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% Project Gutenberg's Theory of Groups of Finite Order, by William Burnside
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% Title: Theory of Groups of Finite Order                                 %
%                                                                         %
% Author: William Burnside                                                %
%                                                                         %
% Release Date: August 2, 2012 [EBook #40395]                             %
% Most recently updated: June 11, 2021                         %
%                                                                         %
% Language: English                                                       %
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% Character set encoding: UTF-8                                           %
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% *** START OF THIS PROJECT GUTENBERG EBOOK THEORY OF GROUPS OF FINITE ORDER ***
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Project Gutenberg's Theory of Groups of Finite Order, by William Burnside

This eBook is for the use of anyone anywhere at no cost and with
almost no restrictions whatsoever.  You may copy it, give it away or
re-use it under the terms of the Project Gutenberg License included
with this eBook or online at www.gutenberg.org


Title: Theory of Groups of Finite Order

Author: William Burnside

Release Date: August 2, 2012 [EBook #40395]
Most recently updated: June 11, 2021

Language: English

Character set encoding: UTF-8     

*** START OF THIS PROJECT GUTENBERG EBOOK THEORY OF GROUPS OF FINITE ORDER ***
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%% -----File: 002.png---Folio ii-------
\newpage
\null\vfill
\begin{center}
\textgoth{London}: \textsc{C. J. CLAY and SONS}, \\[6pt]
\small CAMBRIDGE UNIVERSITY PRESS WAREHOUSE, \\[6pt]
AVE MARIA LANE. \\[6pt]
\footnotesize \textgoth{Glasgow}: 263, ARGYLE STREET. \\[6pt]
%[** CUP device]
\textgoth{Leipzig}: F. A. BROCKHAUS. \\[6pt]
\textgoth{New York}: THE MACMILLAN COMPANY.
\end{center}
\vfill\vfill
%% -----File: 003.png---Folio iii-------
\newpage
\begin{center}
\Huge THEORY OF GROUPS \\[18pt]
\large OF \\[18pt]
\Huge FINITE ORDER
\vfill

\large BY \\[18pt]
\Large W. BURNSIDE, M.A., F.R.S., \\[12pt]
\scriptsize LATE FELLOW OF PEMBROKE COLLEGE, CAMBRIDGE; \\[6pt]
PROFESSOR OF MATHEMATICS AT THE ROYAL NAVAL COLLEGE, GREENWICH.
\vfill\vfill

\large
CAMBRIDGE: \\[6pt]
AT THE UNIVERSITY PRESS. \\[6pt]
1897 \\[6pt]
\footnotesize [\textit{All Rights reserved}.]
\end{center}
%% -----File: 004.png---Folio iv-------
\newpage
\null\vfill
\begin{center}
\textgoth{Cambridge}: \\[6pt]
\scriptsize PRINTED BY J. AND C. F. CLAY, \\[6pt]
AT THE UNIVERSITY PRESS.
\end{center}
\vfill\vfill
%% -----File: 005.png---Folio v-------


\Section{Preface.}

\First{The} theory of groups of finite order may be said to date
from the time of Cauchy. To him are due the first
attempts at classification with a view to forming a theory from
a number of isolated facts. Galois introduced into the theory
the exceedingly important idea of a self-conjugate sub-group,
and the corresponding division of groups into simple and composite.
Moreover, by shewing that to every equation of finite
degree there corresponds a group of finite order on which
all the properties of the equation depend, Galois indicated
how far reaching the applications of the theory might be, and
thereby contributed greatly, if indirectly, to its subsequent
developement.

Many additions were made, mainly by French mathematicians,
during the middle part of the century. The first
connected exposition of the theory was given in the third
edition of M.~Serret's ``\textit{Cours d'Algèbre Supérieure},'' which was
published in 1866. This was followed in 1870 by M.~Jordan's
``\textit{Traité des substitutions et des équations algébriques}.'' The
greater part of M.~Jordan's treatise is devoted to a developement
of the ideas of Galois and to their application to the
theory of equations.

No considerable progress in the theory, as apart from its
applications, was made till the appearance in 1872 of Herr
Sylow's memoir ``\textit{Théorèmes sur les groupes de substitutions}''
in the fifth volume of the \textit{Mathematische Annalen}. Since the
date of this memoir, but more especially in recent years, the
theory has advanced continuously.

In 1882 appeared Herr Netto's ``\textit{Substitutionentheorie und}
%% -----File: 006.png---Folio vi-------
\textit{ihre Anwendungen auf die Algebra},'' in which, as in M.~Serret's
and M.~Jordan's works, the subject is treated entirely from the
point of view of groups of substitutions. Last but not least
among the works which give a detailed account of the subject
must be mentioned Herr Weber's ``\textit{Lehrbuch der Algebra},'' of
which the first volume appeared in 1895 and the second in
1896. In the last section of the first volume some of the more
important properties of substitution groups are given. In the
first section of the second volume, however, the subject is
approached from a more general point of view, and a theory of
finite groups is developed which is quite independent of any
special mode of representing them.

The present treatise is intended to introduce to the reader
the main outlines of the theory of groups of finite order apart
from any applications. The subject is one which has hitherto
attracted but little attention in this country; it will afford
me much satisfaction if, by means of this book, I shall succeed
in arousing interest among English mathematicians in a branch
of pure mathematics which becomes the more fascinating the
more it is studied.

Cayley's dictum that ``a group is defined by means of the
laws of combination of its symbols'' would imply that, in dealing
purely with the theory of groups, no more concrete mode of
representation should be used than is absolutely necessary.
It may then be asked why, in a book which professes to leave
all applications on one side, a considerable space is devoted to
substitution groups; while other particular modes of representation,
such as groups of linear transformations, are not
even referred to. My answer to this question is that while, in
the present state of our knowledge, many results in the pure
theory are arrived at most readily by dealing with properties
of substitution groups, it would be difficult to find a result that
could be most directly obtained by the consideration of groups
of linear transformations.

The plan of the book is as follows. The first Chapter has
been devoted to explaining the notation of substitutions. As
this notation may not improbably be unfamiliar to many
English readers, some such introduction is necessary to make
%% -----File: 007.png---Folio vii-------
the illustrations used in the following chapters intelligible.
Chapters II to VII deal with the more important properties of
groups which are independent of any special form of representation.
The notation and methods of substitution groups
have been rigorously excluded in the proofs and investigations
contained in these chapters; for the purposes of illustration,
however, the notation has been used whenever convenient.
Chapters VIII to X deal with those properties of groups which
depend on their representation as substitution groups. Chapter~XI
treats of the isomorphism of a group with itself. Here,
though the properties involved are independent of the form of
representation of the group, the methods of substitution groups
are partially employed. Graphical modes of representing a
group are considered in Chapters XII and XIII\@. In Chapter~XIV
the properties of a class of groups, of great importance in
analysis, are investigated as a general illustration of the
foregoing theory. The last Chapter contains a series of results
in connection with the classification of groups as simple,
composite, or soluble.

A few illustrative examples have been given throughout
the book. As far as possible I have selected such examples
as would serve to complete or continue the discussion in the
text where they occur.

In addition to the works by Serret, Jordan, Netto and
Weber already referred to, I have while writing this book
consulted many original memoirs. Of these I may specially
mention, as having been of great use to me, two by Herr Dyck
published in the twentieth and twenty-second volumes of the
\textit{Mathematische Annalen} with the title ``\textit{Gruppentheoretische
Studien}''; three by Herr Frobenius in the \textit{Berliner Sitzungsberichte}
for 1895 with the titles, ``\textit{Ueber endliche Gruppen},''
``\textit{Ueber auflösbare Gruppen},'' and ``\textit{Verallgemeinerung des
Sylow'schen Satzes}''; and one by Herr Hölder in the forty-sixth
volume of the \textit{Mathematische Annalen} with the title
``\textit{Bildung zusammengesetzter Gruppen}.'' Whenever a result
is taken from an original memoir I have given a full reference;
any omission to do so that may possibly occur is due to an
oversight on my part.
%% -----File: 008.png---Folio viii-------

To Mr A.~R. Forsyth, Sc.D., F.R.S., Fellow of Trinity
College, Cambridge, and Sadlerian Professor of Mathematics,
and to Mr G.~B. Mathews, M.A., F.R.S., late Fellow of St~John's
College, Cambridge, and formerly Professor of Mathematics in
the University of North Wales, I am under a debt of gratitude
for the care and patience with which they have read the proof-sheets.
Without the assistance they have so generously
given me, the errors and obscurities, which I can hardly hope
to have entirely escaped, would have been far more numerous.
I wish to express my grateful thanks also to Prof.\ O.~Hölder
of Königsberg who very kindly read and criticized parts of
the last chapter. Finally I must thank the Syndics of the
University Press of Cambridge for the assistance they have
rendered in the publication of the book, and the whole Staff
of the Press for the painstaking and careful way in which
the printing has been done.

\Signature{W. BURNSIDE.}{\textit{July}, 1897.}
%% -----File: 009.png---Folio ix-------

\Section{Contents.}

\ToCChap{CHAPTER I.}{ON SUBSTITUTIONS.}

%§§ PAGE

\ToCSect{1}{Object of the chapter}{1}

\ToCSect{2}{Definition of a substitution}{1}

\ToCSect[3]{6}{Notation for substitutions; cycles; products of substitutions}{1-4}

\ToCSect[7]{8}{Identical substitution; inverse substitutions; order of a substitution}{4-6}

\ToCSect[9]{10}{Circular, regular, similar, and permutable substitutions}{7, 8}

\ToCSect{11}{Transpositions; representation of a substitution as a product
of transpositions; odd and even substitutions}{9, 10}


\ToCChap{CHAPTER II.}{THE DEFINITION OF A GROUP.}

\ToCSect{12}{Definition of a group}{11, 12}

\ToCSect{13}{Identical operation}{12, 13}

\ToCSect{14}{Continuous, discontinuous, and mixed groups}{13, 14}

\ToCSect[15]{16}{Order of an operation; products of operations}{14-16}

\ToCSect{17}{Examples of groups of operations; multiplication table of a
group}{17-20}

\ToCSect[18]{19}{Generating operations of a group; defining relations; simply
isomorphic groups}{20-22}

\ToCSect{20}{Dyck's theorem}{22-24}

\ToCSect{21}{Various modes of representing groups}{24}
%% -----File: 010.png---Folio x-------


\ToCChap{CHAPTER III.}{ON THE SIMPLER PROPERTIES OF A GROUP WHICH ARE
INDEPENDENT OF ITS MODE OF REPRESENTATION.}

% §§ PAGE

\ToCSect[22]{23}{Sub-groups; the order of a sub-group divides the order of
the group containing it; symbol for a group}{25-27}

\ToCSect{24}{Transforming one operation by another; conjugate operations
and sub-groups; self-conjugate operations and
sub-groups; simple and composite groups}{27-29}

\ToCSect[25]{26}{The operations of a group which are permutable with a
given operation or sub-group form a group; complete
sets of conjugate operations or sub-groups}{29-33}

\ToCSect{27}{Theorems on self-conjugate sub-groups; maximum sub-groups;
maximum self-conjugate sub-groups}{33-35}

\ToCSect[28]{31}{Multiply isomorphic groups; factor groups; direct product
of two groups}{35-40}

\ToCSect{32}{General isomorphism between two groups}{40, 41}

\ToCSect[33]{35}{Permutable groups; Examples}{41-45}


\ToCChap{CHAPTER IV.}{ON ABELIAN GROUPS.}

\ToCSect[36]{38}{Sub-groups of Abelian groups; every Abelian group is the
direct product of Abelian groups whose orders are
powers of different primes}{46-48}

\ToCSect{39}{Limitation of the discussion to Abelian groups whose orders
are powers of primes}{48}

\ToCSect[40]{44}{Existence of a set of independent generating operations of
such a group; invariance of the orders of the generating
operations; symbol for Abelian group of given
type}{49-55}

\ToCSect[45]{47}{Determination of all types of sub-groups of a given Abelian
group}{55-58}

\ToCSect[48]{49}{Properties of an Abelian group of type $(1, 1, \dots, 1)$}{58-60}

\ToCSect{50}{Examples}{60}
%% -----File: 011.png---Folio xi-------


\ToCChap{CHAPTER V.}{ON GROUPS WHOSE ORDERS ARE POWERS OF PRIMES.}

% §§ PAGE

\ToCSect{51}{Object of the chapter}{61}

\ToCSect{52}{Every group whose order is the power of a prime contains
self-conjugate operations}{61, 62}

\ToCSect[53]{58}{General properties of groups whose orders are powers of
primes}{62-69}

\ToCSect[59]{61}{The number of sub-groups of order~$p^{s}$ of a group of order~$p^{m}$,
where $p$~is a prime, is congruent to unity,
$\pmod{p}$}{69-71}

\ToCSect[62]{63}{Groups of order~$p^{m}$ with a single sub-group of order~$p^{s}$}{71-75}

\ToCSect[64]{67}{Groups of order~$p^{m}$ with a self-conjugate cyclical sub-group
of order~$p^{m-2}$}{75-81}

\ToCSect{68}{Distinct types of groups of orders $p^{2}$ and~$p^{3}$}{81, 82}

\ToCSect[69]{72}{Distinct types of groups of order~$p^{4}$}{82-86}

\ToCSect[73]{74}{Tables of groups of orders $p^{2}$, $p^{3}$, and~$p^{4}$}{86-89}

\ToCSect{75}{Examples}{89}


\ToCChap{CHAPTER VI.}{ON SYLOW'S THEOREM.}

\ToCSect{76}{Object of the chapter}{90, 91}

\ToCSect[77]{78}{Proof of Sylow's theorem}{91-94}

\ToCSect[79]{82}{Direct consequences of Sylow's theorem}{95-100}

\ToCSect[83]{85}{Distinct types of groups of orders~$pq$ ($p$~and $q$ being
different primes), $24$, and~$60$}{100-108}

\ToCSect{86}{Generalization of Sylow's theorem}{108-110}

\ToCSect{87}{Frobenius's theorem}{110-115}

\ToCSect{88}{Groups with properties analogous to those of groups
whose orders are powers of primes}{115-117}


\ToCChap{CHAPTER VII.}{ON THE COMPOSITION-SERIES OF A GROUP.}

\ToCSect{89}{The composition-series, composition-factors, and factor-groups
of a given group}{118, 119}

\ToCSect[90]{91}{Invariance of the composition-series of a group}{119-122}
%% -----File: 012.png---Folio xii-------

\ToCSect{92}{The chief composition-series, or chief series of a group;
its invariance; construction of a composition-series
from a chief series}{122, 123}

\ToCSect[93]{95}{Types of the factor-groups of a chief series; minimum
self-conjugate sub-groups}{124-127}

\ToCSect[96]{97}{Examples of composition-series}{128, 129}

\ToCSect[98]{100}{Soluble groups}{129-132}

\ToCSect{101}{Distinct types of groups of order~$p^{2}q$, where $p$~and $q$
are different primes}{132-137}

\ToCChap{CHAPTER VIII.}{ON SUBSTITUTION GROUPS: TRANSITIVE AND INTRANSITIVE
GROUPS.}

\ToCSect{102}{Substitution groups; degree of a group}{138, 139}

\ToCSect{103}{The symmetric and the alternating groups}{139, 140}

\ToCSect{104}{Transitive and intransitive groups; the degree of a
transitive group is a factor of its order}{140, 141}

\ToCSect{105}{Transitive groups whose substitutions displace all or
all but one of the symbols}{141-144}

\ToCSect[106]{107}{Self-conjugate operations and sub-groups of transitive
groups; transitive groups of which the order is
equal to the degree}{144-148}

\ToCSect[108]{109}{Multiply transitive groups; the order of a $k$-ply transitive
group of degree~$n$ is divisible by $n(n - 1) \dots
(n - k + 1)$; construction of multiply transitive
groups}{148-151}

\ToCSect{110}{Groups of degree~$n$, which do not contain the alternating
group, cannot be more than $(\frac{1}{3}n + 1)$-ply
transitive}{151-153}

\ToCSect{111}{The alternating group of degree~$n$ is simple, except
when $n$ is~$4$}{153, 154}

\ToCSect[112]{113}{Examples of doubly and triply transitive groups}{154-158}

\ToCSect[114]{116}{Properties of intransitive groups}{159-162}

\ToCSect{117}{Intransitive groups of degree~$7$}{163, 164}

\ToCSect[118]{119}{Number of symbols left unchanged by all the substitutions
of a group is the product of the order of
the group and the number of the sets in which
the symbols are interchanged transitively}{165-167}

\ToCLine{Notes to §§~108,~110}{note:1}{167-170}
%% -----File: 013.png---Folio xiii-------


\ToCChap{CHAPTER IX.}{ON SUBSTITUTION GROUPS: PRIMITIVE AND IMPRIMITIVE
GROUPS.}

% §§ PAGE

\ToCSect{120}{Object of the chapter}{171}

\ToCSect{121}{Imprimitive and primitive groups; imprimitive systems}{171, 172}

\ToCSect[122]{125}{Representation of any group in transitive form; primitivity
or imprimitivity of the group so represented}{172-179}

\ToCSect{126}{Number of distinct modes of representing the alternating
group of degree~$5$ in transitive form}{179, 180}

\ToCSect{127}{Imprimitive groups of degree~$6$}{180-183}

\ToCSect[128]{129}{Tests of primitivity: properties of imprimitive systems}{183-186}

\ToCSect{130}{Self-conjugate sub-groups of transitive groups; a self-conjugate
sub-group of a primitive group must be
transitive}{186-188}

\ToCSect{131}{Self-conjugate sub-groups of $k$-ply transitive groups are
in general $(k - 1)$-ply transitive}{188, 189}

\ToCSect[132]{136}{Further properties of self-conjugate sub-groups of primitive
groups}{190-194}

\ToCSect{137}{Examples}{194, 195}


\ToCChap{CHAPTER X.}{ON SUBSTITUTION GROUPS: TRANSITIVITY AND PRIMITIVITY:
(CONCLUDING PROPERTIES).}

\ToCSect{138}{References to tables of primitive groups}{196}

\ToCSect[139]{141}{Primitive groups with transitive sub-groups of smaller
degree: limit to the order of a primitive group of
given degree}{196-199}

\ToCSect{142}{Properties of the symmetric group}{199, 200}

\ToCSect[143]{145}{Further limitations on the orders of primitive groups
of given degree}{200-204}

\ToCSect{146}{Primitive groups whose degrees do not exceed~$8$}{205-211}

\ToCSect[147]{149}{Sub-groups of doubly transitive groups which leave two
symbols unchanged; complete sets of triplets}{212-215}

\ToCSect[150]{151}{The most general groups each of whose substitutions
is permutable with a given substitution, or with
every substitution of a given group}{215-217}

\ToCSect{152}{Transitive groups whose orders are powers of primes}{218, 219}

\ToCSect{153}{Example}{220}
%% -----File: 014.png---Folio xiv-------


\ToCChap{CHAPTER XI.}{ON THE ISOMORPHISM OF A GROUP WITH ITSELF.}

% §§ PAGE

\ToCSect{154}{Object of the chapter}{221}

\ToCSect[155]{156}{Isomorphism of a group with itself; the group of isomorphisms}{222-224}

\ToCSect{157}{Cogredient and contragredient isomorphisms; the group
of cogredient isomorphisms is contained self-conjugately
in the group of isomorphisms}{224-226}

\ToCSect{158}{The holomorph of a group}{226-228}

\ToCSect[159]{161}{Properties of isomorphisms; representation of the group
of isomorphisms in transitive form}{228-232}

\ToCSect{162}{Characteristic sub-groups; groups with no characteristic
sub-groups}{232}

\ToCSect[163]{164}{Characteristic series of a group; its invariance; characteristic
series of a group whose order is the power
of a prime}{232-235}

\ToCSect[165]{167}{Complete groups}{235-239}

\ToCSect[168]{170}{The group of isomorphisms and the holomorph of a
cyclical group}{239-242}

\ToCSect[171]{172}{The group of isomorphisms and the holomorph of an
Abelian group of order~$p^{n}$ and type $(1, 1, \dots, 1)$;
the homogeneous linear group}{242-245}

\ToCSect{173}{The group of isomorphisms of the alternating group}{245, 246}

\ToCSect{174}{The group of isomorphisms of doubly transitive groups
of degree~$p^{n} + 1$ and order~$\frac{1}{2} p^{n} (p^{2n} - 1)$}{246-249}

\ToCSect[175]{178}{Further properties of isomorphisms; the symbols $\vartheta (P)$
and~$\theta (P)$}{249-253}

\ToCSect{179}{Examples}{253, 254}


\ToCChap{CHAPTER XII.}{ON THE GRAPHICAL REPRESENTATION OF A GROUP.}

\ToCSect{180}{Groups with an infinite number of operations}{255}

\ToCSect[181]{182}{The most general discontinuous group that can be
generated by a finite number of operations; relation
of special groups to the general group}{256-259}

\ToCSect{183}{Graphical representation of a cyclical group}{260-262}

\ToCSect[184]{187}{Graphical representation of a general group}{262-267}

\ToCSect[188]{190}{Graphical representation of a special group}{267-272}

\ToCSect[191]{195}{Graphical representation of groups of finite order}{272-280}

\ToCSect{196}{The genus of a group}{280}

\ToCSect[197]{198}{Limitation on the order and on the number of defining
relations of a group of given genus}{281-284}

\ToCLine{Note to §~194}{note:3}{284}
%% -----File: 015.png---Folio xv-------


\ToCChap{CHAPTER XIII.}{ON THE GRAPHICAL REPRESENTATION OF GROUPS: GROUPS OF
GENUS ZERO AND UNITY: CAYLEY'S COLOUR GROUPS.}

% §§ PAGE

\ToCSect[199]{203}{Groups of genus zero; their defining relations and
graphical representation}{285-292}

\ToCSect[204]{209}{Groups of genus unity; their defining relations and
graphical representation}{293-302}

\ToCSect{210}{The graphical representation and the defining relations
of the simple group of order~$168$}{302-305}

\ToCSect[211]{214}{Cayley's colour groups}{306-310}


\ToCChap{CHAPTER XIV.}{ON THE LINEAR GROUP.}

\ToCSect{215}{The homogeneous linear group}{311, 312}

\ToCSect[216]{220}{Its composition-series}{312-317}

\ToCSect{221}{The simple group which it defines}{317-319}

\ToCSect[222]{234}{The fractional linear group; determination of the orders
of its operations and of their distribution in conjugate
sets; determination of all of its sub-groups
and of their distribution in conjugate sets; its
representation as a doubly transitive group}{319-334}

\ToCSect{235}{Generalization of the fractional linear group}{334, 335}

\ToCSect[236]{238}{Representation of the simple group, defined by the
homogeneous linear group, as a doubly transitive
group; special cases}{336-340}

\ToCSect[239]{240}{Generalization of the homogeneous linear group}{340-342}


\ToCChap{CHAPTER XV.}{ON SOLUBLE AND COMPOSITE GROUPS.}

\ToCSect{241}{Object of the chapter}{343, 344}

\ToCSect{242}{Direct applications of Sylow's theorem often shew that
a group of given order must be composite}{344, 345}

\ToCSect[243]{245}{Soluble groups whose orders are $p^{\alpha} q^{\beta}$, where $p$ and~$q$
are primes}{345-352}
%% -----File: 016.png---Folio xvi-------

\ToCSect{246}{Groups whose sub-groups of order~$p^{\alpha}$ are all cyclical,
$p^{\alpha}$~being any power of a prime which divides the
order}{352, 353}

\ToCSect{247}{Groups whose orders contain no square factor}{353, 354}

\ToCSect[248]{249}{Further tests of solubility; groups whose orders contain
no cube factor}{354-360}

\ToCSect[250]{257}{Groups of even order in which the operations of odd
order form a self-conjugate sub-group; either $12$,
$16$, or $56$ must divide the order of a simple group
if it is even}{360-366}

\ToCSect{258}{The simple groups whose orders contain less than $6$
prime factors}{367-370}

\ToCSect[259]{260}{The simple groups whose orders do not exceed~$660$}{370-375}

\ToCSect[261]{263}{Non-soluble composite groups}{375-378}

\ToCLine{Notes to §§~257, 258,~260}{note:4}{379}

%[** TN: Empty optional argument signifies unindented entry]
\ToCLine[]{APPENDIX: On French and German technical terms}{sect:Appendix.}{380-382}

\ToCLine[]{INDEX}{sect:Index.}{383-388}

\iffalse
%[** TN: Applied in the text; q.v. \Erratum macro]
\Section{Errata.}

p.~60, line~18, \emph{for} $7$ \emph{read} $56$.

p.~68, line~10, \emph{after} $p^{r+s}$ \emph{add} where $p$ is an odd prime.

p.~150, line~18, \emph{after} $G$ \emph{add} which displaces $a_{1}$.

p.~150, line~26, \emph{after} $S$, \emph{add} either $ASA$ belongs to $G$ or.

p.~151, line~6, \emph{after} form \emph{add} The reduction is here carried out
on the supposition that $S_{p}$ and $S_{t}$ displace $a_{1}$.
The modification, when this is not the case, is
obvious.
\fi
%% -----File: 017.png---Folio 1-------
\MainMatter
\index{Abelian group@\Topic{Abelian group}|seealso{\Topic{Homogeneous linear group}}}%
\index{Group@\Topic{Group}|seealso{\Topic{Substitution group}}}%
\index{Representation@\Topic{Representation}, graphical|see{\Topic{Graphical representation}}}%
\index{Sub-group@\Topic{Sub-group}|seealso{\Topic{conjugate}, \Topic{self-conjugate} and \Topic{characteristic sub-group}}}%
\index{Type of a group@\Topic{Type of a group}|see{\Topic{Simply isomorphic groups}}}%


\Chapter{I.}{On Substitutions.}
\index{Substitution@\Topic{Substitution}!definition of}%

\Par{1.} \First{Among} the various notations used in the following
pages, there is one of such frequent recurrence that a certain
readiness in its use is very desirable in dealing with the
subject of this treatise. We therefore propose to devote a
preliminary chapter to explaining it in some detail.

\Par{2.} Let $a_{1}$, $a_{2}$,~\dots, $a_{n}$ be a set of $n$~distinct letters. The
operation of replacing each letter of the set by another, which
may be the same letter or a different one, when carried out
under the condition that no two letters are replaced by one and
the same letter, is called a \emph{substitution} performed on the $n$~letters.
Such a substitution will change any given arrangement
\[
a_{1},\ a_{2},\ \dots,\ a_{n}
\]
of the $n$~letters into a definite new arrangement
\[
b_{1},\ b_{2},\ \dots,\ b_{n}
\]
of the same $n$~letters.

\Par{3.} One obvious form in which to write the substitution is
\[
\Sub{a_{1}, a_{2}, \dots, a_{n}}{b_{1}, b_{2}, \dots, b_{n}}
\]
thereby indicating that each letter in the upper line is to be
replaced by the letter standing under it in the lower. The
disadvantage of this form is its unnecessary complexity, each
%% -----File: 018.png---Folio 2-------
of the $n$~letters occurring twice in the expression for the
substitution; by the following process, the expression of the
substitution may be materially simplified.

Let $p$ be any one of the $n$~letters, and $q$ the letter in the
lower line standing under $p$ in the upper. Suppose now that $r$
is the letter in the lower line that stands under $q$ in the upper,
and so on. Since the number of letters is finite, we must arrive
at last at a letter~$s$ in the upper line under which $p$ stands.
If the set of $n$~letters is not thus exhausted, take any letter~$p'$
in the upper line, which has not yet occurred, and let $q'$, $r'$,~\dots\
follow it as $q$, $r$,~\dots\ followed~$p$, till we arrive at~$s'$ in the upper
line with $p'$ standing under it. If the set of $n$~letters is still not
exhausted, repeat the process, starting with a letter~$p''$ which
has not yet occurred. Since the number of letters is finite, we
must in this way at last exhaust them; and the $n$~letters are
thus distributed into a number of sets
\[
\begin{array}{*{5}{l}}
p, & q, & r, & \dots, & s; \\
p', & q', & r', & \dots, & s'; \\
p'', & q'', & r'', & \dots, & s''; \\
\hdotsfor{5}; \\
\end{array}
\]
such that the substitution replaces each letter of a set by the
one following it in that set, the last letter of each set being replaced
by the first of the same set.

If now we represent by the symbol
\[
(pqr \dots s)
\]
the operation of replacing $p$ by~$q$, $q$ by $r$,~\dots, and $s$ by~$p$, the
substitution will be completely represented by the symbol
\[
(pqr \dots s)(p'q'r' \dots s')(p''q''r'' \dots s'')\Dots\Add{.}
\]
The advantage of this mode of expressing the substitution is
that each of the letters occurs only once in the symbol.

\Par{4.} The separate components of the above symbol, such as
\index{Substitution@\Topic{Substitution}!cycles of}%
$(pqr \dots s)$ are called the \emph{cycles} of the substitution. In particular
cases, one or more of the cycles may contain a single letter;
when this happens, the letters so occurring singly are unaltered
by the substitution. The brackets enclosing single letters may
%% -----File: 019.png---Folio 3-------
clearly be omitted without risk of ambiguity, as also may the
unaltered letters themselves. Thus the substitution
\[
\Sub{a, b, c, d, e}{c, b, d, a, e}
\]
may be written $(acd)(b)(e)$, or $(acd)be$, or simply~$(acd)$. If for
any reason it were desirable to indicate that substitutions of the
five letters $a$, $b$, $c$, $d$, $e$ were under consideration, the second of
these three forms would be used.

\Par{5.} The form thus obtained for a substitution is not unique.
The symbol $(qr \dots sp)$ clearly represents the same substitution
as $(pqr \dots s)$, if the letters that occur between $r$ and $s$ in the two
symbols are the same and occur in the same order; so that, as
regards the letters inside the bracket, any one may be chosen
to stand first so long as the cyclical order is preserved unchanged.

Moreover the order in which the brackets are arranged is
clearly immaterial, since the operation denoted by any one
bracket has no effect on the letters contained in the other
brackets. This latter property is characteristic of the particular
expression that has been obtained for a substitution;
it depends upon the fact that the expression contains each
of the letters once only.

\Par{6.} When we proceed to consider the effect of performing
two or more substitutions successively, it is seen at once that
the order in which the substitutions are carried out in general
affects the result. Thus to give a very simple instance, the
substitution~$(ab)$ followed by~$(ac)$ changes $a$ into~$b$, since $b$~is
unaltered by the second substitution. Again, $(ab)$~changes $b$
into~$a$ and $(ac)$~changes $a$ into~$c$, so that the two substitutions
performed successively change $b$ into~$c$. Lastly, $(ab)$ does not
affect~$c$ and $(ac)$~changes $c$ into~$a$. Hence the two substitutions
performed successively change $a$ into~$b$, $b$ into~$c$, $c$ into~$a$, and
affect no other symbols. The result of the two substitutions
performed successively is therefore equivalent to the substitution~$(abc)$;
and it may be similarly shewn that $(ac)$~followed
by $(ab)$ gives~$(acb)$ as the resulting substitution. To avoid
ambiguity it is therefore necessary to assign, once for all, the
%% -----File: 020.png---Folio 4-------
meaning to be attached to such a symbol as~$s_{1}s_{2}$, where $s_{1}$ and
$s_{2}$ are the symbols of two given substitutions. We shall always
understand by the symbol~$s_{1}s_{2}$ \emph{the result of carrying out first
the substitution~$s_{1}$ and then the substitution~$s_{2}$}. Thus the two
simple examples given above may be expressed in the form
\begin{align*}
(ab) (ac) &= (abc), \\
(ac) (ab) &= (acb),
\end{align*}
the sign of equality being used to represent that the substitutions
are equivalent to each other.

If now
\[
s_{1}s_{2} = s_{4} \quad\text{and}\quad s_{2}s_{3} = s_{5},
\]
the symbol $s_{1}s_{2}s_{3}$ may be regarded as the substitution $s_{4}$ followed
by~$s_{3}$ or as $s_{1}$~followed by~$s_{5}$. But if $s_{1}$ changes \emph{any} letter $a$
into~$b$, while $s_{2}$ changes $b$ into~$c$ and $s_{3}$ changes $c$ into~$d$, then $s_{4}$
changes $a$ into~$c$ and $s_{5}$ changes $b$ into~$d$. Hence $s_{4}s_{3}$ and $s_{1}s_{5}$
both change $a$ into~$d$; and therefore, $a$~being any letter operated
upon by the substitutions,
\[
s_{4}s_{3} = s_{1}s_{5}.
\]

Hence the meaning of the symbol $s_{1}s_{2}s_{3}$ is definite; it
\index{Substitution@\Topic{Substitution}!identical}%
\index{Substitution@\Topic{Substitution}!inverse}%
\index{Substitution@\Topic{Substitution}!symbol for the product of two or more}%
\index{Symbol@\Topic{Symbol}!for the product of two or more substitutions}%
depends only on the component substitutions $s_{1}$, $s_{2}$, $s_{3}$ and their
sequence, and it is independent of the way in which they are
associated when their sequence is assigned. And the same
clearly holds for the symbol representing the successive performance
of any number of substitutions. To avoid circumlocution,
it is convenient to speak of the substitution $s_{1}s_{2} \dots s_{n}$
as the \emph{product} of the substitutions $s_{1}$, $s_{2}$,~\dots, $s_{n}$ in the sequence
given. The product of a number of substitutions, thus defined,
always obeys the associative law but does not in general obey
the commutative law of algebraical multiplication.

\Par{7.} The substitution which replaces every symbol by itself
is called the \emph{identical substitution}. The \emph{inverse} of a given
substitution is that substitution which, when performed after
the given substitution, gives as result the identical substitution.
Let $s_{-1}$ be the substitution inverse to~$s$, so that, if
\[
s = \Sub{a_{1}, a_{2}, \dots, a_{n}}{b_{1}, b_{2}, \dots, b_{n}},
\]
then
\[
s_{-1} = \Sub{b_{1}, b_{2}, \dots, b_{n}}{a_{1}, a_{2}, \dots, a_{n}}.
\]
%% -----File: 021.png---Folio 5-------
Let $s_{0}$ denote the identical substitution which can be represented
by
\[
\Sub{a_{1}, a_{2}, \dots, a_{n}}{a_{1}, a_{2}, \dots, a_{n}}.
\]

Then
\[
ss_{-1} = s_{0}\quad \text{and}\quad s_{-1}s = s_{0},
\]
so that $s$ is the substitution inverse to~$s_{-1}$.

Now if
\[
ts = t' s,
\]
then
\[
tss_{-1} = t' ss_{-1},
\]
or
\[
ts_{0} = t' s_{0}.
\]

But $ts_{0}$ is the same substitution as~$t$, since $s_{0}$~produces no
change; and therefore
\[
t = t'.
\]

In exactly the same way, it may be shewn that the relation
\[
st = st'
\]
involves
\[
t = t'.
\]

\Par{8.} The result of performing $r$~times in succession the same
substitution~$s$ is represented symbolically by~$s^{r}$. Since, as
has been seen, products of substitutions obey the associative
law of multiplication, it follows that
\[
s^{\mu} s^{\nu} = s^{\mu + \nu} = s^{\nu} s^{\mu}.
\]

Now since there are only a finite number of distinct
substitutions that can be performed on a given finite set of
symbols, the series of substitutions $s$, $s^{2}$, $s^{3}$,~\dots\ cannot be all
distinct. Suppose that $s^{m + 1}$~is the first of the series which is
the same as~$s$, so that
\[
s^{m + 1} = s.
\]

Then
\[
s^{m} ss_{-1} = ss_{-1},
\]
or
\[
s^{m} = s_{0}.
\]

There is no index~$\mu$ smaller than~$m$ for which this relation
holds. For if
\[
s^{\mu} = s_{0},
\]
then
\[
s^{\mu + 1} = ss_{0} = s,
\]
contrary to the supposition that $s^{m+1}$ is the first of the series
which is the same as~$s$.
%% -----File: 022.png---Folio 6-------

Moreover the $m - 1$ substitutions $s$, $s^{2}$,~\dots, $s^{m - 1}$ must be all
distinct. For if
\[
s^{\mu} = s^{\nu},\quad \nu < \mu < m,
\]
then
\[
s^{\mu - \nu} s^{\nu} (s^{\nu})_{-1} = s^{\nu} (s^{\nu})_{-1},
\]
or
\[
s^{\mu - \nu} = s_{0},
\]
which has just been shewn to be impossible.

The number~$m$ is called the \emph{order} of the substitution~$s$.
\index{Substitution@\Topic{Substitution}!order of}%
In connection with the order of a substitution, two properties
are to be noted. First, if
\[
s^{n} = s_{0},
\]
it may be shewn at once that $n$~is a multiple of $m$ the order of~$s$;
and secondly, if
\[
s^{\alpha} = s^{\beta},
\]
then
\[
\alpha - \beta \equiv 0\ (\mod m).
\]

If now the equation
\[
s^{\mu + \nu} = s^{\mu} s^{\nu}
\]
be assumed to hold, when either or both of the integers $\mu$ and $\nu$
is a negative integer, a definite meaning is obtained for the
symbol~$s^{-r}$, implying the negative power of a substitution; and
a definite meaning is also obtained for~$s^{0}$. For
\[
s^{\mu} s^{-\nu} = s^{\mu - \nu} = s^{\mu - \nu} s^{\nu} (s^{\nu})_{-1} = s^{\mu} (s^{\nu})_{-1},
\]
so that
\[
s^{-\nu} = (s^{\nu})_{-1}.
\]
Similarly it can be shewn that
\[
s^{0} = s_{0}.
\]

Since every power of~$s_{0}$ is the same as~$s_{0}$, and since
wherever $s_{0}$ occurs in the symbol $s_{1} s_{2} \dots s_{n}$ of a compound
substitution it may be omitted without affecting the result, it
is clear that no ambiguity will result from replacing $s_{0}$ everywhere
by~$1$; in other words, we may use~$1$ to represent the
identical substitution which leaves every letter unchanged.
But when this is done, it must of course be remembered that
the equation
\[
s^{m} = 1
\]
is not a reducible algebraical equation, which is capable of being
written in the form
\[
(s - 1) (s^{m - 1} + s^{m - 2} + \dots + 1) = 0.
\]
%% -----File: 023.png---Folio 7-------

Indeed the symbol $s + s'$, where $s$ and $s'$ are any two substitutions,
has no meaning.

\Par{9.} If the cycles of a substitution
\[
s = (pqr\dots s)(p'q'\dots s')(p''q''\dots s'')\dots
\]
contain $m$, $m'$, $m''$,~\dots\ letters respectively, and if
\[
s^{\mu} = 1,
\]
$\mu$ must be a common multiple of $m$, $m'$, $m''$,~\dots. For $s^{\mu}$ changes~$p$
into a letter $\mu$~places from it in the cyclical set $p$,~$q$, $r$,~\dots,~$s$;
and therefore, if it changes $p$ into itself, $\mu$~must be a multiple
of~$m$. In the same way, it must be a multiple of $m'$, $m''$,~\dots.
Hence the order of~$s$ is the least common multiple of $m$, $m'$,
$m''$,~\dots.

In particular, when a substitution consists of a single
\index{Substitution@\Topic{Substitution}!circular}%
\index{Substitution@\Topic{Substitution}!regular}%
\index{Substitution@\Topic{Substitution}!similar}%
cycle, its order is equal to the number of letters which it interchanges.
Such a substitution is called a \emph{circular substitution}.

A substitution, all of whose cycles contain the same
number of letters, is said to be \emph{regular} in the letters which it
interchanges; the order of such a substitution is clearly equal
to the number of letters in one of its cycles.

\Par{10.} Two substitutions, which contain the same number of
cycles and the same number of letters in corresponding cycles,
are called \emph{similar}. If $s$,~$s'$ are similar substitutions, so also
clearly are $s^{r}$,~$s'^{r}$; and the orders of $s$~and $s'$ are the same.

Let now
\[
s = (a_{p}a_{q}\dots a_{s})(a_{p'}a_{q'}\dots a_{s'})\dots
\]
and
\[
t = \Sub{a_{1}, a_{2}, \dots, a_{n}}{b_{1}, b_{2}, \dots, b_{n}}
\]
be any two substitutions. Then
\begin{align*}
t^{-1}st &= \Sub{b_{1}, b_{2}, \dots, b_{n}}{a_{1}, a_{2}, \dots, a_{n}}
  (a_{p}a_{q}\dots a_{s})(a_{p'}a_{q'}\dots a_{s'}) \dots
  \Sub{a_{1}, a_{2}, \dots, a_{n}}{b_{1}, b_{2}, \dots, b_{n}} \\
  &= (b_{p}b_{q}\dots b_{s})(b_{p'}b_{q'}\dots b_{s'})\dots,
\end{align*}
the latter form of the substitution being obtained by actually
carrying out the component substitutions of the earlier form.
Hence $s$ and $t^{-1}st$ are similar substitutions.
%% -----File: 024.png---Folio 8-------

Since
\[
s_{2}s_{1} = s_{1}^{-1}s_{1}s_{2}s_{1},
\]
it follows that $s_{1}s_{2}$ and $s_{2}s_{1}$ are similar substitutions and therefore
that they are of the same order. Similarly it may be shewn
that $s_{1}s_{2}s_{3} \dots s_{n}$, $s_{2}s_{3} \dots s_{n}s_{1}$,~\dots, $s_{n}s_{1} \dots s_{2}s_{3}$ are all similar substitutions.

It may happen in particular cases that $s$ and $t^{-1}st$ are the
\index{Substitution@\Topic{Substitution}!permutable}%
same substitution. When this is so, $t$~and $s$ are \emph{permutable},
that is, $st$ and $ts$ are equivalent to one another; for if
\[
s = t^{-1}st,
\]
then
\[
ts = st.
\]

This will certainly be the case when none of the symbols
that are interchanged by~$t$ are altered by~$s$; but it may happen
when $s$ and $t$ operate on the same symbols. Thus if
\[
s = (ab)(cd), \quad t = (ac)(bd),
\]
then
\[
st = (ad)(bc) = ts.
\]

\begin{Ex}[1.] Shew that every regular substitution is some power of a
circular substitution.
\end{Ex}

\begin{Ex}[2.] If $s$,~$s'$ are permutable regular substitutions of the same
$mn$~letters of orders $m$ and~$n$, these numbers being relatively prime,
shew that $ss'$~is a circular substitution in the $mn$~letters.
\end{Ex}

\begin{Ex}[3\protect\footnotemark.]
\footnotetext{It is often convenient to use digits rather than letters for the purpose of
  illustration.}
If
\begin{alignat*}{3}
&s     &&= (123) &&(456)(789), \\
&s_{1} &&= (147) &&(258)(369), \\
&s_{2} &&=       &&(456)(798),
\end{alignat*}
shew that $s$~is permutable with both $s_{1}$ and~$s_{2}$, and that it can be
formed by a combination of $s_{1}$ and~$s_{2}$.
\end{Ex}

\begin{Ex}[4.] Shew that the only substitutions of $n$ given letters
which are permutable with a circular substitution of the $n$~letters
are the powers of the circular substitution.
\end{Ex}

\begin{Ex}[5.] Determine all the substitutions of the ten symbols
involved in
\[
s = (abcde)(\alpha\beta\gamma\delta\epsilon)
\]
which are permutable with~$s$.
\end{Ex}

The determination of all the substitutions which are
permutable with a given substitution will form the subject
of investigation in Chapter~X\@.
%% -----File: 025.png---Folio 9-------

\Par{11.} A circular substitution of order two is called a \emph{transposition}.
\index{Transpositions@\Topic{Transpositions}!definition of}%
\index{Transpositions@\Topic{Transpositions}!representation of a substitution by means of}%
It may be easily verified that
\[
(pqr\dots s) = (pq)(pr)\dots(ps),
\]
so that every circular substitution can be represented as a
product of transpositions; and thence, since every substitution
is the product of a number of circular substitutions, every
substitution can be represented as a product of transpositions.
It must be remembered, however, that, in general, when a
substitution is represented in this way, some of the letters will
occur more than once in the symbol, so that the order in which
the constituent transpositions occur is essential. There is thus
a fundamental difference from the case when the symbol of a
substitution is the product of circular substitutions, no two of
which contain a common letter.

Since
\[
(p'q') = (pp')(pq')(pp'),
\]
every transposition, and therefore every substitution of $n$~letters,
can be expressed in terms of the $n - 1$ transpositions
\[
(a_{1}a_{2}),\ (a_{1}a_{3}),\ \dots,\ (a_{1}a_{n}).
\]

The number of different ways in which a given substitution
may be represented as a product of transpositions is evidently
unlimited; but it may be shewn that, however the representation
is effected, the number of transpositions is either always even
or always odd. To prove this, it is sufficient to consider the
effect of a transposition on the square root of the discriminant
of the $n$~letters, which may be written
\[
D = \prod_{r = 1}^{r = n - 1} \left\{\prod_{s = r + 1}^{s = n} (a_{r} - a_{s})\right\}.
\]

The transposition $(a_{r}a_{s})$ changes the sign of the factor
$a_{r} - a_{s}$. When $q$~is less than either $r$~or~$s$, the transposition
interchanges the factors $a_{q} - a_{r}$ and $a_{q} - a_{s}$; and when $q$~is
greater than either $r$~or~$s$, it interchanges the factors $a_{r} - a_{q}$ and
$a_{s} - a_{q}$. When $q$~lies between $r$ and~$s$, the pair of factors
$a_{r} - a_{q}$ and $a_{q} - a_{s}$ are interchanged and are both changed in
sign. Hence the effect of the single transposition on~$D$ is
to change its sign. Since any substitution can be expressed as
the product of a number of transpositions, the effect of any
substitution on~$D$ must be either to leave it unaltered or to
%% -----File: 026.png---Folio 10-------
change its sign. If a substitution leaves $D$ unaltered it must,
when expressed as a product of transpositions in any way,
contain an even number of transpositions; and if it changes
the sign of~$D$, every representation of it, as a product of
transpositions, must contain an odd number of transpositions.
Hence no substitution is capable of being expressed both by an
even and by an odd number of transpositions.

A substitution is spoken of as \emph{odd} or \emph{even}, according as the
\index{Substitution@\Topic{Substitution}!even and odd}%
\index{Transpositions@\Topic{Transpositions}!number of, which enter in the representation of a substitution is either always even or always odd}%
transpositions which enter into its representation are odd or
even in number.

Further, an even substitution can always be represented as
a product of circular substitutions of order three. For any
even substitution of $n$~letters can be represented as the product
of an even number of the $n - 1$ transpositions
\[
(a_{1}a_{2}),\ (a_{1}a_{3}),\ \dots,\ (a_{1}a_{n}),
\]
in appropriate sequence and with the proper number of
occurrences; and the product of any consecutive pair of these
$(a_{1}a_{r})(a_{1}a_{s})$ is the circular substitution $(a_{1}a_{r}a_{s})$.

Now
\begin{align*} %[** TN: Set on three lines in the original]
(a_{1}a_{2}a_{s})(a_{1}a_{2}a_{r})(a_{1}a_{2}a_{s})^{2}
  &= (a_{1}a_{2}a_{s})(a_{1}a_{2}a_{r})(a_{1}a_{s}a_{2}) \\
  &= (a_{1}a_{r}a_{s}),
\end{align*}
so that every circular substitution of order three displacing~$a_{1}$,
and therefore every even substitution of $n$~letters, can be
expressed in terms of the $n - 2$ substitutions
\[
(a_{1}a_{2}a_{3}),\ (a_{1}a_{2}a_{4}),\ \dots,\ (a_{1}a_{2}a_{n})
\]
and their powers.

\begin{Ex}[1.] Shew that every even substitution of $n$~letters can be
expressed in terms of
\[
(a_{1}a_{2}a_{3}),\ (a_{1}a_{4}a_{5}),\ \Dots,\ (a_{1}a_{n-1}a_{n}),
\]
when $n$~is odd; and in terms of
\[
(a_{1}a_{2}a_{3}),\ (a_{1}a_{4}a_{5}),\ \Dots,\ (a_{1}a_{n - 2}a_{n - 1}),\ (a_{1}a_{2}a_{n}),
\]
when $n$~is even.
\end{Ex}

\begin{Ex}[2.] If $n + 1$~is odd, shew that every even substitution of
$mn + 1$ letters can be expressed in terms of
\[
(a_{1}a_{2}\Dots a_{n + 1}),\ (a_{1}a_{n + 2}\Dots a_{2n + 1}),\ \Dots,\ (a_{1}a_{(m - 1)n + 2}\Dots a_{mn + 1});
\]
and if $n + 1$~is even, that every substitution of $mn + 1$ letters can
be expressed in terms of this set of $m$~circular substitutions.
\end{Ex}
%% -----File: 027.png---Folio 11-------


\Chapter{II.}{The Definition of a Group.}
\index{Group@\Topic{Group}!definition of}%

\Par{12.} \First{In} the present chapter we shall enter on our main
subject and we shall begin with definitions, explanations and
examples of what is meant by a group.

\begin{Definition}
Let
\[
A,\ B,\ C,\ \dots
\]
represent a set of operations, which can be performed on the
same object or set of objects. Suppose this set of operations
has the following characteristics.

\Inum{($\alpha$)} The operations of the set are all distinct, so that no
two of them produce the same change in every possible application.

\Inum{($\beta$)} The result of performing successively any number of
operations of the set, say $A$, $B$,~\dots, $K$, is another definite
operation of the set, which depends only on the component
operations and the sequence in which they are carried out, and
not on the way in which they may be regarded as associated.
Thus $A$~followed by~$B$ and $B$~followed by~$C$ are operations of
the set, say $D$~and~$E$; and $D$~followed by~$C$ is the same operation
as $A$~followed by~$E$.

\Inum{($\gamma$)} $A$ being any operation of the set, there is always
another operation~$A_{-1}$ belonging to the set, such that $A$
followed by~$A_{-1}$ produces no change in any object.
%% -----File: 028.png---Folio 12-------

The operation~$A_{-1}$ is called the inverse of~$A$.

\index{Group@\Topic{Group}!definition of}%
The set of operations is then said to form a \emph{Group}.
\end{Definition}

From the definition of the inverse of~$A$ given in~\Inum{($\gamma$)}, it
follows directly that $A$~is the inverse of~$A_{-1}$. For if $A$~changes
any object~$\Omega$ into~$\Omega'$, $A_{-1}$~must change $\Omega'$ into~$\Omega$. Hence $A_{-1}$
followed by~$A$ leaves $\Omega'$, and therefore every object, unchanged.

The operation resulting from the successive performance of
the operations $A$, $B$,~\dots, $K$ in the sequence given is denoted by
the symbol $AB \dots K$; and if $\Omega$~is any object on which the
operations may be performed, the result of carrying out this
compound operation on~$\Omega$ is denoted by $\Omega · AB \dots K$.

If the component operations are all the same, say~$A$, and $r$~in
number, the abbreviation~$A^{r}$ will be used for the resultant
operation, and it will be called the $r$th~\emph{power} of~$A$.

\begin{Definition}
Two operations, $A$~and~$B$, are said to be
\index{Permutable operations@\Topic{Permutable operations}!definition of}%
\emph{permutable} when $AB$ and $BA$ are the same operation.
\end{Definition}

\Par{13.} If $AB$ and $AC$ are the same operation, so also are
\index{Symbol@\Topic{Symbol}!for the product of two or more operations}%
$A_{-1}AB$ and~$A_{-1}AC$. But the operation~$A_{-1}A$ produces no
change in any object and therefore $A_{-1}AB$ and~$B$, producing
the same change in every object, are the same operation. Hence
$B$~and~$C$ are the same operation.

This is expressed symbolically by saying that, if
\[
AB = AC,
\]
then
\[
B = C;
\]
the sign of equality being used to imply that the symbols
represent the same operation.

In a similar way, if
\[
BA = CA,
\]
it follows that
\[
B = C.
\]

From conditions \Inum{($\beta$)} and \Inum{($\gamma$)}, $AA_{-1}$~must be a definite
operation of the group. This operation, by definition, produces
no change in any possible object, and it must, by
\index{Identical operation@\Topic{Identical operation}!definition of}%
condition~\Inum{($\alpha$)}, be unique. It is called the \emph{identical} operation.
%% -----File: 029.png---Folio 13-------
If it is represented by~$A_{0}$ and if $A$~be any other operation,
then
\[
A_{0} A = A = AA_{0},
\]
and for every integer~$r$,
\[
A_{0}^{r} = A_{0}.
\]

Hence $A_{0}$ may, without ambiguity, be replaced by~$1$,
wherever it occurs.

\Par{14.} The number of distinct operations contained in a
\index{Group@\Topic{Group}!continuous, discontinuous, or mixed}%
group may be either finite or infinite. When the number is
infinite, the group may contain operations which produce an
infinitesimal change in every possible object or operand.

Thus the totality of distinct displacements of a rigid body
evidently forms a group, for they satisfy conditions \Inum{($\alpha$)},~\Inum{($\beta$)} and~\Inum{($\gamma$)}
of the definition. Moreover this group contains operations
of the kind in question, namely infinitesimal twists; and
each operation of the group can be constructed by the continual
repetition of a suitably chosen infinitesimal twist.

Next, the set of translations, that arise by shifting a cube
parallel to its edges through distances which are any multiples
of an edge, forms a group containing an infinite number
of operations; but this group contains no operation which
effects an infinitesimal change in the position of the cube.

As a third example, consider the set of displacements by
which a complete right circular cone is brought to coincidence
with itself. It consists of rotations through any angle about
the axis of the cone, and rotations through two right angles
about any line through the vertex at right angles to the
axis. Once again this set of displacements satisfies the conditions
\Inum{($\alpha$)},~\Inum{($\beta$)} and~\Inum{($\gamma$)} of the definition and forms a group.

This last group contains infinitesimal operations, namely
rotations round the axis through an infinitesimal angle; and
every finite rotation round the axis can be formed by the
continued repetition of an infinitesimal rotation. There is
however in this case no infinitesimal displacement of the group
by whose continued repetition a rotation through two right
angles about a line through the vertex at right angles to the
axis can be constructed. Of these three groups with an infinite
%% -----File: 030.png---Folio 14-------
number of operations, the first is said to be a \emph{continuous} group,
\index{Group@\Topic{Group}!continuous, discontinuous, or mixed}%
the second a \emph{discontinuous} group, and the third a \emph{mixed} group.

Continuous groups and mixed groups lie entirely outside the
plan of the present treatise; and though, later on, some of the
properties of discontinuous groups with an infinite number of
operations will be considered, such groups will be approached
from a point of view suggested by the treatment of groups
containing a finite number of operations. It is not therefore
necessary here to deal in detail with the classification of infinite
groups which is indicated by the three examples given above;
and we pass on at once to the case of groups which contain a
finite number only of distinct operations.

\Par{15.} \begin{Definition}
If the number of distinct operations contained
\index{Group@\Topic{Group}!order of}%
\index{Order@\Topic{Order}!of a group, definition of}%
in a group be finite, the number is called the \emph{order} of the
group.
\end{Definition}

Let $S$ be an operation of a group of finite order~$N$. Then
the infinite series of operations
\[
S,\ S^{2},\ S^{3},\ \Dots
\]
must all be contained in the group, and therefore a finite
number of them only can be distinct. If $S^{m + 1}$~is the first of
the series which is the same as~$S$, and if $S_{-1}$~is the operation
inverse to~$S$, then
\[
S^{m + 1} S_{-1} = SS_{-1} = 1,
\]
or
\[
S^{m} = 1.
\]

Exactly as in §~8, it may be shewn that, if
\[
S^{\mu} = 1,
\]
$\mu$ must be a multiple of~$m$, and that the operations $S$, $S^{2}$,~\Dots,
$S^{m - 1}$ are all distinct.

Since the group contains only $N$~distinct operations, $m$~must
be equal to or less than~$N$. It will be seen later that, if
$m$~is less than~$N$, it must be a factor of~$N$.

The integer $m$ is called the \emph{order} of the operation~$S$. The
\index{Order@\Topic{Order}!of an operation, definition of}%
order~$m'$ of the operation~$S^{x}$ is the least integer for which
\[
S^{xm'} = 1,
\]
that is, for which
\[
xm' \equiv 0\ (\mod m).
\]
%% -----File: 031.png---Folio 15-------

Hence, if $g$ is the greatest common factor of $x$ and~$m$,
\[
m' = \frac{m}{g};
\]
and, if $m$~is prime, all the powers of~$S$, whose indices are less
than~$m$, are of order~$m$.

Since
\[
S^{x} S^{m-x} = S^{m} = 1,\quad (x < m),
\]
and
\[
S^{x} (S^{x})_{-1} = 1,
\]
it follows that
\[
(S^{x})_{-1} = S^{m-x}.
\]

If now a meaning be attached to~$S^{-x}$, by assuming that
the equation
\[
S^{x+y} = S^{x} S^{y}
\]
holds when either $x$ or $y$ is a negative integer, then
\[
S^{m-x} = S^{m} S^{-x} = S^{-x},
\]
and
\[
(S^{x})_{-1} = S^{-x},
\]
so that $S^{-x}$ denotes the inverse of the operation~$S^{x}$.

\begin{Ex} If $S_{a}$, $S_{b}$,~\Dots, $S_{c}$, $S_{d}$ are operations of a group, shew that
the operation inverse to $S_{a}^{\alpha} S_{b}^{\beta} \Dots S_{c}^{\gamma} S_{d}^{\delta}$ is $S_{d}^{-\delta} S_{c}^{-\gamma} \Dots S_{b}^{-\beta} S_{a}^{-\alpha}$.
\end{Ex}

\Par{16.} If
\[
1,\ S_{1},\ S_{2},\ \Dots,\ S_{N-1}
\]
are the $N$~operations of a group of order~$N$, the set of $N$~operations
\[
S_{r},\ S_{r}S_{1},\ S_{r}S_{2},\ \Dots,\ S_{r}S_{N-1}
\]
are (§~13) all distinct; and their number is equal to the order of
the group. Hence every operation of the group occurs once
and only once in this set.

Similarly every operation of the group occurs once and only
once in the set
\[
S_{r},\ S_{1}S_{r},\ S_{2}S_{r},\ \Dots,\ S_{N-1}S_{r}.
\]

Every operation of the group can therefore be represented
as the product of two operations of the group, and either the
first factor or the second factor can be chosen at will.

A relation of the form
\[
S_{p} = S_{q} S_{r}
\]
between three operations of the group will not in general
involve any necessary relation between the order of~$S_{p}$ and the
%% -----File: 032.png---Folio 16-------
orders of $S_{q}$ and~$S_{r}$. If however the two latter are permutable,
the relation requires that, for all values of~$x$,
\[
S_{p}^{x} = S_{q}^{x} S_{r}^{x};
\]
and in that case the order of~$S_{p}$ is the least common multiple
of the orders of $S_{q}$ and~$S_{r}$.

Suppose now that $S$, an operation of the group, is of order~$mn$,
where $m$~and~$n$ are relatively prime. Then we may shew
that, of the various ways in which $S$~may be represented as the
product of two operations of the group, there is just one in
which the operations are permutable and of orders $m$ and $n$
respectively.

Thus let
\[
S^{n} = M,
\]
and
\[
S^{m} = N,
\]
so that $M$,~$N$ are operations of orders $m$ and~$n$. Since $S^{m}$ and
$S^{n}$ are permutable, so also are $M$ and~$N$, and powers of $M$ and~$N$.

If $x_{0}$,~$y_{0}$ are integers satisfying the equation
\[
xn + ym = 1,
\]
every other integral solution is given by
\[
x = x_{0} + tm,\quad y = y_{0} - tn,
\]
where $t$ is an integer.

Now
\[
M^{x} N^{y} = S^{xn + ym} = S;
\]
and since $x$~and~$m$ are relatively prime, as also are $y$~and~$n$,
$M^{x}$~and $N^{y}$ are permutable operations of orders $m$~and~$n$, so
that $S$~is expressed in the desired form.

Moreover, it is the only expression of this form; for let
\[
S = M_{1} N_{1},
\]
where $M_{1}$ and $N_{1}$ are permutable and of orders $m$ and~$n$.

Then $S^{n} = M_{1}^{n}$, since $N_{1}^{n} = 1$.

Hence
\[
M_{1}^{n} = M,
\]
or
\[
M_{1}^{xn} = M^{x},
\]
or
\[
M_{1}^{1-ym} = M^{x}.
\]
But $M_{1}^{m} = 1$, and therefore $M_{1}^{-ym} = 1$; hence
\[
M_{1} = M^{x}.
\]
In the same way it is shewn that $N_{1}$ is the same as~$N^{y}$. The
representation of~$S$ in the desired form is therefore unique.
%% -----File: 033.png---Folio 17-------

\Par{17.} Two given operations of a group successively performed
give rise to a third operation of the group which, when the
operations are of known concrete form, may be determined by
actually carrying out the two given operations. Thus the set
of finite rotations, which bring a regular solid to coincidence
with itself, evidently form a group; and it is a purely geometrical
problem to determine that particular rotation of the
group which arises from the successive performance of two
given rotations of the group.

When the operations are represented by symbols, the relation
in question is represented by an equation of the form
\[
AB = C;
\]
but the equation indicates nothing of the nature of the actual
operations. Now it may happen, when the operations of two
groups of equal order are represented by symbols,
\begin{alignat*}{2}
&\Inum{(i)}       && 1,\ A,\ B,\ C, \Dots \\
&\Inum{(ii)}\quad && 1,\ A',\ B',\ C', \Dots
\end{alignat*}
that, to every relation of the form
\[
AB = C
\]
between operations of the first group, there corresponds the
relation
\[
A'B' = C'
\]
between operations of the second group. In such a case,
although the nature of the actual operations in the first group
may be entirely different from the nature of those in the
second, the laws according to which the operations of each
group combine among themselves are identical. The following
series of groups of operations, of order six, will at once illustrate
the possibility just mentioned, and will serve as concrete examples
to familiarize the reader with the conception of a group of
operations.

\Example{\Inum{I.} Group of inversions.} Let $P$, $Q$, $R$ be three circles with
a common radical axis and let each pair of them intersect at an
angle~$\frac{1}{3} \pi$. Denote the operations of inversion with respect to
$P$,~$Q$,~$R$ by $C$,~$D$,~$E$; and denote successive inversions at $P$,~$R$
and at $P$,~$Q$ by $A$~and~$B$. The object of operation may be any
point in the plane of the circles, except the two common points
%% -----File: 034.png---Folio 18-------
in which they intersect. Then it is easy to verify, from the
geometrical properties of inversion, that the operations
\[
1,\ A,\ B,\ C,\ D,\ E
\]
are all distinct, and that they form a group. For instance, $DE$~represents
successive inversions at $Q$~and~$R$. But successive
inversions at $Q$ and $R$ produce the same displacement of points
as successive inversions at $P$ and~$Q$, and therefore
\[
DE = B.
\]

\Example{\Inum{II.} Group of rotations.} Let $POP'$, $QOQ'$, $ROR'$ be three
concurrent lines in a plane such that each of the angles $POQ$
and $QOR$ is~$\frac{1}{3} \pi$, and let $IOI'$ be a perpendicular to their plane.
Denote by~$A$ a rotation round~$II'$ through~$\frac{2}{3} \pi$ bringing $PP'$ to~$RR'$;
and by $B$ a rotation round~$II'$ through~$\frac{4}{3} \pi$ bringing $PP'$
to~$Q'Q$. Denote also by $C$,~$D$,~$E$ rotations through two right
angles round $PP'$, $QQ'$,~$RR'$. The object of the rotations may
be any point or set of points in space. Then it may again be
verified, by simple geometrical considerations, that the operations
\[
1,\ A,\ B,\ C,\ D,\ E
\]
are distinct and that they form a group.

\Example{\Inum{III.} Group of linear transformations of a single variable.}
The operation of replacing $x$ by a given function~$f(x)$ of itself
is sometimes represented by the symbol $\bigl(x, f(x)\bigr)$. With this
notation, if
\begin{alignat*}{4}
 A &= \left(x, \frac{1}{1 - x} \right),\quad
&B &= \left(x, \frac{x - 1}{x} \right),\quad
&C &= \left(x, \frac{1}{x} \right),\quad
&D &= (x, 1 - x), \\
%
&&
 E &= \left(x, \frac{x}{x - 1} \right),\quad
&1 &= (x, x),
\end{alignat*}
it may again be verified without difficulty that these six
operations form a group.

\Example{\Inum{IV.} Group of linear transformations of two variables.}
With a similar notation, the six operations
\begin{alignat*}{3}
 A &= \left(x, \frac{y}{x};\ y, \frac{1}{x} \right),\quad
&B &= \left(x, \frac{1}{y};\ y, \frac{x}{y} \right),\quad
&C &= (x, y;\ y, x), \\
%
 D &= \left(x, \frac{1}{x};\ y, \frac{y}{x} \right),\quad
&E &= \left(x, \frac{x}{y};\ y, \frac{1}{y} \right),\quad
&1 &= (x, y;\ x, y)
\end{alignat*}
form a group.
%% -----File: 035.png---Folio 19-------

\Example{\Inum{V.} Group of linear transformations to a prime modulus.}

The six operations defined by
\begin{alignat*}{3}
A &= (x, x + 1),       & B &= (x, x + 2),       & C &= (x, 2x), \\
D &= (x, 2x + 2),\quad & E &= (x, 2x + 1),\quad & 1 &= (x, x),
\end{alignat*}
where each transformation is taken to modulus~$3$, form a group.

\Example{\Inum{VI.} Group of substitutions of $3$~symbols.} The six substitutions
\[
1,\quad A = (xyz),\quad B = (xzy),\quad C = x(yz),\quad D = y(zx),\quad E = z(xy)
\]
are the only substitutions that can be formed with three symbols;
they must therefore form a group.

\Example{\Inum{VII.} Group of substitutions of $6$~symbols.} The substitutions
\begin{gather*}
1,\quad A = (xyz)(abc),\quad B = (xzy)(acb),\quad C = (xa)(yc)(zb), \\
D = \DPtypo{xb(ya)(zc)}{(xb)(ya)(zc)},\quad E = (xc)(yb)(za)
\end{gather*}
may be verified to form a group.

\Example{\Inum{VIII.} Group of substitutions of $6$~symbols.} The substitutions
\begin{gather*}
1,\quad A = (xaybzc),\quad B = (xyz)(abc),\quad C = (xb)(yc)(za), \\
D = (xzy)(acb),\quad E = (xczbya)
\end{gather*}
form a group.

The operations in the first seven of these groups, as well as
the objects of operation, are quite different from one group to
another; but it may be shewn that the laws according to which
the operations, denoted by the same letters in the different
groups, combine together are identical for all seven. There is
no difficulty in verifying that in each instance
\[
A^{3} = 1,\quad C^{2} = 1,\quad B = A^{2},\quad D = AC = CA^{2},\quad E = A^{2}C = CA;
\]
and from these relations the complete system, according to
which the six operations in each of the seven groups combine
together, may be at once constructed. This is given by the
following multiplication table, where the left-hand vertical
column gives the first factor and the top horizontal line the
%% -----File: 036.png---Folio 20-------
\index{Group@\Topic{Group}!multiplication table of}%
\index{Multiplication table@\Topic{Multiplication table} of a group}%
second factor in each product; thus the table is to be read
$A1 = A$, $AB = 1$, $AC = D$, and so on.
\[
\begin{array}{c|*{6}{c}}
\multicolumn{1}{c}{}
  & 1 & A & B & C & D & E \\
\cline{2-7}\Strut
1 & 1 & A & B & C & D & E \\
A & A & B & 1 & D & E & C \\
B & B & 1 & A & E & C & D \\
C & C & E & D & 1 & B & A \\
D & D & C & E & A & 1 & B \\
E & E & D & C & B & A & 1
\end{array}
\]

But, though the operations of the seventh and eighth
groups are of the same nature and though the operands are
identical, the laws according to which the six operations
combine together are quite distinct for the two groups. Thus,
for the last group, it may be shewn that
\[
B = A^{2},\quad C = A^{3},\quad D = A^{4},\quad E = A^{5},\quad A^{6} = 1,
\]
so that the operations of this group may, in fact, be represented
by
\[
1,\ A,\ A^{2},\ A^{3},\ A^{4},\ A^{5}.
\]

\Par{18.} If we pay no attention to the nature of the actual
operations and operands, and consider only the number of the
former and the laws according to which they combine, the first
seven groups of the preceding paragraph are identical with
each other. From this point of view a group, abstractly
considered, is completely defined by its multiplication table;
and, conversely, the multiplication table must implicitly
contain all properties of the group which are independent of
any special mode of representation.

It is of course obvious that this table cannot be arbitrarily
constructed. Thus, if
\[
AB = P \quad\text{and}\quad BC = Q,
\]
the entry in the table for~$PC$ must be the same as that for~$AQ$.
Except in the very simplest cases, the attempt to form a
consistent multiplication table, merely by trial, would be most
laborious.
%% -----File: 037.png---Folio 21-------

The very existence of the table shews that the symbols
\index{Defining relations@\Topic{Defining relations} of a group!definition of}%
\index{Group@\Topic{Group}!defining relations of}%
\index{Group@\Topic{Group}!fundamental or generating operations of}%
denoting the different operations of the group are not all
independent of each other; and since the number of symbols is
finite, it follows that there must exist a set of symbols
$S_{1}$, $S_{2}$,~\dots, $S_{n}$ no one of which can be expressed in terms of the
remainder, while every operation of the group is expressible in
terms of the set. Such a set is called a set of \emph{fundamental} or
\emph{generating} operations of the group. Moreover though no one
of the generating operations can be expressed in terms of the
remainder, there must be relations of the general form
\[
S_{m}^{a} S_{n}^{b} \dots S_{p}^{c} = 1
\]
among them, as otherwise the group would be of infinite order;
and the number of these relations, which are independent
of one another, must be finite. Among them there necessarily
occur the relations
\[
S_{1}^{a_{1}} = 1,\quad S_{2}^{a_{2}} = 1,\ \Dots,\quad S_{n}^{a_{n}} = 1
\]
giving the orders of the fundamental operations.

We thus arrive at a virtually new conception of a group;
it can be regarded as arising from a finite number of fundamental
operations connected by a finite number of independent
relations. But it is to be noted that there is no reason for
supposing that such an origin for a group is unique; indeed,
in general, it is not so. Thus there is no difficulty in verifying
that the group, whose multiplication table is given in §~17, is
completely specified either by the system of relations
\[
A^{3} = 1,\quad C^{2} = 1,\quad (AC)^{2} = 1,
\]
or by the system
\[
C^{2} = 1,\quad D^{2} = 1,\quad (CD)^{3} = 1.
\]

In other words, it may be generated by two operations of
orders $2$~and~$3$, or by two operations of order~$2$. So also the
last group of §~17 is specified either by
\[
A^{6} = 1,
\]
or by
\[
B^{3} = 1,\quad C^{2} = 1,\quad BC = CB.
\]

\Par{19.} \begin{Definition}
Let $G$ and $G'$ be two groups of equal
order. If a correspondence can be established between the
operations of $G$~and~$G'$, so that to every operation of~$G$ there
%% -----File: 038.png---Folio 22-------
\index{Simply isomorphic groups@\Topic{Simply isomorphic groups}!definition of}%
\index{Simply isomorphic groups@\Topic{Simply isomorphic groups}!said to be of the same type}%
corresponds a single operation of~$G'$ and to every operation of~$G'$
there corresponds a single operation of~$G$, while to the
product~$AB$ of any two operations of~$G$ there corresponds the
product~$A'B'$ of the two corresponding operations of~$G$', the
groups $G$ and $G'$ are said to be \emph{simply isomorphic}\footnotemark.
\end{Definition}
\footnotetext{We shall sometimes use the phrase that two groups are of the same \emph{type}
  to denote that they are simply isomorphic.}

Two simply isomorphic groups are, abstractly considered,
identical. In discussing the properties of groups, some definite
mode of representation is, in general, indispensable; and as
long as we are dealing with the properties of a group \textit{per~se},
and not with properties which depend on the form of
representation, the group may, if convenient, be replaced by
any group which is simply isomorphic with it. For the discussion
of such properties, it would be most natural to suppose
the group given either by its multiplication table or by its
fundamental operations and the relations connecting them;
and as far as possible we shall follow this course. Unfortunately,
however, these purely abstract modes of representing a
group are by no means the easiest to deal with. It thus
becomes an important question to determine as far as possible
what different concrete forms of representation any particular
group may be capable of; and we shall accordingly end the
present chapter with a demonstration of the following general
theorem bearing on this question.

\Par{20.} \begin{Theorem}
\index{Dyck's theorem@\Topic{Dyck's theorem} that a group of order~$n$ can be represented as a regular substitution group of degree~$n$|(}%
\index{Representation of a group@\Topic{Representation of a group}!in transitive form; \ie\ as a transitive substitution group}%
\index{Substitution group@\Topic{Substitution group}!representation of any group as a regular}%
Every group of finite order~$N$ is capable of
representation as a group of substitutions of $N$~symbols\footnotemark.
\end{Theorem}
\footnotetext{Dyck, ``Gruppentheoretische Studien,'' \textit{Math.\ Ann.}, Vol.~\VolNo{XX} (1882), p.~30.}
\index{Dyck@\Topic{Dyck}, quoted}%

Let
\[
1,\ S_{1},\ S_{2},\ \dots,\ S_{i},\ \dots,\ S_{N - 1}
\]
be the $N$~operations of the group; and form the complete
multiplication table
\[
\begin{array}{*{7}{c}}
1,     & S_{1},      & S_{2},      & \dots, & S_{i},      & \dots, & S_{N - 1}, \\
S_{1}, & S_{1}^{2},  & S_{2}S_{1}, & \dots, & S_{i}S_{1}, & \dots, & S_{N - 1}S_{1}, \\
\hdotsfor{7} \\
S_{i}, & S_{1}S_{i}, & S_{2}S_{i}, & \dots, & S_{i}^{2},  & \dots, & S_{N - 1}S_{i}, \\
\hdotsfor{7} \\
S_{N - 1}, & S_{1}S_{N - 1}, & S_{2}S_{N - 1}, & \dots, & S_{i}S_{N - 1}, & \dots, & S_{N - 1}^{2},
\end{array}
\]
%% -----File: 039.png---Folio 23-------
that results from multiplying the symbols in the first horizontal
line by $1$,~$S_{1}$, $S_{2}$,~\dots, $S_{N-1}$ in order. Each horizontal line in the
table so obtained contains the same $N$ symbols as the original
line, but any given symbol occupies a different place in each
line; for the supposition that the two symbols $S_{i} S_{p}$ and $S_{i} S_{q}$
were identical would involve
\[
S_{p} = S_{q},
\]
which is not true.

It thus appears that the first line in the table, taken
with the $(i + 1)$th line, defines a substitution
\[
\Sub{\PadTo{S_{i}}{1}, \PadTo{S_{1} S_{i}}{S_{1}}, \PadTo{S_{2} S_{i}}{S_{2}}, \dots, \PadTo{S_{N-1} S_{i}}{S_{N-1}}}
    {S_{i}, S_{1} S_{i}, S_{2} S_{i}, \dots, S_{N-1} S_{i}}
\]
performed on the $N$~symbols $1$,~$S_{1}$, $S_{2}$,~\dots, $S_{N-1}$; and by taking
the first line with each of the others, including itself, a set of $N$~substitutions
performed on these $N$~symbols is obtained. Now
this set of substitutions forms a group simply isomorphic
to the given group. For if
\[
S_{p} S_{i} = S_{q} \quad\text{and}\quad S_{q} S_{j} = S_{r},
\]
the substitutions
\[
\Sub{\PadTo{S_{i}}{1}, \PadTo{S_{1} S_{i}}{S_{1}}, \PadTo{S_{2} S_{i}}{S_{2}}, \dots, \PadTo{S_{N-1} S_{i}}{S_{N-1}}}
    {S_{i}, S_{1} S_{i}, S_{2} S_{i}, \dots, S_{N-1} S_{i}}
\]
and
\[
\Sub{\PadTo{S_{i}}{1}, \PadTo{S_{1} S_{i}}{S_{1}}, \PadTo{S_{2} S_{i}}{S_{2}}, \dots, \PadTo{S_{N-1} S_{i}}{S_{N-1}}}
    {S_{j}, S_{1} S_{j}, S_{2} S_{j}, \dots, S_{N-1} S_{j}},
\]
successively performed, change $S_{p}$ into~$S_{r}$.

But
\[
S_{p} S_{i} S_{j} = S_{r},
\]
and therefore the product of the two substitutions, in the order
given, is the substitution
\[
\Sub{\PadTo{S_{i} S_{j}}{1},
     \PadTo{S_{1} S_{i} S_{j}}{S_{1}},
     \PadTo{S_{2} S_{i} S_{j}}{S_{2}}, \dots,
     \PadTo{S_{N-1} S_{i} S_{j}}{S_{N-1}}}
    {S_{i} S_{j}, S_{1} S_{i} S_{j}, S_{2} S_{i} S_{j}, \dots, S_{N-1} S_{i} S_{j}},
\]
or the product of any two of the $N$~substitutions is again one
of the substitutions of the set. Moreover the above reasoning
shews that the result of performing successively the substitutions
corresponding to the operations $S_{i}$~and~$S_{j}$ gives the
substitution corresponding to the operation~$S_{i} S_{j}$. Hence,
since the number of substitutions is equal to the number of
%% -----File: 040.png---Folio 24-------
operations, the given group and the group of substitutions
are simply isomorphic.

It may also be shewn that each of the $N$~substitutions is
\index{Substitution group@\Topic{Substitution group}!regular, definition of}%
regular (§~9) in the $N$~symbols. For the substitution
\[
\Sub{\PadTo{S_{i}}{1}, \PadTo{S_{1} S_{i}}{S_{1}}, \PadTo{S_{2} S_{i}}{S_{2}}, \dots, \PadTo{S_{N-1} S_{i}}{S_{N-1}}}
    {S_{i}, S_{1} S_{i}, S_{2} S_{i}, \dots, S_{N-1} S_{i}}
\]
changes $S_{r}$ into $S_{r} S_{i}$, $S_{r} S_{i}$ into $S_{r} S_{i}^{2}$, and so on. If then $m$~is
the order of~$S_{i}$, the cycle of the substitution which contains~$S_{r}$
will be
\[
(S_{r},\ S_{r} S_{i},\ S_{r} S_{i}^{2},\ \dots,\ S_{r} S_{i}^{m-1}).
\]

Now $S_{r}$ may be any symbol of the set; hence all the cycles
of the substitution must contain the same number,~$m$, of symbols.

The substitution is therefore regular in the $N$~symbols, and
this can only be the case if $m$~is a factor of~$N$. It follows at
once, as was stated in §~15, that the order of any operation of a
group of order~$N$ must be equal to or a factor of~$N$.

All the substitutions in this form of representing a group
being regular, the group itself is said to be expressed as a
regular substitution group.
\index{Dyck's theorem@\Topic{Dyck's theorem} that a group of order~$n$ can be represented as a regular substitution group of degree~$n$|)}%

\Par{21.} The form, in which it has just been shewn that every
group can be represented, is by no means the only form of
representation possessing this property. Thus Hurwitz\footnote
  {``Algebraische Gebilde mit eindeutigen Transformationen in sich,'' \textit{Math.\
  Ann.}, Vol.~\VolNo{XLI} (1893), p.~421.}
\index{Hurwitz@\Topic{Hurwitz}, quoted}%
has
shewn that every group can be expressed as a group of rational
and reversible transformations which change a suitably chosen
algebraic curve (or Riemann's surface) into itself. We shall see
in Chapter~XIV that every group can also be expressed as a
group of linear transformations of a finite set of variables to a
prime modulus.
%% -----File: 041.png---Folio 25-------


\Chapter{III.}{On the Simpler Properties of a Group which
are Independent of its Mode of Representation.}

\Par{22.} \First{In} this chapter we proceed to discuss some of the
simplest of the properties of groups of finite order which are
independent of their mode of representation.

If among the operations of a group~$G$ a certain set can be
\index{Sub-group@\Topic{Sub-group}!definition of}%
chosen which do not exhaust all the operations of the group~$G$,
yet which at the same time satisfy all the conditions of §~12 so
that they form another group~$H$, this group~$H$ is called a \emph{sub-group}
of the group~$G$. Thus if $S$~be any operation, order~$m$,
of~$G$, the operations
\[
1,\ S,\ S^{2},\ \Dots,\ S^{m-1}
\]
evidently form a group; and when the order of~$G$ is greater
than~$m$, this group is a sub-group of~$G$. A sub-group of this
nature, which consists of the different powers of a single
\index{Group@\Topic{Group}!cyclical}%
operation, is called a \emph{cyclical sub-group}; and a group, which
consists of the different powers of a single operation, is called
a \emph{cyclical group}.

\begin{Theorem}[I.]
If $H$ is a sub-group of~$G$, the order~$n$ of~$H$ is
a factor of the order~$N$ of~$G$.
\end{Theorem}

Let
\[
1,\ T_{1},\ T_{2},\ \Dots,\ T_{n-1}
\]
be the $n$~operations of~$H$; and let $S_{1}$ be any operation of~$G$
which is not contained in~$H$.
%% -----File: 042.png---Folio 26-------
\index{Sub-group@\Topic{Sub-group}!order of a, divides order of group}%

Then the operations
\[
S_{1},\ T_{1} S_{1},\ T_{2} S_{1},\ \Dots,\ T_{n-1} S_{1}
\]
are all distinct from each other and from the operations of~$H$.

For if
\[
T_{p} S_{1} = T_{q} S_{1},
\]
then
\[
T_{p} = T_{q},
\]
contrary to supposition; and if
\[
T_{p} = T_{q} S_{1},
\]
then
\[
S_{1} = T_{q}^{-1} T_{p},
\]
and $S_{1}$~would be contained among the operations of~$H$.

If the $2n$~operations thus obtained do not exhaust all the
operations of~$G$, let $S_{2}$ be any operation of~$G$ not contained
among them.

Then it may be shewn, by repeating the previous reasoning,
that the $n$~operations
\[
S_{2},\ T_{1} S_{2},\ T_{2} S_{2},\ \Dots,\ T_{n-1} S_{2}
\]
are all different from each other and from the previous $2n$
operations. If the group~$G$ is still not exhausted, this process
may be repeated; so that finally the $N$~operations of~$G$ can be
exhibited in the form
\[
\begin{array}{*{5}{c}}
1,     & T_{1},       & T_{2},       & \Dots, & T_{n-1}\DPchg{:}{,} \\
S_{1}, & T_{1} S_{1}, & T_{2} S_{1}, & \Dots, & T_{n-1} S_{1}\DPchg{:}{,} \\
S_{2}, & T_{1} S_{2}, & T_{2} S_{2}, & \Dots, & T_{n-1} S_{2}\DPchg{:}{,} \\
\hdotsfor{5} \\
S_{m-1}, & T_{1} S_{m-1}, & T_{2} S_{m-1}, & \Dots, & T_{n-1} S_{m-1}.
\end{array}
\]
Hence $N = mn$, and $n$~is therefore a factor of~$N$.

\begin{Remark}
When $N$~is a prime~$p$, the group~$G$ can have no sub-group other
than one of order unity consisting of the identical operation alone.
Every operation~$S$ of the group, other than the identical operation,
is of order~$p$, and the group consists of the operations
\[
1,\ S,\ S^{2},\ \Dots,\ S^{p-1}.
\]
A group whose order is prime is therefore necessarily cyclical.
\end{Remark}

\Par{23.} \begin{Theorem}[II.]
The operations common to two groups
$G_{1}$,~$G_{2}$ themselves form a group~$g$, whose order is a factor of the
orders of $G_{1}$ and~$G_{2}$.
\end{Theorem}
%% -----File: 043.png---Folio 27-------

For if $S$,~$T$ are any two operations common to $G_{1}$ and~$G_{2}$,
\index{Group@\Topic{Group}!symbol for a}%
\index{Operations@\Topic{Operations}!common to two groups form a group}%
\index{Symbol@\Topic{Symbol}!for a group generated by given operations}%
$ST$~is also common to both groups; and hence the common
operations satisfy conditions \Inum{($\alpha$)}~and~\Inum{($\beta$)} of the definition in §~12.
But their orders are finite and they must therefore satisfy also
condition~\Inum{($\gamma$)}, and form a group~$g$. Moreover $g$~is a sub-group
of both $G_{1}$ and~$G_{2}$, and therefore by Theorem~I its order is a
factor of the orders of both these groups.

If $G_{1}$ and $G_{2}$ are sub-groups of a third group~$G$, then $g$~is
also clearly a sub-group of~$G$.

The set of operations, that arise by combining in every way
the operations of the groups $G_{1}$~and~$G_{2}$, evidently satisfy the
conditions of §~12 and form a group; but this will not necessarily
or generally be a group of finite order. If however $G_{1}$ and $G_{2}$
are sub-groups of a group~$G$ of finite order, the group~$g'$ that
arises from their combination will necessarily be of finite order;
it may either coincide with~$G$ or be a sub-group of~$G$. In
either case, the order of~$g'$ will be a multiple of the orders of
$G_{1}$~and~$G_{2}$.

It is convenient here to explain a notation that enables
us to avoid an otherwise rather cumbrous phraseology. Let
$S_{1}$,~$S_{2}$, $S_{3}$,~\Dots\ be a given set of operations, and $G_{1}$, $G_{2}$,~\Dots\ a
set of groups. Then the symbol
\[
\{S_{1}, S_{2}, S_{3}, \Dots, G_{1}, G_{2}, \Dots\}
\]
will be used to denote the group that arises by combining in
every possible way the given operations and the operations of
the given groups.

Thus, for instance, the group~$g'$ above would be represented
by
\[
\{G_{1}, G_{2}\};
\]
the cyclical group that arises from the powers of an operation~$S$
by
\[
\{S\};
\]
and, as a further example, the sixth group of §~17 may be
represented by
\[
\{(xy), (xz)\}.
\]

\Par{24.} \begin{Definition}
If $S$ and $T$ are any two operations of a
\index{Conjugate operations@\Topic{Conjugate operations}!definition of}%
\index{Transforming an operation@\Topic{Transforming an operation}!definition of}%
group, the operations $S$ and $T^{-1}ST$ are called \emph{conjugate} operations;
while $T^{-1}ST$~is spoken of as the result of \emph{transforming}
the operation~$S$ by~$T$.
\end{Definition}
%% -----File: 044.png---Folio 28-------

The two operations $S$ and $T^{-1}ST$ are identical only when $S$
and $T$ are permutable. For if
\[
S = T^{-1}ST,
\]
then
\[
TS = ST.
\]

Two conjugate operations are always of the same order.
For
\begin{align*}
(T^{-1}ST)^{n}
  &= T^{-1}ST · T^{-1}ST \Dots T^{-1}ST \\
  &= T^{-1}S^{n}T.
\end{align*}

Therefore, if
\begin{align*}
S^{n} &= 1, \\
(T^{-1}ST)^{n} &= T^{-1}T = 1;
\end{align*}
and conversely, if
\[
(T^{-1}ST)^{n} = 1,
\]
then
\[
S^{n} = T · T^{-1} S^{n}T · T^{-1} = T(T^{-1}ST)^{n} T^{-1} = TT^{-1} = 1.
\]

The operations $ST$ and $TS$ are always conjugate and
therefore of the same order; for
\[
ST = T^{-1}T · ST = T^{-1} · TS · T.
\]

\begin{Ex} Shew that the operations $S_{1}S_{2}\Dots S_{n - 1}S_{n}$ and $S_{r}S_{r + 1}\Dots$
$S_{n}S_{1}\Dots S_{r - 1}$ are conjugate within the group $\{S_{1}, S_{2}, \Dots, S_{n}\}$.
\end{Ex}

\begin{Definition}
An operation~$S$ of a group~$G$, which is
\index{Self-conjugate operation@\Topic{Self-conjugate operation}!definition of}%
identical with all its conjugate operations, is called a \emph{self-conjugate}
operation. Such an operation must evidently be
permutable with each of the operations of~$G$.
\end{Definition}

\begin{Remark}
In every group the identical operation is self-conjugate; and in
a group, whose operations are all permutable, every operation is
self-conjugate. A simple example of a group, which contains self-conjugate
operations other than the identical operation, while at the
same time its operations are not all self-conjugate, is given by
\[
\{(1234), (13)\}.
\]
It is easy to shew that the order of this group is~$8$, and that $(13)(24)$
is a self-conjugate operation.
\end{Remark}

If all the operations of a group be transformed by a given
operation, the set of transformed operations form a group. For
if $T_{1}$ and $T_{2}$ are any two operations of the group, so that $T_{1}T_{2}$~is
also an operation of the group, then
\[
S^{-1} T_{1}S · S^{-1} T_{2}S = S^{-1} T_{1} T_{2}S;
\]
%% -----File: 045.png---Folio 29-------
hence the product of any two operations of the transformed
set is another operation belonging to the transformed set, and
the set therefore forms a group. Moreover the preceding
equation shews that the new group is simply isomorphic
to the original group. If $G$ is the given group, the symbol
$S^{-1}GS$ will be used for the new group. When $S$~belongs to the
group~$G$, the groups $G$ and $S^{-1}GS$ are evidently the same.

Now unless $S$ is a self-conjugate operation of~$G$, the pairs of
operations $T$ and $S^{-1}TS$ will not all be identical when for $T$ the
different operations of~$G$ are put in succession. Hence the
process of transforming all the operations of a group by one of
themselves is equivalent to establishing a correspondence
between the operations of the group, which exhibits it as
simply isomorphic with itself.

\begin{Definitions}
When $H$~is a sub-group of~$G$ and $S$~is any
\index{Composite groups@\Topic{Composite groups}!definition of}%
\index{Conjugate sub-groups@\Topic{Conjugate sub-groups}!definition of}%
\index{Self-conjugate sub-group@\Topic{Self-conjugate sub-group}!definition of}%
operation of~$G$, the groups $H$ and $S^{-1}HS$ are called \emph{conjugate}
sub-groups of~$G$.

If $H$ and $S^{-1}HS$ are identical, $S$~is said to be permutable
with the sub-group~$H$. This does not necessarily involve that
$S$~is permutable with each of the operations of~$H$.

If $H$ and $S^{-1}HS$ are identical, whatever operation $S$ is of~$G$,
$H$~is said to be a \emph{self-conjugate} sub-group of~$G$.

A group is called \emph{composite} or \emph{simple}, according as it does
\index{Group@\Topic{Group}!simple and composite}%
\index{Simple groups@\Topic{Simple groups}!definition of}%
or does not possess at least one self-conjugate sub-group other
than that formed of the identical operation alone.
\end{Definitions}

\Par{25.} \begin{Theorem}[III.]
The operations of a group~$G$, which
are permutable with a given operation~$T$, form a sub-group~$H$;
and the order of~$G$ divided by the order of~$H$ is the number of
operations conjugate to~$T$\footnotemark.
\end{Theorem}
\footnotetext{Among these $T$ of course occurs.}

If $R_{1}$ and $R_{2}$ are any two operations permutable with~$T$, so
\index{Operations@\Topic{Operations}!of a group, which are permutable with a given operation or sub-group, form a group}%
\index{Permutability@\Topic{Permutability} of an operation with a group!definition of}%
that
\[
R_{1}T = TR_{1} \quad\text{and}\quad R_{2}T = TR_{2};
\]
then
\[
R_{1} R_{2}T = R_{1} TR_{2} = TR_{1} R_{2},
\]
and therefore $R_{1}R_{2}$ is permutable with~$T$. The operations
permutable with~$T$ therefore form a group~$H$. Let $n$ be its
order and
\[
1,\ R_{1},\ R_{2},\ \Dots,\ R_{n - 1}
\]
%% -----File: 046.png---Folio 30-------
its operations. Then if $S$~is any operation of~$G$ not contained
in~$H$, the operations
\[
S,\ R_{1}S,\ R_{2}S,\ \Dots,\ R_{n - 1}S
\]
all transform $T$ into the same operation~$T'$.

For
\[
(R_{i}S)^{-1} TR_{i}S = S^{-1} R_{i}^{-1} TR_{i}S = S^{-1}TS.
\]

Also the $n$~operations thus obtained are the only operations
which transform $T$ into~$T'$; for if
\[
S'^{-1}TS' = T',
\]
then
\[
SS'^{-1} TS'S^{-1} = ST'S^{-1} = T;
\]
and therefore $S'S^{-1}$ belongs to~$H$. The number of operations
which transform $T$ into any operation conjugate to it is
therefore equal to the number that transform $T$ into itself, that
is, to the order of~$H$. If then $N$~is the order of~$G$, the
operations of~$G$ may be divided into $\dfrac{N}{n}$~sets of $n$~each, such
that the operations of each set transform $T$ into a distinct
operation, those of the first set, namely the operations of~$H$,
transforming $T$ into itself. The number of operations conjugate
to~$T$, including itself, is therefore~$\dfrac{N}{n}$.

Since
\[
T = ST'S^{-1},
\]
therefore
\[
R_{i}^{-1} TR_{i} = R_{i}^{-1} ST' S^{-1}R_{i};
\]
hence
\[
T'= S^{-1}TS = S^{-1} R_{i}^{-1} TR_{i}S = S^{-1} R_{i}^{-1} S · T' · S^{-1}R_{i}S,
\]
so that every operation of the form $S^{-1}R_{i}S$ is permutable with~$T'$.
Hence if $H$~is the group of operations permutable with~$T$,
and if
\[
S^{-1}TS = T',
\]
then $S^{-1}HS$ is the group of operations permutable with~$T'$.

It is convenient to have a symbol to represent the set of
operations
\[
S,\ R_{1}S,\ R_{2}S,\ \dots,\ R_{n - 1}S,
\]
where
\[
1,\ R_{1},\ R_{2},\ \dots,\ R_{n - 1}
\]
form a group~$H$. We shall in future represent this set of
operations by~$HS$; and we shall use $SH$ to represent the set
\[
S,\ SR_{1},\ SR_{2},\ \dots,\ SR_{n - 1}.
\]
%% -----File: 047.png---Folio 31-------

\begin{Theorem}[IV.]
The operations of a group~$G$ which are
permutable with a sub-group~$H$ form a sub-group~$I$, which is
either identical with~$H$ or contains~$H$ as a self-conjugate sub-group.
The order of~$G$ divided by the order of~$I$ is the number
of sub-groups conjugate to~$H$\footnotemark.
\end{Theorem}
\footnotetext{Among these sub-groups $H$~itself occurs.}

If $S_{1}$,~$S_{2}$ are any two operations of~$G$ which are permutable
\index{Operations@\Topic{Operations}!of a group, which are permutable with a given operation or sub-group, form a group}%
with~$H$, then
\[
S_{1}^{-1} HS_{1} = H,\quad S_{2}^{-1} HS_{2} = H,
\]
and therefore
\[
S_{2}^{-1} S_{1}^{-1} HS_{1} S_{2} = H,
\]
so that $S_{1}S_{2}$ is permutable with~$H$. The operations of~$G$
which are permutable with~$H$ therefore form a group~$I$, which
may be identical with~$H$ and, if not identical with~$H$, must
contain it. Also, if $S$~is any operation of~$I$,
\[
S^{-1}HS = H,
\]
and therefore $H$ is a self-conjugate sub-group of~$I$.

If now $\Sigma$ is any operation of~$G$ not contained in~$I$, it may
be shewn, exactly as in the proof of Theorem~III, that the
operations $I\Sigma$ and no others transform $H$ into a conjugate
sub-group~$H'$ which is not identical with~$H$; and therefore
that the number of sub-groups in the conjugate set to which
$H$~belongs is the order of~$G$ divided by the order of~$I$.

The operations of~$G$ which are permutable with~$H'$ may
also be shewn to form the group~$\Sigma^{-1}I\Sigma$.

It is perhaps not superfluous to point out that two
distinct conjugate sub-groups may have some operations in
common with one another.

\Par{26.} Let $S_{1}$ be any operation of~$G$, and
\[
S_{1},\ S_{2},\ \Dots,\ S_{m}
\]
the distinct operations obtained on transforming $S_{1}$ by every
operation of~$G$. The number,~$m$, of these operations is, by
Theorem~III, a factor of~$N$, the order of~$G$. Moreover if,
instead of transforming~$S_{1}$, we transform any other operation of
the set,~$S_{r}$, by every operation of~$G$, the same set of $m$~distinct
operations of~$G$ will result. Such a set of operations we call a
\index{Conjugate operations@\Topic{Conjugate operations}!complete set of}%
\emph{complete set} of conjugate operations. If $T$~is any operation of~$G$
%% -----File: 048.png---Folio 32-------
\index{Conjugate sub-groups@\Topic{Conjugate sub-groups}!complete set of}%
\index{Isomorphism of two groups@\Topic{Isomorphism of two groups}!simple, definition of}%
which does not belong to this complete set of conjugate
operations, no operation that is conjugate to~$T$ can belong to
the set. Hence the operations of $G$ may be distributed into a
number of distinct sets such that every operation belongs to
one set and no operation belongs to more than one set; while
any set forms by itself a complete set of conjugate operations.
If $m_{1}$, $m_{2}$,~\Dots, $m_{s}$ are the numbers of operations in the different
sets, then
\[
N = m_{1} + m_{2} + \Dots + m_{s};
\]
and, since the identical operation is self-conjugate, one at least
\PageLabel{32}%
of the $m$'s must be unity.

Similarly, if $H_{1}$ is any sub-group of~$G$, and
\[
H_{1},\ H_{2},\ \Dots,\ H_{p}
\]
the distinct sub-groups obtained on transforming $H_{1}$ by every
operation of~$G$, we call the set a \emph{complete set} of conjugate
sub-groups. If $K$~is a sub-group of~$G$ not contained in the set,
no sub-group conjugate to~$K$ can belong to the set. If the
operation~$S_{1}$ belongs to one or more of a complete set of
conjugate sub-groups, $\Sigma^{-1} S_{1}\Sigma$~must also belong to one or more
sub-groups of the set, $\Sigma$~being any operation of~$G$. Hence
among the operations contained in the complete set of conjugate
sub-groups, the complete set of conjugate operations
\[
S_{1},\ S_{2},\ \Dots,\ S_{m}
\]
occurs.

No sub-group of~$G$ can contain operations belonging to
every one of the complete sets of conjugate operations of~$G$.
For if such a sub-group~$H$ existed, the complete set of
conjugate sub-groups, to which $H$ belongs, would contain all
the operations of~$G$. Let $m$ be the order of~$H$ and $n$~($\geq m$) the
order of the sub-group~$I$ formed of those operations of~$G$ which
are permutable with~$H$. Then $H$~is one of $\dfrac{N}{n}$~conjugate sub-groups,
each of which contains $m$~operations. The identical
operation is common to all these sub-groups, and they therefore
cannot contain more than
\[
1 + \frac{N(m - 1)}{n}
\]
%% -----File: 049.png---Folio 33-------
distinct operations in all. This number is less than~$N$, and
therefore the complete set of conjugate sub-groups cannot
contain all the operations of~$G$.

\Par{27.} If a group contains self-conjugate operations, it must
\index{Conjugate sub-groups@\Topic{Conjugate sub-groups}!operations common to or permutable with a complete set of, form a self-conjugate sub-group}%
\index{Operations@\Topic{Operations}!common to or permutable with each of a complete set of conjugate sub-groups form a self-conjugate sub-group}%
contain self-conjugate sub-groups. For the cyclical sub-group
generated by any self-conjugate operation must be self-conjugate.
The only exception to this statement is the case of
the cyclical groups of prime order. Every operation of such a
group is clearly self-conjugate; but since the cyclical sub-group
generated by any operation coincides with the group itself,
there can be no self-conjugate sub-group\footnotemark.
\footnotetext{Strictly speaking, this statement should be qualified by the addition
  ``except that formed by the identical operation alone.'' No real ambiguity
  however will be introduced by always leaving this exception unexpressed.}

If every operation of a group is not self-conjugate, or, in
other words, if the operations of a group are not all permutable
with each other, the totality of the self-conjugate operations
forms a self-conjugate sub-group. For, if $S_{1}$~and~$S_{2}$ are
permutable with every operation of the group, so also is~$S_{1}S_{2}$.

\begin{Theorem}[V.]
The operations common to a complete set of
conjugate sub-groups form a self-conjugate sub-group.
\end{Theorem}

It is an immediate consequence of Theorem~II that the
operations common to a complete set of conjugate sub-groups
form a sub-group. Also the set of conjugate sub-groups, when
transformed by any operation of the group, is changed into
itself. Hence their common sub-group must be self-conjugate.

It may of course happen that the identical operation is the
only one which is common to every sub-group of the set.

\begin{Corollary}
The operations permutable with each of a
complete set of conjugate sub-groups form a self-conjugate
sub-group.
\end{Corollary}

For, if the operations permutable with the sub-group~$H$
form a sub-group~$I$, the operations permutable with every
sub-group of the conjugate set to which $H$ belongs are the
operations common to every sub-group of the conjugate set to
which $I$ belongs.
%% -----File: 050.png---Folio 34-------

Further, the operations which are permutable with every
operation of a complete set of conjugate sub-groups form a
self-conjugate sub-group.

\begin{Theorem}[VI.]
If $T_{1}$, $T_{2}$,~\dots, $T_{r}$ are a complete set of conjugate
\index{Self-conjugate sub-group@\Topic{Self-conjugate sub-group}!generated by a complete set of conjugate operations}%
operations of~$G$, the group $\{T_{1}, T_{2}, \dots, T_{r}\}$, if it does not coincide
with~$G$, is a self-conjugate sub-group of~$G$; and it is the self-conjugate
sub-group of smallest order that contains~$T_{1}$.
\end{Theorem}

Since the operations $T_{1}$, $T_{2}$,~\dots, $T_{r}$ are merely rearranged
in a new sequence when the set is transformed by any operation
of~$G$, it follows that
\[
S^{-1} \{T_{1}, T_{2}, \dots, T_{r}\}S = \{T_{1}, T_{2}, \dots, T_{r}\},
\]
whatever operation of~$G$ may be represented by~$S$. Hence
$\{T_{1}, T_{2}, \dots, T_{r}\}$ is a self-conjugate sub-group. Also any self-conjugate
sub-group of~$G$ that contains~$T_{1}$ must contain
$T_{2}$, $T_{3}$,~\dots, $T_{r}$; and therefore any self-conjugate sub-group of~$G$
which contains~$T_{1}$ must contain $\{T_{1}, T_{2}, \dots, T_{r}\}$.

In exactly the same way it may be shewn that, if
\[
H_{1},\ H_{2},\ \dots,\ H_{s}
\]
{\Loosen are a complete set of conjugate sub-groups of~$G$, the group
$\{H_{1}, H_{2}, \dots, H_{s}\}$,} if it does not coincide with~$G$, is the smallest
self-conjugate sub-group of~$G$ which contains the sub-group~$H_{1}$.

The theorem just proved suggests a process for determining
whether any given group is simple or composite. To this end,
the groups $\{T_{1}, T_{2}, \dots, T_{r}\}$ corresponding to each set of conjugate
operations in the group are formed. If any one of them
differs from the group itself, it is a self-conjugate sub-group
and the group is composite; but if each group so formed
coincides with the original group, the latter is simple. If
the order of~$T_{1}$ contains more than one prime factor and
if $T_{1}^{m}$~is of prime order, it is easy to see that the distinct
operations of the set $T_{1}^{m}$, $T_{2}^{m}$,~\dots, $T_{r}^{m}$ form a complete set of
conjugate operations, and that the group $\{T_{1}, T_{2}, \dots, T_{r}\}$ contains
the group $\{T_{1}^{m}, T_{2}^{m}, \dots, T_{r}^{m}\}$. Hence practically it is sufficient
to form the groups $\{T_{1}, T_{2}, \dots, T_{r}\}$ for all conjugate sets of operations
whose orders are prime.

\begin{Remark}
With the notation of §~26 (\PageRef{p.}{32}), the order of any self-conjugate
sub-group of~$G$ must be of the form $m_{\alpha} + m_{\beta} + \Dots$;
%% -----File: 051.png---Folio 35-------
for if the sub-group contains any given operation, it must contain all
the operations conjugate with it. Moreover one at least of the
numbers $m_{\alpha}$, $m_{\beta}$,~\Dots\ must be unity, since the sub-group must
contain the identical operation. It may happen that the numbers~$m_{r}$
are such that the only factors of~$N$ of the form $m_{\alpha} + m_{\beta} + \Dots$,
one of the~$m$'s being unity, are $N$~itself and unity. When this is
\PageLabel{35}%
the case, $G$~is necessarily a simple group. It must not however
be inferred that, if $N$~has factors of this form, other than $N$~itself
and unity, then $G$~is necessarily composite.
\end{Remark}

If $G_{1}$ and $G_{2}$ are sub-groups of~$G$, it has already been seen
(§~23) that the operations common to $G_{1}$ and $G_{2}$ form a sub-group~$g$
of~$G$; and it is now obvious that, when $G_{1}$ and $G_{2}$ are
self-conjugate sub-groups, so also is~$g$. Moreover the group
$\{G_{1}, G_{2} \}$ is a self-conjugate sub-group unless it coincides with~$G$.
For
\[
S^{-1} \{G_{1}, G_{2}\} S = \{S^{-1}G_{1}S, S^{-1}G_{2}S\} = \{G_{1}, G_{2}\}.
\]

Again, with the same notation, if $T_{1}$ is an operation of~$G$
not contained in the self-conjugate sub-group~$G_{1}$, and if
$T_{1}$, $T_{2}$,~\dots, $T_{r}$ is a complete set of conjugate operations, the
group $\{G_{1}, T_{1}, T_{2}, \dots, T_{r}\}$ is a self-conjugate sub-group, unless
it coincides with~$G$.

\begin{Definitions}
If $G_{1}$, a self-conjugate sub-group of~$G$, is
\index{Maximum sub-group@\Topic{Maximum sub-group}, definition of}%
\index{Maximum self-conjugate sub-group@\Topic{Maximum self-conjugate sub-group}!definition of}%
such that the group
\[
\{G_{1}, T_{1}, T_{2}, \dots, T_{r} \}
\]
coincides with~$G$, when $T_{1}$, $T{2}$,~\dots, $T_{r}$ is any complete set of
conjugate operations not contained in~$G_{1}$, then $G_{1}$~is said to be
a \emph{maximum} self-conjugate sub-group of~$G$. This does not
imply that $G_{1}$~is the self-conjugate sub-group of~$G$ of absolutely
greatest order; but that there is no self-conjugate sub-group
of~$G$, distinct from $G$~itself, which contains $G_{1}$ and is of
greater order than~$G_{1}$.

If $H$ is a sub-group of~$G$, and if, for every operation~$S$ of~$G$
which does not belong to~$H$, the group $\{H, S\}$ coincides with~$G$,
$H$~is said to be a \emph{maximum} sub-group of~$G$.
\end{Definitions}

\Par{28.}
\begin{Definition}
When a correspondence can be established
between the operations of a group~$G$ and the operations
of a group~$G'$, whose order is smaller than the order of~$G$, such
that to each operation~$S$ of~$G$ there corresponds a single
%% -----File: 052.png---Folio 36-------
\index{Isomorphism of two groups@\Topic{Isomorphism of two groups}!multiple, definition of}%
\index{Multiply isomorphic groups@\Topic{Multiply isomorphic groups}!definition of}%
operation~$S'$ of~$G'$, while to the operation $S_{p}S_{q}$ there corresponds
the operation $S_{p}'S_{q}'$, the group~$G$ is said to be \emph{multiply isomorphic}
with the group~$G'$.
\end{Definition}

\begin{Theorem}[VII.]
If a group $G$ is multiply isomorphic with
a group~$G'$, then \Inum{(i)}~the operations of~$G$, which correspond to
the identical operation of~$G'$, form a self-conjugate sub-group of~$G$;
\Inum{(ii)}~to each operation of~$G'$ there correspond the same number
of operations of~$G$; and \Inum{(iii)}~the order of~$G$ is a multiple of the
order of~$G'$.
\end{Theorem}

Let
\[
S_{0},\ S_{1},\ S_{2},\ \dots,\ S_{n-1}
\]
be the set of operations of~$G$ which correspond to the identical
operation of~$G'$. These operations must form a group, since to
$S_{p}S_{q}$ corresponds the operation~$1 · 1$, \ie\ the identical operation
of~$G'$; and therefore $S_{p}S_{q}$ must belong to the set.

Again, to the operation $T^{-1}S_{p}T$ of~$G$ corresponds the operation
$T'^{-1} · 1 · T'$, that is, the identical operation of~$G'$. Hence,
whatever operation of~$G$ is taken for~$T$,
\[
T^{-1} \{S_{0}, S_{1}, \dots, S_{n-1}\} T = \{S_{0}, S_{1}, \dots, S_{n-1}\}.
\]

The sub-group~$\Gamma$ of~$G$ formed of the operations
\[
S_{0},\ S_{1},\ S_{2},\ \dots,\ S_{n-1}
\]
is therefore self-conjugate.

Again, if $T$ and $T_{1}$ are two operations of~$G$ which correspond
to the operation~$T'$ of~$G'$, the operation $T^{-1}T_{1}$ corresponds to
the identical operation of~$G'$, and therefore belongs to~$\Gamma$.
Hence the operations that correspond to~$T'$ are all contained
in the set~$T\Gamma$. The operations of this set are all distinct and
equal in number to the order of~$\Gamma$. Hence if $n$~is the order of~$\Gamma$,
to each operation of~$G'$ there correspond $n$~operations of~$G$.

Finally, since to each operation of~$G$ there corresponds only
one of~$G'$, while to each operation of~$G'$ there correspond $n$ of~$G$,
the order of~$G$ is $n$~times the order of~$G'$.

To any sub-group of~$G'$ of order~$\mu$, there corresponds a
sub-group of~$G$ of order~$\mu n$. For if $T_{p}'T_{q}'$ forms one of the
set
\[
1,\ T_{1}',\ \dots,\ T_{p}',\ \dots,\ T_{q}',\ \dots,\ T_{\mu - 1}',
\]
%% -----File: 053.png---Folio 37-------
at least one, and therefore all, of the set $T_{p} T_{q} \Gamma$ must occur
among
\[
\Gamma,\ T_{1} \Gamma,\ \dots,\ T_{p}\Gamma,\ \dots,\ T_{q}\Gamma,\ \dots,\ T_{\mu - 1}\Gamma,
\]
and hence these operations form a group. Moreover, if the
sub-group of~$G'$ is self-conjugate, so also is the corresponding
sub-group of~$G$.

It should be noticed that no correspondence is thus
established between a sub-group of~$G$ which does not contain~$\Gamma$
and any sub-group of~$G'$.

\Par{29.} The relation of multiple isomorphism between two
groups can be presented in a manner rather different from that
of the last paragraph. Let $G$~be any composite group, and
$\Gamma$~a self-conjugate sub-group of~$G$ consisting of the operations
\[
1,\ T_{1},\ T_{2},\ \dots,\ T_{n - 1}.
\]

Then, as in §~22, the operations of~$G$ can be arranged in the
scheme
\[
\begin{array}{*{6}{c}}
1,         & T_{1},          & T_{2},          & \dots, & T_{n - 1}\DPchg{;}{,} \\
S_{1},     & T_{1}S_{1},     & T_{2}S_{1},     & \dots, & T_{n - 1}S_{1}\DPchg{;}{,} \\
\hdotsfor{5} \\
S_{i},     & T_{1}S_{i},     & T_{2}S_{i},     & \dots, & T_{n - 1}S_{i}\DPchg{;}{,} \\
\hdotsfor{5} \\
S_{m - 1}, & T_{1}S_{m - 1}, & T_{2}S_{m - 1}, & \dots, & T_{n - 1}S_{m - 1}.
\end{array}
\]

Now $\Gamma$ being a self-conjugate sub-group of~$G$, it follows that
\[
S_{i}^{-1} \Gamma S_{i} = \Gamma,
\]
and therefore the two sets of operations $\Gamma S_{i}$ and $S_{i} \Gamma$ coincide
except as regards arrangement. Hence
\[
T_{\alpha} S_{i} T_{\beta} S_{j} = T_{\alpha} T_{\beta'} S_{i} S_{j} = T_{\gamma} S_{i} S_{j},
\]
where
\[
S_{i} T_{\beta} S_{i}^{-1} = T_{\beta'},
\]
and
\[
T_{\alpha}T_{\beta'} = T_{\gamma};
\]
so that both $T_{\beta'}$ and $T_{\gamma}$ belong to~$\Gamma$. Now all the operations of
the set~$\Gamma S_{i}S_{j}$ occur in a single line, say the $(k + 1)$th, of the
above scheme. Hence if any operation of the $(i + 1)$th~line be
followed by any operation of the $(j + 1)$th~line, the result is
some operation of the $(k + 1)$th~line. If then we regard the
set of operations contained in each line of the scheme as a
%% -----File: 054.png---Folio 38-------
single entity, they will by their laws of combination define a
new group of order~$m$. In fact, if we denote these entities by
\[
S_{0}',\ S_{1}',\ \dots,\ S_{m-1}',
\]
a relation of the form
\[
S_{i}' S_{j}' = S_{k}'
\]
has been proved to hold for every pair. Moreover, these relations
necessarily obey the associative law; for if
\[
S_{j}' S_{p}' = S_{q}',
\]
then the relation
\[
S_{k}' S_{p}' = S_{i}' S_{q}'
\]
follows in consequence of the symbols of $G$ itself obeying the
associative law. The symbol~$S_{0}'$, corresponding to the first line
of the scheme, is clearly the symbol of the identical operation in
the new group thus defined.

If now $G$ and $\Gamma$ coincide with the groups of the preceding
paragraph which are represented by the same symbols, then $G'$
of the preceding paragraph must be simply isomorphic with the
group whose operations are
\[
1,\ S_{1}',\ S_{2}',\ \dots,\ S_{m-1}'.
\]

It follows that a group~$G'$ with which a group~$G$ is
\index{Factor groups@\Topic{Factor groups}!definition of}%
multiply isomorphic, in such a way that to the identical operation
of~$G'$ there corresponds a given self-conjugate sub-group~$\Gamma$
of~$G$, is completely defined (as an abstract group) when $G$
and $\Gamma$ are given. This being so it is natural to use a symbol
to denote directly the group thus defined in terms of $G$ and~$\Gamma$.
%[** In footnote, no comma or "Vol." after journal title]
Herr Hölder\footnote
  {``Zur Reduction der algebraischen Gleichungen,'' \textit{Math.\ Ann.}\ \VolNo{XXXIV} (1889),
  p.~31.}
\index{Holder@\Topic{Hölder}, quoted}%
has introduced the symbol
\[
\frac{G}{\Gamma}
\]
to represent this group; he calls it the \emph{quotient} of $G$ by~$\Gamma$,
and a \emph{factor-group} of~$G$. We shall in the sequel make constant
use both of the symbol and of the phrase thus defined.

It may not be superfluous to notice that the symbol~$\dfrac{G}{\Gamma}$ has
%% -----File: 055.png---Folio 39-------
no meaning\footnotemark,
\index{Frobenius@\Topic{Frobenius}, quoted}%
\footnotetext{Herr Frobenius has extended the use of the symbol to the case in which $\Gamma$~is
  any group, whether contained in~$G$ or not, with which every operation of~$G$ is
  permutable: ``Ueber endliche Gruppen,'' \textit{Berliner Sitzungsberichte}, 1895, p.~169.
  We shall always use the symbol in the sense defined in the text.}%
unless $\Gamma$~is a self-conjugate sub-group of~$G$.
Moreover, it may happen that $G$ has two simply isomorphic
self-conjugate sub-groups $\Gamma$~and~$\Gamma'$. When this is
the case, there is no necessary relation between the factor-groups
$\dfrac{G}{\Gamma}$ and $\dfrac{G}{\Gamma'}$ (except of course that their orders are equal);
in other words, the type of the factor-group $\dfrac{G}{\Gamma}$ depends on the
actual self-conjugate sub-group of~$G$ which is chosen for~$\Gamma$ and
not merely on the type of~$\Gamma$.

Further, though in relation to its definition by means of $G$
and $\Gamma$ we call $\dfrac{G}{\Gamma}$ a factor-group of~$G$, we may without ambiguity,
since the symbol represents a group of definite type, omit the
word factor and speak of the group~$\dfrac{G}{\Gamma}$.

It is also to be observed that $G$ has not necessarily a
sub-group simply isomorphic with~$\dfrac{G}{\Gamma}$. This may or may not be
the case.

\Par{30.} If $G$ is multiply isomorphic with $G'$ so that the self-conjugate
sub-group~$\Gamma$ of~$G$ corresponds to the identical
operation of~$G'$, it was shewn, at the end of §~28, that to any
self-conjugate sub-group of~$G'$ there corresponds a self-conjugate
sub-group of~$G$ containing~$\Gamma$. Hence, unless $\dfrac{G}{\Gamma}$ is a simple
group, $\Gamma$~cannot be a maximum self-conjugate sub-group of~$G$.
If $g_{1}$~is any self-conjugate sub-group of~$\dfrac{G}{\Gamma}$, and $G_{1}$~the
corresponding (necessarily self-conjugate) sub-group of~$G$, containing~$\Gamma$,
we may form the factor-group~$\dfrac{G}{G_{1}}$, and determine
again whether this group is simple or composite. By continuing
this process a maximum self-conjugate sub-group of~$G$, containing~$\Gamma$,
must at last be reached.
%% -----File: 056.png---Folio 40-------

\Par{31.} Though $\dfrac{G}{\Gamma}$ is completely defined by $G$~and~$\Gamma$, where $\Gamma$~is
any given self-conjugate sub-group of~$G$, the reader will
easily verify that $G$~is not in general determined when $\Gamma$ and
$\dfrac{G}{\Gamma}$ are given.

We shall have in the sequel to consider the solution of
this problem in various particular cases. There is, however, in
every case one solution of it which is immediately obvious.
We may take any two groups $G_{1}$ and $G_{2}$, simply isomorphic
with the given groups $\Gamma$ and $\dfrac{G}{\Gamma}$, such that $G_{1}$ and $G_{2}$ have
no common operation except identity, while each operation
of one is permutable with each operation of the other. The
group $\{G_{1}, G_{2}\}$, formed by combining these two, is clearly such
that $\dfrac{\{G_{1}, G_{2}\}}{G_{1}}$ is simply isomorphic with~$\dfrac{G}{\Gamma}$; it therefore gives a
solution of the problem.

\begin{Definition}
If two groups $G_{1}$,~$G_{2}$ have no common operation
\index{Direct product@\Topic{Direct product} of two groups!definition of}%
except identity, and if each operation of~$G_{1}$ is permutable
with each operation of~$G_{2}$, the group $\{G_{1}, G_{2}\}$ is called the
\emph{direct product} of $G_{1}$~and~$G_{2}$.
\end{Definition}

\begin{Ex} If $H$,~$h$ are self-conjugate sub-groups of~$G$, and if $h$~is
contained in~$H$, so that $\dfrac{H}{h}$~is a self-conjugate sub-group of~$\dfrac{G}{h}$, shew
that the quotient of~$\dfrac{G}{h}$ by~$\dfrac{H}{h}$ is simply isomorphic with~$\dfrac{G}{H}$.
\end{Ex}

\begin{Remark}
\Par{32.} If $H$ is a self-conjugate sub-group of~$G$, of order~$n$, and if
$H'$~is a self-conjugate sub-group of~$G'$, of order~$n'$, and if $\dfrac{G}{H}$ and $\dfrac{G'}{H'}$
are simply isomorphic, a correspondence of the most general kind
may be established between the operations of $G$~and~$G'$. To every
operation of~$G$ (or~$G'$) there will correspond $n'$ (or~$n$) operations of~$G'$
(or~$G$), in such a way that to the product of any two operations
of~$G$ (or~$G'$) there corresponds a definite set of $n'$ (or~$n$) operations
of~$G'$ (or~$G$). Let
\[
G = H,\ S_{1}H,\ S_{2}H,\ \Dots,\ S_{m - 1}H,
\]
and
\[
G' = H',\ S_{1}'H',\ S_{2}'H',\ \Dots,\ S_{m - 1}'H';
\]
{\Loosen and in the simple isomorphism between $\dfrac{G}{H}$ and~$\dfrac{G'}{H'}$, let $S_{r}$~and~$S_{r}'$
%% -----File: 057.png---Folio 41-------
($r = 0, 1, \dots, m - 1$)} be corresponding operations. Then if we take
the set~$S_{r}' H'$ as the $n'$~operations of~$G'$ that correspond to any
operation of the set~$S_{r}H$ of~$G$, and the set~$S_{r}H$ as the $n$~operations
of~$G$ that correspond to any operation of the set~$S_{r}' H'$ of~$G'$, the
correspondence is, in fact, established.

For, if $h_{1}'$ and $h_{2}'$ are any two operations of~$H'$, the set of operations
\index{Isomorphism of two groups@\Topic{Isomorphism of two groups}!general, definition of}%
$S_{r}' h_{1}' S_{s}' h_{2}'$ includes $n'$~distinct operations only, namely those of
the set~$S_{r}' S_{s}' H'$. Hence to the product of any given operation of
the set~$S_{r}H$ by any given operation of the set~$S_{s}H$, there corresponds
the set of $n'$~operations $S_{r}' S_{s}' H'$; at the same time the product of the
two given operations belongs $\Bigl($in consequence of the isomorphism
between $\dfrac{G}{H}$ and~$\dfrac{G'}{H'}${}$\Bigr)$ to the set~$S_{r} S_{s} H$. The same statements clearly
hold when we interchange accented and unaccented symbols.

We still speak of $G$ and $G'$ as isomorphic groups, and the
correspondence between their operations is said to give an $n$-to-$n'$
isomorphism of the two groups. We shall return to this general
form of isomorphism in dealing with intransitive substitution groups.
\end{Remark}

\Par{33.} \begin{Definition}
Two groups $G$ and $G'$ are said to be
\index{Permutable groups@\Topic{Permutable groups}!definition of}%
\DPchg{permutable}{\emph{permutable}} with each other when the distinct operations of
the set~$S_{i} S_{j}'$, where for $S_{i}$ every operation of the group~$G$ is
put in turn and for $S_{j}'$ every operation of the group~$G'$, coincide
with the distinct operations of the set~$S_{j}' S_{i}$ except possibly
as regards arrangement.
\end{Definition}

If the two groups $G$ and $G'$ are permutable, the group
$\{G, G'\}$ must be of finite order. For, by the definition, every
operation
\[
\dots S_{p} S_{q}' S_{r} S_{t}' \dots
\]
can be reduced to the form~$S_{i} S_{j}'$; and therefore the number of
distinct operations of the group $\{G, G'\}$ cannot exceed the
product of the orders of $G$~and~$G'$. Let $g$~be the group formed
of the common operations of $G$~and~$G'$. Divide the operations
of these groups into the sets
\[
g,\ \Sigma_{1}g,\ \Sigma_{2}g,\ \dots,\ \Sigma_{m - 1}g
\]
and
\[
g,\ g \Sigma_{1}',\ g \Sigma_{2}',\ \DPchg{\Dots\Dots}{\Dots},\ g \Sigma_{m'-1}'.
\]

Then every operation of the set~$S_{i} S_{j}'$ can clearly be expressed
in the form
\[
\Sigma_{p} \gamma \Sigma_{q}',
\]
%% -----File: 058.png---Folio 42-------
where $\gamma$~is some operation of~$g$. And no two operations of this
form can be identical, for if
\[
\Sigma_{p} \gamma_{1} \Sigma_{q}' = \Sigma_{r} \gamma_{2} \Sigma_{t}',
\]
then
\[
\gamma_{2}^{-1} \Sigma_{r}^{-1} \Sigma_{p} \gamma_{1} = \Sigma_{t}' \Sigma_{q}'^{-1};
\]
so that $\Sigma_{t}' \Sigma_{q}'^{-1}$ belongs to~$g$. But this is only possible if
\[
\Sigma_{t}' = \Sigma_{q}',
\]
which leads to
\[
\Sigma_{r} = \Sigma_{p},
\]
and
\[
\gamma_{2} = \gamma_{1}.
\]
The order of $\{G, G'\}$ is therefore the product of the orders of $G$
and~$G'$, divided by the order of~$g$.

If every operation of~$G$ is permutable with~$G'$, then $g$~must
be a self-conjugate sub-group of~$G$. For $G$ and~$G'$ are transformed,
each into itself, by any operation of~$G$; and therefore
their common sub-group~$g$ must be transformed into itself by
any operation of~$G$.

Moreover those operations of~$G$, which are permutable with
every operation of~$G'$, form a self-conjugate sub-group of~$G$.
For if $T$~is an operation of~$G$, which is permutable with every
operation~$S'$ of~$G'$, so that
\[
T^{-1} S' T = S',
\]
and if $S$~is any operation of~$G$, then
\[
S^{-1} T^{-1} S · S_{-1} S'S · S_{-1} TS = S^{-1} S'S,
\]
so that $S_{-1} TS$ is permutable with every operation of~$G'$. Hence
every operation of~$G$, which is conjugate to~$T$, is permutable
with every operation of~$G'$; and the operations of~$G$, which are
permutable with every operation of~$G'$, therefore form a self-conjugate
sub-group.

If $G$ is a simple group, $g$~must consist of the identical
operation only; and either all the operations of~$G$, or none of
them, must be permutable with every operation of~$G'$.

A special case is that in which the two groups $G'$ and~$G$ are
respectively a self-conjugate sub-group~$\Gamma$ and any sub-group~$H$
of some third group; for then every operation of~$H$ is permutable
with~$\Gamma$. If $H$~is a cyclical sub-group generated by an
operation~$S$ of order~$n$, and if $S^{m}$~is the lowest power of~$S$
%% -----File: 059.png---Folio 43-------
which occurs in~$\Gamma$, then $m$~must be a factor of~$n$. For if $m'$~is
the greatest common factor of $m$ and~$n$, integers $x$ and~$y$ can be
found such that
\[
xm + yn = m'.
\]

Now
\[
S^{m'} = S^{xm + yn} = S^{xm},
\]
and therefore $S^{m'}$ belongs to~$\Gamma$. Hence $m'$ cannot be less than~$m$,
and therefore $m$~is a factor of~$n$. Moreover, since $\{S^{m}\}$~is
a sub-group of~$\Gamma$, the order of~$\Gamma$ must be divisible by~$\dfrac{n}{m}$.
Hence:---

\begin{Theorem}[VIII.]
If an operation~$S$, of order~$n$, is permutable
with a group~$\Gamma$, and if $S^{m}$~is the lowest power of~$S$ which occurs
in~$\Gamma$; then $m$~is a factor of~$n$, and $\dfrac{n}{m}$~is a factor of the order
of~$\Gamma$.
\end{Theorem}

The operations of~$\{\Gamma, S\}$ can clearly be distributed in the
sets
\[
\Gamma,\ \Gamma S,\ \Gamma S^{2},\ \dots,\ \Gamma S^{m-1};
\]
and no two of the operations $S$, $S^{2}$,~\dots, $S^{m-1}$ are conjugate in
$\{\Gamma, S\}$.

\Par{34.} A still more special case, but it is most important, is
that in which the two groups are both of them self-conjugate
sub-groups of some third group. If in this case the two groups
are $G$ and~$H$, while $S$~and~$T$ are any operations of the two
groups respectively, then
\[
S^{-1}HS = H,
\]
and
\[
T^{-1}GT = G;
\]
so that every operation of~$G$ is permutable with~$H$ and every
operation of~$H$ is permutable with~$G$.

Consider now the operation $S^{-1} T^{-1} ST$. Regarded as the
product of $S^{-1}$ and~$T^{-1}ST$ it belongs to~$G$, and regarded as the
product of $S^{-1}T^{-1}S$ and~$T$ it belongs to~$H$. Every operation of
this form therefore belongs to the common group of $G$ and~$H$.
If $G$ and~$H$ have no common operation except identity, then
\[
S^{-1} T^{-1} ST = 1,
\]
%% -----File: 060.png---Folio 44-------
or
\[
ST = TS;
\]
and $S$~and~$T$ are permutable. Hence:---

\begin{Theorem}[IX\protect\footnotemark.]
If every operation of~$G$ transforms~$H$ into
itself and every operation of~$H$ transforms~$G$ into itself, and
if $G$~and~$H$ have no common operation except identity; then
every operation of~$G$ is permutable with every operation of~$H$.
\end{Theorem}
%[** No comma after journal title]
\footnotetext{Dyck, ``Gruppentheoretische Studien,'' \textit{Math.\ Ann.}\ Vol.~\VolNo{XXII} (1883), p.~97.}%
\index{Dyck@\Topic{Dyck}, quoted}%

\begin{Corollary}
If every operation of~$G$ transforms~$H$ into itself
and every operation of~$H$ transforms~$G$ into itself, and if either
$G$ or~$H$ is a simple group; then $G$ and~$H$ have no common
operation except identity, and every operation of~$G$ is permutable
with every operation of~$H$.
\end{Corollary}

For, by §~33, if $G$ and~$H$ had a common sub-group, it would
be a self-conjugate sub-group of both of them; and neither of
them could then be simple, contrary to hypothesis. Consequently,
the only sub-group common to $G$ and~$H$ is the
identical operation.

\begin{Remark}
\Par{35.} If $N$ and~$N'$ are the orders of two permutable groups $G$ and~$G'$,
the $NN'$~equations expressing every operation of the form~$SS'$,
where $S$~belongs to~$G$ and $S'$ to~$G'$, in the form~$\Sigma'\Sigma$, where $\Sigma$ and~$\Sigma'$
belong to $G$ and $G'$ respectively, are never all independent. In
particular, if
\[
S_{1},\ S_{2},\ \Dots,\ S_{r}
\]
are a set of independent generating operations of~$G$, and
\[
m_{1},\ m_{2},\ \Dots,\ m_{r}
\]
their orders, it is clearly sufficient that each operation of the form
\[
S'S_{p}^{\alpha}\quad (\alpha = 1, 2, \Dots, m_{p} - 1;\ p = 1, 2, \Dots, r)
\]
should be capable of expression in the form~$\Sigma'\Sigma$.

When, in fact, these conditions are satisfied, it is always possible
by a series of steps to express any operation~$S'S$ in the form~$\Sigma\Sigma'$.
\end{Remark}

\begin{Ex}[1.] Shew that, in the group whose defining relations are
\[
A^{4} = 1,\quad B^{3} = 1,\quad (AB)^{2} = 1,
\]
the three operations $A^{2}$, $B^{-1} A^{2} B$, $BA^{2} B^{-1}$ are permutable and that
they form a complete set of conjugate operations. Hence shew that
$\{A^{2}, B\}$ is a self-conjugate sub-group, and that the order of the group
is~$24$.
\end{Ex}
%% -----File: 061.png---Folio 45-------

\begin{Ex}[2.] Shew that the cyclical group generated by the substitution
$(1234567)$ is permutable with the group
\[
\{(267)(345),\ (23)(47)\};
\]
and that the order of the group resulting from their product is~$168$.
\end{Ex}

\begin{Ex}[3.] If $g_{1}$ and $g_{2}$ are the orders of the groups $G_{1}$ and~$G_{2}$, $\gamma$~the
order of their greatest common sub-group and $g$~the order of
$\{G_{1}, G_{2}\}$, shew that
\[
g\gamma \geq g_{1}g_{2},
\]
\index{Frobenius@\Topic{Frobenius}, quoted}%
and that, if $g\gamma = g_{1}g_{2}$, then $G_{1}$ and $G_{2}$ are permutable. \Attrib{Frobenius.}
\end{Ex}

\begin{Ex}[4.] If $G_{1}$ and $G_{2}$ are two sub-groups of~$G$ of orders $g_{1}$ and~$g_{2}$,
and $S$~any operation of~$G$, prove that the number of distinct operations
of~$G$ contained in the set~$S_{1}SS_{2}$, when for $S_{1}$ and $S_{2}$ are put in
turn every pair of operations of $G_{1}$ and $G_{2}$ respectively, is~$\dfrac{g_{1}g_{2}}{\gamma}$; $\gamma$~being
the order of the greatest sub-group common to $S^{-1}G_{1}S$ and~$G_{2}$.

If $T$~is any other operation of~$G$, shew also that the sets $S_{1}SS_{2}$
and $S_{1}TS_{2}$ are either identical or have no operation in common. \Attrib{Frobenius.}
\end{Ex}

\begin{Ex}[5.] If a group~$G$ of order~$mn$ has a sub-group~$H$ of order~$n$,
and if $n$~has no prime factor which is less than~$m$, shew that $H$~must
be a self-conjugate sub-group. \Attrib{Frobenius.}
\end{Ex}
%% -----File: 062.png---Folio 46-------


\Chapter{IV.}{On Abelian Groups.}
\index{Abelian group@\Topic{Abelian group}!definition of}%
\index{Group@\Topic{Group}!Abelian}%

\Par{36.} \First{We} shall now apply the general results, that have
been obtained in the last chapter, to the study of two special
classes of groups; in the present chapter we shall deal particularly
with those groups whose operations are all permutable
with each other.

\begin{Definition}
A group, whose operations are all permutable
\index{Abel@\Topic{Abel}, quoted}%
\index{Frobenius and Stickelberger@\Topic{Frobenius and Stickelberger}, quoted}%
\index{Jordan@\Topic{Jordan}, quoted}%
\index{Weber@\Topic{Weber}, quoted}%
with each other, is called an \emph{Abelian}\footnote
  {On Abelian groups, the reader may consult Frobenius and Stickelberger,
  ``Ueber Gruppen vertauschbarer Elemente,'' \textit{Crelle's Journal}, Vol.~\VolNo{LXXXVI} (1879),
  p.~217; and a very complete discussion in the second volume of Herr Weber's
  recently published \textit{Lehrbuch der Algebra}. In the proof of the existence of a
  set of independent generating operations (§~41) we have directly followed Herr
  Weber.

  The name ``Abelian group'' has been applied by M.~Jordan (\textit{Traité des substitutions~etc.}\
  pp.~171 et~seq.)\ to an entirely different class of groups, whose
  operations are not permutable. Most writers, we believe, have used the phrase
  in the sense defined in the text.

  The connection of Abel's name with groups of permutable operations is due
  to his having been the first to investigate, with complete generality, the application
  of such groups to the theory of equations, ``Mémoire sur une classe
  particulière d'équations résolubles algébriquement,'' \textit{Crelle's Journal}, Vol.~\VolNo{IV}
  (1829), p.~131; or Collected Works, 1881 edition, Vol.~\VolNo{I}, p.~478.}
group.
\end{Definition}

It is to be expected (and it will be found) that the theory of
Abelian groups is much simpler than that of groups in general;
for the process of multiplication of the operations of such groups
is commutative as well as associative.

Every sub-group of an Abelian group is itself an Abelian
group, since its operations are necessarily all permutable. For
%% -----File: 063.png---Folio 47-------
\index{Abelian group@\Topic{Abelian group}!sub-groups of}%
the same reason, every operation and every sub-group of an
Abelian group is self-conjugate both in the group itself and in
any sub-group in which it is contained.

If $G$ is an Abelian group and $H$~any sub-group of~$G$, then
since $H$~is necessarily self-conjugate, there exists a factor-group~$\dfrac{G}{H}$,
and this again must be an Abelian group. (The
reader must not however infer that, if $H$ and $\dfrac{G}{H}$ are both
Abelian, then $G$~is also Abelian. It is indeed clear that this
is not necessarily the case.)

\Par{37.} Let now $G$ be any Abelian group, and let $p^{m}$ be the
highest power of a prime~$p$ that divides its order. We shall
first shew that $G$~has a single sub-group of order~$p^{m}$, consisting
of all the operations of~$G$ whose orders are powers of~$p$.

If $S_{1}$,~$S_{2}$ are any two operations of~$G$, it follows from §~33,
that, because $S_{1}$ and $S_{2}$ are permutable, the order of~$\{S_{1}, S_{2}\}$ is
equal to or is a factor of the product of the orders of $S_{1}$~and~$S_{2}$.
So again, if $S_{3}$~is an operation of~$G$ not contained in~$\{S_{1}, S_{2}\}$, the
order of $\{S_{1}, S_{2}, S_{3}\}$ is equal to or is a factor of the product of
the orders of $\{S_{1}, S_{2}\}$ and~$S_{3}$. Now by continually including a
fresh operation, not contained in the group already arrived at,
we must in this way after a finite number of steps arrive at the
group~$G$, whose order is divisible by~$p$. Hence, among the
operations $S_{1}$,~$S_{2}$, $S_{3}$,~\Dots, there must be at least one whose
order is divisible by~$p$, and some power of this, say~$S$, will be
an operation of order~$p$. Now, if $m$~is greater than unity, the
order of the factor-group $\dfrac{G}{\{S\}}$, which is also Abelian, is divisible
by~$p$, and therefore this factor-group must have an operation of
order~$p$. Hence $G$~will (§~28) contain a sub-group of order~$p^{2}$.
If $m$~is greater than~$2$, the same reasoning may be
repeated to shew that $G$~has a sub-group of order~$p^{3}$, and so on.
Hence, finally, $G$~has a sub-group of order~$p^{m}$. Let $H$~be this
sub-group; and suppose if possible that $G$~contains an operation~$T$,
whose order is a power of~$p$, which does not belong to~$H$.
Then $\{H, T\}$ is a sub-group of~$G$, whose order is a power of~$p$,
greater than~$p^{m}$; and this is impossible (§~22). The sub-group~$H$
%% -----File: 064.png---Folio 48-------
\index{Abelian group@\Topic{Abelian group}!sub-groups of}%
must therefore contain all the operations of~$G$ whose orders
are powers of~$p$. Hence:---

\begin{Theorem}[I.]
If $p^{m}$ is the highest power of a prime~$p$ that
divides the order of an Abelian group~$G$, then $G$~contains a
single sub-group of order~$p^{m}$, which consists of all the operations
of~$G$ whose orders are powers of~$p$.
\end{Theorem}

\Par{38.} Let the order of~$G$ be
\[
N = p_{1}^{m_{1}} p_{2}^{m_{2}} \Dots p_{n}^{m_{n}},
\]
where $p_{1}$, $p_{2}$,~\Dots, $p_{n}$ are distinct primes; and let
\[
H_{1},\ H_{2},\ \Dots,\ H_{n}
\]
be the sub-groups of~$G$ of orders
\[
p_{1}^{m_{1}},\ p_{2}^{m_{2}},\ \Dots,\ p_{n}^{m_{n}}.
\]

Since the orders of $H_{1}$ and $H_{2}$ are relatively prime, they
can have no common operation except identity; and therefore
the order of $\{H_{1}, H_{2}\}$ is $p_{1}^{m_{1}} p_{2}^{m_{2}}$. This sub-group contains all
the operations of~$G$ whose orders are relatively prime to
$\dfrac{N}{p_{1}^{m_{1}} p_{2}^{m_{2}}}$. For if $T$~were an operation of~$G$ of order~$p_{1}^{\alpha} p_{2}^{\beta}$, not
contained in $\{H_{1}, H_{2}\}$, then $\{H_{1}, H_{2}, T\}$ would be a sub-group
of~$G$, of order $p_{1}^{n_{1}} p_{2}^{n_{2}}$, where $n_{1} > m_{1}$ if $\alpha > 0$ and $n_{2} > m_{2}$ if $\beta > 0$;
and this is impossible (§~22).

This process may clearly be continued to shew that, if
$N = \mu\nu$, where $\mu$~and~$\nu$ are relatively prime, then $G$~contains a
single sub-group of order~$\mu$, consisting of all the operations of~$G$
whose orders are relatively prime to~$\nu$. Moreover $G$~itself is
the direct product (§~31) of $H_{1}$, $H_{2}$,~\Dots, $H_{n}$.

\Par{39.} The first problem of pure group-theory that presents
itself in connection with Abelian groups is the determination
of all distinct Abelian groups of given order~$N$. Let $H_{1}$
and $H_{1}'$ be two distinct Abelian groups of order~$p_{1}^{m_{1}}$, \ie\ two
groups which are not simply isomorphic. Then two Abelian
groups of order~$N$, whose sub-groups of order~$p_{1}^{m_{1}}$ are simply
isomorphic with $H_{1}$ and $H_{1}'$ respectively, are necessarily distinct.
Since then $G$~is the direct product of $H_{1}$, $H_{2}$,~\Dots, $H_{n}$, the
general problem for any composite order~$N$ will be completely
solved when we have determined all distinct types of Abelian
%% -----File: 065.png---Folio 49-------
groups of order~$p_{1}^{m_{1}}$, $p_{2}^{m_{2}}$,~\Dots, $p_{n}^{m_{n}}$. We may therefore, for the
purpose of this problem, confine our attention to those Abelian
groups whose orders are powers of primes.

\Par{40.} Suppose then that $G$~is an Abelian group whose order
is the power of a prime. If among the operations of~$G$ we
choose at random a set
\[
P,\ P',\ P'',\ \Dots
\]
from which the group can be generated, they will not in
general be independent of each other.

\begin{Remark}
As an instance, we may take the group whose multiplication
\index{Group@\Topic{Group}!multiplication table of}%
\index{Multiplication table@\Topic{Multiplication table} of a group}%
table is:---
\[
\begin{array}{c|*{8}{c}}
\multicolumn{1}{c}{}
      & 1     & P_{1} & P_{2} & P_{3} & P_{4} & P_{5} & P_{6} & P_{7} \\
\cline{2-9}
\Strut
1     & 1     & P_{1} & P_{2} & P_{3} & P_{4} & P_{5} & P_{6} & P_{7} \\
P_{1} & P_{1} & P_{2} & P_{3} & 1     & P_{5} & P_{6} & P_{7} & P_{4} \\
P_{2} & P_{2} & P_{3} & 1     & P_{1} & P_{6} & P_{7} & P_{4} & P_{5} \\
P_{3} & P_{3} & 1     & P_{1} & P_{2} & P_{7} & P_{4} & P_{5} & P_{6} \\
P_{4} & P_{4} & P_{5} & P_{6} & P_{7} & 1     & P_{1} & P_{2} & P_{3} \\
P_{5} & P_{5} & P_{6} & P_{7} & P_{4} & P_{1} & P_{2} & P_{3} & 1     \\
P_{6} & P_{6} & P_{7} & P_{4} & P_{5} & P_{2} & P_{3} & 1     & P_{1} \\
P_{7} & P_{7} & P_{4} & P_{5} & P_{6} & P_{3} & 1     & P_{1} & P_{2}
\end{array}
\]

Here $P_{1}$ and $P_{5}$ are two operations from which every operation
of the group may be generated: an inspection of the table will
shew that they are connected by the relation
\[
P_{1}^{2} P_{5}^{2} = 1.
\]

On the other hand, if we choose $P_{1}$ and $P_{4}$ as generating
operations, we find that every operation of the group can be expressed
in the form
\[
P_{1}^{\alpha} P_{4}^{\beta},\quad (\alpha = 0, 1, 2, 3;\ \beta = 0, 1),
\]
while the only conditions to which the permutable operations $P_{1}$
and $P_{4}$ are submitted are
\[
P_{1}^{4} = 1,\quad P_{4}^{2} = 1.
\]
\end{Remark}

The question then arises as to whether $G$~can be generated
by a set of permutable and independent operations, \ie\ by a
set of permutable operations which are connected by no
relations except those that give their orders. That this
%% -----File: 066.png---Folio 50-------
question is always to be answered in the affirmative may be
proved in the following manner.

\Par{41.} Let the order of~$G$ be~$p^{m}$; and let $P_{1}$ be an operation
of~$G$ whose order~$p^{m_{1}}$ is not less than that of any other operation
of the group. Then every operation of the group satisfies
the equation
\[
S^{p^{m_{1}}} = 1.
\]

If $m_{1} = m$, the order of $\{P_{1}\}$ is equal to the order of~$G$; the
latter is then a cyclical group generated by the operation~$P_{1}$.

If $m_{1} < m$, $G$~must contain other operations besides those of~$\{P_{1}\}$.
Denoting $\{P_{1}\}$ by~$G_{1}$, let $Q$ be any operation of~$G$ not
contained in~$G_{1}$, and let $Q^{a}$ be the lowest power of~$Q$ that is
contained in~$G_{1}$. Then (§~33) $a$~must be a power of~$p$; and
when for $Q$ each operation of~$G$ that is not contained in~$G_{1}$ is
taken in turn, $a$~must have some maximum value, say~$p^{m_{2}}$.
Since no operation of~$G$ is of greater order than~$p^{m_{1}}$ it follows
that $m_{2} \leq m_{1}$. We may suppose then $Q$~to be an operation of~$G$,
such that $Q^{p^{m_{2}}}$ is the lowest power of~$Q$ which is contained
in~$G_{1}$.

Then
\[
Q^{p^{m_{2}}} = P_{1}^{\lambda_{1}},
\]
and
\[
Q^{p^{m_{1}}} = 1 = P_{1}^{\DPtypo{\lambda}{\lambda_{1}} p^{m_{1} - m_{2}}},
\]
so that $\lambda_{1}$~is divisible by~$p^{m_{2}}$ and may be expressed in the form~$xp^{m_{2}}$.
The case $\lambda_{1} = 0$ forms no exception to this statement,
since $\lambda_{1}$~is congruent to zero, $\pmod{p^{m_{1}}}$; but we actually take
$x = 0$ in this case.

If now we write
\[
Q P_{1}^{-x} = P_{2},
\]
then
\[
P_{2}^{p^{m_{2}}} = 1,
\]
and $P_{2}^{p^{m_{2}}}$ is the lowest power of~$P_{2}$ which is contained in~$G_{1}$.
Let the sub-group $\{P_{1}, Q\}$ be denoted by~$G_{2}$. Then $G_{2}$~is
generated by the two independent operations $P_{1}$ and $P_{2}$ of
orders $p^{m_{1}}$ and~$p^{m_{2}}$; and every operation~$S$ of~$G$ is such that
\[
S^{p^{m_{2}}} = P_{1}^{\alpha_{1}},
\]
where $\alpha_{1}$~is divisible by~$p^{m_{2}}$.

This process may now be continued step by step. That
it will ultimately lead to a representation of~$G$ as generated
by a set of independent operations may be shewn by induction.
%% -----File: 067.png---Folio 51-------
To this end, we will suppose that a sub-group~$G_{n}$ has been
arrived at which is generated by the $n$ independent operations
\[
P_{1},\ P_{2},\ \Dots,\ P_{n},
\]
of orders
\[
p^{m_{1}},\ p^{m_{2}},\ \Dots,\ \DPtypo{p^{m}}{p^{m_{n}}},
\]
where
\[
m_{1} \geq m_{2} \geq \Dots \geq m_{n}.
\]

We will also suppose that, if $T$~is any operation of~$G$,
\[
T^{p^{m_{n}}} = P_{1}^{\alpha_{1}} P_{2}^{\alpha_{2}} \Dots P_{n-1}^{\alpha_{n-1}},
\]
where $\alpha_{1}$, $\alpha_{2}$,~\Dots, $\alpha_{n-1}$ are all divisible by~$p^{m_{n}}$. In the special
case $n = 2$ these suppositions have been justified.

Let $Q'$ be any operation of~$G$, and let $Q'^{a}$ be the lowest
power of~$Q'$ that occurs in~$G_{n}$. Then when for $Q'$ each operation
of~$G$ that is not contained in~$G_{n}$ is taken in turn, $a$~will have
a certain maximum value, say~$p^{m_{n+1}}$; and from the suppositions
made above, $m_{n+1}$~is not greater than~$m_{n}$. We may therefore
write
\[
Q'^{p^{m_{n+1}}}
  = P_{1}^{\beta_{1}}
    P_{2}^{\beta_{2}} \Dots
    P_{n-1}^{\beta_{n-1}}
    P_{n}^{\beta_{n}},
\]
and hence
\[
Q'^{p^{m_{n}}}
  = P_{1}^{\beta_{1} p^{m_{n}-m_{n+1}}}
    P_{2}^{\beta_{2} p^{m_{n}-m_{n+1}}} \Dots
    P_{n-1}^{\beta_{n-1} p^{m_{n}-m_{n+1}}}
    P_{n}^{\beta_{n} p^{m_{n}-m_{n+1}}}.
\]

Now $P_{1}$, $P_{2}$,~\Dots, $P_{n}$ are independent; it follows therefore
from the second assumption made above, that the
exponents of $P_{1}$, $P_{2}$,~\Dots, $P_{n}$ in this last equation are divisible
by~$p^{m_{n}}$; and hence that $\beta_{1}$, $\beta_{2}$,~\Dots, $\beta_{n}$ are divisible by~$p^{m_{n+1}}$.

If we now suppose that $Q'$ itself is chosen so that $Q'^{p^{m_{n+1}}}$ is
the lowest power of~$Q'$ that occurs in~$G_{n}$, then
\[
Q'^{p^{m_{n+1}}}
  = P_{1}^{x_{1} p^{m_{n+1}}}
    P_{2}^{x_{2} p^{m_{n+1}}} \Dots
    P_{n}^{x_{n} p^{m_{n+1}}};
\]
hence, if
\[
Q'P_{1}^{-x_{1}} P_{2}^{-x_{2}} \Dots P_{n}^{-x_{n}} = P_{n+1},
\]
then
\[
P_{n+1}^{p^{m_{n+1}}} = 1;
\]
and $P_{n+1}^{p^{m_{n+1}}}$ is the lowest power of~$P$ that occurs in~$G_{n}$. Hence
finally $\{P_{1}, P_{2}, \Dots, P_{n}, Q'\}$ is generated by the $n + 1$ independent
operations $P_{1}$, $P_{2}$,~\Dots, $P_{n+1}$ of orders $p^{m_{1}}$, $p^{m_{2}}$,~\Dots,
$p^{m_{n+1}}$ ($m_{1} \geq m_{2} \geq \Dots \geq m_{n+1}$); and $S$~being any operation of~$G$,
\[
S^{p^{m_{n+1}}} = P_{1}^{\gamma_{1}} P_{2}^{\gamma_{2}} \Dots P_{n}^{\gamma_{n}},
\]
where $\gamma_{1}$, $\gamma_{2}$,~\Dots, $\gamma_{n}$ are divisible by~$p^{m_{n+1}}$.
%% -----File: 068.png---Folio 52-------

This completes the proof by induction, and we may now
state the result in the form of the following theorem:---

\begin{Theorem}[II.]
An Abelian group of order~$p^{m}$, where $p$~is a
\index{Abelian group@\Topic{Abelian group}!existence of independent generating operations of}%
prime, can be generated by $r$ \($\leq m$\) independent operations $P_{1}$,
$P_{2}$,~\Dots, $P_{r}$, of orders $p^{m_{1}}$, $p^{m_{2}}$,~\Dots, $p^{m_{r}}$, where
\[
m_{1} + m_{2} + \dots + m_{r} = m,
\]
and $m_{1} \geq m_{2} \geq \Dots \geq m_{r}$. Moreover if $\mu$~is any positive integer
such that $m_{s} \geq \mu \geq m_{s+1}$, while $Q$~is any operation of the group, then
\[
Q^{p^{\mu}} = P_{1}^{\alpha_{1}} P_{2}^{\alpha_{2}} \Dots P_{s}^{\alpha_{s}},
\]
where $\alpha_{1}$, $\alpha_{2}$,~\Dots, $\alpha_{s}$ are divisible by~$p^{\mu}$.
\end{Theorem}

The actual existence of the set of independent generating
operations is demonstrated by the above inductive proof.
The other inferences in the theorem may be established as
follows. A set of independent permutable operations of orders
$p^{m_{1}}$, $p^{m_{2}}$,~\Dots, $p^{m_{r}}$ generate a group of order $p^{m_{1} + m_{2} + \dots + m_{r}}$, and
therefore
\[
m_{1} + m_{2} + \Dots + m_{r} = m.
\]

Since each number in this equation is a positive integer,
the number of terms on the left-hand side cannot be greater
than~$m$. Hence
\[
r \leq m;
\]
and, if $r = m$, every operation of the group except identity is of
order~$p$.

Finally, from the above inductive proof it follows that, if $Q$~is
any operation of the group, then
\[
Q^{p^{m_{s}+1}} = P_{1}^{\beta_{1}} P_{2}^{\beta_{2}} \Dots P_{s}^{\beta_{s}},
\]
where $\beta_{1}$, $\beta_{2}$,~\Dots, $\beta_{s}$ are all divisible by~$p^{m_{s}+1}$.

Hence
\begin{align*}
Q^{p^{\mu}}
  &= P_{1}^{\beta_{1} p^{\mu - m_{s + 1}}}
     P_{2}^{\beta_{2} p^{\mu - m_{s + 1}}} \Dots
     P_{s}^{\DPtypo{\beta}{\beta_{s}} p^{\mu - m_{s + 1}}} \\
  &= P_{1}^{\alpha_{1}}
     P_{2}^{\alpha_{2}} \Dots
     P_{s}^{\alpha_{s}},
\end{align*}
where $\alpha_{1}$, $\alpha_{2}$,~\Dots, $\alpha_{s}$ are all divisible by~$p^{\mu}$.

\Par{42.} It is clear, from the synthetic process by which it has
been proved that an Abelian group of order~$p^{m}$ can be generated
by a set of independent operations, that a considerable latitude
exists in the choice of the actual generating operations; and
the question arises as to the relations between the orders of the
distinct sets of independent generating operations.

The discussion of this question is facilitated by a consideration
of certain special sub-groups of~$G$. If $A$ and $B$ are two
%% -----File: 069.png---Folio 53-------
operations of~$G$, and if the order of~$A$ is not less than that of~$B$,
the order of~$AB$ is equal to, or is a factor of, the order of~$A$.
Hence the totality of those operations of~$G$ whose orders do not
exceed~$p^{\mu}$, or in other words of those operations which satisfy
the relation
\[
S^{p^{\mu}} = 1,
\]
form a sub-group~$G_{\mu}$. The order of~$G_{\mu}$ clearly depends on the
orders of the various operations of~$G$ and in no way on a special
choice of generating operations. Now if
\[
P_{1}^{\alpha_{1}} P_{2}^{\alpha_{2}} \Dots P_{r}^{\alpha_{r}}
\]
belongs to~$G_{\mu}$, then
\[
P_{1}^{\alpha_{1}p^{\mu}} P_{2}^{\alpha_{2}p^{\mu}} \Dots P_{r}^{\alpha_{r}p^{\mu}} = 1.
\]

Hence if $m_{s+1}$ is the first of the series
\[
m_{1},\ m_{2},\ \Dots,\ m_{r},
\]
which is less than~$\mu$, then $\alpha_{s+1}$,~\Dots, $\alpha_{r}$ may have any values
whatever; but $\alpha_{t}$ ($t = 1, 2, \Dots, s$) must be a multiple of~$p^{m_{t}-\mu}$.

It follows from this that $G_{\mu}$~is generated by the $r$~independent
operations
\[
P_{1}^{p^{m_{1}-\mu}},\
P_{2}^{p^{m_{2}-\mu}},\ \Dots,\
P_{s}^{p^{m_{s}-\mu}},\
P_{s+1},\ \Dots,\ P_{r}.
\]
If then the order of~$G_{\mu}$ is~$p^{\nu}$, we have
\[
\nu = \mu s + \sum_{s+1}^{r} m_{t}.
\]

The order of~$G_{1}$, the sub-group formed of all operations of~$G$
whose order is~$p$, is clearly~$p^{r}$.

\Par{43.} Suppose now that by a fresh choice of independent
generating operations, it were found that $G$ could be generated
by the $r'$~independent operations
\[
P_{1}',\ P_{2}',\ \Dots,\ P_{r'}'
\]
of orders
\[
p^{m_{1}'},\ p^{m_{2}'},\ \Dots,\ \DPtypo{p^{m_{r'}}}{p^{m_{r'}'}},
\]
where
\[
m_{1}' \geq m_{2}' \geq \Dots \geq \DPtypo{m_{r'}}{m_{r'}'}.
\]

If $m_{s'+1}'$ is the first of this series which is less than~$\mu$, the
order of~$G_{\mu}$ will be~$p^{\nu'}$, where
\[
\nu' = \mu s' + \sum_{s'+1}^{r'} m_{t}'.
\]

The order of~$G_{\mu}$ is independent of the choice of generating
operations; so that for all values of~$\mu$
\[
\nu = \nu'.
\]
%% -----File: 070.png---Folio 54-------

Hence, by taking $\mu = 1$,
\index{Abelian group@\Topic{Abelian group}!invariance of the orders of a set of independent generating operations of}%
\[
r = r',
\]
or the number of independent generating operations is independent
of their choice.

If now
\[
m_{t} = m_{t}'\quad (t = s + 1, s + 2, \Dots, r),
\]
and
\[
m_{s} > m_{s}',
\]
and if we choose $\mu$ so that
\[
m_{s} \geq \mu > m_{s}';
\]
then
\[
\nu = \mu s + \sum_{s + 1}^{r} m_{t},
\]
and
\[
\nu' = \mu (s - 1) + m_{s}' + \sum_{s + 1}^{r} m_{t}.
\]

The condition
\[
\nu = \nu'
\]
gives
\[
\mu = m_{s}',
\]
in contradiction to the assumption just made.

Similarly we can prove that the assumption $m_{s}' > m_{s}$ cannot
be maintained; hence
\[
m_{s} = m_{s}';
\]
and therefore, however the independent generating operations
of~$G$ are chosen, their number is always~$r$, and their orders are
\[
p^{m_{1}},\ p^{m_{2}},\ \Dots,\ p^{m_{r}}.
\]

\Par{44.} If $G'$ is a second Abelian group of order~$p^{m}$, simply
isomorphic with~$G$, and if
\[
P_{1}',\ P_{2}',\ \Dots,\ \DPtypo{P_{r'}}{P_{r'}'}
\]
of orders
\[
p^{m_{1}'},\ p^{m_{2}'},\ \Dots,\ \DPtypo{p^{m_{r'}}}{p^{m_{r'}'}},
\]
where
\[
m_{1}' \geq m_{2}' \geq \Dots \geq \DPtypo{m_{r'}}{m_{r'}'},
\]
are a set of independent generating operations of~$G'$, exactly the
same process as that of the last paragraph may be used to shew
that
\[
r = r',
\]
and
\[
m_{s} = m_{s}'\quad (s = 1, 2, \Dots, r).
\]

In fact, since corresponding operations of two simply isomorphic
groups have the same order, the order of~$G_{\mu}$ must be
equal to the order of~$G_{\mu}'$; and this is the condition that has
been used to obtain the result of the last paragraph.

Two Abelian groups of order~$p^{m}$ cannot therefore be simply
%% -----File: 071.png---Folio 55-------
\index{Abelian group@\Topic{Abelian group}!sub-groups of|(}%
\index{Abelian group@\Topic{Abelian group}!symbol for, of given type}%
isomorphic unless the series of integers $m_{1}$, $m_{2}$,~\Dots, $m_{r}$ is the
same for each. On the other hand when this condition is satisfied,
it is clear that the two groups are simply isomorphic, since
by taking $P_{s}$ and $P_{s}'$ ($s = 1, 2, \Dots, r$) as corresponding operations,
the isomorphism is actually established.

The number of distinct types of Abelian groups of order~$p^{m}$,
where $p$~is a prime, \ie\ the number of such groups no one
of which is simply isomorphic with any other, is therefore equal
to the number of partitions of~$m$. When the prime~$p$ is given,
each type of group may be conveniently, and without ambiguity,
represented by the symbol of the corresponding partition.
Thus the typical group~$G$ that we have been dealing with
would be represented by the symbol $(m_{1}, m_{2}, \Dots, m_{r})$.

\Par{45.} Having thus determined all distinct types of Abelian
groups of order~$p^{m}$, a second general problem in this connection
is the determination of all possible types of sub-group
when the group itself is given. This will be facilitated by the
consideration of a second special class of sub-groups in addition
to the sub-groups~$G_{\mu}$ already dealt with.

If $S$ and $S'$ are any two operations of~$G$, then
\[
S^{p^{\mu}} S'^{p^{\mu}} = (SS')^{p^{\mu}};
\]
and therefore the totality of the distinct operations obtained by
raising every operation of~$G$ to the power~$p^{\mu}$ will form a
sub-group~$H_{\mu}$.

If
\[
m_{s} \geq \mu > m_{s + 1},
\]
then (Theorem~II, §~41)
\[
S^{p^{\mu}}
  = P_{1}^{\alpha_{1} p^{\mu}}
    P_{2}^{\alpha_{2} p^{\mu}} \Dots
    P_{s}^{\alpha_{s} p^{\mu}},
\]
$S$~being any operation of~$G$. Hence $H_{\mu}$~is generated by the $s$~independent
operations
\[
P_{1}^{p^{\mu}},\ P_{2}^{p^{\mu}},\ \Dots,\ P_{s}^{p^{\mu}};
\]
and the order of~$H_{\mu}$ is~$p^{\esum_{1}^{s} m_{t} - \mu s}$.

\Par{46.} Let now $\Gamma$ of type $(n_{1}, n_{2}, \Dots, n_{s})$ be any sub-group of~$G$.
The order of the group~$\Gamma_{1}$, formed of all the operations of~$\Gamma$
which satisfy the equation
\[
S^{p} = 1,
\]
%% -----File: 072.png---Folio 56-------
is~$p^{s}$ (§~42). This group must be identical with or be a sub-group
of~$G_{1}$ whose order is~$p^{r}$. Hence
\[
s \leq r,
\]
\ie\ the number of independent generating operations of any
sub-group of~$G$ is equal to or is less than the number of independent
generating operations of $G$~itself.

If $p^{n}$ is the order of~$\Gamma$, there must be one or more sub-groups
of~$G$ of order~$p^{n + 1}$ which contain~$\Gamma$; for $\dfrac{G}{\Gamma}$~has operations
of order~$p$. Hence $\Gamma$~must be contained in one or more
sub-groups of~$G$ of order~$p^{m - 1}$. We may begin then by
considering all possible types of sub-groups of~$G$ of order~$p^{m - 1}$.
Let $g$ be such a sub-group and $(m_{1}', m_{2}', \Dots, m_{r'}')$ its type; and
suppose that
\[
m_{t} = m_{t}'\quad (t = 1, 2, \Dots, k - 1),
\]
and
\[
m_{k} \neq m_{k}'.
\]

Then if
\[
m_{k} < m_{k}',
\]
the order of~$h_{m_{k}}$, the sub-group of~$g$ which results by raising all
its operations to the power~$p^{m_{k}}$, is greater than the order of~$H_{m_{k}}$.
This is impossible, since $h_{m_{k}}$~must coincide with~$H_{m_{k}}$ or with one
of its sub-groups. Hence
\[
m_{k}' < m_{k}.
\]

Now
\[
\sum_{1}^{r'} m_{t}' = \sum_{1}^{r} m_{t} - 1;
\]
therefore
\[
m_{k}' = m_{k} - 1,
\]
and
\[
m_{t}' = m_{t}\quad (t = k + 1, \Dots, r).
\]

In the particular case in which $m_{r}$~is unity if we take $k$
equal to~$r$, $m_{r}'$~is zero; and in this case $g$ would have $r - 1$
generating operations.

It is easy to see that sub-groups of order~$p^{m - 1}$ and of all
the types just determined actually exist. For
\[
P_{1},\ P_{2},\ \Dots,\ P_{k - 1},\ P_{k^{p}},\ P_{k + 1},\ \Dots,\ P_{r}
\]
are a set of independent operations which generate a sub-group
of the type $(m_{1}, m_{2}, \dots, m_{k - 1}, m_{k} - 1, m_{k + 1}, \dots, m_{r})$.
%% -----File: 073.png---Folio 57-------

\Par{47.} Returning to the general case, we will assume for all
orders not less than~$p^{n}$ the existence of a sub-group~$\Gamma$ of type
$(n_{1}, n_{2}, \Dots, n_{s})$, when the inequalities
\[
\sum_{1}^{t} n_{u} \leq \sum_{1}^{t} m_{u}\quad (t = 1, 2, \Dots\Add{,} r)
\]
are satisfied, where if $t > s$, $n_{t}$~is zero. When $n$ is equal to $m - 1$,
the truth of this assumption has been established.

If $\Gamma'$ of type $(n_{1}', n_{2}', \Dots, n_{s'}')$ is a sub-group of order~$p^{n - 1}$, it
must be contained in some sub-group of order~$p^{n}$, say $\Gamma$ of type
$(n_{1}, n_{1}, \Dots, n_{s})$. Then, by the preceding discussion,
\[
\sum_{1}^{t} n_{u}' \leq \sum_{1}^{t} n_{u}\quad (t = 1, 2, \Dots, s);
\]
and therefore
\[
\sum_{1}^{t} n_{u}' \leq \sum_{1}^{t} m_{u}\quad (t = 1, 2, \Dots, r),
\]
so that the conditions assumed for sub-groups of order not less
than~$p^{n}$ are necessary for sub-groups of order~$p^{n - 1}$. Moreover if
the conditions
\[
\sum_{1}^{t} n_{u}' \leq \sum_{1}^{t} m_{u}\quad (t = 1, 2, \Dots, r)
\]
are satisfied, and if $n_{k}'$ is the first of the series
\[
n_{1}',\ n_{2}',\ \Dots,
\]
which is less than the corresponding term of the series
\[
m_{1},\ m_{2},\ \Dots,
\]
we may take
\begin{align*}
n_{u} &= n_{u}',\quad u \neq k, \\
n_{k} &= n_{k}' + 1;
\end{align*}
and then the conditions
\[
\sum_{1}^{t} n_{u} \leq \sum_{1}^{t} m_{u}\quad (\DPtypo{k}{t} = 1, 2, \Dots, r);
\]
are satisfied. But these conditions being satisfied it follows,
from the assumption made, that $G$ contains a sub-group of order~$p^{n}$
and type $(n_{1}, n_{2}, \Dots, n_{s})$ and therefore also a sub-group of
order~$p^{n-1}$ and type $(n_{1}', n_{2}', \Dots, \DPtypo{n_{s'}}{n_{s'}'})$. If the conditions assumed
are sufficient for the existence of a sub-group of order~$p^{n}$, they
are thus proved to be sufficient for the existence of one of
order~$p^{n-1}$. Hence since they are sufficient in the case of
sub-groups of order~$p^{m-1}$, they are so generally.
%% -----File: 074.png---Folio 58-------

We may summarize the results obtained in the last three
paragraphs as follows:---

\begin{Theorem}[III.]
The number of distinct types of Abelian
groups of order~$p^{m}$, where $p$~is a prime, is equal to the number of
partitions of~$m$; and each type may be completely represented
by the symbol $(m_{1}, m_{2}, \Dots, m_{r})$ of the corresponding partition.
If the numbers in the partition are written in descending order,
a group of type $(m_{1}, m_{2}, \Dots, m_{r})$ will have a sub-group of type
$(n_{1}, n_{2}, \Dots, n_{s})$, when the conditions
\begin{gather*}
s \leq r, \\
\sum_{1}^{t} n_{u} \leq \sum_{1}^{t} m_{u}\quad (t = 1, 2, \Dots, r),
\end{gather*}
are satisfied; and the type of every sub-group must satisfy these
conditions.
\end{Theorem}

\Par{48.} It will be seen later that the Abelian group of order~$p^{m}$
\index{Abelian group@\Topic{Abelian group}!of order~$p^{m}$ and type $(1, 1, \dots, 1)$}%
and type $(1, 1, 1, \Dots, \text{with $m$ units})$ is of special importance
in the general theory, and we shall here discuss one or two of
its simpler properties.

Since the generating operations of the group are all of order~$p$,
every operation except identity is of order~$p$; and therefore
the type of any sub-group of order~$p^{s}$ is $(1, 1, 1, \Dots, \text{with $s$
units})$. If the group be denoted by~$G$, every sub-group~$G_{\mu}$
coincides with~$G$; while of the sub-groups~$H_{\mu}$, the first coincides
with~$G$ and all the rest consist of the identical operation only.

In choosing a set of independent generating operations, we
may take for the first,~$P_{1}$, any one of the $p^{m} - 1$ operations of
the group, other than identity. The sub-group~$\{P_{1}\}$ is of order~$p$;
and therefore $G$ has $p^{m} - p$ operations which are not contained
in~$\{P_{1}\}$. If we choose any one of these,~$P_{2}$, it is necessarily
independent of~$P_{1}$, and may be taken as a second generating
operation. The sub-group $\{P_{1}, P_{2}\}$ is of order~$p^{2}$ and type
$(1, 1)$; and $G$ has $p^{m} - p^{2}$ operations which are not contained in
this sub-group. If $P_{3}$ be any one of these, no power of~$P_{3}$
other than identity is contained in $\{P_{1}, P_{2}\}$; and $P_{1}$,~$P_{2}$,~$P_{3}$ are
therefore three independent operations which generate a sub-group
of order~$p^{3}$. This process may clearly be continued till
all $m$ generating operations have been chosen. If then the
%% -----File: 075.png---Folio 59-------
position which each generating operation occupies in the set of~$m$,
when they are written in order, be taken into account, there are
\[
(p^{m} - 1) (p^{m} - p) (p^{m} - p^{2}) \Dots (p^{m} - p^{m - 1})
\]
distinct ways in which a set may be chosen. If on the other
hand the sets of generating operations which consist of the
same operations written in different orders be regarded as
identical, the number of distinct sets is
\[
\frac{(p^{m} - 1) (p^{m} - p) \Dots (p^{m} - p^{m - 1})}{m!}.
\]

\Par{49.} No operation $P$ of the group can belong to two distinct
\index{Abelian group@\Topic{Abelian group}!of order~$p^{m}$ and type $(1, 1, \dots, 1)$!number of distinct ways of choosing a set of independent generating operations of}%
sub-groups of order~$p$ except the identical operation. Hence
since every sub-group of order~$p$ contains $p - 1$ operations
besides identity, $G$~must contain $\dfrac{p^{m} - 1}{p - 1}$ sub-groups of order~$p$.

Let $N_{m, r}$ be the number of sub-groups of~$G$ of order~$p^{r}$, so
that
\[
N_{m, 1} = \frac{p^{m} - 1}{p - 1}.
\]

There are, in~$G$, $p^{m} - p^{r}$ operations not contained in any given
sub-group of order~$p^{r}$. If $P$~occurs among these operations, so
also do $P^{2}$, $P^{3}$,~\Dots, $P^{p - 1}$. Hence there are $\dfrac{p^{m} - p^{r}}{p - 1}$ sub-groups of
order~$p$ in~$G$ which are not contained in a given sub-group of
order~$p^{r}$. Each of these may be combined with the given
sub-group to give a sub-group of order~$p^{r + 1}$. When every
sub-group of order~$p^{r}$ is treated in this way, every sub-group of
order~$p^{r + 1}$ will be formed and each of them the same number,~$x$,
of times. Hence
\[
xN_{m, r + 1} = N_{m, r} \frac{p^{m} - p^{r}}{p - 1}.
\]

Now a sub-group of order~$p^{r + 1}$ contains $N_{r + 1, r}$ sub-groups of
order~$p^{r}$, and $\dfrac{p^{r + 1} - p^{r}}{p - 1}$ sub-groups of order~$p$ which are not
contained in any given sub-group of order~$p^{r}$. Hence
\[
x = \frac{p^{r + 1} - p^{r}}{p - 1}  N_{r + 1, r},
\]
and therefore
\[
N_{m, r + 1} = \frac{N_{m, r}}{N_{r + 1\DPtypo{\,}{,} r}} · \frac{p^{m - r} - 1}{p - 1}.
\]
\index{Abelian group@\Topic{Abelian group}!sub-groups of|)}%
%% -----File: 076.png---Folio 60-------

We will now assume that
\[
N_{m, t} = \frac{(p^{m} - 1)(p^{m - 1} - 1) \Dots (p^{m - t + 1} - 1)}{(p - 1)(p^{2} - 1) \Dots (p^{t} - 1)},
\]
for all values of~$m$ and for values of $t$ not exceeding~$r$. This
has been proved for $r = 1$. Then it follows, from the above
relation, that
\[
N_{m, r + 1} = \frac{(p^{m} - 1)(p^{m - 1} - 1) \Dots (p^{m - r} - 1)}{(p - 1)(p^{2} - 1) \Dots (p^{r + 1} - 1)},
\]
that is to say, if the result is true for values of $t$ not exceeding~$r$,
it is also true when $t = r + 1$. Hence the formula is true
generally.

It may be noticed that
\[
N_{m, t} = N_{m, m - t}.
\]

\Par{50.} \begin{Ex}[1.] Shew that a group whose operations except
identity are all of order~$2$ is necessarily an Abelian group.
\index{Abelian group@\Topic{Abelian group}!of order~$p^{m}$ and type $(1, 1, \dots, 1)$!number of sub-groups of, whose order is given}%
\end{Ex}

\begin{Ex}[2.] Prove that in a group of order~$16$, whose operations
except identity are all of order~$2$, the $15$~operations of order~$2$ may
be divided into $5$~sets of $3$ each so that each set of~$3$ with identity
forms a sub-group of order~$4$; and that this division into sets may
be carried out in $\Erratum{7}{56}$ distinct ways.
\end{Ex}

\begin{Ex}[3.] If $G$ is an Abelian group and $H$ a sub-group of~$G$,
shew that $G$~contains one or more sub-groups simply isomorphic
with~$\dfrac{G}{H}$.
\end{Ex}

\begin{Ex}[4.] If the symbols in the successive rows of a determinant
of $n$~rows are derived from those of the first row by performing on
them the substitutions of a regular Abelian group of order~$n$, prove
that the determinant is the product of $n$~linear factors.
\Attrib{\textit{Messenger of Mathematics}, Vol.~\VolNo{XXIII} p.~112.}
\end{Ex}

\begin{Ex}[5.] Discuss the number of ways in which a set of
independent generating operations of an Abelian group of order~$p^{m}$
and given type may be chosen. Shew that, for a group of type
$(m_{1}, m_{2}, \Dots, m_{r})$, where $m_{1} > m_{2} > \Dots > m_{r}$, the number of ways is
of the form $p^{\alpha}(p - 1)^{r}$; and in particular that for a group of order
$p^{\efrac{1}{2} n(n + 1)}$ and type $(n, n - 1, \Dots, 2, 1)$, the number of ways is
$p^{\nu}(p - 1)^{n}$, where $\nu = \frac{1}{6} n(n + 1)(2n + 1) - n$.
\end{Ex}

\begin{Ex}[6.] Shew that for any Abelian group a set of independent
generating operations
\[
S_{1},\ S_{2},\ \dots,\ S_{r - 1},\ S_{r},\ \dots,\ S_{n}
\]
can be chosen such that, for each value of~$r$, the order of~$S_{r}$ is equal
to, or is a factor, of the order of~$S_{r - 1}$.
\end{Ex}
%% -----File: 077.png---Folio 61-------


\Chapter{V.}{On Groups Whose Orders Are the Powers of
Primes.}

\Par{51.} \First{Having} in the last chapter dealt in some detail with
\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime|EtSeq}%
Abelian groups of order~$p^{m}$, where $p$~is a prime, we shall now investigate
some of the more important properties of groups, which
have the power of a prime for their order but are not necessarily
Abelian. Besides illustrating and leading to many interesting
applications of the general theorems of Chapter~III, the discussion
of groups, whose order is the power of a prime, will be
found in many ways to facilitate the subsequent discussion of
other groups, whose order is not thus limited.

\Par{52.} If $G$~is a group whose order is~$p^{m}$, where $p$~is a prime,
the order of every sub-group of~$G$ must also be a power of~$p$; and
therefore (§~25) the total number of operations of~$G$ which are
conjugate with any given operation must be a power of~$p$. The
identical operation of~$G$ is self-conjugate. Hence if the operations
of~$G$, other than the identical operation, are distributed
in conjugate sets containing $p^{\alpha}$, $p^{\beta}$,~\Dots\ operations, the total
number of operations of~$G$ is $1 + p^{\alpha} + p^{\beta} + \Dots$; and therefore
\[
p^{m} = 1 + p^{\alpha} + p^{\beta} + \Dots\Add{.}
\]

This equation can only be true if $p^{r_{1}} - 1$ of the indices
$\alpha$, $\beta$,~\Dots\ are zero, $r_{1}$~being some integer not less than unity.
%% -----File: 078.png---Folio 62-------
\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!always contain self-conjugate operations}%
\index{Self-conjugate operation@\Topic{Self-conjugate operation}!of a group whose order is the power of a prime}%
Hence $G$~must contain $p^{r_{1}}$ self-conjugate operations, which form
(§~27) a self-conjugate sub-group\footnotemark.
\footnotetext{Sylow, \textit{Math.\ Ann.}\ \VolNo{V}.\ (1872), p.~588.} %[** No comma or "Vol."]
\index{Sylow@\Topic{Sylow}, quoted}%
Hence:---

\begin{Theorem}[I.]
Every group whose order is the power of a
prime contains self-conjugate operations, other than the identical
operation; and no such group can be simple.
\end{Theorem}

\Par{53.} Let $H_{1}$ be the sub-group of~$G$ of order~$p^{r_{1}}$, formed of
its self-conjugate operations. Then $\dfrac{G}{H_{1}}$~is a group of order~$p^{m - r_{1}}$,
and it must contain self-conjugate operations other
than identity, forming a self-conjugate sub-group of order~$p^{r_{2}}$.
Corresponding to this self-conjugate sub-group of~$\dfrac{G}{H_{1}}$, $G$~must
have a self-conjugate sub-group~$H_{2}$ of order~$p^{r_{1} + r_{2}}$, which
contains~$H_{1}$. If $r_{1} + r_{2} < m$, $\dfrac{G}{H_{2}}$~is a group of order~$p^{m - r_{1} - r_{2}}$; and
it again will have self-conjugate operations forming a sub-group
of order~$p^{r_{3}}$. Corresponding to this, $G$~will have a self-conjugate
sub-group~$H_{3}$ of order~$p^{r_{1} + r_{2} + r_{3}}$, which contains~$H_{2}$.
If the process thus indicated is continued, till $G$~itself is arrived
at, a series of groups
\[
H_{1},\ H_{2},\ \Dots,\ H_{n},\ G
\]
is formed, each of which is self-conjugate in~$G$ and contains all
that precede it\footnotemark.
\footnotetext{Let $P$~be an operation or \DPchg{subgroup}{sub-group} of~$G$, and let $H$ be a \DPchg{subgroup}{sub-group} of~$G$
  which contains~$P$. If every operation of~$H$ is permutable with~$P$, we shall
  speak of $P$ as ``self-conjugate in~$H$.'' The phrase is to be regarded as
  emphasizing the relation of~$P$ to a sub-group of~$G$ which contains it; and not
  as implying that $H$ is the greatest, or the only, \DPchg{subgroup}{sub-group} of~$G$, within which $P$
  is a self-conjugate operation or \DPchg{subgroup}{sub-group}.}
Moreover $\dfrac{H_{r + 1}}{H_{r}}$~is the group formed of the
self-conjugate operations of~$\dfrac{G}{H_{r}}$, and it is therefore an Abelian
group.

The series of sub-groups thus arrived at is of considerable
importance. From the mode of formation, it is clear that a
second group~$G'$ of order~$p^{m}$ cannot be simply isomorphic with~$G$,
unless $\dfrac{H_{r + 1}}{H_{r}}$ and $\dfrac{H_{r + 1}'}{H_{r}'}$ are simply isomorphic for each value
%% -----File: 079.png---Folio 63-------
of~$r$. It by no means however follows that, when this latter
condition is satisfied, $G$~and $G'$ are always simply isomorphic.
This is indeed not necessarily the case; instances
to the contrary will be found among the groups of order
$p^{3}$ and $p^{4}$ in §§~69--72. While in general there is no limitation
on the type of any Abelian factor-group $\dfrac{H_{r + 1}}{H_{r}}$ formed from
two consecutive terms of the series, it is to be noted that
the last factor-group $\dfrac{G}{H_{n}}$ cannot be cyclical\footnotemark,
\footnotetext{Young, \textit{American Journal}, vol.~\VolNo{XV} (1893), p.~132.}%
\index{Young@\Topic{Young}, quoted}%
a restriction that
can be established as follows. If $\dfrac{G}{H_{n}}$ were cyclical, then, because
$\dfrac{H_{n}}{H_{n - 1}}$ is the group formed of the self-conjugate operations of
$\dfrac{G}{H_{n - 1}}$, the operations of $\dfrac{G}{H_{n - 1}}$ could be arranged in the sets
\[
\dfrac{H_{n}}{H_{n - 1}}, \quad
P\dfrac{H_{n}}{H_{n - 1}}, \quad
P^{2} \dfrac{H_{n}}{H_{n - 1}}, \quad
\dots,\quad
P^{p^{\alpha - 1}} \dfrac{H_{n}}{H_{n - 1}},
\]
$p^{\alpha}$ being the order of $\dfrac{G}{H_{n}}$ and $P^{p^{\alpha}}$~being an operation of~$\dfrac{H_{n}}{H_{n - 1}}$.
But since $P$~is permutable with its own powers and with every
operation of~$\dfrac{H_{n}}{H_{n - 1}}$, these operations would form an Abelian
group. Now the self-conjugate operations of~$\dfrac{G}{H_{n - 1}}$ form the
group $\dfrac{H_{n}}{H_{n - 1}}$, and therefore $\dfrac{G}{H_{n - 1}}$ cannot be Abelian. Hence
the assumption, that $\dfrac{G}{H_{n}}$~is cyclical, is incorrect.

\begin{Remark}
One immediate consequence of this result is that a group~$G$ of
order~$p^{2}$ is necessarily Abelian. For if it were not, its self-conjugate
operations would form a sub-group~$H$ of order~$p$, and $\dfrac{G}{H}$~being of
order~$p$ would be cyclical.

A second consequence is that if $G$, of order~$p^{m}$, is not Abelian,
the order of the sub-group~$H_{1}$, formed of the self-conjugate operations
of~$G$, cannot exceed $p^{m - 2}$.
\end{Remark}
%% -----File: 080.png---Folio 64-------

\Par{54.} It was proved in the last chapter that an Abelian
group has one or more sub-groups the order of which is any
given factor of the order of the group itself. Hence, since $\dfrac{H_{r + 1}}{H_{r}}$
is Abelian, $H_{r + 1}$~must have one or more sub-groups which contain~$H_{r}$,
of any given order greater than that of~$H_{r}$ and less than its
own.

In particular, if $p^{s}$ and $p^{t}$ are the orders of $H_{r}$ and $H_{r + 1}$,
then $\dfrac{H_{r + 1}}{H_{r}}$ must have a series of sub-groups of orders
\[
p^{t - s - 1},\ p^{t - s - 2},\ \Dots,\ p,\ 1,
\]
each of which contains those that follow it, while each one is of
necessity self-conjugate in~$\dfrac{G}{H_{r}}$. Hence between $H_{r}$ and~$H_{r + 1}$, a
series of groups $h_{s + 1}$, $h_{s + 2}$,~\Dots, $h_{t - 1}$ of orders
$p^{s + 1}$, $p^{s + 2}$,~\Dots, $p^{t - 1}$
can be chosen, so that each of the series
\[
H_{r},\ h_{s + 1},\ h_{s + 2},\ \Dots,\ h_{t - 1},\ H_{r + 1}
\]
is self-conjugate in~$G$ and contains all that precede it. We can
therefore always form, in at least one way, a series of sub-groups
of~$G$ of orders
\[
1,\ p,\ p^{2},\ \Dots,\ p^{m - 1},\ p^{m},
\]
such that each is self-conjugate in~$G$ and contains all that
precede it.

It will be shewn in the next paragraph that every sub-group
of order~$p^{s}$ of~$G$ is contained self-conjugately in some
sub-group of order~$p^{s + 1}$, and that every sub-group of order~$p^{s}$ of
a group of order~$p^{s + 1}$ is self-conjugate. Assuming this result, it
follows at once that a series of sub-groups of orders
\[
1,\ p,\ p^{2},\ \Dots,\ p^{m - 1},\ p^{m}
\]
can always be formed, which shall contain any one given
sub-group and which shall be such that each group of the
series contains the previous one self-conjugately. It will not
however, in general, be the case, as in the previous series, that
each group of the series is self-conjugate in~$G$.

\Par{55.} Let $G_{s}$ be any sub-group of~$G$ of order~$p^{s}$, and let $H_{r + 1}$
be the first of the series of sub-groups
\[
H_{1},\ H_{2},\ \Dots,\ H_{n}
\]
%% -----File: 081.png---Folio 65-------
which is not contained in~$G_{s}$. Then $\dfrac{G_{s}}{H_{r}}$~is a sub-group of $\dfrac{G}{H_{r}}$
which does not contain all the operations of~$\dfrac{H_{r + 1}}{H_{r}}$. Every
operation of $\dfrac{H_{r + 1}}{H_{r}}$ is self-conjugate in~$\dfrac{G}{H_{r}}$; and therefore $\dfrac{G_{s}}{H_{r}}$ is
self-conjugate in $\left\{\dfrac{G_{s}}{H_{r}}, \dfrac{H_{r + 1}}{H_{r}}\right\}$, a group of order greater than its
own. Hence $G_{s}$~must be contained self-conjugately in some
group $G_{s + t}$ of order~$p^{s + t}$, where $t$~is not less than unity.
Moreover since $\dfrac{G_{s + t}}{G_{s}}$ must contain operations of order~$p$, there
must be one or more groups of order~$p^{s + 1}$ which contain~$G_{s}$
self-conjugately. Hence:---

\begin{Theorem}[II.]
If $G_{s}$ of order~$p^{s}$ is a sub-group of~$G$, which
\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!every sub-group of, is contained self-conjugately in a greater sub-group}%
is of order~$p^{m}$, then $G$~must contain a sub-group of order~$p^{s + t}$,
$t \nless 1$, within which $G_{s}$~is self-conjugate. In particular, every
sub-group of order~$p^{m - 1}$ of~$G$ is a self-conjugate sub-group\footnotemark.
\end{Theorem}
\index{Burnside@\Topic{Burnside}, quoted}%
\index{Frobenius@\Topic{Frobenius}, quoted}%
\footnotetext{Frobenius, ``Ueber endliche Gruppen,'' \textit{Berliner Sitzungsberichte} (1895),
  p.~173: Burnside, ``Notes on the theory of groups of finite order,'' \textit{Proc.\ London
  Mathematical Society}, Vol.~\VolNo{XXVI} (1895), p.~209.}

Suppose now that $G_{s + t}$, of order~$p^{s + t}$, is the greatest sub-group
of~$G$ which contains a given sub-group~$G_{s}$, of order~$p^{s}$,
self-conjugately; so that $G_{s}$~is one of $p^{m - s - t}$ conjugate sub-groups.
Suppose also that $G_{s + t + u}$ of order~$p^{s + t + u}$, ($u \nless 1$), is
the greatest sub-group of~$G$ that contains $G_{s + t}$ self-conjugately.
Every operation of~$G_{s + t + u}$ transforms $G_{s + t}$ into itself; and no
operation of $G_{s + t + u}$ that is not contained in $G_{s + t}$ transforms $G_{s}$
into itself. Hence, in~$G_{s + t + u}$, $G_{s}$~is one of $p^{u}$~conjugate sub-groups
and each of these is self-conjugate in~$G_{s + t}$.

The $p^{m - s - t}$ sub-groups conjugate with~$G_{s}$ may therefore be
divided into $p^{m - s - t - u}$ sets of $p^{u}$~each, ($u \nless 1$), such that any
operation of one of the sets transforms each sub-group of that
set into itself.

Similarly if $G_{r}$, of order~$p^{r}$, is the greatest sub-group of~$G$
that contains a given operation $P$ self-conjugately, and if $G_{r + s}$,
$s \nless 1$, is the greatest sub-group that contains $G_{r}$ self-conjugately,
then $G_{r}$~must contain self-conjugately $p^{s}$~operations of the
%% -----File: 082.png---Folio 66-------
conjugate set to which $P$ belongs, and therefore any two of
these $p^{s}$~operations are permutable. Hence the $p^{m - r}$ conjugate
operations of the set to which $P$ belongs can be divided into
$p^{m - r - s}$ sets of $p^{s}$~each, ($s \nless 1$), such that all the operations of
any one set are permutable with each other. In particular, if
$P$~is one of a set of $p$ conjugate operations, all the operations of
the set are permutable.

\Par{56.} If $P$ is an operation of~$G$, of order~$p^{n}$, which is conjugate
to one of its own powers~$P^{\alpha}$, there must be some other
operation~$Q$ such that
\[
Q^{-1}PQ = P^{\alpha}.
\]

From this equation it follows that
\[
Q^{-1}P^{x}Q = (Q^{-1}PQ)^{x} = P^{x \alpha}.
\]

Again
\begin{alignat*}{2}
Q^{-2}PQ^{2} &= Q^{-1}P^{\alpha}Q &&= P^{\alpha^{2}}, \\
Q^{-3}PQ^{3} &= Q^{-1}P^{\alpha^{2}}Q &&= P^{\alpha^{3}},
\end{alignat*}
and so on. Hence
\[
Q^{-y}P^{x}Q^{y} = P^{x \alpha^{y}}.
\]

If $Q^{\beta}$ is the lowest power of~$Q$ which is permutable with~$P$,
then in~$\{P, Q\}$\DPnote{* comma?} $P$~will be one of $\beta$ conjugate operations, and
therefore $\beta$~must be a power of~$p$, say~$p^{r}$, less than~$p^{n}$. Now
\[
Q^{-p^{r}}PQ^{p^{r}} = P^{\alpha^{p^{r}}};
\]
and therefore
\[
\alpha^{p^{r}} \equiv 1\ (\mod p^{n}),
\]
while
\[
\alpha^{p^{r - 1}} \nequiv 1\ (\mod p^{n}).
\]

First, we will suppose that $p$~is an odd prime. Then since
\[
x^{p^{r}} = x\ (\mod p),
\]
whatever integer $x$ may be, we may assume that $\alpha = 1 + kp^{s}$,
where $k$~is not a multiple of~$p$; and then
\begin{gather*}
\alpha^{p^{r}} = 1 + kp^{r + s} + \Dots \\
\alpha^{p^{r - 1}} = 1 + kp^{r + s - 1} + \Dots
\end{gather*}

Hence
\[
r + s = n,
\]
and
\[
\alpha \equiv 1\ (\mod p^{n - r}).
\]

In particular, we see that no operation of order~$p$ can be
conjugate to one of its powers. Hence if $P$~and~$P'$ are two
%% -----File: 083.png---Folio 67-------
conjugate operations of order~$p$, $\{P\}$~and $\{P'\}$ have no operation
in common except identity. Also, if $\{P\}$ be a self-conjugate
sub-group of order~$p$, each of its operations is self-conjugate.

If $p$ is~$2$, we must take
\[
\alpha = ±1 + k2^{s},
\]
where $k$~is odd, and we are led by the same process to the
result
\[
\alpha \equiv ±1\ (\mod 2^{n - r}).
\]

\Par{57.} If $P$ is an operation of~$G$ which belongs to the sub-group~$H_{r + 1}$
(§~53) and to no previous sub-group in the set
\[
1,\ H_{1},\ H_{2},\ \Dots,\ H_{n},\ G,
\]
then $\{P, H_{r}\}$ is a sub-group of~$H_{r + 1}$; and therefore every
operation of $\dfrac{\{P, H_{r}\}}{H_{r}}$ is self-conjugate in~$\dfrac{G}{H_{r}}$. The set of
operations~$PH_{r}$ is therefore transformed into itself by every
operation of~$G$, and hence every operation that is conjugate to~$P$
is contained in the set~$PH_{r}$. Suppose now, if possible, that
all the operations conjugate with~$P$ are contained in the set~$PH_{r - 1}$.
Then every operation of~$G$ transforms this set into
itself, and therefore every operation of $\dfrac{\{P, H_{r - 1}\}}{H_{r - 1}}$ is self-conjugate
in~$\dfrac{G}{H_{r - 1}}$. This however cannot be the case, since by supposition
$P$~does not belong to~$H_{r}$. Hence among the operations
conjugate to~$P$, there must be some which belong to~$PH_{r}$
and not to~$PH_{r - 1}$. In particular, if $P$~is one of $p$ conjugate
operations, no operation conjugate to~$P$ can belong to the set~$PH_{r - 1}$.
For there must be an operation~$Q$ such that $Q^{-1}PQ$
belongs to $PH_{r}$ and not to~$PH_{r - 1}$; and this being the case, no
one of the operations $Q^{-x}PQ^{x}$, ($x = 1, 2, \Dots, p - 1$), can belong
to~$PH_{r - 1}$. But these operations with~$P$ constitute the conjugate
set.

It follows from the above reasoning that, if $P$~belongs to~$H_{r + 1}$,
then each operation of the set $P^{-\alpha} Q^{-1} P^{\alpha}Q$, where $Q$~is any
operation of~$G$, belongs to~$H_{r}$. If $Q$~belongs to~$H_{s + 1}$, ($s < r$), the
operation $P^{-\alpha} Q^{-1} P^{\alpha}Q$, regarded as the product of $P^{-\alpha} Q^{-1} P^{\alpha}$ and~$Q$,
must similarly belong to~$H_{s}$. Hence unless every operation
%% -----File: 084.png---Folio 68-------
of~$H_{s + 1}$ is permutable with~$P$, there must be operations conjugate
to~$P$, which belong to the set~$PH_{s}$. Moreover those
operations of~$G$ which transform $P$ into operations of the set~$PH_{s}$
form a sub-group. For if
\[
S^{-1}PS = Ph_{s},
\]
and
\[
S'^{-1}PS' = Ph_{s}',
\]
then
\[
S'^{-1} S^{-1} PSS' = Ph_{s}' · S'^{-1} h_{s}S,
\]
and when both $h_{s}$ and $h_{s}'$ belong to~$H_{s}$, so also does $h_{s}' S'^{-1} h_{s}S'$.

\begin{Remark}
\Par{58.} In illustration of the foregoing paragraph, we will consider
a group~$G$ of order~$p^{r + s}$\Erratum{}{ where $p$~is an odd prime} in which the group~$H_{1}$, is of type~$(r)$, while
$\dfrac{G}{H_{1}}$~is an Abelian group of type~$(1, 1, \Dots \text{with $s$ units})$.

Let $Q$~be an operation of order~$p^{r}$ that generates~$H_{1}$; and let
$P_{1}$, $P_{2}$,~\Dots, $P_{s}$ be $s$~operations no one of which is a power of any
other, such that, with $Q$, they generate~$G$.

If $P_{1}$ and $P_{2}$ are not permutable, then
\[
P_{2}^{-1} P_{1} P_{2} = P_{1} Q^{\alpha},
\]
and
\[
P_{2}^{-p} P_{1} P_{2}^{p} = P_{1} Q^{p\alpha}.
\]

Now $P_{2}^{p}$ belongs to~$H_{1}$, and therefore
\[
Q^{p\alpha} = 1,
\]
so that
\[
\alpha = kp^{r-1}.
\]

Hence the only operations which are conjugate to~$P_{1}$ are
\[
P_{1},\ P_{1} Q^{p^{r - 1}},\ P_{1} Q^{2p^{r - 1}},\ \Dots,\ P_{1} Q^{(p-1)p^{r - 1}};
\]
and $P_{1}$ is one of a set of $p$ conjugate operations, which are transformed
among themselves by~$P_{2}$. Similarly $P_{2}$~is one of a set of $p$
conjugate operations which are transformed among themselves by~$P_{1}$.
Hence both $P_{1}$ and $P_{2}$ are permutable with each of the
operations $P_{3}$, $P_{4}$,~\Dots, $P_{s}$. Again, $P_{3}$~is not self-conjugate, and
there must therefore be another, say~$P_{4}$, of the set $P_{3}$, $P_{4}$,~\Dots, $P_{s}$
which, with its powers, transforms $P_{3}$ into a set of $p$ conjugate
operations. Then $P_{3}$~will similarly transform~$P_{4}$; and $P_{3}$,~$P_{4}$ will
be permutable with each of the set $P_{5}$,~\Dots, $P_{s}$, $P_{1}$,~$P_{2}$. Hence
finally the set of $s$ operations $P_{1}$, $P_{2}$,~\Dots, $P_{s}$ must be divisible
into sets of two, such that each pair are permutable with all the
remaining operations, but are not permutable with each other. The
group can therefore only exist if $s$~is even\footnotemark.
\footnotetext{Young, ``On groups whose order is the power of a prime,'' \textit{American
  Journal of Mathematics}, Vol.~\VolNo{XV} (1893), p.~171.}
\index{Young@\Topic{Young}, quoted}%
%% -----File: 085.png---Folio 69-------

The sub-group $\{P_{1}, P_{2}, Q\}$ is a group which satisfies the same
conditions as~$G$, when $s = 2$. Its order is~$p^{r + 2}$, and it contains a
self-conjugate operation of order~$p^{r}$. Now we shall see in §§~65,~66
that such a group is necessarily of one of two types\footnotemark,
\footnotetext{Let
  \begin{gather*}
  S_{1}^{\alpha_{1}} = 1,\quad S_{2}^{\alpha_{2}} = 1,\ \Dots,\quad S_{n}^{\alpha_{n}} = 1, \\
  f_{1}(S_{i}) = 1,\quad f_{2}(S_{i}) = 1,\ \Dots,\quad f_{k}(S_{i}) = 1,
  \end{gather*}
  where $f_{j}(S_{i})$ is an abbreviation for an expression of the form $S_{p}^{a} S_{q}^{b} \Dots S_{r}^{c}$, be a
  set of relations, such as was considered in §~18, which completely specify a
  group. We may then, without altering the sense in which the word ``type''
  has been used (§§~19,~44), speak of \emph{the type of group} defined by these relations.
  It is essential however that the relations should completely specify the group,
  as otherwise they will define more than one type. For instance, it is clear, from
  §~56, that the equations
  \[
  Q^{p} = 1,\quad P^{p^{2}} = 1,\quad Q^{-1}PQ = P^{\alpha},
  \]
  where $p$~is a given prime, but $\alpha$~is not given, will define more than one type of
  group. Any data in fact, which completely specify a group, may be said to
  define a type of group. Thus in dealing with Abelian groups of order~$p^{m}$, a type
  is defined by a partition of~$m$.}%
the generating
operations of which satisfy one or the other of the sets of equations
\[
Q^{p^{r}} = 1,\quad
P_{1}^{p} = 1,\quad
P_{2}^{p} = 1,\quad
P_{2}^{-1} P_{1}P_{2} = P_{1} Q^{p^{r - 1}}:
\]
or
\[
Q^{p^{r}} = 1,\quad
P_{1}^{p} = Q,\quad
P_{2}^{p} = 1,\quad
P_{2}^{-1} P_{1}P_{2} = P_{1} Q^{p^{r - 1}}.
\]

The same is true for the sub-groups $\{P_{3}, P_{4}, Q\}$, etc.; and all the
operations of any one of these sub-groups are permutable with each
operation of $\{P_{1}, P_{2}, Q\}$.

Hence finally, since the equations to be satisfied by each pair of
operations, such as $P_{1}$~and~$P_{2}$, may be chosen in two distinct ways,
there are in all $2^{\efrac{1}{2}s}$ distinct types of group of order~$p^{r + s}$, for which
$H_{1}$~is a cyclical group of order~$p^{r}$, and $\dfrac{G}{H_{1}}$~is an Abelian group of type
$(1, 1, \Dots\DPchg{\,}{,} \text{to $s$ units})$.
\end{Remark}

\Par{59.} It has been seen in §~55 that every sub-group~$G'$ of
order~$p^{m - 1}$ of a group~$G$ of order~$p^{m}$ is self-conjugate. Suppose
now that $G$~contains two such sub-groups $G'$~and~$G''$. Then
since $G'$ and $G''$ are permutable with each other, while the
order of~$\{G', G''\}$ is~$p^{m}$, the order of the greatest group~$g'$
common to them must (§~33) be~$p^{m - 2}$; and since $g'$~is the
greatest common sub-group of two self-conjugate sub-groups of~$G$,
it must itself be a self-conjugate sub-group of~$G$. The factor
group $\dfrac{G}{g'}$ of order~$p^{2}$ contains the two distinct sub-groups $\dfrac{G'}{g'}$
and $\dfrac{G''}{g'}$, which are of order~$p$ and permutable with each other.
%% -----File: 086.png---Folio 70-------
Hence $\dfrac{G}{g'}$ must be an Abelian group of type $(1, 1)$, and
it therefore contains (§~49) $p + 1$ sub-groups of order~$p$. Hence,
besides~$G'$, $G$~must contain $p$~other sub-groups of order~$p^{m-1}$
which have in common with~$G'$ the sub-group~$g'$. If the $p + 1$
sub-groups thus obtained do not exhaust the sub-groups of~$G$
of order~$p^{m - 1}$, let $G'''$ be a new one. Then, as before, $G'$ and $G'''$
must have a common sub-group~$g''$, of order~$p^{m - 2}$, which is
self-conjugate in~$G$. If $g''$ were the same as~$g'$, $\dfrac{G'''}{g'}$~would be
contained in~$\dfrac{G}{g'}$, which by supposition is not the case. It may
now be shewn as above that there are, in addition to~$G'$, $p$~sub-groups
of order~$p^{m - 1}$ which have in common with~$G'$ the
group~$g''$. These are therefore necessarily distinct from those
before obtained. This process may clearly be repeated till all
the sub-groups of order~$p^{m - 1}$ are exhausted. Hence finally, if
the number of sub-groups of~$G$, of order~$p^{m - 1}$ be $r_{m - 1}$, we have
$r_{m - 1} \equiv 1\ (\mod p)$.

\Par{60.} The self-conjugate operations of a group~$G$ of order~$p^{m}$,
whose orders are~$p$, form with identity a self-conjugate
sub-group whose order is some power of~$p$; and therefore their
number must be congruent to $-1, \pmod{p}$. On the other hand,
if $P$~is any operation of~$G$ of order~$p$ which is not self-conjugate,
the number of operations in the conjugate set to which $P$
belongs is a power of~$p$. Hence the total number of operations
of~$G$, of order~$p$, is congruent to $-1, \pmod{p}$. Now if $r_{1}$~is the
total number of sub-groups of~$G$ of order~$p$, the number of
operations of order~$p$ is $r_{1}(p - 1)$, since no two of these sub-groups
can have a common operation, except identity. It
follows that
\[
r_{1}(p - 1) \equiv -1,\ (\mod p),
\]
and therefore
\[
r_{1} \equiv 1,\ (\mod p).
\]

If now $G_{s}$ is any sub-group of~$G$ of order~$p^{s}$, and if $G_{s + t}$~is the
greatest sub-group of~$G$ in which $G_{s}$ is contained self-conjugately,
then every sub-group of~$G$ which contains $G_{s}$ self-conjugately is
contained in~$G_{s + t}$. But every sub-group of order~$p^{s + 1}$, which
contains~$G_{s}$, contains $G_{s}$ self-conjugately; and therefore every
%% -----File: 087.png---Folio 71-------
\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!number of sub-groups of given order is congruent to unity, $\pmod{p}$}%
sub-group of order~$p^{s + 1}$, which contains~$G_{s}$, is itself contained in~$G_{s + t}$.
By the preceding result, the number of sub-groups of
$\dfrac{G_{s + t}}{G_{s}}$ of order~$p$ is congruent to unity, $\pmod{p}$. Hence the
number of sub-groups of~$G$ of order~$p^{s + 1}$, which contain~$G_{s}$
of order~$p^{s}$, is congruent to unity,~$\pmod{p}$.

\Par{61.} Let now $r_{s}$ represent the total number of sub-groups of
order~$p^{s}$ contained in a group~$G$ of order~$p^{m}$. If any one of
them is contained in $a_{x}$~sub-groups of order~$p^{s + 1}$, and if any one
of the sub-groups of order~$p^{s + 1}$ contains $b_{y}$~sub-groups of order~$p^{s}$\DPchg{;}{,}
then
\[
\sum_{x = 1}^{x = r} a_{x} = \sum_{y = 1}^{y = r_{s + 1}} b_{y};
\]
for the numbers on either side of this equation are both equal
to the number of sub-groups of order~$p^{s + 1}$, when each of the
latter is reckoned once for every sub-group of order~$p^{s}$ that it
contains. It has however been shewn, in the two preceding
paragraphs, that for all values of $x$~and~$y$
\[
a_{x} \equiv 1,\quad b_{y} \equiv 1\ (\mod p).
\]

Hence
\[
r_{s} \equiv r_{s + 1}\ (\mod p).
\]

Now it has just been proved that
\[
r_{1} \equiv 1 \quad \text{and}\quad r_{m - 1} \equiv 1\ (\mod p);
\]
and therefore finally, for all values of~$s$,
\[
r_{s} \equiv 1\ (\mod p).
\]
We may state the results thus obtained as follows:---

\begin{Theorem}[III.]
The number of sub-groups of any given
%[** TN: "mod" italicized in the original]
order~$p^{s}$ of a group of order~$p^{m}$ is congruent to unity,~$\pmod{p}$\footnotemark.
\end{Theorem}
\index{Frobenius@\Topic{Frobenius}, quoted}%
\footnotetext{Frobenius, ``Verallgemeinerung des Sylow'schen Satzes,'' \textit{Berliner Sitzungsberichte},
  (1895), p.~989.}

\begin{Corollary}
The number of self-conjugate sub-groups of
order~$p^{s}$ of a group of order~$p^{m}$ is congruent to unity, \DPtypo{mod~$p$}{$\pmod{p}$}.
\end{Corollary}

This is an immediate consequence of the theorem, since the
number of sub-groups in any conjugate set is a power of~$p$.

\Par{62.} Having shewn that the number of sub-groups of~$G$ of
order~$p^{s}$ is of the form $1 + kp$, we may now discuss under what
%% -----File: 088.png---Folio 72-------
\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!case in which there is only one sub-group of a given order|(}%
circumstances it is possible for $k$ to be zero, so that $G$ then
contains only one sub-group~$G_{s}$ of order~$p^{s}$.

If this is the case, and if $P$~is any operation of~$G$ not contained
in~$G_{s}$, the order of~$P$ must be not less than~$p^{s + 1}$; for if
it were less, $G$~would have some sub-group of order~$p^{s}$ containing~$P$
and this would necessarily be different from~$G_{s}$. If the order
of~$P$ is~$p^{s + t}$, then $\{P^{p^{t}}\}$~is a cyclical sub-group of order~$p^{s}$; and it
must coincide with~$G_{s}$. Hence, if $G_{s}$ is the only sub-group of
order~$p^{s}$, it must be cyclical.

Suppose now that $G$~contains operations of order~$p^{r}$ ($r > s$),
but no operations of order~$p^{r + 1}$; and let $P$ be an operation of~$G$
of order~$p^{r}$. Then $\{P\}$ must be contained self-conjugately in a
non-cyclical sub-group of order~$p^{r + 1}$.

We will take first the case in which $p$~is an odd prime.
Then (§~55) $G$~must contain an operation~$P'$ which does not
belong to~$\{P\}$, such that
\[
P'^{-1} PP' = P^{\alpha},\quad P'^{p} = P^{\beta}.
\]
If $\alpha$ were unity, $\{P, P'\}$ would be an Abelian group of order~$p^{r + 1}$
containing no operation of order~$p^{r + 1}$. Its type would
therefore be $(r, 1)$, and it would necessarily contain an operation
of order~$p$ not occurring in~$\{P\}$. It has been shewn that this is
impossible if $G_{s}$ is the only sub-group of order~$p^{s}$, and therefore
$\alpha$~cannot be unity.

We may then without loss of generality (§~56) assume that
\[
\alpha = 1 + p^{r - 1}.
\]
Moreover if $\beta$ were not divisible by~$p$, the order of~$P'$ would
be~$p^{r + 1}$, contrary to supposition. Hence we must have the
relations
\[
P'^{-1} PP' = P^{1 + p^{r - 1}},\quad P'^{p} = P^{\gamma p}.
\]

By successive applications of the first of these equations, we
get
\[
P'^{-y} P^{x} P'^{y} = P^{x(1 + yp^{r - 1})},
\]
for all values of $x$ and~$y$; and from this it immediately follows
that
\begin{align*}
(P^{x}P')^{p} &= P'^{p} P^{x\{p + \efrac{1}{2} p(p + 1)p^{r - 1}\}} \\
              &= P^{(x + \gamma)p}.
\end{align*}
%% -----File: 089.png---Folio 73-------

Hence if
\[
x = p^{r - 1} - \gamma,
\]
the order of $P^{x}P'$, an operation not contained in~$\{P\}$, is~$p$. This
is impossible if $G_{s}$ is the only sub-group of order~$p^{s}$. If then
$r < m$, $G$~must contain operations of order greater than~$p^{r}$; and
$G$~is therefore a cyclical group. Hence:---

\begin{Theorem}[IV.]
If $G$, of order~$p^{m}$, where $p$~is an odd prime,
contains only one sub-group of order~$p^{s}$, $G$~must be cyclical.
\end{Theorem}

\Par{63.} When $p = 2$, the result is not so simple.

Let $Q$ be an operation that transforms $\{P\}$ into itself, and
suppose that the lowest power of~$Q$ that occurs in~$\{P\}$ is~$Q^{4}$.
Then $\{P, Q\}$ must be defined by
\[
Q^{-1}PQ = P^{\alpha},\quad Q^{4} = P^{\beta},
\]
where
\[
\alpha = ±1 + k2^{r - 2}.
\]
Moreover $\beta$ must be a multiple of~$4$, as otherwise the order of~$Q$
would be greater than~$2^{r}$. A simple calculation now gives
\[
(P^{x}Q^{2})^{2} = P^{\beta + x(1 + \alpha^{2})};
\]
and since
\[
\beta \equiv 0\ (\mod 4),
\]
and
\[
\alpha^{2} \equiv 1\ (\mod 2^{r - 1}),
\]
$x$ can always be chosen so that
\[
\beta + x(1 + \alpha^{2}) \equiv 0\ (\mod 2^{r}).
\]
When $x$ is thus chosen, $P^{x}Q^{2}$~is an operation of order~$2$ which is
not contained in~$\{P\}$. But this is inconsistent with $G_{s}$ being
the only sub-group of order~$2^{s}$; hence $\{P\}$~must contain~$Q^{2}$.

If now $\{P\}$ is not a self-conjugate sub-group, there must
(§~55) be some operation~$P'$ of order~$p^{r}$, conjugate with~$P$ and
not contained in~$\{P\}$, which transforms $\{P\}$ into itself; and
then $\{P, P'\}$ is defined by
\[
P'^{-1}PP' = P^{\alpha},\quad P'^{2} = P^{2\beta},
\]
where $\alpha$ is $-1$ or $±1 + 2^{r - 1}$, and $\beta$~is odd.

If
\[
\alpha = 1 + 2^{r - 1},
\]
then
\[
(P^{x}P')^{2} = P^{2\{\beta + x(1 + 2^{r - 2})\}};
\]
and $x$ can be chosen so that $P^{x}P'$~is of order~$2$. Hence this
case cannot occur.
%% -----File: 090.png---Folio 74-------

If $\alpha$ is either $-1$ or $-1 + 2^{r - 1}$, the defining relations of
$\{P, P'\}$ are not self-consistent unless $r$~is~$2$; for they lead to
\[
P'^{-1}P^{2\beta}P' = P^{-2\beta},
\]
and
\[
P'^{-1}P^{2\beta}P' = P^{2\beta}.
\]

If $r$ is~$2$, and if $P''$~is an operation that transforms $P$ into~$P'$,
so that
\[
P''^{-1} PP'' = P',
\]
then
\[
(PP'')^{2} = PP''^{2}P' = P^{-1}P',
\]
since
\[
P^{2} = P'^{2} = P''^{2}.
\]

Hence $PP''$ would be of order~$8$; and therefore this case
cannot occur.

Hence, finally, $\{P\}$~must be self-conjugate.

If $r + 1 < m$, $\{P\}$~must be transformed into itself by some
operation~$Q'$ not contained in~$\{P,\ Q\}$, so that
\[
Q^{-1}PQ = P^{-1 + k2^{r - 1}},
\]
and
\[
Q'^{-1}PQ' = P^{-1 + k'2^{r - 1}},
\]
where $k$ and $k'$ are each either zero or unity. If both are zero
or both unity, $QQ'$~is permutable with~$P$; and if one is zero and
the other unity,
\[
(QQ')^{-1} P(QQ') = P^{1 + 2^{r - 1}}.
\]
In either case, the group contains an operation of order~$2$ that
does not belong to~$\{P\}$, and this is inadmissible. Hence, lastly,
$r$~must not be less than~$m - 1$.

Suppose now that $r$~is equal to~$m - 1$, and that of the
operations of~$G$, not contained in~$\{P\}$, $Q$~has as small an order
as possible, say $2^t$ ($t \nless s + 1$). Then $Q^{2}$ of order~$2^{t - 1}$ is contained
in~$\{P\}$; without loss of generality we may assume
\[
Q^{2} = P^{2^{m - t}}.
\]

Moreover
\[
Q^{-1}PQ = P^{-1}\quad \text{or}\quad P^{-1 + 2^{m - 2}},
\]
and hence
\[
Q^{-1} P^{2^{m - t}}Q = P^{-2^{m - t}}.
\]
Now $Q$ is permutable with~$P^{2^{m - t}}$, one of its own powers, and
therefore $P^{2^{m - t}}$~is an operation of order~$2$. Hence $t$~is~$2$ and $s$~is
unity. If then $s$~is greater than unity, $G$~must contain
%% -----File: 091.png---Folio 75-------
operations of order~$2^{m}$ and must therefore be cyclical. Moreover
if $s$ is unity, the relations
\[
Q^{-1}PQ = P^{-1 + 2^{m - 2}}, \quad Q^{2} = P^{2^{m - 2}},
\]
lead to
\[
(PQ)^{2} = 1,
\]
and they are therefore inadmissible.

On the other hand, the relations
\[
Q^{-1}PQ = P^{-1}, \quad Q^{2} = P^{2^{m - 2}},
\]
give
\[
(P^{x}Q)^{2} = P^{2^{m - 2}};
\]
and every operation of~$G$, not contained in~$\{P\}$, is of order~$4$.
Also these relations are clearly self-consistent, and they define
a group of order~$2^{m}$. Hence finally:---

\begin{Theorem}[V.]
If a group~$G$, of order~$2^{m}$, has a single
sub-group of order~$2^{s}$, \($s > 1$\), it must be cyclical; if it has a
single sub-group of order~$2$, it is either cyclical or of the type
defined by
\[
P^{2^{m - 1}} = 1,\quad Q^{2} = P^{2^{m - 2}},\quad Q^{-1}PQ = P^{-1}.
\]
\end{Theorem}

\Par{64.} We shall now proceed to discuss, in application of the
foregoing theorems and for the importance of the results themselves,
the various types of groups of order~$p^{m}$ which contain
self-conjugate cyclical sub-groups of orders $p^{m - 1}$ and $p^{m - 2}$
respectively. It is clear from Theorem~V that the case $p = 2$
requires independent investigation; we shall only deal in detail
with the case in which $p$~is an odd prime, and shall state the
results for the case when $p = 2$.

The types of Abelian groups of order~$p^{m}$ which contain
operations of order~$p^{m - 2}$ are those corresponding to the symbols
$(m)$, $(m - 1, 1)$, $(m - 2, 2)$, and $(m - 2, 1, 1)$. We will assume
that the groups which we consider in the following paragraphs
are not Abelian.

\Par{65.} We will first consider a group $G$, of order~$p^{m}$, which
contains an operation~$P$ of order~$p^{m - 1}$. The cyclical sub-group~$\{P\}$
is self-conjugate and contains a single sub-group $\{P^{p^{m - 2}}\}$ of
order~$p$. By Theorem~IV, since $G$~is not cyclical, it must
contain an operation~$Q'$, of order~$p$, which does not occur in~$\{P\}$.
\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!case in which there is only one sub-group of a given order|)}%
%% -----File: 092.png---Folio 76-------
Since $\{P\}$ is self-conjugate and the group is not Abelian, $Q'$~must
transform $P$ into one of its own powers. Hence
\[
Q'^{-1}PQ' = P^{\alpha},
\]
and since $Q'^{p}$ is permutable with~$P$ it follows, from §~56, that
\[
\alpha = 1 + kp^{m - 2}.
\]

Since the group is not Abelian, $k$~cannot be zero; but it may
have any value from $1$ to~$p - 1$. If now
\[
kx \equiv 1\ (\mod p),
\]
then
\[
Q'^{-x} PQ'^{x} = P^{1 + p^{m - 2}};
\]
and therefore, writing $Q$ for~$Q'^{x}$, the group is defined by
\[
P^{p^{m - 1}} = 1,\quad Q^{p} = 1,\quad Q^{-1}PQ = P^{1 + p{m - 2}}.
\]

These relations are clearly self-consistent, and they define a
group of order~$p^{m}$.

There is therefore a single type of non-Abelian group of
order~$p^{m}$ which contains operations of order~$p^{m - 1}$, because, for
any such group, a pair of generating operations may be chosen
which satisfy the above relations.

From the relation
\[
Q^{-1}PQ = P^{1 + p^{m - 2}},
\]
it follows by repetition and multiplication that
\[
Q^{-y} P^{x} Q^{y} = P^{x(1 + yp^{m - 2})},
\]
and therefore that
\[
(P^{x} Q^{y})^{z} = P^{xz\{1 + \efrac{1}{2}(z + 1)yp^{m - 2}\}(1 - yzp^{m - 2})} Q^{yz},
\]
and
\[
(P^{x} Q^{y})^{p} = P^{xp}.
\]

Hence $G$ contains $p$~cyclical sub-groups of order~$p^{m - 1}$ of
which $P$ and~$PQ^{y}$ ($y = 1, 2, \Dots, p - 1$) may be taken as the
generating operations. Since $Q$ and $P^{p}$ are permutable, $G$~also
contains an Abelian non-cyclical sub-group $\{Q, P^{p}\}$ of
order~$p^{m - 1}$ It is easy to verify that the $1 + p$ sub-groups thus
obtained exhaust the sub-groups of order~$p^{m - 1}$; and that, for any
other order~$p^{s}$, there are also exactly $p + 1$ sub-groups of which
$p$~are cyclical and one is Abelian of type~$(s - 1, 1)$.

\begin{Remark}
The reader will find it an instructive exercise to verify the
\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!number of types which contain self-conjugate cyclical sub-groups of order~$p^{m - 2}$|(}%
results of the corresponding case where $p$~is~$2$; they may be stated
thus. There are four distinct types of non-Abelian group of order
%% -----File: 093.png---Folio 77-------
$2^{m}$, which contain operations of order~$2^{m - 1}$, when $m > 3$. Of these,
one is the type given in Theorem~V, and the remaining three are
defined by
\begin{alignat*}{3}
P^{2^{m - 1}} &= 1,\quad &Q^{2} &= 1,\quad &QPQ &= P^{1 + 2^{m - 2}}; \\
P^{2^{m - 1}} &= 1,      &Q^{2} &= 1,      &QPQ &= P^{-1 + 2^{m - 2}}; \\
P^{2^{m - 1}} &= 1,      &Q^{2} &= 1,      &QPQ &= P^{-1}.
\end{alignat*}

When $m = 3$, there are only two distinct types. In this case, the
second and the fourth of the above groups are identical, and the
third is Abelian.
\end{Remark}

\Par{66.} Suppose next that $G$, a group of order~$p^{m}$, has a
self-conjugate cyclical sub-group~$\{P\}$ of order~$p^{m - 2}$, and that
no operation of~$G$ is of higher order than~$p^{m - 2}$. We may
at once distinguish two cases for separate discussion; viz.\
\Inum{(i)}~that in which $P$~is a self-conjugate operation, \Inum{(ii)}~that in
which $P$~is not self-conjugate.

Taking the first case, there can be no operation~$Q'$ in~$G$ such
that $Q'^{p^{2}}$~is the lowest power of~$Q'$ contained in~$\{P\}$, for if there
were, $\{Q', P\}$ would be Abelian and, its order being~$p^{m}$, it would
necessarily coincide with~$G$. Hence any operation~$Q'$, not
contained in~$\{P\}$, generates with~$P$ an Abelian group of type
$(m - 2, 1)$, and we may choose $P$~and~$Q$ as independent
generators of this sub-group, the order of~$Q$ being~$p$. If now
$R'$~is any operation of~$G$ not contained in~$\{Q, P\}$, $R'^{p}$~must
occur in this sub-group, and therefore
\[
R'^{p} = P^{\alpha} Q^{\beta}.
\]

If $\beta$ were not zero, $R^{p^{2}}$~would be the lowest power of~$R$
that occurs in~$\{P\}$, and we have just seen that this cannot be
the case. Hence
\[
R'^{p} = P^{\alpha},
\]
and $\{R', P\}$ is again an Abelian group of type $(m - 2, 1)$. If
$P$~and~$R$ are independent generators of this group, the latter
cannot occur in~$\{Q, P\}$. Now since $Q$~is not self-conjugate,
\[
R^{-1}QR = QP^{\beta};
\]
and since $R^{p}$ is permutable with~$Q$
\[
P^{p\beta} = 1,
\]
so that
\[
\beta \equiv 0\ (\mod p^{m - 3}).
\]

Hence
\[
R^{-1}QR = QP^{kp^{m - 3}},
\]
%% -----File: 094.png---Folio 78-------
where $k$~is not a multiple of~$p$. If finally, $P^{k}$~be taken as a
generating operation in the place of~$P$, the group is defined by
\begin{gather*}
P^{p^{m - 2}} = 1,\quad Q^{p} = 1,\quad R^{p} = 1,\quad R^{-1}QR = QP^{p^{m - 3}}, \\
PQ = QP,\quad PR = RP.
\end{gather*}
There is therefore a single type of group of order~$p^{m}$, which
contains a self-conjugate operation of order~$p^{m - 2}$ and no operation
of order~$p^{m - 1}$.

\Par{67.} Taking next the second case, we suppose that $G$
contains a cyclical self-conjugate sub-group~$\{P\}$ of order~$p^{m - 2}$,
but no self-conjugate operation of this order.

If $G$ contains an operation~$Q'$ such that $Q'^{p^{2}}$~is the lowest
power of~$Q'$ occurring in~$\{P\}$, it follows (§~56), since $\{P\}$~is
self-conjugate, that
\[
Q'^{-1}PQ' = P^{1 + kp^{m - 4}}.
\]

Moreover, since the order of~$Q'$ cannot exceed~$p^{m - 2}$, we have
\[
Q'^{p^{2}} = P^{\alpha p^{2}}.
\]

A simple calculation, similar to that of §~65, will now shew
that
\[
(P^{x}Q')^{p^{2}} = P^{(x + \alpha)p^{2}},
\]
while
\[
(P^{x}Q')^{p} = Q'^{p} P^{x \{p + \efrac{1}{2}k(p + 1)p^{m - 3}\}}.
\]

Hence $P^{-\alpha}Q'$ is an operation of order~$p^{2}$, none of whose
powers, except identity, is contained in~$\{P\}$. If this operation
be represented by~$Q$, $G$~is defined by
\[
P^{p^{m - 2}} = 1,\quad Q^{p^{2}} = 1,\quad Q^{-1}PQ = P^{1 + kp^{m - 4}}.
\]

If $k$ is a multiple of~$p^{2}$, the group is Abelian.

If $k$ is~$k'p$, and if $Q^{\alpha}$~is taken for a new generating operation,
where $\alpha$~is chosen so that
\[
\alpha k' \equiv 1\ (\mod p),
\]
we have a single type defined by
\[
P^{p^{m - 2}} = 1,\quad Q^{p^{2}} = 1,\quad Q^{-1}PQ = P^{1 + p^{m - 3}}.
\]

If $k$ is not a multiple of~$p$, we may take it equal to $k_{1} + k_{2}p$,
where $k_{1}$~and~$k_{2}$ are both less than~$p$, and $k_{1}$~is not zero. (This
case cannot occur if $m < 5$.) Then
\[
Q^{-x_{1} -x_{2}p} PQ^{x_{1} + x_{2}p}
  = P^{1 + k_{1}x_{1}p^{m - 4} + (k_{1}x_{2} + k_{2}x_{1})p^{m - 3}},
\]
%% -----File: 095.png---Folio 79-------
when $m > 5$; while if $m = 5$, the coefficient of $p^{m - 3}$, viz.\ of~$p^{2}$, in
the index of~$P$ must be increased by $\frac{1}{2} x_{1}(x_{1} - 1)k_{1}^{2}$.

If now we choose $x_{1}$ so that
\[
k_{1}x_{1} = 1 + y_{1}p,
\]
and then $x_{2}$ so that
\[
y_{1} + k_{1}x_{2} + k_{2}x_{1} \equiv 0\ (\mod p),
\]
(with the suitable modification when $m = 5$), then $Q^{x_{1} + x_{2}p}$
transforms $P$ into $P^{1 + p^{m - 4}}$. We therefore again in this case
get a single type of group, which, if $Q^{x_{1} + x_{2}p}$ is represented by~$R$,
is defined by
\[
P^{p^{m - 2}} = 1,\quad R^{p^{2}} = 1,\quad R^{-1}PR = P^{1 + p^{m - 4}}.
\]
As already stated, this type exists only when $m > 4$.

If every operation of~$G$ is such that its $p$th~power occurs in~$\{P\}$,
$P$~must be one of a set of $p$ conjugate operations; for if $P$~is
one of $p^{2}$ conjugate operations, there must be an operation~$Q$
which transforms $P$ into $P^{1 + kp^{m - 4}}$, where $k$~is not a multiple of~$p$,
and then $Q^{p^{2}}$ is the lowest power of~$Q$ which is permutable
with~$P$. Since $P$ is one of $p$ conjugate operations, it must be
self-conjugate in an Abelian group $\{Q, P\}$ of type $(m - 2, 1)$.
Again, since $P$ and $P^{1 + p^{m - 3}}$ are conjugate operations, $P$~must be
contained in a group of order~$p^{m - 1}$, defined by
\[
P^{p^{m - 2}} = 1,\quad R^{p} = 1,\quad R^{-1}PR = P^{1 + p^{m - 3}}.
\]

If $Q$ is not a self-conjugate operation, it must be one of $p$
conjugate operations and (§~57) these must be
\[
Q, \quad QP^{p^{m - 3}}, \quad QP^{2p^{m - 3}},\ \Dots,\quad QP^{(p - 1)p^{m - 3}}.
\]

Now if
\[
R^{-1}QR = QP^{\alpha p^{m - 3}},
\]
then
\[
R^{-1} P^{-\alpha} QR = P^{-\alpha}Q.
\]
Hence, if $Q$ is not self-conjugate, $P^{-\alpha}Q$~is a self-conjugate
operation of order~$p^{m - 2}$: and the group is that determined
in the last paragraph. Therefore $Q$~must be self-conjugate,
and the group is defined by
\begin{gather*}
P^{p^{m - 2}} = 1,\quad Q^{p} = 1,\quad R^{p} = 1,\quad R^{-1}PR = P^{1 + p^{m - 3}}, \\
PQ = QP,\quad RQ = QR.
\end{gather*}
%% -----File: 096.png---Folio 80-------

It is clear from these relations that the group thus arrived
at is the direct product of the groups $\{P, R\}$ and~$\{Q\}$\footnotemark.
\footnotetext{In each of the five distinct cases to which we have been led in the discussion
  contained in §§~65--67, we have arrived at a set of defining relations, containing
  no indeterminate symbols, such that in each case a set of generating operations
  can be chosen to satisfy these relations. To justify the statement, in each
  particular case, that such a set of relations gives a distinct type of group, it is
  finally necessary to verify that the relations actually define a group of order~$p^{m}$.
  In the cases dealt with in the text, this verification is implicitly contained in the
  process by which the relations have been arrived at. We have therefore omitted
  the direct verification, which moreover is extremely simple. We shall similarly
  omit the corresponding verifications in the discussion of groups of orders $p^{3}$ and~$p^{4}$,
  as in none of these cases does it present any difficulty.

  To illustrate the necessity of such a verification in general, we may consider
  a simple case. The relations
  \[
  P^{3} = 1,\quad Q^{9} = 1,\quad P^{-1}QP = Q^{\alpha},
  \]
  where $\alpha$~is any given integer, certainly define a group whose order is equal to or
  is a factor of~$27$, since they indicate that $\{P\}$~and~$\{Q\}$ are permutable. They will
  not however give a type of group of order~$27$, unless $\alpha$~is $1$,~$4$ or~$7$. For instance,
  if $\alpha = 5$, the relations involve
  \[
  Q = P^{-3}QP^{3} = Q^{8},\quad\text{or}\quad Q^{7} = 1.
  \]
  Hence
  \[
  Q = Q^{4 · 7 - 3 · 9} = 1,
  \]
  and the relations hold only for a group of order~$3$.

  Again, if $\alpha = 3$, the relations give
  \[
  P^{-1}Q^{3}P = Q^{9} = 1,\quad\text{or}\quad Q^{3} = 1,
  \]
  and as before they define a group of order~$3$.}

\begin{Remark}
It is to be expected, from the result of the corresponding case at
the end of §~65, that the number of distinct types when $p = 2$ is
much greater than when $p$~is an odd prime. There are, in fact,
when $m > 5$, fourteen distinct types of groups of order~$2^{m}$, which
contain a self-conjugate cyclical sub-group of order~$2^{m - 2}$ and no
operation of order~$2^{m - 1}$. They may be classified as follows.

Suppose first that the group has a self-conjugate operation of
order~$2^{m - 2}$. There is then a single type defined by the relations
\[
\Inum{(i)}\quad A^{2^{m - 2}} = 1,\quad B^{2} = 1,\quad C^{2} = 1,\quad CBC = BA^{2^{m - 3}}.
\]

Suppose next that the group~$G$ has no self-conjugate operation
of order~$2^{m - 2}$, and let $\{A\}$~be a self-conjugate cyclical sub-group of
order~$2^{m - 2}$. If $\dfrac{G}{\{A\}}$~is cyclical, there are, when $m > 5$, five distinct
types. The common defining relations of these are
\[
A^{2^{m - 2}} = 1,\quad B^{4} = 1,\quad B^{-1}AB = A^{\alpha};
\]
and the five distinct types are
\begin{gather*}
\Inum{(ii)}\quad \alpha = -1,\qquad
\Inum{(iii)}\quad \alpha = 1 + 2^{m - 3},\qquad
\Inum{(iv)}\quad \alpha = -1 + 2^{m - 3}, \\
\Inum{(v)}\quad \alpha = 1 + 2^{m - 4},\qquad
\Inum{(vi)}\quad \alpha = -1 + 2^{m - 4}.
\end{gather*}
%% -----File: 097.png---Folio 81-------
If $m = 5$, then \Inum{(iv)} and \Inum{(v)} are identical, and \Inum{(vi)}~is Abelian; so that
there are only three distinct types. If $m = 4$, there is a single type;
it is given by~\Inum{(ii)}.

When $\dfrac{G}{\{A\}}$ is not cyclical, the square of every operation of~$G$ is
contained in~$\{A\}$. If all the self-conjugate operations of~$G$ are not
contained in~$\{A\}$, there must be a self-conjugate operation~$B$, of order~$2$,
which does not occur in~$\{A\}$. If $C$~is any operation of~$G$, not
contained in~$\{A, B\}$, then $\{A, C\}$ is a self-conjugate sub-group of
order~$2^{m - 1}$, which has no operation except identity in common with~$\{B\}$.
Hence $G$~is a direct product of a group of order~$2$ and a group
of order~$2^{m - 1}$. There are therefore, for this case, four types \Inum{(vii)},
\Inum{(viii)}, \Inum{(ix)}, \Inum{(x)}, when $m > 4$, corresponding to the four groups of
order~$2^{m - 1}$ of §~65. If $m = 4$, there are two types.

Next, let all the self-conjugate operations of~$G$ be contained in~$\{A\}$;
and suppose that $A$~is one of two conjugate operations. Then
$G$~must contain an Abelian sub-group of type $(m - 2, 1)$, in which $A$
occurs; and it may be shewn that, when $m > 4$, there are two types
defined by the relations
\[
\lintertext{\Inum{(xi)} and \Inum{(xii)}}
\begin{gathered}
A^{2^{m - 2}} = 1,\quad B^{2} = 1,\quad BAB = A,\quad C^{2} = 1, \\
CBC = BA^{2^{m - 3}},\quad CAC = A^{-1}\quad \text{or}\quad A^{-1 + 2^{m - 3}}.
\end{gathered}
\]
When $m = 4$, there is, for this case, no type.

Lastly, suppose that $A$ is one of four conjugate operations. Then
$G$~must contain sub-groups of order~$2^{m - 1}$, of the second and the
third types of §~65, and a sub-group of order~$2^{m - 1}$ of either the
first or fourth type (\lc). In this last case, there are two distinct
types defined by
\[
\lintertext{\Inum{(xiii)} and \Inum{(xiv)}}
\begin{aligned}
&A^{2^{m - 2}} = 1,\quad B^{2} = 1,\quad BAB = A^{1 + 2^{m - 3}}, \\
&\rlap{$C^{2} = 1,\quad CAC = A^{-1 + 2^{m - 3}},\quad CBC = B\quad \text{or}\quad BA^{2^{m - 3}}$.}
\end{aligned}
\]
These two types exist only when $m > 4$.
\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!number of types which contain self-conjugate cyclical sub-groups of order~$p^{m - 2}$|)}%

\Par{68.} We shall now, as a final illustration, determine and
\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!determination of distinct types of orders $p^{2}$, $p^{3}$, $p^{4}$, where $p$~is an odd prime|(}%
tabulate all types of groups of orders $p^{2}$,~$p^{3}$ and~$p^{4}$. It has been
already seen that when $p = 2$ the discussion must, in part at least,
be distinct from that for an odd prime; for the sake of brevity
we shall not deal in detail with this case, but shall state the results
only and leave their verification as an exercise to the reader.

It has been shewn (§~53) that all groups of order~$p^{2}$ are Abelian;
and hence the only distinct types are those represented by $(2)$ and
$(1, 1)$.

For Abelian groups of order~$p^{3}$, the distinct types are $(3)$, $(2, 1)$
and $(1, 1, 1)$.
%% -----File: 098.png---Folio 82-------

If a non-Abelian group of order~$p^{3}$ contains an operation of
order~$p^{2}$, the sub-group it generates is self-conjugate; hence (§~65)
in this case there is a single type of group defined by
\[
P^{p^{2}} = 1,\quad Q^{p} = 1,\quad Q^{-1}PQ = P^{1 + p}.
\]

If there is no operation of order~$p^{2}$, then since there must be a
self-conjugate operation of order~$p$, the group comes under the head
discussed in §~66; there is again a single type of group defined by
\begin{gather*}
P^{p} = 1,\quad Q^{p} = 1,\quad R^{p} = 1,\quad R^{-1}QR = QP, \\
R^{-1}PR = P,\quad Q^{-1}PQ = P.
\end{gather*}

These two types exhaust all the possibilities for non-Abelian
groups of order~$p^{3}$.

\Par{69.} For Abelian groups of order~$p^{4}$, the possible distinct types
are $(4)$, $(3, 1)$, $(2, 2)$, $(2, 1, 1)$ and $(1, 1, 1, 1)$.

For non-Abelian groups of order~$p^{4}$ which contain operations of
order~$p^{3}$ there is a single type, namely that given in §~65 when $m$~is
put equal to~$4$.

For non-Abelian groups which contain a self-conjugate cyclical
sub-group of order~$p^{2}$ and no operation of order~$p^{3}$, there are three
distinct types, obtained by writing $4$ for~$m$ in the group of §~66 and
in the first and the last groups of §~67. The defining relations of
these need not be here repeated, as they will be given in the summarizing
table (§~73).

It remains now to determine all distinct types of groups of
order~$p^{4}$, which contain no operation of order~$p^{3}$ and no self-conjugate
cyclical sub-group of order~$p^{2}$. We shall first deal with groups
which contain operations of order~$p^{2}$.

Let $S$ be an operation of order~$p^{2}$ in a group~$G$ of order~$p^{4}$. The
cyclical sub-group~$\{S\}$ must be self-conjugate in a non-cyclical sub-group
$\{S, T\}$ of order~$p^{3}$, defined by
\[
S^{p^{2}} = 1,\quad T^{p} = 1,\quad T^{-1}ST = S^{1 + kp}.
\]

If $R$ is any operation of~$G$, not contained in~$\{S, T\}$, then since
$\{S\}$~is not self-conjugate, we must have (§~57)
\[
R^{-1}SR = S^{1 + \alpha p} T^{\beta},
\]
and therefore
\[
R^{-1} S^{p}R = S^{p}.
\]

Now
\[
T^{-1}S^{p}T = S^{p},
\]
and therefore the $p$th~power of every operation of order~$p^{2}$ in~$G$ is a
self-conjugate operation.

First let us suppose that $G$~contains other self-conjugate
operations besides those of~$\{S^{p}\}$. Every such operation must occur
%% -----File: 099.png---Folio 83-------
in the group that contains $\{S\}$ self-conjugately; hence in this case
$T$~must be self-conjugate.

We now therefore have
\begin{gather*}
S^{p^{2}} = 1,\quad T^{p} = 1,\quad T^{-1}ST = S, \\
R^{-1}SR = S^{1 + \alpha p}T^{\beta},\quad R^{-1}TR = T, \\
R^{p} = S^{\gamma p}T^{\delta}.
\end{gather*}

These equations give
\[
(S^{x}R)^{p} = R^{p}S^{xp} = S^{(x + \gamma)p}T^{\delta}.
\]

Hence, if $\delta = 0$, $S^{-\gamma}R$~is an operation of order~$p$. Denoting this
by~$R'$ and $S^{\alpha p} T^{\beta}$ by~$T'$, the group is defined by
\begin{gather*}
S^{p^{2}} = 1,\quad T'^{p} = 1,\quad R'^{p} = 1,\quad R'^{-1}SR' = ST', \\
T'^{-1}ST' = S,\quad R'^{-1}T'R' = T'.
\end{gather*}

On the other hand, if $\delta$ is not zero, $S^{\efrac{\delta\alpha}{\beta} - \gamma}R$ is an operation of
order~$p^{2}$ such that $R$ transforms it into a power of itself. This is
contrary to the supposition that the group contains no cyclical self-conjugate
sub-group of order~$p^{2}$. Hence $\delta$~cannot be different from
zero: we therefore have only one type of group.

\Par{70.} Next, let $\{S^{p}\}$ contain all the self-conjugate operations of~$G$;
and as before, let $\{S, T\}$ be the group that contains $\{S\}$ self-conjugately.
If $G$ contains an operation~$S'$ of order~$p^{2}$ which
does not occur in~$\{S, T\}$, there must also be a non-cyclical sub-group
$\{S', T'\}$ of order~$p^{3}$ which contains $\{S'\}$ self-conjugately.
Now $\{S, T\}$ and $\{S', T'\}$ must have a common sub-group of order~$p^{2}$;
since this is self-conjugate in~$G$, it cannot be cyclical. The only
non-cyclical sub-groups of orders\DPnote{** [sic]}~$p^{2}$ that $\{S, T\}$ and $\{S', T'\}$ contain
are $\{S^{p}, T\}$ and $\{S'^{p}, T'\}$. Hence these must be identical, and
therefore $T$~must occur in~$\{S', T'\}$. If now $\{S, T\}$ and $\{S', T'\}$ were
both Abelian, $T$~would be permutable both with~$S$ and with~$S'$, and
would therefore, contrary to supposition, be a self-conjugate
operation. Hence either \Inum{(i)}~$G$~must contain a non-Abelian group
$\{S, T\}$ of order~$p^{3}$, in which $S$ is an operation of order~$p^{2}$; or \Inum{(ii)}~the
Abelian group $\{S, T\}$, in which $S$ is an operation of order~$p^{2}$, must
contain all the operations of~$G$ of order~$p^{2}$.

In the case~\Inum{(i)}, the group is defined by
\begin{gather*}
S^{p^{2}} = 1,\quad T^{p} = 1,\quad T^{-1}ST = S^{1 + p}, \\
R^{-1}SR = S^{1 + \alpha p}T^{\beta},\quad R^{-1}TR = S^{\gamma p}T, \\
R^{p} = S^{\delta p}.
\end{gather*}
%% -----File: 100.png---Folio 84-------

These equations give
\[
(S^{x}R)^{p} = R^{p} S^{xp\{1 + \efrac{1}{6} (p - 1)p(p + 1)\beta (\gamma - 2)\}},
\]
and therefore, except when $p = 3$, $S^{-\delta}R$~is an operation of order~$p$.
Hence, if $p > 3$, we may, without loss of generality, put zero for~$\delta$.
In this case a simple calculation gives
\[
R^{-x}SR^{x} = S^{1 + \alpha_{x} p}T^{\beta_{x}},\quad
R^{-x}TR^{x} = S^{\gamma_{x} p}T,
\]
where
\[
\alpha_{x} = \tfrac{1}{2} \beta\gamma x(x - 1) + x\alpha,\quad
\beta_{x} = x\beta,\quad
\gamma_{x} = x\gamma.
\]

Hence if $R^{x} = R'$, $S^{y} = S'$, $S^{zp}T = T'$, where $x$,~$y$ and~$z$ are not
multiples of~$p$, then
\begin{gather*}
S'^{p^{2}} = 1,\quad T'^{p} = 1,\quad T'^{-1}S'T' = S'^{1 + p}, \\
R'^{-1}S'R' = S'^{1 + (\alpha_{x} - z\beta_{x})p} T'^{y\beta}, \\
R'^{-1}T'R' = S'^{\efrac{\gamma_{x}p}{y}}T',\quad R'^{p} = 1.
\end{gather*}

If now we take
\[
z \equiv \frac{\alpha_{x}}{\beta_{x}},\quad y \equiv \frac{1}{\beta_{x}}\ (\mod p),
\]
then
\[
R'^{-1}S'R' = S'T',\quad R'^{-1}T'R' = S'^{x^{2}\beta\gamma p}T'.
\]

Dropping the accents, the defining relations now become
\begin{gather*}
S^{p^{2}} = 1,\quad T^{p} = 1,\quad R^{p} = 1,\quad T^{-1}ST = S^{1 + p}, \\
R^{-1}SR = ST,\quad R^{-1}TR = S^{\alpha p}T;
\end{gather*}
where $\alpha$ is either zero, unity, or any given non-residue. Since the
sub-group $\{R, T, S^{p}\}$ contains all the operations of order~$p$ of the
group, it follows at once that these three cases give three distinct
types.

When $p = 3$, it will be found that the defining relations again
give three distinct types. The operation~$R$ may always be chosen so
that $\alpha$~and~$\gamma$ are zero. If it is so chosen, the three types correspond
to the values $1$,~$0$; $-1$,~$1$; and $-1$,~$-1$; of $\beta$~and~$\delta$.

In case~\Inum{(ii)}, the group is defined by
\begin{gather*}
S^{p^{2}} = 1,\quad T^{p} = 1,\quad T^{-1}ST = S, \\
R^{-1}SR = S^{1 + \alpha p}T^{\beta},\quad R^{-1}TR = S^{\gamma p}T, \\
R^{p} = 1;
\end{gather*}
with the condition that all operations of~$G$, not contained in~$\{S, T\}$,
are of order~$p$.

The formulæ for $R^{-x}SR^{x}$ and $R^{-x}TR^{x}$ enable us to calculate
directly the power of any given operation of~$G$. Thus they give
\[
(S^{x}R)^{p} = S^{px\{1 + \efrac{1}{6} \beta\gamma(p + 1)p(p - 1) + \efrac{1}{2} \alpha p(p + 1)\}}.
\]
%% -----File: 101.png---Folio 85-------

If $p > 3$, this gives
\[
(S^{x}R)^{p} = S^{px},
\]
so that, if $x$ is not a multiple of~$p$, the order of~$S^{x}R$ is~$p^{2}$. Hence
the type of group under consideration can only occur when $p = 3$.

In this case
\[
(S^{x}R)^{3} = S^{3x(1 + \beta \gamma)}.
\]

Hence if $p = 3$ and $\beta\gamma \equiv -1\ (\mod 3)$, we obtain a new type. A
reduction similar to that in the previous case may now be effected;
and taking unity for~$x$, the group is defined by
\begin{gather*}
S^{9} = 1,\quad T^{3} = 1,\quad R^{3} = 1,\quad T^{-1}ST = S, \\
R^{-1}SR = ST,\quad R^{-1}TR = S^{-3}T.
\end{gather*}

\Par{71.} It only remains to determine the distinct types which
contain no operation of order~$p^{2}$.

Suppose first that the self-conjugate operations of~$G$ form a
group of order~$p^{2}$. This must be generated by two independent
operations $P$~and~$Q$ of order~$p$.

If now $R$ is any other operation of the group, $\{P, Q, R\}$ must be
an Abelian group of type $(1, 1, 1)$. If again $S$~is any operation not
contained in $\{P, Q, R\}$, it cannot be permutable with~$R$; for if it
were, $R$~would be self-conjugate. There must therefore be a relation
of the form (§~57)
\[
S^{-1}RS = RP^{\alpha}Q^{\beta}.
\]

Since any operation of $\{P, Q\}$ may be taken for one of its
generating operations, we may take $P^{\alpha}Q^{\beta}$ or $P'$ for one. If then $Q'$
is an independent operation of~$\{P, Q\}$, $G$~will be defined by
\[
P'^{p} = 1,\quad Q'^{p} = 1,\quad R^{p} = 1,\quad S^{p} = 1,\quad S^{-1}RS = RP',
\]
in addition to the relations expressing that $P'$~and~$Q'$ are self-conjugate.
There is then in this case a single type. That all the
operations of~$G$ in this case are actually of order~$p$ follows from the
obvious fact that every sub-group of order~$p^{3}$ is either Abelian or of
the second type of non-Abelian groups of order~$p^{3}$.

\Par{72.} Suppose, secondly, that the self-conjugate operations of~$G$
form a sub-group of order~$p$, generated by~$P$. There must then be
some operation~$Q$ which belongs to a set of $p$ conjugate operations;
for if every operation of~$G$ which is not self-conjugate were one of
a set of $p^{2}$ conjugate operations, the total number of operations in
the group would be congruent to $p\ (\mod p^{2})$. It follows that $Q$~must
be self-conjugate in a group of order~$p^{3}$; and since $P$~is also
self-conjugate in this group, it must be Abelian. Let $P$,~$Q$,~$R$ be
generators of this group and $S$~any operation of~$G$ not contained in
%% -----File: 102.png---Folio 86-------
it. We may now assume that $Q$~belongs to the sub-group~$H_{2}$
(§~53), and therefore that
\[
S^{-1}QS = QP^{\alpha},
\]
while
\[
S^{-1}RS = RQ^{\beta}P^{\gamma}.
\]

If $\beta$ were zero, $Q^{\efrac{1}{\alpha}} R^{-\efrac{1}{\gamma}}$ would be a self-conjugate operation not
contained in~$\{P\}$; and therefore $\beta$~must be different from zero. We
may now put
\[
Q^{\beta} P^{\gamma} = Q',\quad P^{\alpha\beta} = P';
\]
and the group is then defined by the relations
\begin{gather*}
%[** TN: Set on one line in the original]
P'^{p} = 1,\quad
Q'^{p} = 1,\quad
R^{p} = 1,\quad
S^{p} = 1,\\
S^{-1}RS = RQ',\quad
S^{-1}Q'S = Q'P',
\end{gather*}
together with the relations expressing that $P$ is self-conjugate.
There is thus again in this case, at most, a single type. It remains
to determine whether the operations are all actually of order~$p$.

The defining relations give
\[
S^{-1} P^{\alpha_{x}} Q^{\beta_{x}} R^{\gamma_{x}}S
  = P^{\alpha_{x + 1}} Q^{\beta_{x + 1}} R^{\gamma_{x + 1}},
\]
where
\[
\alpha_{x + 1} = \alpha_{x} + \beta_{x},\quad
\beta_{x + 1} = \beta_{x} + \gamma_{x},\quad
\gamma_{x + 1} = \gamma_{x};
\]
and therefore
\[
S^{-x} P^{\alpha} Q^{\beta} R^{\gamma} S^{x}
  = P^{\alpha_{x}} Q^{\beta_{x}} R^{\gamma_{x}},
\]
where
\[
\alpha_{x} = \alpha + x\beta + \tfrac{1}{2} x(x - 1)\gamma,\quad
\beta_{x} = \beta + x\gamma,\quad
\gamma_{x} = \gamma.
\]

Hence
\begin{align*}
(P^{\alpha} Q^{\beta} R^{\gamma}S)^{p}
  &= P^{\esum_{1}^{p} \alpha_{x}} Q^{\esum_{1}^{p}\beta_{x}} R^{\esum_{1}^{p} \gamma_{x}} \\
  &= P^{p\alpha + \efrac{1}{2} p(p - 1)\beta + \efrac{1}{6} (p + 1)p(p - 1)\gamma}
     Q^{p\beta  + \efrac{1}{2} p(p - 1) \gamma} R^{p\gamma}.
\end{align*}

If $p$ is a greater prime than~$3$, the indices of $P$,~$Q$,~$R$ are all
multiples of~$p$; hence $P^{\alpha} Q^{\beta} R^{\gamma}S$ is of order~$p$, $S$~being any operation
not contained in~$G$. If however $p = 3$, then
\[
(P^{\alpha} Q^{\beta} R^{\gamma}S)^{3} = P^{\gamma},
\]
so that, if $\gamma$ is not a multiple of~$3$, $P^{\alpha} Q^{\beta} R^{\gamma}S$ is an operation of
order~$9$. Hence this last type of group exists as a distinct type for
all primes greater than~$3$; but for $p = 3$, it is not distinct from one
of the previous types containing operations of order~$9$.
\end{Remark}

\Par{73.} In tabulating, as follows, the types of group thus
obtained, we give with each group~$G$ a symbol of the form
\[
(a, b, \Dots)(a', b', \Dots)(a'', b'', \Dots) \Dots,
\]
indicating the types of $H_{1}$, $\dfrac{H_{2}}{H_{1}}$, $\dfrac{H_{3}}{H_{2}}$,~\Dots, where
\[
H_{1},\ H_{2},\ H_{3},\ \Dots,\ H_{n},\ G
\]
is the series of self-conjugate sub-groups defined in §~53. This
%% -----File: 103.png---Folio 87-------
symbol is to be read from the left so that $(a, b, \Dots)$ is the
type of~$H_{1}$.

Moreover in each group there is no operation of higher
order than that denoted by~$P$.


\Subsection{\textit{Table of groups of order $p^{n}$, $p$ an odd prime}\footnotemark.}
\footnotetext{On groups of orders $p^{3}$ and~$p^{4}$, the reader may consult, in addition to
  Young's memoir already referred to, Hölder, ``Die Gruppen der Ordnungen
  $p^{3}$, $pq^{2}$, $pqr$,~$p^{4}$,'' \textit{Math.\ Ann.},\DPnote{** no "Vol."} \VolNo{XLIII}, (1893), in particular, pp.~371--410.}%
\index{Defining relations@\Topic{Defining relations} of a group!for groups of orders $p^{2}$, $p^{3}$, $p^{4}$}%
\index{Types of group@\Topic{Types of group}, distinct, whose order is!$p^{2}$, $p^{3}$, or $p^{4}$, where $p$~is an odd prime}%
\index{Holder@\Topic{Hölder}, quoted}%
\index{Young@\Topic{Young}, quoted}%

%[** TN: Layout reformatted]
\Class{\Inum{I.}}{$n = 2$, two types.}

\Item{(i)} $(2)$;\qquad \Inum{(ii)}\quad $(1, 1)$.

\Class{\Inum{II.}}{$n = 3$, five types.}

\Item{(i)} $(3)$;\qquad \Inum{(ii)} $(2, 1)$;\qquad \Inum{(iii)} $(1, 1, 1)$;

\Item{(iv)} $P^{p^{2}} = 1$,\quad $Q^{p} = 1$,\quad $Q^{-1}PQ = P^{1 + p}$,\quad $(1) (11)$;

\Item{(v)} $\begin{gathered}[t]
P^{p} = 1,\quad Q^{p} = 1,\quad R^{p} = 1,\quad R^{-1}QR = QP, \\
R^{-1}PR = P,\quad Q^{-1}PQ = P,\quad (1) (11).
\end{gathered}$

\Class{\Inum{III.}}{$n = 4$, fifteen types.}

\Item{(i)} $(4)$;\qquad
\Inum{(ii)}\quad $(3, 1)$;\qquad
\Inum{(iii)}\quad $(2, 2)$;

\Item{(iv)} $(2, 1, 1)$;\qquad
\Inum{(v)}\quad $(1, 1, 1, 1)$;

\Item{(vi)} $P^{p^{3}} = 1$,\quad $Q^{p} = 1$,\quad $Q^{-1}PQ = P^{1 + p^{2}}$,\quad $(2) (11)$;

\Item{(vii)} $\begin{gathered}[t]
P^{p^{2}} = 1,\quad Q^{p} = 1,\quad R^{p} = 1,\quad R^{-1}QR = QP^{p}, \\
Q^{-1}PQ = P,\quad R^{-1}PR = P,\quad (2) (11);
\end{gathered}$

\Item{(viii)} $P^{p^{2}} = 1$,\quad $Q^{p^{2}} = 1$,\quad $Q^{-1}PQ = P^{1 + p}$,\quad $(11)(11)$;

\Item{(ix)} $\begin{gathered}[t]
P^{p^{2}} = 1,\quad Q^{p} = 1,\quad R^{p} = 1,\quad R^{-1}PR = P^{1 + p}, \\
P^{-1}QP = Q,\quad R^{-1}QR = Q,\quad (11) (11),
\end{gathered}$ \\
this group~\Inum{(ix)} being the direct product of $\{Q\}$ and~$\{P, R\}$;

\Item{(x)} $\begin{gathered}[t]
P^{p^{2}} = 1,\quad Q^{p} = 1,\quad R^{p} = 1,\quad R^{-1}PR = PQ, \\
Q^{-1}PQ = P,\quad R^{-1}QR = Q,\quad (11) (11);
\end{gathered}$
%% -----File: 104.png---Folio 88-------
\index{Defining relations@\Topic{Defining relations} of a group!for groups of orders $p^{2}$, $p^{3}$, $p^{4}$}%
\index{Types of group@\Topic{Types of group}, distinct, whose order is!$p^{2}$, $p^{3}$, or $p^{4}$, where $p$~is an odd prime}%

\Item[]{(xi),} \Inum{(xii)}, and \Inum{(xiii)} $p > 3$,
\begin{gather*}
P^{p^{2}} = 1,\quad Q^{p} = 1,\quad R^{p} = 1,\quad Q^{-1}PQ = P^{1 + p}, \\
R^{-1}PR = PQ,\quad R^{-1}QR = P^{\alpha p}Q,\quad (1) (1) (11),
\end{gather*}
where for~\Inum{(xi)} $\alpha = 0$, for~\Inum{(xii)} $\alpha = 1$, for~\Inum{(xiii)} $\alpha =$~any non-residue,
$(\mod p)$;

\Item[]{(xi),} \Inum{(xii)}, and \Inum{(xiii)} $p = 3$,
\begin{gather*}
P^{9} = 1,\quad Q^{3} = 1,\quad R^{3} = P^{3 \alpha},\quad Q^{-1}PQ = P^{4}, \\
R^{-1}PR = PQ^{\beta},\quad R^{-1}QR = Q,\quad (1) (1) (11),
\end{gather*}
where for~\Inum{(xi)} $\alpha = 0$, $\beta = 1$, for~\Inum{(xii)} $\alpha = 1$, $\beta = -1$, for~\Inum{(xiii)}
$\alpha = -1$, $\beta = -1$.

\Item{(xiv)} $\begin{gathered}[t]
P^{p} = 1,\quad Q^{p} = 1,\quad R^{p} = 1,\quad S^{p} = 1,\quad S^{-1}RS = RP, \\
S^{-1}QS = Q,\quad S^{-1}PS = P,\quad R^{-1}QR = Q, \\
R^{-1}PR = P,\quad Q^{-1}PQ = P,\quad (11) (11),
\end{gathered}$ \\
this group~\Inum{(xiv)} being the direct product of $\{Q\}$ and $\{P, R, S\}$;

\Item{(xv)} $p > 3$,
\begin{gather*}
P^{p} = 1,\ Q^{p} = 1,\ R^{p} = 1,\ S^{p} = 1,\ S^{-1}RS = RQ,\  S^{-1}QS = QP, \\
S^{-1}PS = P,\ R^{-1}QR = Q,\ R^{-1}PR = P, \\
Q^{-1}PQ = P,\ (1) (1) (11);
\end{gather*}

\Item{(xv)} $p = 3$,
\begin{gather*}
P^{9} = 1,\ Q^{3} = 1,\ R^{3} = 1,\ Q^{-1}PQ = P,\ R^{-1}PR = PQ, \\
R^{-1}QR = P^{-3}Q,\ (1) (1) (11).
\end{gather*}

\Par{74.} To complete the list, we add, as was promised in §~68,
\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!table of distinct types of orders $8$ and~$16$}%
\index{Types of group@\Topic{Types of group}, distinct, whose order is!$8$ or~$16$}%
the types of non-Abelian groups of orders $2^{3}$ and~$2^{4}$; the
possible types of Abelian groups being the same as for an
odd prime.

\Class{}{\textit{Non-Abelian groups of order~$2^{3}$}; two types.}

\Item[]{(i)} identical with \Inum{II}~\Inum{(iv)}, writing $2$ for~$p$;

\Item[]{(ii)} $P^{4} = 1$,\quad $Q^{4} = 1$,\quad $Q^{-1}PQ = P^{-1}$,\quad $Q^{2} = P^{2}$,\quad $(1)(11)$.

\Class{}{\textit{Non-Abelian groups of order~$2^{4}$}; nine types.}

\Item[]{(i)}, \Inum{(ii)}, \Inum{(iii)}, \Inum{(iv)} and \Inum{(v)} identical with \Inum{III}~\Inum{(vi)}, \Inum{(vii)},
\Inum{(viii)}, \Inum{(ix)} and~\Inum{(x)}, writing $2$ for~$p$;
\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!determination of distinct types of orders $p^{2}$, $p^{3}$, $p^{4}$, where $p$~is an odd prime|)}%
%% -----File: 105.png---Folio 89-------
\index{Defining relations@\Topic{Defining relations} of a group!for groups of orders $p^{2}$, $p^{3}$, $p^{4}$}%
\index{Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime!table of distinct types of orders $8$ and~$16$}%
\index{Types of group@\Topic{Types of group}, distinct, whose order is!$8$ or~$16$}%

\Item{(vi)} $\begin{gathered}[t]
P^{4} = 1,\quad Q^{4} = 1,\quad R^{2} = 1,\quad Q^{-1}PQ = P^{-1}, \\
Q^{2} = P^{2},\quad R^{-1}QR = Q,\quad R^{-1}PR = P,\quad (11) (11),
\end{gathered}$ \\
this group~\Inum{(vi)} being the direct product of $\{R\}$ and $\{P, Q\}$;

\Item{(vii)} $P^{8} = 1$,\quad $Q^{2} = 1$,\quad $Q^{-1}PQ = P^{-1}$,\quad $(1) (1) (11)$;

\Item{(viii)} $P^{8} = 1$,\quad $Q^{2} = 1$,\quad $Q^{-1}PQ = P^{3}$,\quad $(1) (1) (11)$;

\Item{(ix)} $P^{8} = 1$,\quad $Q^{4} = 1$,\quad $Q^{-1}PQ = P^{-1}$,\quad $Q^{2} = P^{4}$,\quad $(1) (1) (11)$.

\begin{Remark}
\Par{75.} In the following examples, $G$~is a non-Abelian group whose
order is the power of a prime~$p$; and the sub-groups $H_{1}$, $H_{2}$,~\Dots, $H_{n}$
referred to are the series of self-conjugate sub-groups of §~53.
\end{Remark}

\begin{Ex}[1.] If $P$ and $Q$ are any two operations of~$G$, the totality of
operations of the form $P^{-1}Q^{-1}PQ$ generate a self-conjugate sub-group
identical with or contained in~$H_{n}$.
\end{Ex}

\begin{Ex}[2.] Shew that, if every sub-group of~$G$ is Abelian, $\dfrac{G}{H_{1}}$~is an
Abelian group of type $(1, 1)$; and any two operations of~$G$, neither
of which belongs to~$H_{1}$ and neither of which is a power of the other,
generate~$G$.
\end{Ex}

\begin{Ex}[3.] If $\dfrac{G}{H_{n}}$, $\dfrac{H_{n}}{H{n-1}}$, $\dfrac{H_{n-1}}{H_{n-2}}$,~\Dots\ are of types $(1, 1)$, $(1)$,
$(1)$,~\Dots, shew that $G$~can be generated by two operations.
\end{Ex}

\begin{Ex}[4.] If $G$, of order~$p^{m}$, is not Abelian, and if every sub-group
of~$G$ is self-conjugate, shew that $p$~must be~$2$. \Attrib{Dedekind.}
\index{Dedekind@\Topic{Dedekind}, quoted}%
\end{Ex}

\begin{Ex}[5.] If $G$ is of order~$p^{m}$, and $H_{1}$ of type $(1, 1, \Dots\DPchg{\,}{,} \text{with $m - 2$
units})$, and if $m > 5$, $G$~is the direct product of two groups. If
$m = 5$, there is one type for which $G$ is not a direct product, viz.\
\begin{gather*}
Q_{1}^{p^{2}} = 1,\quad
Q_{2}^{p^{2}} = 1,\quad
P^{p} = 1,\quad
Q_{2}^{-1}Q_{1}Q_{2} = Q_{1}P, \\
%
Q_{1}^{-1}PQ_{1} = P,\quad
Q_{2}^{-1}PQ_{2} = P.
\end{gather*}
\end{Ex}

\begin{Ex}[6.] A group~$G$, of order~$p^{m}$, ($m > 4$), where $p$~is an odd
prime, contains an operation~$P$ of order~$p^{m-2}$, and no cyclical self-conjugate
sub-group of order~$p^{m-2}$. Shew that, if $P$ is one of a set
of $p^{2}$ conjugate operations, the defining relations of the group are of
the form
\begin{gather*}
P^{p^{m-2}} = 1,\quad Q^{p} = 1,\quad Q^{-1}PQ = P^{1 + p^{m - 3}},\quad R^{p} = P^{\alpha p} Q^{\beta}, \\
R^{-1}QR = Q,\quad R^{-1}PR = P^{1 + \beta p^{m - 4}}Q;
\end{gather*}
and that, if $P$ is one of a set of $p$ conjugate operations, the defining
relations are of the form
\begin{gather*}
P^{p^{m-2}} = 1,\quad Q^{p} = 1,\quad Q^{-1}PQ = P,\quad R^{p} = Q^{\alpha}, \\
R^{-1}QR = P^{\beta p^{m - 3}}Q,\quad R^{-1}PR = PQ.
\end{gather*}
Determine in each case the number of distinct types.
\end{Ex}
%% -----File: 106.png---Folio 90-------


\Chapter{VI.}{On Sylow's Theorem.}
\index{Sylow@\Topic{Sylow}, quoted}%

\Par{76.} \First{It} has been proved (§~22) that the order of any sub-group
of a group~$G$ is a factor of the order of~$G$; and it
results at once from the investigation of §§~45--47 that, in
an Abelian group, there is always at least one sub-group
whose order is any given factor of the order of the group.
The latter result is not however generally true for groups
which are not Abelian; and for factors of the order of a group
which contain more than one distinct prime, no general law is
known as to the existence or non-existence of corresponding
sub-groups. If however $p^{m}$, where $p$~is a prime, divides the
order of the group, it may be shewn that the group will always
contain a sub-group of order~$p^{m}$. The special form of this
theorem, that a group whose order is divisible by a prime~$p$
contains operations of order~$p$, is due originally to Cauchy\footnotemark.
\index{Cauchy@\Topic{Cauchy}, quoted}%
\footnotetext{Cauchy, \textit{Exercises d'analyse},~\VolNo{III}, (1844), p.~250.}%
The more general result was first established by Sylow. He
has shewn\footnote
  {Sylow, Théorèmes sur les groupes de substitutions, \textit{Math.\ Ann.},\DPnote{** No "Vol." here, below} \VolNo{V} (1872)
  pp.~584 et~seq. Compare also Frobenius, Neuer Beweis des Sylow'schen Satzes,
  \textit{Crelle}, \VolNo{C.}\ (1886), p.~179.}
\index{Frobenius@\Topic{Frobenius}, quoted}%
that, if $p^{\alpha}$~is the highest power of a prime~$p$ which
divides the order of a group, the group contains a single conjugate
set of $kp + 1$ sub-groups of order~$p^{\alpha}$.

\begin{Remark}
For the special case $p = 2$, the following proof of Cauchy's
theorem is perhaps worth giving for its simplicity. Arrange the
operations of a group of even order in pairs $S$ and $S^{-1}$ of inverse
operations. If at any stage no further pairs can be formed, the
remaining operations must all, except the identical operation, be
%% -----File: 107.png---Folio 91-------
of order~$2$. The number of these remaining operations, which
include the identical operation, is even. Hence there must be at
least one operation of order~$2$, and the total number of such operations
is odd.
\end{Remark}

We shall devote the present chapter to the proof of Sylow's
theorem; and a consideration of some of its more immediate
consequences. These constitute, as will be seen later on, a most
important set of results.

\Par{77.} We shall divide the proof of Sylow's theorem into two
parts. First we shew that, if $p^{\alpha}$~is the highest power of a prime~$p$
which divides the order of a group, the group must have a
sub-group of order~$p^{\alpha}$; and secondly that the sub-groups of
order~$p^{\alpha}$ form a single conjugate set and that their number is
congruent to unity, $\pmod{p}$.

\begin{Lemma}
If $p^{\alpha}$ is the highest power of a prime~$p$ by which
the order of a group~$G$ is divisible, $G$~must contain a sub-group
whose order is divisible by~$p^{\alpha}$.
\end{Lemma}

If the group~$G$ is Abelian, this has been already proved in
§~37. Suppose then that $G$ of order~$N$ ($= p^{\alpha}m$, where $m$~is
prime to~$p$) is not Abelian; and let $G'$ of order~$N'$ be the sub-group
of~$G$ formed of its self-conjugate operations. If $S$~is any
operation of~$G$ which is not self-conjugate, and if $n$~is the order
of the greatest sub-group within which $S$~is permutable, $S$~forms
one of a set of $\dfrac{N}{n}$ conjugate operations; hence, by equating
the order of the group to the sum of the numbers of
operations contained in the different conjugate sets, we obtain
the equation
\[
N = N' + \sum \frac{N}{n},
\]
where the sign of summation is extended to all the different
conjugate sets of those operations which are not self-conjugate.

If $N'$ is not divisible by~$p$, or in other words if $G$ contains
no self-conjugate operations of order~$p$, this equation requires
that at least one of the symbols~$n$ should be divisible by~$p^{\alpha}$;
therefore in this case the lemma is true.

If $N'$ is divisible by~$p^{\alpha'}$, the group~$G'$, being Abelian, must
%% -----File: 108.png---Folio 92-------
contain a sub-group~$g'$ of order~$p^{\alpha'}$; and this sub-group is self-conjugate
in~$G$. Consider now the group $\dfrac{G}{g'}$ of order~$p^{\alpha - \alpha'}m$.
If it has no self-conjugate operations of order~$p$, it must
(by the result just obtained) contain a sub-group whose order
is divisible by~$p^{\alpha - \alpha'}$; and hence $G$~contains a sub-group whose
order is divisible by~$p^{\alpha}$. If, on the other hand, $\dfrac{G}{g'}$~has a sub-group
of self-conjugate operations of order~$p^{\alpha''}$, then $G$~must
contain a self-conjugate sub-group of order~$p^{\alpha' + \alpha''}$, say~$g''$.
We may now repeat the same reasoning with the group $\dfrac{G}{g''}$ of
order~$p^{\alpha - \alpha' - \alpha''}m$. In this way we must in any case be ultimately
led to a sub-group of~$G$ whose order is divisible by~$p^{\alpha}$. The
lemma is therefore established.

We have as an immediate inference:---
\index{Sylow's theorem@\Topic{Sylow's theorem}|(}%

\begin{Theorem}[I.]
If $p^{\alpha}$~is the highest power of a prime~$p$ which
divides the order of a group, the group contains a sub-group of
order~$p^{\alpha}$.
\end{Theorem}

For the group contains a sub-group whose order is divisible
by~$p^{\alpha}$; this sub-group must itself contain a sub-group whose
order is divisible by~$p^{\alpha}$, and so on. Hence at last a sub-group
must be arrived at whose order is~$p^{\alpha}$.

\begin{Corollary}
Since it has been seen in §~53 that a group of
order~$p^{\alpha}$ contains sub-groups of every order~$p^{\beta}$ ($\beta < \alpha$), it follows
that, if the order of a group is divisible by~$p^{\beta}$, the group must
contain a sub-group of order~$p^{\beta}$. In particular, a group whose
order is divisible by a prime~$p$ has operations of order~$p$.
\end{Corollary}

\Par{78.} \begin{Theorem}[II.]
If $p^{\alpha}$~is the highest power of a prime~$p$
which divides the order of a group~$G$, the sub-groups of~$G$ of
order~$p^{\alpha}$ form a single conjugate set, and their number is congruent
to unity, $\pmod{p}$.
\end{Theorem}

If $H$~is a sub-group of~$G$ of order~$p^{\alpha}$, the only operations of~$G$,
which are permutable with~$H$ and have powers of~$p$ for
their orders, are the operations of $H$ itself. For if $P$ is an
operation of order~$p^{r}$, not contained in~$H$ and permutable with
it, and if $p^{s}$ is the order of the greatest group common to~$\{P\}$
%% -----File: 109.png---Folio 93-------
and~$H$, the order of $\{H, P\}$ is~$p^{\alpha + r + s}$. But $G$ can have no such
sub-group, since $p^{\alpha + r - s}$ is not a factor of the order of~$G$.

Suppose now that $H'$~is any sub-group conjugate to~$H$;
and let $p^{\beta}$ be the order of the group~$h$ common to $H$ and~$H'$.
When $H'$~is transformed by all the operations of~$H$, the operations
of~$h$ are the only ones which transform $H'$ into itself.
Hence the operations of~$H$ can be divided into $p^{\alpha - \beta}$ sets of $p^{\beta}$
each, such that the operations of each set transform $H'$ into a
distinct sub-group. In this way, $p^{\alpha - \beta}$~sub-groups are obtained
distinct from each other and from~$H$ and conjugate to~$H$. If
these sub-groups do not exhaust the set of sub-groups conjugate
to~$H$, let $H''$ be a new one. From $H''$ another set of $p^{\alpha - \beta'}$
sub-groups can be formed, distinct from each other and from~$H$
and conjugate to~$H$. Moreover no sub-group of this latter set
can coincide with one of the previous set. For if
\[
P_{1}^{-1} H''P_{1} = P_{2}^{-1} H'P_{2},
\]
where $P_{1}$ and $P_{2}$ are operations of~$H$, then
\[
H'' = P_{3}^{-1} H'P_{3},
\]
where $P_{3}$ ($= P_{2}P_{1}^{-1}$) is an operation of~$H$; and this is contrary to
the supposition that $H''$~is different from each group of the
previous set. By continuing this process, it may be shewn that
the number of sub-groups in the conjugate set containing~$H$ is
\[
1 + p^{\alpha - \beta} + p^{\alpha - \beta'} + \Dots,
\]
where no one of the indices $\alpha - \beta$, $\alpha - \beta'$,~\Dots\ can be less than
unity. The number of sub-groups in the conjugate set containing~$H$
is therefore congruent to unity, $\pmod{p}$.

If now $G$ contains another sub-group~$H_{1}$, of order~$p^{\alpha}$, it must
belong to a different conjugate set. The number of sub-groups
in the new set may be shewn, as above, to be congruent
to unity, $\pmod{p}$. But on transforming $H_{1}$ by the operations of~$H$,
a set of $p^{\alpha - \gamma}$ conjugate sub-groups is obtained, where $p^{\gamma}$~is
the order of the sub-group common to $H$ and~$H_{1}$. A further
sub-group of the set, if it exists, gives rise to $p^{\alpha - \gamma'}$ additional
conjugate sub-groups, distinct from each other and from the
previous $p^{\alpha - \gamma}$. Proceeding thus we shew that the number of
sub-groups in the conjugate set is a multiple of~$p$; and as it
cannot be at once a multiple of~$p$ and congruent to unity, $\pmod{p}$,
%% -----File: 110.png---Folio 94-------
the set does not exist. The sub-groups of order~$p^{\alpha}$ therefore
form a single conjugate set and their number is congruent to
unity, $\pmod{p}$.

\begin{Corollary}[I.]
If $p^{\alpha}m$~is the order of the greatest group~$I$,
within which the group~$H$ of order~$p^{\alpha}$ is contained self-conjugately,
the order of the group~$G$ must be of the form
\[
p^{\alpha}m(1 + kp).
\]
\end{Corollary}

\begin{Corollary}[II.]
The number of groups of order~$p^{\alpha}$ contained
in~$G$, \ie\ the factor $1 + kp$ in the preceding expression for the
order of the group, can be expressed in the form
\[
1 + k_{1}p + k_{2}p^{2} + \dots + k_{\alpha}p^{\alpha},
\]
where $k_{r}p^{r}$ is the number of groups having with a given group~$H$
of the set greatest common sub-groups of order~$p^{\alpha - r}$.
\end{Corollary}

This follows immediately from the arrangement of the set
of groups given in the proof of the theorem. Thus each of the
$p^{\alpha - \beta}$ groups, obtained on transforming $H'$ by the operations of~$H$,
has in common with $H$ a greatest common sub-group of
order~$p^{\beta}$. It may of course happen that any one or more of the
numbers $k_{1}$, $k_{2}$,~\dots, $k_{\alpha}$ is zero. If no two sub-groups of the set
have a common sub-group whose order is greater than~$p^{r}$, then
$k_{1}$, $k_{2}$,~\dots, $k_{\alpha - r - 1}$ all vanish; and the number of sub-groups in the
set is congruent to unity, $\pmod{p^{\alpha - r}}$. Conversely, if $p^{s}$~is the
highest power of~$p$ that divides~$kp$, some two sub-groups of the
set must have a common sub-group whose order is not less than
$p^{\alpha - s}$; for if there were no such common sub-groups, the number
of sub-groups in the set would be congruent to unity, $\pmod{p^{s + 1}}$.

\begin{Corollary}[III.]
Every sub-group of~$G$ whose order is~$p^{\beta}$,
($\beta < \alpha$), must be contained in one or more sub-groups of order~$p^{\alpha}$.
\end{Corollary}

For if the sub-group of order~$p^{\beta}$ is contained in no sub-group
of order~$p^{\beta + 1}$, the only operations whose orders are powers of~$p$
that transform it into itself are its own. In this case, the
preceding method may be used to shew that the number of
sub-groups in the conjugate set to which the given sub-group
belongs must be congruent to unity, $\pmod{p}$. But this is
impossible, as the number of such sub-groups must, on the
assumption made, be a multiple of $p^{\alpha - \beta}$. The sub-group of
%% -----File: 111.png---Folio 95-------
order~$p^{\beta}$ is therefore contained in one of order~$p^{\beta + 1}$, and hence
repeating the same reasoning in one of order~$p^{\alpha}$.

\Par{79.} We shall refer to Theorems I and II together as Sylow's
theorem. In discussing in this and the following paragraphs
some of the results that follow from Sylow's theorem, we shall
adhere to the notation that has been used in establishing the
theorem itself. Thus $p^{\alpha}$ will always denote the highest power
of a prime~$p$ which divides the order of~$G$; the sub-groups of~$G$
of order~$p^{\alpha}$ will be denoted by $H$, $H_{1}$,~\dots, and the greatest sub-groups
of~$G$ that contain these self-conjugately by $I$, $I_{1}$,~\dots.
These latter form a single conjugate set of sub-groups of~$G$,
whose orders are~$p^{\alpha}m$, the order of $G$ itself being $p^{\alpha}m(1 + kp)$.
Moreover the number of groups in this conjugate set is~$1 + kp$.

Suppose now that $S$ is any operation of~$G$ whose order is a
power of~$p$. When the $1 + kp$ sub-groups
\[
H,\ H_{1},\ H_{2},\ \dots,\ H_{kp}
\]
are transformed by~$S$, each one that contains~$S$ is transformed
into itself, while the remainder are interchanged in sets, the
number in any set being a power of~$p$. Hence the number of
these groups which contain~$S$ must be congruent to unity,
$\pmod{p}$. In precisely the same way, it may be shewn that the
number of sub-groups of order~$p^{\alpha}$, which contain a given sub-group
of order~$p^{\beta}$, is congruent to unity, $\pmod{p}$.
\index{Sylow's theorem@\Topic{Sylow's theorem}|)}%

A sub-group (or operation), having a power of~$p$ for its order
and occurring in $1 + lp$ sub-groups of order~$p^{\alpha}$, will not necessarily
be one of the same number of conjugate sub-groups (or
operations) in each of these $1 + lp$ sub-groups. If then $h$~is a sub-group,
of order~$p^{r}$, that occurs in $1 + lp$ sub-groups of order~$p^{\alpha}$,
we may choose one of these, say~$H$, in which $h$~is one of as small
a number as possible of conjugate sub-groups. Let this number
be $p^{\alpha - r - s}$, so that, in~$H$, $h$~is self-conjugate in a group of order~$p^{r + s}$.
Then in~$G$, the order of the greatest group~$i$, in which
$h$~is self-conjugate, must be $p^{r + s}n$, where $n$~is relatively prime to~$p$.
The number of sub-groups of~$i$ of order~$p^{r + s}$ must be
congruent to unity, $\pmod{p}$. No two of these groups of order~$p^{r + s}$
can occur in the same group of order~$p^{\alpha}$; for if they did,
they would generate a group of order~$p^{r + s + t}$, ($t \nless 1$), and this is
%% -----File: 112.png---Folio 96-------
impossible, $p^{r + s}$~being the highest power of~$p$ that divides the
order of~$i$. Moreover the number of groups of order~$p^{\alpha}$ in which
any one of these groups of order~$p^{r + s}$ enters is congruent to
unity, $\pmod{p}$. Hence the number of groups of order~$p^{\alpha}$, in
which $h$~is one of $p^{\alpha - r - s}$ (the least possible number) conjugate
sub-groups, is congruent to unity, $\pmod{p}$. Suppose now that,
in~$H'$, $h$~is one of $p^{\alpha - r - s'}$ ($s' < s$) conjugate sub-groups; so that
in~$H'$ it is self-conjugate in a sub-group of order~$p^{r + s'}$ and
in no greater sub-group. Then the highest power of $p$ that
divides the order of the group common to $I'$ and $i$ is~$p^{r + s'}$; and
therefore, when $H'$ is transformed by all the operations of~$i$, the
number of groups of order~$p^{\alpha}$ formed is a multiple of~$p^{s - s'}$. In
each of these groups, $h$~is one of $p^{\alpha - r - s'}$ conjugate sub-groups.
If this does not exhaust all the groups of order~$p^{\alpha}$ in which $h$ is
one of $p^{\alpha - r - s'}$ conjugate sub-groups, let $H''$ be another. Then
from this another set of groups, whose number is a multiple of~$p^{s - s'}$,
may be formed, which are distinct from each other and
from the previous set, such that in each of them $h$~is one of
$p^{\alpha - r - s'}$ conjugate sub-groups. This process may clearly be
continued till all such groups are exhausted. Hence the
number of groups of order~$p^{\alpha}$, in which $h$ is one of $p^{\alpha - r - s'}$ ($s' < s$)
conjugate sub-groups, is a multiple of~$p^{s - s'}$. If $h$~enters as one
of $p^{\alpha - r - s'}$ conjugate sub-groups in~$H'$, a sub-group conjugate
to~$h$ must enter as one of $p^{\alpha - r - s'}$ conjugate sub-groups in~$H$.
Hence if $p^{r + s}$, $p^{r + s'}$, $p^{r + s''}$,~\dots\ are the orders of the greatest
sub-groups that contain $h$ self-conjugately in the various sub-groups
of order~$p^{\alpha}$ in which it appears, then $H$~must contain
sub-groups of the conjugate set (in~$G$) to which $h$ belongs in
conjugate sets of $p^{r + s}$, $p^{r + s'}$, $p^{r + s''}$,~\dots\ only. The total number~$x$
of such sub-groups contained in~$H$ is clearly connected with
the total number $1 + lp$ of sub-groups of order~$p^{\alpha}$, in which $h$~appears,
by the relation
\[
x(1 + kp) = y(1 + lp),
\]
$y$ being the number of sub-groups in the conjugate set. For
every sub-group of order~$p^{\alpha}$ will contain $x$ of the set: and the
two sides of the equation represent two distinct ways of reckoning
all the sub-groups of the set $h$ contained in the sub-groups
of the set~$H$, when all repetitions are counted.
%% -----File: 113.png---Folio 97-------

It is to be noticed that, with the above notation,
\[
y = \frac{p^{\alpha - r - s} m(1 + kp)}{n};
\]
and therefore
\[
x = \frac{p^{\alpha - r - s} m(1 + lp)}{n}.
\]

\Par{80.} \begin{Theorem}[III.]
\index{Sylow's theorem@\Topic{Sylow's theorem}!some direct consequences of|(}%
Let $p^{\alpha}$ be the highest power of a prime~$p$
which divides the order of a group~$G$, and let $H$ be a sub-group
of~$G$ of order~$p^{\alpha}$. Let $h$ be a sub-group common to $H$ and some
other sub-group of order~$p^{\alpha}$, such that no sub-group, which contains~$h$
and is of greater order, is common to any two sub-groups of
order~$p^{\alpha}$. Then there must be some operation of~$G$, of order
prime to~$p$, which is permutable with~$h$ and not with~$H$\footnotemark.
\end{Theorem}
\index{Burnside@\Topic{Burnside}, quoted}%
\index{Frobenius@\Topic{Frobenius}, quoted}%
\footnotetext{Frobenius, ``Ueber endliche Gruppen,'' \textit{Berliner Sitzungsberichte} (1895),
  p.~176: and Burnside, ``Notes on the theory of groups of finite order,'' \textit{Proc.\
  London Mathematical Society}, Vol.~\VolNo{XXVI} (1895), p.~209.}

Suppose that $H$ and $H'$ are two groups of order~$p^{\alpha}$ to
which $h$ is common; and let $h_{1}$ and $h_{1}'$ be sub-groups of $H$ and~$H'$,
of greater order than~$h$, in which $h$~is self-conjugate. If $h_{1}$
and $h_{1}'$ generate a group whose order is a power of~$p$, it must
occur in some group~$H''$ of order~$p^{\alpha}$; and then $H$ and $H''$ have
a common group~$h_{1}$, which contains~$h$ and is of greater order.
This is contrary to supposition, and therefore the order of the
group generated by $h_{1}$ and $h_{1}'$ is not a power of~$p$. Hence $h$ is
permutable with some operation whose order is prime to~$p$.

Let $p^{r}$ be the order of~$h$, and $p^{r + s}n$ be the order of the
greatest sub-group~$i$ of~$G$ that contains $h$ self-conjugately. If $i$~contained
a self-conjugate sub-group of order~$p^{r + s}$, $h_{1}$~and $h_{1}'$
would be sub-groups of it and they would generate a group
whose order is a power of~$p$. This is not the case, and therefore
$i$~must contain $1 + k'p$ sub-groups of order~$p^{r + s}$, so that we may
write $m'(1 + k'p)$ for~$n$; and then, in~$i$, a sub-group of order~$p^{r + s}$
is self-conjugate in a sub-group of order~$p^{r + s}m'$. No sub-group
of~$i$ of order~$p^{r + t}$ ($t > 0$) can occur in more than one sub-group
of order~$p^{\alpha}$; and the $1 + k'p$ sub-groups of~$i$ of order~$p^{r + s}$ belong
therefore to $1 + k'p$ distinct sub-groups of order~$p^{\alpha}$. Moreover
$h$~occurs in no sub-groups of order~$p^{\alpha}$ other than these $1 + k'p$.
For if $h$ occurred in another sub-group~$H_{1}$, it would in this
sub-group be self-conjugate in a group of order~$p^{r + s'}$ ($s' > 0$); and
%% -----File: 114.png---Folio 98-------
this group would occur in~$i$. This group would then be common
to two sub-groups of order~$p^{\alpha}$, contrary to supposition.

An operation of~$i$, which transforms one of its sub-groups
of order~$p^{r + s}$ into another, must transform the sub-group of order~$p^{\alpha}$
containing the one into that containing the other. Hence $i$
must contain operations which are not permutable with~$H$.
The greatest common sub-group of $i$~and~$I$ is that sub-group
of~$i$ of order~$p^{r + s}m'$ which contains the sub-group of order~$p^{r + s}$
belonging to $H$ self-conjugately. For every operation, that
transforms this sub-group of order~$p^{r + s}$ into itself, must transform
$H$ into itself; and no operation can transform $H$ into itself
which transforms this sub-group of order~$p^{r + s}$ into another.

\Par{81.} Let $P$ be an operation, or sub-group, which is self-conjugate
in~$H$; and let $Q$~be another operation, or sub-group,
of~$H$, which is conjugate to~$P$ in~$G$, but not conjugate to~$P$
in~$I$. Suppose first that, if possible, $Q$~is self-conjugate in~$H$.
There must be an operation~$S$ which transforms $P$ into~$Q$ and
$H$ into some other sub-group~$H'$, so that
\begin{align*}
S^{-1}PS &= Q, \\
S^{-1}HS &= H'.
\end{align*}

Now in the sub-group which contains $Q$ self-conjugately,
the sub-groups of order~$p^{\alpha}$ form a single conjugate set, and
$H$~must occur among them. Hence this sub-group must contain
an operation~$T$ such that
\[
T^{-1}QT = Q,
\]
and
\[
T^{-1}H'T = H.
\]
It follows that
\begin{align*}
T^{-1}S^{-1}PST &= Q, \\
T^{-1}S^{-1}HST &= H,
\end{align*}
or that, contrary to supposition, $P$~and~$Q$ are conjugate in~$I$.
Hence:---

\begin{Theorem}[IV.]
Let $G$ and $H$ be defined as in the previous
theorem, and let $I$ be the greatest sub-group of~$G$ which contains
$H$ self-conjugately. Then if $P$ and $Q$ are two self-conjugate
operations or sub-groups of~$H$, which are not conjugate in~$I$,
they are not conjugate in~$G$.
\end{Theorem}
%% -----File: 115.png---Folio 99-------

\begin{Corollary}
If $H$ is Abelian, no two operations of~$H$ which
are not conjugate in~$I$ can be conjugate in~$G$. Hence the
number of distinct sets of conjugate operations in~$G$, which
have powers of~$p$ for their orders, is the same as the number of
such sets in~$I$.
\end{Corollary}

\Par{82.} Suppose next that $Q$ is not self-conjugate in~$H$.
Then every operation that transforms $Q$ into~$P$ must transform
$H$ into a sub-group of order~$p^{\alpha}$ in which $P$~is not self-conjugate.
Of the sub-groups of order~$p^{\alpha}$, to which $P$ belongs and in which
$P$~is not self-conjugate, choose $H'$ so that, in~$H'$, $P$~forms one
of as small a number of conjugate operations or sub-groups as
possible. Let $g$ be the greatest sub-group of~$H'$ that contains
$P$ self-conjugately. Among the sub-groups of order~$p^{\alpha}$ that
contain $P$ self-conjugately, there must be one or more to which $g$
belongs. Let $H$ be one of these; and suppose that $h$~and~$h'$
are the greatest sub-groups of $H$ and $H'$ respectively that
contain $g$ self-conjugately. The orders of both $h$ and $h'$ must
(Theorem~II, §~55) be greater than the order of~$g$; and in consequence
of the assumption made with respect to~$H'$, every
sub-group, having a power of~$p$ for its order and containing~$h$,
must contain $P$ self-conjugately.

Now consider the sub-group $\{h, h'\}$. Since it does not
contain $P$ self-conjugately, its order cannot be a power of~$p$.
Also if $p^{\beta}$ is the highest power of~$p$ that divides its order, it
must contain more than one sub-group of order~$p^{\beta}$. For any
sub-group of order~$p^{\beta}$, to which $h$ belongs, contains $P$ self-conjugately;
and any sub-group of order~$p^{\beta}$, to which $h'$ belongs,
does not. Suppose now that $S$~is an operation of~$\{h, h'\}$, having
its order prime to~$p$ and transforming a sub-group of~$\{h, h'\}$ of
order~$p^{\beta}$, to which $h$ belongs, into one to which $h'$ belongs.
Then $S$~cannot be permutable with~$P$; for if it were, $P$~would
be self-conjugate in each of these sub-groups of order~$p^{\beta}$.

When $P$ is an operation, we may reason in the same way
with respect to~$\{P\}$.

Since $g$ is self-conjugate in both $h$ and~$h'$, $S$~must transform
$g$ into itself. Now $P$ is self-conjugate in~$g$, and therefore
%% -----File: 116.png---Folio 100-------
$S^{-r}PS^{r}$ is also self-conjugate in~$g$ for all values of~$r$. If then
$S^{t}$~is the first power of~$S$ which is permutable with~$P$, the
series of groups $P$, $S^{-1}PS$,~\Dots, $S^{-t+1} PS^{t-1}$ are all distinct and
each is a self-conjugate sub-group of~$g$. Every group in this
series is therefore permutable with every other. Hence:---

\begin{Theorem}[V.]
If $G$ and $H$ are defined as in the two preceding
theorems, and if $P$ is a self-conjugate sub-group or operation of~$H$,
then either \Inum{(i)}~$P$~must be self-conjugate in every sub-group
of~$G$, of order~$p^{\alpha}$, in which it enters, or \Inum{(ii)}~there must be an
operation~$S$, of order~$q$ prime to~$p$, such that the set of sub-groups
$S^{-r} \{P\} S^{r}$ \($r = 0, 1, \Dots, q - 1$\) are all distinct and permutable
with each other.
\end{Theorem}

\begin{Remark}
\Par{83.} In illustration of Sylow's theorem and its consequences, we
\index{Defining relations@\Topic{Defining relations} of a group!for groups of order~$pq$}%
\index{Types of group@\Topic{Types of group}, distinct, whose order is!$pq$, where $p$~and $q$ are different primes}%
will now consider certain special types of group; and we will deal
first with a \emph{group whose order is the product of two different primes}.

If $p_{1}$ and $p_{2}$ ($p_{1} < p_{2}$) are distinct primes, a group of order~$p_{1}p_{2}$
must, by Sylow's theorem, have a self-conjugate sub-group of order~$p_{2}$.
If $p_{2}$ is not congruent to unity, $\pmod{p_{1}}$, the group must also
have a self-conjugate sub-group of order~$p_{1}$. The two self-conjugate
sub-groups of orders $p_{1}$ and $p_{2}$ can have no common operation
except the identical operation; and therefore (Theorem~IX, §~34)
every operation of one must be permutable with every operation of
the other. Hence if $p_{2}$ is not congruent to unity, $\pmod{p_{1}}$, a group
of order~$p_{1}p_{2}$ must be Abelian, and therefore also cyclical.

If $p_{2}$ is congruent to unity, $\pmod{p_{1}}$, there may be either $1$ or $p_{2}$
conjugate sub-groups of order~$p_{1}$.

If there is one, it is self-conjugate and the group is cyclical.

If there are $p_{2}$ sub-groups of order~$p_{1}$, let $P_{1}$ denote an operation
which generates one of them. Then if $P_{2}$ is an operation of order~$p_{2}$,
\[
P_{1}^{-1} \{P_{2}\} P_{1} = \{P_{2}\};
\]
therefore
\[
P_{1}^{-1} P_{2} P_{1} = P_{2}^{\alpha},
\]
so that
\[
P_{2} = P_{1}^{-p_{1}} P_{2} P_{1}^{p_{1}} = P_{2}^{\alpha^{p_{1}}}.
\]
Hence
\[
\alpha^{p_{1}} \equiv 1\ (\mod p_{2}).
\]
If $\alpha$ were unity, $P_{2}$~would be permutable with~$P_{1}$, and $\{P_{1}\}$~would not
be one of $p_{2}$ conjugate sub-groups. Hence $\alpha$~must be a primitive
root of the above congruence.

A group of order~$p_{1}p_{2}$, which has $p_{2}$ conjugate sub-groups of order~$p_{1}$,
must therefore, when it exists, be defined by the relations
\[
P_{1}^{p_{1}} = 1,\quad P_{2}^{p_{2}} = 1,\quad P_{1}^{-1} P_{2} P_{1} = P_{2}^{\alpha},
\]
\index{Sylow's theorem@\Topic{Sylow's theorem}!some direct consequences of|)}%
%% -----File: 117.png---Folio 101-------
\index{Types of group@\Topic{Types of group}, distinct, whose order is!$pq$, where $p$~and $q$ are different primes}%
where $\alpha$~is a primitive root of the congruence\footnote
  {It should be noticed that, if $\alpha$ is not a root of the congruence, the relations
  define a group of order~$p_{1}$. Thus from
  \[
  P_{1}^{-1} P_{2} P_{1} = P_{2}^{\alpha},
  \]
  we have
  \[
  P_{1}^{-p_{1}} P_{2} P_{1}^{p_{1}} = P_{2}^{\alpha^{p_{1}}}.
  \]
  Hence
  \[
  P_{2}^{\alpha^{p_{1} - 1}} = 1,\quad \text{and}\quad P_{2}^{p_{2}} = 1,
  \]
  so that $P_{2} = 1$, and the group reduces to~$\{P_{1}\}$.}
\[
\alpha^{p_{1}} \equiv 1\ (\mod p_{2}).
\]

It follows from §~33 that the order of the group defined by these
relations cannot exceed~$p_{1} p_{2}$.  But also from the given relations it
is clearly impossible to deduce new relations of the form
\[
P_{1}^{x} P_{2}^{y} = P_{1}^{x'} P_{2}^{y'},
\]
where $x$ and $x'$ are less than~$p_{1}$, and $y$ and $y'$ are less than~$p_{2}$; hence
the order cannot be less than~$p_{1} p_{2}$.

Again, let $\beta$ be a primitive root of the congruence, distinct from~$\alpha$,
so that
\[
\beta^{x} \equiv \alpha\ (\mod p_{2}).
\]

Then if, in the group defined by
\[
P_{1}^{p_{1}} = 1,\quad P_{2}^{p_{2}} = 1,\quad P_{1}^{-1} P_{2} P_{1} = P_{2}^{\beta},
\]
we represent $P_{1}^{x}$ by~$P_{1}'$, the defining relations become
\[
P_{1}'^{p_{1}} = 1,\quad P_{2}^{p_{2}} = 1,\quad P_{1}'^{-1} P_{2} P_{1}' = P_{2}^{\alpha};
\]
and the group is simply isomorphic with the previous group.

Hence finally, when $p_{2}$~is congruent to unity, $\pmod{p_{1}}$, there is a
single type of group of order~$p_{1} p_{2}$, which contains $p_{2}$ conjugate
sub-groups of order~$p_{1}$.

\Par{84.} We will next deal with the problem of determining all
\index{Types of group@\Topic{Types of group}, distinct, whose order is!$24$|(}%
distinct types of group of order~$24$.

A group of order~$24$ must contain either $1$ or $3$~sub-groups of
order~$8$, and either $1$ or $4$~sub-groups of order~$3$. If it has one
sub-group of order~$8$ and one sub-group of order~$3$, the group must,
since each of these sub-groups is self-conjugate, be their direct
product. We have seen (§§~68, 74) that there are five distinct types
of group of order~$8$; there are therefore five distinct types of group
of order~$24$, which are obtained by taking the direct product of any
group of order~$8$ and a group of order~$3$.

If there are $3$~groups of order~$8$, some two of them must
(Theorem~II, Cor.~II, §~78) have a common sub-group of order~$4$;
and (Theorem~III, §~80) this common sub-group must be a self-conjugate
sub-group of the group of order~$24$. Moreover if, in
this case, a sub-group of order~$8$ is Abelian, each operation of the
%% -----File: 118.png---Folio 102-------
self-conjugate sub-group of order~$4$ must (Theorem~IV, Cor.\ §~81)
be a self-conjugate operation of the group of order~$24$.

With the aid of these general considerations, it now is easy to
determine for each type of group of order~$8$, the possible types of
group of order~$24$, in addition to the five types already obtained.

\Inum{(i)} Suppose a group of order~$8$ to be cyclical, and let $A$ be an
operation that generates it. If $\{A\}$~is self-conjugate and $B$~is an
operation of order~$3$, then
\[
B^{-1}AB = A^{\alpha},
\]
and therefore
\[
B^{-3}AB^{3} = A^{\alpha^{3}}.
\]
Hence
\[
\alpha^{3} \equiv 1\ (\mod 8),
\]
and therefore
\[
\alpha \equiv 1\ (\mod 8);
\]
so that $A$ and~$B$ are permutable. This is one of the types already
obtained. Hence for a new type, $\{A\}$~cannot be self-conjugate, and
$A^{2}$~must be a self-conjugate operation; $B$~is therefore one of two
conjugate operations, while $\{B\}$~is self-conjugate. Hence the only
possible new type in this case is given by
\[
A^{-1}BA = B^{-1}.
\]

\Inum{(ii)} Next, let a group of order~$8$ be an Abelian group defined by
\[
A^{4} = 1,\quad B^{2} = 1,\quad AB = BA.
\]

If this is self-conjugate, then, by considerations similar to those
of the preceding case, we infer that the group is the direct product
of groups of orders $8$~and~$3$. Hence there is not in this case a new
type.

If the group of order~$8$ is not self-conjugate, the self-conjugate
group of order~$4$ may be either $\{A\}$ or $\{A^{2}, B\}$. In either case, if $C$~is
an operation of order~$3$, it must be one of two conjugate operations
while $\{C\}$~is self-conjugate. Hence there are two new types respectively
given by
\[
C^{3} = 1,\quad BCB = C^{-1},\quad A^{-1}CA = C;
\]
and
\[
C^{3} = 1,\quad A^{-1}CA = C^{-1},\quad BCB = C.
\]

\Inum{(iii)} Let a group of order~$8$ be an Abelian group defined by
\[
A^{2} = 1,\quad B^{2} = 1,\quad C^{2} = 1,\quad  AB = BA,\quad BC = CB,\quad CA = AB.
\]

If it is self-conjugate, and if the group of order~$24$ is not the
direct product of groups of orders $8$~and~$3$, an operation~$D$ of order~$3$
must transform the $7$~operations of order~$2$ among themselves; and
it must therefore be permutable with one of them. Now the relations
\[
D^{-1}AD = A,\quad D^{-1}BD = AB,
\]
are not self-consistent, because they give
\[
D^{-2}BD^{2} = B.
\]
%% -----File: 119.png---Folio 103-------
Hence, since the group of order~$8$ is generated by $A$,~$B$ and any
other operation of order~$2$ except~$AB$, we may assume, without loss
of generality, that
\[
D^{-1}AD = A,\quad  D^{-1}BD = C,\quad  D^{-1}CD = A^{x}B^{y}C^{z}.
\]
These relations give
\[
B = D^{-3}BD^{3} = D^{-1}A^{x}B^{y}C^{z}D = A^{x(1 + z)}B^{yz}C^{y + z^{2}},
\]
and therefore
\[
y = z = 1.
\]

Now if
\[
D^{-1}CD = ABC,
\]
and if
\[
AB = B',\quad AC = C',
\]
then
\[
D^{-1}B'D = C',\quad  D^{-1}C'D = B'C';
\]
so that the two alternatives $x = 0$ and $x = 1$ lead to simply
isomorphic groups.

Hence there is in this case a single type. It is the direct
product of $\{A\}$ and $\{D, B, C\}$, where
\[
D^{-1}BD = C,\quad D^{-1}CD = BC.
\]

If the group of order~$8$ is not self-conjugate, the self-conjugate
group of order~$4$ may be taken to be~$\{A, B\}$; and $D$~being an operation
of order~$3$, there is a single new type given by
\[
D^{3} = 1,\quad CDC = D^{-1},\quad ADA = D,\quad BDB = D.
\]

\Inum{(iv)} Let a group of order~$8$ be a non-Abelian group defined by
\[
A^{4} = 1,\quad B^{4} = 1,\quad A^{2} = B^{2},\quad B^{-1}AB = A^{-1};
\]
and let $C$ be an operation of order~$3$. If the group of order~$8$ is
self-conjugate, and the group of order~$24$ is not a direct product of
groups of orders $8$~and~$3$, $C$~must transform the $3$~sub-groups of
order~$4$, $\{A\}$,~$\{B\}$ and~$\{AB\}$, among themselves. Hence we may take
\[
C^{-1}AC = B,
\]
and
\[
C^{-1}BC = AB\quad \text{or}\quad (AB)^{3}.
\]
If $C$ transforms $B$ into~$(AB)^{3}$, then
\[
C^{-3}AC^{3} = A^{-1},
\]
and $C$ cannot be an operation of order~$3$. Hence in this case there
is only one new type, given by
\[
C^{3} = 1,\quad C^{-1}AC = B,\quad C^{-1}BC = AB.
\]

If the sub-group of order~$8$ is not self-conjugate, the self-conjugate
sub-group of order~$4$ is cyclical, and each of its operations must be
permutable with~$C$. Hence again we get a single new type, given by
\[
C^{3} = 1,\quad A^{-1}CA = C,\quad B^{-1}CB = C^{-1}.
\]

\Inum{(v)} Lastly, let a sub-group of order~$8$ be a non-Abelian group
defined by
\[
A^{4} = 1,\quad B^{2} = 1,\quad BAB = A^{-1}.
\]
%% -----File: 120.png---Folio 104-------

This contains one cyclical and two non-cyclical sub-groups of
order~$4$. If it is self-conjugate, the group of order~$24$ must therefore
be the direct product of groups of orders $8$~and~$3$; and there is no
new type.

If the sub-group of order~$8$ is not self-conjugate, and the self-conjugate
sub-group of order~$4$ is the cyclical group~$\{A\}$, then $A$
must be permutable with an operation~$C$ of order~$3$, and there is a
single new type given by
\[
C^{3} = 1,\quad A^{-1}CA = C,\quad B^{-1}CB = C^{-1}.
\]

If the self-conjugate sub-group of order~$4$ is not cyclical, it may
be taken to be $\{1, A^{2}, B, A^{2}B\}$. If $C$ is permutable with each
operation of this sub-group, there is a single type given by
\[
C^{3}= 1,\quad A^{-1}CA = C^{-1},\quad B^{-1}CB = C.
\]

If $C$ is not permutable with every operation of the self-conjugate
sub-group, it must transform $A^{2}$,~$B$,~$A^{2}B$ among themselves and we
may take
\[
C^{-1}A^{2}C = B,\quad C^{-1}BC = A^{2}B.
\]

Now $\{C, A^{2}, B\}$ is self-conjugate, and therefore $A$ must transform~$C$
into another operation of order~$3$ contained in this sub-group.
Hence
\[
A^{-1}CA = C^{x}A^{2y}B^{z}.
\]

The only values of $x$,~$y$,~$z$ which are consistent with the previous
relation
\[
A^{2}CA^{2} = CA^{2}B,
\]
are
\[
x = 2,\quad y = z = 1.
\]

The last new type is therefore defined by
\begin{gather*}
A^{4} = 1,\quad B^{2} = 1,\quad BAB = A^{-1}, \\
C^{3} = 1,\quad C^{-1}A^{2}C = B,\quad C^{-1}BC = A^{2}B, \\
A^{-1}CA = C^{2}A^{2}B.
\end{gather*}

When $B$ is eliminated between these relations, it will be found
that the only independent relations remaining are
\[
A^{4} = 1,\quad C^{3} = 1,\quad (AC)^{2} = 1.
\]

It is a good exercise to verify that these form a complete set of
defining relations for the group. (Compare Ex.~1, §~35.)

There are therefore, in all, fifteen distinct types of group of order~$24$.
The last of these is the only type, which has neither a self-conjugate
sub-group of order~$8$, nor one of order~$3$. The reader
should satisfy himself, as an exercise, that, in the ten cases where
the group is not a direct product of groups of orders $8$~and~$3$, the
defining relations which we have given are self-consistent. This is
of course an essential part of the investigation, and it may, in
more complicated cases, involve some little difficulty. We have
omitted the verification here, where it is very easy, for the sake of
brevity.
\index{Types of group@\Topic{Types of group}, distinct, whose order is!$24$|)}%
%% -----File: 121.png---Folio 105-------

It is to be noticed that the last type obtained gives an example,
and indeed the simplest possible, of Theorem~V, §~82. Thus in
$\{A, B\}$ of order~$8$, $A^{2}$~is a self-conjugate operation and $B$~is not. In
the group of order~$24$, the operations $A^{2}$ and $B$ are conjugate; and
$C$~is an operation, of order prime to~$2$, such that $A^{2}$, $C^{-1}A^{2}C$, $C^{-2}A^{2}C^{2}$
generate three mutually permutable sub-groups.

A discussion similar to that of the present section (but simpler,
since in each case the number of types is smaller), will verify the
following table\footnotemark:---
\footnotetext{Miller, \textit{Comptes Rendus}, \VolNo{CXXII} (1896), p.~370.}
\index{Miller@\Topic{Miller}, quoted}%
\[
\begin{array}{l|*{11}{c|}}
\text{Order} & 6 & 10 & 12 & 14 & 15 & 18 & 20 & 22 & 26 & 28 & 30 \\
\hline
\Strut
\text{Number} & 2 & 2 & 5 & 2 & 1 & 5 & 5 & 2 & 2 & 4 & 4 \\
\hline
\end{array}
\]

This table, taken with the results of Chapter~V, gives the number
of distinct types of groups for all orders less than~$32$.

\Par{85.} As a second example, we will discuss the various distinct
\index{Types of group@\Topic{Types of group}, distinct, whose order is!$60$|(}%
types of group of order~$60$.

A group of order~$60$ must, by Sylow's theorem, contain either $1$
or $6$ cyclical sub-groups of order~$5$.

We will first suppose that a group~$G$ of order~$60$ contains a
single cyclical sub-group of order~$5$, which is necessarily self-conjugate.
There will then be four operations of order~$5$ in~$G$; and
we may deal with two sub-cases according as these operations are or
are not self-conjugate.

\Inum{(i)} Suppose that each operation of order~$5$ is self-conjugate.
There must then be either $1$ or $3$ sub-groups of order~$4$. If there is
only one, it must be permutable with an operation of order~$3$;
and then $G$~contains a sub-group of order~$12$. If there are three,
it follows, by Theorem~II, Cor.~II (§~78), that some pair of them
must have a common sub-group of order~$2$. But (Theorem~III, §~80)
this sub-group of order~$2$ must be permutable with some operation
of order prime to~$2$, which is not permutable with a sub-group of
order~$4$. This operation must be of order~$3$; hence in this case also
there must be a sub-group of order~$12$. Thus then the group is,
with either alternative, the direct product of a group of order~$5$ and
a group of order~$12$. Now there are five distinct types of group of
order~$12$; there are therefore five distinct types of group of order~$60$
which contain self-conjugate operations of order~$5$.

\Inum{(ii)} Suppose that $S$ is an operation of order~$5$ which is not self-conjugate.
If $T$~is an operation which is not permutable with~$S$,
then
\[
T^{-1}ST = S^{\alpha},
\]
%% -----File: 122.png---Folio 106-------
where $\alpha$~is not unity. Also, if $T^{x}$ is the lowest power of~$T$ which is
permutable with~$S$, then
\[
T^{-x}ST^{x} = S^{\alpha^{x}} = S,
\]
and therefore
\[
\alpha^{x} \equiv 1\ (\mod 5).
\]

It follows that $x$ must be either $2$ or~$4$. If $x$~is~$2$ for every
operation~$T$ which is not permutable with~$S$, then $S$ and $S^{4}$ form a
complete set of conjugate operations; as also do $S^{2}$ and~$S^{3}$. If $x$~is~$4$
for any operation~$T$, the four operations $S$, $S^{2}$, $S^{3}$, $S^{4}$ form a
single conjugate set.

First, let $S$ be one of two conjugate operations; it must then be
self-conjugate in a sub-group of order~$30$, and by Sylow's theorem
this sub-group must contain a single sub-group of order~$3$. It will
therefore be given by
\begin{alignat*}{2}
&\Inum{($\alpha$)}\quad & A^{2} &= 1,\quad B^{3} = 1,\quad S^{5} = 1,\quad ABA = B, \\
\lintertext{or}
&\Inum{($\beta$)}\quad & A^{2} &= 1,\quad B^{3} = 1,\quad S^{5} = 1,\quad ABA = B^{2},
\end{alignat*}
according as $B$ is or is not a self-conjugate operation; $S$~in either
case being permutable with both $A$ and~$B$.

If the sub-groups of order~$4$ are cyclical, $G$~must contain an
operation~$A_{1}$ of order~$4$ whose square is~$A$; and $A_{1}$~must transform $S$
into its inverse and $\{B\}$ into itself. The latter condition clearly
cannot be satisfied if the self-conjugate sub-group of order~$30$ is of
type~\Inum{($\beta$)}. Hence we have two types given by
\begin{gather*}
A_{1}^{4} = 1,\quad B^{3} = 1,\quad S^{5} = 1,\quad B^{-1}SB = S, \\
A_{1}^{-1}SA_{1} = S^{4},\quad \text{and}\quad A_{1}^{-1}BA_{1} = B\quad \text{or}\quad B^{2}.
\end{gather*}

If the sub-groups of order~$4$ are not cyclical, $G$~must contain an
operation~$A'$ of order~$2$, which is permutable with~$A$; and $A'$~transforms
$S$ into its inverse and $\{B\}$~into itself. In this case, if the self-conjugate
sub-group is of type~\Inum{($\alpha$)}, there are two types given by
\begin{gather*}
A'^{2} = 1,\quad A^{2} = 1,\quad B^{3} = 1,\quad S^{5} = 1,\quad ASA = S,\quad B^{-1}SB = S, \\
A'AA' = A,\quad A'SA' = S^{4},\quad ABA = B,\quad A'BA' = B\quad \text{or}\quad B^{2}.
\end{gather*}

If the self-conjugate sub-group is of type~\Inum{($\beta$)}, there is a single
type in which the last two of the preceding equations are replaced
by
\[
ABA = B^{2},\quad A'BA' = B.
\]

Secondly, let the operations of order~$5$ form a single conjugate
set. The sub-groups of order~$4$ must then be cyclical since $G$~contains
an operation~$T$, such that $T^{4}$~is the lowest power of~$T$ which
is permutable with~$S$. Also $S$~is permutable in a sub-group of order~$15$.
This sub-group must be self-conjugate; and therefore $G$~contains
a self-conjugate sub-group of order~$3$. Let
\[
B^{3} = 1,\quad S^{5} = 1,\quad B^{-1}SB = S
\]
define the self-conjugate sub-group of order~$15$; and let $A_{1}$ be an
%% -----File: 123.png---Folio 107-------
operation of order~$4$, none of whose powers is permutable with~$S$.
We may then take
\[
A_{1}^{-1}SA_{1} = S^{2};
\]
since $\{B\}$ is self-conjugate, $A_{1}$~must transform this sub-group into
itself. There are therefore two types given by
\[
A_{1}^{4} = 1,\quad B^{3} = 1,\quad S^{5} = 1,\quad B^{-1}SB = S,\quad A_{1}^{-1}SA_{1} = S^{2},
\]
and
\[
A_{1}^{-1}BA_{1} = B\quad \text{or\quad} B^{2}.
\]
Hence there are in all twelve distinct groups of order~$60$, each of
which has a self-conjugate sub-group of order~$5$.

\Inum{(iii)} Next, suppose that $G$~contains $6$~conjugate sub-groups of
order~$5$. No operation of order~$3$ can be permutable with an
operation of order~$5$, and therefore by Sylow's theorem there must
be $10$~conjugate sub-groups of order~$3$. Hence $G$~contains $24$~operations
of order~$5$ and $20$~operations of order~$3$. If any one
operation of order~$5$ were permutable with an operation of order~$2$,
all its powers would be permutable with the same operation, and
therefore, since the sub-groups of order~$5$ form a single conjugate
set, every operation of order~$5$ would be permutable with an
operation of order~$2$. The group would then contain at least $24$~operations
of order~$10$. This is clearly impossible, since the sum of
the numbers of operations of orders $3$,~$5$ and~$10$ would be greater
than the order of the group. Hence the sub-group of order~$10$,
which contains self-conjugately a sub-group of order~$5$, must be of
the type
\[
A^{2} = 1,\quad S^{5} = 1,\quad ASA = S^{4}.
\]

In a similar way, we shew that a sub-group of order~$6$, which
contains self-conjugately a sub-group of order~$3$, is of the type
\[
A^{2} = 1,\quad B^{3} = 1,\quad ABA = B^{-1}.
\]

Since no operation of order $3$~or~$5$ is permutable with an operation
of order~$2$, it follows (Theorem~III, §~80) that no two sub-groups of
order~$4$ can have a common operation other than identity. Hence
there must be $5$~sub-groups of order~$4$; for if there were $3$~or~$15$,
some of them would necessarily have common operations. Each
sub-group of order~$4$ is therefore contained self-conjugately in a
sub-group of order~$12$. Such a sub-group of order~$12$ can contain no
self-conjugate operation of order~$2$, since $G$~contains no operation of
order~$6$. Hence the sub-groups of order~$4$ are non-cyclical, and the
$3$~operations of order~$2$ in any sub-group of order~$4$ are conjugate
operations in the sub-group of order~$12$ containing it. This sub-group
must therefore be of the type
\[
B^{3} = 1,\quad B^{-1}A_{1}B = A_{2},\quad B^{-1}A_{2}B = A_{1}A_{2},
\]
where $A_{1}$ and $A_{2}$ are two permutable operations of order~$2$.

The $5$~sub-groups of order~$4$ contain therefore $15$~distinct operations
of order~$2$; and these form a conjugate set. We have already
%% -----File: 124.png---Folio 108-------
seen that the $20$~operations of order~$3$ form a conjugate set, and
that the $24$~operations of order~$5$ form two conjugate sets of $12$
each. Hence the $60$~operations of the group are distributed in $5$~conjugate
sets, containing respectively $1$,~$12$, $12$,~$15$ and $20$~operations.
It follows at once (§~27, \PageRef{p.}{35}) that the group, when it exists,
is simple.

A sub-group of order~$12$, the existence of which has been proved,
must be one of $5$~conjugate sub-groups; and, since the group is
simple, no operation can transform each of these into itself. Hence
if the $5$~conjugate sub-groups
\[
H_{1},\ H_{2},\ H_{3},\ H_{4},\ H_{5}
\]
are transformed by any operation of the group into
\[
H_{1}',\ H_{2}',\ H_{3}',\ H_{4}',\ H_{5}',
\]
and if we regard
\[
%[** TN: Left delimiter missing in the original]
\Sub{H_{1}, H_{2}, H_{3}, H_{4}, H_{5}}
    {H_{1}', H_{2}', H_{3}', H_{4}', H_{5}'}
\]
as a substitution performed on $5$~symbols, the group is simply
isomorphic with a substitution group of $5$~symbols. In other words,
the group can be represented as a group of substitutions of $5$~symbols.
Now there are just $60$~even substitutions of $5$~symbols; and it is
easy to verify that the group they form satisfy all the conditions
above determined. Moreover it will be formally proved in Chapter~VIII,
and it is indeed almost obvious, that no group of substitutions
can be simple if it contains odd substitutions. Hence finally, there
is one and only one type of group of order~$60$ which contains $6$~sub-groups
of order~$5$.
\end{Remark}
\index{Types of group@\Topic{Types of group}, distinct, whose order is!$60$|)}%

\begin{Ex}[1.] If $p$,~$q$,~$r$ are distinct primes, shew that a group of order~$p^{2}qr$,
which contains a self-conjugate operation of order~$r$, must be
the direct product of two groups of orders $p^{2}q$ and $r$ respectively.
\end{Ex}

\begin{Ex}[2.] Shew that there is a single type of group of order~$84$
which contains $28$~sub-groups of order~$3$; and determine its defining
relations.
\end{Ex}

\begin{Ex}[3.] If $p^{\alpha}$ ($\alpha > 1$) is the highest power of~$p$ which divides the
order of~$G$, and if $1 + kp$ be the number of sub-groups of~$G$ of order~$p^{\alpha}$,
shew that \Inum{(i)}~if $1 + kp < p^{2}$, \Inum{(ii)}~if a group of order~$p^{x}$ is cyclical
and $1 + kp < p^{\alpha}$, $G$~is composite.
\index{Maillet@\Topic{Maillet}, quoted}%
\Attrib{Maillet, \textit{Comptes Rendus}, \VolNo{CXVIII}, (1894), p.~1188.}
\end{Ex}

\Par{86.} A remarkable and important extension of Sylow's
theorem has recently been given by Herr Frobenius\footnotemark.
\index{Frobenius@\Topic{Frobenius}, quoted}%
\footnotetext{Frobenius: \textit{Berliner Sitzungsberichte} (1895), p.~988.}%
In
the theorem, as stated above, $p^{\alpha}$~is the highest power of a
%% -----File: 125.png---Folio 109-------
prime~$p$ that divides the order of a group. Herr Frobenius
shews that, if $p^{k}$~is any power of a prime~$p$ that divides the
order of a group, the number of sub-groups of order~$p^{k}$ is
congruent to unity, $\pmod{p}$. These groups do not however, in
general, form a single conjugate set.

For a group whose order is a power of~$p$ higher than~$p^{k}$, this
result has been already proved in §~61. Suppose now that
\[
G_{k},\ G_{k}',\ G_{k}'',\ \Dots,
\]
are the sub-groups of order~$p^{k}$ contained in a group~$G$, and that
$p^{\alpha}$~is the highest power of~$p$ dividing the order of~$G$. It has
been seen, in the proof of Sylow's theorem, that $G$~contains at
least one sub-group~$G_{\alpha}$ of order~$p^{\alpha}$. The above set of sub-groups
of order~$p^{k}$ may then be divided into two classes, those
namely which are contained in~$G_{\alpha}$, and those which are not.
The result of §~61 shews that the number of the sub-groups
contained in the first of these classes is congruent to unity,
$\pmod{p}$; and if $G_{\alpha}$ is a self-conjugate sub-group of~$G$, all the
sub-groups must be contained in this class. If $G_{\alpha}$ is not
self-conjugate in~$G$, and if $G_{k}$, any one of the sub-groups
belonging to the second class, be transformed by all the
operations of~$G_{\alpha}$, a set of $p^{x}$ sub-groups of order~$p^{k}$ will result.

Now it is easy to see that $x$~is not less than unity, and that
every one of these $p^{x}$~sub-groups belongs to the second class.
For suppose first that $x$~is zero, so that $G_{k}$~is transformed into
itself by every operation of~$G_{\alpha}$. Then $\{G_{\alpha}, G_{k}\}$ is a sub-group
of~$G$ whose order is a power of~$p$; and since $G_{k}$~is not contained
in~$G_{\alpha}$, the order of this sub-group is not less than $p^{\alpha + 1}$. This is
impossible, since $p^{\alpha + 1}$~is not a factor of the order of~$G$. Secondly,
if $P^{-1}G_{k}P$ were contained in~$G_{\alpha}$, $P$~being some operation of~$G_{\alpha}$,
then $G_{k}$~would be contained in~$PG_{\alpha}P^{-1}$ or in~$G_{\alpha}$, contrary
to supposition. If the sub-groups of the second class are
not thus exhausted, and if $G_{k}'$ is a new one, we may on transforming~$G_{k}'$
by the operations of~$G_{\alpha}$ form a fresh set of $p^{x'}$~sub-groups
of the second class which are distinct from each
other and from the previous set; and this process may be
continued till the second class is exhausted. The number of
sub-groups in the second class is therefore a multiple of~$p$.
Hence:---
%% -----File: 126.png---Folio 110-------

\begin{Theorem}[VI.]
If $p^{k}$, where $p$~is prime, divides the order of
\index{Sylow's theorem@\Topic{Sylow's theorem}!extension of}%
a group~$G$, the number of sub-groups of~$G$ of order~$p^{k}$ is congruent
to unity, $\pmod{p}$.
\end{Theorem}

\Par{87.} If $p$~is a prime which divides the order of a group~$G$,
it immediately follows from the foregoing theorem that the
number of operations of~$G$, which satisfy the relation
\[
S^{p} = 1,
\]
is a multiple of~$p$. For the number of sub-groups of~$G$ of order~$p$
is $k_{1}p + 1$, and no two of these sub-groups can contain a
common operation except identity. There are therefore
\[
(k_{1}p + 1)(p - 1)
\]
distinct operations of order~$p$ in~$G$; these, together with the
identical operation, which also satisfies
\[
S^{p} = 1,
\]
give in all $(k_{1}p - k_{1} + 1)p$ operations.

This result has been generalized by Herr Frobenius\footnote
  {``Verallgemeinerung des Sylow'schen Satzes,'' \textit{Berliner Sitzungsberichte}
  (1895), pp.~984,~985.}
\index{Frobenius@\Topic{Frobenius}, quoted}%
in the
following form:---
\index{Frobenius's theorem@\Topic{Frobenius's theorem} that, if $n$~is a factor of the order of a group, the number of operations of the group whose orders divide~$n$ is a multiple of~$n$|(}%

\begin{Theorem}[VII.]
If $n$~is a factor of the order~$N$ of a group~$G$,
the number of operations of~$G$, including identity, whose
orders are factors of~$n$, is a multiple of~$n$.
\end{Theorem}

It is easy to verify the truth of this theorem directly for
small values of~$N$; and we may therefore assume it true for
every group whose order is less than that of the given group~$G$.
Again, when $n$~is equal to~$N$, the theorem is obviously true.
If then, on the assumption that the theorem is true for all
factors of~$N$ which are greater than~$n$, we shew that it is true
for~$n$, the general truth of the theorem will follow by induction.

If $p$ is any factor of~$\dfrac{N}{n}$, we assume that the number of
operations of~$G$ whose orders divide~$np$ is a multiple of~$np$, and
therefore also of~$n$; and we have to shew that the number of
operations of~$G$, whose orders divide~$np$ and do not divide~$n$, is
%% -----File: 127.png---Folio 111-------
also a multiple of~$n$. Let this set of operations be denoted by~$A$.
If $p^{\lambda - 1}$ is the highest power of~$p$ that divides~$n$, the order of
every operation of the set~$A$ must be equal to or be a multiple
of~$p^{\lambda}$. Let $P$ be one of these operations and $m$~its order. Then,
if $\phi(m)$ is the number of integers less than and prime to~$m$,
the cyclical sub-group~$\{P\}$ contains $\phi(m)$~operations of order~$m$,
and each of these belongs to the set~$A$. If these do not exhaust
the set, let $P'$ of order~$m'$ be another operation belonging to it.
Then no one of the $\phi(m')$~operations of order~$m'$, contained in
the cyclical sub-group~$\{P'\}$, can be identical with any of the
preceding set of $\phi(m)$~operations, since $P'$~itself is not contained
in that set; at the same time, the new set of $\phi(m')$~operations
all belong to~$A$. This process may be continued till
the set~$A$ is exhausted. Now $m$, $m'$,~\Dots\ are all divisible by~$p^{\lambda}$,
and therefore $\phi(m)$, $\phi(m')$,~\Dots\ are all divisible by~$p^{\lambda - 1}(p - 1)$.
Hence the number of operations in the set~$A$ is a multiple
of~$p^{\lambda - 1}$.

If now $n = p^{\lambda - 1}s$, where by supposition $s$~is relatively prime
to~$p$, it remains to shew that the number of operations in the
set~$A$ is a multiple of~$s$. For this purpose, let $P$ be any
operation of~$G$ of order~$p^{\lambda}$; and let those operations of~$G$, which
are permutable with~$P$, form a sub-group~$H$ of order~$p^{\lambda}r$. The
number of operations of~$H$ whose orders divide~$s$ is the same as
the number whose orders divide~$t$, where $t$~is the greatest common
measure of $r$~and~$s$. Now the order of~$\dfrac{H}{\{P\}}$ is less than~$N$, and
therefore we may assume that the number of operations of~$\dfrac{H}{\{P\}}$
whose orders divide~$t$ is a multiple of~$t$, say~$kt$. $H$~therefore
contains $kt$~operations of the form~$PT$, where $P$ and $T$ are
permutable and the order of~$T$ divides~$s$. Now $P$ is one of
a set of $\dfrac{N}{p^{\lambda}r}$ conjugate operations in~$G$; and corresponding to
each of these, there is a similar set of $kt$~operations. Moreover
no two of these operations can be identical; for we have
seen (§~16) that, if $m$~and~$n$ are relatively prime, an operation
of order~$mn$ of a group~$G$, can be expressed in only one way
as the product of two permutable operations of~$G$ of orders $m$~and~$n$.
%% -----File: 128.png---Folio 112-------

The complete set of $\dfrac{N}{p^{\lambda} r} kt$ operations of the form~$PT$ belongs
to~$A$; if $A$ is not thus exhausted, its remaining operations can
be divided into similar sets. Now $N$ is divisible by both $r$
and $s$, and therefore by their least common multiple~$\dfrac{rs}{t}$. Hence
$\dfrac{Nkt}{p^{\lambda} r}$ is divisible by~$s$, and therefore also the number of operations
in the set~$A$ is divisible by~$s$. The number of operations
in~$A$, being divisible both by~$p^{\lambda - 1}$ and by~$s$, is therefore divisible
by~$n$. Hence finally, the number of operations of~$G$, whose
orders divide~$n$, is a multiple of~$n$; and the theorem is proved.

\begin{Corollary}[I.]
Let $N_{n}$ be the number of operations of~$G$
whose orders divide~$n$, and suppose that
\[
N_{n} = n.
\]
Let $p$~be the smallest prime factor of~$n$, and $p^{\alpha}$~the highest
power of~$p$ that divides~$n$, so that
\[
n = p^{\alpha} n_{1},
\]
where every prime factor of~$n_{1}$ is greater than~$p$. If the order
of~$G$ is divisible by a higher power of~$p$ than~$p^{\alpha}$, then
\[
N_{pn} = kpn,
\]
where $k$~is an integer. Now if the order~$m$ of any operation
divides~$pn$ and does not divide~$n$, then $m$~must be a
multiple of~$p^{\alpha + 1}$; and therefore among the operations, whose
orders divide~$n$, there must be operations whose orders are
equal to or are multiples of~$p^{\alpha}$. Hence
\[
N_{p^{\alpha - 1} n_{1}} = \lambda p^{\alpha - 1} n_{1},
\]
where
\[
\lambda < p.
\]
Now $N_{p^{\alpha} n_{1}} - N_{p^{\alpha - 1} n_{1}}$ is the number of operations whose orders
are factors of~$n$ and multiples of~$p^{\alpha}$; and it has been shewn, in
the proof of the theorem, that this number is a multiple of~$p^{\alpha - 1} (p - 1)$.
Hence
\[
(p - \lambda) p^{\alpha - 1} n_{1} = \mu p^{\alpha - 1} (p - 1).
\]
Since every prime factor of~$n_{1}$ is greater than~$p$, this equation
requires that $\lambda$~should be unity; therefore
\[
N_{p^{\alpha - 1} n_{1}} = p^{\alpha - 1} n_{1}.
\]
\index{Frobenius's theorem@\Topic{Frobenius's theorem} that, if $n$~is a factor of the order of a group, the number of operations of the group whose orders divide~$n$ is a multiple of~$n$|)}%
%% -----File: 129.png---Folio 113-------

This process may be repeated to shew that, for each value
of~$\beta$ which is not greater than~$\alpha$, we have
\[
N_{p^{\alpha - \beta} n_{1}} = p^{\alpha - \beta}n_{1},
\]
so that, finally,
\[
N_{n_{1}} = n_{1}.
\]

Moreover it is easy to see that the reasoning holds when
$p^{\alpha}$~is the highest power of~$p$ that divides the order of~$G$, provided
that then the sub-groups of~$G$, of order~$p^{\alpha}$, are cyclical.

Suppose now that
\[
n = p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \Dots p_{n}^{\alpha_{n}},
\]
where $p_{1}$, $p_{2}$,~\Dots, $p_{n}$ are primes in ascending order; and either
that, for each value of~$r$ from $1$ to~$n - 1$, $p_{r}^{\alpha_{r}}$~is not the highest
power of~$p_{r}$ which divides the order of~$G$: or that, if $p_{r}^{\alpha_{r}}$~is the
highest power of~$p_{r}$ that divides the order of~$G$, the sub-groups
of order~$p_{r}^{\alpha_{r}}$ are cyclical.

Then we may prove as above, first, that $G$~contains operations
of each of the orders $p_{1}^{\alpha_{1}}$, $p_{2}^{\alpha_{2}}$,~\Dots, $p_{n - 1}^{\alpha_{n - 1}}$; and secondly
that, for each value of~$r$ from $2$ to~$n$,
\[
N_{p_{r}^{\alpha_{r}} \Dots p_{n}^{\alpha_{n}}}
 = p_{r}^{\alpha_{r}} \Dots p_{n}^{\alpha_{n}}.
\]

The equation
\[
N_{p_{n}^{\alpha_{n}}} = p_{n}^{\alpha_{n}}
\]
implies that $G$ has a self-conjugate sub-group of order~$p_{n}^{\alpha_{n}}$; for
$G$~has a sub-group of this order, and if $G$ had more than one, it
would necessarily contain more than $p_{n}^{\alpha_{n}}$~operations whose
orders divide~$p_{n}^{\alpha_{n}}$.

Again, since $G$ contains a self-conjugate sub-group of order~$p_{n}^{\alpha_{n}}$
and also operations of order~$p_{n - 1}^{\alpha_{n - 1}}$, it must contain a sub-group of
order $p_{n - 1}^{\alpha_{n - 1}}p_{n}^{\alpha_{n}}$. If it had more than one sub-group of this order,
it would contain more than $p_{n - 1}^{\alpha_{n - 1}}p_{n}^{\alpha_{n}}$ operations whose orders
divide $p_{n - 1}^{\alpha_{n - 1}}p_{n}^{\alpha_{n}}$.  But since
\[
N_{p_{n - 1}^{\alpha_{n - 1}}p_{n}^{\alpha_{n}}}
 = p_{n - 1}^{\alpha_{n - 1}}p_{n}^{\alpha_{n}},
\]
this is impossible. Hence $G$~contains a single sub-group of
%% -----File: 130.png---Folio 114-------
order $p_{n - 1}^{\alpha_{n -1}}p_{n}^{\alpha_{n}}$, which is necessarily self-conjugate. In the
same way we shew that, for each order
\[
p_{r}^{\alpha_{r}} \Dots p_{n}^{\alpha_{n}}\quad (r = n - 1, n - 2, \Dots, 1),
\]
$G$~contains a single sub-group.

Finally then, under the conditions stated above, the equation
\[
N_{n} = n
\]
involves the property that $G$~has a self-conjugate sub-group of
order~$n$ and no other sub-group of the same order.
\end{Corollary}

\begin{Corollary}[II\protect\footnotemark.]
\footnotetext{Frobenius, \textit{Berliner Sitzungsberichte}, (1895), p.~987.}%
\index{Frobenius@\Topic{Frobenius}, quoted}%
If $m$~and~$n$ are relatively prime factors of
the order of~$G$, and if the numbers of operations of~$G$ whose
orders divide $m$~and~$n$ respectively are equal to $m$~and~$n$; then
every operation whose order is a factor of~$m$ is permutable with
every operation whose order is a factor of~$n$, and the number of
operations of~$G$ whose orders divide~$mn$ is equal to~$mn$.
\end{Corollary}

Every operation of~$G$ whose order divides~$mn$ can be expressed
as the product of two permutable operations whose
orders divide $m$~and $n$ respectively; and therefore the number
of operations whose orders divide~$mn$ cannot be greater than~$mn$.
On the other hand, it follows from the theorem that the
number of such operations cannot be less than~$mn$; and therefore
every operation whose order divides~$m$ must be permutable
with every operation whose order divides~$n$.

\begin{Corollary} [III\protect\footnotemark.]
\footnotetext{Frobenius: ``Ueber endliche Gruppen,'' \textit{Berliner Sitzungsberichte} (1895),
  p.~170.}%
If a group of order~$mn$, where $m$~and~$n$ are
relatively prime, contains a self-conjugate sub-group of order~$n$,
the group contains exactly $n$ operations whose orders divide~$n$.
\end{Corollary}

If the group~$G$ has an operation~$S$ whose order divides~$n$,
and if $S$ is not contained in the self-conjugate sub-group~$H$ of
order~$n$, $\{S, H\}$~would be a sub-group of~$G$, whose order is
greater than~$n$ and at the same time contains no factor in
common with~$m$. This is impossible, and therefore $H$~must
contain all the operations of~$G$ whose orders divide~$n$.
%% -----File: 131.png---Folio 115-------

\begin{Corollary} [IV\protect\footnotemark.]
\footnotetext{Frobenius, \textit{loc.~cit.}\ p.~170.}%
If $G$~has a self-conjugate sub-group~$H$ of
order~$mn$, where $m$ and $n$ are relatively prime, and if $H$ has a
self-conjugate sub-group~$K$ of order~$n$, then $K$~is a self-conjugate
sub-group of~$G$.
\end{Corollary}

For by the preceding Corollary, $H$~contains exactly $n$~operations
whose orders divide~$n$, namely the operations forming the
sub-group~$K$; and every operation that transforms $H$ into itself
must interchange these $n$~operations among themselves. Hence
every operation of~$G$ is permutable with~$K$.

\begin{Ex}[1.]
Shew that if, the conditions of Corollary~II being satisfied,
the order of~$G$ is~$mn$, then either $G$~is the direct product of two
groups of orders $m$~and~$n$, or $G$~contains an Abelian self-conjugate
sub-group.
\end{Ex}

\begin{Ex}[2.]
If $p$~and~$q$ are distinct primes, there cannot be more than
one type of group of order~$p^{\alpha}q^{\beta}$ which contains no operation of order~$pq$.
\end{Ex}

\Par{88.} We have seen in §~53 that a group~$G$, whose order is
the power of a prime, contains a series of self-conjugate sub-groups
\[
H_{1},\ H_{2},\ \Dots,\ H_{n - 1},\ H_{n},\ G,
\]
such that in $\dfrac{G}{H_{r}}$ every operation of~$\dfrac{H_{r + 1}}{H_{r}}$ is self-conjugate. We
shall conclude the present chapter by shewing that any group
which has such a series of self-conjugate sub-groups is the
direct product of two or more groups whose orders are powers
of primes.

\begin{Theorem}[VIII.]
If a group~$G$, of order~$p^{\alpha}q^{\beta} \Dots r^{\gamma}$, where
$p$, $q$,~\Dots, $r$ are distinct primes, has a series of self-conjugate
sub-groups
\[
H_{1},\ H_{2},\ \Dots,\ H_{n - 1},\ H_{n},\ G,
\]
such that in $\dfrac{G}{H_{r}}$ every operation of~$\dfrac{H_{r + 1}}{H_{r}}$ is self-conjugate, then
$G$~is the direct product of groups of order~$p^{\alpha}$, $q^{\beta}$,~\Dots, $r^{\gamma}$.
\end{Theorem}

Suppose, if possible, that $p$~divides the order of~$H_{2}$ and does
not divide the order of~$H_{1}$. If $P$ is an operation of order~$p$
%% -----File: 132.png---Folio 116-------
contained in~$H_{2}$, $\{P, H_{1}\}$~is a self-conjugate sub-group of~$G$;
and every operation conjugate to~$P$ is contained in the set~$PH_{1}$.
But the only operation of this set, whose order is~$p$, is~$P$. Hence
$P$~must be a self-conjugate operation, contrary to the supposition
that has been made. Hence if the order of~$H_{1}$ is not
divisible by~$p$, neither is the order of~$H_{2}$. Suppose, next, that
$p^{x}$~is the highest power of~$p$ that divides the orders of both $H_{r}$
and~$H_{r - 1}$. Then the order of the sub-group $\dfrac{H_{r}}{H_{r - 1}}$ of~$\dfrac{G}{H_{r - 1}}$,
formed of the self-conjugate operations of the latter, is not
divisible by~$p$; and therefore the order of~$\dfrac{H_{r + 1}}{H_{r - 1}}$ is not divisible
by~$p$. Hence $p^{x}$~is the highest power of~$p$ that divides the
order of~$H_{r + 1}$. This reasoning may be repeated to shew that
$p^{x}$~is the highest power of~$p$ that divides the order of each of the
groups $H_{r + 2}$, $H_{r + 3}$,~\Dots. Hence $x$~must be equal to~$\alpha$; and
therefore the order of~$H_{1}$ must be divisible by each of the
primes $p$, $q$,~\Dots,~$r$.

Suppose now that, for each prime~$p$ which divides the order
of~$G$, every operation of~$H_{r}$, whose order is a power of~$p$, is
permutable with every operation of~$G$ whose order is relatively
prime to~$p$. Let $P$ be any operation, whose order is a power of~$p$,
belonging to~$H_{r + 1}$ and not to~$H_{r}$; and let $Q$ be any operation
of~$G$ whose order is relatively prime to~$p$. If $Q$~is not permutable
with~$P$, then
\[
Q^{-1}PQ = Ph_{r},
\]
where $h_{r}$~is some operation of~$H_{r}$. The order of~$h_{r}$ must be a
power of~$p$. For let $h_{r} = h_{r}' h_{r}''$, where the order of~$h_{r}'$ is a power
of~$p$ and the order of~$h_{r}''$ is relatively prime to~$p$. Then, from
the supposition made with regard to the sub-group~$H_{r}$, the
operation~$Ph_{r}$ is the product of the permutable operations $Ph_{r}'$
and~$h_{r}''$. But, since the order of~$Ph_{r}$ is a power of~$p$, this
is impossible unless $h_{r}''$~is identity. If the order of~$h_{r}$ is~$p^{\beta}$,
then
\[
Q^{-p^{\beta}} PQ^{p^{\beta}} = Ph_{r}^{p^{\beta}} = P;
\]
and this equation implies that $Q$ is permutable with~$P$, since
$p^{\beta}$~and the order of~$Q$ are relatively prime. Hence if the
%% -----File: 133.png---Folio 117-------
supposition that has been made holds for~$H_{r}$, it also holds for~$H_{r + 1}$.
But it certainly holds for~$H_{1}$, and therefore it is true for~$G$.
Hence every operation of~$G$ whose order is a power of~$p$ is
permutable with every operation of~$G$ whose order is relatively
prime to~$p$. The group therefore contains self-conjugate sub-groups
of each of the orders $p^{\alpha}$, $q^{\beta}$,~\Dots, $r^{\gamma}$; and it follows, from
the definition of §~31, that $G$~is the direct product of these
groups.

\begin{Remark}
We add here two examples in further illustration of the applications
of Sylow's theorem.
\end{Remark}

\begin{Ex}[1.] If $p$~is a prime, greater than~$3$, shew that the number of
distinct types of group of order~$6p$ is $6$~or~$4$, according as $p$~is congruent
to $1$~or $5$, $\pmod{6}$.
\end{Ex}

\begin{Ex}[2.] If $p$~is a prime, greater than~$5$, shew that the number of
distinct types of group of order~$12p$ is $18$,~$12$, $15$ or~$10$, according
as $p$~is congruent to $1$,~$5$, $7$ or $11$, $\pmod{12}$.
\end{Ex}
%% -----File: 134.png---Folio 118-------


\Chapter{VII.}{On the Composition-Series of Group.}
\index{Composition factors@\Topic{Composition factors}!definition of}%
\index{Composition-series@\Topic{Composition-series}!definition of}%
\index{Factor groups@\Topic{Factor groups}!set of, for a given group}%
\index{Factor groups@\Topic{Factor groups}!invariance of}%

\Par{89.} \First{Let} $G_{1}$~be a maximum self-conjugate sub-group (§~27)
of a given group~$G$, $G_{2}$~a maximum self-conjugate sub-group of~$G_{1}$,
and so on. Since $G$ is a group of finite order, we must, after
a finite number of sub-groups, arrive in this way at a sub-group
$G_{n - 1}$, whose only self-conjugate sub-group is that formed of the
identical operation alone, so that $G_{n - 1}$ is a simple group.

\begin{Definitions} The series of groups
\[
G,\ G_{1},\ G_{2},\ \Dots,\ G_{n - 1},\ 1,
\]
obtained in the manner just described is called a \emph{composition-series}
of~$G$.

The set of groups
\[
\frac{G}{G_{1}},\ \frac{G_{1}}{G_{2}},\ \Dots,\ \frac{G_{n - 2}}{G_{n - 1}},\ G_{n - 1},
\]
is called a set of \emph{factor-groups} of~$G$, and the orders of these
groups are said to form a set of \emph{composition-factors} of~$G$.
\end{Definitions}

Each of the set of factor-groups is necessarily (§~30) a
simple group.

The set of groups forming a composition-series of $G$ is not,
in general, unique. Thus $G$ may have more than one maximum
self-conjugate sub-group, in which case the second term in the
series may be taken different from~$G_{1}$. Moreover the groups
succeeding $G_{1}$ are not all necessarily self-conjugate in~$G$; and
when some of them are not so, we obtain a new composition-series
%% -----File: 135.png---Folio 119-------
on transforming the whole set by a suitably chosen
operation of~$G$. That the new set thus obtained is again a
composition-series is obvious; for if $G_{r + 1}$ is a maximum self-conjugate
sub-group of~$G_{r}$, so also is $S^{-1} G_{r + 1} S$ of~$S^{-1} G_{r} S$. We
proceed to prove that, if a group has two different composition-series,
the number of terms in them is the same and the
factor-groups derived from them are identical except as regards
the sequence in which they occur.

This result, which is of great importance in the subsequent
theory, is due to Herr Hölder\footnotemark;
\index{Holder@\Topic{Hölder}, quoted}%
\footnotetext{``Zurückführung einer beliebigen algebraischen Gleichung auf eine Kette
  von Gleichungen,'' \textit{Math.\ Ann.}\ \VolNo{XXXIV}, (1889), p.~33.}%
and the proof we here give does
not differ materially from his.

The less general result, that, however the composition-series
may be chosen, the composition-factors are always the same
except as regards their sequence, had been proved by M.~Jordan\footnote
  {``Traité des substitutions,'' (1870), p.~42.}
\index{Jordan@\Topic{Jordan}, quoted}%
some years before the date of Herr Hölder's memoir.

\Par{90.} \begin{Theorem}[I.]
If $H$ is any self-conjugate sub-group of a
group~$G$; and if $K$,~$K'$ are two self-conjugate sub-groups of~$G$ contained
in~$H$, such that there is no self-conjugate sub-group of~$G$
contained in~$H$ and containing either $K$ or $K'$ except $H$,~$K$ and
$K'$ themselves; and if $L$ is the greatest common sub-group of $K$
and~$K'$, so that $L$ is necessarily self conjugate in~$G$; then the
groups $\dfrac{H}{K}$ and $\dfrac{K'}{L}$ are simply isomorphic, as also are the groups
$\dfrac{H}{K'}$ and~$\dfrac{K}{L}$.
\end{Theorem}

Since $K$ and $K'$ are self-conjugate sub-groups of~$G$ contained
in~$H$, $\{K, K'\}$~must also be a self-conjugate sub-group of~$G$
contained in~$H$; and since, by supposition, there is in~$H$ no self-conjugate
sub-group of~$G$ other than $H$ itself, which contains
either $K$ or~$K'$, $\{K, K'\}$~must coincide with~$H$. Hence (§~33)
the product of the orders of $K$ and $K'$ is equal to the product
of the orders of $H$ and~$L$.

If the order of~$\dfrac{K}{L}$ is~$m$, the operations of~$K$ may be divided
into the $m$~sets
\[
L,\ S_{1}L,\ S_{2}L,\ \Dots,\ S_{m - 1}L,
\]
%% -----File: 136.png---Folio 120-------
such that any operation of one set multiplied by any operation
of a second gives some operation of a definite third set, and the
group $\dfrac{K}{L}$ is defined by the laws according to which the sets
combine.

Consider now the $m$~sets of operations
\[
K',\ S_{1}K',\ S_{2}K',\ \Dots,\ S_{m - 1}K'.
\]

No two operations of any one set can be identical. If
operations from two different sets are the same, say
\[
S_{p}k_{1}' = S_{q}k_{2}',
\]
where $k_{1}'$ and $k_{2}'$ are operations of~$K'$, then
\[
S_{q}^{-1} S_{p} = k_{2}' k_{1}'^{-1},
\]
some operation of~$K'$. But $S_{q}^{-1}S_{p}$ is an operation of~$K$; hence,
as it belongs both to $K$ and~$K'$, it must belong to~$L$, so that
\[
S_{p} = S_{q}l,
\]
where $l$ is some operation of~$L$. This however contradicts the
supposition that the operations $S_{p}L$ and $S_{q}L$ are all distinct. It
follows that the operations of the above $m$~sets are all distinct.

Now they all belong to the group~$H$; and their number,
being the order of~$K'$ multiplied by the order of~$\dfrac{K}{L}$, is equal to
the order of~$H$. Hence in respect of the self-conjugate sub-group~$K'$,
which $H$ contains, the operations of the group~$H$ can
be divided into the sets
\[
K',\ S_{1}K',\ S_{2}K',\ \Dots,\ S_{m - 1}K',
\]
and the group $\dfrac{H}{K'}$ is defined by the laws according to which
these sets combine. But if
\[
S_{p}L · S_{q}L = S_{r}L,
\]
then necessarily
\[
S_{p}K' · S_{q}K' = S_{r}K'.
\]
Hence the groups $\dfrac{H}{K'}$ and $\dfrac{K}{L}$ are simply isomorphic. In precisely
the same way it is shewn that $\dfrac{H}{K}$ and $\dfrac{K'}{L}$ are simply
isomorphic.
%% -----File: 137.png---Folio 121-------

\begin{Corollary}
If $H$ coincides with~$G$, $K$~and $K'$ are maximum
self-conjugate sub-groups of~$G$. Hence if $K$ and $K'$ are maximum
self-conjugate sub-groups of~$G$, and if $L$ is the greatest group
common to $K$ and~$K'$, then $\dfrac{G}{K}$ and $\dfrac{K'}{L}$ are simply isomorphic; as
also are $\dfrac{G}{K'}$ and~$\dfrac{K}{L}$.
\end{Corollary}

Now $\dfrac{G}{K}$ and $\dfrac{G}{K'}$ are simple groups; and therefore, $\dfrac{K}{L}$ and $\dfrac{K'}{L}$
being simple groups, $L$~must be a maximum self-conjugate
sub-group of both $K$ and~$K'$.

\Par{91.} We may now at once proceed to prove by a process of
induction the properties of the composition-series of a group
stated at the end of §~89. Let us suppose that, for groups
whose orders do not exceed a given number~$n$, it is already
known that any two composition-series contain the same
number of groups and that the factor-groups defined by them
are the same except as regards their sequence. If $G$, a group
whose order does not exceed~$2n$, has more than one composition-series,
let two such series be
\begin{align*}
&G,\ G_{1},\ G_{2},\Dots,\ 1; \\
\lintertext{and}
&G,\ G_{1}',\ G_{2}',\Dots,\ 1.
\end{align*}
If $H$ is the greatest common sub-group of $G_{1}$ and~$G_{1}'$, and if
\[
H,\ I,\ J,\ \Dots,\ 1
\]
is a composition-series of~$H$, then, by the Corollary in the
preceding paragraph,
\begin{align*}
G,\ G_{1},\ H,\ I,\ &J,\Dots,\ 1, \\
\lintertext{and}
G,\ G_{1}',\ H,\ I,\ &J,\Dots,\ 1
\end{align*}
are two composition-series of~$G$ which contain the same number
of terms and give the same factor-groups. For it has there
been shewn that $\dfrac{G}{G_{1}}$ and $\dfrac{G_{1}'}{H}$ are simply isomorphic; as also are
$\dfrac{G}{G_{1}'}$ and $\dfrac{G_{1}}{H}$. Now the order of~$G_{1}$, being a factor of the order of~$G$,
cannot exceed~$n$. Hence the two composition-series
\begin{alignat*}{3}
&G_{1},&&\ G_{2}, &&\Dots,\ 1, \\
\lintertext{and}
&G_{1},&&\ H,\ I, &&\Dots,\ 1,
\end{alignat*}
%% -----File: 138.png---Folio 122-------
\index{Composition-series@\Topic{Composition-series}!invariance of}%
by supposition contain the same number of groups and give the
same factor-groups; and the same is true of the two composition-series
\begin{alignat*}{2}
&G_{1}',\ G_{2}',&&\Dots,\ 1, \\
\lintertext{and}
&G_{1}',\ H,\ I,&&\Dots,\ 1.
\end{alignat*}
Hence finally, the two original series are seen, by comparing
them with the two new series that have been formed, to have
the same number of groups and to lead to the same factor-groups.
The property therefore, if true for groups whose order
does not exceed~$n$, is true also for groups whose order does not
exceed~$2n$. Now the simplest group, which has more than one
composition-series, is that defined by
\[
A^{2} = 1,\quad B^{2} = 1,\quad AB = BA.
\]
For this group there are three distinct composition-series, viz.\
\begin{alignat*}{2}
&\{A,\, B\},\ \{A\}, &&1; \\
&\{A,\, B\},\ \{B\}, &&1; \\
\lintertext{and}
&\{A,\, B\},\ \{AB\},\ &&1:
\end{alignat*}
and for these the theorem is obviously true. It is therefore
true generally. Hence:---

\begin{Theorem}[II.]
Any two composition-series of a group consist
of the same number of sub-groups, and lead to two sets of factor-groups
which, except as regards the sequence in which they occur,
are identical with each other.
\end{Theorem}

The definite set of simple groups, which we thus arrive at
from whatever composition-series we may start, are essential
constituents of the group: the group is said to be \emph{compounded}
from them. The reader must not, however, conclude either that
the group is defined by its set of factor-groups, or that it necessarily
contains a sub-group simply isomorphic with any given
one of them.

\Par{92.} It has been already pointed out that the groups in a
composition-series of~$G$ are not necessarily, all of them, self-conjugate
groups of~$G$.

Suppose now that a series of groups
\[
G,\ H_{1},\ H_{2},\ \Dots,\ H_{m - 1},\ 1
\]
%% -----File: 139.png---Folio 123-------
are chosen so that each one is a self-conjugate sub-group of~$G$,
while there is no self-conjugate sub-group of~$G$ contained in any
one group of the series and containing the next group.

\begin{Definition}
The series of groups, obtained in the manner
\index{Chief composition series@\Topic{Chief composition series}, or \Topic{Chief series}!definition of}%
\index{Chief composition series@\Topic{Chief composition series}, or \Topic{Chief series}!invariance of}%
just described, is called a \emph{chief composition-series}, or a \emph{chief-series}
of~$G$.
\end{Definition}

It should be noticed that such a series is not necessarily
obtained by dropping out from a composition-series those of its
groups which are not self-conjugate in the original group.
Thus it follows immediately, from the results of §~54, that the
composition-series of a group whose order is the power of a
prime can be chosen, either \Inum{(i)}~so that every group of the series
is a self-conjugate sub-group, or \Inum{(ii)}~so as to contain any given
sub-group, self-conjugate or not.

A chief composition-series of a group is not necessarily
unique; and when a group has more than one, the following
theorem, exactly analogous to Theorem~II, holds:---

\begin{Theorem}[III.]
Any two chief-composition-series of a group
consist of the same number of terms and lead to two sets of
factor-groups, which, except as regards the sequence in which they
occur, are identical with each other.
\end{Theorem}

The formal proof of this theorem would be a mere repetition
of the proof of §~91, Theorem~I itself being used to start from
instead of its Corollary; it is therefore omitted.

Although it is not always possible to pass from a composition-series
to a chief-series, the process of forming a composition-series
on the basis of a given chief-series can always be carried
out. Thus if, in a chief-series, $H_{r + 1}$~is not a maximum self-conjugate
sub-group of~$H_{r}$, the latter group must have a
maximum self-conjugate sub-group $G_{r, 1}$ which contains~$H_{r + 1}$.
If $H_{r + 1}$ is not a maximum self-conjugate sub-group of~$G_{r, 1}$, then
such a group, $G_{r, 2}$, may be found still containing~$H_{r + 1}$; and this
process may be continued till we arrive at a group~$G_{r, s - 1}$, of
which $H_{r + 1}$~is a maximum self-conjugate sub-group. A similar
process may be carried out for each pair of consecutive terms
in the chief-series; the resulting series so obtained is a composition-series
of the original group.
%% -----File: 140.png---Folio 124-------

\Par{93.} The factor-groups $\dfrac{H_{r}}{H_{r + 1}}$ arising from a chief-series are
not necessarily simple groups. If between $H_{r}$ and $H_{r + 1}$ no
groups of a corresponding composition-series occur, the group
$\dfrac{H_{r}}{H_{r + 1}}$ is simple; but when there are such intermediate groups,
$\dfrac{H_{r}}{H_{r + 1}}$ cannot be simple. We proceed to discuss the nature of
this group in the latter case.

Let $G$ be multiply isomorphic with~$G'$, so that the self-conjugate
sub-group $H_{r + 1}$ of~$G$ corresponds to the identical
operation of~$G'$. Also let
\[
H_{1}',\ H_{2}',\ \Dots,\ H_{p}',\ H'_{p + 1},\ \Dots,\ H_{r}',\ 1
\]
be the sub-groups of~$G'$ which correspond to the sub-groups
\[
H_{1},\ H_{2},\ \Dots,\ H_{p},\ H_{p + 1},\ \Dots,\ H_{r},\ H_{r + 1}
\]
of~$G$. Since $H_{p}$ contains~$H_{p + 1}$, $H_{p}'$~must contain~$H'_{p + 1}$; and
since $H_{p + 1}$ is self-conjugate in~$G$, $H'_{p + 1}$~is self-conjugate in~$G'$.
Also if $G'$ had a self-conjugate sub-group contained in~$H_{p}'$ and
containing~$H'_{p + 1}$, $G$~would have a self-conjugate sub-group
contained in~$H_{p}$ and containing~$H_{p + 1}$. This is not the case,
and therefore
\[
G',\ H_{1}',\ H_{2}',\ \Dots,\ H_{r}',\ 1
\]
is a chief-series of~$G'$. Hence $\dfrac{H_{r}}{H_{r + 1}}$ is simply isomorphic with~$H_{r}'$,
the last group but one in the chief-series of~$G'$.

\begin{Definition}
If $\Gamma$ is a self-conjugate sub-group of~$G$, and
\index{Minimum self-conjugate sub-group@\Topic{Minimum self-conjugate sub-group}!definition of}%
if $G$~has no self-conjugate sub-group, contained in~$\Gamma$, whose
order is less than that of~$\Gamma$, then $\Gamma$~is called a \emph{minimum} self-conjugate
sub-group of~$G$.
\end{Definition}

Making use of the phrase thus defined, the discussion of the
factor-groups $\dfrac{H_{r}}{H_{r + 1}}$ of a chief-series is the same as that of the
minimum self-conjugate sub-groups of a given group.

\Par{94.} To simplify the notation as much as possible, let $I$ be
a minimum self-conjugate sub-group of~$G$; and, if $I$ is not a
simple group, suppose that in a composition-series of~$G$ the
%% -----File: 141.png---Folio 125-------
term following~$I$ is~$g_{1}$. Since $g_{1}$~cannot be self-conjugate in~$G$,
it must be one of a set of conjugate sub-groups
\[
%[** TN: Item tag set like an equation number in the original; only instance]
\Inum{($\alpha$)}\quad g_{1},\ g_{2},\ \Dots,\ g_{n}.
\]

Now, if $g_{r} = S^{-1}g_{1}S$, then as $g_{1}$~is a maximum self-conjugate
sub-group of~$I$, $S^{-1}g_{1}S$~or $g_{r}$ is a maximum self-conjugate
sub-group of $S^{-1}IS$ or~$I$; and hence every one of the above set
of conjugate sub-groups is a maximum self-conjugate sub-group
of~$I$. If $g_{rs}$ represents the greatest sub-group common to $g_{r}$ and~$g_{s}$,
then, by Theorem~I of the present chapter (§~90),
\begin{gather*}
\Dots,\ I,\ g_{1},\ g_{1r},\ \Dots \\
\Dots,\ I,\ g_{r},\ g_{1r},\ \Dots
\end{gather*}
are composition-series of~$G$. Hence $\dfrac{g_{1}}{g_{1r}}$  is simply isomorphic
with~$\dfrac{I}{g_{r}}$. Now $\dfrac{I}{g_{r}}$ and $\dfrac{SIS^{-1}}{Sg_{r}S^{-1}}$, or~$\dfrac{I}{g_{1}}$ are simply isomorphic;
therefore $\dfrac{g_{1}}{g_{1r}}$ and $\dfrac{I}{g_{1}}$ are simply isomorphic. Similarly we may
shew that, whatever $r$~and~$s$ may be, $\dfrac{g_{r}}{g_{rs}}$~and $\dfrac{I}{g_{1}}$ are simply
isomorphic.

If $g_{1r}$ and $g_{1s}$ are the same group, whatever $r$~and~$s$
may be, this group is common to the whole set of conjugate
groups~\Inum{($\alpha$)}. If these groups have any common sub-group,
except identity, it is (Theorem~V, §~27) a self-conjugate sub-group
of~$G$, and this self-conjugate sub-group would be contained
in~$I$, contrary to supposition. Hence if $g_{1r}$ and $g_{1s}$ are
the same group, for all values of $r$~and~$s$, this group consists of
the identical operation alone, and a composition-series is
\[
\Dots,\ I,\ g_{1},\ 1.
\]

If $g_{1r}$ and $g_{1s}$ are distinct and if $g_{1rs}$ is their greatest common
sub-group, then
\begin{gather*}
\Dots,\ I,\ g_{1},\ g_{1r},\ g_{1rs},\ \Dots \\
\Dots,\ I,\ g_{1},\ g_{1s},\ g_{1rs},\ \Dots
\end{gather*}
are two composition-series, and $\dfrac{g_{1r}}{g_{1rs}}$~is simply isomorphic with with~$\dfrac{I}{g_{1}}$
for all values of $r$~and~$s$.
%% -----File: 142.png---Folio 126-------

This reasoning may be repeated. If $g_{1rs}$ and $g_{1rt}$ are the
same group whatever $s$~and $t$ may be, they must each be the
identical operation. If not, $g_{1rst}$~is another term of the composition-series,
and $\dfrac{g_{1rs}}{g_{1rst}}$~is simply isomorphic with~$\dfrac{I}{g_{1}}$.

Hence however the composition-series from $I$ onwards be
constructed, the corresponding factor-groups are all simply
isomorphic with each other. Moreover every group after $I$ in
the composition-series is self-conjugate in~$I$. For $g_{1}$ and $g_{r}$
being self-conjugate sub-groups of~$I$, so also is their greatest
common sub-group~$g_{1r}$; and $g_{1r}$,~$g_{1s}$ being self-conjugate sub-groups
of~$I$, $g_{1rs}$~is also self-conjugate; and so on. Let the
composition-series thus arrived at be now written
\[
\Dots,\ I,\ \gamma_{1},\ \gamma_{2},\ \Dots,\ \gamma_{s-1},\ 1;
\]
where, as has been proved, $\dfrac{I}{\gamma_{1}}$, $\dfrac{\gamma_{1}}{\gamma_{2}}$,~\Dots, $\gamma_{s-1}$, are simply isomorphic.

The final group~$\gamma_{s-1}$ must be one of a conjugate set of, say,
$\nu$~groups in~$G$, no one of which has any operation except
identity in common with any other. Since each of these $\nu$~groups
is self-conjugate in~$I$, and since no two of them have a
common operation except identity, it follows by Theorem~IX,
§~34, that every operation of any one of them is permutable
with every operation of the remaining $\nu - 1$. The group generated
by the $\nu$ groups conjugate to~$\gamma_{s-1}$, being self-conjugate in~$G$
and contained in~$I$, must coincide with~$I$. Now, if $\gamma_{s-1}^{1}$ and
$\gamma_{s-1}^{2}$ are any two of this set of $\nu$~groups, $\{\gamma_{s-1}^{1}, \gamma_{s-1}^{2}\}$~is their
direct product, and it is a self-conjugate sub-group of~$I$. If
$s > 2$, $\{\gamma_{s-1}^{1}, \gamma_{s-1}^{2}\}$ does not coincide with~$I$; and therefore there
must be another sub-group~$\gamma_{s-1}^{3}$, of the set to which $\gamma_{s-1}$
belongs, which is not contained in~$\{\gamma_{s-1}^{1}, \gamma_{s-1}^{2}\}$. Since both
the latter group and $\gamma_{s-1}^{3}$ are self-conjugate in~$I$, while $\gamma_{s-1}^{3}$
is a simple group, no operation of~$\gamma_{s-1}^{3}$ except identity can be
contained in~$\{\gamma_{s-1}^{1}, \gamma_{s-1}^{2}\}$. Hence $\gamma_{s-1}^{1}$,~$\gamma_{s-1}^{2}$ and~$\gamma_{s-1}^{3}$ are independent,
\ie\ no operation of one of these groups can be expressed
in terms of operations of the other two. The group
$\{\gamma_{s-1}^{1}, \gamma_{s-1}^{2}, \gamma_{s-1}^{3}\}$ is the direct product of $\gamma_{s-1}^{1}$,~$\gamma_{s-1}^{2}$ and~$\gamma_{s-1}^{3}$,
and it is self-conjugate in~$I$. If $s > 3$, the same reasoning
%% -----File: 143.png---Folio 127-------
may be repeated. Finally, from the set of $\nu$~groups conjugate
to~$\gamma_{s - 1}$, it must be possible to choose $s$~independent groups
\[
\gamma_{s - 1}^{1},\ \gamma_{s - 1}^{2},\ \Dots,\ \gamma_{s - 1}^{s},
\]
such that no operation of any one of them can be expressed in
terms of the operations of the remaining $s - 1$; and $I$~will then
be the direct product of these $s$~groups. Hence:---

\begin{Theorem}[IV.]
If between two consecutive terms $H_{r}$ and $H_{r+1}$
in the chief-composition-series of a group there occur the groups
$G_{r, 1}$, $G_{r, 2}$,~\Dots, $G_{r, s - 1}$ of a composition-series; then \Inum{(i)}~the factor groups
\[
\frac{H_{r}}{G_{r, 1}},\ \frac{G_{r, 1}}{G_{r, 2}},\ \Dots,\ \frac{G_{r, s - 1}}{H_{r + 1}}
\]
are all simply isomorphic, and \Inum{(ii)}~$\dfrac{H_{r}}{H_{r + 1}}$ is the direct product of
$s$~groups of the type~$\dfrac{H_{r}}{G_{r, 1}}$.
\end{Theorem}

\begin{Corollary}
If the order of $\dfrac{H_{r}}{H_{r+1}}$ is a power,~$p^{s}$, of a prime,
$\dfrac{H_{r}}{H_{r + 1}}$ must be an Abelian group of type $(1, 1, \Dots\DPchg{\,}{,} \text{to $s$ units})$.
\end{Corollary}

\Par{95.} A chief-series of a group~$G$ can always be constructed
\index{Minimum self-conjugate sub-group@\Topic{Minimum self-conjugate sub-group}!is a simple group or the direct product of simply isomorphic simple groups}%
which shall contain among its terms any given self-conjugate
sub-group of~$G$. For if $\Gamma$ is a self-conjugate sub-group of~$G$,
and if $\dfrac{G}{\Gamma}$~is simple, we may take $\Gamma$ for the group which follows~$G$
in the chief-series. If on the other hand $\dfrac{G}{\Gamma}$~is not simple, it
must contain a minimum self-conjugate sub-group. Then $\Gamma_{1}$,
the corresponding self-conjugate sub-group of~$G$, contains~$\Gamma$;
and if there were a self-conjugate sub-group of~$G$ contained in~$\Gamma_{1}$,
and containing~$\Gamma$, the self-conjugate sub-group of~$\dfrac{G}{\Gamma}$, which
corresponds to~$\Gamma_{1}$, would not be a minimum self-conjugate sub-group.
We may now repeat the same process with~$\Gamma_{1}$, and so
on; the sub-groups thus introduced will, with $G$~and~$\Gamma$,
clearly form the part of a chief-series extending from $G$ to~$\Gamma$.
The series may be continued from~$\Gamma$, till we arrive at the
identical operation, in the usual way.
%% -----File: 144.png---Folio 128-------

\begin{Remark}
\Par{96.} It will perhaps assist the reader if we illustrate the foregoing
\index{Composition-series@\Topic{Composition-series}!examples of}%
theory by one or two simple examples. We take first a group
of order~$12$, defined by the relations
\begin{gather*}
A^{2} = 1,\quad B^{2} = 1,\quad AB = BA, \\
R^{3} = 1,\quad R^{-1}AR = B,\quad R^{-1}BR = AB\footnotemark.
\end{gather*}
\footnotetext{The reader will notice that $B$~can be eliminated from these relations, and
  that the group can be defined by $A^{2} = 1$, $R^{3} = 1$, $(AR)^{3} = 1$. The structure of the
  group however is given, at a glance, by the equations in the text.}%
From the last two equations, it follows that
\[
R^{-1}ABR = A,
\]
and therefore $R$~transforms the sub-group $\{A, B\}$ of order~$4$ into
itself; so that this sub-group is self-conjugate, and the order of the
group is $12$ as stated. The self-conjugate sub-group $\{A, B\}$ thus
determined is clearly a maximum self-conjugate sub-group. Also it is
the only one. For if there were another its order would be~$6$, and
it would contain all the operations of order~$3$ in the group. Now
since $\{R\}$~is only permutable with its own operations, the group
contains $4$~sub-groups of order~$3$, and therefore there can be no
self-conjugate sub-group of order~$6$. The three cyclical sub-groups
$\{A\}$,~$\{B\}$ and~$\{AB\}$ of order~$2$ are transformed into each other by~$R$,
and therefore no one of them is self-conjugate.

Hence the only chief-series is
\[
\{R, A, B\},\quad \{A, B\},\quad 1,
\]
and there are three composition-series, viz.\
\begin{alignat*}{4}
&\{R, A, B\}, &&\{A, B\}, &&\{A\}, &&1; \\
&\{R, A, B\}, &&\{A, B\}, &&\{B\}, &&1; \\
\lintertext{and}
&\{R, A, B\},\quad &&\{A, B\},\quad &&\{AB\},\quad &&1.
\end{alignat*}

The orders of the factor-groups in the chief-series are $3$ and~$2^{2}$,
and the group of order~$2^{2}$ is, as it should be, an Abelian group whose
operations are all of order~$2$. The composition-factors are $3$,~$2$,~$2$ in
the order written.

\Par{97.} As a rather less simple instance, we will now take a group
generated by four permutable independent operations $A$,~$B$, $P$,~$Q$, of
orders $2$,~$2$, $3$,~$3$ respectively and an operation~$R$ of order~$3$, for
which
\[
R^{-1}AR = B,\quad R^{-1}BR = AB,\quad R^{-1}PR = P,\quad R^{-1}QR = QP\footnotemark.
\]

\footnotetext{Here again the group can clearly be defined in terms of $A$,~$Q$ and~$R$.}%
The sub-group $\{A, B, P, Q\}$, of order~$36$, is clearly a maximum
self-conjugate sub-group, and therefore the order of the group is~$108$.
Since $A$,~$B$ and~$AB$ are conjugate operations, every self-conjugate
sub-group that contains~$A$ must contain~$B$; and since $Q$~and $QP$ are
%% -----File: 145.png---Folio 129-------
conjugate, every self-conjugate sub-group that contains~$Q$ must
contain~$P$. Hence the only other possible maximum self-conjugate
sub-groups are those of the form $\{A, B, P, RQ^{\alpha}\}$; and since
\[
Q^{-1} RQ^{\alpha}Q = RQ^{\alpha} P^{-1},
\]
these groups actually are self-conjugate. The same reasoning shews
that the only maximum self-conjugate sub-group of $\{A, B, P, Q\}$ or
of $\{A, B, P, RQ^{\alpha}\}$, which is self-conjugate in the original group, is
$\{A, B, P\}$; and the only maximum self-conjugate sub-groups of the
latter, which are self-conjugate in the original group, are $\{A, B\}$ and~$\{P\}$.
Hence all the chief-series of the group are given by
\[
\{R, A, B, P, Q\},\quad
\begin{gathered}
\{A, B, P, Q\}, \\
\text{or} \\
\{A, B, P, RQ^{\alpha}\},
\end{gathered} \quad
\{A, B, P\}, \quad
\begin{gathered}
\{A, B\}, \\
\text{or} \\
\{P\},
\end{gathered}
\quad 1.
\]
Since $\{A, B, P, Q\}$ is an Abelian group, all of its sub-groups are
self-conjugate. Hence if $G_{1}$,~$G_{2}$, and~$G_{3}$ are \emph{any} maximum sub-groups
of $\{A, B, P, Q\}$, $G_{1}$~and $G_{2}$ respectively, then
\[
\{R, A, B, P, Q\},\quad \{A, B, P, Q\},\quad G_{1},\quad G_{2},\quad G_{3},\quad 1
\]
is a composition-series.

Again, since $A$~and~$B$ are conjugate in $\{A, B, P, RQ^{\alpha}\}$, the only
maximum self-conjugate sub-groups of this group are those of the
form $\{A, B, P^{x}(RQ^{\alpha})^{y}\}$. If $y$~is zero, this sub-group is Abelian; and
we may take for the next term in the composition series any
maximum sub-group~$g_{2}$ of this Abelian sub-group, and for the last
term but one any sub-group~$g_{3}$ of~$g_{2}$. But if $y$~is not zero,
$\{A, B, P^{x}(RQ^{\alpha})^{y}\}$ can only be followed by~$\{A, B\}$. Hence the
remaining composition-series are of the forms:---
\[
\{R, A, B, P, Q\},\quad \{A, B, P, RQ^{\alpha}\},\quad \{A, B, P\},\quad g_{2},\quad g_{3},\quad 1,
\]
and
\begin{gather*}
\{R, A, B, P, Q\},\quad \{A, B, P, RQ^{\alpha}\},\quad \{A, B, P^{x}(RQ^{\alpha})^{y}\}, \\
\{A, B\},\quad \{A\}\ \text{or}\ \{B\}\ \text{or}\ \{AB\},\quad 1.
\end{gather*}
It should be noticed that if, in the last of these series, we drop out
the terms which are not self-conjugate in the original group, here
the third and fifth terms, we do not arrive at a chief-series. This
illustrates a remark made in §~92.
\end{Remark}
\index{Composition-series@\Topic{Composition-series}!examples of}%

\Par{98.} \begin{Theorem}[V.]
If $H$~is a sub-group of~$G$, each composition-factor
of~$H$ must be equal to or be a factor of some
composition-factor of~$G$.
\end{Theorem}

If $G$ is simple, its only composition-factor is equal to its
order: the theorem in this case is obvious.
%% -----File: 146.png---Folio 130-------

If $G$ is not simple, let $G_{r + 1}$ be the first term in a composition-series
of~$G$ which does not contain~$H$; and let $G_{r}$ be
the term preceding~$G_{r + 1}$. If $H_{1}$ is the greatest common sub-group
of $G_{r + 1}$ and~$H$, then $H_{1}$~is a self-conjugate sub-group
of~$H$. For every operation of~$H$ transforms both $H$ and $G_{r + 1}$
into themselves; and therefore every operation of $H$ transforms~$H_{1}$,
the greatest common sub-group of $H$ and~$G_{r + 1}$, into itself.
Now the order of $\{H, G_{r + 1}\}$ is equal to the product of the
orders of $H$ and $G_{r + 1}$ divided by the order of~$H_{1}$; and $\{H, G_{r + 1}\}$
is contained in~$G_{r}$. Hence the order of $\dfrac{H}{H_{1}}$ is equal to or is
a factor of the order of~$\dfrac{G_{r}}{G_{r + 1}}$. If then a composition-series of~$H$
be taken, containing the term~$H_{1}$, the orders of the factor-groups,
formed by those terms of the series terminating with~$H_{1}$,
are equal to or are factors of the order of~$\dfrac{G_{r}}{G_{r + 1}}$. The same
reasoning may now be used for $H_{1}$ that has been applied to~$H$;
and the theorem is therefore true.

\begin{Corollary}
If all the composition-factors of~$G$ are primes,
so also are the composition-factors of every sub-group of~$G$.
\end{Corollary}

\Par{99.} \begin{Definition} A group, all whose composition-factors
\index{Group@\Topic{Group}!soluble}%
\index{Soluble groups@\Topic{Soluble groups}!definition of}%
are primes, is called a \emph{soluble} group.
\end{Definition}

The Corollary of the last theorem may be stated in the
form:---if a group is soluble, so also are all its sub-groups.

A soluble group of order $p^{\alpha} q^{\beta} \Dots r^{\gamma}$, where $p$, $q$,~\Dots, $r$
are distinct primes, has $\alpha + \beta + \Dots + \gamma$ composition-factors;
these are capable of $\dfrac{(\alpha + \beta + \Dots + \gamma)!}{\alpha!\, \beta!\, \Dots\, \gamma!}$ distinct arrangements.

For a specified group the composition-factors may, as we
have already seen, occur in two or more distinct arrangements;
but it is immediately obvious that two groups of the same
order cannot be of the same type, \ie\ simply isomorphic, unless
the distinct arrangements, of which the composition-factors are
capable, are the same for both. A first step therefore towards
the enumeration of all distinct types of soluble groups of a
given order, will be to classify them according to the distinct
%% -----File: 147.png---Folio 131-------
\index{Soluble groups@\Topic{Soluble groups}!properties of}%
arrangements of which the composition-factors are capable; for
no two groups belonging to different classes can be of the same
type.

The case, in which the composition-factors are capable of all
possible arrangements, is one which will always occur. Taking
in this case $\beta$~$q$'s followed by $\alpha$~$p$'s for the last $\alpha + \beta$ composition-factors,
the group contains a sub-group~$G'$ of order~$p^{\alpha} q^{\beta}$. In
the composition-series of this group, with the composition-factors
taken as proposed, there is a sub-group~$H$ of order~$p^{\alpha}$
contained self-conjugately in a sub-group~$H_{1}$ of order~$p^{\alpha}q$. This
sub-group~$H_{1}$ is contained self-conjugately in a group~$H_{2}$ of
order~$p^{\alpha}q^{2}$. Hence (Theorem~VII, Cor.~IV, §~87) $H$~is
contained self-conjugately in~$H_{2}$. Again, $H_{2}$~is contained self-conjugately
in a group~$H_{3}$ of order~$p^{\alpha}q^{3}$, and therefore again $H$~is
self-conjugate in~$H_{3}$. Proceeding thus, we shew that $H$~is
self-conjugate in~$G'$. It follows that $n$, the number of
conjugate sub-groups of order~$p^{\alpha}$ contained in the group, is
not a multiple of~$q$. Now $q$~may be any one of the distinct
primes other than~$p$ that divide the order of the group. Hence
finally the group contains a self-conjugate sub-group of order~$p^{\alpha}$.
In the same way we shew that it contains self-conjugate
sub-groups of orders $q^{\beta}$,~\Dots, $r^{\gamma}$. The group must therefore
be the direct product of groups whose orders are $p^{\alpha}$, $q^{\beta}$,~\Dots,~$r^{\gamma}$.

Hence:---

\begin{Theorem}[VI.]
A soluble group, the composition-factors
of which may be taken in any order, is the direct product of
groups whose orders are powers of primes.
\end{Theorem}

\Par{100.} \begin{Theorem}[VII.]
If $G$ is a soluble group of order~$p^{\alpha} m$,
where $p$~is prime and does not divide~$m$, and if every operation
of~$G$ whose order is a power of~$p$ is permutable with every
operation whose order is relatively prime to~$p$; then $G$~is the
direct product of two groups of orders $p^{\alpha}$ and~$m$.
\end{Theorem}

Let $H$ be a sub-group of~$G$ of order~$p^{\alpha}$. This sub-group,
from the conditions of the theorem, is necessarily self-conjugate.
Let a chief-series of~$G$ be formed which contains~$H$, say
\[
G,\ \Dots,\ H',\ H,\ \Dots,\ 1.
\]
%% -----File: 148.png---Folio 132-------

The order of~$H'$ must be of the form~$p^{\alpha}q^{\beta}$; and $H'$~must
contain a single sub-group~$h$ of order~$q^{\beta}$. This sub-group~$h$ is
self-conjugate in~$G$, since $H'$~is self-conjugate in~$G$; and $h$~is the
only sub-group of order~$q^{\beta}$ contained in~$H'$.

Consider now the group~$\dfrac{G}{h}$. Every operation of this group
whose order is a power of~$p$ is permutable with every operation
whose order is relatively prime to~$p$. Hence we may repeat
the same reasoning to shew that $\dfrac{G}{h}$~contains a self-conjugate
sub-group of order~$r^{\gamma}$ where $r$~is a prime distinct from~$p$, but
possibly the same as~$q$. It follows that $G$~has a self-conjugate
sub-group~$h_{1}$ of order~$q^{\beta}r^{\gamma}$. We may now repeat the same
reasoning with~$\dfrac{G}{h_{1}}$; and in this way we must at last reach
a self-conjugate sub-group of~$G$ of order~$m$. Hence, since $G$~contains
self-conjugate sub-groups of orders $p^{\alpha}$ and~$m$, which
are relatively prime, $G$~must be the direct product of these
sub-groups.

It should be noticed that the above reasoning does not
necessarily hold if $G$ is not soluble; for then the order
of~$H'$ may be of the form~$p^{\alpha} \mu^{\beta}$, where~$\mu$, the order of a simple
group, contains more than one prime factor. In that case it
would not be necessarily true that $H'$~contains a group of order~$\mu^{\beta}$.

\begin{Remark}
\Par{101.} Though the actual determination of all types of soluble
\index{Soluble groups@\Topic{Soluble groups}!properties of}%
groups of a given order more properly forms part of the subject of
Chapter~XI, we will here, as a further illustration of the subject of
the present Chapter, deal with the problem for groups whose orders
are of the form~$p^{2}q$, $p$~and~$q$ being distinct primes.

Every such group must be soluble. In fact, if $p > q$, the group
must contain a self-conjugate sub-group of order~$p^{2}$; and if $p < q$, there
must be a self-conjugate sub-group of order~$q$ unless $p = 2$, $q = 3$;
while in this case if there are $4$~sub-groups of order~$3$, there must be
a self-conjugate sub-group of order~$4$. These statements all follow
immediately from Sylow's theorem.

The Abelian groups of order~$p^{2}q$ may be specified immediately;
and therefore in what follows we will assume that the group is not
Abelian. There are $3$~possible arrangements for the composition-factors,
viz.\
\[
p,\ p,\ q; \quad p,\ q,\ p; \quad q,\ p,\ p.
\]
%% -----File: 149.png---Folio 133-------

If the factors are capable of all three arrangements, the group is
the direct product of groups of orders $p^{2}$ and~$q$; it is therefore an
Abelian group.

If the two arrangements $p$,~$p$,~$q$ and $q$,~$p$,~$p$ are possible, there are
self-conjugate sub-groups of orders $p^{2}$ and~$q$; the group again is
Abelian, and all three arrangements are possible.

There are now five other possibilities.

\Class{\Inum{I.}}{$p$,~$p$,~$q$ and $p$,~$q$,~$p$, the only possible arrangements.}
\index{Defining relations@\Topic{Defining relations} of a group!for groups of order~$p^{2}q$|(}%
\index{Types of group@\Topic{Types of group}, distinct, whose order is!$p^{2}q$, where $p$~and $q$ are different primes|(}%

There must be here a sub-group of order~$pq$, containing self-conjugate
sub-groups of orders $p$ and $q$ and therefore Abelian. Let
this be generated by operations $P$~and~$Q$, of orders $p$~and~$q$. Since the
group has sub-groups of order~$p^{2}$, there must be operations of orders
$p$ or~$p^{2}$, not contained in the sub-group of order~$pq$, and permutable
with~$P$. Let $R$ be such an operation, so that $R^{p}$~belongs to the
sub-group of order~$pq$. $R$~cannot be permutable with~$Q$, as the
group would be then Abelian; hence
\[
R^{-1}QR = Q^{\alpha},
\]
so that
\[
R^{-p} QR^{p} = Q^{\alpha^{p}},
\]
and
\[
\alpha^{p} \equiv 1\ (\mod q).
\]
This case can therefore only occur if $p$~is a factor of~$q - 1$. There are
two distinct types, according as $R$ is of order~$p$ or of order~$p^{2}$; \ie\
according as the sub-groups of order~$p^{2}$ are non-cyclical or cyclical.
If $\alpha$~and~$\beta$ are any two distinct primitive roots of the congruence
\[
\alpha^{p} \equiv 1\ (\mod q),
\]
the relations
\[
R^{-1}QR = Q^{\alpha},
\]
and
\[
R^{-1}QR = Q^{\beta},
\]
do not lead to distinct types, since the latter may be reduced to the
form by replacing $R$ by~$R^{x}$, where
\[
\beta^{x} \equiv \alpha\ (\mod q).
\]

The two types are respectively defined by the relations
\begin{gather*}
Q^{q} = 1,\quad P^{p} = 1,\quad R^{p} = 1,\quad P^{-1}QP = Q, \\
R^{-1}PR = P,\quad R^{-1}QR = Q^{\alpha};
\end{gather*}
and
\[
Q^{q} = 1,\quad R^{p^{2}} = 1,\quad R^{-1}QR = Q^{\alpha}.
\]

In each case, $\alpha$~is a primitive root of the congruence $\alpha^{p} \equiv 1\ (\mod q)$.

\Class{\Inum{II.}}{$p$,~$q$,~$p$ and $q$,~$p$,~$p$, the only possible arrangements.}

There must be a self-conjugate sub-group of order~$pq$, in which
the sub-group of order~$q$ is not self-conjugate, and a self-conjugate
sub-group of order~$p^{2}$. The sub-group of order~$pq$ must be given by
\begin{gather*}
P^{p} = 1,\quad Q^{q} = 1,\quad Q^{-1}PQ = P^{\alpha}, \\
\alpha^{q} \equiv 1\ (\mod p);
\end{gather*}
%% -----File: 150.png---Folio 134-------
so that in this case $q$~must be a factor of~$p - 1$. If the sub-group of
order~$p^{2}$ is not cyclical, there must be an operation~$R$ of order~$p$, not
contained in the sub-group of order~$pq$. Any such operation must
be permutable with~$P$. Moreover since the sub-group of order~$pq$ is
self-conjugate and contains only $p$~sub-groups of order~$q$, the
sub-group~$\{Q\}$ must be permutable with some operation of order~$p$.
Hence we may assume that $R$~is permutable with~$\{Q\}$, and, since
$p > q$, with~$Q$.

We thus obtain a single type defined by
\begin{alignat*}{3}
P^{p} &= 1,      &R^{p} &= 1,       &       PR &= RP, \\
Q^{q} &= 1,\quad &   QR &= RQ,\quad & Q^{-1}PQ &= P^{\alpha}.
\end{alignat*}

If the sub-group of order~$p^{2}$ is cyclical, all the operations, which
have powers of~$p$ for their orders and are not contained in the sub-group
of order~$pq$, must be of order~$p^{2}$. There can therefore be no
operation of order~$p$, which is permutable with~$\{Q\}$; therefore there
is no corresponding type.

\Class{\Inum{III.}}{$p$,~$p$,~$q$, the only possible arrangement.}

There must be a self-conjugate sub-group of order~$pq$, which has
no self-conjugate sub-group of order~$p$; it is therefore defined by
\begin{gather*}
P^{p} = 1,\quad Q^{q} = 1, \quad P^{-1}QP = Q^{\alpha}, \\
\alpha^{p} \equiv 1\ (\mod q);
\end{gather*}
so that here $p$~must be a factor of~$q - 1$.

If the sub-groups of order~$p^{2}$ are not cyclical, there must be an
operation~$R'$ of order~$p$, not contained in this sub-group and permutable
with~$P$. Hence
\[
R'^{-1}QR' = Q^{\beta};
\]
and if
\[
\beta \equiv \alpha^{x}\ (\mod q),
\]
then $R'P^{-x}$ is an operation of order~$p$, which is not contained in the
sub-group of order~$pq$ and is permutable with~$Q$. It is therefore a
self-conjugate operation of order~$p$. Hence $p$,~$q$,~$p$ is a possible
arrangement of the composition-factors, and there is in this case
no type.

If the sub-groups of order~$p^{2}$ are cyclical, there must be an operation~$R$
of order~$p^{2}$, such that
\[
R^{p} = P.
\]
Hence
\[
R^{-1}QR = Q^{\beta},
\]
where $\beta$ is a primitive root of the congruence
\[
\beta^{p^{2}} = 1\ (\mod q).
\]
This case then can only occur when $p^{2}$~is a factor of~$q - 1$; and we
again have a single type defined by
\[
R^{p^{2}} = 1, \quad Q^{q} = 1,\quad R^{-1}QR = Q^{\beta}.
\]
%% -----File: 151.png---Folio 135-------

\Class{\Inum{IV.}}{$p$,~$q$,~$p$, the only possible arrangement.}

Here the self-conjugate sub-group of order~$pq$ must be given by
\begin{gather*}
P^{p} = 1,\quad Q^{q} = 1,\quad Q^{-1}PQ = P^{\alpha}, \\
\alpha^{q} \equiv 1\ (\mod p),
\end{gather*}
and $q$ must be a factor of~$p - 1$. As in~\Inum{II}, there must be an
operation~$R$ of order~$p$, permutable with~$\{Q\}$ and therefore with~$Q$;
and since $R$~transforms $\{P, Q\}$ into itself, it must be permutable
with~$P$. This however makes the sub-group $\{P, R\}$ self-conjugate,
which requires $q$,~$p$,~$p$ to be a possible arrangement of the composition-factors.
Hence there is no type corresponding to this case.

\Class{\Inum{V.}}{$q$,~$p$,~$p$, the only possible arrangement.}

If the sub-group of order~$p^{2}$ is cyclical, and is generated by~$P$,
while $Q$~is an operation of order~$q$, we must have
\[
Q^{-1}PQ = P^{\alpha},
\]
where
\[
\alpha^{q} \equiv 1\ (\mod p^{2}).
\]
Here $q$ must be a factor of~$p - 1$; since the congruence has just
$q - 1$ primitive roots, there is a single type of group.

If the sub-group of order~$p^{2}$ is not cyclical, it can be generated
by two permutable operations $P_{1}$ and $P_{2}$ of order~$p$. Since a
sub-group of order~$q$ is not self-conjugate, either $p$ or $p^{2}$ must be
congruent to unity $(\mod q)$; and therefore $q$~must be a factor of
either $p - 1$ or~$p + 1$.

Suppose first, that $q$ is a factor of~$p - 1$.

There are $p + 1$ sub-groups of order~$p$. When these are
transformed by any operation~$Q$ of order~$q$, those which are not
permutable with~$Q$ must be interchanged in sets of~$q$. Hence at
least two of these sub-groups must be permutable with~$Q$, and we
may take the generating operations of two such sub-groups for $P_{1}$
and~$P_{2}$. Therefore
\[
Q^{-1}P_{1}Q = P_{1}^{\alpha},\quad Q^{-1}P_{2}Q = P_{2}^{\beta}.
\]

Now if either $\alpha$ or~$\beta$, say~$\beta$, were unity, then $\{Q, P_{1}\}$ would be a
self-conjugate sub-group and $p$,~$q$,~$p$ would be a possible arrangement
of the composition-factors. Hence neither $\alpha$ nor~$\beta$ can be unity, and
we may take
\[
Q^{-1}P_{1}Q = P_{1}^{\alpha},\quad Q^{-1}P_{2}Q = P_{2}^{\alpha^{x}},
\]
where $\alpha$ is a primitive root of
\[
\alpha^{q} \equiv 1\ (\mod p),
\]
and $x$ is not zero.

It remains to determine how many distinct types these equations
contain. When $q = 2$, the only possible value of~$x$ is unity; and there
is a single type. When $q$~is an odd prime, and we take
\[
Q^{y} = Q',\quad P_{1} = P_{2}',\quad P_{2} = P_{1}',\quad xy \equiv 1\ (\mod q),
\]
%% -----File: 152.png---Folio 136-------
the equations become
\[
Q'^{-1} P_{1}' Q' = P_{1}'^{\alpha},\quad Q'^{-1} P_{2}' Q' = P_{2}'^{\alpha^{y}},
\]
and therefore the values $x$~and~$y$ of the index of~$\alpha$, where
\[
xy \equiv 1\ (\mod q),
\]
give the same type. Now the only way, in which the two equations
can be altered into two equations of the same form, is by replacing~$Q$
by some other operation of the group whose order is~$q$ and by
either interchanging $P_{1}$ and~$P_{2}$ or leaving each of them unchanged.
Moreover the other operations of the group whose orders are~$q$ are
those of the form $Q^{l} P_{1}^{m} P_{2}^{n}$, where $l$ is~not zero, and this operation
transforms $P_{1}$ and~$P_{2}$ in the same way as~$Q^{l}$. Hence finally, the
values $x$~and~$y$ of the index will only give groups of the same type
when
\[
xy \equiv 1\ (\mod q).
\]
There are therefore $\frac{1}{2} (q + 1)$ distinct types, when $q$~is an odd prime;
they are given by the above equations.

Suppose next, that $q$~is a factor of~$p + 1$.

Any operation~$Q$, of order~$q$, will transform the sub-groups of
order~$p$, with which it is not permutable, so as to interchange them
in sets of~$q$; and hence if it is permutable with any sub-group, it
must be permutable with $q$ at least. This, by the last case, is
clearly impossible, and hence $Q$~is permutable with no sub-group of
order~$p$. We may therefore, by suitably choosing the generating
operations of the group of order~$p^{2}$, assume that
\[
Q^{-1} P_{1} Q = P_{2},\quad Q^{-1} P_{2} Q = P_{1}^{\alpha} P_{2} ^{\beta}.
\]

If now
\[
Q^{-x - 1} P_{1} Q^{x + 1} = P_{1}^{\alpha_{x}} P_{2}^{\beta_{x}},
\]
then
\[
\alpha_{x + 1} \equiv \alpha \beta_{x},\quad
\beta_{x + 1}  \equiv \alpha_{x} + \beta \beta_{x}\ (\mod p),
\]
and therefore
\[
\beta_{x + 1} - \beta \beta_{x} - \alpha \beta _{x - 1} \equiv 0\ (\mod p).
\]
Hence if $\iota_{1}$ and $\iota_{2}$ are the roots of the congruence
\[
\iota^{2} - \beta \iota - \alpha \equiv 0\ (\mod p),
\]
then
\[
\beta_{x} \equiv \frac{\iota_{2}^{x + 1} - \iota_{1}^{x + 1}}{\iota_{2} - \iota_{1}}.
\]

Now since $Q^{q}$ is the lowest power of~$Q$ that is permutable with~$P_{1}$,
$\beta_{q - 1}$~must be the first term of the series $\beta_{1}$, $\beta_{2}$,~\dots\ which vanishes.
Hence $q$~is the least value of~$z$ for which
\[
\iota_{2}^{z} \equiv \iota_{1}^{z},
\]
and therefore the congruence
\[
\iota^{2} - \beta \iota - \alpha \equiv 0
\]
is irreducible. Moreover $\alpha_{q - 1}$ must be congruent to unity, and
therefore
\[
1 \equiv -\iota_{1}\iota_{2} \frac{\iota_{2}^{q - 1} - \iota_{1}^{q - 1}}{\iota_{2} - \iota_{1}} \equiv \iota_{1}^{q}.
\]
%% -----File: 153.png---Folio 137-------

From the quadratic congruence satisfied by~$\iota$, it follows that
\[
\alpha \equiv -\iota^{p + 1} \equiv -1,\quad \beta \equiv \iota^{p} + \iota\ (\mod p);
\]
and thence
\[
\alpha_{x} \equiv -\frac{\iota^{px} - \iota^{x}}{\iota^{p} - \iota},\quad
\beta_{x}  \equiv  \frac{\iota^{p(x + 1)} - \iota^{x + 1}}{\iota^{p} - \iota}.
\]

Finally, we may shew that, when $q$~is a factor of~$p + 1$, the
equations
\begin{gather*}
P_{1}^{p} = 1,\quad P_{2}^{p} = 1,\quad Q^{q} = 1,\quad P_{1}P_{2} = P_{2}P_{1}, \\
Q^{-1}P_{1}Q = P_{2},\quad  Q^{-1}P_{2}Q = P_{1}^{-1}P_{2}^{\iota^{p} + \iota},
\end{gather*}
where $\iota$ is a primitive root of the congruence
\[
\iota^{q} \equiv 1\ (\mod p),
\]
define a single type of group, whatever primitive root of the congruence
is taken for~$\iota$.

Thus from the given equations it follows that
\begin{align*}
Q^{-x}P_{1}Q^{x} &= P_{1}^{\alpha_{x - 1}}P_{2}^{\beta_{x - 1}} = P_{3},\text{ say,} \\
\intertext{and}
Q^{-x}P_{2}'Q^{x} &= (P_{1}^{\alpha_{x - 1}}P_{2}^{\beta_{x - 1}})^{\alpha_{x - 1}} (P_{1}^{\alpha_{x}}P_{2}^{\beta_{x}})^{\beta_{x - 1}}, \\
&= P_{1}^{\alpha_{x}\beta_{x - 1} - \alpha_{x - 1}\beta_{x}} P_{3}^{\alpha_{x - 1} + \beta_{x}} \\
&= P_{1}^{-1}P_{3}^{\iota^{px} + \iota^{x}}.
\end{align*}
If then we take $P_{1}$,~$P_{3}$ and~$Q^{x}$ as generating operations in the place
of $P_{1}$,~$P_{2}$ and~$Q$, the defining relations are reproduced with $\iota^{x}$ in the
place of~$\iota$. The relations therefore define a single type of group\footnotemark.
\index{Cole and Glover@\Topic{Cole} and \Topic{Glover}, quoted}%
\index{Holder@\Topic{Hölder}, quoted}%
\footnotetext{On groups whose order is of the form~$p^{2}q$ the reader may consult; Hölder,
  ``Die Gruppen der Ordnungen $p^{3}$, $pq^{2}$, $pqr$,~$p^{4}$,'' \textit{Math.\ Ann.}\ \DPnote{** No "Vol."}\VolNo{XLIII} (1893), in
  particular pp.~335--360; and Cole and Glover, ``On groups whose orders are
  products of three prime factors,'' \textit{Amer.\ Journal}, \VolNo{XV} (1893), pp.~202--214.}%

We have, for the sake of brevity, in each case omitted the
verification that the defining relations actually give a group of
order~$p^{2}q$. This presents no difficulty, even for the last type; for
the previous types it is immediately obvious.
\end{Remark}
\index{Defining relations@\Topic{Defining relations} of a group!for groups of order~$p^{2}q$|)}%
\index{Types of group@\Topic{Types of group}, distinct, whose order is!$p^{2}q$, where $p$~and $q$ are different primes|)}%
%% -----File: 154.png---Folio 138-------


\Chapter{VIII.}{On Substitution-Groups: Transitive and
Intransitive Groups.}

\Par{102.} \First{It} has been proved, in the theorem in §~20, that every
group is capable of being represented as a group of substitutions
performed on a number of symbols equal to the order of the
group. For applications to Algebra, and in particular to the
Theory of Equations, the presentation of a group as a group of
substitutions is of special importance; and we shall now proceed
to consider the more important properties of this special mode
of representing groups\footnotemark.
\footnotetext{When it is necessary to call attention directly to the fact that the group
  we are dealing with is supposed to be presented as a group of substitutions, the
  group will be spoken of as a \emph{substitution-group}.}

\begin{Definition}
When a group is represented by means of
\index{Degree@\Topic{Degree} of a substitution group!definition of}%
\index{Substitution group@\Topic{Substitution group}!degree of, definition of}%
substitutions performed on a finite set of $n$ distinct symbols,
the integer~$n$ is called the \emph{degree} of the group.
\end{Definition}

It is obvious, by a consideration of simple cases, that a
group can always be represented in different forms as a group
of substitutions, the number of symbols which are permuted
in two forms not being necessarily the same; examples have
already been given in Chapter~II\@. The ``degree of a group''
is therefore only an abbreviation of ``the degree of a special
representation of the group as a substitution-group.''

The $n!$ substitutions, including the identical substitution,
that can be performed upon $n$~distinct symbols, clearly form a
%% -----File: 155.png---Folio 139-------
group; for they satisfy the conditions of the definition (§~12).
Moreover they form the greatest group of substitutions that can
be performed on the $n$~symbols, because every possible substitution
occurs among them. When a group then is spoken of as of
degree~$n$, it is implicitly being regarded as a sub-group of this
most general group of order~$n!$ which can be represented by
substitutions of the $n$~symbols; and therefore (Theorem~I,
§~22) the order of a substitution-group of degree~$n$ must be
a factor of~$n!$.

\Par{103.} It has been seen in §~11 that any substitution performed
on $n$~symbols can be represented in various ways as
the product of transpositions; but that the number of transpositions
entering in any such representation of the substitution
is either always even or always odd. In particular, the identical
substitution can only be represented by an even number of
transpositions. Hence if $S$~and~$S'$ are any two even (§~11)
substitutions of $n$~symbols, and $T$~any substitution at all of $n$~symbols,
then $SS'$ and $T^{-1}ST$ are even substitutions. The
even substitutions therefore form a self-conjugate sub-group~$H$
of the group~$G$ of all substitutions.

If now $T$ is any odd substitution, the set of substitutions~$TH$
are all odd and all distinct. Moreover they give all the
odd substitutions; for if $T'$~is any odd substitution distinct
from~$T$, then $T^{-1}T'$~is an even substitution and must be
contained in~$H$. Hence the number of even substitutions is
equal to the number of odd substitutions: and the order of~$G$
is twice that of~$H$.

\begin{Definitions}
The group of order~$n!$ which consists of all
\index{Alternating group@\Topic{Alternating group}!definition of}%
\index{Group@\Topic{Group}!symmetric}%
\index{Group@\Topic{Group}!alternating}%
\index{Symmetric group@\Topic{Symmetric group}!definition of}%
the substitutions that can be performed on $n$ symbols is called
the \emph{symmetric} group of degree~$n$.

The group of order~$\frac{1}{2} n!$ which consists of all the even
substitutions of $n$~symbols is called the \emph{alternating}\footnote
  {The symmetric group has been so called because the only functions of the
  $n$~symbols which are unaltered by all the substitutions of the group are the
  symmetric functions.

  All the substitutions of the alternating group leave the square root of the
  discriminant unaltered (§~11).}
group of
degree~$n$.
\end{Definitions}
%% -----File: 156.png---Folio 140-------

If the substitutions of a group of degree~$n$ are not all even,
the preceding reasoning may be repeated to shew that its even
substitutions form a self-conjugate sub-group whose order is
half the order of the group; and this sub-group is a sub-group
of the alternating group of the $n$~symbols.

\Par{104.} \begin{Definition}
A substitution-group is called \emph{transitive}
\index{Degree@\Topic{Degree} of a substitution group!is a factor of the order, if the group is transitive}%
\index{Intransitive groups@\Topic{Intransitive groups}!definition of}%
\index{Substitution group@\Topic{Substitution group}!transitive and intransitive, definition of}%
\index{Substitution group@\Topic{Substitution group}!degree of transitive, is a factor of its order}%
\index{Transitive group@\Topic{Transitive group}!definition of}%
when, by means of its substitutions, a given symbol~$a_{1}$ can be
changed into every other symbol $a_{2}$, $a_{3}$,~\Dots, $a_{n}$ operated on
by the group. When it has not this property, the group is
called \emph{intransitive}.
\end{Definition}

A transitive group contains substitutions changing any one
symbol into any other. For if $S$~and~$T$ respectively change~$a_{1}$
into $a_{s}$~and~$a_{t}$, then $S^{-1}T$~changes $a_{s}$ into~$a_{t}$.

\begin{Theorem}[I.]
The substitutions of a transitive group~$G$, which
leave a given symbol~$a_{1}$ unchanged, form a sub-group; and the
number of substitutions, which change $a_{1}$ into any other symbol~$a_{r}$,
is equal to the order of this sub-group.
\end{Theorem}

The substitutions which leave $a_{1}$ unchanged must form a
sub-group~$H$ of~$G$; for if $S$~and~$S'$ both leave~$a_{1}$ unchanged, so
also does~$SS'$.

Let the operations of~$G$ be divided into the sets
\[
H,\ HS_{1},\ HS_{2},\ \Dots,\ HS_{m - 1}.
\]

No operation of the set~$HS_{p}$ leaves $a_{1}$ unchanged; and each
operation of the set~$HS_{p}$ changes $a_{1}$ into~$a_{p}$, if $S_{p}$~does so. If
the operations of any other set~$HS_{q}$ also changed $a_{1}$ into~$a_{p}$, then
$S_{p}S_{q}^{-1}$ would leave $a_{1}$ unchanged and would belong to~$H$, which
it does not. Hence each set changes~$a_{1}$ into a distinct symbol.
The number of sets must therefore be equal to the number of
symbols, while from their formation each set contains the same
number of substitutions. If $N$~is the order and $n$ the degree
of the transitive group~$G$, then $\dfrac{N}{n}$~is the order of the sub-group
that leaves any symbol~$a_{1}$ unchanged; and there are $\dfrac{N}{n}$
substitutions changing $a_{1}$ into any other given symbol~$a_{p}$.

\begin{Corollary}
The order of a transitive group must be
divisible by its degree.
\end{Corollary}
%% -----File: 157.png---Folio 141-------

Every group conjugate to~$H$ leaves one symbol unchanged.
\index{Substitution group@\Topic{Substitution group}!transitive, whose substitutions, except identity, displace all or all but one of the symbols|(}%
For if $S$ changes $a_{1}$ into~$a_{p}$, then $S^{-1}HS$~leaves $a_{p}$ unchanged.
The sub-groups which leave the different symbols unchanged
form therefore a conjugate set.

{\Loosen A transitive group of degree~$n$ and order~$mn$ has, as we
have just seen, $m - 1$ substitutions other than identity which
leave a given symbol~$a_{1}$ unchanged. Hence there must be at
least $mn - 1 - n(m - 1)$, \ie\ $n - 1$, substitutions in the group
that displace every symbol. If the $m - 1$ substitutions, other
than identity, that leave $a_{1}$ unchanged are all distinct from the
$m - 1$ that leave $a_{p}$ unchanged, whatever other symbol~$a_{p}$ may
be, $n- 1$~will be the actual number of substitutions that
displace all the symbols; and no operation other than identity
will displace less than $n - 1$ symbols. If however the sub-groups
that leave $a_{1}$~and~$a_{p}$ unchanged have other substitutions
besides identity in common, these substitutions must displace
less than $n - 1$ symbols; and there will be more than $n - 1$
substitutions which displace all the symbols.}

\begin{Ex} If the substitutions of two transitive groups of degree~$n$
which displace all the symbols are the same, the groups can only
differ in the substitutions that keep just one symbol unchanged.
\index{Netto@\Topic{Netto}, quoted}%
\Attrib{Netto.}
\end{Ex}

\begin{Remark}
\Par{105.} We have seen that every group can be represented as a
substitution-group whose degree is equal to its order. A reference
to the proof of this theorem (§~20) will shew that such a substitution-group
is transitive, and that the identical substitution is the only
one which leaves any symbol unchanged.

We will now consider some of the properties of a transitive
group of degree~$n$, whose operations, except identity, displace all or
all but one of the $n$~symbols. It has just been seen that such a group
has exactly $n - 1$ operations which displace all the $n$~symbols.
If these $n - 1$ operations, with identity, form a sub-group, the sub-group
must clearly be self-conjugate.

Suppose now that $nm$ is the order of the group. Then the order of
the sub-group, that leaves one symbol~$a_{1}$ unchanged, is~$m$; and if $a_{2}$~is
any other of the $n$~symbols, no two operations of this sub-group
can change $a_{2}$ into the same symbol. For if $S$,~$S'$ were two operations
of the sub-group both changing $a_{2}$ into~$a_{r}$, then $SS'^{-1}$~would be
an operation, distinct from identity, which would keep both $a_{1}$ and
$a_{2}$ unchanged. Let now $P$ be any operation that displaces all the
symbols. Then the set of $m$~operations $S^{-1}PS$, where for $S$ is put in
%% -----File: 158.png---Folio 142-------
turn each operation of the sub-group that keeps $a_{1}$ unchanged, are all
distinct; for each of them changes $a_{1}$ into a different symbol. If this
set does not exhaust the operation conjugate to~$P$, and if $P_{1}$~is another
such operation, then the set of $m$~operations $S^{-1}P_{1}S$ are all distinct
from each other and from those of the previous set. This process
may be continued till the operations conjugate to~$P$ are exhausted.
The number of operations conjugate to~$P$ is therefore a multiple
of~$m$. Since $P$~itself does not belong to any one of the $n$~conjugate
sub-groups that each keep one symbol fixed, no operation conjugate
to~$P$ can belong to any of them. Hence each of the $km$~operations
of the conjugate set to which $P$ belongs displaces all the symbols.
The $n - 1$ operations that displace all the symbols can therefore be
divided into sets, so that the number in each set is a multiple of~$m$;
and hence $m$~must be a factor of~$n - 1$.

Suppose now that $p$ is any prime factor of~$n$, and that $p^{\alpha}$~is the
highest power of~$p$ which divides~$n$. If $P$~is an operation whose
order is a power of~$p$, and if $p^{\beta}\mu$~is the order of the greatest sub-group~$h$
that contains $P$ self-conjugately, then $P$~is one of $\dfrac{nm}{p^{\beta}\mu}$
conjugate operations. Now (§~87) the sub-group~$h$ contains~$k\mu$ ($k \nless 1$)
operations whose orders are relatively prime to~$p$; and therefore
there are $k\mu$~operations of the form~$PQ$, where $Q$~is permutable with~$P$
and the order of~$Q$ is relatively prime to the order of~$P$. If $P'$~is
any operation conjugate to~$P$, there are similarly $k\mu$~operations of
the form~$P'Q'$; and (§~16) no one of these operations can be
identical with any one of those of the previous set. The group
therefore will contain $\dfrac{nmk}{p^{\beta}}$ distinct operations, which are conjugate
to the various operations of the set~$PQ$. Moreover since $P$~displaces
all the symbols, each one of these operations must displace all the
symbols.

If then the group has $r$~distinct sets of conjugate operations
whose orders are powers of~$p$, the number of operations, whose orders
are divisible by~$p$, is equal to
\[
\sum_{s=1}^{s=r} \frac{nmk_{s}}{p^{\beta_{s}}}.
\]

Also the number of operations, which displace all the symbols
and the orders of which are not divisible by~$p$, is of the form $\dfrac{nk_{0}}{p^{\alpha}} - 1$
(§~87).

Hence finally
\begin{align*}
m \sum_{s=1}^{s=r} \frac{k_{s}}{p^{\beta_{s}}}
  &= \frac{1}{n} \left\{n - 1 - \left(\frac{nk_{0}}{p^{\alpha}} - 1\right)\right\} \\
  &= \frac{p^{\alpha} - k_{0}}{p^{\alpha}}.
\end{align*}
%% -----File: 159.png---Folio 143-------
The greatest possible value of~$m$ will correspond to the suppositions
\[
r = 1,\quad k_{1} = 1,\quad k_{0} = 1;
\]
and these give
\[
m = p^{\alpha} - 1.
\]

Hence if $n = p^{\alpha} q^{\beta} \dots r^{\gamma}$, where $p$, $q$,~\dots, $r$ are distinct primes, $m$~cannot
be greater than the least of the numbers
\[
p^{\alpha} - 1,\quad q^{\beta} - 1,\quad \dots,\quad r^{\gamma} - 1.
\]

Certain particular cases may be specially noticed. First, a group
of degree~$n$ and order~$n(n - 1)$, whose operations other than
identity displace all or all but one of the symbols, can exist only
when $n$~is the power of a prime\footnotemark.
\index{Jordan@\Topic{Jordan}, quoted}%
\footnotetext{Jordan, ``Récherches sur les substitutions,'' \textit{Liouville's Journal}, \Ord{2}{me}~sér.\
  Vol.~\VolNo{XVII} (1872), p.~355.}%
Groups which satisfy these
conditions will be discussed in §~112.

Similarly, a group of degree~$n$ and order~$nm$, where $m$~is not less
than~$\sqrt{n}$, whose operations other than identity displace all or all but
one of the symbols, can only exist when $n$~is the power of a prime.

If $n$ is equal to twice an odd number, a transitive group of
degree~$n$, none of whose operations except identity leave two symbols
unchanged, must be of order~$n$.

Lastly we may shew that, if $m$~is even, a group of degree~$n$ and
order~$nm$, none of whose operations except identity leave two symbols
unchanged, must contain a self-conjugate Abelian sub-group of order
and degree~$n$.

A sub-group that keeps one symbol fixed must, if $m$~is even,
contain an operation of order~$2$. If it contained $r$~such operations,
the group would contain~$nr$; and each of these could be expressed
as the product of $\frac{1}{2}(n - 1)$ independent transpositions. Now from $n$~symbols
$\frac{1}{2}n(n - 1)$ transpositions can be formed. If then $r$ were
greater than~$1$, among the operations of order~$2$ that keep one
symbol fixed there would be pairs of operations with a common
transposition; and the product of two such operations would be an
operation, distinct from identity, which would keep two symbols at
least fixed. This is impossible; therefore $r$~must be unity. Now let
\[
A_{1},\ A_{2},\ \Dots,\ A_{n}
\]
be the $n$~operations of order~$2$ belonging to the group. Since no
two of these operations contain a common transposition,
\[
A_{1}A_{r},\ A_{2}A_{r},\ \Dots,\ A_{r - 1}A_{r},\ A_{r + 1}A_{r},\ \Dots, A_{n}A_{r}
\]
are the $n - 1$ operations which displace all the symbols. These
operations may also be expressed in the form
\[
A_{r}A_{1},\ A_{r}A_{2},\ \Dots,\ A_{r}A_{r - 1},\ A_{r}A_{r + 1},\ \Dots, A_{r}A_{n};
\]
and since
\[
A_{p}A_{r} · A_{r}A_{q} = A_{p}A_{q},
\]
%% -----File: 160.png---Folio 144-------
the product of any two of these operations is either identity or
another operation which displaces all the symbols. Hence the $n - 1$
operations which displace all the symbols, with identity, form a self-conjugate
sub-group. Now
\[
A_{r} · A_{p}A_{r} · A_{r} = A_{r}A_{p},
\]
so that $A_{r}$~transforms every operation of this sub-group into its
inverse. Hence
\[
A_{r}A_{p} · A_{q}A_{r} = A_{q}A_{p} = A_{q}A_{r} · A_{r}A_{q};
\]
\ie\ every two operations of this sub-group are permutable, and the
sub-group is therefore Abelian.
\end{Remark}

\Par{106.} If
\[
S = (a_{1}a_{2} \Dots a_{i})(a_{i + 1}a_{i + 2} \Dots a_{j})\Dots,
\]
and
\[
T = \Sub{a_{1}, a_{2}, \Dots, a_{n}}{b_{1}, b_{2}, \Dots, b_{n}},
\]
are any two substitutions of a group, then (§~10)
\[
T^{-1}ST = (b_{1}b_{2} \Dots b_{i})(b_{i + 1}b_{i + 2} \Dots b_{j})\Dots.
\]

Hence every substitution of the group, which is conjugate
\index{Self-conjugate operation@\Topic{Self-conjugate operation}!of a transitive substitution group must be a regular substitution}%
\index{Substitution group@\Topic{Substitution group}!conjugate substitutions of, are similar}%
to~$S$, is also similar to~$S$. It does not necessarily or generally
follow that two similar substitutions of a group are conjugate.
That this is true however of the symmetric group is obvious,
for then the substitution~$T$ may be chosen so as to replace the
$n$~symbols by any permutation of them whatever.

A self-conjugate substitution of a transitive group of degree~$n$
must be a regular substitution (§~9) changing all the $n$~symbols.
For if it did not change all the $n$~symbols, it would belong to
one of the sub-groups that keep a symbol unchanged. Hence,
since it is a self-conjugate substitution, it would belong to each
sub-group that keeps a symbol unchanged, which is impossible
unless it is the identical substitution. Again if it were not
regular, one of its powers would keep two or more symbols
unchanged, and this cannot be the case since every power of a
self-conjugate substitution must be self-conjugate. On the
other hand, a self-conjugate sub-group of a transitive group
need not contain any substitution which displaces all the
symbols. Thus if
\begin{align*}
S &= (12)(34), \\
T &= (135)(246),
\end{align*}
then $\{S, T\}$ is a transitive group of degree~$6$. The only
substitutions conjugate to~$S$ are
\[
T^{-1}ST = (34)(56) \quad \text{and} \quad TST^{-1} = (12)(56);
\]
\index{Substitution group@\Topic{Substitution group}!transitive, whose substitutions, except identity, displace all or all but one of the symbols|)}%
%% -----File: 161.png---Folio 145-------
and these, with $S$ and identity, form a self-conjugate sub-group
of order~$4$, none of whose substitutions displace more than $4$
symbols. The form of a self-conjugate sub-group of a transitive
group will be considered in greater detail in the next Chapter.

\Par{107.} Since a self-conjugate substitution of a transitive
\index{Substitutions@\Topic{Substitutions} which are permutable!with every substitution of a given group}%
substitution-group~$G$ of degree~$n$ must be a regular substitution
which displaces all the symbols, the self-conjugate sub-group~$H$
of~$G$ which consists of all its self-conjugate operations must
have $n$ or some submultiple of~$n$ for its order. For if $S$ and $S'$
are two self-conjugate substitutions of~$G$, so also is~$S^{-1}S'$; and
therefore $S$ and $S'$ cannot both change $a$ into~$b$. The order of~$H$
therefore cannot exceed~$n$; and if the order is~$n'$, the
substitutions of~$H$ must interchange the $n$~symbols in sets of~$n'$,
so that $n'$~is a factor of~$n$. Let now $S$, some substitution
performed on the $n$~symbols of the transitive substitution-group~$G$
of degree~$n$, be permutable with every substitution of~$G$.
Then $S$ is a self-conjugate operation of the transitive
substitution-group $\{S, G\}$ of degree~$n$, and it is therefore a
regular substitution in all the $n$~symbols. The totality of the
substitutions~$S$, which are permutable with every substitution
of~$G$, form a group (not necessarily Abelian); and the order of
this group is~$n$ or a factor of~$n$.

The special case, in which $G$ is a group whose order~$n$ is
equal to its degree, is of sufficient importance to merit particular
attention. If
\[
S_{1} (= 1),\ S_{2},\ \Dots,\ S_{n}
\]
are the operations of~$G$, it has been seen (§~20) that the
substitution-group can be expressed as consisting of the $n$~substitutions
\[
\Sub{\PadTo{S_{1} S_{x}}{S_{1}}, \PadTo{S_{2} S_{x}}{S_{2}}, \Dots, \PadTo{S_{n} S_{x}}{S_{n}}}
    {S_{1} S_{x}, S_{2} S_{x}, \Dots, S_{n}S_{x}},\quad (x = 1, 2, \Dots, n),
\]
performed upon the symbols of the operations of the group.

Now if pre-multiplication be used in the place of post-multiplication,
it may be verified exactly as in §~20 that the $n$~substitutions
\[
\Sub{\PadTo{S_{x} S_{1}}{S_{1}}, \PadTo{S_{x} S_{2}}{S_{2}}, \Dots, \PadTo{S_{x} S_{n}}{S_{n}}}
    {S_{x} S_{1}, S_{x} S_{2}, \Dots, S_{x} S_{n}},\quad (x = 1, 2, \Dots, n)
\]
%% -----File: 162.png---Folio 146-------
form a group~$G'$ simply isomorphic with~$G$; and the substitution
of~$G'$ which is given corresponds to the operation~$S_{x}^{-1}$ of~$G$.
Moreover
\begin{multline*}
  \Sub{\PadTo{S_{x} S_{1}}{S_{1}}, \Dots, \PadTo{S_{x} S_{n}}{S_{n}}}
      {S_{x} S_{1}, \Dots, S_{x} S_{n}}^{-1}
  \Sub{\PadTo{S_{1} S_{y}}{S_{1}}, \Dots, \PadTo{S_{n} S_{y}}{S_{n}}}
      {S_{1} S_{y}, \Dots, S_{n} S_{y}}
  \Sub{\PadTo{S_{x} S_{1}}{S_{1}}, \Dots, \PadTo{S_{x} S_{n}}{S_{n}}}
      {S_{x} S_{1}, \Dots, S_{x} S_{n}} \\
  \begin{aligned}
    &= \Sub{S_{x} S_{1}, \Dots, S_{x} S_{n}}
           {S_{1} S_{y}, \Dots, S_{1} S_{n}}
       \Sub{\PadTo{S_{x} S_{1} S_{y}}{S_{1} S_{y}}, \Dots,
            \PadTo{S_{x} S_{n} S_{y}}{S_{n} S_{y}}}
           {S_{x} S_{1} S_{y}, \Dots, S_{x} S_{n} S_{y}} \\
    &= \Sub{\PadTo{S_{x} S_{1} S_{y}}{S_{x} S_{1}}, \Dots,
            \PadTo{S_{x} S_{n} S_{y}}{S_{x} S_{n}}}
           {S_{x} S_{1} S_{y}, \Dots, S_{x} S_{n} S_{y}}
      = \Sub{\PadTo{S_{1} S_{y}}{S_{1}}, \Dots, \PadTo{S_{n} S_{y}}{S_{n}}}
            {S_{1} S_{y}, \Dots, S_{n} S_{y}},
  \end{aligned}
\end{multline*}
so that every operation of~$G'$ is permutable with every operation
of~$G$, while $G$~and $G'$ can be expressed as transitive
substitution-groups in the same $n$~symbols. Hence:---

\begin{Theorem}[II.]
Those substitutions of $n$~symbols which are
\index{Substitutions@\Topic{Substitutions} which are permutable!with every substitution of a given group}%
\index{Substitutions@\Topic{Substitutions} which are permutable!with every substitution of a group, whose degree is equal to its order, form a simply isomorphic group}%
permutable with every substitution of a substitution-group~$G$ of
order~$n$, transitive in the $n$~symbols, form a group~$G'$ of order
and degree~$n$, simply isomorphic with~$G$\footnotemark.
\end{Theorem}
\footnotetext{Jordan, \textit{Traité des Substitutions} (1870), p.~60.}%
\index{Jordan@\Topic{Jordan}, quoted}%

It has, in fact, been seen that there is a substitution-group~$G'$
of order and degree~$n$, every one of whose substitutions is
permutable with every substitution of~$G$; and also that the
order of any such group cannot exceed~$n$.

If $S_{x}$ is a self-conjugate substitution of~$G$, the substitutions
\[
\Sub{\PadTo{S_{1} S_{x}}{S_{1}}, \Dots, \PadTo{S_{n} S_{x}}{S_{n}}}
      {S_{1} S_{x}, \Dots, S_{n} S_{x}}
\quad \text{and} \quad
\Sub{\PadTo{S_{x} S_{1}}{S_{1}}, \Dots, \PadTo{S_{x} S_{n}}{S_{n}}}
      {S_{x} S_{1}, \Dots, S_{x} S_{n}}
\]
are the same. Hence $G$~and $G'$ have for their greatest common
sub-group, that which is constituted by the self-conjugate
substitutions of either; and if $n'$ is the order of this sub-group,
the order of $\{G, G'\}$ is~$\dfrac{n^{2}}{n'}$. In particular, if $G$ is Abelian, $G$ and
$G'$ coincide; and if $G$ has no self-conjugate operation except
identity, $\{G, G'\}$~is the direct product of $G$ and~$G'$.
\index{Direct product@\Topic{Direct product} of two simply isomorphic groups of order~$n$ represented as a transitive group of degree~$n$}%

The sub-group of $\{G, G'\}$ which leaves one symbol, say~$S_{1}$,
unchanged, is formed of the distinct substitutions of the set
\[
\Sub{\PadTo{S_{x}^{-1} S_{1} S_{x}}{S_{1}},
     \PadTo{S_{x}^{-1} S_{2} S_{x}}{S_{2}}, \Dots,
     \PadTo{S_{x}^{-1} S_{n} S_{x}}{S_{n}}}
    {S_{x}^{-1} S_{1} S_{x},
     S_{x}^{-1} S_{2} S_{x}, \Dots,
     S_{x}^{-1} S_{n} S_{x}},\quad (x = 1, 2, \Dots, n).
\]
%% -----File: 163.png---Folio 147-------

When $G$ has no self-conjugate operation except identity,
\index{Direct product@\Topic{Direct product} of two simply isomorphic groups of order~$n$ represented as a transitive group of degree~$n$}%
the order of this sub-group is~$n$, and it is simply isomorphic
with~$G$. In fact, in this case the order of $\{G, G'\}$, a transitive
group of degree~$n$, is~$n^{2}$, and therefore the order of a sub-group
that keeps one symbol unchanged is~$n$. Again
\begin{multline*}
\Sub{\PadTo{S_{x}^{-1} S_{1} S_{x}}{S_{1}},
     \PadTo{S_{x}^{-1} S_{2} S_{x}}{S_{2}}, \Dots,
     \PadTo{S_{x}^{-1} S_{n} S_{x}}{S_{n}}}
    {S_{x}^{-1} S_{1} S_{x},
     S_{x}^{-1} S_{2} S_{x}, \Dots,
     S_{x}^{-1} S_{n} S_{x}}
\Sub{\PadTo{S_{y}^{-1} S_{1} S_{y}}{S_{1}},
     \PadTo{S_{y}^{-1} S_{2} S_{y}}{S_{2}}, \Dots,
     \PadTo{S_{y}^{-1} S_{n} S_{y}}{S_{n}}}
    {S_{y}^{-1} S_{1} S_{y},
     S_{y}^{-1} S_{2} S_{y}, \Dots,
     S_{y}^{-1} S_{n} S_{y}} \\
%
\begin{aligned}
&= \Sub{\PadTo{S_{x}^{-1} S_{1} S_{x}}{S_{1}},
     \PadTo{S_{x}^{-1} S_{2} S_{x}}{S_{2}}, \Dots,
     \PadTo{S_{x}^{-1} S_{n} S_{x}}{S_{n}}}
    {S_{x}^{-1} S_{1} S_{x},
     S_{x}^{-1} S_{2} S_{x}, \Dots,
     S_{x}^{-1} S_{n} S_{x}} \\
&\qquad\qquad
  \Sub{\PadTo{S_{y}^{-1} S_{x}^{-1} S_{1} S_{x} S_{y}}{S_{x}^{-1} S_{1} S_{x}},
       \PadTo{S_{y}^{-1} S_{x}^{-1} S_{2} S_{x} S_{y}}{S_{x}^{-1} S_{2} S_{x}}, \Dots,
       \PadTo{S_{y}^{-1} S_{x}^{-1} S_{n} S_{x} S_{y}}{S_{x}^{-1} S_{n} S_{x}}}
      {S_{y}^{-1} S_{x}^{-1} S_{1} S_{x} S_{y},
       S_{y}^{-1} S_{x}^{-1} S_{2} S_{x} S_{y}, \Dots,
       S_{y}^{-1} S_{x}^{-1} S_{n} S_{x} S_{y}} \\
%
&= \Sub{\PadTo{S_{y}^{-1} S_{x}^{-1} S_{1} S_{x} S_{y}}{S_{1}},
       \PadTo{S_{y}^{-1} S_{x}^{-1} S_{2} S_{x} S_{y}}{S_{2}}, \Dots,
       \PadTo{S_{y}^{-1} S_{x}^{-1} S_{n} S_{x} S_{y}}{S_{n}}}
      {S_{y}^{-1} S_{x}^{-1} S_{1} S_{x} S_{y},
       S_{y}^{-1} S_{x}^{-1} S_{2} S_{x} S_{y}, \Dots,
       S_{y}^{-1} S_{x}^{-1} S_{n} S_{x} S_{y}};
\end{aligned}
\end{multline*}
thus giving a direct verification that the sub-group is isomorphic
with the group whose operations are
\[
S_{1},\ S_{2},\ \Dots,\ S_{n}.
\]

When $G$ contains self-conjugate operations, it will be
multiply isomorphic with the sub-group~$K_{1}$ of~$\{G, G'\}$ which
keeps the symbol $S_{1}$ fixed; and if $g$ is the group constituted by
the self-conjugate operations of~$G$ (or~$G'$), then $K_{1}$~is simply
isomorphic with~$\dfrac{G}{g}$.

If $K_{1}$ is not a maximum sub-group of~$\{G, G'\}$, let $I$ be a
greater sub-group containing~$K_{1}$. Then $I$ and $G'$ (or~$G$) must
contain common substitutions. For every substitution of
$\{G, G'\}$ is of the form
\[
\Sub{\PadTo{S_{y} S_{1} S_{x}}{S_{1}},
     \PadTo{S_{y} S_{2} S_{x}}{S_{2}}, \Dots,
     \PadTo{S_{y} S_{n} S_{x}}{S_{n}}}
    {S_{y} S_{1} S_{x}, S_{y} S_{2} S_{x}, \Dots, S_{y} S_{n} S_{x}};
\]
and if this substitution belongs to~$I$, then
\[
%[** TN: Set on two lines in the original]
\Sub{\PadTo{S_{y} S_{1} S_{x}}{S_{1}},
     \PadTo{S_{y} S_{2} S_{x}}{S_{2}}, \Dots,
     \PadTo{S_{y} S_{n} S_{x}}{S_{n}}}
    {S_{y} S_{1} S_{x}, S_{y} S_{2} S_{x}, \Dots, S_{y} S_{n} S_{x}}
\Sub{\PadTo{S_{x} S_{1} S_{x}^{-1}}{S_{1}},
     \PadTo{S_{x} S_{2} S_{x}^{-1}}{S_{2}}, \Dots,
     \PadTo{S_{x} S_{n} S_{x}^{-1}}{S_{n}}}
    {S_{x} S_{1} S_{x}^{-1}, S_{x} S_{2} S_{x}^{-1}, \Dots, S_{x} S_{n} S_{x}^{-1}},
\]
or
\[
\Sub{\PadTo{S_{x} S_{y} S_{1}}{S_{1}},
     \PadTo{S_{x} S_{y} S_{2}}{S_{2}}, \Dots,
     \PadTo{S_{x} S_{y} S_{n}}{S_{n}}}
    {S_{x} S_{y} S_{1}, S_{x} S_{y} S_{2}, \Dots, S_{x} S_{y} S_{n}},
\]
%% -----File: 164.png---Folio 148-------
is a substitution of~$G'$ which belongs to~$I$. Moreover, since $G'$
is a self-conjugate sub-group of~$\{G, G'\}$, the substitutions of~$G'$
which belong to~$I$ form a self-conjugate sub-group of~$I$: this
sub-group we will call~$H'$.

Now every substitution of the group can be represented as
the product of a substitution of~$K_{1}$ by a substitution of~$G$: and
therefore all the sub-groups conjugate to~$I$ will be obtained on
transforming $I$ by the operations of~$G$. Hence, because every
substitution of~$G$ transforms $H'$ into itself, $H'$~is common to
the complete set of conjugate sub-groups to which $I$ belongs;
and $H'$~is therefore a self-conjugate sub-group of~$\{G, G'\}$.
Finally then, $K_{1}$~is a maximum sub-group of~$\{G, G'\}$, if and only
if $G$~is a simple group.

\Par{108.} \begin{Definition}
A substitution-group, that contains one
\index{Substitution group@\Topic{Substitution group}!multiply transitive, definition of}%
or more substitutions changing $k$ given symbols $a_{1}$, $a_{2}$,~\dots, $a_{k}$
into any other $k$~symbols, is called \emph{$k$-ply transitive}.
\end{Definition}

Such a group clearly contains substitutions changing any
set of $k$~symbols into any other set of~$k$; and the order of the
sub-group keeping any~$j$ ($\ngtr k$) symbols unchanged is independent
of the particular $j$~symbols chosen.

\begin{Theorem}[III.]
The order of a $k$-ply transitive group of
\index{Substitution group@\Topic{Substitution group}!order of a $k$-ply transitive, whose degree is~$n$, is a multiple of $n(n - 1)\Dots (n - k + 1)$}%
degree~$n$ is $n(n - 1) \Dots (n - k + 1)m$, where $m$~is the order of
the sub-group that leaves any $k$~symbols unchanged. This
sub-group is contained self-conjugately in a sub-group of order~$k!m$.
\end{Theorem}

If $N$ is the order of the group, the order of the sub-group
which keeps one symbol fixed is~$\dfrac{N}{n}$, by Theorem~I (§~104).
Now this sub-group is a transitive group of degree~$n - 1$; and
therefore the order of the sub-group that keeps two symbols
unchanged is~$\dfrac{N}{n(n - 1)}$. If $k > 2$, this sub-group again is a
transitive group of degree~$n - 2$; and so on. Proceeding thus,
the order of the sub-group which keeps $k$~symbols unchanged
is seen to be
\[
\frac{N}{n(n - 1) \Dots (n - k + 1)},
\]
which proves the first part of the theorem.
%% -----File: 165.png---Folio 149-------

Let $a_{1}$, $a_{2}$,~\Dots, $a_{k}$ be the $k$~symbols which are left unchanged
by a sub-group~$H$ of order~$m$. Since the group is
$k$-ply transitive, it must contain substitutions of the form
\[
\Sub{a_{1}, a_{2}, \Dots, a_{k}, \PadTo[l]{b'}{b}, \PadTo[l]{c'}{c}, \Dots}
    {a_{1}', a_{2}', \Dots, a_{k}', b', c', \Dots},
\]
where $a_{1}'$, $a_{2}'$,~\Dots, $a_{k}'$ are the same $k$~symbols as $a_{1}$, $a_{2}$,~\Dots, $a_{k}$
arranged in any other sequence. Also every substitution of this
form is permutable with~$H$, since it interchanges among themselves
the symbols left unchanged by~$H$. Further, if $S_{1}$ and
$S_{2}$ are any two substitutions of this form, $S_{1}^{-1} S_{2}$~will belong to~$H$
if, and only if, $S_{1}$~and $S_{2}$ give the same permutation of the
symbols $a_{1}$, $a_{2}$,~\Dots, $a_{k}$. Hence finally, since $k!$~distinct
substitutions can be performed on the $k$~symbols, the order of
the sub-group that contains $H$ self-conjugately is~$k!m$.

If $m$ is unity, the identical substitution is the only one that
keeps any $k$~symbols fixed, and there is just one substitution
that changes $k$~symbols into any other~$k$. In the same case, the
group contains substitutions which displace $n - k + 1$ symbols
only, and there are none, except the identical substitution,
which displace less.

If $m > 1$, the group will contain $m - 1$ substitutions besides
identity, which leave unchanged any $k$ given symbols, and
therefore displace $n - k$ symbols at most.

It follows from §~105 that a $k$-ply transitive group of degree~$n$
and order $n(n - 1) \Dots (n - k + 1)$ can exist only if $n - k + 2$
is the power of a prime. For such a group must contain
sub-groups of order $(n - k + 2)(n - k + 1)$, which keep $k - 2$
symbols unchanged and are doubly transitive in the remaining
$n - k + 2$. When $k$~is~$n$, the group is the symmetric group; and
when $k$~is~$n - 2$, we shall see (in §~110) that the group is the
alternating group. If $k$ is less than~$n - 2$, M.~Jordan\footnote
  {``Récherches sur les substitutions,'' \textit{Liouville's Journal}, \Ord{2}{me}~sér., Vol.~\VolNo{XVII}
  (1872), pp.~357--363.}
\index{Jordan@\Topic{Jordan}, quoted}%
\PageLabel{149}%
has
shewn that, with two exceptions for $n = 11$ and $n = 12$, the
value of~$k$ cannot exceed~$3$. The actual existence of triply
transitive groups of order $(p^{n} + 1) p^{n} (p^{n} - 1)$, for all prime
values of~$p$, will be established in §~113.
%% -----File: 166.png---Folio 150-------

\Par{109.} A $k$-ply transitive group, of degree~$n$ and order~$N$,
\index{Substitution group@\Topic{Substitution group}!construction of multiply transitive}%
is not generally contained in some $(k + 1)$-ply transitive
group of degree~$n + 1$ and order~$N(n + 1)$. To determine
whether this is the case for any given group, M.~Jordan\footnote
  {\textit{Traité des Substitutions} (1870), pp.~31,~32.}
\index{Jordan@\Topic{Jordan}, quoted}%
has
suggested the following tentative process, which for moderate
values of~$n$ is always practicable.

Let $G$ be a transitive group of order~$N$ in the $n$~symbols
\[
a_{1},\ a_{2},\ \Dots,\ a_{n};
\]
and suppose that $G$~is that sub-group of the transitive group~$\Gamma$
in the $n + 1$ symbols
\[
a_{1},\ a_{2},\ \Dots,\ a_{n},\ a_{n + 1},
\]
which leaves the symbol $a_{n + 1}$ unchanged. Then $\Gamma$ must be at
least doubly transitive, and it therefore contains a substitution
of order~$2$ which interchanges the two symbols $a_{1}$~and~$a_{n + 1}$.
Let $A$ be such a substitution. Then
\[
\Gamma = \{G, A\};
\]
for $\{G, A\}$ is contained in~$\Gamma$, and its order cannot be less than
the order of~$\Gamma$. Also if $S$ is any substitution of~$G$\Erratum{}{ which displaces~$a_{1}$}, two other
substitutions $S'$ and $S''$ of~$G$ can always be found such that
\[
ASA = S'AS''.
\]

In fact, if $ASA$ changes $a_{r}$ into~$a_{n + 1}$, and if $S'$~changes $a_{r}$
into~$a_{1}$, then $AS'^{-1} ASA$ leaves $a_{n + 1}$ unchanged, and it therefore
belongs to~$G$.

Conversely, if $A$~is any operation of order~$2$ which changes
$a_{1}$ into $a_{n + 1}$ and permutes the remaining~$a$'s among themselves;
and if, whatever substitution of~$G$ is represented by~$S$,\Erratum{ either $ASA$~belongs to~$G$ or} two other
substitutions of~$G$ can be found such that
\[
ASA = S'AS'';
\]
then $\{G, A\}$ is a group with the required properties. In fact,
when these conditions are satisfied, every substitution of the
group $\{G, A\}$ can be expressed in one of the two forms
\[
S_{1},\ S_{2}AS_{3},
\]
%% -----File: 167.png---Folio 151-------
where $S_{1}$,~$S_{2}$,~$S_{3}$ are substitutions of~$G$. For instance, the
substitution
\begin{align*}
AS_{p}AS_{q}AS_{r}
  &= S_{p}'AS_{p}'' · S_{q}AS_{r}, \\
  &= S_{p}'AS_{t}AS_{r}, \quad\text{if}\quad S_{p}''S_{q} = S_{t}, \\
  &= S_{p}'S_{t}'AS_{t}''S_{r},
\end{align*}
which is of the second form.
\Erratum{}{The reduction is here carried out on supposition that $S_{p}$~and~$S_{t}$ displace~$a_{1}$.
The modification, when this is not the case, is obvious.}

Moreover every operation of the form $S_{2}AS_{3}$ displaces~$a_{n + 1}$,
and therefore the sub-group of~$\{G, A\}$ which leaves $a_{n + 1}$ unchanged
is~$G$.

It is clearly sufficient that the conditions
\[
ASA = S'AS''
\]
should be satisfied, when each of a set of independent generating
operations of~$G$ is taken for~$S$.

\begin{Ex} Construct a doubly transitive group of degree~$12$ of which
the sub-group that keeps one symbol unchanged is
\[
\{(123456789ab),\ (256a4)(39b87)\}.
\]
\end{Ex}

\Par{110.} Let
\[
S = (a_{1}a_{2} \Dots a_{i}) \Dots (\Dots a_{j - 1}a_{j})
    (a_{j + 1} \Dots a_{k - 1}a_{k} \Dots) \Dots
\]
be a substitution of a $k$-ply transitive group displacing $s$ ($> k$)
symbols. If $j < k - 1$, take
\[
T = \Sub{a_{1}, a_{2}, \Dots, a_{k - 1}, a_{k}, \Dots}
        {a_{1}, a_{2}, \Dots, a_{k - 1}, b_{k}, \Dots},
\]
where $b_{k}$~is some other symbol occurring in~$S$. Since the
group is $k$-ply transitive, it must contain a substitution such as~$T$.
Now
\[
T^{-1}ST = (a_{1}a_{2} \Dots a_{i}) \Dots (\Dots a_{j - 1}a_{j})
           (a_{j + 1} \Dots a_{k - 1}b_{k} \Dots) \Dots;
\]
and this is certainly not identical with~$S$, so that $T^{-1}STS^{-1}$
cannot be the identical substitution. Moreover $a_{1}$, $a_{2}$,~\Dots, $a_{k - 2}$
are not affected by $T^{-1}STS^{-1}$; and therefore this substitution
will displace at most $2s - 2k + 2$ symbols.

If $j = k - 1$, take
\[
T = \Sub{a_{1}, a_{2}, \Dots, a_{k - 1}, a_{k}, \Dots}
        {a_{1}, a_{2}, \Dots, a_{k - 1}, c_{k}, \Dots},
\]
where $c_{k}$ is a symbol that does not occur in~$S$. Then
\[
T^{-1}ST = (a_{1}a_{2} \Dots a_{i}) \Dots (\Dots a_{k - 2}a_{k - 1})(c_{k} \Dots),
\]
%% -----File: 168.png---Folio 152-------
and this cannot coincide with~$S$. Now in this case, $a_{1}$, $a_{2}$,~\Dots, $a_{k-1}$
are not affected by $T^{-1}STS^{-1}$; and therefore this operation
will again displace at most $2s - 2k + 2$ symbols.

If then
\[
2s - 2k + 2 < s,
\]
or
\[
s < 2k - 2,
\]
the group must contain a substitution affecting fewer symbols
than~$S$. This process may be repeated till we arrive at a
substitution
\[
\Sigma = (\alpha_{1}\alpha_{2} \Dots \alpha_{i})
         (\alpha_{i+1} \Dots \alpha_{j}) \Dots
         (\alpha_{j+1} \Dots \alpha_{k}),
\]
which affects exactly $k$~symbols; and if this substitution be
transformed by
\[
T = \Sub{\alpha_{1}, \alpha_{2}, \Dots, \alpha_{k-1}, \alpha_{k}, \Dots}
        {\alpha_{1}, \alpha_{2}, \Dots, \alpha_{k-1},  \beta_{k}, \Dots},
\]
then
\[
T^{-1} \Sigma T = (\alpha_{1}\alpha_{2} \dots \alpha_{i})
                  (\alpha_{i+1} \dots \alpha_{j}) \dots
                  (\alpha_{j+1} \dots a_{k-1} \beta_{k}),
\]
and
\[
\Sigma^{-1} T^{-1} \Sigma T = (\alpha_{k} \beta_{k} \alpha_{j+1}).
\]

Thus in the case under consideration the group contains
one, and therefore every, circular substitution of three symbols;
and hence (§~11) it must contain every even substitution. It
is therefore either the alternating or the symmetric group. If
then a $k$-ply transitive group of degree~$n$ does not contain the
alternating group of $n$~symbols, no one of its substitutions,
except identity, must displace fewer than $2k - 2$ symbols.
It has been shewn that such a group contains substitutions
displacing not more than $n - k + 1$ symbols; and therefore, for
a $k$-ply transitive group of degree~$n$, other than the alternating
or the symmetric group, the inequality
\[
n - k + 1  \nless 2k- 2,
\]
or
\PageLabel{152}%
\[
k \ngtr \tfrac{1}{3} n + 1,
\]
must hold. Hence:---

\begin{Theorem}[IV.]
A group of degree~$n$, which does not contain
\index{Substitution group@\Topic{Substitution group}!limit to the degree of transitivity of}%
the alternating group of $n$~symbols, cannot be more than
$(\frac{1}{3}n + 1)$-ply transitive.
\end{Theorem}

The symmetric group is $n$-ply transitive; and, since of the
two substitutions
\[
\Sub{a_{1}, a_{2}, \Dots, a_{n-2}, a_{n-1}, a_{n}}
    {b_{1}, b_{2}, \Dots, b_{n-2}, b_{n-1}, b_{n}}
\quad\text{and}\quad
\Sub{a_{1}, a_{2}, \Dots, a_{n-2}, a_{n-1}, a_{n}}
    {b_{1}, b_{2}, \Dots, b_{n-2}, b_{n}, b_{n-1}},
\]
%% -----File: 169.png---Folio 153-------
one is evidently even and the other odd, the alternating group
is $(n - 2)$-ply transitive. The discussion just given shews that
no other group of degree~$n$ can be more than $(\frac{1}{3}n + 1)$-ply
transitive\footnotemark.
\index{Bochert@\Topic{Bochert}, quoted}%
\index{Jordan@\Topic{Jordan}, quoted}%
\footnotetext{For a further discussion of the limits of transitivity of a substitution-group,
  compare Jordan, \textit{Traité des Substitutions}, pp.~76--87; and Bochert,
  \textit{Math.\ Ann.},\DPnote{** No "Vol."} \VolNo{xxix}, (1886) pp.~27--49; \VolNo{XXXIII}, (1888) pp.~573--583.}%

\Par{111.} The process used in the preceding paragraph may be
applied to shew that, unless $n = 4$, the alternating group of $n$~symbols
is simple. It has just been shewn that the alternating
group is $(n - 2)$-ply transitive. Therefore, if $S$ is a substitution
of the alternating group displacing fewer than $n - 1$~symbols, a
substitution $T^{-1}ST$ can certainly be found such that $S^{-1}T^{-1}ST$
is a circular substitution of three symbols. In this case, the
self-conjugate group generated by~$S$ and its conjugate substitutions
contains all the circular substitutions of three symbols,
and therefore it coincides with the alternating group itself. If
$S$~displaces $n - 1$ symbols, then $T^{-1}ST$ can be taken so that
$S^{-1}T^{-1}ST$ displaces not more than $2(n - 1) - 2(n - 2) + 2$, or
$4$~symbols; and if $S$~displaces $n$~symbols, $S^{-1}T^{-1}ST$~can be found
to displace not more than $2n - 2(n - 2) + 2$, or $6$~symbols.

It is therefore only necessary to consider the case $n = 5$, when
$S$~displaces $n - 1$ symbols; and the cases $n = 4$, $5$,~$6$, when $S$~displaces
$n$~symbols; in all other cases, the group generated by~$S$
and its conjugate substitutions must contain circular substitutions
of $3$~symbols.

When $n = 5$, and $S$ is an even substitution displacing $4$~symbols,
we may take
\[
S = (12) (34).
\]
If
\[
T = (12) (35),
\]
then
\[
T^{-1}ST = (12) (45),
\]
and
\[
S^{-1}T^{-1}ST = (345).
\]
Hence, in this case again, we are led to the alternating group
itself.
%% -----File: 170.png---Folio 154-------

When $n = 6$, and $S$~is an even substitution displacing all
the symbols, we may take
\[
S = (12) (3456),
\]
or
\[
S' = (123) (456).
\]
If now
\[
T = (12) (3645),
\]
then
\[
S^{-1}T^{-1}ST = (356),
\]
and
\[
S'^{-1}T^{-1}S'T = (14263);
\]
and, in either case, we are led to the alternating group.

When $n = 5$, and $S$~is an even substitution displacing all
the symbols, we may put
\[
S = (12345).
\]
If
\[
T = (345),
\]
then
\[
S^{-1}T^{-1}ST = (134);
\]
and again the alternating group is generated.

When $n = 4$, and $S$~is an even substitution displacing all the
symbols, we may take
\[
S = (12) (34).
\]
Here the only two substitutions conjugate to~$S$ are clearly
$(13) (24)$ and $(14) (23)$, which are permutable with each other
and with~$S$. Hence the alternating group of $4$~symbols, which
is of order~$12$, has a self-conjugate sub-group of order~$4$.

Finally when $n = 3$, the alternating group, being the group
$\{(123)\}$, is a simple cyclical group of order~$3$. Hence:---

\begin{Theorem}[V.]
The alternating group of $n$~symbols is a simple
\index{Alternating group@\Topic{Alternating group}!is simple, except for degree~$4$}%
group except when $n = 4$.
\end{Theorem}

\begin{Remark}
\Par{112.} It has been seen in §~108 that the order of a doubly transitive
group of degree~$n$ is equal to or is a multiple of~$n(n - 1)$. If it
is equal to this number, every substitution of the group, except
identity, must displace either all or all but one of the symbols; for
a sub-group of order~$n - 1$ which keeps one symbol fixed is transitive
in the remaining $n - 1$ symbols, and therefore all its substitutions,
except identity, displace all the $n - 1$ symbols.

Now it has been shewn in §~105 that a transitive group of degree~$n$
and order~$n(n - 1)$, whose operations displace all or all but one of
the symbols can exist only if $n$~is the power of a prime~$p$. The
$n - 1$ operations displacing all the symbols are the only operations
of the group whose orders are powers of~$p$; and therefore with
%% -----File: 171.png---Folio 155-------
\index{Doubly transitive groups@\Topic{Doubly transitive groups}!of degree~$n$ and order~$n(n - 1)$|(}%
\index{Substitution group@\Topic{Substitution group}!doubly transitive, of degree~$n$ and order~$n(n - 1)$|(}%
identity they form a self-conjugate sub-group of order~$n$. Moreover
it also follows from §~105 that the $n - 1$ operations of this sub-group
other than identity form a single conjugate set. Hence this sub-group
must be Abelian, and all its operations are of order~$p$.

Suppose first that $n$~is a prime~$p$, and that $P$~is any operation of
the group of order~$p$. If $\alpha$~is a primitive root of~$p$, there must be
an operation~$S$ in the group such that
\[
S^{-1}PS = P^{\alpha};
\]
then $S^{p-1}$ is the lowest power of~$S$ which is permutable with~$P$.
Now $S$~must belong to a sub-group of order~$p - 1$ that keeps one
symbol fixed, and we have just seen that the order of~$\{S\}$ is not less
than~$p - 1$. The sub-group of order~$p - 1$ is therefore cyclical, and
\[
S^{p-1} = 1.
\]
Hence the group, if it exists, must be defined by
\[
P^{p} = 1,\quad S^{p-1} = 1,\quad S^{-1}PS = P^{\alpha}.
\]

It is an immediate result of a theorem, which will be proved in the
next chapter~(§~123), that this group can be actually represented as
a transitive substitution-group of degree~$p$; this may be also verified
directly as follows.

Let
\[
P = (a_{1}a_{2} \dots a_{p}),
\]
so that
\[
P^{\alpha} = (a_{1}a_{\alpha + 1}a_{2\alpha + 1} \dots a_{(p - 1)\alpha + 1}),
\]
where the suffixes are to be reduced $(\mod p)$; and suppose that $S$~is
a substitution that keeps unchanged. Then since
\[
S^{-1}PS = P^{\alpha},
\]
$S$ must change $a_{2}$ into~$a_{\alpha + 1}$, $a_{3}$ into~$a_{2 \alpha + 1}$, and generally, $a_{r}$ into
$a_{(r - 1)\alpha + 1}$. Hence
\[
S = (a_{2} a_{\alpha + 1} a_{\alpha^{2} + 1} \dots) \dots;
\]
and since $\alpha$~is a primitive root of~$p$, there is only a single cycle; so
that
\[
S = (a_{2} a_{\alpha + 1} a_{\alpha^{2} + 1} \dots a_{\alpha^{p - 2} + 1}).
\]

The substitutions $P$ and~$S$ thus constructed actually generate a
doubly transitive substitution-group of degree~$p$ and order~$p(p - 1)$.

Without making a complete investigation of the case in which $n$~is
the power of a prime, we go on to shew that, $p$~being any prime,
there is always a doubly transitive group of degree~$p^{m}$ and order
$p^{m}(p^{m} - 1)$, in which a sub-group of order~$p^{m} - 1$ is cyclical\footnotemark.
\index{Mathieu@\Topic{Mathieu}, quoted}%
\footnotetext{On the subject of this and the following paragraph, the reader  should
\PageLabel{155}%
  consult the memoirs by Mathieu, \textit{Liouville's Journal}, \Ord{2}{me} Sér., t.~\VolNo{V} (1860),
  pp.~9--42; \textit{ib.}\ t.~\VolNo{VI} (1861), pp.~241--323; where the groups here considered were
  first shewn to exist.}
%% -----File: 172.png---Folio 156-------

Let $i$ be a primitive root of the congruence
\[
i^{p^{m} - 1} \equiv 1\ (\mod p),
\]
so that the distinct roots of the congruence are
\[
i,\ i^{2},\ i^{3},\ \Dots,\ i^{p^{m} - 1}.
\]
Every rational function of~$i$ with real integral coefficients satisfies
the same congruence; and therefore every such function is congruent
$(\mod p)$ to some power of~$i$ not exceeding the $(p^{m} - 1)$th.

Consider now a set of transformations of the form
\[
x' \equiv \alpha x + \beta\ (\mod p),
\]
where $\alpha$ is a power of~$i$, and $\beta$~is either a power of~$i$ or zero. Two
such transformations, performed successively, give another transformation
of the same form; and since $\alpha$~cannot be zero, the inverse
of each transformation is another definite transformation; so that
the totality of transformations of this form constitute a group.
Moreover
\[
\begin{aligned}
x' &\equiv \alpha x + \beta, \\
\lintertext{and}
x' &\equiv \alpha' x + \beta'
\end{aligned}
\ (\mod p),
\]
are not the same transformation unless
\[
\alpha \equiv \alpha' \quad\text{and}\quad \beta \equiv \beta'\ (\mod p).
\]
{\Loosen Hence, since $\alpha$ can take $p^{m} - 1$ distinct values and $\beta$ can take $p^{m}$
distinct values, the order of the group, formed of the totality of
these transformations, is $p^{m}(p^{m} - 1)$.}

The transformations for which $\alpha$ is unity clearly form a sub-group.
If $S$~and~$T$ represent
\[
x' \equiv \alpha x + \beta \quad\text{and}\quad x' \equiv x + \gamma
\]
respectively, $S^{-1}TS$ represents
\[
x' \equiv x + \alpha \gamma.
\]
Hence the transformations for which $\alpha$ is unity form a self-conjugate
sub-group whose order is~$p^{m}$. Every two transformations of this
sub-group are clearly permutable; and the order of each of them
except identity is~$p$.

Again, the transformations for which $\beta$~is zero form a sub-group.
Since every one of them is a power of the transformation
\[
x' \equiv ix,
\]
this sub-group is a cyclical sub-group of order~$p^{m} - 1$. If the transformation
just written be denoted by~$I$, then $S^{-1}IS$ is
\[
x' \equiv ix + \beta (1 - i).
\]
Hence the only operations permutable with~$\{I\}$ are its own operations,
and therefore $\{I\}$~is one of $p^{m}$ conjugate sub-groups.
%% -----File: 173.png---Folio 157-------

The set of transformations
\[
x' \equiv \alpha x + \beta
\]
therefore forms a group of order~$p^{m}(p^{m} - 1)$. This group contains a
self-conjugate Abelian sub-group of order~$p^{m}$ and type $(1, 1, \dots, 1)$,
and $p^{m}$~conjugate cyclical sub-groups of order~$p^{m} - 1$, none of whose
operations are permutable with any of the operations of the self-conjugate
sub-group.

Now if the operation
\[
x' \equiv \alpha x + \beta
\]
be performed on each term of the series
\[
0,\ i,\ i^{2},\ \Dots,\ i^{p^{m}-1},
\]
it will, since every rational integral function of~$i$ with real integral
coefficients is congruent $(\mod p)$ to some power of~$i$, change the
term into another of the same series; and since the congruence
\[
\alpha i^{x} + \beta \equiv \alpha i^{y} + \beta
\]
gives
\[
x \equiv y\ (\mod p^{m-1}),
\]
no two terms of the series can thus be transformed into the same
term. Moreover the only operation that leaves every term of the
series unchanged is clearly the identical operation.

To each operation of the form
\[
x' \equiv \alpha x + \beta
\]
therefore will correspond a single substitution performed on the $p^{m}$
symbols just written, so that to the product of two operations will
correspond the product of the two homologous substitutions. The
group is therefore simply isomorphic with a substitution group of
degree~$p^{m}$. Moreover since the linear congruence
\[
x \equiv \alpha x + \beta\ (\mod p)
\]
has only a single solution when $\alpha$ is different from unity, and none
when $\alpha$~is unity, every substitution except identity must displace all
or all but one of the symbols. The substitution group is therefore
doubly transitive\footnotemark.
\index{Burnside@\Topic{Burnside}, quoted}%
\footnotetext{The author has shewn (\textit{Messenger of Mathematics}, Vol.~\VolNo{XXV} (1896), pp.~147--153)
  that the type of group considered in the text is the only type of doubly
  transitive group of degree~$p^{m}$ and order~$p^{m}(p^{m} - 1)$ when $m = 3$; and that, when
  $m = 2$ and $p > 3$, the same is true. When $m = 2$ and $p = 3$, there is one other type.}
\end{Remark}

\begin{Ex}[1.]
Apply the method just explained to the actual construction
of a doubly transitive group of degree~$8$ and order~$56$.
\end{Ex}

\begin{Ex}[2.] Shew that the equations
\[
A^{2} = 1,\quad S^{2^{m}-1} = 1,\quad AS^{-1}AS = S^{-n}AS^{n},
\]
where $n$~is such that a primitive root of the congruence,
\[
i^{2^{m}-1} - 1 \equiv 0\ (\mod 2),
\]
\index{Doubly transitive groups@\Topic{Doubly transitive groups}!of degree~$n$ and order~$n(n - 1)$|)}%
\index{Substitution group@\Topic{Substitution group}!doubly transitive, of degree~$n$ and order~$n(n - 1)$|)}%
%% -----File: 174.png---Folio 158-------
satisfies the congruence
\[
i^{n} + i + 1 \equiv 0\ (\mod 2),
\]
suffice to define a group which can be expressed as a doubly transitive
group of degree~$2^{m}$ and order~$2^{m}(2^{m} - 1)$.
\Attrib{\textit{Messenger of Mathematics}, Vol.~\VolNo{XXV}, p.~189.}
\end{Ex}

\begin{Remark}
\Par{113.} A slight extension of the method of the preceding paragraph
\index{Substitution group@\Topic{Substitution group}!triply transitive, of degree~$n$ and order~$n(n - 1)(n - 2)$}%
will enable us to shew that, for every prime~$p$, it is possible to
construct a triply transitive group of degree~$p^{m} + 1$ and order
\index{Triply transitive@\Topic{Triply transitive} groups of degree~$n$ and order $n(n - 1)(n - 2)$}%
$(p^{m} + 1)p^{m}(p^{m} - 1)$. The analysis of this group will form the
subject of investigation in Chapter~XIV; here we shall only demonstrate
its existence.

In the place of the operations of the last paragraph, we now
consider those of the form
\[
x' \equiv \frac{\alpha x + \beta}{\gamma x + \delta}\ (\mod p),
\]
where again $\alpha$, $\beta$, $\gamma$, $\delta$ are powers of~$i$, limited now by the condition
that $\alpha\delta - \beta\gamma$ is not congruent to zero $(\mod p)$. When this relation
is satisfied, the set of operations again clearly form a group. Moreover
if we represent $\dfrac{i^{x}}{0}$ by~$\infty$ for all values of~$x$, any operation of this
group, when carried out on the set of quantities
\[
\infty,\ 0,\ i,\ i^{2},\ \Dots,\ i^{p^{m} - 1},
\]
will change each of them into another of the set; while no operation
except identity will leave each symbol of the set unchanged. Hence
the group can be represented as a substitution group of degree
$p^{m} + 1$.

Now
\[
\frac{x' - i^{a'}}{x'- i^{b'}}\, \frac{i^{c'} - i^{b'}}{i^{c'} - i^{a'}} \equiv
\frac{x - i^{a}}{x - i^{b}}\, \frac{i^{c} - i^{b}}{i^{c} - i^{a}}
\]
is an operation of the above form, which changes the three symbols
$i^{a}$,~$i^{b}$,~$i^{c}$ into $i^{a'}$,~$i^{b'}$,~$i^{c'}$ respectively; and it is easy to modify this form
so that it holds when $0$~or~$\infty$ occurs in the place of~$i^{a}$, etc. Hence
the substitution group is triply transitive, since it contains an
operation transforming any three of the $p^{m} + 1$ symbols into any
other three.

On the other hand, if the typical operation keeps the symbol~$x$
unchanged, then
\[
\gamma x^{2} + (\delta - \alpha) x - \beta \equiv 0\ (\mod p),
\]
and this congruence cannot have more than two roots among the
set of $p^{m} + 1$ symbols. Hence no substitution of the group, except
identity, keeps more than two symbols fixed.

Finally then, since the group is triply transitive and since it
contains no operation, except identity, that keeps more than two
symbols fixed, its order must (§~108) be $(p^{m} + 1) p^{m} (p^{m} - 1)$.
\end{Remark}
%% -----File: 175.png---Folio 159-------

\Par{114.} An intransitive substitution group, as defined in §~104,
is one which does not contain substitutions changing $a_{1}$ into
each of the other symbols $a_{2}$, $a_{3}$,~\Dots, $a_{n}$ operated on by the
group. Let us suppose that the substitutions of such a group
change $a_{1}$ into $a_{2}$, $a_{3}$,~\Dots, $a_{k}$ only. Then all the substitutions
of the group must interchange these $k$~symbols among themselves;
for if the group contained a substitution changing $a_{2}$
into~$a_{k + 1}$, then the product of any substitution changing $a_{1}$ into~$a_{2}$
by this latter substitution would change $a_{1}$ into~$a_{k + 1}$. Hence
the $n$~symbols operated on by the group can be divided into a
number of sets, such that the substitutions of the group change
the symbols of any one set transitively among themselves, but
do not interchange the symbols of two distinct sets. It follows
immediately that the order of the group must be a common
multiple of the numbers of symbols in the different sets.

Suppose now that $a_{1}$, $a_{2}$,~\Dots, $a_{k}$ is a set of symbols which
are interchanged transitively by all the substitutions of a group~$G$
of degree~$n$. If for a time we neglect the effect of the
substitutions of~$G$ on the remaining $n - k$ symbols, the group~$G$
will reduce to a transitive group~$H$ of degree~$k$. The group~$G$
is isomorphic with the group~$H$; for if we take as the substitutions
of~$G$, that correspond to a given substitution of~$H$, those
which produce the same permutation of the symbols $a_{1}$, $a_{2}$,~\Dots, $a_{k}$
as that produced by the substitution of~$H$, then to the product
of any two substitutions of~$G$ will correspond the product of
the corresponding two substitutions of~$H$. The isomorphism
thus shewn to exist may be simple or multiple. In the former
case, the order of~$H$ is the same as that of~$G$; in the
latter case, the substitutions of~$G$ which correspond to the
identical substitution of~$H$, \ie\ those substitutions of~$G$ which
change none of the symbols $a_{1}$, $a_{2}$,~\Dots, $a_{k}$ form a self-conjugate
sub-group.

We will consider in particular an intransitive group~$G$
\index{Intransitive groups@\Topic{Intransitive groups}!properties of|(}%
\index{Intransitive groups@\Topic{Intransitive groups}!transitive sets of symbols in}%
which interchanges the symbols in \emph{two} transitive sets; these
we will refer to as the $\alpha$'s and the~$\beta$'s. Let $G_{\alpha}$~and $G_{\beta}$ be the
two groups transitive in the $\alpha$'s and $\beta$'s respectively, to which
$G$~reduces when we alternately leave out of account the effect
of the substitutions on the $\beta$'s and the~$\alpha$'s. Also let $g_{\alpha}$ and~$g_{\beta}$
%% -----File: 176.png---Folio 160-------
be the self-conjugate sub-groups of~$G$, which keep respectively
all the $\beta$'s and all the $\alpha$'s unchanged; and denote the group
$\{g_{\alpha}, g_{\beta}\}$ by~$g$. This last group~$g$, which is the direct product
of $g_{\alpha}$ and~$g_{\beta}$, is self-conjugate in~$G$, since it is generated by the
two self-conjugate groups $g_{\alpha}$ and~$g_{\beta}$. Now $g_{\alpha}$~is self-conjugate
not only in~$G$ but also in~$G_{\alpha}$; for $G_{\alpha}$~permutes the~$\alpha$'s in the
same way that $G$ does, while any substitution of~$g_{\alpha}$, not
affecting the~$\beta$'s, is necessarily permutable with every substitution
performed on the~$\beta$'s. The group~$G_{\alpha}$ is simply
isomorphic with the group~$\dfrac{G}{g_{\beta}}$, and $G_{\beta}$ with~$\dfrac{G}{g_{\alpha}}$; hence, using $n_{H}$
to denote the order of a group~$H$,
\[
n_{G} = n_{G_{\alpha}} n_{g_{\beta}} = n_{G_{\beta}} n_{g_{\alpha}}.
\]
Let the substitutions of~$G$ be now divided into sets in respect
of the self-conjugate sub-group~$g$, so that
\[
G = g,\ Sg,\ Tg,\ \Dots.
\]

The group $\dfrac{G}{g}$ is defined by the laws according to which
these sets combine among themselves, the sets being such that,
if any substitution of one set be multiplied by any substitution
of a second set, the resulting substitution will belong to a
definite third set.

If we now neglect the effect of the substitutions on the
symbols~$\beta$, the group~$G$ reduces to~$G_{\alpha}$ and $g$~reduces to~$g_{\alpha}$, and
hence
\[
G_{\alpha} = g_{\alpha},\ S_{\alpha} g_{\alpha},\ T_{\alpha} g_{\alpha},\ \Dots,
\]
where $S_{\alpha}$, $T_{\alpha}$,~\Dots\ represent the substitutions $S$, $T$,~\Dots, so far
as they affect the~$\alpha$'s. Moreover the substitutions in the different
sets into which $G_{\alpha}$ is thus divided must be all distinct since,
by the preceding relations between the orders of the groups,
their number is just equal to the order of~$G_{\alpha}$. Hence $\dfrac{G_{\alpha}}{g_{\alpha}}$ is
defined by the laws according to which these sets of substitutions
combine. But if
\[
Sg · Tg = Ug,
\]
then necessarily
\[
S_{\alpha} g_{\alpha} · T_{\alpha} g_{\alpha} = U_{\alpha} g_{\alpha},
\]
%% -----File: 177.png---Folio 161-------
and therefore, finally, the three groups $\dfrac{G}{g}$,~$\dfrac{G_{\alpha}}{g_{\alpha}}$, and~$\dfrac{G_{\beta}}{g_{\beta}}$ are simply
isomorphic.

\Par{115.} The relation of simple isomorphism between $\dfrac{G_{\alpha}}{g_{\alpha}}$ and $\dfrac{G_{\beta}}{g_{\beta}}$
thus arrived at establishes between the groups $G_{\alpha}$ and $G_{\beta}$ an
isomorphism of the most general kind (§~32).

To every operation of~$G_{\alpha}$ correspond $n_{g_{\beta}}$~operations of~$G_{\beta}$,
and to every operation of~$G_{\beta}$ correspond $n_{g_{\alpha}}$~operations
of~$G_{\alpha}$; so that to the product of any two operations of~$G_{\alpha}$ (or~$G_{\beta}$)
there corresponds a definite set of $n_{g_{\beta}}$~operations of~$G_{\beta}$ (or~$n_{g_{\alpha}}$
operations of~$G_{\alpha}$).

Returning now to the intransitive group~$G$, its genesis from
the two transitive groups $G_{\alpha}$ and~$G_{\beta}$, with which it is isomorphic,
may be represented as follows. The $n_{g_{\alpha}}$ to $n_{g_{\beta}}$ correspondence,
such as has just been described, having been established
between the groups $G_{\alpha}$ and~$G_{\beta}$, each substitution of~$G_{\alpha}$ is
multiplied by the $n_{g_{\beta}}$~substitutions that correspond to it in~$G_{\beta}$.
The set of $n_{G_{\alpha}}n_{g_{\beta}}$ substitutions so obtained form a group, for
\[
S_{\alpha} S_{\beta} · S_{\alpha}' S_{\beta}'
  = S_{\alpha} S_{\alpha}' · S_{\beta} S_{\beta}'
  = S_{\alpha}'' S_{\beta}'',
\]
where, if $S_{\beta}$,~$S_{\beta}'$ are substitutions corresponding to $S_{\alpha}$,~$S_{\alpha}'$, then
$S_{\beta}''$~is a substitution corresponding to~$S_{\alpha}''$. Moreover, this
group may be equally well generated by multiplying every one
of the substitutions of~$G_{\beta}$ by the $n_{g_{\alpha}}$~corresponding substitutions
of~$G_{\alpha}$; and by a reference to the representations of $G$,~$G_{\alpha}$ and~$G_{\beta}$,
as divided into sets of substitutions given above, it is immediately
obvious that all these substitutions occur in~$G$. Hence,
as their number is equal to the order of~$G$, the group thus
formed coincides with~$G$.

\Par{116.} The general result for any intransitive group, the
simplest case of which has been considered in the two last
paragraphs, may be stated in the following form:---

\begin{Theorem}[VI.]
If $G$ is an intransitive group of degree~$n$
which permutes the $n$~symbols in $s$~transitive sets, and if \Inum{(i)}~$G_{r}$~is
what $G$~becomes when the substitutions of~$G$ are performed on
%% -----File: 178.png---Folio 162-------
the $r$th~set of symbols only, \Inum{(ii)}~$\Gamma_{r}$~is what $G$ becomes when the
substitutions of~$G$ are performed on all the sets except the~$r$th,
\Inum{(iii)}~$g_{r}$~is that sub-group of~$G$ which changes the symbols of the
$r$th~set only, \Inum{(iv)}~$\gamma_{r}$~is that sub-group of~$G$ which keeps all the
symbols of the $r$th~set unchanged: then the groups $\dfrac{G_{r}}{g_{r}}$ and $\dfrac{\Gamma_{r}}{\gamma_{r}}$
are simply isomorphic, and $n_{r}$,~$\nu_{r}$ being the orders of $g_{r}$,~$\gamma_{r}$, an
$n_{r}$~to~$\nu_{r}$ correspondence is thus established between the substitutions
of the groups $G_{r}$ and~$\Gamma_{r}$. Moreover, the substitutions of~$G$
are given, each once and once only, by multiplying each
substitution of~$G_{r}$ by the $\nu_{r}$~substitutions of~$\Gamma_{r}$ that correspond to
it\footnotemark.
\end{Theorem}
\index{Bolza@\Topic{Bolza}, quoted}%
\index{Klein and Fricke@\Topic{Klein} and \Topic{Fricke}, quoted}%
\footnotetext{On intransitive groups, reference may be made to Bolza, ``On the
  construction of intransitive groups,'' \textit{Amer.\ Journal}, Vol.~\VolNo{XI}, (1889), pp.~195--214.
  The general isomorphism which underlies the construction of these
  groups is considered by Klein and Fricke, \textit{Vorlesungen über die Theorie der
  elliptischen Modulfunctionen}, Vol.~\VolNo{I}, (1890), pp.~402--406.}

It is not necessary to give an independent proof of this
theorem, since if, in the discussion of the two preceding paragraphs,
$G_{\alpha}$,~$G_{\beta}$, $g_{a}$,~$g_{b}$,~$g$ be replaced by $G_{r}$,~$\Gamma_{r}$, $g_{r}$,~$\gamma_{r}$, $\{g_{r}, \gamma_{r}\}$, it
will be found each step of the process there carried out may
be repeated without alteration.

If we regard $G_{\alpha}$ and $G_{\beta}$ as two given transitive groups
in distinct sets of symbols, the determination of all the
intransitive groups in the combined symbols, which reduce
to $G_{\alpha}$ or $G_{\beta}$ when the symbols of the second or first set
are neglected, involves a knowledge of the composition of
the two groups. To each distinct $m$~to~$n$ isomorphism, that
can be established between the two groups, there will correspond
a distinct intransitive group. If $G_{\alpha}$ is a simple group,
containing therefore only itself and identity self-conjugately,
then to each substitution of~$G_{\beta}$ there must correspond either
one or all of the substitutions of~$G_{\alpha}$; and the former can be
the case only when $G_{\beta}$~contains a self-conjugate sub-group~$H$,
such that $\dfrac{G_{\beta}}{H}$~is simply isomorphic with~$G_{\alpha}$. Hence, if the
order of~$G_{\beta}$ is less than twice the order of~$G_{\alpha}$, the only possible
intransitive group is the direct product of $G_{\alpha}$ and~$G_{\beta}$, unless $G_{\beta}$~is
simply isomorphic with~$G_{\alpha}$.
\index{Intransitive groups@\Topic{Intransitive groups}!properties of|)}%
%% -----File: 179.png---Folio 163-------

\begin{Remark}
\Par{117.} In illustration of the preceding paragraphs, we will
\index{Intransitive groups@\Topic{Intransitive groups}!of degree~$7$ with transitive sets of $3$ and~$4$}%
determine the number of distinct intransitive groups of degree~$7$,
when the symbols are interchanged in transitive sets of $4$ and $3$
respectively. The four symbols will be referred to as the~$\alpha$'s, and
the three symbols as the~$\beta$'s.

If a transitive group of degree~$4$ contains operations of order~$3$,
it must be either the symmetric or the alternating group. If it
contains no operation of order~$3$, its order must be either $8$ or~$4$.
By Sylow's theorem, the symmetric group of degree~$4$, being of
order~$24$, contains a single set of conjugate sub-groups of order~$8$, so
that there is only one type of transitive group of degree~$4$ and order~$8$.
This is given by
\[
S = (1234),\quad T=(13),
\]
so that
\[
S^{4} = 1,\quad T^{2} = 1,\quad TST = S^{-1}.
\]

This group contains three self-conjugate sub-groups of order~$4$,
namely,
\[
\begin{array}{l*{4}{c}}
\Inum{(i)}   & 1, & (1234),   & (13)(24), & (1432), \\
\Inum{(ii)}  & 1, & (12)(34), & (13)(24), & (14)(23), \\
\Inum{(iii)} & 1, & (13),     & (13)(24), & (24).
\end{array}
\]

The two latter are simply isomorphic; but as substitution groups,
they are of distinct form. Hence for $G_{\alpha}$ in the construction of the
intransitive group, we may take either $G_{24}$, the symmetric group of
the~$\alpha$'s, or $G_{12}$, the alternating group of the~$\alpha$'s, or $G_{8}$, the above
group of order~$8$, or finally $G_{4}$,~$G_{4}'$,~$G_{4}''$, the above three groups of
order~$4$. The only transitive groups of $3$~symbols are $G_{6}$, the symmetric
group, and $G_{3}$, the alternating group: one of these must be
taken for~$G_{\beta}$.

\Class{\Inum{I.}}{$G_{\alpha} = G_{24}$.}

The only self-conjugate sub-groups of~$G_{24}$ are $G_{12}$ and~$G_{4}'$.
Also $\dfrac{G_{24}}{G_{12}}$ is a group of order~$2$; and it may easily be verified
that $\dfrac{G_{24}}{G_{4}'}$ is simply isomorphic with the symmetric group of degree~$3$.

Hence \Inum{(i)}~if $G_{\beta}$ is~$G_{6}$, we may take
\begin{alignat*}{4}
g_{\alpha} &= G_{24},\quad & g_{\beta} &= G_{6}, &&\text{ so that } & n_{G} &= 144, \\
g_{\alpha} &= G_{12},      & g_{\beta} &= G_{3}, &&\Ditto{ so that } & n_{G} &= 72, \\
g_{\alpha} &= G_{4}',      & g_{\beta} &= 1,     &&\Ditto{ so that } & n_{G} &= 24; \\
\intertext{and \Inum{(ii)}~if $G_{\beta}$ is~$G_{3}$, we must take}
g_{\alpha} &= G_{24},      & g_{\beta} &= G_{3}, &&\text{ giving } & n_{G} &= 72.
\end{alignat*}
%% -----File: 180.png---Folio 164-------
\index{Intransitive groups@\Topic{Intransitive groups}!of degree~$7$ with transitive sets of $3$ and~$4$}%

\Class{\Inum{II.}}{$G_{\alpha} = G_{12}$.}

The only self-conjugate sub-group of~$G_{12}$ is~$G_{4}'$; and $\dfrac{G_{12}}{G_{4}'}$~is a
cyclical group of order~$3$.

Hence \Inum{(i)}~if $G_{\beta}$ is~$G_{6}$, we must take
\begin{alignat*}{4}
g_{\alpha} &= G_{12},\quad & g_{\beta} &= G_{6}, &&\text{ giving } & n_{G} &= 72; \\
\intertext{and \Inum{(ii)}~if $G_{\beta}$ is~$G_{3}$, we may take}
g_{\alpha} &= G_{12},      & g_{\beta} &= G_{3}, &&\text{ giving } & n_{G} &= 36, \\
\lintertext{or}
g_{\alpha} &= G_{4}',      & g_{\beta} &= 1,  &&\Ditto{ giving }  & n_{G} &= 12.
\end{alignat*}

\Class{\Inum{III.}}{$G_{\alpha} = G_{8}$.}

The self-conjugate sub-groups of~$G_{8}$, of order~$4$, are determined
above.

If \Inum{(i)}~$G_{\beta}$ is~$G_{6}$, we may take
\begin{alignat*}{3}
g_{\alpha} &= G_{8},   & g_{\beta} &= G_{6}, & \text{ giving }  & n_{G} = 48, \\
g_{\alpha} &= G_{4},   & g_{\beta} &= G_{3}, & \Ditto{ giving } & n_{G} = 24, \\
g_{\alpha} &= G_{4}',  & g_{\beta} &= G_{3}, & \Ditto{ giving } & n_{G} = 24, \\
g_{\alpha} &= G_{4}'', \quad & g_{\beta} &= G_{3}, & \Ditto{ giving } & n_{G} = 24.
\end{alignat*}

The two latter groups are simply isomorphic; but regarded as
substitution groups, they are of distinct forms.

If \Inum{(ii)}~$G_{\beta}$ is~$G_{3}$, we must take
\[
g_{\alpha} = G_{8},\quad g_{\beta} = G_{3}, \text{ giving } n_{G} = 24.
\]

\Class{\Inum{IV.}}{$G_{\alpha} = G_{4}$.}

If \Inum{(i)}~$G_{\beta}$ is~$G_{6}$, we may take
\[
g_{\alpha} = G_{4},\quad g_{\beta} = G_{6}, \text{ giving } n_{G} = 24,
\]
or
\[
g_{\alpha} = G_{2},\quad g_{\beta} = G_{3}, \Ditto{ giving } n_{G} = 12;
\]
$G_{2}$ representing the group $\{1, (13)(24)\}$.

If \Inum{(ii)}~$G_{\beta}$ is~$G_{3}$, we must take
\[
g_{\alpha} = G_{4},\quad g_{\beta} = G_{3}, \text{ giving } n_{G} = 12.
\]

\Class{\Inum{V.}}{$G_{\alpha} = G_{4}'$.}

There are, exactly as in the last case, three possibilities: $G_{4}'$~taking
the place of~$G_{4}$. These groups are not however simply isomorphic
with the preceding three.

\Class{\Inum{VI.}}{$G_{\alpha} = G_{4}''$.}

There are in this case four possibilities. Of these three correspond
to those of~\Inum{IV}, $G_{4}''$~taking the place of~$G_{4}$. The remaining
one is given by $G_{\beta} = G_{6}$, $g_{\beta} = G_{3}$, $g_{\alpha} = G_{2}'$, where $G_{2}'$~represents the
group $\{1, (13)\}$. Regarded as substitution groups all these are of
distinct form from the groups of~\Inum{V}\@.
%% -----File: 181.png---Folio 165-------

There are thus $22$~distinct intransitive substitution groups of
degree~$7$, in which the symbols are interchanged transitively in two
sets of $4$~and~$3$ respectively.
\end{Remark}

\Par{118.} Let $\nu_{r}$ ($r = 0, 1, 2, \Dots, n$) be the number of substitutions
\index{Number of symbols@\Topic{Number of symbols} unchanged by all the substitutions of a group}%
of a group of degree~$n$ and order~$N$ which leave exactly
$r$ symbols unchanged, so that
\[
N = \sum_{r = 0}^{r = n} \nu_{r}.
\]

Suppose first that the group is transitive; and in a sub-group,
which keeps one symbol unchanged, let $\nu_{r}'$ ($r = 1,
2, \Dots, n$) be the number of substitutions that leave exactly $r$
symbols unchanged, so that
\[
\frac{N}{n} = \sum_{r = 1}^{r = n} \nu_{r}'.
\]

Each of the $n$~sub-groups, which leave a single symbol
unchanged, have $\nu_{r}'$~substitutions which leave exactly $r$ symbols
unchanged; and each of these substitutions belong to $r$~sub-groups
which leave one symbol unchanged. Hence
\[
n\nu_{r}' = r\nu_{r},
\]
and therefore
\[
N = \sum_{r = 1}^{r = n} r\nu_{r};
\]
or the number of unchanged symbols in all the substitutions
of a transitive group is equal to the order of the group.

Suppose, next, that the group is intransitive; and consider a
set of $s$~symbols among the~$n$, which are permuted transitively
among themselves by the operations of the group. Let $N_{1}$ be
the order of the self-conjugate sub-group~$H_{1}$, which leaves
unchanged each of this set of $s$ symbols. Then if we consider
the effect of the substitutions on this set of $s$ symbols only, the
group reduces to a transitive group of order~$\dfrac{N}{N_{1}}$ with which the
original group is multiply isomorphic. If $S'$ is any substitution
of this group of order~$\dfrac{N}{N_{1}}$, and if $SH_{1}$~denote the corresponding $N_{1}$
operations of the original group, then every substitution of the
set~$SH_{1}$ produces the same effect on the $s$~symbols that $S'$
%% -----File: 182.png---Folio 166-------
\index{Intransitive groups@\Topic{Intransitive groups}!transitive sets of symbols in}%
\index{Number of symbols@\Topic{Number of symbols} unchanged by all the substitutions of a group}%
produces. Now the number of unchanged symbols in all the
substitutions of the transitive group of degree~$s$ and order~$\dfrac{N}{N_{1}}$
is~$\dfrac{N}{N_{1}}$; therefore, in all the substitutions of the original group,
the number of symbols of the set of~$s$ that remain unchanged is~$N$.
The same reasoning applies to each separate set of the $n$~symbols,
which are permuted transitively among themselves by
the operations of the group. Hence if there are $t_{1}$ such transitive
sets, the total number of symbols which remain unchanged
in all the substitutions of the group is~$Nt_{1}$; or
\[
Nt_{1} = \sum_{r = 1}^{r = n} r\nu_{r}.
\]

\Par{119.} The formula just obtained is the first of a series of
similar \DPchg{formulae}{formulæ}, due to Herr Frobenius,\footnote
  {``Ueber die Congruenz nach einem aus zwei endlichen Gruppen gebildeten
  Doppelmodul,'' \textit{Crelle}, t.~\VolNo{CI}, (1887), p.~288.}
\index{Frobenius@\Topic{Frobenius}, quoted}%
which are capable of
many useful applications.

{\Loosen Consider the symbol $(a_{1}, a_{2}, \Dots, a_{p})$, where the letters are
$p$~distinct letters chosen from the $n$ which are operated on by
the group, the sequence in which the $p$~letters are arranged
being regarded as essential. The number of such symbols
that can be formed from the $n$~letters is $n(n - 1)\Dots (n - p + 1)$.
Every substitution of the group interchanges this set of
symbols among themselves; and no substitution can leave one
of the symbols unchanged unless it leaves each of the letters
forming it unchanged. Moreover no substitution of the group,
other than identity, can leave each of the set of symbols
unchanged. Hence the group can be expressed as a substitution
group in this set of $n(n - 1)\Dots (n - p + 1)$ symbols.
Every substitution of the group which leaves exactly $r$ letters
unchanged will, if $r < p$, leave none of the set of symbols
unchanged; while if $r \geq p$, it will leave exactly}
\[
r(r - 1)\Dots (r - p + 1)
\]
unchanged. Hence, if the set of symbols are interchanged by
the substitutions of the group in $t_{p}$~transitive sets, then
\[
Nt_{p} = \sum_{r = p}^{r = n} \frac{r!}{r - p!} \nu_{r}.
\]
%% -----File: 183.png---Folio 167-------

From the symbol $(a_{1}, a_{2}, \Dots, a_{p})$, containing $p$~letters,
may be formed $n - p$ symbols containing $p + 1$ letters, by
adding any one of the remaining $n - p$ letters in the last place.
If the symbol $(a_{1}, a_{2}, \Dots, a_{p})$ is one of a transitive set of
$s$~symbols, to these there will correspond $s(n - p)$ symbols of
$p + 1$ letters. No symbol of $p + 1$ letters, which is not included
among these $s(n - p)$ symbols, can enter into a transitive set
with any one of the $s(n - p)$; since if it did, the $s$~symbols of $p$~letters
would not form a transitive set. Hence the $s(n - p)$
symbols must form, by themselves, a number of transitive sets
of the symbols of $p + 1$ letters; and this number clearly cannot
be less than~$1$ nor greater than $n - p$. Accordingly, to every
transitive set of the symbols of $p$~letters, there correspond
$x$ ($1 \leq x \leq n - p$) transitive sets of the symbols of $p + 1$ letters;
and therefore
\[
t_{p} \leq t_{p + 1} \leq (n - p)t_{p}.
\]

\begin{Ex}[1.] If $t_{p}$ and $t_{p + 1}$ are equal and if $p < n - 1$, shew that $t_{p + 1}$~is
unity and that the group is $(p + 1)$-ply transitive.
\end{Ex}

\begin{Ex}[2.] Apply the method of §~119 to shew that no substitution,
except identity, of a $k$-ply transitive group, which does not contain
the alternating group, can displace less than $2k - 2$ letters.
\end{Ex}


\Note{\textit{Note to} §~108.}

\begin{Remark}
The results of M.~Jordan, stated on \PageRef{p.}{149}, may be established
as follows. Let $G$~be a quadruply transitive group of degree~$n$ and
order $n(n - 1)(n - 2)(n - 3)$, so that $n - 2$ is the power of a prime.
There is a single operation of~$G$ which changes any four symbols
into any other four symbols. Let $a$,~$b$, $c$,~$d$ be four of the symbols
operated on by~$G$. The operations of~$G$ which permute these
symbols among themselves form a sub-group~$H$, simply isomorphic
with the symmetric group of degree~$4$. Suppose that the remaining
$n - 4$ symbols are permuted by~$H$ in transitive sets of $n_{1}$, $n_{2}$,~\dots\
symbols each.

The only groups with which $H$~is multiply isomorphic are
\Inum{(i)}~the symmetric group of degree~$3$, \Inum{(ii)}~a group of order~$2$. If
then, when we consider the effect of~$H$ on a set of $n_{1}$~symbols which
are permuted transitively by it, the group of degree~$n_{1}$ so obtained
is one with which $H$~is multiply isomorphic, this group must be
%% -----File: 184.png---Folio 168-------
either the symmetric group of degree~$3$ or a group of order~$2$.
Hence $n_{1}$ must be either $6$,~$3$ or~$2$. Since in this case $H$ will
contain operations leaving all the $n_{1}$~symbols unchanged, the value~$6$
for~$n_{1}$ is inadmissible. If $n_{1}$ is~$3$, the operations of~$H$, which
give the substitutions
\[
1,\quad (ab)\,(cd),\quad (ad)\,(bc),\quad (ac)(bd),
\]
of $a$,~$b$, $c$,~$d$, leave the $3$~symbols unchanged; and if $n_{1}$ is~$2$, the
operations of~$H$ which give all the even substitutions of $a$,~$b$, $c$,~$d$,
leave the $2$~symbols unchanged.

If, on the other hand, when we consider the effect of~$H$ on the
set of $n_{1}$ symbols only, the transitive group of degree~$n_{1}$ so obtained
is simply isomorphic with~$H$, then $n_{1}$~must be either $4$,~$6$, $8$,~$12$ or~$24$.
In this case, the value~$4$ for~$n_{1}$ is inadmissible. In fact, if
$n_{1}$ were~$4$ the operation of~$H$ which leaves $c$~and~$d$ unchanged
would also leave two of the $n_{1}$ symbols unchanged; and this is
impossible since no operation of~$H$, except identity, can leave more
than $3$ symbols unchanged. For a similar reason, the value~$12$ for~$n_{1}$
is inadmissible; while, if $n_{1}$ is~$6$, the sub-group that keeps one of
the $n_{1}$~symbols unchanged must be cyclical.

The only other possible value for $n_{1}$ is unity.

Suppose, first, that $n_{1}$ is~$2$. If any of the remaining numbers
$n_{2}$, $n_{3}$,~\dots\ differ from~$24$, it may be shewn immediately that
$H$~contains operations which keep more than $3$ symbols fixed.
Hence in this case $n$~is congruent to $6\ (\mod 24)$; and since $n - 2$
cannot then be a power of a prime unless $n = 6$, this case gives the
alternating group of degree~$6$.

Next, suppose that $n_{1}$ is~$3$. The only admissible values for
$n_{2}$, $n_{3}$,~\dots\ are $8$~and~$24$. In this case, the sub-group of~$H$ which
keeps all the $n_{1}$ ($= 3$) symbols unchanged is a non-cyclical sub-group
of order~$4$. But we have seen in §~105 that, if $n - 3$ is even, a
sub-group of order~$n - 3$ which keeps $3$~symbols unchanged can only
have a single operation of order~$2$. Hence this case cannot occur.

Next, suppose that $n_{1}$ is~$8$. The only admissible values for
$n_{2}$, $n_{3}$,~\dots\ are again~$24$. This case cannot occur since no number
congruent to $10, \pmod{24}$, can be a power of a prime.

The only remaining possibilities are:
%[** TN: Reformatted; ad hoc dot spacing]
\begin{alignat*}{3}
&\Inum{(i)}   & n_{1} &= n_{2} = \ \dots\dots\dots\dots\dots &&= 24, \\
&\Inum{(ii)}  & n_{1} &= 1,\quad n_{2} = n_{3} = \ \dots\dots &&= 24, \displaybreak[0]\\
&\Inum{(iii)}\quad &  n_{1} &= 6,\quad n_{2} = n_{3} = \ \dots\dots &&= 24, \displaybreak[0]\\
&\Inum{(iv)}  & n_{1} &= 1,\quad n_{2} = 6,\quad n_{3} = \dots &&= 24, \displaybreak[0]\\
&\Inum{(v)}   & n_{1} &= 1,\quad n_{2} = 6,\quad n_{3} = \dots &&= 0, \\
&\Inum{(vi)}  & n_{1} &= 1,\quad n_{2} = n_{3} = \ \dots\dots &&= 0.
\end{alignat*}

The first of them cannot occur, since then $n - 2$ is not the power
%% -----File: 185.png---Folio 169-------
of a prime. In the second case, an operation of~$H$ which leaves $c$
and $d$ unchanged would be of the form
\[
(ab) (\alpha\beta) (\gamma\delta) \Dots,
\]
where there are $\frac{1}{2}(n - 3)$ independent transpositions; while an
operation of~$H$, of order~$2$, which leaves none of the symbols $a$,~$b$,
$c$,~$d$ unchanged, will consist of the product of $\frac{1}{2}(n - 1)$ independent
transpositions. Now the operation
\[
(ab) (\alpha\beta) (\gamma\delta) \Dots
\]
occurs in the group, conjugate to~$H$, which permutes $a$,~$b$, $\alpha$,~$\beta$,
among themselves; and as an operation of this group, it must be
the product of $\frac{1}{2}(n - 1)$ independent transpositions. This is a
contradiction: hence this case cannot occur.

In case~\Inum{(iii)}, let $1$,~$2$, $3$, $4$, $5$,~$6$ be the six symbols that are
permuted transitively. The self-conjugate sub-group of~$H$ of order~$4$
will consist of identity and the three substitutions
\begin{gather*}
(ab) (cd) (34) (56) (\alpha\beta) (\gamma\delta) \Dots, \\
(ac) (bd) (56) (12) (\alpha\gamma) (\beta\delta) \Dots, \\
(ad) (bc) (12) (34) (\alpha\delta) (\beta\gamma) \Dots.
\end{gather*}

This sub-group will also occur as the self-conjugate sub-group of
order~$4$ of the group, conjugate to~$H$, which permutes $\alpha$,~$\beta$, $\gamma$,~$\delta$
among themselves. If $S$,~$S'$ are two operations, of order~$3$, of these
sub-groups which give the same permutation of $1$,~$2$, $3$, $4$, $5$,~$6$, then
$SS'^{-1}$~is an operation of~$G$, distinct from identity, which keeps the
six symbols unchanged. Hence this case cannot occur. Precisely
the same reasoning applies to case~\Inum{(iv)}.

Finally then, cases \Inum{(v)} and~\Inum{(vi)}, in which $n$~is respectively $11$ and
$5$ are the only remaining possibilities.

When $n$ is~$5$, $G$~is the symmetric group of degree~$5$.

When $n$ is $11$, the group~$G$, if it exists, is of degree~$11$ and
order $11 · 10 · 9 · 8$. A sub-group of order~$8$ which keeps three
symbols fixed must contain a single operation of order~$2$; hence
it must either be cyclical or of the type given in Theorem~V, §~63.
When expressed in $8$~symbols, this is generated by
\[
(1254) (3867)\quad \text{and}\quad (1758) (2643),
\]
while we may take a cyclical group of order~$8$ to be generated by
\[
(12835476).
\]

It is easy to verify that each of these substitutions transforms the
group of order~$9$, generated by
\[
(123) (456) (789)\quad \text{and}\quad (147) (258) (369),
\]
into itself.

If we now apply the method of §~109, we find that each of
these doubly transitive groups of degree~$9$ is contained in a triply
%% -----File: 186.png---Folio 170-------
transitive group of degree~$10$; a repetition of the same process of
trial shews that, while the group
\[
\{(123) (456) (789),\ (12835476)\}
\]
is not contained in a quadruply transitive group of degree~$11$ and
order~$11·10·9·8$, the group
\[
\{(123) (456) (789),\ (1254) (3867),\ (1758) (2643)\}
\]
and the two substitutions
\[
(a2) (58) (64) (79),\ (ab) (57) (68) (49)
\]
actually generate such a group.

A further trial will shew that this group and the substitution
\[
(bc) (47) (58) (69)
\]
generate a group of degree~$12$ and order $12·11·10·9·8$; but that
there is no group of degree~$13$ which contains the last group as the
sub-group that keeps one symbol fixed.

The three substitutions, of order two, just given generate that
sub-group of order~$24$, of the transitive group of degree~$11$ and
order $11·10·9·8$ in the symbols
\[
a,\ b,\ c,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9,
\]
which permutes $a$,~$b$, $c$,~$2$ among themselves.

{\Loosen If $k \nless 4$, a $k$-ply transitive group, of degree~$n$ and order
$n(n - 1)\Dots(n - k + 1)$, must contain a quadruply transitive group of
degree $n - k + 4$ and order $(n - k + 4) (n - k + 3) (n - k + 2) (n - k + 1)$.
Hence, if the group is neither the alternating group nor the symmetric
group of degree~$n$, it must be either the group of degree~$11$
or the group of degree~$12$ that have been determined above.}


\Note{\textit{Note to} §~110.}

It may be shewn that, with a single exception when $n = 12$, the
inequality (\PageRef{p.}{152})
\[
k \ngtr \tfrac{1}{3} n + 1
\]
may be replaced by
\[
k < \tfrac{1}{3} n + 1.
\]

Since $k$ is an integer, it is only necessary to consider the case in
\index{Substitution group@\Topic{Substitution group}!quintuply transitive, of degree~$12$}%
which $n$~is a multiple of~$3$, so that we may write $3m$ for~$n$. If, in
this case, $k = m + 1$, then $2k - 2 = 2m$; and a $(m + 1)$-ply transitive
group of degree~$3m$, which does not contain the alternating group,
can therefore have no substitution which displaces fewer than $2m$
symbols. Its order must therefore be $3m (3m - 1)\Dots (2m + 1) 2m$.
From M.~Jordan's results, which have just been proved, such a
group exists only when $n = 12$; and therefore when $n$~is not~$12$, a
$k$-ply transitive group of degree~$n$, which does not contain the
alternating group, can only exist if $\DPtypo{k\frac{1}{3} < n + 1}{k < \frac{1}{3}n + 1}$.
\end{Remark}
%% -----File: 187.png---Folio 171-------


\Chapter{IX.}{On Substitution Groups: Primitive and Imprimitive Groups.}

\Par{120.} \First{We} have seen that the symbols permuted by the
operations of an intransitive substitution group may be divided
into sets, such that every substitution of the group permutes
the symbols of each set among themselves. For a transitive
group the symbols must, from this point of view, be regarded
as forming a single set. It may however in particular cases
be possible to divide the symbols permuted by a transitive
group into sets in such a way, that every substitution of the
group either interchanges the symbols of any set among themselves
or else changes them all into the symbols of some other
set. That this may be possible, it is clearly necessary that each
set shall contain the same number of symbols.

In the present Chapter we shall discuss those properties
of transitive groups which depend on their possessing or not
possessing the property indicated.

\Par{121.} \begin{Definition}
When the symbols operated on by a
\index{Imprimitive groups@\Topic{Imprimitive groups}!definition of}%
\index{Primitive groups@\Topic{Primitive groups}!definition of}%
\index{Substitution group@\Topic{Substitution group}!primitive and imprimitive, definition of}%
transitive substitution group can be divided into sets, each set
containing the same number of distinct symbols and no symbol
occurring in two different sets, and when the sets are such
that all the symbols of any set are either interchanged among
themselves or changed into the symbols of another set by every
substitution of the group, the group is called \emph{imprimitive}.
When no such division into sets is possible, the group is called
%% -----File: 188.png---Folio 172-------
\index{Imprimitive systems@\Topic{Imprimitive systems}!definition of}%
\emph{primitive}. The sets of symbols which are interchanged by
an imprimitive group are called \emph{imprimitive systems}.
\end{Definition}

A simple example of an imprimitive group is given by
group~\Inum{VII} of §~17. An examination of the substitutions of this
group will shew that they all either transform the systems of
symbols $xyz$ and $abc$ into themselves or else interchange them,
and that the same is true of the systems $xa$,~$yb$,~$zc$; so that, in
this case, the symbols may be divided into two distinct sets of
imprimitive systems.

It follows at once, from the definition, that an imprimitive
group cannot be more than simply transitive. For if it were
doubly transitive, it would contain substitutions changing any
two symbols into any other two, and of these the first pair
might be chosen from the same imprimitive system and the
second pair from distinct systems.

The question as to whether a given group can be expressed
as a transitive group of given degree, and the further question
as to whether such a representation of the group, when possible,
is imprimitive or primitive, finds its complete solution in the
following investigation due to Herr Dyck\footnotemark.
\index{Dyck@\Topic{Dyck}, quoted}%
\footnotetext{Dyck, ``Gruppentheoretische Studien,~II,'' \textit{Math.\ Ann.}\DPnote{** No "Vol."} \VolNo{XXII}, (1883), pp.~86--95.}

\Par{122.} In §~20 it was shewn how any group~$G$ of order~$N$
could be represented as a substitution group of $N$~symbols.
This form of the group, defined as the regular form, is simply
transitive; for all its substitutions except identity displace all
the symbols, and therefore there must be just one substitution
changing a given symbol into any other. Let us now suppose
that $N = \mu\nu$, and that $G$~has a sub-group~$H$ of order~$\mu$, consisting
of the operations
\[
S_{1} (= 1),\ S_{2},\ S_{3},\ \Dots,\ S_{\mu},
\]
so that every operation of the group can be represented uniquely
in the form
\[
S_{m}T_{n}\quad (m = 1, 2, 3, \Dots, \mu;\ n = 1, 2, 3, \Dots, \nu);\ T_{1} = 1.
\]

The tableau representing the group as a substitution group
of $N$~symbols will, in terms of these symbols, take the form
%% -----File: 189.png---Folio 173-------
\index{Imprimitive systems@\Topic{Imprimitive systems}!of a regular group}%
\index{Representation of a group@\Topic{Representation of a group}!in transitive form; \ie\ as a transitive substitution group|(}%
given on the following page. Every symbol in this tableau is
of the form
\[
S_{m} T_{n} S_{m'} T_{n'};
\]
and such a symbol will belong to the column headed by $S_{m}T_{n}$
and to the line beginning with~$S_{m'}T_{n'}$. The symbols in any
line (or column) differ in arrangement only from those in the
leading line (or column); hence $T_{k}$~must occur in the line
beginning with~$S_{m'} T_{n'}$. We may therefore suppose that
\[
S_{m} T_{n} S_{m'} T_{n'} = T_{k};
\]
and then
\begin{align*}
  S_{2} S_{m} T_{n} S_{m'} T_{n'} &= S_{2} T_{k}; \\
  S_{3} S_{m} T_{n} S_{m'} T_{n'} &= S_{3} T_{k}, \\
  \multispan{3}{\dotfill} \\
  S_{\mu} S_{m} T_{n} S_{m'} T_{n'} &= S_{\mu} T_{k}.
\end{align*}
Since
\[
1,\ S_{2},\ S_{3},\ \Dots,\ S{\mu}
\]
form a group, these symbols differ from
\[
S_{m},\ S_{2}S_{m},\ S_{3}S_{m},\ \Dots,\ S_{\mu}S_{m}
\]
only in the sequence in which they are written; and therefore
the set of symbols
\[
S_{p} T_{n} S_{m'} T_{n'}\quad (p = 1, 2, \Dots, \mu)
\]
is identical, except as regards arrangement, with the set
\[
S_{p}T_{k}\quad (p = 1, 2, \Dots, \mu).
\]

Hence the substitution of~$G$, represented by the line beginning
with~$S_{m'} T_{n'}$, changes the set of symbols $S_{p} T_{n}$ into the
set~$S_{p} T_{k}$ in some sequence or other.

Every substitution of the group therefore changes the
symbols of each of the $\nu$~sets, into which the first line of the
tableau is divided, either into themselves or into the symbols of
some other of the $\nu$ sets. Hence:---

\begin{Theorem}[I.]
If a group of order~$\mu\nu$ contains a sub-group
of order~$\mu$, the regular form of the group will be imprimitive in
such a way that the $\mu\nu$~symbols may be divided into $\nu$~imprimitive
systems of $\mu$~symbols each.
\end{Theorem}

The converse of this theorem is also true. For if the $\mu\nu$~symbols
\[
1,\ S_{2},\ S_{3},\ \Dots,\ S_{\mu\nu},
\]
%% -----File: 190.png---Folio 174-------
\begin{sidewaystable}
\setlength{\arraycolsep}{2pt}
\PageLabel{174}%
\ifthenelse{\boolean{ForPrinting}}{}{\Squeeze[0.75]}{$
\begin{array}{@{}*{5}{c}|c|*{5}{c}|c|*{5}{c}@{}}
1  & S_{2} & S_{3} & \dots & S_{\mu} & \dots &
T_{a} & S_{2}T_{a} & S_{3}T_{a} & \dots & S_{\mu}T_{a} & \dots &
T_{\nu} & S_{2}T_{\nu} & S_{3}T_{\nu} & \dots & S_{\mu}T_{\nu} \\
%
S_{2} & S_{2}S_{2} & S_{3}S_{2} & \dots & S_{\mu}S_{2} & \dots &
T_{a}S_{2} & S_{2}T_{a}S_{2} & S_{3}T_{a}S_{2} & \dots & S_{\mu}T_{a}S_{2} & \dots &
T_{\nu}S_{2} & S_{2}T_{\nu}S_{2} & S_{3}T_{\nu}S_{2} & \dots & S_{\mu}T_{\nu}S_{2} \\
%
\multicolumn{5}{c|}{\dotfill} & \dots &
\multicolumn{5}{c|}{\dotfill} & \dots &
\multicolumn{5}{c}{\dotfill} \\
%
S_{\mu} & S_{2}S_{\mu}   & S_{3}S_{\mu}  & \dots & S_{\mu}S_{\mu}  & \dots & T_{a}S_{\mu}  & S_{2}T_{a}S_{\mu}  & S_{3}T_{a}S_{\mu}  & \dots & S_{\mu}T_{a}S_{\mu}  & \dots & T_{\nu}S_{\mu}  & S_{2}T_{\nu}S_{\mu}  & S_{3}T_{\nu}S_{\mu}  & \dots & S_{\mu}T_{\nu}S_{\mu} \\
  \hline
\multicolumn{5}{c|}{\dotfill} & \Strut\dots &
\multicolumn{5}{c|}{\dotfill} & \dots &
\multicolumn{5}{c}{\dotfill} \\
%
\hline
\Strut
T_{b} & S_{2}T_{b} & S_{3}T_{b} & \dots & S_{\mu}T_{b} & \dots &
T_{a}T_{b} & S_{2}T_{a}T_{b} & S_{3}T_{a}T_{b} & \dots & S_{\mu}T_{a}T_{b} & \dots &
T_{\nu}T_{b} & S_{2}T_{\nu}T_{b} & S_{3}T_{\nu}T_{b} & \dots & S_{\mu}T_{\nu}T_{b} \\
%
S_{2}T_{b} & S_{2}S_{2}T_{b} & S_{3}S_{2}T_{b} & \dots & S_{\mu}S_{2}T_{b} & \dots &
T_{a}S_{2}T_{b} & S_{2}T_{a}S_{2}T_{b} & S_{3}T_{a}S_{2}T_{b} & \dots & S_{\mu}T_{a}S_{2}T_{b} & \dots &
T_{\nu}S_{2}T_{b} & S_{2}T_{\nu}S_{2}T_{b} & S_{3}T_{\nu}S_{2}T_{b} & \dots & S_{\mu}T_{\nu}S_{2}T_{b} \\
\multicolumn{5}{c|}{\dotfill} & \dots &
\multicolumn{5}{c|}{\dotfill} & \dots &
\multicolumn{5}{c}{\dotfill} \\
%
S_{\mu}T_{b} & S_{2}S_{\mu}T_{b} & S_{3}S_{\mu}T_{b} & \dots & S_{\mu}S_{\mu}T_{b} & \dots &
T_{a}S_{\mu}T_{b} & S_{2}T_{a}S_{\mu}T_{b} & S_{3}T_{a}S_{\mu}T_{b} & \dots & S_{\mu}T_{a}S_{\mu}T_{b} & \dots &
T_{\nu}S_{\mu}T_{b} & S_{2}T_{\nu}S_{\mu}T_{b} & S_{3}T_{\nu}S_{\mu}T_{b} & \dots & S_{\mu}T_{\nu}S_{\mu}T_{b} \\
\hline
\multicolumn{5}{c|}{\dotfill} & \dots &
\multicolumn{5}{c|}{\dotfill} & \dots &
\multicolumn{5}{c}{\dotfill} \\
%
\hline
\Strut
T_{\nu} & S_{2}T_{\nu} & S_{3}T_{\nu} & \dots & S_{\mu}T_{\nu} & \dots &
T_{a}T_{\nu} & S_{2}T_{a}T_{\nu} & S_{3}T_{a}T_{\nu} & \dots & S_{\mu}T_{a}T_{\nu} & \dots &
T_{\nu}T_{\nu} & S_{2}T_{\nu}T_{\nu} & S_{3}T_{\nu}T_{\nu} & \dots & S_{\mu}T_{\nu}T_{\nu} \\
%
S_{2}T_{\nu} & S_{2}S_{2}T_{\nu} & S_{3}S_{2}T_{\nu} & \dots & S_{\mu}S_{2}T_{\nu} & \dots &
T_{a}S_{2}T_{\nu} & S_{2}T_{a}S_{2}T_{\nu} & S_{3}T_{a}S_{2}T_{\nu} & \dots & S_{\mu}T_{a}S_{2}T_{\nu} & \dots &
T_{\nu}S_{2}T_{\nu} & S_{2}T_{\nu}S_{2}T_{\nu} & S_{3}T_{\nu}S_{2}T_{\nu} & \dots & S_{\mu}T_{\nu}S_{2}T_{\nu} \\
%
\multicolumn{5}{c|}{\dotfill} & \Strut\dots &
\multicolumn{5}{c|}{\dotfill} & \dots &
\multicolumn{5}{c}{\dotfill} \\
%
S_{\mu}T_{\nu} & S_{2}S_{\mu}T_{\nu} & S_{3}S_{\mu}T_{\nu} & \dots & S_{\mu}S_{\mu}T_{\nu} & \dots &
T_{a}S_{\mu}T_{\nu} & S_{2}T_{a}S_{\mu}T_{\nu} & S_{3}T_{a}S_{\mu}T_{\nu} & \dots & S_{\mu}T_{a}S_{\mu}T_{\nu} & \dots &
T_{\nu}S_{\mu}T_{\nu} & S_{2}T_{\nu}S_{\mu}T_{\nu} & S_{3}T_{\nu}S_{\mu}T_{\nu} & \dots & S_{\mu}T_{\nu}S_{\mu}T_{\nu}
\end{array}
$}
\end{sidewaystable}
%% -----File: 191.png---Folio 175-------
by whose permutations the group can be expressed in regular
form, are divisible into $\nu$ imprimitive systems of $\mu$ each, let
\[
1,\ S_{2},\ S_{3},\ \Dots,\ S_{\mu}
\]
be that system which contains identity. Then this system
and the systems
\[
S_{m},\ S_{2}S_{m},\ S_{3}S_{m},\ \Dots,\ S_{\mu}S_{m},\quad m = 2,\ 3,\ \Dots,\ \mu,
\]
having the symbol~$S_{m}$ in common, must have all their symbols
in common; therefore the product of any two of the operations
of this set of $\mu$ operations is another operation of the set. The
set therefore forms a group.

\Par{123.} Let us now represent the imprimitive system
\[
T_{m},\ S_{2}T_{m},\ S_{3}T_{m},\ \Dots,\ S_{\mu}T_{m}
\]
by the single symbol~$U_{m}$, for all values of~$m$. If we then, in
the preceding tableau representing the group, pay attention only
to the way in which the systems are permuted among themselves,
without regarding the permutations of the symbols within the individual
systems, we obtain a substitution group of the $\nu$ symbols
\[
U_{1},\ U_{2},\ U_{3},\ \Dots,\ U_{\nu}.
\]

This group is isomorphic with the original group~$G$; and if
no substitution of the original group interchanges among
themselves the symbols of each imprimitive system, the isomorphism
must be simple. Now a substitution of~$G$, which
does not change any imprimitive system into another, must, if
it exists, be a substitution~$S'$ of the sub-group~$H$, which is
constituted by
\[
1,\ S_{2},\ S_{3},\ \Dots,\ S_{\mu},
\]
such that $T_{m}S'$ belongs to the set~$U_{m}$ for each suffix~$m$; and
therefore, for each suffix~$m$, we must have an equation
\[
T_{m}S' = S''T_{m},
\]
where $S''$ is another substitution of~$H$. The sub-group~$H$
must therefore contain every substitution of~$G$ which is conjugate
to~$S'$: in other words, it must contain a self-conjugate
sub-group of~$G$. Hence:---

\begin{Theorem}[II.]
If $H$, of order~$\mu$, is a sub-group of~$G$ of
\index{Transitive group@\Topic{Transitive group}!representation of any group as a|(}%
order~$\mu\nu$, and if no self-conjugate sub-group of~$G$ is contained in~$H$,
then $G$~can be expressed as a transitive group of degree~$\nu$.
\end{Theorem}
%% -----File: 192.png---Folio 176-------

That the converse of this theorem is true is immediately
obvious.

\Par{124.} If the sub-group~$H$ of~$G$ is contained in a greater
sub-group~$K$ of order~$\mu\nu'$, where $\nu = \nu'\nu''$, the operations of~$K$
consist of the sets
\[
H,\ HT_{2},\ HT_{3},\ \Dots,\ HT_{\nu'},
\]
and the operations of~$G$ of the sets
\[
K,\ KR_{2},\ KR_{3},\ \Dots,\ KR_{\nu''};
\]
while the set~$KR_{m}$ is made up of the sets
\[
HR_{m},\ HT_{2} R_{m},\ HT_{3} R_{m},\ \Dots,\ HT_{\nu'} R_{m}.
\]

The regular form of the group has $\nu'\nu''$~imprimitive systems
\index{Imprimitive systems@\Topic{Imprimitive systems}!of any transitive group}%
corresponding to the sub-group~$H$, and $\nu''$~imprimitive systems
corresponding to the sub-group~$K$; the above method of representing~$K$
shews that each of the latter systems contain $\nu'$~complete
systems of the former set.

Now it has just been proved that, if $H$~contains no self-conjugate
sub-group of~$G$, the group can be represented as a
transitive substitution group of the symbols
\[
HT_{n} R_{m}\quad (n = 1, 2, \Dots, \nu',\ m = 1, 2, \Dots, \nu'').
\]

But from the division of the symbols in the regular form of
the group into $\nu''$~imprimitive systems, it follows that the set of
symbols
\[
HT_{n} R_{m}\quad (n = 1, 2, \Dots, \nu')
\]
must either be permuted among themselves or be changed into
another set
\[
HT_{n} R_{m'}\quad (n = 1, 2, \Dots, \nu')
\]
by every substitution of the group. The representation of the
group as a transitive group of $\nu\nu''$~symbols is therefore imprimitive.
Hence:---

\begin{Theorem}[III.]
If a group~$G$ of order~$\mu\nu'\nu''$ has a sub-group~$H$
of order~$\mu$, which contains no self-conjugate sub-group of~$G$;
and if $H$~is contained in a sub-group~$K$ of~$G$ of order~$\mu\nu'$;
then the representation of~$G$ as a transitive group of degree~$\nu'\nu''$
\(in respect of~$H$\) is imprimitive, the $\nu'\nu''$~symbols being
divisible into $\nu''$~systems of $\nu'$~symbols each.
\end{Theorem}
%% -----File: 193.png---Folio 177-------

\begin{Corollary}[I.]
A transitive group of order~$\mu\nu$ and degree~$\nu$
\index{Representation of a group@\Topic{Representation of a group}!in primitive form}%
will be primitive if, and only if, a sub-group of order~$\mu$ that
keeps one symbol fixed is a maximum sub-group.
\end{Corollary}

\begin{Corollary}[II.]
A group, which contains other self-conjugate
operations besides identity, cannot be represented in primitive
form.
\end{Corollary}

For if a sub-group~$H$ of order~$\mu$ contains a self-conjugate
operation, the group (of order~$\mu\nu$) cannot be represented as
a transitive group of degree~$\nu$ in respect of~$H$; and if $H$
contains none of the self-conjugate operations, and is not a
self-conjugate sub-group, it cannot be a maximum sub-group.

In particular, a group whose order is the power of a prime
cannot be represented as a primitive group.

\begin{Corollary}[III.]
An Abelian group when represented as a
transitive substitution group, must be in regular form.
\end{Corollary}

\begin{Corollary}[IV.]
A simple group can always be represented
in primitive form.
\end{Corollary}

\Par{125.} Every possible representation of a group as a
transitive substitution group is given by the method of the
preceding paragraphs. There is another method of dealing
with the same problem which we may shortly consider here in
view of its utility in many special cases, though it does not in
general lead to all possible modes of representation. Let
\[
H_{1},\ H_{2},\ \Dots,\ H_{\nu}
\]
be a conjugate set of sub-groups (or operations) of a given
group~$G$, and let
\[
I_{1},\ I_{2},\ \Dots,\ I_{\nu}
\]
be a set of sub-groups of~$G$, such that $I_{r}$~is the greatest sub-group
containing $H_{r}$ self-conjugately. The latter set of groups
are not necessarily all distinct; in fact, we have seen in §~55
that, when the order of~$G$ is the power of a prime, they cannot
be all distinct.

If $S$ is any operation of~$G$, then
\[
\Sub{\PadTo{S^{-1}H_{1}S}{H_{1}},
     \PadTo{S^{-1}H_{2}S}{H_{2}}, \Dots,
     \PadTo{S^{-1}H_{\nu}S}{H_{\nu}}}
    {S^{-1}H_{1}S, S^{-1}H_{2}S, \Dots, S^{-1}H_{\nu}S}
\]
%% -----File: 194.png---Folio 178-------
is a substitution performed on the set of symbols $H_{1}$, $H_{2}$,~\dots, $H_{\nu}$;
and if for $S$ each substitution of the group is written in turn,
these substitutions form a transitive substitution group of
degree~$\nu$. The substitution corresponding to~$S$, followed by
the substitution
\[
\Sub{\PadTo{T^{-1}H_{1}T}{H_{1}},
     \PadTo{T^{-1}H_{2}T}{H_{2}}, \Dots,
     \PadTo{T^{-1}H_{\nu}T}{H_{\nu}}}
    {T^{-1}H_{1}T, T^{-1}H_{2}T, \Dots, T^{-1}H_{\nu}T}
\]
corresponding to~$T$, gives the substitution
\[
\Sub{\PadTo{T^{-1}S^{-1}H_{1}ST}{H_{1}},
     \PadTo{T^{-1}S^{-1}H_{2}ST}{H_{2}}, \Dots,
     \PadTo{T^{-1}S^{-1}H_{\nu}ST}{H_{\nu}}}
    {T^{-1}S^{-1}H_{1}ST, T^{-1}S^{-1}H_{2}ST, \Dots, T^{-1}S^{-1}H_{\nu}ST};
\]
and therefore the substitutions form a group isomorphic with~$G$,
since the product of the substitutions corresponding to $S$ and
$T$ is the substitution corresponding to~$ST$.

Moreover, since there are operations of~$G$ which transform
$H_{1}$ into each of the other sub-groups (or operations) of the
conjugate set, the substitution group is transitive in the $\nu$~symbols.
The substitution group will be simply isomorphic
with~$G$ if, and only if, there is no operation of~$G$ which
transforms each of the $\nu$~sub-groups into itself. Now the only
operations of~$G$ which transform $H_{1}$ into itself are the operations
of~$I_{1}$; and hence the substitution group will be simply
isomorphic with~$G$ if, and only if, the conjugate set of sub-groups
\[
I_{1},\ I_{2},\ \Dots,\ I_{\nu},
\]
has no common sub-group except identity. This will be the
case only when $I_{1}$~contains no self-conjugate sub-group of~$G$.

It has been seen (§~123) that, when this condition is satisfied,
$G$~can be represented as a transitive substitution group whose
degree is~$\nu$, the ratio of the orders of $G$~and~$I_{1}$. That the
form there obtained is identical with the form obtained in the
present paragraph may be easily verified. Thus in the earlier
form, the substitution corresponding to~$S$ is
\[
\Sub{\PadTo{I_{1}T_{1}S}{I_{1}T_{1}},
     \PadTo{I_{1}T_{2}S}{I_{1}T_{2}}, \Dots,
     \PadTo{I_{1}T_{\nu}S}{I_{1}T_{\nu}}}
    {I_{1}T_{1}S, I_{1}T_{2}S, \Dots, I_{1}T_{\nu}S}, \quad (T_{1} = 1),
\]
%% -----File: 195.png---Folio 179-------
or in abbreviated form
\[
\Sub{I_{1}T_{x}}{I_{1}T_{x'}},
\]
if
\[
T_{x}S = iT_{x'},
\]
$i$ being some operation of~$I_{1}$.

In the present mode of representation, the substitution
corresponding to~$S$ can be written in the form
\[
\Sub{\PadTo{S^{-1} T_{1}^{-1} H_{1} T_{1}S}{T_{1}^{-1} H_{1} T_{1}},
     \PadTo{S^{-1} T_{2}^{-1} H_{1} T_{2}S}{T_{2}^{-1} H_{1} T_{2}}, \Dots,
     \PadTo{S^{-1} T_{\nu}^{-1} H_{1} T_{\nu}S}{T_{\nu}^{-1} H_{1} T_{\nu}}}
    {S^{-1} T_{1}^{-1} H_{1} T_{1}S,
     S^{-1} T_{2}^{-1} H_{1} T_{2}S, \Dots,
     S^{-1} T_{\nu}^{-1} H_{1} T_{\nu}S},
\]
since each operation of the set~$I_{1} T_{n}$ will transform $H_{1}$ into the
same sub-group. Now, if a corresponding abbreviated form be
used, this substitution may be written
\[
\Sub{T_{x}^{-1} H_{1} T_{x}}{T_{x'}^{-1} H_{1} T_{x'}},
\]
and therefore the symbols~$I_{1} T_{x}$ in the one form are permuted
by the substitutions in identically the same manner as the
corresponding symbols $T_{x}^{-1} H_{1} T_{x}$ in the other form.

It should be noticed that, if $G$~contains self-conjugate operations
other than identity, these operations necessarily occur
in~$I_{1}$; and therefore in such a case the present method cannot
lead to a substitution group which is simply isomorphic with~$G$.
In any case, if $K$~is the greatest self-conjugate sub-group of~$G$
contained in~$I_{1}$, the substitution group is simply isomorphic
with~$\dfrac{G}{K}$.
\index{Representation of a group@\Topic{Representation of a group}!in transitive form; \ie\ as a transitive substitution group|)}%
\index{Transitive group@\Topic{Transitive group}!representation of any group as a|)}%

\begin{Remark}
\Par{126.} As an illustration of the preceding paragraphs, we will
\index{Transitive group@\Topic{Transitive group}!number of distinct modes of representing the alternating group of degree~$5$ as a}%
determine the different modes in which the alternating group of
degree~$5$ can be represented as a transitive group.

The only cyclical sub-groups contained in~$G_{60}$, the alternating
group of degree~$5$, are groups of orders $2$,~$3$ and~$5$; and of each of
these cyclical sub-groups there is a single conjugate set.

The non-cyclical sub-groups may be determined as follows. The
lowest possible order for such a sub-group is~$4$; since this is the
highest power of~$2$ that divides~$60$, there is a single conjugate set of
sub-groups of order~$4$. The next lowest possible order is~$6$. Now
no operation of order~$3$ is permutable with an operation of order~$2$,
as the group contains no operations of order~$6$; on the other hand,
every sub-group of order~$3$ is permutable with an operation of order~$2$;
thus
\[
(12) (45) (123) (12) (45) = (132).
\]
%% -----File: 196.png---Folio 180-------
\index{Transitive group@\Topic{Transitive group}!number of distinct modes of representing the alternating group of degree~$5$ as a}%
There is therefore a single set of conjugate sub-groups of order~$6$.
The next lowest possible order is~$10$. The group contains no operation
of order~$10$; but every sub-group of order~$5$ is permutable with
an operation of order~$2$; thus
\[
(14) (23) (12345) (14) (23) = (15432).
\]
There is therefore a single conjugate set of sub-groups of order~$10$.
The next lowest possible order is~$12$. If the group contains a
sub-group of this order, it must be transitive in $4$~symbols. Now the
alternating group of $4$~symbols is of order~$12$. Hence $G_{60}$~must
contain a single conjugate set of sub-groups of order~$12$. The only
other possible orders are $15$,~$20$ and~$30$. The reader will readily
verify directly that there are no sub-groups of these orders. This
can also be seen indirectly, since $G_{60}$~is a simple group and therefore,
if there were a sub-group of order~$15$, the group could be expressed
transitively in $4$~symbols. Since the group contains operations of
order~$5$, this is clearly impossible.

Hence finally, since each of the sub-groups leads to a transitive
representation of the group, $G_{60}$~can be represented as a transitive
substitution group in $30$,~$20$, $15$, $12$, $10$, $6$ and $5$~symbols, and in
one distinct form in each case. The second method, as given in
§~125, does not lead to all these modes of representation. The
group will be found to contain $15$~conjugate operations (or sub-groups)
of order~$2$: $10$~conjugate sub-groups and $20$~conjugate
operations of order~$3$: $5$~conjugate sub-groups of order~$4$: $6$~conjugate
sub-groups and two sets of $12$~conjugate operations, each of
order~$5$: $10$~conjugate sub-groups of order~$6$: $6$~conjugate sub-groups
of order~$10$: and $5$~conjugate sub-groups of order~$12$. Hence, by
using the second method, the representation of the group as transitive
in $30$~symbols would be missed.

Since a sub-group of order~$2$ is contained in sub-groups of orders
$4$,~$6$,~$10$ and~$12$, the $30$~symbols permuted by~$G_{60}$, when it is expressed
as a transitive group of degree~$30$, can be divided into sets
of imprimitive systems, containing respectively $2$,~$3$,~$5$ and $6$~symbols
each. Similarly, when $G_{60}$~is represented as a transitive group of
degree $20$,~$15$ or~$12$, it is imprimitive. When expressed as a group
of order $10$,~$6$ or~$5$, it is primitive.

\Par{127.} As a further illustration, we shall determine all the
distinct forms of imprimitive groups of degree~$6$. Let $G$ be such
a group, and $H$ that sub-group of~$G$ which interchanges among
themselves the symbols of each imprimitive system.

We will first suppose that there are two systems of three symbols
each, viz.\
\[
1,\ 2,\ 3 \quad\text{and}\quad 4,\ 5,\ 6.
\]
%% -----File: 197.png---Folio 181-------
\index{Imprimitive groups@\Topic{Imprimitive groups}!of degree~$6$}%

In this case, $H$~cannot consist of the identical operation only;
for there must be a substitution changing $1$ into~$2$, and this must
permute $1$,~$2$ and~$3$ among themselves, and therefore also $4$,~$5$ and~$6$
among themselves.

Let $H$~contain substitutions which leave $4$,~$5$ and~$6$ unchanged.
These must (§~114) form a self-conjugate sub-group of~$H$, which will
be either
\[
\{(123),\ (12)\}\quad \text{or}\quad \{(123)\}.
\]

Since $G$~has substitutions interchanging the systems, $H$~must
similarly contain
\[
\{(456),\ (45)\}\quad \text{or}\quad \{(456)\}.
\]

In the first alternative, $H$~is the group
\[
\lintertext{\Inum{(i)}} \{(123),\ (12),\ (456),\ (45)\};
\]
for $H$~contains this group, and on the other hand, this is the most
general group that interchanges the six symbols in two intransitive
systems of three each. The order of this group is~$2^{2} · 3^{2}$.

In the second alternative, $H$~contains the self-conjugate sub-group
\[
\{(123),\ (456)\}.
\]

Now if $H$~is of order~$2^{2} · 3^{2}$, it is necessarily of the form~\Inum{(i)}. If
it is of order~$2 · 3^{2}$, it must contain a substitution of order~$2$ which
transforms the sub-group just given into itself. This may be taken,
without loss of generality, to be~$(12) (45)$; and then $H$~is the
group
\[
\lintertext{\Inum{(ii)}} \{(123),\ (456),\ (12) (45)\}.
\]

If $H$ is of order~$3^{2}$, it is the group
\[
\lintertext{\Inum{(iii)}} \{(123),\ (456)\}.
\]

Next, let $H$~contain no substitutions which leave $4$,~$5$ and~$6$
unchanged. Then $H$~is simply isomorphic with a group of degree~$3$,
and therefore it must be of the form
\[
\lintertext{\Inum{(iv)}} \{(123) (456),\ (12) (45)\},
\]
or
\[
\lintertext{\Inum{(v)}} \{(123) (456)\}.
\]

Now $\dfrac{G}{H}$~is of order~$2$; therefore $G$~must have a substitution of
order $2$ or~$4$, which interchanges the systems. If the order of~$H$ is
odd, this substitution must be of order~$2$. When the substitution
is of order~$2$, we may, without loss of generality, take it to be
$(14) (25) (36)$ or $(14) (26) (35)$. If $H$ is of the form \Inum{(i)},~\Inum{(ii)}, or~\Inum{(iv)},
we get for~$G$, in each case, the same group whichever of these substitutions
we take. When $H$ is of the form~\Inum{(iii)}, we get two groups which
are easily seen to be conjugate in the symmetric group. These we
do not regard as distinct. When $H$~is of the form~\Inum{(v)}, we get two
%% -----File: 198.png---Folio 182-------
\index{Imprimitive groups@\Topic{Imprimitive groups}!of degree~$6$}%
distinct groups, one of which is simply isomorphic with the symmetric
group of three symbols, while the other is a cyclical group.
In these two latter cases, the symbols can be divided into three imprimitive
systems of two each.

If the substitution which interchanges the systems is of order~$4$,
its square must occur in~$H$; we may therefore take it to be
$(1425) (36)$. When $H$~is of form~\Inum{(i)}, this gives the same group as
before; but when $H$~is of either of the forms \Inum{(ii)} or~\Inum{(iv)}, we get new
forms for~$G$. There are therefore eight distinct forms of groups of
degree~$6$, in which the symbols form two imprimitive systems of
three symbols each.

Secondly, suppose that there are three systems of symbols,
containing two each, viz.\
\[
1,\ 2;\quad 3,\ 4;\quad \text{and}\quad 5,\ 6.
\]

The self-conjugate sub-group~$H$ is of order $2^{3}$,~$2^{2}$, $2$ or~$1$.
Corresponding to the first three cases, the forms of~$H$ are easily
seen to be
\begin{alignat*}{2}
&\Inum{(i)}       && \{(12),\ (34),\ (56)\}, \\
&\Inum{(ii)}      && \{(12) (34),\ (34) (56),\ (12) (56)\}, \\
\lintertext{and}
&\Inum{(iii)}\quad&& \{(12) (34) (56)\}.
\end{alignat*}

Again, since $\dfrac{G}{H}$~interchanges the three systems, it must be
simply isomorphic with a group of degree~$3$, and its order is therefore
either $3$ or~$6$. First, let its order be~$3$. It must then contain a
substitution of order~$3$ which, without loss of generality, may be
taken to be $(135)(246)$; this gives, with the three above forms of~$H$,
three distinct forms for~$G$. The form of~$G$ corresponding to the form~\Inum{(iii)}
of~$H$ is, however, the same as one of those already determined.

If $\dfrac{G}{H}$~is of order~$6$, $G$~must contain as a self-conjugate sub-group
one of the three groups just obtained. Also if $\dfrac{G}{H}$~were cyclical,
there would be a substitution in~$G$, not belonging to~$H$ and
permutable with $(135) (246)$. This is clearly impossible, and therefore
$G$~must contain a substitution which transforms $(135) (246)$
into its inverse. We may take this to be $(13) (24)$ or $(14) (23) (56)$.
With the form~\Inum{(i)} for~$H$, these two substitutions lead to the same
group. When $H$~is of the form~\Inum{(ii)}, they give two distinct forms
for~$G$. When $H$~is of the form~\Inum{(iii)}, $G$~admits the imprimitive
systems $1$,~$3$,~$5$, and $2$,~$4$,~$6$.

Lastly, if $H$~is the identical operation, $G$~is necessarily of order~$6$;
no new forms can arise.
%% -----File: 199.png---Folio 183-------

There are therefore five distinct forms of groups of degree~$6$, in
which the symbols form three imprimitive systems of two symbols
each but do not at the same time form two imprimitive systems of
three symbols each.
\end{Remark}

\Par{128.} An actual test to determine whether any transitive
group is primitive or imprimitive may be applied as follows.
Consider the effect of the substitutions of the group~$G$ on $r$ of
the symbols which are permuted transitively by it. Those
substitutions, which permute the $r$~symbols, say
\[
a_{1},\ a_{2},\ \Dots,\ a_{r}
\]
among themselves, form a sub-group~$H$. Now suppose that
every substitution, which changes $a_{1}$ into one of the $r$~symbols,
belongs to~$H$. Then if $S$~is a substitution, which does not
permute the $r$~symbols among themselves, it must change
them into a new set
\[
b_{1},\ b_{2},\ \Dots,\ b_{r},
\]
which has no symbol in common with the previous set; and
every operation of the set~$HS$ changes all the $a$'s into~$b$'s.
Moreover, since $G$~is transitive, $H$~must permute the $a$'s
transitively; and therefore the set~$HS$ must contain substitutions
changing $a_{1}$ into each one of the~$b$'s.

Suppose now, if possible, that the group contains a substitution~$S'$,
which changes some of the $a$'s into~$b$'s, and the
remainder into new symbols. We may assume that $S'$~changes
$a_{1}$ into~$b_{1}$, and $a_{2}$~into a new symbol~$c_{2}$. Among the set~$HS$
there is at least one substitution,~$T$, which changes $a_{1}$ into~$b_{1}$.
Hence $TS'^{-1}$ changes $a_{1}$ into itself and some new symbol into~$a_{2}$.
This however contradicts the supposition that every
substitution, which changes $a_{1}$ into one of the set of~$a$'s,
belongs to~$H$. Hence no substitution such as~$S'$ can belong to~$G$;
and every substitution, which changes one of the $a$'s into
one of the~$b$'s, must change all the $a$'s into~$b$'s.

If the substitutions of the group are not thus exhausted,
there must be another set of $r$~symbols
\[
c_{1},\ c_{2},\ \dots,\ c_{r},
\]
which are all distinct from the previous sets, such that some
substitution changes all the $a$'s into~$c$'s. We may now repeat
%% -----File: 200.png---Folio 184-------
the previous reasoning to shew that every substitution, which
changes an $a$ into a~$c$, must change all the $a$'s into~$c$'s. By
continuing this process, we finally divide the symbols into a
number of distinct sets of $r$~each, such that every substitution of
the group must change the~$a$'s either into themselves or into
some other set: and therefore also must change every set either
into itself or into some other set. The group must therefore be
imprimitive. Hence:---

\begin{Theorem}[IV.]
If, among the symbols permuted by a transitive
\index{Primitivity@\Topic{Primitivity}, test of}%
group, it is possible to choose a set such that every
substitution of the group, which changes a chosen symbol of the
set either into itself or into another of the set, permutes all the
symbols of the set among themselves; then the group is imprimitive,
and the set of symbols forms an imprimitive system.
\end{Theorem}

\begin{Corollary}[I.]
If $a_{1}$, $a_{2}$,~\Dots, $a_{r}$ are a part of the symbols
permuted by a primitive group, there must be substitutions of
the group, which replace some of this set of symbols by others
of the set, and the remainder by symbols not belonging to the
set.
\end{Corollary}

\begin{Corollary}[II.]
A sub-group of a primitive group, which
keeps one symbol unchanged, must contain substitutions which
displace any other symbol.
\end{Corollary}

If the sub-group~$H$, that leaves $a_{1}$ unchanged, leaves every
symbol of the set $a_{1}$, $a_{2}$,~\Dots, $a_{r}$ unchanged, then $H$~must be
transformed into itself by every substitution which changes any
one of these symbols into any other. Every substitution, which
changes one of the set into another, must therefore permute
the set among themselves; and the group, contrary to supposition,
is imprimitive.

\Par{129.} It has already been seen that, in particular cases, it
may be possible to distribute the symbols, which are permuted by
an imprimitive group, into imprimitive systems in more than
one way. When this is possible, suppose that two systems
which contain~$a_{1}$ are
\begin{align*}
&a_{1},\ a_{2},\ \Dots,\ a_{r},\ a_{r+1},\ \Dots,\ a_{m}, \\
\lintertext{and}
&a_{1},\ a_{2},\ \Dots,\ a_{r},\ a_{r+1}',\ \Dots\Dots\Dots,\ a_{n}';
\end{align*}
%% -----File: 201.png---Folio 185-------
and that the symbols common to the two systems are
\[
a_{1},\ a_{2},\ \Dots,\ a_{r}.
\]

A substitution of the group, which changes $a_{1}$ into~$a_{r+1}'$,
must change $a_{1}$, $a_{2}$,~\Dots, $a_{r}$ into $r$~symbols of that system of
the first set which contains~$a_{r+1}'$, while it changes the system of
the second set that contains~$a_{1}$ into itself. Hence the latter
system contains at least $r$~symbols of that system of the first
set in which $a_{r+1}'$ occurs. By considering the effect of the
inverse substitution, it is clear that the system
\[
a_{1},\ a_{2},\ \Dots,\ a_{r},\ a_{r+1}',\ \Dots,\ a_{n}'
\]
cannot have more than $r$~symbols in common with the system
of the first set that contains~$a_{r+1}'$. Hence the $n$~symbols of this
system can be divided into sets of~$r$, such that each set is
contained in some system of the first set. It follows that~$n$, and
therefore also~$m$, must be divisible by~$r$.

Suppose now that $b_{1}$ is any symbol which is not contained
in either of the above systems. A substitution that changes
$a_{1}$ into $b_{1}$ must change the two systems into two others, which
have $r$~symbols
\[
b_{1},\ b_{2},\ \Dots,\ b_{r}
\]
in common; and since no two systems of either set have a
common symbol, these $r$~symbols must be distinct from
\[
a_{1},\ a_{2},\ \Dots,\ a_{r}.
\]

Further, from the mode in which the set $b_{1}$, $b_{2}$,\ \Dots, $b_{r}$ has
been obtained, any operation, which changes one of the symbols
$a_{1}$, $a_{2}$,~\Dots, $a_{r}$ into one of the symbols $b_{1}$, $b_{2}$,~\Dots, $b_{r}$, must
change all the symbols of the first set into those of the second.
Hence the symbols operated on by the group can be divided
into systems of $r$ each, by taking together the sets of $r$~symbols
which are common to the various pairs of the two
given sets of imprimitive systems; and the group is imprimitive
in regard to this new set of systems of $r$~symbols each.
Hence:---

\begin{Theorem}[V.]
If the symbols permuted by a transitive group
\index{Imprimitive systems@\Topic{Imprimitive systems}!properties of}%
can be divided into imprimitive systems in two distinct ways, $m$~being
the number of symbols in each system of one set and $n$~in
%% -----File: 202.png---Folio 186-------
each system of the other; and if some system of the first set has
$r$~symbols in common with some system of the second set; then
\Inum{(i)}~$r$~is a factor of both $m$ and~$n$, and \Inum{(ii)}~the symbols can be
divided into a set of systems of $r$~each, in respect of which the
group is imprimitive.
\end{Theorem}

This result may also be regarded as an immediate consequence
of Theorem~III, §~124. For if $H_{1}$~and~$H_{2}$ are two
sub-groups of~$G$ each of which contains the sub-group~$R$, and
if $K$~is the greatest common sub-group of $H_{1}$ and~$H_{2}$, then $K$
contains~$R$. Now if $R$~contains no self-conjugate sub-group
of~$G$, then $G$~can be represented as a transitive group whose
degree is the order of~$G$ divided by the order of~$R$. If the
respective orders of $H_{1}$ and $H_{2}$ are $m$~times and $n$~times the
order of~$R$, the symbols can be divided in two distinct ways
into sets of imprimitive systems, the systems containing $m$
and $n$~symbols respectively. Also, if the order of~$K$ is $r$~times
the order of~$R$, then $r$~is a factor of~$m$ and of~$n$; by considering
the sub-group~$K$, the symbols may be divided into a set of
systems which contain $r$~symbols each.

\begin{Remark}
It might be expected that, just as we can form a new set of
imprimitive systems by taking together the symbols which are
common to pairs of systems of two given sets, so we might form
another new set of systems by combining all the systems of one set
which have any symbols in common with a single system of the
other set. A very cursory consideration will shew however that
this is not in general the case. In fact, it is sufficient to point out
that, with the notation already used, the number of symbols in such
a new system would be~$\dfrac{mn}{r}$; and this number is not necessarily a
factor of the degree of the group. Also, even if this number is a
factor of the degree of the group, it will not in general be the case
\index{Imprimitive systems@\Topic{Imprimitive systems}!properties of}%
that the symbols so grouped together form an imprimitive system.
\end{Remark}

\Par{130.} We may now discuss, more fully than was possible in
§~106, the form of a self-conjugate sub-group of a given transitive
group. Such a sub-group must clearly contain one or more
operations displacing every symbol operated on by the group.
For if every operation of the sub-group keeps the symbol $a_{1}$
unchanged, then since it is self-conjugate, every operation will
keep $a_{2}$, $a_{3}$,~\Dots, unchanged: and the sub-group must reduce
to the identical operation only.
%% -----File: 203.png---Folio 187-------

Suppose now, if possible, that $H$~is an intransitive self-conjugate
\index{Self-conjugate sub-group@\Topic{Self-conjugate sub-group}!of a primitive group must be transitive}%
\index{Self-conjugate sub-group@\Topic{Self-conjugate sub-group}!of an imprimitive group}%
sub-group of a transitive group~$G$; and that $H$
permutes the $n$~symbols of~$G$ in the separate transitive sets
$a_{1}$, $a_{2}$,~\Dots, $a_{n_{1}}$; $b_{1}$, $b_{2}$,~\Dots, $b_{n_{2}}$;~\Dots. If $S$~is any operation
of~$G$ which changes $a_{1}$ into~$b_{1}$, then, since
\[
S^{-1}HS = H,
\]
it must change all the $a$'s into~$b$'s; and since
\[
SHS^{-1} = H,
\]
$S^{-1}$ must change all the $b$'s into~$a$'s. Hence the number of
symbols in the two sets, and therefore the number of symbols
in each of the sets, must be the same.

Moreover every operation of~$G$, since it transforms $H$ into
itself, must either permute the symbols of any set among
themselves, or it must change them all into the symbols of
some other set. Hence $G$~must be imprimitive, and $H$~must
consist of those operations of~$G$ which permute the symbols of
each imprimitive system among themselves.

Conversely, when $G$~is imprimitive, it is immediately obvious
that those operations of~$G$, if any such exist, which permute the
symbols of each of a set of imprimitive systems among themselves,
form a self-conjugate sub-group. Hence:---

\begin{Theorem}[VI.]
A self-conjugate sub-group of a primitive
group must be transitive; and if an imprimitive group has an
intransitive self-conjugate sub-group, it must consist of the
operations which permute among themselves the symbols of each
of a set of imprimitive systems.
\end{Theorem}

If $G$~is an imprimitive group of degree~$mn$, and if there are
$n$ imprimitive systems of $m$~symbols each, then we have seen
in §~123 that $G$~is isomorphic with a group~$G'$ of degree~$n$. In
particular instances, it may at once be evident, from the order of~$G$,
that this isomorphism cannot be simple. For example, if
the order of~$G$ has a factor which does not divide~$n!$, this is
certainly the case: and more generally, if it is known independently
that $G$~cannot be expressed as a transitive group of
degree~$n$, then $G$~must certainly be multiply isomorphic with~$G'$.
In such instances the self-conjugate sub-group of~$G$, which
%% -----File: 204.png---Folio 188-------
corresponds to the identical operation of~$G'$, is that intransitive
self-conjugate sub-group, which interchanges among themselves
the symbols of each imprimitive system.

If $G$~is soluble, a minimum self-conjugate sub-group of~$G$
must have for its order a power of a prime. Also, if $G$~has an
intransitive self-conjugate sub-group, it must have an intransitive
minimum self-conjugate sub-group. Hence if $G$~is soluble
and has intransitive self-conjugate sub-groups, the symbols
permuted by~$G$ must be capable of division into imprimitive
systems, such that the number in each system is the power of
\index{Self-conjugate sub-group@\Topic{Self-conjugate sub-group}!of an imprimitive group}%
a prime.

\Par{131.} Let $G$ be a $k$-ply transitive group of degree~$n$ ($k > 2$),
and let $G_{r}$~be that sub-group of~$G$ which keeps $r$ ($< k$) given
symbols unchanged, so that $G_{r}$~is $(k - r)$-ply transitive in the
remaining $n - r$ symbols. Also, let $H$ be a self-conjugate
sub-group of~$G$, and let $H_{r}$ be that sub-group of~$H$ which keeps
the same $r$~symbols unchanged; so that $H_{r}$ is the common
sub-group of $H$ and~$G_{r}$. Since every operation of~$G_{r}$ transforms
both $H$ and $G_{r}$ into themselves, every operation of~$G_{r}$ must be
permutable with~$H_{r}$; \ie\ $H_{r}$~is a self-conjugate sub-group of~$G_{r}$.
Now, if $r = k - 2$, $G_{k - 2}$~is doubly transitive in the $n - k + 2$
symbols on which it operates; it is therefore primitive. Hence,
unless $H_{k - 2}$ consists of the identical operation only, it must be
transitive in the $n - k + 2$ symbols.

If $H_{k - 2}$ is the identical operation, $H$~contains no operation,
except identity, which displaces less than $n - k + 3$ symbols.
Suppose, first, that $H$~contains operations, other than identity,
which leave one or more symbols unchanged. Then, since $H$~is
a self-conjugate sub-group and $G$~is $k$-ply transitive, it may
be shewn, exactly as in §~110, that $H$~must contain operations
displacing not more than $2k - 2$ symbols. Hence $H_{k - 2}$ can
consist of the identical operation alone, only if
\[
n - k + 3 \ngtr 2k - 2,
\]
or
\[
k \nless \tfrac{1}{3}n + \tfrac{5}{3}.
\]

When this inequality holds, we have seen (\PageRef{p.}{152}) that $G$~contains
the alternating group. Hence in this case, if $G$~does
not contain the alternating group, it follows that $H_{k - 2}$~is transitive
in the $n - k + 2$ symbols on which it operates.
%% -----File: 205.png---Folio 189-------

Since $H$~is self-conjugate and $G$~is $k$-ply transitive, $H$~must
\index{Self-conjugate sub-group@\Topic{Self-conjugate sub-group}!of a $k$-ply transitive group is, in general, $(k - 1)$-ply transitive}%
contain a sub-group conjugate to~$H_{k - 2}$ which keeps any other
$k - 2$ symbols unchanged. Hence $H_{k - 3}$~must be doubly transitive
in the $n - k + 3$ symbols on which it operates; and so on.
Finally, if $G$~is not the symmetric group (the alternating group,
being simple, contains no self-conjugate sub-group) $H$~must be
$(k - 1)$-ply transitive.

Suppose, next, that $H$~contains no operation, except identity,
which leaves any symbol unchanged. Then if, with the notation
of §~110, $j = k - 1$ for every operation of~$H$, the argument there
used does not apply. For it is impossible to choose the operation~$T$
so that $c_{k}$~is a symbol which does not occur in~$S$.

The self-conjugate sub-group~$H$ contains a single operation
changing a given symbol~$a_{1}$ into any other symbol~$a_{r}$. Also $G$~contains
operations which leave $a_{1}$ unchanged and change $a_{r}$
into any other symbol~$a_{s}$. Hence the operations of~$H$, other
than identity, form a single conjugate set in~$G$; and therefore
$H$~must be an Abelian group of order~$p^{m}$ and type $(1, 1, \Dots,
\text{to~$m$ units})$; $p$~being a prime. Further, since $G$~is by supposition
at least triply transitive, it must contain operations which
transform any two operations of~$H$, other than identity, into
any other two. If $p$~were an odd prime, and $P_{1}$~and~$P_{2}$ were
two of the generating operations of~$H$, it follows that $G$~would
have an operation~$S$ such that
\[
S^{-1} P_{1} S = P_{1}, \quad S^{-1} P_{2} S = P_{1}^{\alpha};
\]
and this is impossible. Hence $p$~must be~$2$. Further if $G$~were
more than triply transitive, and if $A$,~$B$,~$C$ were three independent
generating operations of~$H$, then $G$~would have an operation~$\Sigma$
such that
\[
\Sigma^{-1} A\Sigma = A, \quad
\Sigma^{-1} B\Sigma = B, \quad
\Sigma^{-1} C\Sigma = AB.
\]
This again is impossible, and therefore $k$~must be~$3$. Hence:---

\begin{Theorem}[VII.]
A self-conjugate sub-group of a $k$-ply
transitive group of degree~$n$ \($2 < k < n$\), is in general at least
$(k - 1)$-ply transitive\footnotemark.
\index{Jordan@\Topic{Jordan}, quoted}%
\footnotetext{Jordan, \textit{Traité des Substitutions}, p.~65; where, however, the exceptional
  case is overlooked.}%
The only exception is that a triply
transitive group of degree~$2^{m}$ may have a self-conjugate sub-group
of order~$2^{m}$.
\end{Theorem}
%% -----File: 206.png---Folio 190-------

\Par{132.} For the further discussion of the self-conjugate
\index{Direct product@\Topic{Direct product} of two groups!represented as a transitive group}%
sub-groups of a primitive group, it is necessary to consider in
what forms the direct product of two groups can be represented
as a transitive group.

Let $G$~be the direct product of two groups $H_{1}$ and~$H_{2}$, and
suppose that $G$~can be represented as a transitive group
of degree~$n$. When $G$ is thus represented, we will suppose
that $H_{1}$~is transitive in the $n$~symbols that $G$~permutes. We
have seen in §~107 that every substitution of $n$~symbols, which
is permutable with each of the substitutions of a group
transitive in the $n$~symbols, must displace all the $n$~symbols.
It follows that every substitution of~$H_{2}$ must displace all the $n$~symbols
on which $G$~operates; and that the order of~$H_{2}$ is
equal to or is a factor of~$n$.

If the order of~$H_{2}$ is equal to~$n$, then $H_{2}$~is transitive in the
$n$~symbols, so that the order of~$H_{1}$ cannot be greater than~$n$.
In this case, $H_{1}$~and~$H_{2}$ must (§~107) be two simply isomorphic
groups of order~$n$, which have no self-conjugate operations
except identity. Further, if $H_{1}$~and $H_{2}$ in this case are not
simple groups, let $K$ be a self-conjugate sub-group of~$H_{1}$.
Since every operation of~$H_{2}$ is permutable with every operation
of~$H_{1}$, $K$~is a self-conjugate sub-group of~$G$. Now the order of~$K$
is less than~$n$, the degree of~$G$; therefore $K$~is intransitive
and $G$~is imprimitive. On the other hand, we have seen
(\textit{loc.~cit.})\ that, if $H_{1}$~and $H_{2}$ are simple, the sub-group of~$G$
that keeps one symbol fixed is a maximum sub-group: and
therefore $G$~is primitive. Hence:---

\begin{Theorem}[VIII.]
If the direct product~$\Gamma$ of $H_{1}$ and $H_{2}$ can
be represented as a transitive group of degree~$n$, in such a way
that $H_{1}$~and $H_{2}$ are transitive sub-groups of~$\Gamma$, then $H_{1}$~and $H_{2}$
must be simply isomorphic groups of order~$n$, which have no
self-conjugate operations except identity. When this condition
is satisfied, $\Gamma$~will be primitive if, and only if, $H_{1}$~and $H_{2}$ are
simple.
\end{Theorem}

\Par{133.} Suppose now that a primitive group~$G$, of degree~$n$,
has two distinct minimum self-conjugate sub-groups $H_{1}$ and~$H_{2}$.
Then every operation of~$H_{1}$ (or~$H_{2}$) is permutable with~$H_{2}$
%% -----File: 207.png---Folio 191-------
(or~$H_{1}$), and $H_{1}$,~$H_{2}$ have no common operation except
identity. Hence (§~34) the group $\{H_{1}, H_{2}\}$, which we will call~$\Gamma$,
is the direct product of $H_{1}$ and~$H_{2}$. Now $\Gamma$~is a self-conjugate
sub-group of~$G$: it is therefore transitive in
the $n$~symbols which $G$~permutes. Also $H_{1}$ and~$H_{2}$, being
self-conjugate sub-groups of~$G$, are transitive. Hence, by
Theorem~VIII (§~132), $H_{1}$~and $H_{2}$ are simply isomorphic, and $n$~is
equal to the order of~$H_{1}$. Moreover, since $H_{1}$~is a minimum
self-conjugate sub-group of~$G$ which contains no self-conjugate
operations except identity, it must (Theorem~IV, §~94) be either
a simple group of composite order, or the direct product of
several simply isomorphic simple groups of composite order. It
follows that $G$~cannot have two distinct minimum self-conjugate
sub-groups unless the degree of~$G$ is equal to or is a power
of the order of some simple group of composite order.

\Par{134.} Let now $\Gamma$ be a minimum self-conjugate sub-group
of a doubly transitive group~$G$, and suppose that $\Gamma$~is the direct
product of the $\alpha$~simply isomorphic simple groups
\[
H_{1},\ H_{2},\ \Dots,\ H_{\alpha}.
\]

Since $G$~is primitive, $\Gamma$~is transitive. If $H_{1}$~is a cyclical
group of prime order~$p$, the order of~$\Gamma$ is~$p^{\alpha}$; therefore the
degree of~$\Gamma$, or what is the same thing, the degree of~$G$, is~$p^{\alpha}$.

If $H_{1}$~is a simple group of composite order, and if $\alpha > 2$,
then (§~132) $H_{1}$~cannot be transitive. The intransitive systems
of~$H_{1}$, since they form a set of imprimitive systems for~$\Gamma$, must
each contain the same number $m$ of symbols. If $m$~is less than
the order of~$H_{1}$, a sub-group of~$H_{1}$ which leaves unchanged one
symbol of one intransitive system will leave unchanged one
symbol of each intransitive system. Now we shall see, in §~136,
that the operations of an imprimitive self-conjugate sub-group
of a doubly transitive group must displace all or all but one of
the symbols. Hence $m$~cannot be less than the order of~$H_{1}$.
We may similarly shew that, if $m$~is equal to the order of~$H_{1}$,
some of the operations of~$\Gamma$ must keep more than one symbol
fixed; and therefore, if $\alpha > 2$, the group assumed cannot exist.
If $\alpha = 2$, $H_{1}$~may be transitive. But in this case $\{H_{1}, H_{2}\}$
certainly contains operations which leave more than one symbol
unchanged; and again the group assumed cannot exist. Hence
%% -----File: 208.png---Folio 192-------
finally no doubly transitive group can contain a minimum self-conjugate
sub-group of the type assumed.

No general law can be stated regarding self-conjugate
sub-groups of simply transitive primitive groups; but for
groups which are at least doubly transitive the preceding
results may be summed up as follows:---

\begin{Theorem}[IX.]
A group~$G$ which is at least doubly transitive
\index{Doubly transitive groups@\Topic{Doubly transitive groups}!generally contain simple self-conjugate sub-groups}%
\index{Primitive groups@\Topic{Primitive groups}!when soluble, have a power of a prime for degree}%
either must be simple or must contain a simple group~$H$ as a
self-conjugate sub-group. In the latter case no operation of~$G$,
except identity, is permutable with every operation of~$H$. The
only exceptions to this statement are that a triply transitive group
of degree~$2^{m}$ may have a self-conjugate sub-group of order~$2^{m}$;
and that a doubly transitive group of degree~$p^{m}$, where $p$~is a
prime, may have a self-conjugate sub-group of order~$p^{m}$.
\end{Theorem}

\begin{Corollary}
If a primitive group is soluble, its degree
must be the power of a prime\footnotemark.
\end{Corollary}
\index{Galois@\Topic{Galois}, quoted}%
\footnotetext{This result, stated in a somewhat different form, is given, among many
\PageLabel{192}%
  others, in the letter written by Galois to his friend Chevalier on the evening
  of May~29th, 1832, the day before the duel in which he was killed. The letter
  was first printed in the \textit{Revue Encyclopédique} (1832), p.~568; it was reprinted
  in the collection of Galois's mathematical writings in \textit{Liouville's Journal}, t.~\VolNo{XI}
  (1846), pp.~381--444}
%[** TN: See http://www.gutenberg.org/ebooks/40213]

In fact, if a group is soluble, so also is its minimum
self-conjugate sub-group. The latter must be therefore an
Abelian group of order~$p^{\alpha}$: and since this group must be
transitive, its order is equal to the degree of the primitive
group.

\begin{Remark}
\Par{135.} As illustrating the occurrence of an imprimitive self-conjugate
sub-group in a primitive group, we will construct a
primitive group of degree~$36$ which has an imprimitive self-conjugate
sub-group. For this purpose, let\footnote
  {The commas in the symbols for the substitutions are here used to prevent
  confusion among the one-digit and two-digit numbers.}
\begin{multline*}
S = (1, 2, 3)(4, 5, 6)(7, 8, 9)(10, 11, 12)(13, 14, 15) \\
(16, 17, 18)(19, 20, 21)(22, 23, 24)(25, 26, 27) \\
(28, 29, 30)(31, 32, 33)(34, 35, 36),
\end{multline*}
and
\begin{multline*}
A = (3, 4)(5, 6)(9, 10)(11, 12)(15, 16)(17, 18) \\
(21, 22)(23, 24)(27, 28)(29, 30)(33, 34)(35, 36);
\end{multline*}
so that $\{S, A\}$ is an intransitive group of degree~$36$, the symbols
being interchanged in $6$~transitive systems containing $6$~symbols
each. This group is simply isomorphic with
\[
\{(123) (456),\ (34) (56)\};
\]
%% -----File: 209.png---Folio 193-------
and it may be easily verified that this group is simply isomorphic with
the alternating group of $5$~symbols, the order of which is~$60$.

Also let
\begin{multline*}
J = (2, 7) (3, 13) (4, 19) (5, 25) (6, 31) (9, 14) (10, 20) (11, 26) \\
(12, 32) (16, 21) (17, 27) (18, 33) (23, 28) (24, 34) (30, 35).
\end{multline*}

Then
\begin{multline*}
JSJ = (1, 7, 13) (19, 25, 31) (2, 8, 14) (20, 26, 32) \\
  (3,  9, 15) (21, 27, 33) (4, 10, 16) (22, 28, 34) \\
  (5, 11, 17) (23, 29, 35) (6, 12, 18) (24, 30, 36),
\end{multline*}
and
\begin{multline*}
JAJ = (13, 19) (25, 31) (14, 20) (26, 32) (15, 21) (27, 33) \\
  (16, 22) (28, 34) (17, 23) (29, 35) (18, 24) (30, 36);
\end{multline*}
and $\{JSJ, JAJ\}$ is similar to and simply isomorphic with~$\{S, A\}$.

Now it may be directly verified that $S$~and~$A$ are, each of them,
permutable with $JSJ$ and~$JAJ$; and therefore every operation of
the group $\{S, A\}$ is permutable with every operation of the group
$\{JSJ, JAJ\}$. Also these two groups can have no common operation,
since the symbols into which $\{S, A\}$ changes any given symbol are
all distinct from those into which $\{JSJ, JAJ\}$ change it. Hence
$\{S, A, JSJ, JAJ\}$ is the direct product of $\{S, A\}$ and~$\{JSJ, JAJ\}$;
it is therefore a group of order~$3600$. It is also, from its mode of
formation, a transitive group of degree~$36$; and it interchanges the
symbols in two and only two distinct sets of imprimitive systems, of
which
\[
1,\ 2,\ 3,\ 4,\ 5,\ 6 \quad\text{and}\quad 1,\ 7,\ 13,\ 19,\ 25,\ 31
\]
may be taken as representatives.

Now $J$~does not interchange the $36$~symbols in either of these
systems, and therefore it cannot occur in $\{S, A, JSJ, JAJ\}$.
Further
\[
J\{S, A, JSJ, JAJ\}J = \{S, A, JSJ, JAJ\};
\]
and therefore $\{J, S, A\}$ is a transitive group of degree~$36$ and order~$7200$.
Also, since $J$~does not interchange the symbols in either of
the two sets of imprimitive systems of $\{S, A, JSJ, JAJ\}$, it follows
that $\{J, S, A\}$ is primitive.

\Par{136.} We have seen that, with a single exception, a $k$-ply
\index{Imprimitive self-conjugate sub-group@\Topic{Imprimitive self-conjugate sub-group} of a doubly transitive group}%
transitive group ($k > 2$) cannot have an imprimitive self-conjugate
sub-group; while the example in the preceding section illustrates
the occurrence of an imprimitive self-conjugate sub-group in a simply
transitive primitive group. We will now consider, from a rather
different point of view, the possibility of an imprimitive self-conjugate
sub-group in a doubly transitive group. Let $G$~be a
doubly transitive group of degree~$mn$, and let $H$~be an imprimitive
self-conjugate sub-group of~$G$. Suppose that $m$~is the smallest
number, other than unity, of symbols which occur in an imprimitive
system; and let
\[
a_{1},\ a_{2},\ \Dots,\ a_{m}
\]
%% -----File: 210.png---Folio 194-------
form an imprimitive system. Since $G$~is doubly transitive, it must
contain a substitution~$S$, which leaves $a_{1}$ unchanged and changes $a_{2}$
into~$a_{m + 1}$, a symbol not contained in the given set. If $S$~changes
the given set into
\[
a_{1},\ a_{m + 1},\ \Dots,\ a_{2m - 1},
\]
then, since
\[
S^{-1}HS = H,
\]
this new set must form an imprimitive system for~$H$. Also, since
$m$~is the smallest number of symbols that can occur in an imprimitive
system, the two sets have no symbol in common except~$a_{1}$.

Now $a_{m + 1}$ may be any symbol not contained in the original
system. Hence it must be possible to distribute the $mn$~symbols
into sets of imprimitive systems of $m$~each, such that every pair of
symbols occurs in one system and no pair in more than one system.
This implies that $mn - 1$ is divisible by $m - 1$, or that $n - 1$ is
divisible by~$m - 1$.

Consider now a substitution of~$H$ which leaves $a_{1}$ unchanged.
It must permute among themselves the remaining $m - 1$ symbols
of each of the $\dfrac{mn - 1}{m - 1}$ systems in which $a_{1}$~occurs. If $a_{r}$~is any
other symbol, a similar statement applies to it. Now no two
systems have more than one symbol in common. Hence every
substitution, which leaves both $a_{1}$ and $a_{r}$ unchanged, must leave all
the symbols unchanged. The sub-group~$H$ is therefore such that
each of its substitutions displaces all or all but one of the symbols.
Moreover the substitutions, which leave $a_{1}$ unchanged, permute
among themselves the remaining symbols of each system in which
$a_{1}$~occurs; therefore the order of~$H$ must be~$mn\mu$, where $\mu$~is a
factor of~$m - 1$.

It has been seen in §~112 that, for certain values of~$mn$, groups
satisfying these conditions actually exist. The doubly transitive
groups, of degree~$p^{m}$ and order~$p^{m}(p^{m} - 1)$ there obtained, have
\index{Imprimitive self-conjugate sub-group@\Topic{Imprimitive self-conjugate sub-group} of a doubly transitive group}%
imprimitive self-conjugate sub-groups of orders $p^{m}(p - 1)$ and~$p^{m}$.
\end{Remark}

\Par{137.} \begin{Ex}[1.] Shew that a transitive group of order~$N$ and
degree~$2n$, which is imprimitive in respect of two systems of $n$~symbols
each, must have a self-conjugate sub-group of order~$\frac{1}{2}N$.
\end{Ex}

\begin{Ex}[2.] Shew that an imprimitive group of order~$Np^{2}$ and
degree~$np$, where $p$~is an odd prime, which has $p$~imprimitive
systems of $n$~symbols each, must contain a self-conjugate sub-group
whose order is a multiple of~$p$.
\end{Ex}

\begin{Ex}[3.] Shew that, if $n$~is the smallest number of symbols in
which a group~$G$ can be represented as a transitive group, then $n^{\alpha}$~is
the smallest number of symbols in which the direct product of $\alpha$
groups, simply isomorphic with~$G$, can be represented as a transitive
group.
\end{Ex}
%% -----File: 211.png---Folio 195-------

\begin{Ex}[4.] Prove that if a group~$G$, of order~$N$, represented as a
transitive substitution group, is imprimitive only when its degree is~$N$;
then either $G$~is Abelian, or $N$~is the product of two distinct
primes.
\index{Dyck@\Topic{Dyck}, quoted}%
\Attrib{Dyck.}
\end{Ex}

\begin{Ex}[5.] Shew that if, in the tableau of \PageRef{p.}{174}, the sub-group
\[
1,\ S_{2},\ S_{3},\ \Dots,\ S_{\mu}
\]
is a self-conjugate sub-group~$H$ of the given group~$G$, then, in each
square compartment of the tableau, every horizontal line contains
the same $\mu$~symbols.

Shew also that, if each square compartment of the tableau is
regarded as a single symbol, the permutations of these symbols
given by the tableau represents the group~$\dfrac{G}{H}$ in regular form.
\Attrib{Dyck.}
\end{Ex}
%% -----File: 212.png---Folio 196-------


\Chapter{X.}{On Substitution Groups: Transitivity and
Primitivity: (Concluding Properties.)}

\Par{138.} \First{The} determination of all distinct transitive sub-groups
of the symmetric group of $n$ symbols has been carried
out for values of~$n$ up to~$12$\footnotemark.
\index{Cole@\Topic{Cole}, quoted}%
\index{Jordan@\Topic{Jordan}, quoted}%
\index{Miller@\Topic{Miller}, quoted}%
\footnotetext{Jordan, \textit{Comptes Rendus}, t.~\VolNo{LXXIII} (1871), pp.~853--857; \textit{ib.}\ t.~\VolNo{LXXV}
  (1872), pp.~1754--1757.

  And see the \textit{Bulletin of the New York Mathematical Society}, Cole, 1st~series,
  Vol.~\VolNo{II}, pp.~184--190; 250--258: Miller, 1st~series, Vol.~\VolNo{III}, pp.~168,~169; 242--245:
  2nd series, Vol.~\VolNo{I}, pp.~67--72; 255--258: Vol.~\VolNo{II}, pp.~138--145. \textit{Quarterly
  Journal of Mathematics}, Cole, Vol.~\VolNo{XXVI}, pp.~372--388: Vol.~\VolNo{XXVII}, pp.~35--50:
  Miller, Vol.~\VolNo{XXVII}, pp.~99--118; Vol.~\VolNo{XXVIII}, pp.~193--231. These memoirs
  were all published between 1892 and~1896; in them further references will be
  found.}
This investigation, if carried out
for sufficiently great values of~$n$, would involve the expression
of all types of group in all possible transitive forms.

From the point of view of one of the chief problems of pure
group-theory, namely, the determination of all distinct types
of group of a given order, this analysis of the symmetric group
of $n$~symbols is not a succinct process, as it continually involves
the redetermination of groups which have been already obtained.
Thus a simple group of degree~$mn$, in which the symbols
are permuted in $m$~imprimitive systems of $n$~each, would in
this analysis have been already obtained as a group of degree~$m$.
With reference then to the more restricted problem of determining
all types of simple groups, it would certainly be
sufficient to find all primitive sub-groups of the symmetric
group.

\Par{139.} We shall proceed to determine a superior limit to the
order of a primitive group of degree~$n$, other than the alternating
or the symmetric group.
%% -----File: 213.png---Folio 197-------

Let $G$~be a primitive group of degree~$n$, and suppose that $G$~contains
\index{Primitive groups@\Topic{Primitive groups}!with a transitive sub-group of smaller degree}%
a sub-group~$H$ which leaves $n - m$ symbols unchanged
and is transitive in the remaining~$m$. Since $G$~is primitive, $H$
and the sub-groups conjugate to it must generate a transitive
self-conjugate sub-group of~$G$; and therefore there must be
some sub-group~$H'$, conjugate to~$H$, such that the $m$~symbols
operated on by~$H$ and the $m$ operated on by~$H'$ are not all
distinct. Suppose $H'$ is chosen so that these two sets of $m$~symbols
have as great a number in common as possible, say~$s$;
and represent by
\begin{alignat*}{2}
&\alpha_{1},\ \alpha_{2},\ \Dots,\ \alpha_{r},\ &&\gamma_{1},\ \gamma_{2},\ \Dots,\ \gamma_{s}, \\
\lintertext{and}
&\beta_{1},\ \beta_{2},\ \Dots,\ \beta_{r},\ &&\gamma_{1},\ \gamma_{2},\ \Dots,\ \gamma_{s},
\end{alignat*}
where $r + s = m$, the symbols operated on by $H$~and~$H'$ respectively.
Then $\{H, H'\}$ is a transitive group in the $2r + s$ symbols
$\alpha$,~$\beta$, and~$\gamma$, which leaves unaltered all the remaining symbols
of~$G$.

If $S$~is an operation of~$H$ which changes $\alpha_{2}$ into~$\alpha_{1}$, $S^{-1}H'S$
does not affect~$\alpha_{1}$. Hence, unless $S$ interchanges the $\alpha$'s among
themselves, the $m$~symbols operated on by~$H'$ and the $m$
operated on by~$S^{-1}H'S$ will have more than~$s$ in common.
Every operation of~$H$ which changes one~$\alpha$ into another must
therefore interchange all the $\alpha$'s among themselves; hence $H$~must
be imprimitive.

If then $H$~is primitive, $s$~must be equal to~$m - 1$. In any
case, if $s = m - 1$, $\{H, H'\}$~is a doubly transitive and therefore
primitive group of degree~$m + 1$, which leaves the remaining
$n - m - 1$ symbols of~$G$ unchanged. We may reason about this
sub-group as we have done about~$H$. Among the sub-groups
conjugate to~$\{H, H'\}$, there must be one at least which operates
on $m$ of the symbols displaced by~$\{H, H'\}$. This, with $\{H, H'\}$,
generates a triply transitive group of degree~$m + 2$, which
leaves $n - m - 2$ symbols unchanged. Proceeding thus, we find
finally that $G$~itself must be $(n - m + 1)$-ply transitive.

\Par{140.} If $s$ is less than $m - 1$, we may again deal with the
sub-group $\{H, H'\}$, or~$H_{1}$, exactly as we have dealt with~$H$. It
is a transitive group of degree~$m_{1}$ ($> m$), which leaves $n - m_{1}$
symbols unchanged. If, among the sub-groups conjugate to~$H_{1}$,
%% -----File: 214.png---Folio 198-------
none operates on more than $s_{1}$ of the symbols affected by~$H_{1}$,
and if $H_{1}'$~is a suitably chosen conjugate sub-group, then
$\{H_{1}, H_{1}'\}$ is a transitive group of degree $2m_{1} - s_{1}$, which leaves
$n - 2m_{1} + s_{1}$ symbols unchanged. Continuing this process, we
must, before arriving at a group of degree~$n$, reach a stage at
which the number~$s_{r}$ is equal to~$m_{r} - 1$.

For suppose, if possible, that among the groups conjugate
to~$K$, of degree~$\rho + \sigma$, none displaces more than $\sigma$ of the
symbols acted on by~$K$, while at the same time $2\rho + \sigma = n$. If
\begin{alignat*}{2}
&\alpha_{1},\ \alpha_{2},\ \Dots,\ \alpha_{\rho},\ &&\gamma_{1},\ \gamma_{2},\ \Dots,\ \gamma_{\sigma}, \\
\lintertext{and}
&\beta_{1},\ \beta_{2},\ \Dots,\ \beta_{\rho},\ &&\gamma_{1},\ \gamma_{2},\ \Dots,\ \gamma_{\sigma}
\end{alignat*}
are the symbols affected by $K$ and $K'$ respectively, then since $G$~is
primitive, it must contain an operation~$S$ which changes $\alpha_{1}$ into~$\alpha_{2}$
without at the same time changing all the $\alpha$'s into~$\alpha$'s. If
then we transform~$K'$ by~$S$, the two groups $K'$ and $S^{-1}K'S$ must
operate on more than $\sigma$ common symbols, contrary to supposition.

Hence $G$ must in this case certainly contain a transitive
sub-group of degree~$n - 1$, and therefore is itself at least
doubly transitive\footnotemark.
\index{Jordan@\Topic{Jordan}, quoted}%
\index{Netto@\Topic{Netto}, quoted}%
\index{Marggraff@\Topic{Marggraff}, quoted}%
\footnotetext{The results contained in §§~139,~140 are due to Jordan (\textit{Liouville's Journal},
  Vol.~\VolNo{XVI}, 1871) and Netto (\textit{Crelle's Journal}, Vol.~\VolNo{CIII}, 1889). They have been
  extended by Marggraff: ``Ueber primitiven Gruppen mit transitiven Untergruppen
  geringeren Grades,'' (\textit{Inaugural Dissertation}, Giessen, 1892). With the
  notation used in the text, Marggraff shews that, unless the symbols affected by~$H$
  can be divided into imprimitive systems of $r$~symbols each, in at least $r + 1$
  distinct ways, $G$~will be $(n - m + 1)$-ply transitive. In particular, if $H$~is a
  cyclical group of degree~$m$, $G$~is $(n - m + 1)$-ply transitive. He also shews that
  in any case $m \geq \frac{1}{2}n$.}%

\Par{141.} Returning to the case in which $H$~is primitive and $G$
\index{Primitive groups@\Topic{Primitive groups}!with a transitive sub-group of smaller degree}%
therefore $(n - m + 1)$-ply transitive, we at once obtain an
inferior limit for~$m$. We have seen, in fact, in Theorem~IV,
§~110, that a group of degree~$n$, other than the alternating or
the symmetric group, cannot be more than $(\frac{1}{3}n + 1)$-ply transitive.
Hence
\[
n - m + 1 \ngtr \tfrac{1}{3}n + 1,
\]
or
\[
m \nless \tfrac{2}{3}n.
\]
We may sum up these results as follows:---
%% -----File: 215.png---Folio 199-------

\begin{Theorem}[I.]
A primitive group~$G$ of degree~$n$, which has a
\index{Primitive groups@\Topic{Primitive groups}!limit to the order of, for a given degree}%
sub-group~$H$ that keeps $n - m$ symbols unchanged and is transitive
in the remaining $m$~symbols, is at least doubly transitive. If
$H$~is primitive and $G$~does not contain the alternating group, $m$~cannot
be less than~$\frac{2}{3}n$, and $G$~is $(n - m + 1)$-ply transitive.
\end{Theorem}

\begin{Corollary}
The order of a primitive group of degree~$n$
cannot exceed $\dfrac{n!}{2 · 3 \Dots p}$, where $2$, $3$,~\Dots\ $p$ are the distinct
primes which are less than~$\frac{2}{3}n$.
\end{Corollary}

If $q^{\alpha}$~is the highest power of a prime~$q$ that divides~$n!$, the
sub-groups of order~$q^{\alpha}$ of the symmetric group form a single
conjugate set, and each of them must contain circular substitutions
of order~$q$. Hence if $q < \frac{2}{3}n$, it follows by the theorem
that no primitive group of degree~$n$, other than the alternating
or the symmetric group, can contain a sub-group of order~$q^{\alpha}$;
and therefore $q^{\alpha - 1}$ is the highest power of~$q$ that can divide
the order of the group.

\Par{142.} The ratio of $2 · 3 \Dots p$ to~$n$ increases rapidly as $n$~increases,
and it is at once obvious that, when $n > 7$, this ratio is
greater than unity; hence for values of~$n$ greater than~$7$, the
symmetric group can have no primitive sub-group of order~$(n - 1)!$.

The order of the greatest imprimitive sub-group of the
symmetric group is $a!\left(\dfrac{n}{a}!\right)^{\alpha}$, where $a$~is the smallest factor of~$n$.
When $n > 4$, this is less than~$(n - 1)!$.

The order of the greatest intransitive sub-group of the
symmetric group, other than the sub-groups that keep one
symbol fixed, is $2!(n - 2)!$. This is always less than $(n - 1)!$.

Hence when $n > 7$, the only sub-groups of order~$(n - 1)!$ of
the symmetric group are the sub-groups which each keep one
symbol fixed; and these form a conjugate set of $n$~sub-groups.

When $n = 7$, a sub-group of order~$(n - 1)!$ must be intransitive,
and therefore the same result holds in this case; this also
is true when $n$~is $3$,~$4$, or~$5$.
%% -----File: 216.png---Folio 200-------

Lastly, when $n = 6$, there may, by the foregoing theorem,
\index{Symmetric group@\Topic{Symmetric group}!of degree~$n$ has a single set of conjugate sub-groups of order~$\Factorial{(n - 1)}$ except when $n$ is~$6$}%
\index{Symmetric group@\Topic{Symmetric group}!of degree~$6$ has $12$~simply isomorphic sub-groups of order~$\Factorial{5}$ which form two distinct conjugate sets}%
be primitive sub-groups of order~$5!$. That such sub-groups
actually exist may be verified at once by considering the
symmetric group of $5$~symbols. This group contains $6$~cyclical
sub-groups of order~$5$, and each of them is self-conjugate in a
sub-group of order~$20$. Hence, since the only self-conjugate
sub-group contained in the symmetric group of degree~$5$ is the
corresponding alternating group, the symmetric group of degree~$5$
can be expressed as a doubly transitive group of degree~$6$.
The symmetric group of degree~$6$ therefore contains a set of $6$
conjugate doubly transitive sub-groups of order~$5!$, which are
simply isomorphic with the intransitive sub-groups that each
keep one symbol fixed. Finally, if the $12$~sub-groups of order~$5!$,
which are thus accounted for, do not exhaust all the sub-groups
of this order, any other would have in common with
each of the~$12$ a sub-group of order~$20$; and therefore the
operations of order~$3$ contained in it would be distinct from
those in the previous~$12$. But these clearly contain all the
operations of order~$3$ of the symmetric group, and therefore
there can be no other sub-groups of order~$5!$. Hence:---

\begin{Theorem}[II.]
The symmetric group of degree~$n$ \($n \neq 6$\)
contains $n$~and only~$n$ sub-groups of order~$(n - 1)!$, which form a
single conjugate set. The symmetric group of degree~$6$ contains
$12$~sub-groups of order~$5!$, which are simply isomorphic with one
another and form two conjugate sets of $6$ each.
\end{Theorem}

\Par{143.} We shall now discuss certain further limitations on
the order of a primitive group of given degree. Though it will
be seen that these do not lead to general results, similar to that
given by Theorem~I, §~141, yet in many special cases they are
of considerable assistance in determining the possible existence
of groups of given orders and degrees.

We consider first a group~$G$ of order~$N$ and of prime degree~$p$.
If $G$~is not cyclical, it must contain substitutions which keep
only one symbol unchanged. For let $P$ be a substitution of~$G$ of
order~$p$. The only substitutions permutable with~$P$ are its own
powers (§~107); and the only substitutions permutable with~$\{P\}$
are substitutions which keep one symbol unchanged and are regular
%% -----File: 217.png---Folio 201-------
in the remaining $p - 1$ symbols (§~112)\footnotemark.
\footnotetext{It is shewn in §~112 that $\{P\}$~is permutable with a circular substitution of
\PageLabel{201}%
  $p - 1$ symbols, which leaves one symbol~$a_{1}$ unchanged. If there are other substitutions
  which leave $a_{1}$ unchanged and are permutable with~$\{P\}$, some such
  substitution will leave two symbols unchanged. This is clearly impossible.
  Hence the group of order~$p(p - 1)$ is the greatest group of the $p$~symbols in which
  $\{P\}$~is self-conjugate.}%
Now if the only substitutions
permutable with~$\{P\}$ are its own, then $\{P\}$~is one of $\dfrac{N}{p}$
conjugate sub-groups; and these contain $N\left(1 - \dfrac{1}{p}\right)$ substitutions
of order~$p$. In this case, $G$~would contain exactly exactly $\dfrac{N}{p}$
substitutions whose orders are not divisible by~$p$. But this is
clearly impossible, since $\dfrac{N}{p}$~is the order of a sub-group which
keeps one symbol unchanged, and there are $p$~such sub-groups.
Hence there must be substitutions in~$G$, other than those of~$\{P\}$,
which are permutable with~$\{P\}$; and each of these substitutions
keeps one symbol unchanged.

It follows from §~134 that $G$, if soluble, must contain a
\index{Primitive groups@\Topic{Primitive groups}!of prime degree}%
self-conjugate sub-group of order~$p$: therefore no group of
prime degree~$p$, which contains more than one sub-group of
order~$p$, can be soluble.

If $1 + kp$ is the number of sub-groups of order~$p$ contained
in~$G$, then
\[
N = p \frac{p - 1}{d} (1 + kp),
\]
where $d$ is a factor of $p - 1$; and a sub-group of order~$p$ is
transformed into itself by every substitution of a cyclical
sub-group of order~$\dfrac{p - 1}{d}$. When $d$~is odd, a substitution which
generates this cyclical sub-group is an odd substitution; and
$G$~then contains a self-conjugate sub-group of order~$\frac{1}{2}N$.

If both $p$ and $\frac{1}{2}(p - 1)$ are primes, the order of a group of
degree~$p$, which contains more than one sub-group of order~$p$,
must be divisible by~$\frac{1}{2}(p - 1)$. For if the order is not divisible
by~$\frac{1}{2}(p - 1)$, the order of the sub-group, within which a sub-group
of order~$p$ is self-conjugate, must be~$2p$. Now the
%% -----File: 218.png---Folio 202-------
substitutions of order~$2$ in this sub-group consist of $\frac{1}{2}(p - 1)$
transpositions, so that they are odd substitutions. The group
must therefore contain a self-conjugate sub-group in which
these operations of order~$2$ do not occur. In such a sub-group,
the only operations permutable with those of a sub-group of
order~$p$ are its own; and we have seen that no such group can
exist. The order of the group must therefore, as stated above,
be divisible by~$\frac{1}{2}(p - 1)$.

\Par{144.} Let $G$ be a primitive group of degree~$n$ and order~$N$;
and let $p$ be a prime, which is a factor of~$N$ but not of either $n$
or $n - 1$. Moreover, suppose that $n$~is congruent to $\nu, \pmod{p}$;
$\nu$~being less than~$p$. If $n < p^{2}$, and if $p^{\alpha}$ is the highest power of~$p$
which divides~$N$, the sub-groups of order~$p^{\alpha}$ must be Abelian
groups of type $(1, 1, \dots \text{ to $\alpha$ units})$. In fact, such a sub-group
must be intransitive, and, since $n < p^{2}$, the number of symbols
in each transitive system of the sub-group must be~$p$. In
any case the number of symbols left unchanged by a sub-group
of order~$p^{\alpha}$ is of the form $kp + \nu$.

Suppose now that, in a sub-group of order~$\dfrac{N}{n}$ which leaves
one symbol unchanged, a sub-group~$H$ of order~$p^{\alpha}$ is one of
$\dfrac{N}{p^{\alpha}mn}$ conjugate sub-groups. Then each of the $n$~sub-groups
that keep one symbol unchanged contains $\dfrac{N}{p^{\alpha}mn}$ sub-groups of
order~$p^{\alpha}$; and each sub-group of order~$p^{\alpha}$ belongs to $kp + \nu$
sub-groups that keep one symbol unchanged. Hence $G$~contains
$\dfrac{N}{p^{\alpha}m (kp + \nu)}$ sub-groups of order~$p^{\alpha}$, and any one of
them, say~$H$, is contained self-conjugately in a sub-group~$I$ of
order~$p^{\alpha}m (kp + \nu)$. This sub-group~$I$ must interchange transitively
among themselves the $kp + \nu$ symbols left unchanged by~$H$.
For let $a$~and~$b$ be any two of these symbols; and let $S$ be
an operation which changes $a$ into $b$ and transforms $H$ into~$H'$.
There must be an operation~$T$ which keeps $b$ unchanged and
transforms $H'$ into~$H$, since in the sub-group that keeps $b$
unchanged there is only one conjugate set of sub-groups of
order~$p^{\alpha}$. Then $ST$~changes $a$ into $b$ and transforms~$H$ into
%% -----File: 219.png---Folio 203-------
itself; and therefore $I$~contains substitutions which change $a$
into~$b$. Now it may happen that the existence of a sub-group
such as~$I$ requires that $G$ is either the alternating or the symmetric
group. When this is the case, we infer that the order
of a primitive group of degree~$n$, other than the alternating or
the symmetric group, cannot be divisible by~$p$.

\begin{Remark}
\Par{145.} As a simple example, we will shew that the order of
a group of degree~$19$ cannot be divisible by~$7$, unless it contains
the alternating group. It follows from Theorem~I, Corollary,
§~141, that the order of a group of degree~$19$, which does not
contain the alternating group, cannot be divisible by a power of~$7$
higher than the first, and that if the group contains a substitution
of order~$7$, the substitution must consist of two cycles of $7$~symbols
each. The sub-group of order~$7$ must therefore leave $5$~symbols
unchanged; hence, by §~144, it must be contained self-conjugately in
a sub-group whose order is divisible by~$5$. Now (§~83) a group of
order~$35$ is necessarily Abelian; so that the group of degree~$19$ must
contain a substitution of order~$5$ which is permutable with a substitution
of order~$7$. Such a substitution of order~$5$ must clearly
consist of a single cycle, and its presence in a group of degree~$19$
requires that the latter should contain the alternating group. It
follows that, if a group of degree~$19$ does not contain the
alternating group, its order is not divisible by~$7$.

As a second example, we will determine the possible forms for
the order of a group of degree~$13$, which does not contain the alternating
group. It follows, from Theorem~I, Corollary, §~141, that
the order of such a group must be of the form $2^{\alpha} · 3^{\beta} · 5^{\gamma} · 11^ {\delta} · 13$;
where $\alpha$,~$\beta$, $\gamma$,~$\delta$ do not exceed $9$,~$4$, $1$,~$1$ respectively.

Suppose, first, that $\gamma$~is unity, if possible. A substitution of
order~$5$ must consist of two cycles of $5$~symbols each; and a sub-group
of order~$5$ must therefore be self-conjugate in a sub-group of
order~$15$. There is then a substitution of order~$3$ which is permutable
with a substitution of order~$5$. Such a substitution must,
as in the last example, consist of a single cycle; and its existence
would imply that the group contains the alternating group. It
follows that, for the group as specified, $\gamma$~must be zero.

Suppose, next, that $\delta$~is unity, if possible. The group is then
(Theorem~I, §~141) triply transitive; and the order of the sub-group,
that keeps two symbols fixed and is transitive in the remaining~$11$,
is $2^{\alpha} · 3^{\beta} · 11$; and this sub-group must contain more than one sub-group
of order~$11$. We have seen in §~143 that no such group can
exist. Therefore $\delta$~must be zero.

The two smallest numbers of the form $2^{m} 3^{n}$ which are congruent
to unity, $\pmod{13}$, are $3^{3}$ and~$2^{4} 3^{2}$; and every number of this form,
%% -----File: 220.png---Folio 204-------
which is congruent to unity, $\pmod{13}$, can be written $(2^{4}3^{2})^{x}3^{3y}$.
Hence the order of every group of order~$13$, which contains no odd
substitution, must be of the form $(2^{4}3^{2})^{x} · 3^{3y} · z · 13$, where $z$~is $2$,~$3$
or~$6$. Since $3^{4}$~is the highest power of~$3$ that can divide the order
of the group, the only admissible values of $x$~and~$y$ are \Inum{(i)}~$x = 0$,
$y = 1$: \Inum{(ii)}~$x = 2$, $y = 0$: \Inum{(iii)}~$x = 1$, $y = 0$.

Suppose, first, that $x = 0$, $y = 1$. The order of the group is
$2 · 3^{3} · 13$, $3^{4} · 13$, or $2 · 3^{4} · 13$. There must be $13$~sub-groups of order~$3^{4}$
(or~$3^{3}$), and since $13$~is not congruent to unity, $\pmod{9}$, there must
be sub-groups of order~$3^{3}$ (or~$3^{2}$) common to some two sub-groups of
order~$3^{4}$ (or~$3^{3}$). Such a sub-group must be self-conjugate (Theorem~III,
§~80). This case therefore cannot occur.

Next, suppose that $x = 2$, $y = 0$. Then $z$~must be~$2$, and the
order of the group is $2^{9} · 3^{4} · 13$. Now it is easy to verify that $2^{9} · 3^{4}$
is not a possible order either for an intransitive or for an imprimitive
group of degree~$12$. The order of a sub-group of the group of degree~$12$
which keeps one symbol fixed is $2^{7} · 3^{3}$. This sub-group can have
no substitution consisting of a single cycle of $3$~symbols, since no
such substitution can occur in the original group. Hence it must
permute the $11$ symbols in two transitive sets of $9$~and $2$~symbols
respectively. It must therefore contain a self-conjugate sub-group
of order $2^{6} · 3^{3}$ which keeps $3$~of the $12$~symbols unchanged; and this
sub-group must occur self-conjugately in $3$~of the $12$~sub-groups which
keep one symbol unchanged. This however makes the group of
degree~$12$ imprimitive, contrary to supposition. Hence this case
cannot occur.

Finally, then, the only possible values of $x$~and~$y$ are $x = 1$,
$y = 0$. The order of a group of degree~$13$, which has more than
one sub-group of order~$13$ and no odd substitutions, is $2^{5} · 3^{2} · 13$,
$2^{4} · 3^{3} · 13$, or $2^{5} · 3^{3} · 13$. The order of a group of order~$13$ with odd
substitutions will be twice one of the preceding three numbers.

A further and much more detailed examination would be necessary
to determine whether groups of degree~$13$ correspond to any or
all of these orders. We shall see in Chapter~XIV that there is a
group of degree~$13$ and order~$2^{4} · 3^{3} · 13$; and M.~Jordan\footnote
  {\textit{Comptes Rendus}, t.~\VolNo{LXXV} (1872), p.~1757.}
\index{Jordan@\Topic{Jordan}, quoted}%
states
that this is the only group of degree~$13$ which contains more than
one sub-group of order~$13$.
\end{Remark}

\begin{Ex} If $n$ ($> 3$) and $2n + 1$ are primes, shew that there is no
triply transitive group of degree~$2n + 3$ which does not contain the
alternating group.
\end{Ex}

\Par{146.} As a further illustration, and for the actual value of
the results themselves, we proceed to determine all types of
primitive groups for degrees not exceeding~$8$.
%% -----File: 221.png---Folio 205-------

\Class{\Inum{(i)}}{$n = 3$.}
\index{Primitive groups@\Topic{Primitive groups}!of degrees $3$,~$4$ and~$5$}%

The symmetric group of $3$~symbols has a single sub-group,
viz.\ the alternating group. Both these groups are necessarily
primitive.

\Class{\Inum{(ii)}}{$n = 4$.}

Since a group whose order is the power of a prime cannot
(§~124) be represented in primitive form, a primitive group of
degree~$4$ must contain $3$ as a factor of its order. Hence the
only primitive groups of degree~$4$ are the symmetric and the
alternating groups.

\Class{\Inum{(iii)}}{$n = 5$.}

Since $5$~is a prime, every group of degree~$5$ is a primitive
group. The symmetric group of degree~$5$ contains $6$~cyclical
sub-groups of order~$5$; therefore, by Sylow's theorem, every
group of degree~$5$ must contain either $1$~or $6$ sub-groups of
order~$5$. Since the alternating group is simple, every sub-group
that contains $6$~sub-groups of order~$5$ must contain the
alternating group. Hence, besides the alternating and the
symmetric groups, we have only sub-groups which contain a
sub-group of order~$5$ self-conjugately. In such a group, an
operation of order~$5$ can be permutable with its own powers
only. Hence (§~112) the only sub-groups of the type in question
other than cyclical sub-groups, are groups of orders $20$ and~$10$.
These are defined by
\begin{align*}
&\{(12345),\ (2354)\}, \\
\lintertext{and}
&\{(12345),\ (25) (34)\}.
\end{align*}

\Class{\Inum{(iv)}}{$n = 6$.}
\index{Primitive groups@\Topic{Primitive groups}!of degree~$6$}%

If the order of a primitive group of degree~$6$ is not divisible
by~$5$, the order must (§~141) be equal to or be a factor of~$2^{3} · 3$.
The order of a sub-group that keeps one symbol fixed is equal to
or is a factor of~$2^{2}$. Hence the sub-group must keep two symbols
fixed, and therefore (§~128) the group cannot be primitive. Hence
the order of every primitive group of degree~$6$ is divisible by~$5$,
and every such group is at least doubly transitive. The
symmetric group contains $36$~sub-groups of order~$5$; and hence,
since no transitive group of degree~$6$ can contain a self-conjugate
sub-group of order~$5$, every primitive group of degree~$6$,
%% -----File: 222.png---Folio 206-------
\index{Primitive groups@\Topic{Primitive groups}!of degree~$6$}%
which does not contain the alternating group, must have $6$~sub-groups
of order~$5$.

If $G$ is such a group, the sub-group of $G$ that keeps one
symbol fixed is a transitive group of degree~$5$ which has a
self-conjugate sub-group of order~$5$. If this transitive group
of degree~$5$ were cyclical, every operation of the doubly transitive
group~$G$ of order~$30$ would displace all or all but one of the
symbols. Since $6$~is not the power of a prime, this is impossible
(§~105). Hence the sub-group of~$G$ which keeps one symbol
fixed must be of one of the two types given above; and the order
of~$G$ must be $120$ or~$60$. Now we have seen, in §~142, that the
symmetric group of degree~$6$ has a single conjugate set of primitive
sub-groups of order~$120$ and a single set of order~$60$. Hence
there is a single type of primitive group of degree~$6$, corresponding
to each of the orders $120$ and~$60$. These are defined by
\begin{align*}
&\{(126)(354),\ (12345),\ (2354)\}, \\
\lintertext{and}
&\{(126)(354),\ (12345),\ (25)(34)\}:
\end{align*}
where the last two substitutions in each case generate a sub-group
that keeps one symbol unchanged.

\Class{\Inum{(v)}}{$n = 7$.}
\index{Primitive groups@\Topic{Primitive groups}!of degree~$7$|(}%

Every transitive group of degree~$7$ is primitive; and if it
does not contain the alternating group, its order must (§~141) be
equal to or be a factor of $7 · 6 · 5 · 4$. A cyclical sub-group of order~$7$
must (footnote, \PageRef{p.}{201}), in a group of degree~$7$ that contains
more than one such sub-group, be self-conjugate in a group of
order $21$ or~$42$. Now neither $20$ nor $40$ is congruent to unity,
$\pmod{7}$; and therefore $5$~cannot be a factor of the order of such
a group. Hence the order of a transitive group of degree~$7$,
that does not contain the alternating group, is equal to or is a
factor of $7 · 6 · 4$. But $8$~is the only factor of $7 · 6 · 4$ which is
congruent to unity, $\pmod{7}$; and therefore, if the group contains
more than one sub-group of order~$7$, its order must be equal to
$7 · 6 · 4$ and it must contain $8$~sub-groups of order~$7$.

Such a group must be doubly transitive; for if a sub-group
of order~$24$ interchanges the symbols in two intransitive
systems, it is easily shewn that the group would contain
substitutions displacing three symbols only, and therefore that
it would contain the alternating group. A sub-group of degree~$24$,
%% -----File: 223.png---Folio 207-------
transitive in $6$~symbols, must contain $4$~sub-groups of order~$3$.
For if it contained only one, it would necessarily have
circular substitutions of $6$~symbols, and the group of order
$7 · 6 · 4$ would have a self-conjugate sub-group of order $7 · 6 · 2$;
which is not the case. Hence the sub-groups of order~$24$ must
be simply isomorphic with the symmetric group of $4$~symbols.

The actual construction of the group is now reduced to
a limited number of trials. A group of degree~$6$, simply
isomorphic with the symmetric group of $4$~symbols, and containing
no odd substitutions, may always be represented in the
form
\[
\{(234)(567),\ (2763)(45)\};
\]
and we have to find a circular substitution of the seven symbols
$1$,~$2$, $3$, $4$, $5$, $6$,~$7$ such that the group generated by it shall be
permutable with this group. Moreover since, in the required
group, every operation of order~$3$ transforms some operation of
order~$7$ into its square, we may assume without loss of generality
that the circular substitution of order~$7$ contains the sequence
\dots$12$\dots\ and is transformed into its own square by~$(234)(567)$.
There are only three circular substitutions satisfying these
conditions, viz.\
\begin{align*}
&(1235476), \\
&(1236457), \\
\lintertext{and}
&(1237465).
\end{align*}

It appears on trial that the group generated by the first of
these is not permutable with the sub-group of order~$24$, while
the groups generated by the other two are. There are therefore
just two groups of order $7 · 6 · 4$ which contain the given group
of order~$24$. Now in the symmetric group of $7$~symbols, a
sub-group of order $7 · 6 · 4$ must, from the foregoing discussion,
be one of a set of $30$ conjugate sub-groups. These all enter in
the alternating group; and therefore, in that group, they must
form two sets of $15$ conjugate sub-groups each. Each of these
contains $7$~sub-groups of the type
\[
\{(234)(567),\ (2763)(45)\};
\]
and the alternating group contains a conjugate set of $105$ such
such sub-groups. Hence each sub-group of this set will enter
in two, and only in two, sub-groups of the alternating group of
%% -----File: 224.png---Folio 208-------
order $7 · 6 · 4$; and in the symmetric group these two sub-groups
are conjugate. Finally then, the sub-groups of order $7 · 6 · 4$
form a single conjugate set in the symmetric group. They are
defined by
\[
\{(1236457),\ (234)(567),\ (2763)(45)\},
\]
the two latter substitutions giving a sub-group that keeps
one symbol fixed.

These groups are simple; for since they are expressed as
transitive groups of degree~$7$, there can be no self-conjugate
sub-group whose order divides~$24$, while it is evident that a
self-conjugate sub-group that contains an operation of order~$7$
must coincide with the group itself. Also since there are $8$~sub-groups
of order~$7$, these groups can be expressed as doubly
transitive groups of eight symbols.

A group of degree~$7$, which has only one sub-group of order~$7$,
must either be cyclical or be contained in the group of order
$7 · 6$ given by §~112. Such groups are defined by
\begin{align*}
&\{(1234567),\ (243756)\}, \\
\lintertext{or}
&\{(1234567),\ (235)(476)\}, \\
\lintertext{or}
&\{(1234567),\ (27)(45)(36)\}.
\end{align*}

The simple group of order~$168$, which here occurs as a
\PageLabel{208}%
transitive group of degree~$7$, is the only simple group of that
order. For, if possible, let there be a simple group~$G$ of order~$168$
and of a distinct type from the above. It certainly cannot
be expressed as a group of degree~$7$; and therefore it must
have $21$~sub-groups of order~$8$. If two of these sub-groups
have a common sub-group of order~$4$, it must be contained
self-conjugately (§~80) in a sub-group of order $24$ or~$56$; and
this is inconsistent with the suppositions made. If on the
other hand, $2$~is the order of the greatest sub-group common
to two sub-groups of order~$8$, such a common sub-group of order~$2$
must, on the suppositions made, be self-conjugate in a sub-group
of order~$12$. But a group of order~$12$, which has a self-conjugate
operation of order~$2$, must have a self-conjugate
sub-group of order~$3$; and therefore $G$ would only contain $7$~sub-groups
of order~$3$, and could be expressed as a group of
degree~$7$; contrary to supposition. No other supposition is
possible with regard to the sub-groups of order~$8$, since $21$~is
\index{Primitive groups@\Topic{Primitive groups}!of degree~$7$|)}%
%% -----File: 225.png---Folio 209-------
not congruent to unity, $\pmod{8}$. Hence, finally, there is no
simple group of order~$168$ distinct from the group of degree~$7$.

\Class{\Inum{(vi)}}{$n = 8$.}
\index{Primitive groups@\Topic{Primitive groups}!of degree~$8$|(}%

The order of a primitive group of degree~$8$, which does not
contain the alternating group, cannot (§~141) be divisible by~$5$.
Suppose, if possible, that the order of such a group is $2^{\alpha + 3} · 3$
($\alpha = 0, 1, 2, 3$). A substitution of order~$3$ must consist of two
cycles; and therefore the sub-group of order~$\DPchg{2^{\alpha}3}{2^{\alpha} · 3}$, which keeps
one symbol fixed, must interchange the others in two intransitive
systems of $3$ and $4$ respectively. In this sub-group, a
sub-group of order~$3$ must be one of four conjugate sub-groups,
and therefore $\alpha$~is either $2$ or~$3$. Now a group of order~$2^{5} · 3$
or $2^{6} · 3$ is soluble, as is seen at once by considering the sub-groups
of order $2^{5}$ or~$2^{6}$. Hence a primitive group of order~$2^{5} · 3$
or $2^{6} · 3$ must contain a transitive self-conjugate sub-group of
order~$8$, whose operations are all of order~$2$.

If $7$ is a factor of the order of the group, the group must be
doubly transitive; and from the case of $n = 7$, it follows that
the possible orders are $8 · 7$, $8 · 7 · 2$, $8 · 7 · 3$, $8 · 7 · 6$, and
$8 · 7 · 6 · 4$. Moreover, for the orders $8 · 7 · 2$ and $8 · 7 · 6$, the
group contains odd substitutions and therefore it contains self-conjugate
sub-groups of order $8 · 7$ and $8 · 7 · 3$ respectively.

A simple group of order~$8 · 7 · 3$ is necessarily identical in
type with the group of this order determined on \PageRef{p.}{208};
and a group of order~$8 · 7 · 3$, which is not simple, is certainly
soluble. Hence a composite group of order~$8 · 7 · 3$, and
a group of order $8 · 7 · 6$ which does not contain a simple
sub-group of order~$8 · 7 · 3$, must both, if expressible as primitive
groups of degree~$8$, contain transitive self-conjugate
sub-groups of order~$8$ whose operations are all of order~$2$.
With the possible exception then of groups of order $8 · 7 · 6 · 4$,
the only primitive groups of degree~$8$, which do not contain a
self-conjugate sub-group of order~$8$, are the simple group of
order~$8 · 7 · 3$ and any group of $8 · 7 · 6$ which contains this self-conjugately.
We have seen that the simple group of order
$8 · 7 · 3$ contains a single set of $8$ conjugate sub-groups of order~$21$,
and therefore it can be expressed in one form only as a
group of degree~$8$. A group of degree~$8$ and order~$8 · 7 · 6$,
%% -----File: 226.png---Folio 210-------
which contains this self-conjugately, can occur only in one
form, if at all; for, if it exists, it must be triply transitive, and
it must be given by combining the simple group with any
operation of order~$2$ which transforms one of its operations of
order~$7$ into its own inverse. That such a group does exist has
been shewn in §~113. These two groups are actually given by
\begin{align*}
&\{(15642378),\ (1234567),\ (243756)\}, \\
\lintertext{and}
&\{(1627) (5438),\ (1234567),\ (235) (476)\};
\end{align*}
where in each case the last two substitutions give a sub-group
that keeps one symbol fixed.

To avoid unnecessary prolixity we shall, in dealing with the
primitive groups of degree~$8$, which contain a transitive self-conjugate
sub-group of order~$8$ whose operations are all of
order~$2$, anticipate some of the results of the next Chapter. It
will there be shewn that, if $G$~is an Abelian group of order~$N$,
and $L$~a group of isomorphisms (§~156) of~$G$, a group of degree~$N$
may be formed which has a transitive self-conjugate sub-group
simply isomorphic with $G$, while at the same time the sub-group
that keeps one symbol fixed is isomorphic with~$L$ (§~158).
The group of isomorphisms of a group of order~$8$, whose
operations are all of order~$2$, will be shewn in Chapter~XIV to
be identical with the simple group of order~$168$. This group
has a single set of conjugate sub-groups of each of the orders $7$
and $21$, but no sub-group of order $14$ or~$42$. When expressed
as a group of degree~$7$, it has a single set of conjugate sub-groups
of order~$12$ (or~$24$) which leave no symbols unchanged.
There are therefore primitive groups of degree~$8$ containing
transitive self-conjugate sub-groups of order~$8$ corresponding
to each of the orders $8 · 7$, $8 · 7 · 3$, $2^{5} · 3$, $2^{6} · 3$, and $8 · 7 · 6 · 4$;
and in each case there is a single type of such group.

It remains to determine whether there can be any type
of group, of degree~$8$ and order $8 · 7 · 6 · 4$, other than that just
obtained. Such a group must be one of $15$~conjugate sub-groups
in the alternating group of degree~$8$, and can therefore
itself be expressed as a group of degree~$14$. Since it
certainly cannot be expressed as a group of degree~$7$, the
group of degree~$14$ must be transitive. The order of the sub-group,
in this form, that keeps one symbol fixed is~$2^{5} · 3$. If
%% -----File: 227.png---Folio 211-------
this keeps only one symbol unchanged, it must interchange the
remaining symbols in four intransitive systems of $3$,~$3$, $3$ and~$4$
respectively, since a substitution of order~$3$ must clearly consist
of $4$~cycles. A group of order~$2^{5} · 3$ cannot however be so
expressed; and therefore the sub-group that keeps one symbol
fixed must keep two fixed. The group of degree~$14$ is therefore
imprimitive, and the group must contain a sub-group of order~$2^{6} · 3$.
Moreover, since the group cannot be expressed as a
transitive group of degree~$7$, this sub-group of order~$2^{6} · 3$ must
contain (§~123) a sub-group which is self-conjugate in the group
itself. The order of this sub-group must be a power of~$2$;
since the group is primitive, it cannot be less than~$2^{3}$. On the
other hand, the order cannot be greater than~$2^{3}$ since the group
contains a simple sub-group of order~$7 · 6 · 4$. Hence finally,
there is no type of primitive group of degree~$8$ and order
$8 · 7 · 6 · 4$ other than that already obtained.

There is no difficulty now in actually constructing the
primitive groups of degree~$8$ which have a self-conjugate sub-group
of order~$8$. They are all contained in the group of order
$8 · 7 · 6 · 4$; and this may be constructed from a sub-group
keeping one symbol fixed, which is given on \PageRef{p.}{208}, by the
method of §~109. It will thus be found that the group in
question is given by
\[
\{(81)(26)(37)(45),\ (1236457),\ (234)(567),\ (2763)(45)\};
\]
while the groups of orders $8 · 7 · 3$ and $8 · 7$ are given by omitting
respectively the last and the two last of the four generating
operations.

The construction of the two remaining groups, of order~$2^{5} · 3$
and $2^{6} · 3$, is left as an exercise for the reader.

It may be noticed that it has been shewn incidentally, in
discussing above the possibility of a second type of group of
degree~$8$ and order $8 · 7 · 6 · 4$, that the alternating group of
degree~$8$ can be expressed as a doubly transitive group of
degree~$15$.

It may similarly be shewn that the alternating group of
degree~$7$ can be expressed as a doubly transitive group of
degree~$15$, and the alternating group of degree~$6$ as a simply
transitive and primitive group of degree~$15$.
\index{Primitive groups@\Topic{Primitive groups}!of degree~$8$|)}%
%% -----File: 228.png---Folio 212-------

\Par{147.} We have seen in §~105 that a doubly transitive group,
\index{Doubly transitive groups@\Topic{Doubly transitive groups}!the sub-groups of, which keep two symbols fixed}%
of degree~$n$ and order~$n (n - 1)$, can exist only when $n$~is the
power of a prime. For such a group, the identical operation is
the only one which keeps more than one symbol unchanged.
We shall now go on to consider the sub-groups of a doubly
transitive group, of degree~$n$ and order~$n(n - 1)m$, which keep
two symbols fixed. The order of any such sub-group is~$m$;
since the group contains operations changing any two symbols
into any other two, the sub-groups which keep two symbols
fixed must form a single conjugate set.

Suppose first that the sub-group, which keeps two symbols
unchanged, displaces all the other symbols. The sub-group that
keeps $a$ and $b$ unchanged cannot then be identical with that
which keeps $c$ and $d$ unchanged, unless the symbols $c$ and $d$ are
the same pair as $a$ and~$b$. Since there are $\frac{1}{2}n(n - 1)$ pairs
of $n$~symbols, the conjugate set contains $\frac{1}{2}n(n - 1)$ sub-groups;
and each sub-group of order~$m$ keeping two symbols fixed
must be self-conjugate in a sub-group of order~$2m$, which
consists of the operations of the sub-group of order~$m$ and of
those operations interchanging the two symbols that the sub-group
of order~$m$ keeps fixed.

Suppose next that all the operations of a sub-group~$H$,
which keeps two symbols fixed, keep $x$~symbols fixed, while
none of the remaining $n - x$ symbols are unchanged by all the
operations of~$H$. From $x$~symbols $\frac{1}{2}x(x - 1)$ pairs can be
formed, and therefore the sub-group that keeps one pair unchanged
must keep $\frac{1}{2}x(x - 1)$ pairs unchanged. In this case,
the conjugate set contains $\dfrac{n(n - 1)}{x(x - 1)}$ distinct sub-groups of order~$m$,
and $H$~is therefore self-conjugate in a group~$K$ of order
$x(x - 1)m$. The operations of this sub-group which do not
belong to~$H$ interchange among themselves the $x$~symbols that
are left unchanged by~$H$. Now since the group itself is doubly
transitive, there must be operations which change any two of
these $x$~symbols into any other two; and any such operation
being permutable with~$H$ must belong to~$K$. Hence if we
consider the effect of~$K$ on the $x$~symbols only which are left
unchanged by~$H$, $K$~reduces to a doubly transitive group of
%% -----File: 229.png---Folio 213-------
degree~$x$ and order~$x(x - 1)$. It follows that $x$ must be a prime
or the power of a prime.

\begin{Remark}
\Par{148.} The preceding paragraph suggests the combinatorial
problem of forming from $n$~distinct symbols $\dfrac{n(n - 1)}{x(x - 1)}$ sets of $x$~symbols,
such that every pair of symbols occurs in one set of~$x$ and
no pair occurs in more than one.

There is one class of cases in which a solution of this problem is
given immediately by the theory of Abelian groups. Let $G$ be an
Abelian group of order~$p^{m}$, where $p$~is a prime, and type $(1, 1, \Dots\DPchg{\,}{,}
\text{to $m$ units})$. We have seen, in §~49, that $G$ has $\dfrac{p^{m} - 1}{p - 1}$ sub-groups of
order~$p$, and $\dfrac{p^{m} - 1 · p^{m - 1} - 1}{p - 1 · p^{2} - 1}$ sub-groups of order~$p^{2}$. Now any pair
of sub-groups of order~$p$ generates a sub-group of order~$p^{2}$, and therefore
every pair of sub-groups of order~$p$ occurs in one and only one
sub-group of order~$p^{2}$. Moreover, every sub-group of order~$p^{2}$ contains
$p + 1$ sub-groups of order~$p$. When $p$~is a prime and $m$~any integer,
it is therefore always possible to form from $\dfrac{p^{m} - 1}{p - 1}$ symbols
$\dfrac{p^{m} - 1 · p^{m - 1} - 1}{p - 1 · p^{2} - 1}$ sets of $p + 1$ symbols each, such that every pair of
the symbols occurs in one set of $p + 1$ and no pair occurs in more
than one set.

Supposing that, for given values of $n$ and~$x$, such a distribution
is possible, it is still of course an open question as to whether there
is a doubly transitive substitution group of the $n$~symbols, such that
every substitution which keeps any two symbols unchanged keeps
also unchanged the whole set of~$x$ in which they occur. When $x$~is
greater than~$3$, the question as to the existence of such groups is
one which still remains to be investigated. There is however an
important class of groups, to be considered later (Chapter~XIV),
that possess a closely analogous property. These groups are doubly
transitive; and from the $n$~symbols upon which they operate, we can
form $\dfrac{n(n - 1)}{x(x - 1)}$ sets of~$x$, that are interchanged transitively by the
substitutions of the group: the sets being such that every pair
occurs in one set and no pair in more than one set.

If $n(n - 1)m$ is the order of such a group, and if $H$~is a
sub-group of order~$m$ which keeps a given pair fixed, then $H$~must
interchange among themselves the remaining $x - 2$ symbols of that
set of~$x$ which contains the pair kept unchanged by~$H$. $H$~contains,
as a self-conjugate sub-group, the group~$h$ which leaves every symbol
of the set of~$x$ unchanged; and if $m''$~is the order of this sub-group,
while $m = m'm''$, then $m'$~is the order of the group to which $H$
%% -----File: 230.png---Folio 214-------
reduces when we consider its effect only on the $x - 2$ symbols. Now
$h$~is self-conjugate in the group~$K$ that interchanges all the symbols
of the set of~$x$ among themselves. But since the original group is
doubly transitive, it must contain substitutions which change any
two of the set of~$x$ into any other two, and every such substitution
must belong to~$K$. Hence $K$~must be doubly transitive in the $x$~symbols,
and therefore finally the order of the group, to which $K$~reduces
when we consider its effect on the $x$~symbols only, is
$x(x - 1)m'$. Since the order of~$h$, which keeps unchanged each of
the $x$~symbols is~$m''$, the order of~$K$ is~$x(x - 1)m$.

\Par{149.} When $x = 3$, $n$~must be of the form $6k + 1$ or $6k + 3$, since
\index{Doubly transitive groups@\Topic{Doubly transitive groups}!with a complete set of triplets}%
otherwise $\dfrac{n(n - 1)}{x(x - 1)}$ would not be an integer. The substitutions of
a doubly transitive group of degree~$n$, which possesses a complete
set of $\frac{1}{6}n(n - 1)$ triplets, must be such that every substitution which
leaves two given symbols unchanged also leaves a third definite
symbol unchanged.

The smallest possible value of~$n$ is~$7$; and the group of Ex.~2,
§~35, which is one of the groups obtained in §~146, satisfies all the
conditions. In fact, the group is clearly a doubly transitive group
of degree~$7$; and since its order is $7 · 6 · 4$, the order of the sub-group
which keeps two symbols fixed is~$4$. Now in the sub-group
\[
\{(267) (345),\ (23) (47)\},
\]
which keeps one symbol fixed, the only operations that keep $1$~and~$2$
unchanged are
\[
(35) (67),\ (36) (57),\ (37) (56).
\]

These with identity form a sub-group of order~$4$, which must
therefore be the sub-group that keeps three symbols fixed. The
complete set of triplets in this case is
\[
124,\ 137,\ 156,\ 235,\ 267,\ 346,\ 457.
\]

The next smallest value of~$n$ is~$9$, and in this case again, a group
with the required properties exists.
\end{Remark}

\begin{Ex} Shew that the group
\[
\{(26973854),\ (456) (798)\}
\]
is an imprimitive group of order~$48$, each imprimitive system
containing two symbols; and that the sub-group, which keeps the
symbols of one imprimitive system unchanged, is isomorphic with
the symmetric group of three symbols. Prove that this group is
permutable with
\[
\{(123) (456) (789),\ (147) (258) (369)\},
\]
and thence that
\[
\{(123) (456) (789),\ (26973854),\ (456) (798)\},
\]
is a doubly transitive group of degree~$9$, which possesses a complete
set of $12$~triplets.
\end{Ex}
%% -----File: 231.png---Folio 215-------

\begin{Remark}
The reader is not to infer from the examples given that, when
$n$~is of the form $6k + 1$ or $6k + 3$, there is always a doubly transitive
group of degree~$n$ which possesses a complete set of triplets. It is
a good exercise to verify that there is no such group when $n$~is~$13$.

The case $n = 13$, $x = 4$ is the simplest case that can occur of the
division of $n$~symbols into sets of~$x$ in the manner of §~148 when $x$~is
greater than~$3$. We shall see in Chapter~XIV that there is a
doubly transitive group of degree~$13$ such that from the $13$~symbols
permuted by the group a complete set of $13$ quartets can be
formed, which are themselves permuted by the operations of the
group. Of the operations forming a sub-group that keeps two given
symbols fixed, half will keep fixed the two other symbols, which form
a quartet with the two given symbols, and half will permute them.

On the question of the independent formation of a complete set
of triplets of $n$~symbols, and in certain cases of the group of
degree~$n$ which interchanges the triplets among themselves, reference
may be made to the memoirs mentioned in the subjoined footnote\footnotemark.
\end{Remark}
\index{Heffter@\Topic{Heffter}, quoted}%
\index{Moore@\Topic{Moore}, quoted}%
\index{Netto@\Topic{Netto}, quoted}%
\footnotetext{Netto: ``Substitutionentheorie,'' pp.~220--235; ``Zur Theorie der Tripelsysteme,''
  \textit{Math.\ Ann.}\ Vol.~\VolNo{XLII}, (1892), pp.~143--152. Moore: ``Concerning
  triple systems,'' \textit{Math.\ Ann.}\ Vol.~\VolNo{XLIII}, (1893), pp.~271--285. Heffter: ``Ueber
  Tripelaysteme,'' \textit{Math.\ Ann.}, Vol.~\VolNo{XLIX}, (1897), pp.~101--112.}

\Par{150.} We shall conclude the present Chapter with some
applications of substitution groups, which enable us to complete
and extend certain earlier results.

We have seen in §~107 that the substitutions of $n$~symbols,
\index{Substitutions@\Topic{Substitutions} which are permutable!with a given substitution}%
which are permutable with each of the substitutions of a
regular substitution group~$G$ of order~$n$ of the same $n$ symbols,
form another regular substitution group of order~$n$; and that,
if $G$~is Abelian, the latter group coincides with~$G$. Hence the
only substitutions of $n$~symbols, which are permutable with a
circular substitution of the $n$~symbols, are the powers of the
circular substitution.

Let now $S$ be a regular substitution of order~$m$, in $mn$~symbols.
It must permute the symbols in $n$~cycles of $m$~symbols
each; and so we may take
\[
S = (a_{11} a_{12} \Dots a_{1m})
    (a_{21} a_{22} \Dots a_{2m}) \Dots
    (a_{n1} a_{n2} \Dots a_{nm}).
\]

If $T$ is permutable with~$S$, and if it changes $a_{rp}$ into~$a_{rq}$, it
clearly must permute the $m$~symbols
\[
a_{r1},\ a_{r2},\ \Dots,\ a_{rm}
\]
%% -----File: 232.png---Folio 216-------
among themselves; and therefore, so far as regards its effect on
these $m$~symbols, $T$~must be a power of
\[
(a_{r1} a_{r2}\Dots a_{rm}).
\]

Again, if $T$~changes $a_{rp}$ into~$a_{sq}$, it must change the set
\[
a_{r1},\ a_{r2},\ \Dots,\ a_{rm},
\]
into the set
\[
a_{s1},\ a_{s2},\ \Dots,\ a_{sm};
\]
as otherwise it would not be permutable with~$S$.

Now the totality of the substitutions of the $mn$~symbols,
\index{Substitutions@\Topic{Substitutions} which are permutable!with a given substitution}%
which are permutable with~$S$, form a group~$G_{S}$. This group
must, from the properties of~$T$ just stated, be imprimitive,
interchanging the symbols in $n$ imprimitive systems of $m$
symbols each; and the symbols in any cycle of~$S$ will form an
imprimitive system. Moreover, the self-conjugate sub-group~$H_{S}$
of this group, which permutes the symbols of each system
among themselves, is the group of order~$m^{n}$ generated by
\[
(a_{11} a_{12}\Dots a_{1m}),\
(a_{21} a_{22}\Dots a_{2m}),\ \Dots,\
(a_{n1} a_{n2}\Dots a_{nm}).
\]

In fact, every substitution of this group is clearly permutable
with~$S$; and conversely, every substitution of the $mn$~symbols,
which does not permute the systems, must belong to
this group.

Now $\dfrac{G_{S}}{H_{S}}$ is capable of representation as a group of degree~$n$,
for none of its operations changes every one of the $n$~systems
into itself. Hence $n!$~is the greatest possible order of~$\dfrac{G_{S}}{H_{S}}$. On
the other hand, every operation of the group, generated by
\[
(a_{11} a_{21}\Dots a_{n1})
(a_{12} a_{22}\Dots a_{n2})\Dots
(a_{1m} a_{2m}\dots a_{nm})
\]
and
\[
(a_{11} a_{21}) (a_{12} a_{22}) \Dots (a_{1m} a_{2m}),
\]
is clearly permutable with~$S$; and this group, being simply
isomorphic with the group
\[
\{(a_{1} a_{2}\Dots a_{n}),\ (a_{1} a_{2})\},
\]
\ie\ with the symmetric group of $n$~symbols, is of order~$n!$.

Hence, finally, the order of~$G_{S}$ is~$m^{n} · n!$; and $G_{S}$~is generated
by
\begin{gather*}
(a_{11} a_{12}\Dots a_{n1})
(a_{12} a_{22}\Dots a_{n2}) \Dots
(a_{1m} a_{2m}\Dots a_{nm}), \\
(a_{11} a_{21})(a_{12} a_{22})\Dots (a_{1m} a_{2m}),
\end{gather*}
and
\[
(a_{11} a_{12}\Dots a_{1m}).
\]
%% -----File: 233.png---Folio 217-------

\Par{151.} Let $h_{r}$ be a regular substitution group of order~$m$ in
the $m$~symbols
\[
a_{r1},\ a_{r2},\ \Dots,\ a_{rm},
\]
and let $S_{rt}$ be one of its substitutions. Then if for $r$ we write
in turn $1$, $2$,~\Dots, $n$, and if for each value of~$t$ from $1$ to~$m$ we
form the substitution
\[
S_{1t} S_{2t} \Dots S_{nt},
\]
the set of $m$~substitutions so formed constitute an intransitive
group~$H$ in the $mn$~symbols, simply isomorphic with~$h_{r}$.

The method of §~150 can be applied directly to determine
\index{Substitutions@\Topic{Substitutions} which are permutable!with every substitution of a given group}%
the group~$G_{H}$ of degree~$mn$, each of whose substitutions are\DPnote{[** [sic]]}
permutable with every substitution of~$H$. The order of this
group is~$m^{n} · n!$; and it can be generated by
\begin{gather*}
(a_{11} a_{21} \Dots a_{n1})
(a_{12} a_{22} \Dots a_{n2}) \Dots
(a_{1m} a_{2m} \Dots a_{nm}), \\
(a_{11} a_{21}) (a_{12} a_{22}) \Dots (a_{1m} a_{2m}),
\end{gather*}
and
\[
h_{1}';
\]
where $h_{1}'$~is the regular group in the symbols
\[
a_{11},\ a_{12},\ \Dots,\ a_{1m},
\]
each of whose substitutions is permutable with every substitution
of~$h_{1}$.

This group will contain~$H$ if, and only if, $H$~is an Abelian
group. Moreover, the only self-conjugate substitutions of~$G_{H}$
are the substitutions of~$H$ contained in it. For if $G_{H}$
contained other self-conjugate substitutions $S_{1}$, $S_{2}$,~\dots, every
operation of~$G_{H}$ would be permutable with every operation of
the group $\{H,\ S_{1},\ S_{2},\ \dots\}$. Now $G_{H}$~is transitive, so that
$S_{1}$, $S_{2}$,~\dots\ must displace all the symbols; and therefore
$\{H,\ S_{1},\ S_{2},\ \dots\}$ has all its substitutions regular in the $mn$~symbols.
If its order is~$mn_{1}$ where $n = n_{1}n_{2}$, the order of the
group formed of all the substitutions of $mn$~symbols, which are
permutable with each of its operations, is $(mn_{1})^{n_{2}} · n_{2}!$; and this
number is less than~$m^{n} · n!$. Thus the supposition, that $G_{H}$~has
self-conjugate operations other than the operations of~$H$ which
it contains, leads to an impossibility.

By means of this and the preceding section, the reader will
have no difficulty in forming the group of $n$~symbols, which is
permutable with every operation of any given group in the $n$~symbols.
%% -----File: 234.png---Folio 218-------

\Par{152.} If a group, whose order is a power of a prime~$p$, be
\index{Substitution groups@\Topic{Substitution groups} whose orders are powers of primes}%
expressed as a transitive substitution group, its degree must
also be a power of~$p$ (§~123). Moreover such a group, since it
has self-conjugate operations, must necessarily be imprimitive.

The greatest value of~$m$, for which a group of order~$p^{m}$ can
be expressed as a transitive group of degree~$p^{n}$, where $n$~is
regarded as given, is determined at once by considering the
symmetric group of degree~$p^{n}$. The highest power of~$p$ that
divides~$p^{n}!$ is~$p^{\nu}$, where
\[
\nu = p^{n - 1} + p^{n - 2} + \Dots + p + 1.
\]

Hence the symmetric group of degree~$p^{n}$ contains a set of
conjugate sub-groups of order~$p^{\nu}$ and it contains no groups whose
order is a higher power of~$p$. Also, these groups are transitive in
the $p^{n}$~symbols; for any one of them must contain a circular
substitution of order~$p^{n}$. There are therefore groups of order~$p^{\nu}$
which can be expressed as transitive groups of degree~$p^{n}$; but
no group of order~$p^{\nu'}$ ($\nu' > \nu$) can be so expressed. Moreover, in
order that a group of order~$p^{m}$ may be capable of representation
as a transitive group of degree~$p^{n}$, it must be simply isomorphic
with a transitive sub-group of the above substitution group of
order~$p^{\nu}$.

This group may be constructed synthetically as follows.
Since the sub-group of order~$p^{\nu - n}$, that leaves one symbol
unchanged, is contained in a sub-group of order~$p^{\nu - n + 1}$ (§~55),
there must be imprimitive systems containing $p$~symbols each.
If then we distribute the $p^{n}$~symbols into $p^{n - 1}$~sets of $p$~each,
and with each set of~$p$ form a circular substitution, the $p^{n - 1}$
permutable and independent circular substitutions will generate
an intransitive group of order~$p^{p^{n - 1}}$. It will be the self-conjugate
sub-group of the group of order~$p^{\nu}$, which permutes the
symbols of each system among themselves.

Again, the systems of $p$~symbols each may be taken in sets
of~$p$ to form systems of $p^{2}$~symbols each, since the previously
considered sub-group of order~$p^{n - \nu + 1}$ must be contained in a sub-group
of order~$p^{n - \nu + 2}$. Hence we may form systems of $p^{2}$~symbols
by combining the previous systems in sets of~$p$; and then with
each $p^{2}$~symbols we can form a circular substitution, whose \DPchg{$p^{th}$}{$p$th}~power
is the product of the $p$~circular substitutions of order~$p$,
%% -----File: 235.png---Folio 219-------
which have been previously formed from the $p^{2}$~symbols. The
symbols of any set of~$p^{2}$ will then be interchanged by a transitive
group of order~$p^{p + 1}$; and since there are $p^{n - 2}$ such sets, we obtain
in this way an intransitive group of $p^{p^{n - 1} + p^{n - 2}}$. The group thus
formed is that self-conjugate sub-group of the original group,
which interchanges among themselves the symbols of each
system of~$p^{2}$. This process may be continued, taking greater
and greater systems, till at the last step we combine the $p$~systems
of $p^{n - 1}$~symbols each into a single system by means of
a circular substitution of order~$p^{n}$. The order of the resulting
group is clearly~$p^{\nu}$, as it should be.

The self-conjugate operations of this group form a sub-group
of order~$p$.

For suppose, if possible, they form a sub-group of order~$p^{r}$.
Every operation of this sub-group displaces all the symbols;
and therefore, when expressed as a substitution group in the $p^{n}$~symbols,
it must interchange them transitively in $p^{n - r}$~sets of
$p^{r}$ each.

Now (§~151) those substitutions of the $p^{n}$~symbols, which
are permutable with every operation of this sub-group, form a
group of order~$p^{rp^{n - r}} · p^{n - r}!$; this number is only divisible by~$p^{\nu}$,
\index{Substitution groups@\Topic{Substitution groups} whose orders are powers of primes}%
as it must be, when $r = 1$.

\begin{Ex} Shew that, for the group of degree~$p^{2}$ and order~$p^{p + 1}$, the
factor groups $\dfrac{H_{r + 1}}{H_{r}}$ (of §~53) are all of type~$(1)$ except the last,
which is of type~$(1, 1)$.
\end{Ex}

The fact that $\nu$~is a function of~$p$ when $n$~is given, explains
why, in classifying all groups of order~$p^{n}$, some of the lower
primes may behave in an exceptional manner. Thus we saw, in
§~73, that for certain groups of order~$p^{4}$ it was necessary to
consider separately the case $p = 3$. The present article makes
it clear that, while there may be more than one type of group
of order~$p^{4}$ ($p > 3$), which can be expressed as a transitive group
of degree~$p^{2}$, there is only a single type of group of order~$3^{4}$
which can be expressed transitively in $9$~symbols.

\begin{Remark}
\Par{153.} In the memoirs referred to in the footnote on \PageRef{p.}{155},
M.~Mathieu has demonstrated the existence of a remarkable group,
%% -----File: 236.png---Folio 220-------
of degree~$12$ and order~$12 · 11 · 10 · 9 · 8$\DPtypo{.}{}, which is quintuply transitive.
The generating operations of this group have been given in
the note to §~108, at the end of Chapter~VIII\@. The verification of
some of the more important properties of this group, as stated in
the succeeding example, forms a good exercise on the results of this
and the two preceding Chapters.
\end{Remark}

\begin{Ex} Shew that the substitutions
\begin{gather*}
(1254) (3867),\quad       (1758) (2643), \\
(12) (48) (57) (69),\quad (a2) (58) (46) (79), \\
(ab) (57) (68) (49),\quad (bc) (47) (58) (69),
\end{gather*}
generate a quintuply transitive group of degree~$12$ and order
\[
12 · 11 · 10 · 9 · 8.
\]

Prove that this group is simple; that a sub-group of degree~$11$
and order $11 · 10 · 9 · 8$, which leaves one symbol unchanged, is a
simple group; and that a sub-group of degree~$10$ and order~$10 · 9 · 8$,
which leaves two symbols unchanged, contains a self-conjugate sub-group
simply isomorphic with the alternating group of degree~$6$.

Shew also that the group of degree~$12$ contains \Inum{(i)}~$1728$
sub-groups of order~$11$ each of which is self-conjugate in a group
of order~$55$: \Inum{(ii)}~$2376$ sub-groups of order~$5$, each of which is self-conjugate
in a group of order~$40$: \Inum{(iii)}~$880$ sub-groups of order~$27$,
each of which is self-conjugate in a group of order~$108$: \Inum{(iv)}~$1485$
sub-groups of order~$64$.

Prove further that the group is a maximum sub-group of the
alternating group of degree~$12$.
\end{Ex}
%% -----File: 237.png---Folio 221-------


\Chapter{XI.}{On the Isomorphism of a Group with Itself.}

\Par{154.} \First{It} is shewn in §~24 that, if all the operations of a
group are transformed by one of themselves, which is not self-conjugate,
a correspondence is thereby established among the
operations of the group which exhibits the group as simply
isomorphic with itself.

In an Abelian group every operation is self-conjugate, and
the only correspondence established in the manner indicated is
that in which every operation corresponds to itself. If however
in an Abelian group we take, as the operation which corresponds
to any given operation~$S$, its inverse~$S^{-1}$, then to $ST$ will
correspond $S^{-1}T^{-1}$ or~$(ST)^{-1}$; and the correspondence exhibits
the group as simply isomorphic with itself. For this particular
correspondence, a group of order~$2$ is the only one in which
each operation corresponds to itself.

It is therefore possible for every group, except a group of
order~$2$, to establish a correspondence between the operations of
the group, which shall exhibit the group as simply isomorphic
with itself. Moreover, we shall see that in general there are
such correspondences which cannot be established by either of
the processes above given. We devote the present Chapter to
a discussion of the isomorphism of a group with itself. It will
be seen that, for many problems of group-theory, and in particular
for the determination of the various types of group
which are possible when the factor-groups of the composition-series
are given, this discussion is most important.
%% -----File: 238.png---Folio 222-------

\Par{155.}
\begin{Definition}
A correspondence between the operations
\index{Isomorphisms@\Topic{Isomorphisms} of a group with itself!definition of}%
of a group, such that to every operation~$S$ there
corresponds a single operation~$S'$, while to the product~$ST$ of
two operations there corresponds the product~$S'T'$ of the
corresponding operations, is said to define an \emph{isomorphism of
the group with itself}.
\end{Definition}

That isomorphism in which each operation corresponds to
itself is called the \emph{identical isomorphism}. In every isomorphism
of a group with itself, the identical operation corresponds to
itself; and the orders of two corresponding operations are
the same. For if $1$~and~$S$ were corresponding operations, so
also would be $1 · 1$ and~$S^{2}$; and therefore more than one operation
would correspond to~$1$. Again, if $S$~and~$S'$, of orders $n$~and~$n'$,
are corresponding operations, so also are $S^{n}$ and~$S'^{n}$; and therefore
$n$~must be a multiple of~$n'$. Similarly $n'$~must be a multiple
of~$n$; and therefore $n$~and~$n'$ are equal.

If the operations of a group of order~$N$ are represented by
\[
1,\ S_{1},\ S_{2},\ \Dots,\ S_{N - 1},
\]
and if, for a given isomorphism of the group with itself, $S'_{r}$~is
the operation that corresponds to~$S_{r}$ ($r = 1, 2, \Dots, N - 1$), the
isomorphism will be completely represented by the symbol
\[
\Isom{1, S_{1}, S_{2}, \Dots, S_{N - 1}}
     {1, S_{1}', S_{2}', \Dots, S_{N - 1}'}.
\]

In this symbol, two operations in the same vertical line are
corresponding operations. When no risk of confusion is
thereby introduced, the simpler symbol
\[
\Isom{S}{S'}
\]
is used.

\Par{156.} An isomorphism of a group with itself, thus defined,
is not an operation. The symbol of an isomorphism however
defines an operation. It may, in fact, be regarded as a substitution
performed upon the $N$~letters which represent the
operations of the group. Corresponding to every isomorphism
there is thus a definite operation; and it is obvious that the
operations, which correspond to two distinct isomorphisms, are
%% -----File: 239.png---Folio 223-------
themselves distinct. The totality of these operations form a
group. For let
\[
\Isom{S}{S'}
\quad\text{and}\quad
\Isom{S'}{S''}
\]
be any two isomorphisms of the group with itself. Then if, as
hitherto, we use curved brackets to denote a substitution, we
have
\[
\Sub{S}{S'} \Sub{S'}{S''} = \Sub{S}{S''}.
\]

But since $\Isom{S}{S'}$ is an isomorphism, the relation
\[
S_{p} S_{q} = S_{r}
\]
requires that
\[
S_{p}' S_{q}' = S_{r}'.
\]

And since $\Isom{S'}{S''}$ is an isomorphism, the relation
\[
S_{p}' S_{q}' = S_{r}'
\]
requires that
\[
S_{p}'' S_{q}'' = S_{r}''.
\]

Hence if
\[
S_{p} S_{q} = S_{r},
\]
then
\[
S_{p}'' S_{q}'' = S_{r}'';
\]
and therefore $\Isom{S}{S''}$ is an isomorphism.

The product of the substitutions which correspond to two
isomorphisms is therefore the substitution which corresponds to
some other isomorphism.

The set of substitutions which correspond to the isomorphisms
of a given group with itself, therefore form a group.

\begin{Definition}
A group, which is simply isomorphic with
\index{Group@\Topic{Group}!group of isomorphisms of a}%
the group thus derived from a given group, is called \emph{the
group of isomorphisms} of the given group.
\end{Definition}

It is not, of course, necessary always to regard this group as
a group of substitutions performed on the symbols of the
operations of the given group. But however the group of
isomorphisms may be represented, each one of its operations
corresponds to a definite isomorphism of the given group. To
avoid an unnecessarily cumbrous phrase, we may briefly apply
%% -----File: 240.png---Folio 224-------
the term ``isomorphism'' to the operations of the group of
isomorphisms. So long, at all events, as we are dealing with
the properties of a group of isomorphisms, no risk of confusion
is thereby introduced. Thus we shall use the phrase ``the
isomorphism $\Sub{S}{S'}$'' as equivalent to ``the operation of the
group of isomorphisms which corresponds to the isomorphism
$\Isom{S}{S'}$.''

\Par{157.} If $\Sigma$~is some operation of a group~$G$, while for $S$ each
\index{Isomorphisms@\Topic{Isomorphisms} of a group with itself!cogredient and contragredient, definition of}%
\index{Isomorphisms@\Topic{Isomorphisms} of a group with itself!class of, definition of}%
operation of the group is put in turn, the symbol
\[
\Sub{S}{\Sigma^{-1}S \Sigma}
\]
defines an isomorphism of the group. For if
\[
S_{p} S_{q} = S_{r},
\]
then
\[
\Sigma^{-1} S_{p} \Sigma · \Sigma^{-1} S_{q}\Sigma
  = \Sigma^{-1} S_{p} S_{q}\Sigma
  = \Sigma^{-1} S_{r} \Sigma;
\]
and $\Sigma_{-1} S_{r}\Sigma$ is an operation of the group. An isomorphism of a
group, which is thus formed on transforming the operations of the
group by one of themselves, is called a \emph{cogredient} isomorphism.
All others are called \emph{contragredient} isomorphisms\footnotemark.
\index{Holder@\Topic{Hölder}, quoted}%
\index{Klein@\Topic{Klein}, quoted}%
\footnotetext{Klein, ``Vorlesungen über das Ikosaeder und die Auflösung der Gleichungen
  vom fünften Grade'' (1884), p.~232. Hölder, \textit{Math.\ Ann.}, Vol.~\VolNo{XLIII}
  (1893), p.~314.}%
If $\Sub{S}{S'}$ is
any contragredient isomorphism, the isomorphisms
\[
\Sub{S}{S'} \Sub{S}{\Sigma^{-1}S\Sigma},
\]
when for $\Sigma$ each operation of the group is taken successively,
are said to form a \emph{class} of contragredient isomorphisms.

\begin{Theorem}[I.]
The totality of the cogredient isomorphisms of
a group~$G$ form a group isomorphic with~$G$; this group is
a self-conjugate sub-group of the group of isomorphisms of~$G$\footnotemark.
\end{Theorem}
\footnotetext{Hölder, ``Bildung zusammengesetzter Gruppen,'' \textit{Math.\ Ann.}, Vol.~\VolNo{XLVI}
  (1895), p.~326.}

The product of the isomorphisms
\[
\Sub{S}{\Sigma^{-1}S\Sigma} \quad\text{and}\quad
\Sub{S}{\Sigma'^{-1}S\Sigma'}
\]
%% -----File: 241.png---Folio 225-------
is given by
\begin{align*}
\Sub{S}{\Sigma^{-1}S\Sigma}
\Sub{S}{\Sigma'^{-1}S\Sigma'}
& =
\Sub{S}{\Sigma^{-1}S\Sigma}
\Sub{\Sigma^{-1}S\Sigma}{\Sigma'^{-1}\Sigma^{-1}S\Sigma\Sigma'} \\
& =
\Sub{S}{\Sigma'^{-1}\Sigma^{-1}S\Sigma\Sigma'} \\
& =
\Sub{S}{\Sigma''^{-1}S\Sigma''},
\end{align*}
where
\[
\Sigma \Sigma' = \Sigma''.
\]

The product of two cogredient isomorphisms is therefore another
cogredient isomorphism; hence the cogredient isomorphisms
form a group. Moreover, if we take the isomorphism
\[
\Sub{S}{\Sigma^{-1}S\Sigma}
\]
as corresponding to the operation~$\Sigma$ of the group~$G$, then to
every operation of~$G$ there will correspond a definite cogredient
isomorphism, so that to the product of any two operations
of~$G$ there corresponds the product of the two corresponding
isomorphisms. The group~$G$ and its group of cogredient
isomorphisms are therefore isomorphic. If $G$~contains no
self-conjugate operation, identity excepted, no two isomorphisms
corresponding to different operations of~$G$ can be identical;
and therefore, in this case, $G$~is simply isomorphic with its
group of cogredient isomorphisms. If however $G$~contains self-conjugate
operations, forming a self-conjugate sub-group~$H$,
then to every operation of~$H$ there corresponds the identical
isomorphism; and the group of cogredient isomorphisms is
simply isomorphic with~$\dfrac{G}{H}$.

Let now
\[
\Sub{S}{S'}
\]
be any isomorphism. Then
\begin{align*}
\Sub{S}{S'}^{-1}
\Sub{S}{\Sigma^{-1}S\Sigma}
\Sub{S}{S'}
& =
\Sub{S'}{S}
\Sub{S}{\Sigma^{-1}S\Sigma}
\Sub{S}{S'} \\
& =
\Sub{S'}{\Sigma^{-1}S\Sigma}
\Sub{\Sigma^{-1}S\Sigma}{\Sigma'^{-1}S'\Sigma'} \\
& =
\Sub{S'}{\Sigma'^{-1}S'\Sigma'}.
\end{align*}
%% -----File: 242.png---Folio 226-------

The isomorphism $\Sub{S}{S'}$ therefore transforms every cogredient
\index{Group of isomorphisms@\Topic{Group of isomorphisms}!contains the group of cogredient isomorphisms self-conjugately}%
isomorphism into another cogredient isomorphism. It follows
that the group of cogredient isomorphisms is self-conjugate
within the group of isomorphisms.

\Par{158.} Let $G$ be a group of order~$N$, whose operations are
\[
1,\ S_{1},\ S_{2},\ \Dots,\ S_{N - 1};
\]
and let $L$ be the group of isomorphisms of~$G$. We have seen
in §~20 that $G$~may be represented as a transitive group of
substitutions performed on the $N$~symbols
\[
1,\ S_{1},\ S_{2},\ \Dots,\ S_{N - 1};
\]
and that, when it is so represented, the substitution which
corresponds to the operation~$S_{x}$ is
\[
\Sub{\PadTo{S_{x}}{1},
     \PadTo{S_{1}S_{x}}{S_{1}},
     \PadTo{S_{2}S_{x}}{S_{2}}, \Dots,
     \PadTo{S_{N-1}S_{x}}{S_{N-1}}}
    {S_{x}, S_{1}S_{x}, S_{2}S_{x}, \Dots, S_{N-1}S_{x}},
\]
or more shortly
\[
\Sub{S}{SS_{x}}.
\]

When $G$ is thus represented, we will denote it by~$G'$. We
have already seen that $L$~can be represented as an intransitive
substitution group of the same $N$~symbols; a typical substitution
of~$L$, when it is so represented, is
\[
\Sub{1, S_{1}, S_{2}, \Dots, S_{N-1}}
    {1, S'_{1}, S'_{2}, \Dots, S'_{N-1}},
\]
or more shortly
\[
\Sub{S}{S'}.
\]

When $L$ is thus represented, we will denote it by~$L'$. It is
clear that the two substitution groups $G'$ and $L'$ have no
substitution in common except identity. For every substitution
of~$L'$ leaves the symbol~$1$ unchanged; and no substitution of~$G'$,
except identity, leaves $1$ unchanged.

Now
\begin{align*}
\Sub{S}{S'}^{-1}
\Sub{S}{SS_{x}}
\Sub{S}{S'}
& =
\Sub{S'}{SS_{x}}
\Sub{SS_{x}}{S'S'_{x}} \\
& =
\Sub{S'}{S'S'_{x}} \\
& =
\Sub{S}{SS'_{x}}.
\end{align*}
%% -----File: 243.png---Folio 227-------

Every operation of~$L'$ is therefore permutable with~$G'$.
Hence if $M$~is the order of~$L$, the group~$\{G', L'\}$, which we will
call~$K'$, is a transitive group of degree~$N$ and order~$NM$,
containing $G'$ self-conjugately. Further $\Sub{S}{S'}$ transforms $\Sub{S}{SS_{x}}$
into $\Sub{S}{SS'_{x}}$; and these two substitutions of~$G'$ correspond to
the operations $S_{x}$ and~$S_{x}'$ of~$G$. Hence the isomorphism,
established on transforming the substitutions of~$G'$ by any
substitution $\Sub{S}{S'}$ of~$L'$, is the isomorphism denoted by the
symbol~$\Sub{S}{S'}$.

{\Loosen Since $\Sub{S}{S_{x}SS_{x}^{-1}}$ is a substitution of~$L'$, the substitution
$\Sub{S}{S_{x}SS_{x}^{-1}} \Sub{S}{SS_{x}}$, or $\Sub{S}{S_{x}S}$, belongs to~$K'$. Hence $K'$~contains
the set of substitutions}
\[
\Sub{S}{S_{x}S},\quad (x = 0, 1, 2, \Dots, N - 1).
\]

These form (§~107) a transitive group~$G''$, simply isomorphic
with~$G'$ and such that every substitution of~$G''$ is
permutable with every substitution of~$G'$. Moreover (\lc),
the substitutions of~$G''$ are the only substitutions of the
$N$~symbols which are permutable with each of the substitutions
of~$G'$.

Suppose now that $\Sigma$~is any substitution of the $N$~symbols
which is permutable with~$G'$. When the substitutions of~$G'$
are transformed by~$\Sigma$, the resulting isomorphism is identical
with that given by some substitution, say $\Sub{S}{S'}$, of $L'$.
Hence $\Sigma \Sub{S}{S'}^{-1}$ is a substitution of the $N$~symbols which is
permutable with every substitution of~$G'$. It therefore belongs
to~$G''$; and hence $\Sigma$~belongs to~$K'$. It follows that $K'$~contains
every substitution of the $N$~symbols which is permutable
with~$G'$.
%% -----File: 244.png---Folio 228-------

The only substitutions common to $G'$ and $G''$ are the self-conjugate
substitutions of either. The factor group $\dfrac{K'}{\{G', G''\}}$
is simply isomorphic with~$\dfrac{L}{g}$, where $g$~is the group of cogredient
isomorphisms of $G$ contained in~$L$. The groups $G'$ and $G''$ are
identical only when $G'$~is Abelian; in this case, $g$~consists of the
identical operation alone.

\begin{Definition}
A group~$K$, simply isomorphic with the
\index{Group@\Topic{Group}!holomorph of a}%
\index{Holomorph@\Topic{Holomorph}!definition of}%
substitution group~$K'$ which has just been constructed, we
shall call the \emph{holomorph} of~$G$.
\end{Definition}

\Par{159.} An isomorphism must change any set of operations,
which are conjugate to each other, into another set which are
conjugate. For if
\[
\Sub{S}{S'}
\]
be the isomorphism, and if
\[
S_{z}^{-1} S_{x} S_{z} = S_{y},
\]
then
\[
{S'_{z}}^{-1} S'_{x} S'_{z} = S'_{y},
\]
so that $S'_{x}$ and $S'_{y}$ are conjugate operations when $S_{x}$~and~$S_{y}$ are
conjugate. A~cogredient isomorphism changes every set of
conjugate operations into itself; and all the members of a class
of contragredient isomorphisms permute the conjugate sets in
the same way. If
\[
\Sub{S}{S'}
\]
is an isomorphism which changes every conjugate set of operations
of~$G$ into itself, and if
\[
\Sub{S}{S''}
\]
is any isomorphism of~$G$, then the isomorphism
\[
\Sub{S}{S''}^{-1} \Sub{S}{S'}\Sub{S}{S''}
\]
changes every conjugate set into itself. It follows that those
isomorphisms, which change every conjugate set of operations
into itself, form a self-conjugate sub-group of the complete group
of isomorphisms. This sub-group clearly contains the group of
cogredient isomorphisms and may be identical with it.
%% -----File: 245.png---Folio 229-------

If now $\Sub{S}{S'}$ is any isomorphism of~$G$ of order~$n$, the
substitutions
\[
\Sub{S}{S'}\quad \text{and}\quad \Sub{S}{SS_{x}},\quad(x = 0, 1, \Dots, N - 1),
\]
generate a group of order~$N_{n}$. When $J$~is used to represent
the isomorphism, this group may be denoted by $\{J, G\}$; as shewn
above, it contains $G$ self-conjugately. Suppose that $n$~is prime
and not a factor of~$N$. The operation~$J$ is not permutable
with every operation of~$G$; and therefore there must be operations~$S$
of~$G$ which are permutable with no operation of the
conjugate set to which $J$~belongs. The number of operations
which in $\{J, G\}$ are conjugate to such an operation~$S$ must be a
multiple of~$n$; and since $n$~is not a factor of~$N$, this conjugate
set of operations must be made up of~$n$ distinct conjugate sets
of operations in~$G$. The isomorphism~$J$ must therefore interchange
some of the conjugate sets of~$G$.

The same result is clearly true if the order~$n$ of~$J$ has any
prime factor not contained in~$N$. Hence:---

\begin{Theorem}[II.]
An isomorphism of a group~$G$, whose order
contains a prime factor which does not occur in the order of~$G$,
must interchange some of the conjugate sets of~$G$.
\end{Theorem}

\Par{160.} If the isomorphism $\Sub{S}{S'}$ or~$J$ leaves no operation
except identity unchanged, it must in $\{J, G\}$ be one of $N$
conjugate operations. For if
\[
S_{x}^{-1} J S_{x} = S_{y}^{-1} J S_{y},
\]
$J$~would be permutable with $S_{y}S_{x}^{-1}$, which is not the case.

These $N$~conjugate operations are
\[
J,\ JS_{1},\ JS_{2},\ \Dots,\ JS_{N-1},
\]
and since the first transforms every operation of~$G$, except
identity, into a different one, the same must be true of all the
set. If now $J$~transformed any operation~$S$ into a conjugate
operation $\Sigma^{-1} S \Sigma, J \Sigma^{-1}$ would transform $S$ into itself; hence
$J$~must transform every conjugate set of~$G$ into a different
conjugate set.
%% -----File: 246.png---Folio 230-------

\begin{Remark}
The special case in which the order of~$J$ is two may here be
considered. Representing the $N$~operations conjugate to~$J$ by
\[
J,\ J_{1},\ J_{2},\ \Dots,\ J_{N-1},
\]
the $N$~operations of~$G$ are
\[
J^{2},\ JJ_{1},\ JJ_{2},\ \Dots,\ JJ_{N-1}.
\]

Now
\[
J^{-1} · JJ_{x} · J = J_{x}J = (JJ_{x})^{-1},
\]
so that $J$~transforms every operation of~$G$ into its own inverse.
But if
\[
S' = S^{-1},
\]
and
\[
T' = T^{-1},
\]
then
\[
S'T' = S^{-1}T^{-1} = (TS)^{-1}.
\]

Now as $S'T'$ is the operation into which the isomorphism
transforms~$ST$, it must be~$(ST)^{-1}$, and therefore
\[
ST = TS.
\]

The group~$G$ is therefore an Abelian group of odd order.
\end{Remark}

\Par{161.} Any sub-group~$H$ of~$G$ is transformed by an isomorphism
into a simply isomorphic sub-group~$H'$; but $H$~and~$H'$
are not necessarily conjugate within~$G$. If however the set of
conjugate sub-groups
\[
H_{1},\ H_{2},\ \Dots,\ H_{m},
\]
are the only sub-groups of~$G$ of a given type, every isomorphism
must interchange them among themselves; and if no
isomorphism transforms each one of the set into itself, the
group of isomorphisms can be represented as a transitive group
of degree~$m$.

Suppose now that no operation of~$G$ is permutable with
each of the conjugate sub-groups
\[
H_{1},\ H_{2},\ \Dots,\ H_{m},
\]
so that $G$, or what in this case is the same thing (since $G$~can
have no self-conjugate operation except identity) the group of
cogredient isomorphisms of~$G$, can be expressed as the transitive
group of $m$~symbols that arises on transforming the set of $m$~sub-groups
by each operation of~$G$. Let $J$~be any operation, of
order~$\mu$, that transforms~$G$ and each of the set of $m$ conjugate
sub-groups, into itself. Then $J^{\mu}$~is the lowest power of~$J$ that
can occur in~$G$, since no operation of~$G$ transforms each of the $m$~sub-groups
%% -----File: 247.png---Folio 231-------
into itself. Now in $\{J, G\}$, the greatest sub-group
that contains $H_{r}$ self-conjugately is~$\{J, I_{r}\}$, $I_{r}$~being the greatest
sub-group of~$G$ that contains~$H_{r}$ self-conjugately. Also, in
$\{J, G\}$ the set of sub-groups $\{J, I_{r}\}$, ($r = 1, 2, \dots, m$), is a
complete conjugate set. Now the set of groups
\[
I_{1},\ I_{2},\ \Dots,\ I_{m}
\]
have by supposition no common operation except identity; and
therefore the greatest common sub-group of
\[
\{J, I_{1}\},\ \{J, I_{2}\},\ \Dots,\ \{J, I_{m}\}
\]
is~$\{J\}$. Hence $\{J\}$~is a self-conjugate sub-group of~$\{J, G\}$; and
since $G$~is also a self-conjugate sub-group of~$\{J, G\}$, while $\{J\}$
and~$G$ have no common operation except identity, $J$~must be
permutable with every operation of~$G$. Every operation
therefore which is permutable with~$G$, and with each of the
sub-groups
\[
H_{1},\ H_{2},\ \Dots,\ H_{m},
\]
is permutable with every operation of~$G$. Thus finally, no
contragredient isomorphism can transform each of the sub-groups~$H_{r}$
($r = 1, 2, \Dots, m$) into itself. Hence:---

\begin{Theorem}[III.]
If the conjugate set of $m$ sub-groups
\[
H_{1},\ H_{2},\ \Dots,\ H_{m}
\]
contains all the sub-groups of~$G$ of a given type, and if no
operation of~$G$ is permutable with each sub-group of the set, the
group of isomorphisms of~$G$ can be represented as a transitive
group of degree~$m$.
\end{Theorem}

\begin{Corollary}[I.]
If $G$ contains $kp + 1$ sub-groups of order~$p^{\alpha}$,
where $p^{\alpha}$~is the highest power of a prime~$p$ that divides the
order of~$G$; and if the greatest sub-group~$I$, that contains a
sub-group of order~$p^{\alpha}$ self-conjugately, contains no self-conjugate
sub-group of~$G$; then the group of isomorphisms of~$G$ can
be represented as a transitive group of degree~$kp + 1$.
\end{Corollary}

For it has been seen that the groups of order~$p^{\alpha}$ contained
in~$G$ form a single conjugate set.

\begin{Corollary}[II.]
If the conjugate set of $m$ sub-groups of~$G$
\[
H_{1},\ H_{2},\ \Dots,\ H_{m}
\]
is changed into itself by every isomorphism of~$G$, and if no
%% -----File: 248.png---Folio 232-------
operation of~$G$ is permutable with every one of these sub-groups;
then the group of isomorphisms of~$G$ can be expressed
as a transitive group of degree~$m$.
\end{Corollary}

In fact, under the conditions stated, the reasoning applied
to prove the theorem may be used to shew that no isomorphism
of~$G$ can transform each of the $m$ sub-groups into itself.

\Par{162.} \begin{Definition}
Any sub-group of a group~$G$ which is
\index{Characteristic sub-group@\Topic{Characteristic sub-group}!definition of}%
\index{Characteristic sub-group@\Topic{Characteristic sub-group}!groups with no, are either simple or the direct product of simply isomorphic simple groups}%
transformed into itself by every isomorphism of~$G$, is called\footnote
  {Frobenius, ``Ueber endliche Gruppen,'' \textit{Berliner Sitzungsberichte}, 1895,
  p.~183.}
\index{Frobenius@\Topic{Frobenius}, quoted}%
a \emph{characteristic sub-group} of~$G$.
\end{Definition}

A characteristic sub-group of a group~$G$ is necessarily a
self-conjugate sub-group of~$G$; but a self-conjugate sub-group
is not necessarily characteristic. A simple group, having no
self-conjugate sub-groups, can have no characteristic sub-groups.
Let now $G$ be any group, and let $K$~be the holomorph
of~$G$. A characteristic sub-group of~$G$ is then a self-conjugate
sub-group of~$K$; and conversely, every self-conjugate sub-group
of~$K$ which is contained in~$G$ is a characteristic sub-group of~$G$.

Suppose now a chief-series of~$K$ formed which contains~$G$.
If $G$~has no characteristic sub-group, it must be the
last term but one of this series, the last term being identity.
It follows by §~94 that $G$~must be the direct product of a
number of simply isomorphic simple groups. Hence:---

\begin{Theorem}[IV.]
A group, which has no characteristic sub-group,
must be either a simple group or the direct product of
simply isomorphic simple groups.
\end{Theorem}

The converse of this theorem is clearly true.

\Par{163.} Suppose now that $G$~is a group which has characteristic
\index{Characteristic series@\Topic{Characteristic series}!definition of}%
sub-groups; and let
\[
G,\ G_{1},\ \Dots,\ G_{r},\ G_{r+1},\ \Dots,\ 1
\]
be a series of such sub-groups, each containing all that follow it
and chosen so that, for each consecutive pair $G_{r}$ and $G_{r+1}$, there
is no characteristic sub-group of~$G$ contained in~$G_{r}$ and containing~$G_{r+1}$,
except $G_{r+1}$~itself. Such a series is called\footnote
  {Frobenius, ``Ueber auflösbare Gruppen, II,'' \textit{Berliner Sitzungsberichte}, 1895,
  p.~1027.}
a \emph{characteristic series} of~$G$.
%% -----File: 249.png---Folio 233-------
\index{Characteristic series@\Topic{Characteristic series}!invariance of}%

It may clearly be possible to choose such a series in more
than one way. If
\[
G,\ G_{1}',\ \Dots,\ G_{r}',\ G_{r+1}',\ \Dots,\ 1
\]
be a second characteristic series of~$G$, then
\begin{align*}
&K,\ J,\ \Dots,\ H,\ G,\ G_{1},\ \Dots,\ G_{r},\ G_{r+1},\ \Dots,\ 1 \\
\lintertext{and}
&K,\ J,\ \Dots,\ H,\ G,\ G_{1}',\ \Dots,\ G_{r}',\ G_{r+1}',\ \Dots,\ 1
\end{align*}
are two chief-series of~$K$. In fact, if $K$~had a self-conjugate
sub-group contained in~$G_{r}$ and containing $G_{r+1}$, then $G$~would
have a characteristic sub-group contained in~$G_{r}$ and containing
$G_{r+1}$. The two chief-series of~$K$ coincide in the terms from $K$
to~$G$ inclusive. Hence the two sets of factor-groups
\[
\frac{G}{G_{1}},\ \frac{G_{1}}{G_{2}},\ \Dots,\ \frac{G_{r}}{G_{r+1}},\ \Dots
\]
and
\[
\frac{G}{G_{1}'},\ \frac{G_{1}'}{G_{2}'},\ \Dots,\ \frac{G_{r}'}{G_{r+1}'},\ \Dots
\]
must be equal in number and, except possibly as regards the
sequence in which they occur, identical in type. Moreover, each
factor-group must be either a simple group or the direct product
of simply isomorphic simple groups.

\Par{164.} We will now shew how to determine a characteristic
\index{Characteristic series@\Topic{Characteristic series}!of a group whose order is a power of a prime|(}%
series for a group whose order is the power of a prime\footnotemark.
\footnotetext{Frobenius, \lc, pp.~1028, 1029.}%

{\Loosen First, let the group~$G$ be Abelian; and suppose that it is
generated by a set of independent operations, of which $n_{s}$ are of
the order~$p^{m}$, ($s = 1, 2, \dots, r$), while}
\[
m_{1} > m_{2} > \Dots > m_{r}.
\]

The sub-group~$G_{\mu}$ (§~42), formed of the operations of~$G$
which satisfy the relation
\[
S^{p^{\mu}} = 1,
\]
is clearly a characteristic sub-group. As a first step towards
forming the characteristic series, we may take the set of
groups
\[
G_{m_{1}} (= G),\ G_{m_{1} - 1},\ G_{m_{1} - 2},\ \Dots,\ G_{2},\ G_{1},\ 1;
\]
for this is a set of characteristic sub-groups such that each
contains the one that follows it.
%% -----File: 250.png---Folio 234-------

Now the sub-group~$H_{\nu}$ (§~45), formed of the distinct
operations that remain when every operation of~$G$ is raised
to the power~$p^{\nu}$, is also a characteristic sub-group; and since
the operations common to two characteristic sub-groups also
form a characteristic sub-group, the sub-group~$K_{\mu,\nu}$ (common to
$G_{\mu}$ and~$H_{\nu}$) is characteristic. It follows from this that $G_{1}$~will be
a characteristic sub-group only when $r = 1$. If $r > 1$, $G_{1}$~is not
contained in~$H_{m_{r - 1} - 1}$, and the common sub-group~$K_{1, m_{r - 1} - 1}$ of
these two is characteristic. If $r > 2$, this sub-group again is
not contained in~$H_{m_{r - 2} - 1}$; and the common sub-group~$K_{1, m_{r - 2} - 1}$
of $G_{1}$ and~$H_{m_{r - 2} - 1}$ is a characteristic sub-group contained
in~$K_{1, m_{r - 1} - 1}$. Continuing thus, we form between $G_{1}$ and~$1$ the
series
\[
G_{1},\ K_{1, m_{r - 1} - 1},\ K_{1, m_{r - 2} - 1},\ \Dots,\ K_{1, m_{1} - 1},\ 1.
\]

In a similar way, between $G_{\alpha}$ and~$G_{\alpha-1}$ we introduce such of
the series
\[
\{G_{\alpha - 1}, K_{\alpha, m_{r - 1} - \alpha} \},\
\{G_{\alpha - 1}, K_{\alpha, m_{r - 2} - \alpha} \},\ \Dots,\
\{G_{\alpha - 1}, K_{\alpha, m_{1} - \alpha} \}
\]
as are distinct, the symbol $m_{s} - \alpha$ being replaced by zero where
it is negative.

From the original series we thus form a new one, in which
again each group is characteristic and contains the following.
This series may be shewn to be a characteristic series.

Let
\[
P_{m_{s}, 1},\ P_{m_{s}, 2},\ \Dots,\ P_{m_{s}, n_{s}}
\]
be the $n_{s}$~generating operations of~$G$, whose orders are~$p^{m_{s}}$.
Then if $\{G_{\alpha - 1}, K_{\alpha, m_{s} - \alpha}\}$ and $\{G_{\alpha - 1}, K_{\alpha, m_{s - 1} - \alpha}\}$ are distinct, the
generating operations of the latter differ only from those of the
former in containing the set
\[
P_{m_{s}, x}^{p^{m_{s} - \alpha + 1}},\quad (x = 1, 2, \Dots, n_{s}),
\]
in the place of
\[
P_{m_{s}, x}^{p^{m_{s} - \alpha} },\quad (x = 1, 2, \Dots, n_{s}).
\]

Now any permutation of the $n_{s}$~generating operations
\[
P_{m_{s}, x},\quad (x = 1, 2, \Dots, n_{s}),
\]
among themselves, the remaining generating operations being
unaltered, must clearly give an isomorphism of~$G$ with itself;
and therefore no sub-group of~$G$, contained in $\{G_{\alpha - 1}, K_{\alpha, m_{s} - \alpha} \}$
%% -----File: 251.png---Folio 235-------
and containing $\{G_{\alpha - 1}, K_{\alpha, m_{s - 1} - \alpha}\}$, can be a characteristic sub-group.
This result being true for every pair of distinct groups
which succeed each other in the series that has been formed,
it follows that the series is a characteristic series. It may be
noticed that, if $\Gamma$~and~$\Gamma'$ are any two consecutive sub-groups in
a characteristic series of~$G$, the order of~$\dfrac{\Gamma}{\Gamma'}$ must be~$p^{\nu}$, where $\nu$~is
one of the $r$~numbers~$n_{s}$.

Secondly, suppose that $G$~is not Abelian. We may first
consider the series of sub-groups
\[
G,\ H_{n},\ H_{n - 1},\ \Dots,\ H_{1},\ 1
\]
of §~53. Each of these is clearly a characteristic sub-group, and
each contains the succeeding. Moreover, $\dfrac{H_{r + 1}}{H_{r}}$ is an Abelian
group; and, by the process that we have just investigated, a
characteristic series may be formed for it. To each group in
this series will correspond a characteristic sub-group of~$H_{r + 1}$
containing~$H_{r}$; and the set of groups so obtained forms part of
a complete characteristic series of~$G$. When between each
consecutive pair of groups in the above series the groups thus
formed are interpolated, the resulting series of groups is a
characteristic series for~$G$.

\Par{165.} The isomorphisms of a given group with itself are
closely connected with the composition of every composite group
in which the given group enters as a self-conjugate sub-group.
Let $G$~be any composite group and $H$~a self-conjugate sub-group
of~$G$. Then since every operation of~$G$ transforms~$H$
into itself, to every such operation will correspond an isomorphism
of~$H$ with itself. If $S$~is an operation of~$G$ not
contained in~$H$, and if the isomorphism of~$H$ arising on
transforming its operations by~$S$ is contragredient, so also is
the isomorphism arising from each of the set of operations~$SH$.
In this case, no one of this set of operations is permutable with
every operation of~$H$. If however the isomorphism arising from~$S$
is cogredient, there must be some operation~$h$ of~$H$ which
gives the same isomorphism as~$S$; and then $Sh^{-1}$~is permutable
with every operation of~$H$. In this case, the set of operations~$SH$
will give all the cogredient isomorphisms of~$H$.
\index{Characteristic series@\Topic{Characteristic series}!of a group whose order is a power of a prime|)}%
%% -----File: 252.png---Folio 236-------

Suppose now that $H_{1}$ is that sub-group of~$G$ which is formed
of all the operations of~$G$ that are permutable with every
operation of~$H$. Then to every operation of~$G$, not contained
in~$\{H, H_{1}\}$, must correspond a contragredient isomorphism of~$H$;
and to every operation of the factor group $\dfrac{G}{\{H, H_{1}\}}$ corresponds
a class of contragredient isomorphisms. If then $L$~is the
group of isomorphisms of~$H$, and if $L_{1}$~is that self-conjugate
sub-group of~$L$ which gives the cogredient isomorphisms of~$H$,
$\dfrac{G}{\{H, H_{1}\}}$ must be simply isomorphic with a sub-group of~$\dfrac{L}{L_{1}}$.

If now $H$~contains no self-conjugate operation except
identity, $H$~and~$H_{1}$ can contain no common operation except
identity; and since each of them is a self-conjugate sub-group
of~$G$, every operation of~$H$ is permutable with every operation
of~$H_{1}$. In this case, $\{H, H_{1}\}$ is the direct product of $H$ and~$H_{1}$.

If, further, $L$~coincides with~$H$, so that $H$~admits of no
contragredient isomorphisms, $\dfrac{G}{\{H, H_{1}\}}$ must reduce to identity.
In this case, $G$~is the direct product of $H$ and~$H_{1}$.

\begin{Definition}
A group, which contains no self-conjugate
\index{Complete group@\Topic{Complete group}!definition of}%
\index{Complete group@\Topic{Complete group}!groups which contain a, self-conjugately are direct products}%
\index{Group@\Topic{Group}!complete}%
operation except identity and which admits of no contragredient
isomorphism, is called\footnote
  {Hölder, ``Bildung zusammengesetzter Gruppen,'' \textit{Math.\ Ann.}, Vol.~\VolNo{XLVI}
  (1895), p.~325.}
\index{Holder@\Topic{Hölder}, quoted}%
a \emph{complete} group.
\end{Definition}

The result of the present paragraph may be expressed in
the form:---

\begin{Theorem}[V.]
A group, which contains a complete group as
a self-conjugate sub-group, must be the direct product of the
complete group and some other group\footnotemark.
\end{Theorem}
\footnotetext{\textit{Ibid.}\ p.~325.}%

\Par{166.} \begin{Theorem}[VI.]
If $G$ is a group with no self-conjugate
operations except identity; and if the group of cogredient isomorphisms
of~$G$ is a characteristic sub-group of~$L$, the group of
isomorphisms of~$G$; then $L$~is a complete group\footnotemark.
\end{Theorem}
\footnotetext{Hölder (\textit{loc.~cit.}\ p.~331) gives a theorem which is similar but not quite
  equivalent to Theorem~VI.}%
%% -----File: 253.png---Folio 237-------

With the notation of §~158, the operations of~$L$ may be
represented by the substitutions
\[
\Sub{S}{S'}.
\]
The group of cogredient isomorphisms, which we will call~$G'$, is
given by the substitutions
\[
\Sub{S}{S_{x}^{-1} SS_{x}}\DPtypo{;}{,}\quad (x = 0, 1, \Dots, N - 1)\DPtypo{,}{;}
\]
it is simply isomorphic with~$G$.

Now
\begin{align*}
\Sub{S}{S'}^{-1} \Sub{S}{S_{x}^{-1} SS_{x}} \Sub{S}{S'}
  &= \Sub{S'}{S_{x}'^{-1} S'S_{x}'}, \\
  &= \Sub{S}{S_{x}'^{-1} SS_{x}'}:
\end{align*}
and therefore no operation of~$L$ is permutable with every
operation of~$G'$. Hence every isomorphism of~$G'$ is given on
transforming its operations by those of~$L$. Suppose now
that $J$~is an operation which transforms $L$ into itself. Since
$G'$~is by supposition a characteristic sub-group of~$L$, the
operation~$J$ transforms $G'$ into itself. If $J$~does not belong to~$L$,
we may assume that $J$~is permutable with every operation of~$G'$.
For if it is not, it must give the same isomorphism of~$G'$
as some operation~$S$ of~$L$; and then $JS^{-1}$~is permutable with
every operation of~$G'$, and is not contained in~$L$. Now $J$~being
permutable with every operation of~$G'$, we have
\[
J^{-1} s^{-1} gsJ = s^{-1}gs,
\]
where $s$~is any operation of~$L$, and $g$~any operation of~$G'$.

Moreover
\[
JgJ^{-1} = g,
\]
and therefore
\[
J^{-1} s^{-1} JgJ^{-1} sJ = s^{-1}gs.
\]

Hence $s$ and $J^{-1}sJ$ give the same isomorphism of~$G'$. Now
no two distinct operations of~$L$ give the same isomorphism of~$G'$,
so that $s$ and $J^{-1}sJ$ must be identical; in other words, $J$~is
permutable with every operation of~$L$. Hence $L$~admits of
no contragredient isomorphisms. Moreover, $G'$~has no self-conjugate
operations, and no operation of~$L$ is permutable with
%% -----File: 254.png---Folio 238-------
every operation of~$G'$; hence $L$~has no self-conjugate operations.
It is therefore a complete group.

\begin{Corollary}
If $G$~is a simple group of composite order, or
\index{Complete group@\Topic{Complete group}!group of isomorphisms of a simple group of composite order is a}%
if it is the direct product of a number of isomorphic simple
groups of composite order, the group of isomorphisms~$L$ of~$G$ is
a complete group.
\end{Corollary}

For suppose, if possible, in this case that $G'$~is not a
characteristic sub-group of~$L$; and that, by a contragredient
isomorphism of~$L$, $G$~is transformed into~$G''$. Then $G''$~is a
self-conjugate sub-group of~$L$, and each of the groups $G'$ and~$G''$
transforms the other into itself. Hence (§~34) either every
operation of~$G'$ is permutable with every operation of~$G''$, or $G'$
and~$G''$ must have a common sub-group. The former supposition
is impossible since no operation of~$L$ is permutable with
every operation of~$G$. On the other hand, if $G'$~and~$G''$ have a
common sub-group, it is a self-conjugate sub-group of~$L$ and it
therefore is a characteristic sub-group of $G'$. Now (§~162) $G$~has
no characteristic sub-groups, and therefore the second supposition
is also impossible. It follows that, in this case, $G'$~is a characteristic
sub-group of~$L$, and that $L$~is a complete group.

\Par{167.} \begin{Theorem}[VII.]
If $G$~is an Abelian group of odd order,
and if $K$~is the holomorph of~$G$; then when $G$~is a characteristic
sub-group of~$K$, the latter group is a complete group.
\end{Theorem}

If $N$~is the order of~$G$, then $K$~can be expressed (§~158) as a
transitive group of degree~$N$. When $K$~is so expressed, those
operations of~$K$ which leave one symbol unchanged form a
sub-group~$H$, which is simply isomorphic with the group of
isomorphisms of~$G$. Now (§~160) an Abelian group of odd order
admits of a single isomorphism of order two, which changes every
operation into its own inverse. The corresponding substitution
of~$H$ is a self-conjugate substitution in~$H$, and is one of $N$~conjugate
substitutions in~$K$. These are the only substitutions
of~$K$ which transform every substitution of~$G$ into its inverse.
If $G$~is a characteristic sub-group of~$K$, every isomorphism of~$K$
must transform~$G$, and therefore also the set of $N$ conjugate
substitutions of order two, into itself. Also, no substitution of~$K$
can be permutable with each one of these $N$~substitutions,
since each of them keeps just one symbol unchanged. Hence
%% -----File: 255.png---Folio 239-------
(Theorem~III, Cor.~II, §~161) the group of isomorphisms of~$K$
can be expressed as a transitive group of degree~$N$, which
contains $G$ as a transitive self-conjugate sub-group. But
when the group of isomorphisms of~$K$ is so expressed, $K$~itself
consists of all the substitutions of the $N$~symbols which are
permutable with~$G$; and at the same time, every isomorphism
of~$K$ transforms $G$ into itself. Hence the group of isomorphisms
of~$K$ must coincide with $K$~itself; \ie\ $K$~admits of no
contragredient isomorphisms. Also $K$~obviously contains no
self-conjugate operation except identity; hence it is a complete
group.

\begin{Corollary}
The holomorph of an Abelian group of order~$p^{m}$,
\index{Complete group@\Topic{Complete group}!holomorph of an Abelian group of order~$p^{m}$ and type $(1, 1, \dots, 1)$ is a}%
\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of a cyclical group|(}%
where $p$~is an odd prime, and type $(1, 1, \dots\DPchg{\,}{,} \text{to $m$ units})$, is a
complete group.
\end{Corollary}
% [** TN: Logically (but not typographycally), corollary ends here.]

For if, in this case, $G$~is not a characteristic
sub-group of~$K$, let $G'$~be a sub-group of~$K$ which, in the group
of isomorphisms of~$K$, is conjugate to~$G$. Then $G$~and~$G'$ being
both self-conjugate in~$K$ must have a common sub-group, since
$K$~cannot contain their direct product. But the common sub-group
of $G$ and~$G'$, being self-conjugate in~$K$, is a characteristic
sub-group of~$G$. This is impossible (§~162); hence $G$~is a
characteristic sub-group of~$K$.

\begin{Ex} Shew that the holomorph of an Abelian group of degree~$2^{m}$
and type $(1, 1, \dots\DPchg{\,}{,} \text{to $m$ units})$ is a complete group.
\end{Ex}

\begin{Remark}
\Par{168.} We shall now discuss the groups of isomorphisms of
certain special groups; and we shall begin with the case of a
cyclical group~$G$, of prime order~$p$, generated by an operation~$P$.
Every isomorphism of such a group must interchange among themselves
the $p - 1$ operations
\[
P,\ P^{2},\ \Dots,\ P^{p - 1};
\]
and if any isomorphism replaces $P$ by~$P^{\alpha}$, it must replace $P^{2}$ by~$P^{2\alpha}$,
and so on. Moreover, the symbol
\[
\Sub{\PadTo{P^{\alpha}}{P}\Add{,}
     \PadTo{P^{2\alpha}}{P^{2}}\Add{,} \Dots\Add{,}
     \PadTo{P^{(p - 1)\alpha}}{P^{p - 1}}}
    {P^{\alpha}\Add{,} P^{2\alpha}\Add{,} \Dots\Add{,} P^{(p - 1)\alpha}}
\]
does actually represent an isomorphism whatever number $\alpha$~may be
from $1$ to $p - 1$; for each operation occurs once in the second line,
and the change indicated leaves the multiplication-table of the group
unaltered. If $\alpha = p$, the symbol does not represent an isomorphism:
if $\alpha = p + \alpha'$, the symbol represents the same isomorphism as
\[
\Sub{\PadTo{P^{\alpha'}}{P}\Add{,}
     \PadTo{P^{2\alpha'}}{P^{2}}\Add{,} \Dots\Add{,}
     \PadTo{P^{(p - 1)\alpha'}}{P^{p - 1}}}
    {P^{\alpha'}\Add{,} P^{2\alpha'}\Add{,} \Dots\Add{,} P^{(p - 1)\alpha'}}.
\]
%% -----File: 256.png---Folio 240-------

The group of isomorphisms of a group of prime order~$p$ is
therefore a group of order~$p - 1$. Now the $n$th~power of the
isomorphism
\[
\Sub{P}{P^{\alpha}}
\]
is
\[
\Sub{P}{P^{\alpha^{n}}}.
\]

Hence if $\alpha$~is a primitive root of the congruence
\[
\alpha^{p - 1} - 1 \equiv 0,\ (\mod p),
\]
the group of isomorphisms is a cyclical group generated by the
isomorphism
\[
\Sub{P}{P^{\alpha}}.
\]

Finally, if $S$~is an operation satisfying the relations
\[
S^{p - 1} = 1,\quad S^{-1}PS = P^{\alpha},
\]
where $\alpha$~is a primitive root of~$p$, $\{S, P\}$~is the holomorph of~$G$.

The reader will at once observe that this group of order~$p(p - 1)$
is identical with the doubly transitive group of §~112. It is a
complete group.

\Par{169.} We shall consider next the case of any cyclical group.
\index{Defining relations@\Topic{Defining relations} of a group!for the holomorph of a cyclical group}%
\index{Holomorph@\Topic{Holomorph}!of a cyclical group|(}%

Suppose, first, that $G$~is a cyclical group of order~$p^{n}$, where $p$~is
an odd prime; and let it be generated by an operation~$S$. The group
contains $p^{n - 1}(p - 1)$ operations of order~$p^{n}$; if $S'$~is any one of these,
\[
\Sub{S}{S'}
\]
defines a distinct isomorphism. The group of isomorphisms is therefore
a group of order~$p^{n - 1}(p - 1)$. Moreover, since the congruence
\[
\alpha^{p^{n - 1}(p - 1)} - 1 \equiv 0,\ (\mod p^{n}),
\]
has primitive roots, the group of isomorphisms is a cyclical group.
The holomorph of~$G$ is defined by
\[
S^{p^{n}} = 1,\quad J^{p^{n - 1}(p - 1)} =  1,\quad J^{-1}SJ = S^{\alpha},
\]
where $\alpha$~is a primitive root of the congruence
\[
\alpha^{p^{n - 1}(p - 1)} - 1 \equiv 0,\ (\mod p^{n}).
\]

If $G$~is a cyclical group of order~$2^{n}$, it follows, in the same way,
that the group of isomorphisms is an Abelian group of order~$2^{n - 1}$.
In this case, however, the congruence
\[
\alpha^{2^{n - 1}} - 1 \equiv 0,\ (\mod 2^{n}), \quad n > 2,
\]
has no primitive root, and therefore the group of isomorphisms is not
cyclical. The congruence
\[
\alpha^{2^{n - 2}} - 1 \equiv 0\Add{,}\ (\mod 2^{n})
\]
%% -----File: 257.png---Folio 241-------
however has primitive roots, and a primitive root~$\alpha$ of this congruence
can always be found to satisfy the condition
\[
\alpha^{2^{n - 3}}  \equiv  1 + 2^{n - 1},\ (\mod 2^{n}).
\]

The powers of the isomorphism
\[
\Sub{S}{S^{\alpha}}
\]
then form a cyclical group of order~$2^{n - 2}$; and the only isomorphism
of order~$2$ contained in it is
\[
\Sub{S}{S^{1 + 2^{n - 1}}}.
\]

Hence
\[
\Sub{S}{S^{\alpha}} \quad\text{and}\quad \Sub{S}{S^{-1}},
\]
the latter not being contained in the sub-group generated by the
former, are two permutable and independent isomorphisms of orders
$2^{n - 2}$ and~$2$. They generate an Abelian group of order~$2^{n - 1}$ which is
the group of isomorphisms of~$G$. The corresponding holomorph is
given by
\begin{gather*}
S^{2^{n}} = 1,\quad J_{1}^{2^{n - 2}} = 1,\quad J_{2}^{2} = 1,\quad J_{1}J_{2} = J_{2}J_{1}, \\
J_{1}^{-1}SJ_{1} = S^{\alpha},\quad J_{2}SJ_{2} = S^{-1},
\end{gather*}
where $\alpha$~satisfies the conditions given above.

If $G$~is a cyclical group of order~$4$, its group of isomorphisms is
clearly a group of order~$2$.

\Par{170.} It is now easy to construct the group of isomorphisms of
\index{Complete group@\Topic{Complete group}!holomorph of a cyclical group of odd order is a}%
\index{Defining relations@\Topic{Defining relations} of a group!for the holomorph of a cyclical group}%
any cyclical group~$G$, and the corresponding holomorph. If the order
of~$G$ is $2^{n}p_{1}^{m_{1}} p_{2}^{m_{2}} \Dots$, where $p_{1}$, $p_{2}$,~\Dots\ are odd primes, $G$~is the
direct product of cyclical groups of orders $2^{n}$, $p_{1}^{m_{1}}$, $p_{2}^{m_{2}}$,~\Dots; and
every isomorphism of~$G$ transforms each of these groups into itself.
Hence if the groups of isomorphisms of these cyclical groups be
formed, and their direct product be then constructed, every operation
of the group so formed will give a distinct isomorphism of the group~$G$.
Moreover, the order
\[
2^{n - 1}p_{1}^{m_{1} - 1}(p_{1} - 1)p_{2}^{m_{2} - 1}(p_{2} - 1) \Dots
\]
of this group is equal to the number of operations of~$G$ whose order
is $2^{n}p_{1}^{m_{1}}p_{2}^{m_{2}} \Dots$, or in other words to the number of isomorphisms
of which $G$~is capable. The group thus formed is therefore the group
of isomorphisms of~$G$. The corresponding holomorph is clearly the
direct product of the holomorphs of the cyclical groups of orders $2^{n}$,
$p_{1}^{m_{1}}$,~etc.

When the order of~$G$ is odd, the holomorph~$K$ is easily shewn
to be a complete group. Suppose it to be expressed transitively,
as in §~158, in $N$~symbols, where $N$~is the order of~$G$; if $G$~is
not a characteristic sub-group of~$K$, let $G'$ be a group into which
$G$~is transformed by a contragredient isomorphism of~$K$. Then $G'$~is
%% -----File: 258.png---Folio 242-------
a self-conjugate sub-group of~$K$; and since $G'$~is cyclical, every
sub-group of~$G$' is a self-conjugate sub-group of~$K$. Hence a
generating operation of~$G'$ must be a circular substitution of
$N$~symbols. The $N$~operations of order~$2$ which transform each
operation of~$G'$ into its own inverse therefore each keep one symbol
fixed; hence each of them must transform every operation of~$G$
into its inverse. But there is only one such set of operations
of order~$2$, and therefore $G'$~cannot differ from~$G$. It follows by
Theorem~VII, §~167 that, as $G$~is a characteristic sub-group of~$K$, the
group~$K$ itself is complete.

If the order of~$G$ is even, $K$~must contain a self-conjugate
operation other than identity, namely the operation of order~$2$
contained in~$G$. Moreover, $K$~admits of a contragredient isomorphism
whose square is cogredient. From the mode of formation of~$K$,
it is clearly sufficient to verify this when the order of~$G$ is a
power of~$2$. The holomorph~$K$ is given by
\begin{gather*}
S^{2^{n}} = 1,\quad J_{1}^{2^{n - 2}} = 1,\quad J_{2}^{2} = 1,\quad J_{1}J_{2} = J_{2}J_{1}, \\
J_{1}^{-1}SJ_{1} = S^{\alpha},\quad J_{2}SJ_{2} = S^{-1};
\end{gather*}
where $\alpha$~is a primitive root of
\[
\alpha^{2^{n - 2}} - 1 \equiv 0\ (\mod 2^{n}),
\]
such that
\[
\alpha^{2^{n - 3}} + 1 \nequiv 0\ (\mod 2^{n}).
\]

In this group, $J_{2}$~is one of $2^{n - 1}$ conjugate operations
\[
J_{2},\ J_{2}S^{2},\ J_{2}S^{4},\ \Dots.
\]

Now it may be directly verified that $K$~admits the isomorphism
represented by
\[
\Sub{S, \PadTo{J_{1}S^{\efrac{1}{2} (1 - \alpha)}}{J_{1}}, \PadTo{J_{2}S}{J_{2}}}
    {S, J_{1}S^{\efrac{1}{2} (1 - \alpha)}, J_{2}S},
\]
Moreover, since this isomorphism changes $J_{2}$ into~$J_{2}S$, it cannot be a
cogredient isomorphism. Finally, the square of this isomorphism is
\[
\Sub{S, \PadTo{J_{1}S^{1 - \alpha}}{J_{1}}, \PadTo{J_{2}S^{2}}{J_{2}}}
    {S, J_{1}S^{1 - \alpha}, J_{2}S^{2}},
\]
or
\[
\Sub{S, \PadTo{S^{-1}J_{1}S}{J_{1}}, \PadTo{S^{-1}J_{2}S}{J_{2}}}
    {S, S^{-1}J_{1}S, S^{-1}J_{2}S},
\]
which is a cogredient isomorphism.

\Par{171.} We shall next consider the group of isomorphisms of an
Abelian group of order~$p^{n}$ and type $(1, 1, \dots\DPchg{\,}{,} \text{to $n$ units})$. Such a
group is generated by $n$~independent permutable operations of
\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of a cyclical group|)}%
\index{Holomorph@\Topic{Holomorph}!of a cyclical group|)}%
order~$p$, say
\[
P_{1},\ P_{2},\ \Dots,\ P_{n}.
\]
%% -----File: 259.png---Folio 243-------

Since every operation of the group is self-conjugate and of order~$p$,
\index{Abelian group@\Topic{Abelian group}!of order~$p^{m}$ and type $(1, 1, \dots, 1)$!group of isomorphisms of}%
\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots, 1)$}%
while the group contains no characteristic sub-group, there must
be isomorphisms transforming any one operation of the group into
any other. We may therefore begin by determining under what
conditions the symbol
\[
\Sub{P_{r}}
    {P_{1}^{a_{1r}} P_2^{a_{2r}} \Dots P_{n}^{a_{nr}}} \quad (r = 1, 2, \Dots, n),
\]
defines an isomorphism. This symbol replaces the operation
$P_{1}^{x_{1}} P_{2}^{x_{2}} \Dots P_{n}^{x_{n}}$ by $P_{1}^{y_{1}} P_{2}^{y_{2}} \Dots P_{n}^{y_{n}}$, where
\[
\begin{aligned}
y_{1} &\equiv a_{11}x_{1} + a_{12}x_{2} + \Dots + a_{1n}x_{n}, \\
y_{2} &\equiv a_{21}x_{1} + a_{22}x_{2} + \Dots + a_{2n}x_{n}, \\
\multispan{3}{\dotfill} \\
y_{n} &\equiv a_{n1}x_{1} + a_{n2}x_{2} + \Dots + a_{nn}x_{n},
\end{aligned}
\quad
(\mod p).
\]

Unless the $p^{n}$ operations $P_{1}^{y_{1}} P_{2}^{y_{2}} \Dots P_{n}^{y_{n}}$ thus formed are all
distinct, when for $P_{1}^{x_{1}} P_{2}^{x_{2}} \Dots P_{n}^{x_{n}}$ is put successively each of the $p^{n}$~operations
of the group, the symbol does not represent an isomorphism.
On the other hand, when this condition is satisfied, the symbol
represents a permutation of the operations among themselves which
leaves the multiplication table of the group unchanged; it is therefore
an isomorphism.

If this condition is satisfied, $x_{1}$, $x_{2}$,~\Dots, $x_{n}$ must be definite
numbers $(\mod p)$, when $y_{1}$, $y_{2}$,~\Dots, $y_{n}$ are given; and therefore
the above set of $n$ simultaneous congruences must be capable
of definite solution with respect to the~$x$'s. The necessary and
sufficient condition for this is that the determinant
\[
\begin{vmatrix}
a_{11}, & a_{12}, & \Dots, & a_{1n} \\
a_{21}, & a_{22}, & \Dots, & a_{2n} \\
\hdotsfor{4} \\
a_{n1}, & a_{n2}, & \Dots, & a_{nn}
\end{vmatrix}
\]
should not be congruent to zero $(\mod p)$.

Every distinct set of congruences of the above form, for
which this condition is satisfied, represents a distinct isomorphism
of the group, two sets being regarded as distinct if the
congruence
\[
a_{rs} \equiv a'_{rs}\ (\mod p)
\]
does not hold for each corresponding pair of coefficients. Moreover,
to the product of two isomorphisms will correspond the set of congruences
which results from carrying out successively the operations
indicated by the two sets that correspond to the two isomorphisms.
%% -----File: 260.png---Folio 244-------

The group of isomorphisms is therefore simply isomorphic with
the group of operations defined by all sets of congruences
\[
\begin{aligned}
  y_{1} &\equiv a_{11}x_{1} + a_{12}x_{2} + \Dots + a_{1n}x_{n}, \\
  y_{2} &\equiv a_{21}x_{1} + a_{22}x_{2} + \Dots + a_{2n}x_{n}, \\
  \multispan{3}{\dotfill} \\
  y_{n} &\equiv a_{n1}x_{1} + a_{n2}x_{2} + \Dots + a_{nn}x_{n},
\end{aligned}
\quad (\mod p)
\]
for which the relation
\[
\begin{vmatrix}
  a_{11}, & a_{12}, & \Dots, & a_{1n} \\
  a_{21}, & a_{22}, & \Dots, & a_{2n} \\
  \hdotsfor{4} \\
  a_{n1}, & a_{n2}, & \Dots, & a_{nn} \\
\end{vmatrix}
\nequiv 0\ (\mod p)
\]
is satisfied.

\Par{172.} The group thus defined is of great importance in many
\index{Homogeneous linear group@\Topic{Homogeneous linear group}!definition of}%
branches of analysis. It is known as the \emph{linear homogeneous} group.
In a subsequent Chapter we shall consider some of its more
important properties; we may here conveniently determine its
order. Let this be represented by~$N_{n}$, so that $N_{r}$~represents the
number of distinct solutions of the congruences
\[
\begin{vmatrix}
  a_{11}, & a_{12}, & \Dots, & a_{1r} \\
  a_{21}, & a_{22}, & \Dots, & a_{2r} \\
  \hdotsfor{4} \\
  a_{r1}, & a_{r2}, & \Dots, & a_{rr} \\
\end{vmatrix}
\nequiv 0\ (\mod p)
\]

Since the group of isomorphisms of the Abelian group of order~$p^{n}$
\index{Abelian group@\Topic{Abelian group}!of order~$p^{m}$ and type $(1, 1, \dots, 1)$!group of isomorphisms of}%
\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots, 1)$}%
transforms every one of its operations (identity excepted) into
every other, it can be represented as a transitive substitution group
of $p^{n} - 1$ symbols, and therefore, if $M$~is the order of the sub-group
that keeps $P_{1}$ unchanged,
\[
N_{n} = (p^{n} - 1) M.
\]

Now, in the congruences corresponding to an operation that keeps
$P_{1}$ unchanged, we have
\[
a_{11} \equiv 1,\ a_{12} \equiv a_{13} \equiv \Dots \equiv a_{1n} \equiv 0,\ (\mod p).
\]

Hence $M$~is equal to the number of solutions of the congruences
\[
\begin{vmatrix}
  1, & 0, & \Dots, & 0 \\
  a_{21}, & a_{22}, & \Dots, & a_{2n} \\
  \hdotsfor{4} \\
  a_{n1}, & a_{n2}, & \Dots, & a_{nn} \\
\end{vmatrix}
\nequiv 0\ (\mod p)
\]

The value of the determinant does not depend on the values of
$a_{21}$, $a_{31}$,~\Dots, $a_{n1}$, and therefore
\[
M = p^{n - 1} N_{n - 1}.
\]
%% -----File: 261.png---Folio 245-------

Hence
\[
N_{n} = (p^{n} - 1) p^{n - 1} N_{n - 1},
\]
and therefore immediately
\[
N_{n} = (p^{n} - 1) (p^{n} - p) \Dots (p^{n} - p^{n-1}).
\]

The reader will notice that an independent proof of this result
\index{Holomorph@\Topic{Holomorph}!of an Abelian group of order~$p^{n}$ and type $(1, 1, \dots, 1)$}%
has already been obtained in §~48. The discussion there given of
the number of distinct ways, in which independent generating
operations of an Abelian group of type $(1, 1, \Dots, 1)$ may be
\index{Abelian group@\Topic{Abelian group}!of order~$p^{m}$ and type $(1, 1, \dots, 1)$!holomorph of}%
chosen, is clearly equivalent to a determination of the order of the
corresponding group of isomorphisms.

{\Loosen The holomorph of an Abelian group, of order~$p^{n}$ and type
$(1, 1, \Dots\DPchg{\,}{,} \text{to $n$ units})$, can similarly be represented as a group of
linear transformations to the prime modulus~$p$. Consider, in fact,
the set of transformations}
\[
\begin{aligned}
  y_{1} &\equiv a_{11}x_{1} + a_{12}x_{2} + \Dots + a_{1n}x_{n} + b_{1}, \\
  y_{2} &\equiv a_{21}x_{1} + a_{22}x_{2} + \Dots + a_{2n}x_{n} + b_{2}, \\
  \multispan{3}{\dotfill} \\
  y_{n} &\equiv a_{n1}x_{1} + a_{n2}x_{2} + \Dots + a_{nn}x_{n} + b_{n},
\end{aligned}
\quad (\mod p);
\]
where the coefficients take all integral values $(\mod p)$ consistent
with
\[
\begin{vmatrix}
a_{11}, & a_{12}, &\Dots, &a_{1n} \\
a_{21}, & a_{22}, &\Dots, &a_{2n} \\
\hdotsfor{4} \\
a_{n1}, & a_{n2}, &\Dots, &a_{nn}
\end{vmatrix}
\nequiv 0.
\]

The set of transformations clearly forms a group whose order is~$N_{n} p^{n}$.
The sub-group formed by all the transformations
\[
y_{1} \equiv x_{1} + b_{1},\
y_{2} \equiv x_{2} + b_{2},\ \Dots,\
y_{n} \equiv x_{n} + b_{n},\ (\mod p),
\]
is an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots\DPchg{\,}{,} \text{to $n$ units})$, and it
is a self-conjugate sub-group. Moreover, the only operations of the
group, which are permutable with every operation of this self-conjugate
sub-group, are the operations of the sub-group itself; and,
since the order of the group is equal to the order of the holomorph
of the Abelian group, it follows that the group of transformations
must be simply isomorphic with the holomorph of the Abelian
group. In the simplest instance, where $p^{n}$~is~$2^{2}$, the holomorph is
simply isomorphic with the alternating group of four symbols. In
any case the holomorph, when expressed as in §~158, is a doubly
transitive group of degree~$p^{n}$.
\end{Remark}

\Par{173.} It has been seen in §~142 that, except when $n = 6$, the
symmetric group of degree~$n$ has $n$~and only~$n$ sub-groups
of order~$n - 1!$,\DPnote{** No () in original} which form a conjugate set. Hence by
Theorem~III, §~161, the group of isomorphisms of the symmetric
group of degree~$n$ can be expressed, except when $n = 6$,
%% -----File: 262.png---Folio 246-------
as a transitive group of degree~$n$. The symmetric group of $n$~symbols
however consists of all possible substitutions that can
be performed on the $n$~symbols, and therefore it must coincide
with its group of isomorphisms. Hence\footnotemark:---
\index{Holder@\Topic{Hölder}, quoted}%
\footnotetext{Hölder, \textit{Math.\ Ann.}\ Vol.~\VolNo{XLVI}, (1895), p.~345.}%

\begin{Theorem}[VIII.]
The symmetric group of $n$~symbols is a
\index{Complete group@\Topic{Complete group}!symmetric group is a, except for degree~$6$}%
\index{Symmetric group@\Topic{Symmetric group}!is a complete group, except for degree~$6$}%
complete group, except when $n = 6$.
\end{Theorem}

\begin{Corollary}
Except when $n = 6$, the alternating group of $n$~symbols
\index{Alternating group@\Topic{Alternating group}!group of isomorphisms of}%
\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of the alternating group}%
admits of one and only one class of contragredient
isomorphisms.
\end{Corollary}

For with this exception, the alternating group of degree~$n$
has just $n$ sub-groups of order~$\frac{1}{2} (n - 1)!$.

\begin{Remark}
\Par{174.} The alternating group of degree~$6$ occurs as a special
case of another class of groups of which we will determine the
groups of isomorphisms. These are the doubly and the triply transitive
groups that have been defined in §§~112,~113.

The doubly transitive group of degree~$p^{m}$ and order~$p^{m}(p^{m} - 1)$
\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of doubly transitive groups of degree~$p^{n} + 1$ and order $\frac{1}{2}p^{n}(p^{2n} - 1)$|(}%
there considered has a single set of $p^{m}$ conjugate cyclical sub-groups
of order~$p^{m} - 1$. Its group of isomorphisms can therefore be expressed
as a transitive group of degree~$p^{m}$. Let $p^{m}(p^{m} - 1) \mu$~be the
order of the group of isomorphisms. The order of a sub-group that
keeps two symbols fixed is~$\mu$; and every operation of this sub-group
must transform a cyclical sub-group of order~$p^{m} - 1$ into itself. With
the notation of §~112, we will consider the sub-group which keeps $0$
and $i$ fixed. Every operation of this sub-group must transform the
cyclical sub-group generated by the congruence\footnote
  {This and all subsequent congruences in the present section are to be taken,
  $\pmod{p}$.}
\[
x' \equiv ix
\]
into itself; and no operation of the sub-group can be permutable
with the given operation. If $S$~is an operation of the sub-group
which transforms
\[
x' \equiv ix
\]
into
\[
x' \equiv i^{\alpha}x,
\]
then $S$, when represented as a substitution, is given by
\[
\Sub{\PadTo{i^{\alpha + 1}}{i^{2}},
     \PadTo{i^{2\alpha + 1}}{i^{3}},
     \PadTo{i^{3\alpha + 1}}{i^{4}}, \Dots}
    {i^{\alpha + 1}, i^{2\alpha + 1}, i^{3\alpha + 1}, \Dots}.
\]

Now $S$~must transform the sub-group of order~$p^{m}$ into itself; it
must therefore be permutable with that operation of this sub-group
which changes $0$ into~$i$.
%% -----File: 263.png---Folio 247-------

This operation is given by
\[
x' \equiv x + i,
\]
and if we denote it by~$T$, then $S^{-1}TS$ changes $i^{y + 1}$ into $i^{\alpha z + 1}$, where
\[
i^{\efrac{y}{\alpha} + 1} + i \equiv i^{z + 1}.
\]

Now $T$ changes $i^{y + 1}$ into $i^{y + 1} + i$. Hence, since $S$~and $T$ are
permutable, we must for all values of~$y$ have the simultaneous
congruences
\[
i^{\efrac{y}{\alpha} + 1} + i \equiv i^{z + 1},
\]
and
\[
i^{y + 1} + i \equiv i^{\alpha z + 1}.
\]

Eliminating $z$, the congruence
\[
1 + i^{y} \equiv (1 + i^{\efrac{y}{\alpha}})^{\alpha}
\]
must hold for all values of~$y$ from $0$ to~$p^{m} - 1$. If $\alpha$~is not a power
of~$p$, this congruence involves an identity of the form
\[
a_{1}i^{\alpha_{1}} + a_{2}i^{\alpha_{2}} + \Dots \equiv 0,
\]
where all the indices are less than $p^{m} - 1$; and this is impossible.
Hence the only possible values of~$\alpha$ are $p$, $p^{2}$, $p^{3}$,~\Dots; and the
greatest possible value of~$\mu$ is~$m$.

Now the congruence
\[
x' \equiv x^{p}
\]
defines a substitution performed on the $p^{m}$~symbols permuted by the
group, and this substitution is permutable with the group. For if
we denote this operation by~$J$, and any operation
\[
x' \equiv \alpha x + \beta
\]
of the group by $\Sigma$, then $J^{-1} \Sigma J$ is
\[
x' \equiv \alpha^{p}x + \beta^{p},
\]
another operation of the group. Moreover, $J$~clearly transforms
\[
x' \equiv ix
\]
into its $p$th~power.

The group of isomorphisms of the doubly transitive group of
order~$p^{m}$ is therefore the group defined by
\[
x' \equiv ix,\quad x' \equiv x + i,\quad x' \equiv x^{p};
\]
where $i$~is a primitive root of the congruence
\[
i^{p^{m - 1}} \equiv 1.
\]

From this it immediately follows that the group of isomorphisms
of the triply transitive group, of degree~$p^{m} + 1$ and order
\[
(p^{m} + 1) p^{m} (p^{m} - 1),
\]
%% -----File: 264.png---Folio 248-------
defined by the congruences
\[
x' \equiv \frac{\alpha x + \beta}{\gamma x + \delta},
\]
where $\alpha$,~$\beta$, $\gamma$,~$\delta$ satisfy the conditions of §~113, is the group of order
$(p^{m} + 1) p^{m} (p^{m} - 1) m$ obtained by combining the previous congruences
with
\[
x' \equiv x^{p}.
\]

In fact, it may be immediately verified that the operation given
by this congruence is permutable with the group, and does not give
a cogredient isomorphism of the group. Moreover, by Theorem~III,
§~161, the group of isomorphisms of the given group can be expressed
as a transitive group of degree~$p^{m} + 1$; and therefore, among a class of
contragredient isomorphisms, there must be some transforming into
itself a sub-group which keeps one symbol fixed. Hence the order
of the group of isomorphisms cannot exceed
\[
(p^{m} + 1) p^{m} (p^{m} - 1) m.
\]

When $p$ is an odd prime, the triply transitive group of degree
$p^{m} + 1$, which may be defined by
\[
x' \equiv -\frac{1}{x},\quad x' \equiv ix,\quad x' \equiv x + i,
\]
contains as a self-conjugate sub-group a doubly transitive group
defined by
\[
x' \equiv -\frac{1}{x},\quad x' \equiv i^{2}x,\quad x' \equiv x + i.
\]

It will be shewn in Chapter~XIV that this is a simple group.
When $p^{m}$~is equal to~$3^{2}$, it is easy to verify that this group is simply
isomorphic with the alternating group of degree~$6$.

The group of isomorphisms of this simple group can be expressed
as a transitive group of degree~$p^{m} + 1$, and must clearly contain the
triply transitive group of order $(p^{m} + 1)p^{m}(p^{m} - 1)$ self-conjugately.
It therefore coincides with the group of order $(p^{m} + 1)p^{m}(p^{m} - 1)m$,
which has just been determined. The latter group is therefore
(Theorem~VI, Cor.\ §~166) a complete group. Further, if the simple
group be denoted by~$G$ and the group of isomorphisms by~$L$, the
factor group~$\dfrac{L}{G}$ will be determined by
\[
x' \equiv ix,\quad x' \equiv x^{p},
\]
when all operations of~$G$ are treated as the identical operation.
Denoting these operations by $I$~and~$J$, then $J^{-1}IJ$ is~$I^{p}$, which is
the same as $I$ multiplied by an operation of~$G$. Hence $\dfrac{L}{G}$~is an
Abelian group generated by two independent operations of orders $2$
and~$m$.
\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of doubly transitive groups of degree~$p^{n} + 1$ and order $\frac{1}{2}p^{n}(p^{2n} - 1)$|)}%
%% -----File: 265.png---Folio 249-------

The group of isomorphisms of the alternating group of degree~$6$
is therefore a group of order~$1440$; and the symmetric group of
degree~$6$ admits a single class of contragredient isomorphisms\footnotemark.
\end{Remark}
\footnotetext{Hölder, \textit{loc.~cit.}\ p.~343.}%
\index{Holder@\Topic{Hölder}, quoted}%

\Par{175.} Let $P$~be any group whose order is the power of a
prime, and let
\[
P,\ P_{1},\ P_{2},\ \Dots,\ P_{n},\ 1
\]
of orders
\[
p^{\alpha},\ p^{\alpha_{1}},\ p^{\alpha_{2}},\ \Dots,\ P^{\alpha_{n}},\ 1
\]
be a characteristic series (§~163) of~$P$. Every isomorphism
of~$P$ must transform $P_{r}$ and $P_{r + 1}$ into themselves, and therefore
also $\dfrac{P_{r}}{P_{r + 1}}$ into itself. Suppose now that an isomorphism~$I$
of~$P$ transforms every operation of $P_{r + 1}$ into itself and
every operation of $\dfrac{P_{r}}{P_{r + 1}}$ into itself. If $S$~is any operation of~$P_{r}$,
not contained in~$P_{r + 1}$, $I$~must transform $S$ into~$SA$, where
$A$~is some operation of~$P_{r + 1}$; so that, if $p^{\mu}$~is the order of~$A$,
$I$~transforms the operations of the set~$SP_{r + 1}$ cyclically
among themselves in sets of~$p^{\mu}$. Similarly, if $S'$~is an operation
of~$P_{r}$ not contained in the set~$SP_{r + 1}$, $I$~will transform the
operations of the set $S'P_{r + 1}$ cyclically among themselves in
sets of~$p^{\mu'}$. Hence the order of the isomorphism~$I$ is a multiple
of~$p$; and any isomorphism, that transforms every operation
of~$P_{r + 1}$ into itself and every operation of~$\dfrac{P_{r}}{P_{r + 1}}$ into itself and is
of order prime to~$p$, will transform every operation of~$P_{r}$ into
itself. Therefore, the only isomorphism of~$P$, that transforms
every operation of each of the groups
\[
\frac{P}{P_{1}},\ \frac{P_{1}}{P_{2}},\ \Dots,\ \frac{P_{n - 1}}{P_{n}},\ P_{n}
\]
into itself and is of order prime to~$p$, is the identical isomorphism.
Now each of these groups is an Abelian group, whose
operations are all of order~$p$; and it has been shewn that,
if $p^{m}$~is the order of such a group, the order of its group of
isomorphisms is
\[
(p^{m} - 1)(p^{m} - p)\Dots (p^{m} - p^{m - 1}).
\]
%% -----File: 266.png---Folio 250-------

Every isomorphism of~$P$, whose order is relatively prime to~$p$,
must therefore be such that its order is a factor of one of the
expressions of the above form, obtained by writing
\[
\alpha - \alpha_{1},\ \alpha_{1} - \alpha_{2},\ \Dots,\ \alpha_{n},
\]
in succession for~$m$. If then $k$~is the greatest of these numbers,
the order of any isomorphism of~$P$, whose order is relatively
prime to~$p$, is a factor of
\[
(p - 1) (p^{2} - 1) \Dots (p^{k} - 1).
\]

Herr Frobenius\footnote
  {``Ueber auflösbare Gruppen,~\VolNo{II},'' \textit{Berliner Sitzungsberichte}, 1895, p.~1028.}
\index{Frobenius@\Topic{Frobenius}, quoted}%
\index{Symbol@\Topic{Symbol}!theta@$\vartheta(P)$ and $\theta(P)$, definition of}%
has introduced the symbol~$\vartheta(P)$ to denote
this product. If $G$~is a group of order $p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \Dots p_{n}^{\alpha_{n}}$, where
$p_{1}$, $p_{2}$,~\Dots, $p_{n}$ are distinct primes, and if $P_{1}$, $P_{2}$,~\Dots, $P_{n}$ are
groups of orders $p_{1}^{\alpha_{1}}$, $p_{2}^{\alpha_{2}}$,~\Dots, $p_{n}^{\alpha_{n}}$ contained in~$G$, we shall use
the symbol~$\vartheta(G)$ to denote the least common multiple of
\[
\vartheta(P_{1}),\ \vartheta(P_{2}),\ \Dots,\ \vartheta(P_{n}).
\]

\Par{176.} If $P'$~is a sub-group of a group~$P$ of order~$p^{\alpha}$, $\vartheta(P')$~is
not necessarily a factor of~$\vartheta(P)$. For instance, the group of
order~$p^{4}$, generated by the four operations $A$,~$B$, $C$,~$D$ of order~$p$,
of which $A$~and $B$ are self-conjugate while
\[
D^{-1}CD = CA,
\]
has a characteristic sub-group $\{A, B\}$ of order~$p^{2}$, and it
has no characteristic sub-group of order~$p^{3}$. Hence
\[
\vartheta(P) = (p - 1) (p^{2} - 1).
\]

The sub-group~$P'$ however, which is generated by $A$,~$B$ and~$C$,
is an Abelian group whose operations are all of order~$p$; and
therefore
\[
\vartheta(P') = (p - 1) (p^{2} - 1) (p^{3} - 1).
\]

So again, for the Abelian group~$P$ of order~$p^{3}$, generated by
$A$~and~$B$, where
\[
A^{p^{2}} = 1,\quad B^{p} = 1,\quad AB = BA,
\]
we have
\[
\vartheta(P) = p - 1;
\]
while its sub-group~$P'$, generated by $A^{p}$~and~$B$, is an Abelian
group whose operations are all of order~$p$, so that
\[
\vartheta(P') = (p - 1) (p^{2} - 1).
\]
%% -----File: 267.png---Folio 251-------

Suppose now that $P$~is an Abelian group of order~$p^{\alpha}$,
generated by $\rho$~independent and permutable operations. For
such a group, we define\footnote
  {Frobenius, \textit{loc.~cit.}\ p.~1030.}
\index{Frobenius@\Topic{Frobenius}, quoted}%
\index{Symbol@\Topic{Symbol}!theta@$\vartheta(P)$ and $\theta(P)$, definition of}%
a new symbol~$\theta(P)$ by the equation
\[
\theta(P) = (p - 1) (p^{2} - 1) \Dots (p^{\rho} - 1).
\]

In forming the characteristic series of~$P$ in §~164, we
commenced with the series of groups~$G_{r}$ ($r = 1, 2, \Dots$), such
that $G_{r}$~consists of all the operations of~$P$ satisfying the
relation
\[
S^{p^{r}} = 1.
\]

Since $P$~is generated by $\rho$~independent operations, the order
of~$G_{1}$ is~$p^{\rho}$, and the order of~$\dfrac{G_{r + 1}}{G_{r}}$ cannot be greater than~$p^{\rho}$. If
the generating operations of~$P$ are all of the same order~$\beta$, the
series
\[
P,\ G_{\beta - 1},\ G_{\beta - 2},\ \Dots,\ G_{1},\ 1
\]
is a complete characteristic series of~$P$, and each factor-group
$\dfrac{G_{r + 1}}{G_{r}}$ is of order~$p^{\rho}$. In this case, therefore,
\[
\vartheta(P) = \theta(P).
\]

If however the generating operations are not all of the same
order, $G_{1}$~will not be the last term of the complete characteristic
series; nor will $G_{r + 1}$~and $G_{r}$ be consecutive terms in the series, if
the order of~$\dfrac{G_{r + 1}}{G_{r}}$ is~$p^{\rho}$. Hence, in this case, $\vartheta(P)$~will be a
factor of~$\theta(P)$.

If now $P'$~is any sub-group of~$P$, then it has been seen
(§~46) that the number of independent generating operations of~$P'$
is equal to or less than~$\rho$. Hence $\theta(P')$~is equal to or is a
factor of~$\theta(P)$.

\Par{177.} \begin{Theorem}[IX.]
If a group~$G$, of order~$N$, is transformed
into itself by an operation~$S$, whose order is relatively
prime to~$N\vartheta(G)$, every operation of~$G$ is permutable with~$S$\footnotemark.
\end{Theorem}
\footnotetext{Frobenius, \textit{loc.~cit.}\ p.~1030.}

Let $p_{1}^{\alpha_{1}}$~be the highest power of a prime~$p_{1}$ which divides~$N$.
%% -----File: 268.png---Folio 252-------
\index{Isomorphisms@\Topic{Isomorphisms} of a group with itself!limitation on the order of}%
The number of sub-groups of~$G$, and therefore also of $\{S, G\}$,
whose order is~$p_{1}^{\alpha_{1}}$ is a factor of~$N$; and since the order of~$S$ is
relatively prime to~$N$, one at least of them, say~$P_{1}$, must be
transformed into itself by~$S$. Now the order of~$S$ is relatively
prime both to~$p_{1}$ and to~$\vartheta(P_{1})$; and therefore the isomorphism of~$P$
given on transforming its operations by~$S$ must be the
identical isomorphism. In other words, $S$~is permutable with
every operation of~$P_{1}$. In the same way it follows that, if
$p_{2}^{\alpha_{2}}$, $p_{3}^{\alpha_{3}}$,~\Dots\ are the highest powers of $p_{2}$, $p_{3}$,~\Dots\ that divide~$N$,
there must be sub-groups of orders $p_{2}^{\alpha_{2}}$, $p_{3}^{\alpha_{3}}$,~\Dots\ with every
operation of each of which $S$~is permutable. But these groups
of orders $p_{1}^{\alpha_{1}}$, $p_{2}^{\alpha_{2}}$, $p_{3}^{\alpha_{3}}$,~\Dots\ generate~$G$; therefore $S$~is permutable
with every operation of~$G$.

\begin{Corollary}
If the order of an isomorphism of~$G$ is relatively
prime to~$N$, it must be a factor of~$\vartheta(G)$.
\end{Corollary}

It follows immediately, from the theorem, that there is no
isomorphism of~$G$ whose order contains a prime factor not
occurring in~$N\vartheta(G)$. Suppose, if possible, that the group of
isomorphisms of~$G$ contains an operation~$S$ of order~$q^{m}$, where
$q$~is a prime which does not divide~$N$, and that the highest
power of~$q$ that divides~$\vartheta(G)$ is~$q^{m'}$, where $m' < m$. If $p_{r}^{\alpha_{r}}$~is
the highest power of any prime~$p_{r}$ which divides~$N$, $S$~must
transform some sub-group~$P_{r}$ of order~$p_{r}^{\alpha_{r}}$ into itself. Since
$q^{m}$~is not a factor of~$\vartheta(P_{r})$, some power of~$S$ must be permutable
with every operation of~$P_{r}$. Hence, as in the theorem,
it follows that some power of~$S$, certainly~$S^{q^{m - 1}}$, must be permutable
with every operation of~$G$. But no operation of the
group of isomorphisms of~$G$, except identity, is permutable
with every operation of~$G$. Hence the group of isomorphisms
cannot contain an operation of order~$q^{m}$, if $m > m'$; and therefore
there is no isomorphism of~$G$ whose order contains a
higher power of~$q$ than~$q^{m'}$. If then $q^{r} q'^{r'} \Dots$, a number
relatively prime to~$N$, is the order of an isomorphism of~$G$, all
the numbers $q^{r}$, $q'^{r'}$,~\Dots\ are factors of~$\vartheta(G)$, and so also is
their product.

\Par{178.} The method, by which it has been shewn in §~175 that
the order of any isomorphism of~$P_{r}$ which transforms every
operation of each of the groups $\dfrac{P_{r}}{P_{r + 1}}$ and $P_{r + 1}$ into itself is a
%% -----File: 269.png---Folio 253-------
power of~$p$, may be used to obtain the following more general
result.

\begin{Theorem}[X.]
If $H$~is a self-conjugate sub-group of~$G$, the
order of an isomorphism of~$G$, which transforms every operation
of each of the groups $\dfrac{G}{H}$ and $H$ into itself, is a factor of the order
of~$H$.
\end{Theorem}

If $S$~is any operation of~$G$ not contained in~$H$, the
isomorphism will change $S$ into~$Sh$, where $h$~is some operation
of~$H$. If then $m$~is the order of~$h$, the isomorphism transforms
\[
S,\ Sh,\ Sh^{2},\ \Dots,\ Sh^{m - 1}
\]
cyclically; and therefore it transforms all the operations of the
set~$SH$ in cycles of $m$~each. If $S'$~is any operation of~$G$ not contained
in~$SH$, the isomorphism will interchange the operations
of the set~$S'H$ among themselves in cycles of $m'$~each, where $m'$
again is the order of some operation of~$H$. The isomorphism,
when expressed as a substitution performed on the operations
of~$G$, will consist of a number of cycles of $m$, $m'$,~\Dots\ symbols;
and its order is therefore the least common multiple of
$m$, $m'$,~\Dots. Now if $q$~is any prime that divides the order
of~$H$, and $q^{n}$~the highest power of~$q$ that occurs as the order of
a cyclical operation of~$H$, no power of~$q$ higher than~$q^{n}$ can
occur in any of the numbers $m$, $m'$,~\Dots; and $q^{n}$~is therefore
the highest power of~$q$ that can occur in their least common
multiple. This least common multiple, which is the order of
the isomorphism, must therefore divide the order of~$H$.

\Par{179.} \begin{Ex}[1.] Shew that, for the group of order~$p^{3}$ defined by
\begin{gather*}
P^{p} = 1, \quad Q^{p} = 1, \quad R^{p} = 1, \quad Q^{-1}PQ = PR, \\
RP = PR,\quad RQ = QR,
\end{gather*}
the symbol
\[
\Sub{\PadTo{P^{x} Q^{y} R^{z}}{P},
     \PadTo{P^{\alpha} Q^{\beta} R^{\gamma}}{Q},
     \PadTo{R^{n}}{R}}
    {P^{x} Q^{y} R^{z}, P^{\alpha} Q^{\beta} R^{\gamma}, R^{n}}
\]
gives an isomorphism if
\[
\beta x - \alpha y \equiv n,\ (\mod p).
\]
Hence determine the order of the group of isomorphisms.
\end{Ex}
%% -----File: 270.png---Folio 254-------

\begin{Ex}[2.] Shew that, for the group of order~$p^{3}$ defined by
\[
P^{p^{2}} = 1, \quad Q^{p} = 1,\quad Q^{-1}PQ = P^{1 + p},
\]
the symbol
\[
\Sub{\PadTo{P^{x} Q^{y}}{P}\Add{,} \PadTo{P^{\alpha p}Q}{Q}}
    {P^{x} Q^{y}, P^{\alpha p}Q}
\]
gives an isomorphism if $x$ is not a multiple of~$p$; and determine the
order of the group of isomorphisms.
\end{Ex}

\begin{Ex}[3.] Shew that the group of isomorphisms of the group of
order~$2^{n}$, defined by (§~63)
\[
P^{2^{n - 1}} = 1,\quad Q^{2} = P^{2^{n - 2}}, \quad Q^{-1}PQ = P^{-1},
\]
is of order~$2^{2n - 3}$, when $n > 3$. If $n = 3$, its order is~$24$ and it is
simply isomorphic with the last type but one of §~84.
\end{Ex}

\begin{Ex}[4.] If $G$~is a complete group of order~$N$, shew that the order
of~$K$, the holomorph of~$G$, is~$N_{2}$; and that the order of the holomorph
of~$K$ is~$2N^{4}$.
\end{Ex}
%% -----File: 271.png---Folio 255-------


\Chapter{XII.}{On the Graphical Representation of Group\protect\footnotemark.}
\index{Dyck@\Topic{Dyck}, quoted}%
\footnotetext{The investigations of this Chapter are due to Dyck, ``Gruppentheoretische
  Studien,'' \textit{Math.\ Ann.}, Vol.~\VolNo{XX}, (1882), pp.~1--44. We have followed Dyck's
  memoir closely except in two respects. Firstly, we have used a rather more
  definite geometrical operation than that of the memoir; and secondly, we have
  not specially considered a regular and symmetric division of a closed surface,
  apart from a merely regular division.}

\Par{180.} \First{Our} discussions hitherto have been confined exclusively
to groups of finite order. When however, as we now
propose to do, we consider a group in relation to the operations
that generate it, it becomes almost necessary to deal, incidentally
at least, with groups whose order is not finite; for it is not
possible to say a~priori what must be the number and the nature
of the relations between the given generating operations, which
will ensure that the order of the resulting group is finite.

Many of the definitions given in respect of finite groups
may obviously be extended at once to groups containing an
infinite number of operations. Among these may be specially
mentioned the definitions of a sub-group, of conjugate operations
and sub-groups, of self-conjugate sub-groups, of the
relation of isomorphism between two groups and of the
factor-group given by this relation. In regard to the last of
them, the isomorphism between two groups, one at least of
which is not of finite order, may be such that to one operation
of the one group there correspond an infinitely great number of
operations of the other. On the other hand, all the results
obtained for finite groups, which depend directly or indirectly on
the order of the group, necessarily become meaningless when
the group is not a group of finite order.
%% -----File: 272.png---Folio 256-------

\Par{181.} Suppose that
\index{General discontinuous group@\Topic{General discontinuous group} with a finite number of generating operations}%
\index{Group@\Topic{Group}!general}%
\[
S_{1},\ S_{2},\ \Dots,\ S_{n}
\]
represent any $n$~distinct operations which can be performed,
directly or inversely, on a common object, and that between
these operations no relations exist. Then the totality of the
operations represented by
\[
\Dots S_{p}^{\alpha} S_{q}^{\beta} S_{r}^{\gamma} \Dots,
\]
where the number of factors is any whatever and the indices
are any positive or negative integers, form a group~$G$ of infinite
order, which is generated by the $n$~operations. If, moreover,
whenever such a succession of factors as $S_{p}^{\alpha} S_{p}^{\beta}$ occurs in the
above expression, it is replaced by $S_{p}^{\alpha + \beta}$, each operation of
the group can be expressed in one way and in one way only by
an expression of the above form, which is then called \emph{reduced}.

It will sometimes be convenient to avoid the use of negative
indices in the expression of any operation of the group. To this
end we may write
\[
S_{1}S_{2} \Dots S_{n}S_{n + 1} = 1,
\]
so that $S_{n + 1}$ is a definite operation of the group; then
\[
S_{r}^{-1} = S_{r + 1}S_{r + 2} \Dots S_{n}S_{n + 1}S_{1} \Dots S_{r - 1}, \quad (r = 1, 2, \Dots, n).
\]
By using these relations to replace all negative powers of
operations wherever they occur, we may represent every
operation of the group in a single definite way by means of
the $n + 1$ operations
\[
S_{1},\ S_{2},\ \Dots,\ S_{n},\ S_{n + 1},
\]
with positive indices only.

The group, thus defined and represented, is the most
general group conceivable that is generated by $n$ distinct
operations. Any two such groups, for which $n$~is the same, are
simply isomorphic with each other.

Suppose now that
\[
\Bar{S}_{1},\ \Bar{S}_{2},\ \Dots,\ \Bar{S}_{n}
\]
represent $n$ distinct operations, but that, instead of being entirely
independent, they are connected by a relation of the form
\[
\Bar{S}_{p}^{a} \Bar{S}_{q}^{b} \Dots \Bar{S}_{r}^{c} = 1,
\]
which will be represented by
\[
f(\Bar{S}_{i}) = 1.
\]
%% -----File: 273.png---Folio 257-------
\index{General discontinuous group@\Topic{General discontinuous group} with a finite number of generating operations!relation of special groups to}%
\index{Group@\Topic{Group}!special}%
\index{Isomorphism@\Topic{Isomorphism} of general and special groups}%

If $\Bar{G}$ is the group generated by these operations, an isomorphism
may be established between $G$ and $\Bar{G}$ by taking $\Bar{S}_{i}$
($i = 1, 2, \Dots, n$) as the operation of~$\Bar{G}$ that corresponds to
the operation~$S_{i}$ of~$G$.

Then to every operation of $G$
\[
\Dots S_{p}^{\alpha} S_{q}^{\beta} S_{r}^{\gamma} \Dots
\]
will correspond a single definite operation
\[
\Dots \Bar{S}_{p}^{\alpha} \Bar{S}_{q}^{\beta} \Bar{S}_{r}^{\gamma} \Dots
\]
of~$\Bar{G}$; for the supposition that two distinct operations of~$\Bar{G}$
correspond to the same operation of~$G$ leads to the result that
between the generating operations of~$G$ there is a relation,
which is not the case. On the other hand, to the identical
operation of~$\Bar{G}$ there will correspond an infinite number of
distinct operations of~$G$, namely those which are formed by
combining together in every possible way all operations of~$G$ of
the form
\[
R^{-1} f(S_{i})R,
\]
where $R$~is any operation of~$G$. These operations of~$G$ form a
self-conjugate sub-group~$H$; the corresponding factor-group
$\dfrac{G}{H}$ is simply isomorphic with~$\Bar{G}$.

If between the generating operations of~$\Bar{G}$ there are several
independent relations
\[
f_{1}(\Bar{S}_{i}) = 1,\quad
f_{2}(\Bar{S}_{i}) = 1,\quad \Dots,\quad
f_{m}(\Bar{S}_{i}) = 1,
\]
it may be shewn exactly as before that the groups $G$ and $\Bar{G}$ are
isomorphic in such a way that to the identical operation of~$\Bar{G}$
there corresponds that self-conjugate sub-group of~$G$, which is
formed by combining in every possible way all the operations
of~$G$ of the form
\[
R^{-1} f_{j}(S_{i}) R,\quad (j = 1, 2, \Dots, m).
\]

\Par{182.} We may at once extend the result of the preceding
paragraph in the following way:---

\begin{Theorem}[I.]
If $G$~is the group generated by the $n$~operations
\[
S_{1},\ S_{2},\ \Dots,\ S_{n},
\]
between which the $m$~relations
\[
f_{1}(S_{i}) = 1,\quad
f_{2}(S_{i}) = 1,\quad \Dots,\quad
f_{m}(S_{i}) = 1,
\]
%% -----File: 274.png---Folio 258-------
exist; and if $\Bar{G}$~is the group generated by the $n$~operations
\[
\Bar{S}_{1},\ \Bar{S}_{2},\ \Dots,\ \Bar{S}_{n},
\]
which are connected by the same $m$~relations
\[
f_{1}(\Bar{S}_{i}) = 1,\quad
f_{2}(\Bar{S}_{i}) = 1,\quad \Dots,\quad
f_{m}(\Bar{S}_{i}) = 1,
\]
as hold between the generating operations of~$G$, and by the
further $m'$~relations
\[
g_{1}(\Bar{S}_{i}) = 1,\quad
g_{2}(\Bar{S}_{i}) = 1,\quad \Dots,\quad
g_{m'}(\Bar{S}_{i}) = 1;
\]
then $\Bar{G}$~is simply isomorphic with the factor-group~$\dfrac{G}{H}$; where $H$~is
that self-conjugate sub-group of~$G$, which results from combining
in every possible way all operations of the form
\[
R^{-1} g_{j}(S_{i}) R,\quad (j= 1, 2, \Dots, m'),
\]
\index{General discontinuous group@\Topic{General discontinuous group} with a finite number of generating operations!relation of special groups to}%
\index{Isomorphism@\Topic{Isomorphism} of general and special groups}%
$R$ being any operation of~$G$.
\end{Theorem}

{\Loosen In proving this theorem, it is sufficient to notice that, if we
take $\Bar{S}_{i}$ ($i = 1, 2, \Dots, n$) as the operation of~$\Bar{G}$ which corresponds
to the operation~$S_{i}$ of~$G$, then to each operation of~$G$ a single
definite operation of~$\Bar{G}$ will correspond, while to the identical
operation of~$\Bar{G}$ there corresponds the self-conjugate sub-group~$H$
of~$G$.}

\begin{Remark}
The theorem just stated is of such a general nature that it is
perhaps desirable to illustrate it by considering shortly some simple
examples.

Let us take first the case of a group~$G$, generated by two independent
operations $S_{1}$ and~$S_{2}$, subject to no relations; and let us
suppose that the single relation
\[
\Bar{S}_{1} \Bar{S}_{2} = \Bar{S}_{2} \Bar{S}_{1}
\]
holds between the generating operations of~$\Bar{G}$. The self-conjugate
sub-group~$H$ of~$G$ then consists of all the operations
\[
\Dots S_{1}^{\alpha_{n}} S_{2}^{\beta_{n}} \Dots
\]
of~$G$ which reduce to identity if we regard $S_{1}$~and $S_{2}$ as permutable;
or, in other words, of those operations of~$G$ for which the relations
\[
\sum \alpha_{n} = 0,\quad \sum \beta_{n} = 0
\]
simultaneously hold.

In respect of this sub-group, the operations of~$G$ can be divided
into an infinite number of classes of the form
\[
S_{1}^{p} S_{2}^{q} H.
\]
%% -----File: 275.png---Folio 259-------

For the operations of the class $S_{1}^{p} S_{2}^{q}H$, multiplied by those of
the class $S_{1}^{p'} S_{2}^{q'}H$, give always operations of the class
\[
S_{1}^{p} S_{2}^{q} S_{1}^{p'} S_{2}^{q'}H,
\]
since $H$~is a self-conjugate sub-group; and, because
\[
S_{1}^{p} S_{2}^{q} S_{1}^{p'} S_{2}^{q'}
  = S_{1}^{p + p'} · S_{1}^{-p'} S_{2}^{q} S_{1}^{p'} S_{2}^{-q} · S_{2}^{q + q'},
\]
while $S_{1}^{-p'} S_{2}^{q} S_{1}^{p'} S_{2}^{-q}$ belongs to~$H$, the class $S_{1}^{p} S_{2}^{q} S_{1}^{p'} S_{2}^{q'} H$ is the
same as $S_{1}^{p + p'} S_{2}^{q + q'} H$. Hence the operations of any two given classes,
multiplied in either order, give the same third class; and therefore
the group~$\dfrac{G}{H}$ is an Abelian group generated by two permutable,
but otherwise unrestricted, operations.

As a second illustration, we will choose a case in which $\Bar{G}$~is of
finite order. Let $G$ be generated by the operations $S$~and~$T$, which
satisfy the relations
\[
S^{3} = 1,\quad T^{3} = 1,\quad (ST)^{3} = 1;
\]
and for~$\Bar{G}$, suppose that the generating operations satisfy the \DPtypo{addtional}{additional}
relation
\[
(ST^{2})^{3} = 1.
\]

Then $H$~is formed by combining in all possible ways the
operations
\[
R^{-1} (ST^{2})^{3} R.
\]

Now it may be easily verified that in~$G$, the operation~$ST^{2}$
belongs to a set of three conjugate operations
\[
ST^{2},\quad TST,\quad T^{2}S;
\]
and that these three operations are permutable among themselves,
while their product is identity. Hence $H$~consists of the Abelian
group
\[
(ST^{2})^{3\alpha} (TST)^{3\beta};
\]
and in respect of~$H$, $G$~may be divided into $27$~classes of the form
\[
S^{x} (ST^{2})^{y} (TST)^{z} H,\quad (x, y, z = 0, 1, 2).
\]

The group~$\Bar{G}$ will be defined by the laws according to which
these $27$~classes combine among themselves; and the reader will
have no difficulty in verifying that it is isomorphic with the non-Abelian
group of order~$27$, whose operations are all of order~$3$ (§~73).
\end{Remark}

\begin{Ex} If $S_{1} (= 1)$, $S_{2}$,~\Dots, $S_{N}$ are the operations of a group~$G$
of finite order, prove that the totality of the operations of the form
$S_{p}^{-1} S_{q}^{-1} S_{p} S_{q}$ generate a self-conjugate sub-group~$H$; and that the
group~$\dfrac{G}{H}$ is an Abelian group. Shew also that, if $H'$~is a self-conjugate
sub-group of~$G$ and if $\dfrac{G}{H'}$~is Abelian, then $H'$~contains~$H$.
\index{Miller@\Topic{Miller}, quoted}%
\Attrib{Miller, \textit{Quarterly Journ.\ of Math.}, Vol.~\VolNo{XXVIII}, (1896), p.~266.}
\end{Ex}
%% -----File: 276.png---Folio 260-------

\Par{183.} For the further discussion of a group, as defined by its
\index{Graphical representation@\Topic{Graphical representation}!of a cyclical group}%
generating operations and the relations between them, a suitable
graphical mode of representation becomes of the greatest assistance.
To this we shall now proceed.

In the simple case in which the group is generated by a
single unrestricted operation, such a representation may be
constructed as follows. Let $C_{1}$~and $C_{-1}$ be two circles which
touch each other; $C_{2}$~and $C_{-2}$ the inverses of~$C_{-1}$ in~$C_{1}$ and $C_{1}$
in~$C_{-1}$; $C_{3}$~and $C_{-3}$ the inverses of~$C_{-2}$ in~$C_{1}$ and $C_{2}$ in~$C_{-1}$, and
so on. These circles (fig.~1) divide the plane in which they are
% [Illustration: Fig. 1.]
\Figure{1}{fig01}
drawn into an infinite number of crescent-shaped spaces. Suppose
now that the space between $C_{1}$~and~$C_{-1}$ is left white, and
the spaces between $C_{1}$~and~$C_{2}$ and between $C_{-1}$~and~$C_{-2}$ (on either
side of this white space) are coloured black; the next pair on
either side left white, the next coloured black, and so on. Then
any white space may be transformed into any other (and any
black into any other) by an even number of inversions at the
circles $C_{-1}$ and~$C_{1}$; and if $S$~denote\DPnote{** [sic]} the operation consisting
of an inversion at~$C_{-1}$ followed by an inversion at~$C_{1}$,
the space between $C_{-1}$~and~$C_{1}$ will be transformed into
another perfectly definite white space by the operation~$S^{n}$,
while conversely the operation necessary to transform the space
between $C_{-1}$~and~$C_{1}$ into any other given white space will be a
definite power of~$S$. Hence if one of the white spaces, say
%% -----File: 277.png---Folio 261-------
that between $C_{-1}$~and~$C_{1}$, is taken to correspond to the identical
operation, there is then a unique correspondence between the
white spaces and the operations of the group generated by
the unrestricted operation~$S$; and the figure that has been
constructed gives a graphical representation of the group. It
should be noticed that the actual geometrical process of inversion,
which has been here used to construct the spaces
corresponding to the operations of the group, is in no way
essential to the graphical representation. It is however convenient
as giving definiteness to the construction; and later,
when we deal with the case of a general group, such definiteness
becomes almost a necessity.

In a precisely similar manner, the group generated by a
single relation~$S$, satisfying the relation
\[
S^{n} = 1,
\]
may be treated. In this case, we take two circles $C_{-1}$ and~$C_{1}$
intersecting at an angle~$\dfrac{\pi}{n}$, and from these form, as before,
the circles obtained by successive inversions. This gives a
\index{Graphical representation@\Topic{Graphical representation}!of a cyclical group}%
% [Illustration: Fig. 2.]
\Figure{2}{fig02}
finite series of $n$~circles, each of which intersects the two next to
it on either side at angles~$\dfrac{\pi}{n}$, while the $n$~circles divide the
%% -----File: 278.png---Folio 262-------
plane into $2n$~spaces. If these are left white and coloured
black in alternate succession, and if one of the white is taken
to correspond to the identical operation, there is a unique
correspondence between the white spaces and the operations of
the group generated by~$S$, where $S$~represents the result of
successive inversions first at~$C_{-1}$ and then at~$C_{1}$.

This operation obviously satisfies the relation
\[
S^{n} = 1
\]
and no simpler relation; so that the figure gives a graphical
representation of a cyclical group of order~$n$.

The systems of circles in figures 1~and 2 have a common
\index{Graphical representation@\Topic{Graphical representation}!of a general group|(}%
geometrical property which may be noticed here as it will be of
use in the sequel. Successive inversions at any one of the pairs
$C_{-1}$~and~$C_{1}$, $C_{1}$~and~$C_{2}$, $C_{2}$~and~$C_{3}$ are equivalent to the operation~$S$;
and therefore successive inversions at $C_{-1}$~and~$C_{r}$ are equivalent
to the operation~$S^{r}$. Hence the result of an even
number of inversions at any of the circles in either figure is
equivalent to some operation of the group that the figure
represents.

\Par{184.} We may now proceed to construct a graphical
representation of the group which is generated by $n$~operations
subject to no relations. To this end, suppose $n + 1$ circles
drawn, each of which is external to all the others while each
touches two and only two of the rest. Such a system can be
drawn in an infinite variety of ways: we will suppose, to give
definiteness and simplicity to the resulting figure, that the
$n + 1$ points of contact lie on a circle, which cuts the $n + 1$
circles orthogonally. If these $n + 1$ points taken in order are
$A_{1}$, $A_{2}$,~\Dots, $A_{n + 1}$, the successive circles are $A_{n + 1}A_{1}$, $A_{1}A_{2}$,~\Dots,
$A_{n}A_{n + 1}$. We will suppose that only so much of these
circles is drawn as lies within the common orthogonal circle
$A_{1}A_{2}\Dots A_{n + 1}$. The $n + 1$ circular arcs $A_{n + 1}A_{1}$, $A_{1}A_{2}$,~\Dots\
then bound a finite simply connected plane figure which we
will denote provisionally by~$P$. Suppose now that $P$~is inverted
in each of its sides, that the resulting figures are inverted in
each of their new sides, and so on continually. Then from
their mode of formation no two of the figures thus arising can
%% -----File: 279.png---Folio 263-------
overlap either wholly or in part; and when the process is
continued without limit, every point in the interior of the
orthogonal circle $A_{1}A_{2} \Dots A_{n + 1}$ will lie in one and only one of
the figures thus formed from~$P$ by successive inversions.

If $P'$~is any one of the figures or polygons so formed, the set
of inversions by which it is derived from~$P$ is perfectly definite.
For suppose, if possible, that $P'$~is derived from~$P$ by two
distinct sets of inversions represented by $\Sigma$~and~$\Sigma'$. Then
$\Sigma\Sigma'^{-1}$~is a set of inversions in the sides of~$P$ which transforms~$P$
into itself. But every set of inversions necessarily transforms~$P$
into some polygon lying outside it, and therefore
\[
\Sigma\Sigma'^{-1} = 1;
\]
or the set of inversions composing~$\Sigma$ is identical term for term
%[Illustration: Fig. 3.]
\Figure{3}{fig03}
with the set composing~$\Sigma'$. It immediately follows that the
polygons can be divided into two sets, according as they are
derived from~$P$ by an even or an odd number of inversions. The
%% -----File: 280.png---Folio 264-------
latter we shall suppose coloured black, and the former (including~$P$)
left white. Every white polygon will be surrounded
by black polygons and vice~versâ. Since there is only one
definite set of inversions that will transform~$P$ into any other
white polygon~$P'$, the $n + 1$ corners of~$P'$ will correspond one
by one to the $n + 1$ corners of~$P$; and when the perimeters of
the two polygons are described in the same direction of rotation
with regard to their interiors, the angular points that correspond
will occur in the same cyclical order. On the other hand, in
order that the corresponding angular points of a white and a
black polygon may occur in the same cyclical order, their perimeters
must be described in opposite directions. In consequence
of these results, we may complete our figure (fig.~3) by lettering
every angular point of every polygon with the same letter that
occurs at the corresponding angular point of the polygon~$P$.

\Par{185.} If now $T_{1}$, $T_{2}$,~\Dots, $T_{n + 1}$ represent inversions at
$A_{1}A_{2}$, $A_{2}A_{3}$,~\Dots, $A_{n + 1}A_{1}$, the operation $T_{r - 1}T_{r}$ leaves the
corner~$A_{r}$ of~$P$ unchanged and it transforms~$P$ into the next
white polygon which has the corner~$A_{r}$ in common with~$P$, the
direction of turning round~$A_{r}$ coinciding with the direction
$A_{n + 1}A_{n} \Dots A_{1}$ of describing the perimeter of~$P$. For brevity,
we shall describe this transformation of~$P$ as a positive rotation
round~$A_{r}$. If then we denote the operation $T_{r - 1}T_{r}$ by the
single symbol~$S_{r}$, we may say that $S_{r}$~produces a positive
rotation of~$P$ round~$A_{r}$. Let $P_{1}$~be the new polygon so
obtained; and let $P_{1}'$~be the polygon into which any other
white polygon~$P'$ is changed by a positive rotation round the
corner of~$P'$ that corresponds to~$A_{r}$. Then if $\Sigma$~is the set of
inversions that changes $P$ into~$P'$, it also changes $P_{1}$ into~$P_{1}'$: so
that $\Sigma^{-1}S_{r}\Sigma$ changes $P'$ into~$P_{1}'$, \ie\ produces a positive rotation
round the corner~$A_{r}$ of~$P'$; and $S_{r}\Sigma$~changes $P$ into~$P_{1}'$.

Let us now represent the operations
\[
T_{n + 1}T_{1},\ T_{1}T_{2},\ \Dots,\ T_{n}T_{n + 1},
\]
by
\[
S_{1},\ S_{2},\ \Dots,\ S_{n + 1},
\]
so that
\[
S_{1}S_{2} \Dots S_{n}S_{n + 1} = 1.
\]

Then every operation, consisting of a pair of inversions in
the sides of~$P$, can be represented in terms of
\[
S_{1},\ S_{2},\ S_{3},\ \Dots,\ S_{n}.
\]
%% -----File: 281.png---Folio 265-------
For an inversion at~$A_{r}A_{r + 1}$, followed by an inversion at~$A_{s}A_{s + 1}$,
is given by~$T_{r}T_{s}$; and
\begin{align*}
T_{r}T_{s}
  &= T_{r}T_{r + 1} · T_{r + 1}T_{r + 2} · \Dots · T_{s - 1}T_{s} \\
  &= S_{r + 1}S_{r + 2}\Dots S_{s}.
\end{align*}

If $s > r$, this is of the form required. If $s < r$, the term~$S_{n + 1}$
that then occurs may be replaced by
\[
S_{n}^{-1} S_{n - 1}^{-1} \Dots S_{2}^{-1} S_{1}^{-1}.
\]
Hence finally, every operation consisting of an even number of
inversions in the sides of~$P$ can be expressed in terms of
\[
S_{1},\ S_{2},\ S_{3},\ \Dots,\ S_{n};
\]
and with a restriction to positive indices, every such operation
can be expressed in terms of
\[
S_{1},\ S_{2},\ \Dots,\ S_{n},\ S_{n + 1}.
\]

Now it has been seen that no two operations, each consisting
of a set of inversions in the sides of~$P$, can be identical unless
the component inversions are identical term for term. Hence
no two reduced operations of the form
\[
f(S_{i})
\]
are identical; in other words, the $n$ generating operations
\[
S_{1},\ S_{2},\ \Dots,\ S_{n}
\]
are subject to no relations.

If then we take the polygon~$P$ to correspond to the identical
operation of the group~$G$ generated by
\[
S_{1},\ S_{2},\ \Dots,\ S_{n},
\]
each white polygon may be taken as associated with the
operation which will transform~$P$ into it. The foregoing discussion
makes it clear that in this way a unique correspondence
is established between the operations of~$G$ and the system of
white polygons; or in other words, that the geometrical figure
gives a complete graphical representation of the group.

Moreover, since the operation $\Sigma^{-1}S_{r} \Sigma$ is a positive rotation
round the corner~$A_{r}$ of the polygon~$\Sigma$ (calling now $P$ the
polygon~$1$), a simple rule may be formulated for determining
by a mere inspection of the figure what operation of the group
any given white polygon corresponds to.
%% -----File: 282.png---Folio 266-------

This rule may be stated as follows. Let a continuous line
be drawn inside the orthogonal circle from a point in the white
polygon~$1$ to a point in any other white polygon, so that every
consecutive pair of white polygons through which the line
passes have a common corner, a positive rotation round which
leads from the first to the second of the pair. This is always
possible. Then if the common corners of each consecutive pair
of white polygons through which the line passes, starting from~$1$,
are $A_{p}$, $A_{q}$,~\Dots, $A_{r}$, $A_{s}$, the final white polygon corresponds
to the operation
\[
S_{s}S_{r} \Dots S_{q}S_{p}\footnotemark.
\]

\footnotetext
  {The reader who refers to Prof.~Dyck's memoir should notice that the
  definition of the operation~$S_{r}$ above given is not exactly equivalent to that used
  by Prof.~Dyck. With his notation, the white polygon here considered would
  correspond to the operation $S_{p}S_{q} \Dots S_{r}S_{s}$.}%
\begin{Remark}
\Par{186.} The graphical representation of a general group we have
\index{Graphical representation@\Topic{Graphical representation}!of a special group|(}%
thus arrived at is only one of an infinite number that could be constructed;
and we choose this in preference to others mainly because
the form of the figure and the relative positions of the successive polygons
are readily apprehended by the eye. As regards the mere establishment
of such a representation we might, still using the process of
inversion for the purpose of forming a definite figure, have started
with $n + 1$ circles each exterior to and having no point in common
with any of the others. Taking as the figure~$P$ the space external
to all the circles and inverting it continually in the circles, we should
form a series of black and white spaces of which the latter would
again give a complete picture of the group. It is however only
necessary to begin the construction of such a figure in order to
convince ourselves that it would not appeal to the eye in the same
way as the figure actually chosen.

Moreover, as in the representation of a cyclical group, the process
of inversion is in no way essential to the representation at which we
arrive. Any arbitrary construction, which would give us the series
of white and black polygons having, in the sense of the geometry of
position, the same relative configuration as our actual figure, would
serve our purpose equally well.
\end{Remark}

\Par{187.} If $\Sigma$~is the operation which transforms $P$ into~$P'$, and
if $Q$~is the black polygon which has the side~$A_{r}A_{r + 1}$ in common
with~$P$, then $\Sigma$~transforms $Q$ into the black polygon which has
in common with~$P'$ the side corresponding to~$A_{r}A_{r + 1}$. If then
we take $Q$ to correspond to the identical operation, any black
polygon will correspond to the same operation as that white
\index{Graphical representation@\Topic{Graphical representation}!of a general group|)}%
%% -----File: 283.png---Folio 267-------
polygon with which it has the side~$A_{r}A_{r + 1}$ in common. In this
way we may regard our figure as divided in a definite way into
double-polygons, each of which represents a single operation of
the group.

\Par{188.} We have next to consider how, from the representation
of a general group whose $n$~generating operations are subject
to no relations, we may obtain the representation of a special
group generated by $n$ operations connected by a series of
relations
\[
F_{j}(S_{i}) = 1,\quad (j = 1, 2, \Dots, m).
\]

It has been seen (§~181) that to the identical operation of the
special group there corresponds a self-conjugate sub-group~$H$ of
the general group; or in other words, that the set of operations
$\Sigma H$ of the general group give one and only one operation in the
special group.

Hence, to obtain from our figure for the general group one
that will apply to the special group, we must regard all the
double-polygons of the set~$\Sigma H$ as equivalent to each other;
and if from each such set of double-polygons we choose one as
a representative of the set, the totality of these representative
polygons will have a unique correspondence with the operations
of the special group.

We shall first shew that a set of representative double-polygons
can always be chosen so as to form a single simply
connected figure. Starting with the double-polygon,~$P_{1}$, that
corresponds to the identical operation of the general group as
the one which shall correspond to the identical operation
of the special group, we take as a representative of some
other operation of the special group a double-polygon,~$P_{2}$,
which has a side in common with~$P_{1}$. Next we take as
a representative of some third operation of the special group
a double-polygon which has a side in common with either
$P_{1}$ or~$P_{2}$; and we continue the choice of double-polygons
in this way until it can be carried no further. The set of
double-polygons thus arrived at of necessity forms a single
simply connected figure~$C$, bounded by circular arcs; and no
two of the double-polygons belonging to it correspond to the
same operation of the special group. Moreover, in~$C$ there is
%% -----File: 284.png---Folio 268-------
one double-polygon corresponding to each operation of the
special group. To shew this, let $C'$ be the figure formed by
combining with~$C$ every double-polygon which has a side in
common with~$C$; and form $C''$ from~$C'$, $C'''$~from~$C''$, and so on,
as $C'$~has been formed from~$C$. From the construction of~$C$ it
follows that every polygon in~$C'$ is equivalent, in respect of the
special group, to some polygon in~$C$. Similarly, every polygon
in~$C''$ is equivalent to some polygon in~$C'$ and therefore to some
polygon in~$C$; and so on. Hence finally, every polygon in the
complete figure of the general group is equivalent to some
polygon in~$C$, in respect of the special group; and therefore,
since no two polygons of~$C$ are equivalent in respect of the
special group, the figure~$C$ is formed of a complete set of representative
double-polygons for the special group.

Suppose now that $S$~is a double-polygon outside~$C$, with a
side $A_{r}' A_{r + 1}'$ belonging to the boundary of~$C$. Within~$C$ there
must be just one polygon, say~$ST$, of the set~$SH$. If this
polygon lay entirely inside~$C$, so as to have no side on the
boundary of~$C$, every polygon having a side in common with it
would belong to~$C$. Now since $S$~and $ST$ are equivalent, every
polygon having a side in common with~$S$ is equivalent to some
polygon having a side in common with~$ST$. Hence since $C$~contains
no two equivalent polygons, $ST$~must have a side on
the boundary of~$C$; and if this side is~$A_{r}'' A_{r + 1}''$, the operation~$T$
of~$H$ transforms $A_{r}' A_{r + 1}'$ into $A_{r}'' A_{r + 1}''$. Moreover, no operation
of~$H$ can transform $A_{r}' A_{r + 1}'$ into another side of~$C$; for if this
were possible, $C$~would contain two polygons equivalent to~$S$.
It is also clear that, regarded as sides of polygons within~$C$,
$A_{r}' A_{r + 1}'$~and $A_{r}'' A_{r + 1}''$ belong to polygons of different colours.
Hence a correspondence in pairs of the sides of~$C$ is established:
to each portion $A_{r}' A_{r + 1}'$ of the boundary of~$C$, which forms
a side of a white (or black) polygon of~$C$, there corresponds
another definite portion $A_{r}'' A_{r + 1}''$, forming a side of a black (or
white) polygon of~$C$, such that a certain operation of~$H$ and its
inverse will change one into the other, while no other operation
of~$H$ will change either into any other portion of the boundary
of~$C$.

The system of double-polygons forming the figure~$C$, and
the correspondence of the sides of~$C$ in pairs, will now give a
%% -----File: 285.png---Folio 269-------
complete graphical representation of the group. For the figure
has been formed so that there is a unique correspondence
between the white polygons of~$C$ and the operations of the
group, such that until we arrive at the boundary the previously
obtained rule will apply; and when we arrive at a polygon on
the boundary, the correspondence of the sides in pairs enables
the process to be continued.

\Par{189.} From the mode in which the figure~$C$ has been formed,
no two of the figures~$CH$ can have a polygon in common, when
for $H$~is taken in turn each operation of the self-conjugate
sub-group~$H$ of the general group~$G$; also the complete
set contains every double-polygon of our original figure. This
set of figures, or rather the division of the original figure into
this set, will then represent in a graphical form the self-conjugate
sub-group~$H$ of~$G$. Moreover, the operations which
transform corresponding pairs of sides of~$C$ into each other will,
when combined and repeated, clearly suffice to transform~$C$ into
any one of the figures~$CH$ and will therefore form a set of
generating operations of~$H$.

\begin{Remark}
\Par{190.} A simple example, in which the process described in the
\index{Graphical representation@\Topic{Graphical representation}!examples of|(}%
preceding paragraphs is actually carried out, will help to familiarize
the reader with the nature of the process and will also serve to
introduce a further modification of our figure. The example we
propose to consider is the special group with two generating operations
which are connected by the relations
\[
S_{1}^{3} = 1,\quad S_{2}^{3} = 1,\quad S_{1}S_{2} = S_{2}S_{1}.
\]

As a first step, we will take account only of the relation
\[
S_{1}S_{2} = S_{2}S_{1},
\]
and form for this special group the figure~$C$. All operations
\[
\Dots S_{1}^{\alpha_{n}} S_{2}^{\beta_{n}} \Dots,
\]
for which $\Sigma \alpha_{n}$ and $\Sigma \beta_{n}$ have given values, are in the special group
identical. We may thus select from the figure for the general group
the set of polygons
\[
S_{2}^{\alpha} S_{1}^{\beta}\quad (\alpha, \beta = -\infty\ \text{to}\ +\infty)
\]
as a set of representative polygons; and a reference to the diagram\footnote
  {In fig.~4 the orthogonal circle, which is not shewn, is taken to be a straight
  line.}
(fig.~4) makes it clear that this set of polygons forms a figure with
a single bounding curve. The black polygon which corresponds to
the operation~$\Sigma$ has here been chosen as that which has the side
$A_{1}A_{2}$ in common with the white polygon~$\Sigma$.
\index{Graphical representation@\Topic{Graphical representation}!of a special group|)}%
%% -----File: 286.png---Folio 270-------

Each double-polygon, except those of the set~$S_{1}^{m}$, contributes two
sides to the boundary of~$C$, one belonging to a white polygon and
% [Illustration: Fig. 4.]
\Figure[\textwidth]{4}{fig04}
one to a black. The polygons, which border~$C$ and have sides in
common with $S_{2}^{\alpha} S_{1}^{\beta}$, are $S_{1} S_{2}^{\alpha} S_{1}^{\beta}$ and $S_{1}^{-1} S_{2}^{\alpha} S_{1}^{\beta}$; and these, regarded
as operations of the special group, are equivalent to $S_{2}^{\alpha} S_{1}^{\beta + 1}$ and
$S_{2}^{\alpha} S_{1}^{\beta - 1}$. Hence the correspondence between the sides of~$C$ is such
that

\Inum{(i)}\quad to the side~$A_{1}A_{3}$ of the white polygon $S_{2}^{\alpha} S_{1}^{\beta}$ corresponds the
side~$A_{1}A_{3}$ of the black polygon $S_{2}^{\alpha} S_{1}^{\beta - 1}$;

\Inum{(ii)}\quad to the side~$A_{1}A_{3}$ of the black polygon $S_{2}^{\alpha} S_{1}^{\beta}$ corresponds the
side~$A_{1}A_{3}$ of the white polygon $S_{2}^{\alpha}S_{1}^{\beta + 1}$.

When we now take account of the additional relations
\[
S_{1}^{3} = 1,\quad S_{2}^{3} = 1,
\]
the figure~$C$ is found to reduce to a set of nine double-polygons,
which is completely represented by fig.~5.

In addition to the correspondences between the sides of~$C$ to
which those just written simplify when the indices of $S_{1}$~and~$S_{2}$ are
reduced $(\mod 3)$, we have now also the correspondences, indicated
in the figure by curved lines with \DPchg{arrowheads}{arrow-heads}, which result from the
new relations. Our figure may be further modified in such a way that
its form takes direct account of these four new correspondences.
Thus without in any way altering the configuration of the double-polygons,
from the point of view of geometry of position, we may
continuously deform the figure so that the pairs of corresponding
sides indicated by the curved \DPchg{arrowheads}{arrow-heads} are brought to actual
coincidence. When this is done, the resulting figure will have the
form shewn in fig.~6. The correspondence in pairs of the sides of
the boundary is indicated in the figure by full and dotted lines.
%% -----File: 287.png---Folio 271-------
% Illustration Fig. 5.
\begin{sidewaysfigure}[hp!]
\centering
\Graphic{\textheight}{fig05}
\end{sidewaysfigure}
%\Figure[\textwidth]{5}{fig05}
%% -----File: 288.png---Folio 272-------
The two unmarked portions $A_{1}A_{3}A_{1}$ correspond, as also do the two
similar portions marked with a full line, and the two marked with a
dotted line.
%[Illustration: Fig. 6.]
\Figure{6}{fig06}

It will be noticed that, in this final form of the figure for the
special group, direct account is taken of the finite order of the
generating operations $S_{1}$~and~$S_{2}$ and also of the operation~$S_{1}S_{2}$. The
simplification of the figure that results by thus taking account
directly of the finite order of the generating operations, and the
greater ease with which the eye follows this simplified representation,
are immediately obvious on a comparison of figs.\ 5~and~6.
\index{Graphical representation@\Topic{Graphical representation}!examples of|)}%
\end{Remark}

\Par{191.} In the applications of this graphical representation
of a group that we have specially in view, namely to groups
of finite order, the generating operations themselves are
necessarily of finite order. The generating operations
\[
S_{1},\ S_{2},\ \Dots,\ S_{n},
\]
of such a group may be taken as of orders
\[
m_{1},\ m_{2},\ \Dots,\ m_{n};
\]
%% -----File: 289.png---Folio 273-------
and if
\[
S_{1} S_{2} \Dots S_{n} S_{n + 1} = 1,
\]
then $S_{n + 1}$~will be of finite order~$m_{n + 1}$. We shall therefore next
consider a group generated by $n$~operations which satisfy the
relations
\[
S_{1}^{m_{1}} = 1,\quad S_{2}^{m_{2}} = 1,\ \Dots,\quad S_{n}^{m_{n}} = 1,\quad (S_{1} S_{2} \Dots S_{n})^{m_{n + 1}} = 1.
\]

The simple example we have given makes it clear that,
\index{Graphical representation@\Topic{Graphical representation}!of a group of finite order|(}%
at least in some particular cases, relations of this form may
be directly taken into account in constructing our figure; in
such a way that in the complete figure, consisting of a finite or
an infinite number of double-polygons, the correspondence in
pairs of the sides of the boundary, if any, will depend upon
further relations between the generating operations.

We may, in fact, always take account of relations of the
form in question in the construction of our figure as follows.

Let us take as before $n + 1$ arcs of circles
\[
A_{n + 1}A_{1},\ A_{1}A_{2},\ \Dots,\ A_{n - 1}A_{n},\ A_{n}A_{n + 1},
\]
bounding a polygonal figure~$P$ of $n + 1$ corners; but now,
instead of supposing the circles $A_{r - 1}A_{r}$ and $A_{r}A_{r + 1}$ to touch at~$A_{r}$,
let them cut at an angle (measured inside~$P$) of
\[
\frac{\pi}{m_{r}}, \quad (r = 1, 2, \Dots, n),
\]
while $A_{n}A_{n + 1}$ and $A_{n + 1}A_{1}$ cut at an angle~$\dfrac{\pi}{m_{n + 1}}$. Such a figure
can again be chosen in an infinite variety of ways: we will
suppose that it is drawn so that the $n + 1$ circles have a common
orthogonal circle. This clearly is always possible; but it is not
now necessarily the case that this orthogonal circle is real. Let
the figure~$P$ be now inverted in each of its sides; let the new
figures so formed be inverted in each of their new sides; and so
on continually. Then since the angles of~$P$ are sub-multiples of
two right angles, no two of the figures thus formed can overlap
in part without coinciding entirely. Moreover, when the process
is completely carried out, every point within the orthogonal
circle when it is real, and every point in the plane of the figure
when the orthogonal circle is evanescent or imaginary, will lie
in one and in only one of the polygons thus formed from~$P$ by
successive inversions.
%% -----File: 290.png---Folio 274-------

\Par{192.} Exactly as with the general group, these polygons
are coloured white or black according as they are derivable
from~$P$ by an even or an odd number of inversions. The
corners of any white polygon correspond one by one to the
corners of $P$; so that, when the perimeters of the polygons are
described in the same direction, corresponding corners occur in
the same cyclical order.

If now the operation of successive inversions at $A_{n + 1}A_{1}$ and
$A_{1}A_{2}$ is represented by~$S_{1}$, and that of successive inversions at
$A_{r - 1}A_{r}$ and $A_{r}A_{r + 1}$ by~$S_{r}$, ($r = 2, 3, \Dots, n$); all operations,
consisting of an even number of inversions in the sides of~$P$,
can be represented in terms of
\[
S_{1},\ S_{2},\ \Dots,\ S_{n}.
\]

Moreover, from the construction of the polygon~$P$, these
operations satisfy the relations
\[
S_{1}^{m_{1}} = 1,\quad S_{2}^{m_{2}} = 1,\quad \Dots, \quad S_{n}^{m_{n}} = 1,\quad S_{n + 1}^{m_{n + 1}} = 1,
\]
where
\[
S_{1}S_{2} \Dots S_{n}S_{n + 1} = 1.
\]

Again, if $P'$~is any white polygon of the figure, which can be
derived from~$P$ by the operation~$\Sigma$, a positive rotation (§~185) of~$P'$
round its corner~$A_{r}'$ is effected by the operation $\Sigma^{-1}S_{r}\Sigma$;
and, if $P''$~is the polygon so obtained, $P''$~is derived from~$P$ by
the operation~$S_{r}\Sigma$. It is to be observed that a positive rotation
of a polygon round its $A_{r}$~corner is now an operation of finite
order~$m_{r}$.

Suppose now that two operations $\Sigma$ and $\Sigma'$ transform~$P$ into
the same polygon~$P'$, so that $\Sigma\Sigma'^{-1}$ leaves $P$ unchanged. If
this operation, written at length, is
\[
S_{p}^{\alpha} \Dots S_{q}^{\beta} S_{r}^{\gamma} S_{s}^{\delta},
\]
and if $P$~is transformed into~$P_{1}$ by a positive rotation round~$A_{s}$
repeated $\delta$~times, $P_{1}$~into~$P_{2}$ by a rotation round its
corner~$A_{r}$ repeated $\gamma$~times, and so on; then the operation
may be indicated by a broken line drawn from $P$~to~$P_{1}$, from
$P_{1}$~to~$P_{2}$, and so on, the line returning at last to~$P$. But the
operation indicated by such a line is clearly equivalent to complete
rotations, (\ie\ rotations each of which lead to identity),
%% -----File: 291.png---Folio 275-------
round each of the corners which the broken line includes. In
other words, $\Sigma\Sigma'^{-1}$~reduces to identity when account is taken of
the relations which the generating operations satisfy. Hence
finally, to every white polygon~$P'$ will correspond one and only
one of the operations of the group, namely that operation
which transforms $P$ into~$P'$. The same is clearly true of the
black polygons; and by taking $P$ and a chosen black polygon
which has a side in common with~$P$ as corresponding to the
identical operation, the required unique correspondence is
established between the complete set of double-polygons in the
figure and the operations of the group, the relations which the
generating operations satisfy being directly indicated by the
configuration of the figure. Moreover, as with the general
group (§~185), a simple rule may be stated for determining,
from an inspection of the figure, the polygon that corresponds
to any given operation of the group.

\Par{193.} The number of polygons in the figure and therefore
the order of the group will still, in general, be infinite. We
may now proceed, just as in the previous case of a quite general
group, to derive from the figure representing the group~$G$,
generated by $n$ operations satisfying the relations
\[
S_{1}^{m_{1}} = 1,\quad
S_{2}^{m_{2}} = 1,\quad \Dots,\quad
S_{n}^{m_{n}} = 1,\quad (S_{1} S_{2} \Dots S_{n})^{m_{n + 1}} = 1,
\]
a suitable representation of the more special group~$\Bar{G}$, generated
by $n$~operations which satisfy the above relations and in addition
the further $m$~relations
\[
f_{j}(S_{i}) = 1,\quad (j = 1, 2, \Dots, m).
\]

As has been seen in §~182, if $H$~is the self-conjugate sub-group
of~$G$ which is formed by combining all possible operations
of the form
\[
R^{-1} f_{j}(S_{i})R,
\]
and if $\Sigma$~is any operation of~$G$, then the set of operations~$\Sigma H$,
regarded as operations of~$\Bar{G}$, are all equivalent to each other.
From each set of polygons $\Sigma H$ in the figure of~$G$, we may
therefore choose one to represent the corresponding operation
of~$\Bar{G}$; and, as was shewn with the general group, a complete
set of such representative polygons may be selected to form a
%% -----File: 292.png---Folio 276-------
connected figure, \ie\ a figure which does not consist of two or
more portions which are either isolated or connected only by
corners. Moreover, as in the former case, the sides of this
figure~$C$ will be connected in pairs $A_{r}' A_{r + 1}'$ and $A_{r}'' A_{r + 1}''$, which
are transformed into each other by some operation~$T$ of~$H$ and
its inverse, while no other operation of~$H$ will transform either
$A_{r}' A_{r + 1}'$ or $A_{r}'' A_{r + 1}''$ into any other side of~$C$.

It is not now however necessarily the case that the figure~$C$,
as thus constructed, is simply connected. Let us suppose then
that $C$~has one or more inner boundaries as well as an outer
boundary, and denote one of these inner boundaries by~$L$. If
the sides of~$L$ do not all correspond in pairs, and if $A_{r}' A_{r + 1}'$~is a
side of~$L$ such that the other side $A_{r}'' A_{r + 1}''$ corresponding to it
does not belong to~$L$, we may replace the double-polygon~$P''$ in~$C$
of which $A_{r}'' A_{r + 1}''$~is a side by the double-polygon, not
previously belonging to~$C$, of which $A_{r}' A_{r + 1}'$~is a side. If $P''$~has
a side on the boundary~$L$, the new figure~$C'$ thus obtained
will have one inner boundary less than~$C$; and if $P''$~has no side
on the boundary~$L$, the new inner boundary~$L'$ that is thus
formed from~$L$ will contain one double-polygon less than~$L$,
while the number of inner boundaries is not increased. This
process may be continued till the new inner boundary~$L_{1}$ which
replaces~$L$ is such that all of its sides correspond in pairs.

Let now $A_{s}A_{s + 1}$ and $A_{s}' A_{s + 1}'$ be a pair of corresponding
sides of~$L_{1}$, such that $A_{s}A_{s + 1}$~is transformed into $A_{s}' A_{s + 1}'$ by an
operation~$h$ of the self-conjugate sub-group~$H$. A side~$A_{t}A_{t + 1}$
of another boundary of~$C$ may be chosen such that $A_{s}A_{s + 1}$
and $A_{t}A_{t + 1}$ are sides of a simply connected portion, say~$B$, of~$C$;
while no side of~$L_{1}$ except $A_{s}A_{s + 1}$ forms part of the boundary of~$B$.
The polygons of~$B$ are equivalent, in respect of the special
group, to those of~$Bh$. Moreover, since the sides of~$L_{1}$ correspond
in pairs, no side of~$Bh$, except $A_{s}' A_{s + 1}'$ can coincide with
a side of~$L_{1}$. Hence when $B$~is replaced by~$Bh$, the inner
boundary~$L_{1}$ will be got rid of and no new inner boundary will
be formed. Finally then, $C$~may always be chosen so as to form
a single simply connected figure.

The simply connected plane figure~$C$, which has thus been
constructed, with the correspondence of the sides of its boundary
%% -----File: 293.png---Folio 277-------
in pairs, will now give a complete graphical representation of
the special group. The rule already formulated will determine
the operation of the group to which each white polygon
corresponds; and when, in carrying out this rule, we come to
a polygon on the boundary, the correspondence of the sides of
the boundary in pairs will enable the process to be continued.

The correspondence of the sides of~$C$ in pairs involves a
correspondence of the corners in sets of two or more. Thus if
$A_{r}$~is a corner of~$C$ and if, of the $m_{r}$ white polygons which in
the complete figure have a corner at~$A_{r}$, $n_{1}$~lie within~$C$, there
must within~$C$ be $m_{r} - n_{1}$ white polygons equivalent to the
remainder, and each of these must have an $A_{r}$~corner on the
boundary. If $A_{r}'$~is a corner of~$C$ such that there are $n_{2}$~white
polygons, lying within~$C$ and having a corner at~$A_{r}'$, and if one
of the sides of the boundary with a corner at~$A_{r}'$ corresponds
to one of the sides of the boundary with a corner at~$A_{r}$, these
$n_{2}$~white polygons must be equivalent to $n_{2}$ of the white polygons,
lying outside~$C$ and having a corner at~$A_{r}$. If
\[
n_{1} + n_{2} < m_{r},
\]
there must be a third corner~$A_{r}''$, contributing $n_{3}$~more white
polygons towards the set. With this we proceed as before; and
the process may be continued till the whole of the $m_{r}$~white
polygons surrounding~$A_{r}$ are accounted for. The set of corners
$A_{r}$, $A_{r}'$, $A_{r}''$,~\Dots\ will then form a set of corresponding corners,
which are equivalent to each other in respect of the special
group; and the whole of the corners of~$C$ may be divided into
such sets. At each set of corresponding corners~$A_{r}$ of~$C$ there
must clearly be also $m_{r}$~black polygons belonging to~$C$; and
the sum of the angles of~$C$ at a set of corresponding corners
must be equal to four right angles.

\Par{194.} When the order of the group is finite, we may still
further so modify our figure as to take account of the correspondence
of the sides of the boundary in pairs. We may, in
fact, by a suitable bending and stretching of the figure, bring
corresponding sides of the boundary to actual coincidence.
When this is done, the figure will no longer be a piece of a
plane with a single boundary, but will form a continuous
%% -----File: 294.png---Folio 278-------
\index{Regular division@\Topic{Regular division} of a surface, representation of a group by means of}%
surface, which is unbounded and in general is multiply connected.
Every point~$A_{r}$ on the surface, which in the plane
figure did not lie on the boundary, will be a corner common
to $2m_{r}$~polygons alternately black and white; and, in consequence
of what has just been seen in regard to the correspondence
of corners of the boundary, the same is true
for every point~$A_{r}$ on the surface which in the plane figure
consisted of a set of corresponding corners of the boundary. If
$N$~is the order of the group, the continuous unbounded surface
will be divided into $2N$~polygons, black and white. The configuration
of the set of white polygons with respect to any one
of them will, from the point of view of geometry of position, be
the same as that with respect to any other; and the like is
true for the black polygons. Such a division of a continuous
unbounded surface is described as a \emph{regular} division; and we
have finally, as a graphical representation of any group of finite
order~$N$, a division of a continuous surface into $2N$~polygons,
half black and half white, which is regular with respect to each
% [Illustration: Fig. 7.]
\Figure{7}{fig07}
\index{Graphical representation@\Topic{Graphical representation}!of a group of finite order|)}%
%% -----File: 295.png---Folio 279-------
set. The correspondence between the operations of the group
and the white polygons on the surface is given by the rule that
a single positive rotation of the white polygon~$\Sigma$ round its
corner~$A_{r}$ leads to the white polygon~$S_{r}\Sigma$.

\begin{Remark}
\Par{195.} We may again here illustrate this final modification of the
\index{Graphical representation@\Topic{Graphical representation}!examples of}%
graphical representation of a finite group by a simple example. For
this purpose, we choose the group defined by
\begin{gather*}
S_{1}^{4} = 1,\quad S_{2}^{4} = 1,\quad S_{3}^{4} = 1, \\
S_{1}S_{2}S_{3} = 1,\quad S_{2}^{-1}S_{1}S_{2}S_{1} = 1.
\end{gather*}

This group (§~74) is a non-Abelian group of order~$8$, containing
a single operation of order~$2$. The reader will have no difficulty in
verifying that the plane figure for this group is given by fig.~7; and
that opposite sides of the octagonal boundary correspond. The
single operation of order~$2$ is
\[
S_{1}^{2} (= S_{2}^{2} = S_{3}^{2});
\]
this corresponds to a displacement of the triangles among themselves
in which all the six corners remain fixed. If now corresponding
sides of the boundary are brought to coincidence, the continuous
surface formed will be a double-holed anchor-ring, or sphere with
% [Illustration: Fig. 8.]
\Figure[0.6\textwidth]{8}{fig08}
%% -----File: 296.png---Folio 280-------
two holes through it. A view of one half of the surface divided
into black and white triangles, is given in fig.~8. The half of the
surface, not shewn, is divided up in a similar manner; and the
operation of order~$2$ replaces each triangle of one half by the
corresponding triangle of the other, an operation which clearly leaves
the six corners of the polygons undisplaced.
\end{Remark}

\Par{196.} The form of the plane figure~$C$, which with the correspondence
\index{Genus@\Topic{Genus} of a group, definition of}%
\index{Group@\Topic{Group}!genus of a}%
of its bounding sides in pairs represents the group,
is capable of indefinite modification by replacing individual
polygons on the boundary by equivalent polygons. If however
we reckon a pair of corresponding sides of the boundary as a
single side and a set of corresponding corners of the boundary
as a single corner, it is clear that, however the figure may be
modified, the numbers of its corners, sides and polygons remain
each constant. This may be immediately verified on replacing
any single boundary polygon by its equivalent.

If now $A$~be the number of corners, and $E$~the number of
sides in the figure~$C$ when reckoned as above, $2N$~being the
number of polygons, then the connectivity\footnote
  {Forsyth, \textit{Theory of Functions}, p.~325.}
\index{Forsyth@\Topic{Forsyth}, quoted}%
$2p + 1$ of the closed
surface is given by the equation
\[
2p = 2 + E - 2N - A.
\]

When the group and its generating operations are given,
the integer~$p$ is independent of the form of the plane figure~$C$,
which as has been seen is capable of considerable modification.
The plane figure~$C$ however depends directly on the set of
generating operations that is chosen for the group. For a given
group of finite order, such a set is not in general unique; and
the number of generating operations as well as their order will
in general vary from one set to another. It does not necessarily
follow, and in fact it is not generally the case, that the connectivity
of the surface by whose regular division the group is
represented, is independent of the choice of generating operations.
There must however obviously be a lower limit to the
number~$p$ for any given group of finite order, whatever
generating operations are chosen; this we shall call the \emph{genus}
of the group\footnotemark.
\index{Hurwitz@\Topic{Hurwitz}, quoted}%
\footnotetext{Hurwitz, ``Algebraische Gebilde mit eindeutigen Transformationen in
  sich,'' \textit{Math.\ Ann.}\DPnote{** No comma or "Vol."}\ \VolNo{XLI}, (1893), p.~426.}%
%% -----File: 297.png---Folio 281-------

\Par{197.} We shall now shew that there is a limit to the order
of a group which can be represented by the regular division of
a surface of given connectivity $2p + 1$. If $N$~is the order of
such a group, generated by the $n$~operations
\[
S_{1},\ S_{2},\ \Dots,\ S_{n},
\]
which satisfy the relations
\begin{gather*}
S_{1}^{m_{1}} = 1,\quad
S_{2}^{m_{2}} = 1,\quad \Dots, \quad
S_{n}^{m_{n}} = 1, \\
S_{1}S_{2} \Dots S_{n} = 1;
\end{gather*}
the surface will be divided in $2N$~polygons of $n$~sides each.
Let $A_{1}$, $A_{2}$,~\Dots, $A_{n}$ be the angular points of one of these
polygons; and suppose that on the surface there are $C_{1}$~corners
in the set to which $A_{1}$ belongs, $C_{2}$~in the set to which $A_{2}$ belongs,
and so on. Round each corner~$A_{r}$ there are $2m_{r}$~polygons; and
each polygon has one and only one corner of the set to which
$A_{r}$ belongs. Hence
\[
C_{r}m_{r} = N,
\]
and so
\[
C_{1} + C_{2} + \Dots + C_{n} = N \sum_{1}^{n} \frac{1}{m_{r}}.
\]

Again, each side belongs to two and only to two polygons, so
that the number of sides is
\[
Nn.
\]

Using these values for $A$~and~$E$ in the formula of §~196,
we obtain the equation
\[
2(p - 1) = N\left(n - 2 - \sum_{1}^{n} \frac{1}{m_{r}}\right).
\]

A complete discussion of this equation for the cases $p = 0$
and $p = 1$ will be given in the next chapter.

When $p$~is a given integer greater than unity, we can
determine the greatest value that is possible for~$N$ by finding
the least possible positive value of the expression
\[
n - 2 - \sum_{1}^{n} \frac{1}{m_{r}}.
\]

If $n > 4$, this quantity is not less than~$\frac{1}{2}$, since $m_{r}$~cannot be
less than~$2$.

If $n = 4$, the simultaneous values
\[
m_{1} = m_{2} = m_{3} = m_{4} = 2
\]
%% -----File: 298.png---Folio 282-------
are not admissible, since they make the expression zero. Its
least value in this case will therefore be given by
\[
m_{1} = m_{2} = m_{3} = 2,\quad m_{4} = 3;
\]
and the expression is then equal to~$\frac{1}{6}$.

If $n = 3$, we require the least positive value of
\[
K = 1 - \frac{1}{m_{1}} - \frac{1}{m_{2}} - \frac{1}{m_{3}}.
\]

Now the three sets of values
\begin{alignat*}{3}
m_{1} &= 3,\quad & m_{2} &= 3,\quad & m_{3} &= 3, \\
m_{1} &= 2,\quad & m_{2} &= 4,\quad & m_{3} &= 4, \\
\lintertext{and}
m_{1} &= 2,\quad & m_{2} &= 3,\quad & m_{3} &= 6,
\end{alignat*}
each make $K$ zero; and therefore no positive value of~$K$ can be
less than the least of those given by
\begin{alignat*}{3}
m_{1} &= 3,\quad & m_{2} &= 3,\quad & m_{3} &= 4, \\
m_{1} &= 2,\quad & m_{2} &= 4,\quad & m_{3} &= 5, \\
\lintertext{and}
m_{1} &= 2,\quad & m_{2} &= 3,\quad & m_{3} &= 7.
\end{alignat*}

These sets of values give for~$K$ the values $\frac{1}{12}$,~$\frac{1}{20}$ and~$\frac{1}{42}$.
Hence finally, the absolutely least positive value of the expression
is~$\frac{1}{42}$, and therefore the greatest admissible value of~$N$ is
\[
84(p - 1).
\]

Hence\footnotemark:---
\footnotetext{Hurwitz, \textit{loc.~cit.}\ p.~424.}
\index{Hurwitz@\Topic{Hurwitz}, quoted}%

\begin{Theorem}[II.]
The order of a group, that can be represented
\index{Limitation@\Topic{Limitation}!on the order of a group of given genus}%
by the regular division of a surface of connectivity $2p + 1$, cannot
exceed $84(p - 1)$, $p$~being greater than unity.
\end{Theorem}

\Par{198.} If, when a group is represented by the regular division
of an unbounded surface, we draw a line from any point inside
the white polygon~$1$ (or any other polygon) returning after any
path on the surface to the point from which it started, it will
represent a relation between the generating operations of the
group. For in following out along the line so drawn the rule
that determines the operation of the group corresponding to
each white polygon, some operation
\[
F(S_{i})
\]
%% -----File: 299.png---Folio 283-------
will be found to correspond to the final polygon; and this
being the white polygon~$1$, it follows that
\[
F(S_{i}) = 1.
\]

If the surface is simply connected, any such line can be
\index{Defining relations@\Topic{Defining relations} of a group!limitation on the number of, when the genus is given}%
\index{Limitation@\Topic{Limitation}!on the number of defining relations of a group of given genus}%
continuously altered till it shrinks to a point; and therefore the
$n + 1$ relations between the $n$~generating operations completely
define the group, since all other relations can be deduced from
them.

If however the surface is of connectivity $2p + 1$, there are
$2p$~independent closed paths that can be drawn on the surface,
no one of which can by continuous displacement either be
shrunk up to a point or brought to coincidence with another;
and every closed path on the surface can by continuous displacement
either be brought to a point or to coincidence with
a path constructed by combination and repetition of the $2p$~independent
paths\footnotemark.
\footnotetext{Forsyth, \textit{Theory of Functions}, p.~330.}%
\index{Forsyth@\Topic{Forsyth}, quoted}%
Any one of these $2p$~independent paths will
give a relation between the $n$~generating operations of the
group, which cannot be deduced from the $n + 1$ relations on
which the angles of the polygons depend. Moreover, every relation
between the generating operations can be represented by a
closed path on the surface; and therefore there can be no further
relation independent of the original relations and those obtained
from the $2p$~independent paths. There cannot therefore be
more than $2p$~independent relations between the $n$~generating
operations of a group, in addition to the $n + 1$ relations that give
the order of the generating operations and of their product;
$2p + 1$ being the connectivity of the surface by whose regular
division into $n$-sided polygons the group is represented.

The $2p$~relations given by $2p$~independent paths on the
surface are not, however, necessarily independent. In fact we
have already had an example to the contrary in §~195. On the
closed surface, by the regular division of which the group there
considered is represented, four independent closed paths can be
drawn. Any three of the corresponding relations can be derived
from the fourth by transformation.

The only known cases in which the $2p$~relations are independent
are those of a class of groups of genus one (§~205).
%% -----File: 300.png---Folio 284-------

\begin{Ex} Draw the figure of the group generated by $S_{1}$,~$S_{2}$,~$S_{3}$, where
\[
S_{1}^{2} = 1,\quad S_{2}^{3} = 1,\quad S_{3}^{8} = 1,\quad S_{1} S_{2} S_{3} = 1.
\]

Shew from the figure that the special group, given by the
additional relation
\[
(S_{1} S_{3}^{4})^{2} = 1,
\]
is a finite group of order~$48$; and that it can be represented by the
regular division of a surface of connectivity~$5$.
\end{Ex}


\Note{\textit{Note to} §~194.}

\begin{Remark}
If in the process of bending and stretching, described in §~194,
by means of which the plane figure~$C$ is changed into an unbounded
surface, the angles of the polygons all remain unaltered, the circles
of the plane figure will become continuous curves on the surface.
These curves on the surface, which we will still call circles, are
necessarily re-entrant. It is not however necessarily the case that,
on the surface, a circle will not cut itself.

In the plane figure for the general group, an inversion at any
circle of the figure leaves the figure unchanged geometrically but
interchanges the black and white polygons. Each circle is, in fact, a
line of symmetry for the figure such that, in respect of it, there
is corresponding to every white polygon a symmetric black polygon
and vice~versâ.

Similarly on the surface a circle which does not cut itself may
be a line of symmetry, such that a reflection at it is an operation of
order two which leaves the surface and its division into polygons
unchanged, but interchanges black and white polygons. When this
is the case, every circle on the surface will be a line of symmetry and
no circle will cut itself. On the other hand no such operation can
ever be connected with a circle which cuts itself.

When such lines of symmetry exist, Prof.~Dyck speaks of the
division of the surface as regular and symmetric.
\end{Remark}
%% -----File: 301.png---Folio 285-------


\Chapter{XIII.}{On the Graphical Representation of Groups:
Groups of Genus Zero and Unity: Cayley's Colour Groups.}

\Par{199.} \First{We} shall now proceed to a discussion in the cases
$p = 0$ and $p = 1$ of the relation
\[
2(p - 1) = N \left(n - 2 - \sum_{1}^{n} \frac{1}{m_{r}}\right),
\]
which connects the number and the orders of the generating
operations of a group with the order of the group itself; and to
the consideration of the corresponding groups.

For any given value of~$p$, other than $p = 1$, we may regard
this relation as an equation connecting the positive integers
$N$,~$n$, $m_{1}$, $m_{2}$,~\Dots, $m_{n}$. It does not however follow from the
investigations of the last Chapter that there is always a
group or a set of groups corresponding to a given solution of
the equation. In fact, for values of~$p$ greater than~$1$, this is not
necessarily the case. We shall however find that, when $p = 0$,
there is a single type of group corresponding to each solution
of the equation; and that, when $p = 1$, there is an infinite
number of types of group, all characterized by a common
property, corresponding to each solution of the equation.
When $p = 0$, the groups are (§~196) of genus zero; and all
possible groups of genus zero are found by putting $p = 0$
%% -----File: 302.png---Folio 286-------
in the equation. The groups thus obtained are of special
importance in many applications of group-theory; for this
reason, they will be dealt with in considerable detail.

\Par{200.} When $p = 0$, the equation may be written in the form
\index{Groups@\Topic{Groups}!of genus zero|(}%
\[
2\left(1 - \frac{1}{N}\right) = \sum_{1}^{n} \left(1 - \frac{1}{m_{r}}\right);
\]
in this form, it is clear that the only admissible values of~$n$ are
$2$ and~$3$.

First, let $n = 2$. The only possible solution then is
\[
N = m_{1} = m_{2} = n,
\]
$n$~being any integer. The corresponding group is a cyclical
group of order~$n$.

Secondly, let $n = 3$. In this case, one at least of the three
integers $m_{1}$,~$m_{2}$,~$m_{3}$ must be equal to~$2$, as otherwise the right-hand
side of the equation would be not less than~$2$. We may
therefore without loss of generality put $m_{1} = 2$. If now both
$m_{2}$ and $m_{3}$ were greater than~$3$, the right-hand side would still
be not less than~$2$; and therefore we may take $m_{2}$ to be either
$2$ or~$3$. When $m_{1}$~and~$m_{2}$ are both~$2$, the equation becomes
\[
\frac{2}{N} = \frac{1}{m_{3}};
\]
giving
\[
m_{3} = n,\quad N = 2n,
\]
where $n$~is any integer.

When $m_{1}$~is~$2$ and $m_{2}$~is~$3$, the equation is
\[
\frac{2}{N} + \frac{1}{6} = \frac{1}{m_{3}}.
\]

This has three solutions in positive integers; namely,
\begin{alignat*}{2}
m_{3} &= 3,\quad & N &= 12; \\
m_{3} &= 4,\quad & N &= 24; \\
\lintertext{and}
m_{3} &= 5,\quad & N &= 60.
\end{alignat*}
%% -----File: 303.png---Folio 287-------

The solutions of the equation for the case $p = 0$ may therefore
be tabulated in the form:---
\[
\begin{array}{*{5}{c|}}
\TEntry{}  & \TEntry{m_{1}} & \TEntry{m_{2}} & \TEntry{m_{3}} & \TEntry{N} \\
\cline{2-5}
\Strut
\Inum{I}   & n & n &   & n \\
\cline{2-5}
\Strut
\Inum{II}  & 2 & 2 & n & 2n \\
\cline{2-5}
\Strut
\Inum{III} & 2 & 3 & 3 & 12 \\
\cline{2-5}
\Strut
\Inum{IV}  & 2 & 3 & 4 & 24 \\
\cline{2-5}
\Strut
\Inum{V}   & 2 & 3 & 5 & 60 \\
\cline{2-5}
\end{array}
\]

\Par{201.} That a single type of group actually exists, corresponding
to each of these solutions, may be seen at once by
returning to our plane figure. The sum of the internal angles
of the triangle $A_{1} A_{2} A_{3}$ formed by circular arcs is, in each of
these cases, greater than two right angles; and the common
orthogonal circle is therefore imaginary. The complete figure
will therefore divide the whole plane into black and white
triangles, so that there are no boundaries to consider. Moreover,
the number of white triangles in each case must be equal
to the corresponding value of~$N$; for the preceding investigation
shews that this is a possible value, and on the other hand the
process, by which the figure is completed from a given original
triangle, is a unique one. There is therefore a group corresponding
to each solution; and the correspondence which has
been established in any case between the operations of a group
and the polygons of a figure, proves that there cannot be two
distinct types of group corresponding to the same solution.

\Par{202.} The plane figure for $p = 0$ does not, in fact, differ
essentially from the figure drawn on a continuous simply connected
surface in space. The former may be regarded as the
stereographic projection of the latter. The five distinct types
are represented graphically by the following figures.

The first is a cyclical group, and the figure (fig.~9) does not
differ essentially from fig.~2 in §~183.

The group given by the second solution of the equation
\index{Group@\Topic{Group}!dihedral}%
is called the dihedral group. It is represented by fig.~10.
%% -----File: 304.png---Folio 288-------
\index{Graphical representation@\Topic{Graphical representation}!of a cyclical group}%
\index{Graphical representation@\Topic{Graphical representation}!examples of|(}%
%[Illustration: Fig. 9.]
\Figure{9}{fig09}
%[Illustration: Fig. 10.]
\Figure{10}{fig10}
%% -----File: 305.png---Folio 289-------
\index{Group@\Topic{Group}!tetrahedral}%
\index{Group@\Topic{Group}!octohedral}%
\index{Group@\Topic{Group}!icosahedral}%
%[Illustration: Fig. 11.]
\Figure{11}{fig11}

The group given by the third solution of the equation is
represented in fig.~11. It is known as the tetrahedral group.

To the fourth solution of the equation corresponds the group
represented by fig.~12. It is known as the octohedral group.

To the fifth solution of the equation corresponds the group
represented in fig.~13. It is known as the icosahedral group.

The four last groups are identical with the groups of rotations
which will bring respectively a double pyramid on an $n$-sided
base, a tetrahedron, an octohedron, and an icosahedron to coincidence
with itself in each case\footnotemark.
\index{Klein@\Topic{Klein}, quoted}%
\footnotetext{Klein, ``Vorlesungen über das Ikosaeder,'' Chap.~\DPchg{\textsc{i}}{I}.}%

When the figures are drawn on a sphere, and the three circles
of the original triangle and therefore also all the circles of the
figure are taken to be great-circles of the sphere, the actual displacements
of the triangles among themselves which correspond
%% -----File: 306.png---Folio 290-------
\PageLabel{290}%
%[Illustration: Fig. 12.]
\Figure{12}{fig12}
%[Illustration: Fig. 13.]
\Figure{13}{fig13}
\index{Graphical representation@\Topic{Graphical representation}!examples of|)}%
%% -----File: 307.png---Folio 291-------
to the operations of the group can be effected by real rotations
about diameters of the sphere; thus the statement of the preceding
sentence may be directly verified.

\Par{203.} In terms of their generating operations, the five types
\index{Defining relations@\Topic{Defining relations} of a group!for groups of genus zero}%
of group of genus zero are given by the relations:---
\begin{alignat*}{5}
&\Inum{I.}   & S_{1}^{n} &= 1, & S_{2}^{n} &= 1, &           &     & S_{1}S_{2} &= 1; \\
&\Inum{II.}  & S_{1}^{2} &= 1, & S_{2}^{2} &= 1, & S_{3}^{n} &= 1, & S_{1}S_{2}S_{3} &= 1; \\
&\Inum{III.}\quad & S_{1}^{2} &= 1,\quad & S_{2}^{3} &= 1,\quad & S_{3}^{3} &= 1,\quad & S_{1}S_{2}S_{3} &= 1; \\
&\Inum{IV.}  & S_{1}^{2} &= 1, & S_{2}^{3} &= 1, & S_{3}^{4} &= 1, & S_{1}S_{2}S_{3} &= 1; \\
&\Inum{V.}   & S_{1}^{2} &= 1, & S_{2}^{3} &= 1, & S_{3}^{5} &= 1, & S_{1}S_{2}S_{3} &= 1.
\end{alignat*}

The first of these does not require special discussion.

In the dihedral group, we have
\[
S_{1} S_{3} S_{1} = S_{1} S_{2} = S_{3}^{-1}.
\]

The dihedral group of order~$2n$ therefore contains a cyclical
sub-group of order~$n$ self-conjugately; and every operation of
the group which does not belong to this self-conjugate sub-group
is of order~$2$. The operations of the group are given,
each once and once only, by the form
\[
S_{1}^{\alpha} S_{3}^{\beta},\quad (\alpha = 0, 1;\ \beta = 0, 1, 2, \Dots, n - 1).
\]

When $n = 3$, this group is simply isomorphic with the
symmetric group of three symbols.

In the tetrahedral group, since
\begin{gather*}
(S_{1} S_{2})^{3} = 1, \\
S_{2}^{-1} S_{1} S_{2} S_{1} = S_{2} S_{1} S_{2}^{-1},
\end{gather*}
and therefore $S_{1}$, $S_{2}^{-1} S_{1}S_{2}$, $S_{2}S_{1}S_{2}^{-1}$ are permutable with each
other. These operations of order~$2$ (with identity) form a self-conjugate
sub-group of order~$4$; and the $12$~operations of the
group are therefore given by the form
\[
S_{2}^{\alpha} (S_{2}^{-1} S_{1} S_{2})^{\beta} S_{1}^{\gamma},
\]
or
\[
S_{2}^{\alpha} S_{1}^{\beta} S_{2} S_{1}^{\gamma},\quad (\alpha = 0, 1, 2;\ \beta, \gamma = 0, 1).
\]

If
\[
S_{1} = (12) (34),\quad S_{2} = (123),
\]
then
\[
S_{1} S_{2} = (134);
\]
and therefore the tetrahedral group is simply isomorphic with
the alternating group of four symbols.
%% -----File: 308.png---Folio 292-------

If, in the octohedral group, we write
\[
S_{3}^{2} = S',
\]
then
\[
S_{2}S' = S_{1}S_{3} = S_{1} S_{2}^{2} S_{1};
\]
and therefore
\[
(S_{2}S')^{3} = 1.
\]

Hence $S_{2}$ and $S'$ generate a tetrahedral group.

Again
\[
S_{1}S_{2}S_{1} = (S_{2}S')^{2},
\]
and
\[
S_{1}S'S_{1} = S_{2}S' S_{2}^{-1},
\]
so that this is a self-conjugate sub-group. The operations of
the group are given, each once and once only, in the form
\[
S_{1}^{\alpha} S_{2}^{\beta} S_{3}^{2\gamma} S_{2} S_{3}^{2\delta},\quad
  (\alpha, \gamma, \delta = 0, 1;\ \beta = 0, 1, 2).
\]

If
\[
S_{1} = (12),\quad S_{2} = (234),
\]
then
\[
S_{1}S_{2} = (1342);
\]
and therefore the octohedral group is simply isomorphic with
the symmetric group of four symbols.

The icosahedral group is simple. It is, in fact, simply
isomorphic with the alternating group of five symbols which
has been shewn (§~111) to be a simple group. Thus if
\[
S_{1} = (12)(34),\quad S_{2} = (135),
\]
then
\[
S_{1}S_{2} = (12345);
\]
so that the substitutions $S_{1}$ and $S_{2}$ satisfy the relations
\[
S_{1}^{2} = 1,\quad S_{2}^{3} = 1,\quad (S_{1}S_{2})^{5} = 1.
\]

They must therefore generate an icosahedral group or one
of its sub-groups. On the other hand, from the substitutions
$S_{1}$ and $S_{2}$ all the even substitutions of five symbols may be
formed, and these are $60$~in number. The group therefore
cannot be a sub-group of the icosahedral group; the only
alternative is that the two are identical.

As the icosahedral group has no self-conjugate group, we
cannot in this case so easily construct a form which will
represent each operation of the group just once in terms of
the generating operations. It is however not difficult to verify
that this is true of the set of forms\footnote
  {Dyck, ``Gruppentheoretische Studien,'' \textit{Math.\ Ann.}\ \VolNo{XX}, (1882), p.~35, and
  Klein, \textit{loc.~cit.}\ p.~26.}
\index{Dyck@\Topic{Dyck}, quoted}%
\index{Klein@\Topic{Klein}, quoted}%
\[
\begin{gathered}
  S_{3}^{\alpha},\quad S_{3}^{\alpha} S_{1} S_{3}^{\beta},\quad
  S_{3}^{\alpha} S_{1} S_{3}^{2} S_{1} S_{3}^{\beta}, \\
  S_{3}^{\alpha} S_{1} S_{3}^{2} S_{1} S_{3}^{3} S_{1},
\end{gathered}
\quad (\alpha, \beta = 0, 1, 2, 3, 4).
\]
\index{Groups@\Topic{Groups}!of genus zero|)}%
%% -----File: 309.png---Folio 293-------

\Par{204.} We shall next deal with the equation in the case
$p = 1$. In this case alone, the order of the group disappears
from the equation, which merely gives a relation between the
number and order of the generating operations. This may be
written in the form
\[
2 = \sum_{1}^{n} \left(1 - \frac{1}{m_{r}}\right);
\]
and $n$~must therefore be either $4$ or~$3$.

When $n$~is~$4$, the equation becomes
\[
2 = \frac{1}{m_{1}} + \frac{1}{m_{2}} + \frac{1}{m_{3}} + \frac{1}{m_{4}},
\]
and the only solution is clearly
\[
m_{1} = m_{2} = m_{3} = m_{4} = 2.
\]

When $n$~is~$3$, the equation takes the form
\[
1 = \frac{1}{m_{1}} + \frac{1}{m_{2}} + \frac{1}{m_{3}},
\]
and is easily seen to have three solutions, viz.\
\begin{alignat*}{3}
m_{1} &= 3,\quad & m_{2} &= 3,\quad & m_{3} &= 3; \\
m_{1} &= 2, & m_{2} &= 4, & m_{3} &= 4; \\
\lintertext{and}
m_{1} &= 2, & m_{2} &= 3, & m_{3} &= 6.
\end{alignat*}

\Par{205.} Take first the solution
\[
n = 4,\quad m_{1} = m_{2} = m_{3} = m_{4} = 1.
\]
The corresponding general group is defined by the relations
\begin{gather*}
S_{1}^{2} = 1,\quad S_{2}^{2} = 1, \quad S_{3}^{2} = 1,\quad S_{4}^{2} = 1, \\
S_{1}S_{2}S_{3}S_{4} = 1.
\end{gather*}

If we proceed to form the plane figure representing this
group, the sum of the internal angles of the quadrilateral
$A_{1}A_{2}A_{3}A_{4}$ is equal to four right angles, and the four circles
that form it therefore pass through a point. If this point be
taken at infinity, the four circles (and therefore all the circles of
the figure) become straight lines. The plane figure will now
take the form given in fig.~14, and the four generating operations
are actual rotations through two right angles about lines
through $A_{1}$,~$A_{2}$,~$A_{3}$ and~$A_{4}$, perpendicular to the plane of the
figure.
%% -----File: 310.png---Folio 294-------
\index{Graphical representation@\Topic{Graphical representation}!examples of}%

%[Illustration: Fig. 14.]
\Figure{14}{fig14}

Every operation of the group is therefore, in this form of
representation, either a rotation through two right angles
about a corner of the figure or a translation; and it will clearly
be the former or the latter according as it consists of an odd
or an even number of factors, when expressed in terms of the
generating operations. The operations which correspond to
translations form a sub-group; for if two operations each
consist of an even number of factors, so also does their
product. Moreover, this sub-group is self-conjugate, since the
number of factors in $\Sigma^{-1}S\Sigma$ is even if the number in~$S$ is
even. This self-conjugate sub-group is generated by the two
operations
\[
S_{1}S_{2}\quad \text{and}\quad S_{2}S_{3};
\]
for
\begin{gather*}
S_{1}S_{3} = S_{1}S_{2} · S_{2}S_{3}, \\
S_{2}S_{1} = (S_{1}S_{2})^{-1},\quad
S_{3}S_{2} = (S_{2}S_{3})^{-1},\quad
S_{3}S_{1} = (S_{2}S_{3})^{-1} (S_{1}S_{2})^{-1};
\end{gather*}
and therefore every operation containing an even number of
factors can be represented in terms of $S_{1}S_{2}$ and~$S_{2}S_{3}$. Lastly,
these two operations are permutable with each other; for
\[
S_{2}S_{3} · S_{1}S_{2} = S_{2}S_{3} · S_{4}S_{3} = S_{1}S_{3} = S_{1}S_{2} · S_{2}S_{3};
\]
and therefore every operation of the group is contained, once
and only once, in the form
\[
S_{1}^{\alpha} (S_{1}S_{2})^{\beta} (S_{2}S_{3})^{\gamma},\quad (\alpha = 0, 1:\ \beta, \gamma = -\infty, \dots, 0, \dots, \infty).
\]

The results thus arrived at may also be verified very simply
by purely kinematical considerations. If a group generated
%% -----File: 311.png---Folio 295-------
by $S_{1}$,~$S_{2}$,~$S_{3}$ and~$S_{4}$ is of finite order, there must, since it is of
genus~$1$, be either one or two additional relations between the
generating operations; and any such relation is expressible by
equating the symbol of some operation of the general group to
unity. Such a relation is therefore either of the form
\[
S_{1} (S_{1}S_{2})^{b} (S_{2}S_{3})^{c} = 1,
\]
or
\[
(S_{1}S_{2})^{b} (S_{2}S_{3})^{c} = 1.
\]

The operation $S_{1} (S_{1}S_{2})^{b} (S_{2}S_{3})^{c}$ of the general group, consisting
of an odd number of factors, must be a rotation round
some corner of the figure, say a rotation round the corner~$A_{r}$ of
the white quadrilateral~$\Sigma$; it is therefore identical with $\Sigma^{-1}S_{r}\Sigma$.

Now the relation
\[
\Sigma^{-1}S_{r}\Sigma = 1,
\]
gives
\[
S_{r} = 1.
\]

A relation of the first of the two forms is therefore inconsistent
with the supposition that the group is actually generated
by $S_{1}$,~$S_{2}$ and~$S_{3}$. It, in fact, reduces the generating operations
and the relations among them to
\[
S_{1}^{2} = 1,\quad S_{2}^{2} = 1,\quad S_{3}^{2} = 1,\quad S_{1}S_{2}S_{3} = 1,
\]
which define a group of genus zero.

The only admissible relations for a group of genus~$1$ are
\index{Groups@\Topic{Groups}!of genus one|(}%
therefore those of the form
\[
(S_{1}S_{2})^{b} (S_{2}S_{3})^{c} = 1.
\]

A single relation of this form reduces the operations of the
general group to those contained in
\begin{gather*}
S_{1}^{\alpha} (S_{1}S_{2})^{\beta} (S_{2}S_{3})^{\gamma}, \\
(\alpha = 0, 1;\ \beta = 0, 1, \dots, b - 1;\ \gamma = -\infty, \dots, 0, \dots, \infty);
\end{gather*}
and the group so defined is still of infinite order.

Finally, two independent relations
\begin{gather*}
(S_{1}S_{2})^{b} (S_{2}S_{3})^{c} = 1, \\
(S_{1}S_{2})^{b'} (S_{2}S_{3})^{c'} = 1,
\end{gather*}
where
\[
\frac{b}{b'} \neq \frac{c}{c'},
\]
must necessarily lead to a group of finite order. If $m$~is the
greatest common factor of $b$~and~$b'$, so that
\[
b = b_{1}m,\quad b' = b_{1}'m,
\]
%% -----File: 312.png---Folio 296-------
where $b_{1}$ and~$b_{1}'$ are relatively prime; and if
\[
b_{1}x - b_{1}'y = 1;
\]
the two relations give
\[
(S_{2}S_{3})^{cb_{1}' - c'b_{1}} = 1,
\]
and
\[
(S_{1}S_{2})^{m} = (S_{2}S_{3})^{c'y - cx}.
\]

Every operation of the group is now contained, once and
only once, in the form
\begin{gather*}
S_{1}^{\alpha} (S_{1}S_{2})^{\beta} (S_{2}S_{3})^{\gamma}, \\
(\alpha = 0, 1;\ \beta = 0, 1, \dots, m - 1;\ \gamma = 0, 1, \dots, cb_{1}' - b_{1}c' - 1);
\end{gather*}
%[** TN: Minus sign in next line printed as a tilde in the original]
and the order of the group is $2(bc' - b'c)$.

\Par{206.} Corresponding to the solution
\[
n = 3,\quad m_{1} = m_{2} = m_{3} = 3,
\]
we have the general group generated by $S_{1}$,~$S_{2}$,~$S_{3}$, where
\begin{gather*}
S_{1}^{3} = 1,\quad S_{2}^{3} = 1,\quad S_{3}^{3} = 1, \\
S_{1} S_{2} S_{3} = 1.
\end{gather*}

The sum of the three angles of the triangle $A_{1} A_{2} A_{3}$ is two
\index{Graphical representation@\Topic{Graphical representation}!examples of}%
% [Illustration: Fig.~15.]
\Figure{15}{fig15}
%% -----File: 313.png---Folio 297-------
right angles, and therefore again the circles in the plane figure
(fig.~15) may be taken as straight lines. When the figure is
thus chosen, the generating operations are rotations through~$\frac{2}{3}\pi$
about the angles of an equilateral triangle; and every operation
of the group is either a translation or a rotation.

The three operations
\[
S_{1} S_{2}^{2},\quad S_{2} S_{1} S_{2},\quad S_{2}^{2} S_{1},
\]
when transformed by~$S_{2}$, are interchanged among themselves.
When transformed by~$S_{1}$ they become
\[
S_{2}^{2} S_{1},\quad
S_{1}^{2} S_{2} S_{1} S_{2} S_{1},\quad
S_{1}^{2} S_{2}^{2} S_{1}^{2},
\]
and since
\[
(S_{1} S_{2})^{3} = 1,
\]
the two latter are $S_{1} S_{2}^{2}$ and $S_{2}S_{1} S_{2}$ respectively.

Hence the three operations generate a self-conjugate sub-group;
and since
\[
S_{1} S_{2}^{2} · S_{2} S_{1} S_{2} · S_{2}^{2} S_{1} = 1,
\]
this sub-group is generated by $S_{1} S_{2}^{2}$ and $S_{2} S_{1} S_{2}$.

These two operations are permutable; for
\[
S_{1} S_{2}^{2} · S_{2} S_{1} S_{2}
  = S_{1}^{2} S_{2}
  = S_{2} S_{1} S_{2} S_{1} S_{2} · S_{2}
  = S_{2} S_{1} S_{2} · S_{1} S_{2}^{2}.
\]

Hence finally, every operation of the group is represented,
once and only once, by the form
\[
S_{1}^{\alpha} (S_{1}S_{2}^{2})^{\beta} (S_{2}S_{1}S_{2})^{\gamma},\quad
  (\alpha = 0, 1, 2;\ \beta, \gamma = -\infty, \dots, 0, \dots, \infty).
\]

This result might also be arrived at by purely kinematical
considerations; for an inspection of the figure shews that the
two simplest translations are
\[
S_{1} S_{2}^{2} \quad\text{and}\quad S_{2} S_{1} S_{2},
\]
and that every translation in the group can be obtained by the
combination and repetition of these. Every operation in which
the index~$\alpha$ is not zero must be a rotation through $\frac{2}{3}\pi$ or~$\frac{4}{3}\pi$
about one of the angles of the figure; it is therefore necessarily
identical with an operation of the form
\[
\Sigma^{-1} S_{r}^{\alpha} \Sigma.
\]

If now the group generated by $S_{1}$,~$S_{2}$,~$S_{3}$ is of finite order,
there must be either one or two additional relations between
them. A relation of the form
\[
S_{1}^{a} (S_{1} S_{2}^{2})^{b} (S_{2} S_{1} S_{2})^{c} = 1,
\]
%% -----File: 314.png---Folio 298-------
where $a$~is either $1$ or~$2$, is equivalent to
\[
\Sigma^{-1} S_{r}^{a}\Sigma = 1,
\]
so that
\[
S_{r} = 1.
\]
Such a relation would reduce the group to a cyclical group of
order~$3$. This is not admissible, if the group is actually to be
generated by two distinct operations $S_{1}$ and~$S_{2}$.

A relation
\[
(S_{1} S_{2}^{2})^{b} (S_{2} S_{1} S_2)^{c} = 1,
\]
gives, on transformation by~$S_{2}^{-1}$,
\[
(S_{2} S_{1} S_{2})^{b} (S_{2}^{2} S_{1})^{c} = 1.
\]

Now
\[
S_{2}^{2} S_{1} = (S_{1} S_{2}^{2})^{-1} (S_{2} S_{1} S_{2})^{-1},
\]
so that
\[
(S_{1} S_{2}^{2})^{-c} (S_{2} S_{1} S_{2})^{b - c} = 1.
\]

If $m$~is the greatest common factor of $b$~and~$c$, so that
\[
b = b'm,\quad c = c'm,
\]
where $b'$~and~$c'$ are relatively prime; and if
\[
b'x - c'y = 1,
\]
the two relations
\[
(S_{1} S_{2}^{2})^{b} (S_{2} S_{1} S_{2})^{c} = 1,
\]
and
\[
(S_{1} S_{2}^{2})^{-c} (S_{2} S_{1} S_{2})^{b - c} = 1,
\]
lead to
\[
(S_{2} S_{1} S_{2})^{m(b'^{2} - b'c' + c'^{2})} = 1,
\]
and
\[
(S_{1} S_{2})^{m} = (S_{2} S_{1} S_{2})^{m\{(c' - b')y - c'x\}};
\]
and every operation of the group is contained, once and only
once, in the form
\[
S_{1}^{\alpha} (S_{1} S_{2}^{2})^{\beta} (S_{2} S_{1} S_{2})^{\gamma},
\]
where
\[
\alpha = 0, 1, 2;\ \beta = 0, 1, \dots, m - 1;\ \gamma = 0, 1, \dots, m(b'^{2} - b'c' + c'^{2}) - 1.
\]
Thus the group is of finite order $3(b^{2} - bc + c^{2})$. In this case
then, unlike the previous one, a single additional relation is
sufficient to ensure that the group is of finite order. Any
further relation, which is independent, must of necessity reduce
the group to a cyclical group of order~$3$ or to the identical
operation.

\Par{207.} The two remaining solutions may now be treated in
less detail. The general group corresponding to the solution
\[
n = 3,\quad m_{1} = 2,\quad m_{2} = m_{3} = 4,
\]
%% -----File: 315.png---Folio 299-------
is given by
\begin{gather*}
S_{1}^{2} = 1,\quad S_{2}^{4} = 1,\quad S_{3}^{4} = 1, \\
S_{1} S_{2} S_{3} = 1,
\end{gather*}
and is represented graphically by fig.~16.
\index{Graphical representation@\Topic{Graphical representation}!examples of}%

% [Illustration: Fig.~16.]
\Figure{16}{fig16}

All the translations of the group can be generated from the
two operations
\[
S_{1} S_{2}^{2},\quad S_{2} S_{1} S_{2};
\]
and every operation of the group is given, once and only once,
by the form
\[
S_{2}^{\alpha} (S_{1} S_{2}^{2})^{\beta} (S_{2} S_{1} S_{2})^{\gamma},\quad
  (\alpha = 0, 1, 2, 3;\ \beta, \gamma = -\infty, \dots, 0, \dots, \infty).
\]

An additional relation of the form
\[
S_{2}^{a} (S_{1} S_{2}^{2})^{b} (S_{2} S_{1} S_{2})^{c} = 1,
\]
where $a$~is $1$,~$2$ or~$3$, leads either to
\[
S_{1} = 1,\quad S_{2} = 1 \quad\text{or}\quad S_{3} = 1,
\]
and is therefore inconsistent with the supposition that the
group is generated by two distinct operations.

An additional relation
\[
(S_{1} S_{2}^{2})^{b} (S_{2} S_{1} S_{2})^{c} = 1
\]
gives
\[
(S_{1} S_{2}^{2})^{-c} (S_{2} S_{1} S_{2})^{b} = 1:
\]
and if
\begin{gather*}
b = b_{1}m,\quad c = c_{1}m, \\
b_{1}x + c_{1}y = 1,
\end{gather*}
%% -----File: 316.png---Folio 300-------
where $b_{1}$~and~$c_{1}$ are relatively prime, these relations are
equivalent to
\begin{gather*}
(S_{2} S_{1} S_{2})^{m(b_{1}^{2} + c_{1}^{2})} = 1, \\
(S_{1} S_{2}^{2})^{m} = (S_{2} S_{1} S_{2})^{m(b_{1}y - c_{1}x)}.
\end{gather*}

Every operation of the group is then contained, once and
only once, in the form
\begin{gather*}
S_{2}^{\alpha} (S_{1} S_{2}^{2})^{\beta} (S_{2} S_{1} S_{2})^{\gamma}, \\
(\alpha = 0, 1, 2, 3;\ \beta = 0, 1, \dots, m - 1;\ \gamma = 0, 1, \dots, m(b_{1}^{2} + c_{1}^{2}) - 1);
\end{gather*}
and the order of the group is $4(b^{2} + c^{2})$.

\Par{208.} Lastly, the general group corresponding to the
solution
\[
n = 3,\quad m_{1} = 2,\quad m_{2} = 3,\quad m_{3} = 6,
\]
is given by
\begin{gather*}
S_{1}^{2} = 1,\quad S_{2}^{3} = 1,\quad S_{3}^{6} = 1, \\
S_{1} S_{2} S_{3} = 1;
\end{gather*}
and it is represented graphically by fig.~17.
\index{Graphical representation@\Topic{Graphical representation}!examples of}%

% [Illustration: Fig.~17.]
\Figure{17}{fig17}
%% -----File: 317.png---Folio 301-------

Now it may again be verified, either from the generating
relations or from the figure, that the two operations
\[
S_{2}^{2} S_{3}^{2} \quad\text{and}\quad S_{3} S_{2}^{2} S_{3},
\]
which are permutable with each other, generate all the operations
which in the kinematical form of the group are translations;
and that every operation of the group is represented, once and
only once, by the form
\begin{gather*}
S_{3}^{\alpha} (S_{2}^{2} S_{3}^{2})^{\beta} (S_{3} S_{2}^{2} S_{3})^{\gamma}, \\
(\alpha = 0, 1, \dots, 5;\ \beta, \gamma = -\infty, \dots, 0, \dots, \infty).
\end{gather*}

Also as before, any further relation, which does not reduce
the group to a cyclical group, is necessarily of the form
\[
(S_{2}^{2} S_{3}^{2})^{b} (S_{3} S_{2}^{2} S_{3})^{c} = 1.
\]

On transforming this relation by~$S_{3}^{-1}$ we obtain
\[
(S_{3} S_{2}^{2} S_{3})^{b} (S_{3}^{2} S_{2}^{2})^{c} = 1.
\]

Now
\[
S_{3}^{2} S_{2}^{2} = (S_{2}^{2} S_{3}^{2})^{-1} (S_{3} S_{2}^{2} S_{3});
\]
so that
\[
(S_{2}^{2} S_{3}^{2})^{-c} (S_{3} S_{2} S_{3})^{b + c} = 1.
\]

If then
\begin{gather*}
b = b_{1}m,\quad c = c_{1}m, \\
b_{1}x + c_{1}y = 1,
\end{gather*}
where $b_{1}$~and~$c_{1}$ are relatively prime, it follows that
\[
(S_{3} S_{2}^{2} S_{3})^{m(b_{1}^{2} + b_{1}c_{1} + c_{1}^{2})} = 1,
\]
and
\[
(S_{2}^{2} S_{3}^{2})^{m} = (S_{3} S_{2} S_{3})^{m\{b_{1}y + c_{1}(y - x)\}}.
\]

Every operation of the group is then contained, once and
only once, in the form
\begin{gather*}
S_{3}^{\alpha} (S_{2}^{2} S_{3}^{2})^{\beta} (S_{3} S_{2}^{2} S_{3})^{\gamma}, \\
(\alpha = 0, 1, \dots, 5;\ \beta = 0, 1, \dots, m - 1;\ \gamma = 0, 1, \dots, m(b_{1}^{2} + b_{1}c_{1} + c_{1}^{2}) - 1);
\end{gather*}
and the order of the group is $6(b^{2} + bc + c^{2})$.

\Par{209.} There are thus four distinct classes of groups\footnote
  {Dyck, ``Ueber Aufstellung und Untersuchung von Gruppe und Irrationalität
  regulärer Riemann'scher Flächen,'' \textit{Math.\ Ann.}\ \VolNo{XVII}, (1880), pp.~501--509.}
\index{Dyck@\Topic{Dyck}, quoted}%
\index{Defining relations@\Topic{Defining relations} of a group!for groups of genus unity}%
of
genus~$1$, which are defined in terms of their generating operations
by the following sets of relations:---

\Item{\Inum{I.}} $\begin{gathered}[t]
S_{1}^{2} = 1,\quad S_{2}^{2} = 1,\quad S_{3}^{2} = 1, \quad (S_{1} S_{2} S_{3})^{2} = 1, \\
(S_{1} S_{2})^{a} (S_{2} S_{3})^{b} = 1,\quad
(S_{1} S_{2})^{a'} (S_{2} S_{3})^{b'} = 1,\quad (ab' - a'b > 0); \\
N = 2(ab' - a'b).
\end{gathered}$
%% -----File: 318.png---Folio 302-------

\Item{II.} $\begin{gathered}[t]
S_{1}^{3} = 1,\quad S_{2}^{3} = 1,\quad (S_{1} S_{2})^{3} = 1, \\
(S_{1}^{2} S_{2}^{2})^{a} (S_{2} S_{1} S_{2})^{b} = 1; \\
N = 3(a^{2} - ab + b^{2}).
\end{gathered}$

\Item{III.} $\begin{gathered}[t]
S_{1}^{2} = 1,\quad S_{2}^{4} = 1,\quad (S_{1} S_{2})^{4} = 1, \\
(S_{1} S_{2}^{2})^{a} (S_{2} S_{1} S_{2})^{b} = 1; \\
N = 4 (a^{2} + b^{2}).
\end{gathered}$

\Item{IV.} $\begin{gathered}[t]
S_{1}^{6} = 1,\quad S_{2}^{3} = 1,\quad (S_{1} S_{2})^{2} = 1, \\
(S_{1}^{2} S_{2}^{2})^{a} (S_{1} S_{2}^{2} S_{1})^{b} = 1; \\
N = 6 (a^{2} + ab + b^{2}).
\end{gathered}$
\index{Defining relations@\Topic{Defining relations} of a group!for groups of genus unity}%
\index{Groups@\Topic{Groups}!of genus one|)}%

For special values of $a$~and~$b$, some of these groups may be
groups of genus zero; for instance, in Class~\Inum{I}, if $ab' - a'b$ is a
prime, the group is a dihedral group. It is left as an exercise
to the reader to determine all such exceptional cases.

\begin{Ex}
Prove that the number of distinct types of group, of genus two, is
\index{Groups@\Topic{Groups}!of genus two}%
three; viz.\ the groups defined by
\begin{alignat*}{4}
&\Inum{(i)}      & A^{4} &= 1,\quad& B^{2} &= A^{2},\quad&& B^{-1}AB = A^{-1}; \\
&\Inum{(ii)}     & A^{8} &= 1,     & B^{2} &= 1,         && B^{-1}AB = A^{3}; \\
&\Inum{(iii)}\quad&A^{8} &= 1,     & B^{2} &= 1,         && (AB)^{3} = 1,\quad (A^{4}B)^{2} = 1.
\end{alignat*}
\end{Ex}

\Par{210.} As a final illustration of the present method of
\PageLabel{302}%
graphical representation, we will consider the simple group of
order~$168$ (§~146), given by
\[
\{(1236457),\ (234) (567),\ (2763) (45)\}.
\]

The operations of this group are of orders $7$,~$4$, $3$ and~$2$; and
it is easy to verify that three operations of orders $2$,~$3$, and~$7$
can be chosen such that their product is identity.

In fact, if
\[
S_{2} = (16) (34),\quad S_{3} = (253) (476),\quad S_{7} = (1673524);
\]
then
\[
S_{2} S_{3} S_{7} = 1.
\]

Moreover, these three operations generate the group. The
connectivity of the corresponding surface, by the regular division
of which the group can be represented, is $2p + 1$; where
\[
2p - 2 = 168(3 - 2 - \tfrac{1}{2} - \tfrac{1}{3} - \tfrac{1}{7}).
\]
%% -----File: 319.png---Folio 303-------

This gives
\[
p = 3;
\]
it follows from Theorem~II, §~197, that the genus of the
group is~$3$.

The figure for the general group, generated by $S_{2}$,~$S_{3}$, and~$S_{7}$,
where
\begin{gather*}
S_{2}^{2} = 1,\quad S_{3}^{3} = 1,\quad S_{7}^{7} = 1, \\
S_{2} S_{3} S_{7} = 1,
\end{gather*}
acquires as symmetrical a form as possible, by taking the
centre of the orthogonal circle for that angular point of the
triangle~$1$ at which the angle is~$\frac{1}{7}\pi$. In fig.~18 a portion of
the general figure, which is contained between two radii of the
orthogonal circle inclined at an angle~$\frac{2}{7}\pi$, is shewn. The
remainder may be filled in by inversions at the different
portions of the boundary.
\index{Graphical representation@\Topic{Graphical representation}!examples of}%

%[Illustration: Fig. 18.]
\Figure[0.8\textwidth]{18}{fig18}
%% -----File: 320.png---Folio 304-------
The operations, which correspond to the white triangles of
the figure, are given by the following table:---
\[
\begin{array}{cr|cr|cr}
1 &               1  & 9 &          S_{2}S_{7}^{5}S_{2} &17 & S_{2}S_{7}^{5}S_{2}S_{7}^{3}S_{2} \\
2 &          S_{2} &10 &     S_{7}S_{2}S_{7}^{5}S_{2} &18 &      S_{7}^{6}S_{2}S_{7}^{3}S_{2} \\
3 &     S_{7}S_{2} &11 & S_{7}^{2}S_{2}S_{7}^{5}S_{2} &19 &               S_{2}S_{7}^{3}S_{2} \\
4 & S_{7}^{2}S_{2} &12 & S_{7}^{3}S_{2}S_{7}^{5}S_{2} &20 &          S_{7}S_{2}S_{7}^{3}S_{2} \\
5 & S_{7}^{3}S_{2} &13 & S_{7}^{6}S_{2}S_{7}^{4}S_{2} &21 &               S_{2}S_{7}^{2}S_{2} \\
6 & S_{7}^{4}S_{2} &14 &          S_{2}S_{7}^{4}S_{2} &22 &      S_{7}^{6}S_{2}S_{7}^{2}S_{2} \\
7 & S_{7}^{5}S_{2} &15 &     S_{7}S_{2}S_{7}^{4}S_{2} &23 &      S_{7}^{5}S_{2}S_{7}^{2}S_{2} \\
8 & S_{7}^{6}S_{2} &16 & S_{7}^{2}S_{2}S_{7}^{4}S_{2} &24 & S_{2}S_{7}^{4}S_{2}S_{7}^{2}S_{2}.
\end{array}
\]

The representation of the special group is derived from
this general figure by retaining only a set of $168$ white (and
corresponding black) triangles, which are distinct when $S_{2}$,~$S_{3}$
and~$S_{7}$ are replaced by the corresponding substitutions given on
\PageRef{p.}{302}. When each white triangle is thus marked with the
corresponding substitution, it is found that a complete set of
$168$ distinct white (and black) triangles is given by the portion
of the figure actually drawn and the six other distinct portions
obtained by rotating it round the centre of the orthogonal
circle through multiples of~$\frac{2}{7}\pi$.

To complete the graphical representation of the group of
order~$168$, it is necessary to determine the correspondence in
pairs of the sides of the boundary. This is facilitated by
noticing that the angular points $A_{1}$, $A_{2}$, $A_{3}$,~\Dots\ of the
boundary must correspond in sets. Now the white triangle,
which has an angle at~$A_{1}$ and lies inside the polygon, is
given by
\[
S_{2} S_{7}^{4} S_{2} S_{7}^{2} S_{2}.
\]

This must be equivalent to a white triangle, which lies
outside the polygon, and has a side on the boundary and an
angle at one of the points $A_{2}$, $A_{3}$,~\Dots.

The triangle, which satisfies these conditions and has an
angle at~$A_{2n + 1}$, is given by
\[
S_{7} S_{2} S_{7}^{4} S_{2} S_{7}^{2} S_{2} S_{7}^{n};
\]
while the triangle, which satisfies the conditions and has an
angle at~$A_{2n + 2}$, is given by
\[
S_{7}^{6} S_{2} S_{7}^{5} S_{2} S_{7}^{3} S_{2} S_{7}^{n}.
\]
%% -----File: 321.png---Folio 305-------

When $(16) (34)$ and $(1673524)$ are written for $S_{2}$ and~$S_{7}$, we
find that
\[
S_{2} S_{7}^{4} S_{2} S_{7}^{2} S_{2}
  = S_{7} S_{2} S_{7}^{4} S_{2} S_{7}^{2} S_{2} S_{7}^{5}.
\]

The white triangle with an angle at~$A_{1}$ inside the polygon
is therefore equivalent to the white triangle with an angle at~$A_{11}$,
which lies outside the polygon and has a side on the
boundary. It follows, from the continuity of the figure, that
the arcs $A_{1}A_{2}$ and $A_{11}A_{10}$ of the boundary correspond. Since
the operation~$S_{7}$ changes the figure and the boundary into
themselves, it follows that $A_{3}A_{4}$,~$A_{13}A_{12}$; $A_{5}A_{6}$,~$A_{1}A_{14}$; $A_{7}A_{8}$,~$A_{3}A_{2}$;
$A_{9}A_{10}$,~$A_{5}A_{4}$; $A_{11}A_{12}$,~$A_{7}A_{6}$; and $A_{13}A_{14}$,~$A_{9}A_{8}$; are pairs
of corresponding sides. Hence the above single condition is
sufficient to ensure that the general group shall reduce to the
special group of order~$168$.

By taking account of the relation
\[
(S_{7} S_{2})^{3} = 1,\quad \text{or}\quad (S_{2} S_{7})^{3} = 1,
\]
the form of the condition may be simplified. Thus it may be
written
\begin{align*}
S_{7}^{4} S_{2} S_{7}^{2} S_{2}
  &= S_{2} S_{7} S_{2} S_{7} · S_{7}^{3} S_{2} S_{7}^{2} S_{2} S_{7}^{5} \\
  &= S_{7}^{6} S_{2} S_{7}^{3} S_{2} S_{7}^{2} S_{2} S_{7}^{5},
\end{align*}
or
\[
S_{2} S_{7}^{2} S_{2} = S_{7}^{2} S_{2} S_{7}^{3} S_{2} S_{7}^{2} S_{2} S_{7}^{5}.
\]

Now
\begin{align*}
S_{2} S_{7} · S_{7} S_{2}
  &= S_{7}^{6} S_{2} S_{7}^{6} S_{2} · S_{2} S_{7}^{6} S_{2} S_{7}^{6} \\
  &= S_{7}^{6} S_{2} S_{7}^{5} S_{2} S_{7}^{6}.
\end{align*}

Hence
\[
S_{7}^{4} S_{2} S_{7}^{5} S_{2} S_{7} = S_{2} S_{7}^{3} S_{2} S_{7}^{2} S_{2},
\]
or
\[
S_{7}^{4} S_{2} S_{7}^{4} · S_{2} S_{7}^{6} S_{2} = S_{2} S_{7}^{3} S_{2} S_{7}^{2} S_{2},
\]
or
\[
S_{7}^{4} S_{2} S_{7}^{4} S_{2} S_{7}^{4} = S_{2} S_{7}^{3} S_{2},
\]
or finally
\[
(S_{7}^{4} S_{2})^{4} = 1.
\]

The simple group of order~$168$ is therefore defined abstractly
\index{Defining relations@\Topic{Defining relations} of a group!for the simple group of order~$168$}%
by the relations\footnote
  {This agrees with the result as stated by Dyck, ``Gruppentheoretische
  Studien,'' \textit{Math.\ Ann.}\ Vol.~\VolNo{XX}, (1882), p.~41.}
\index{Dyck@\Topic{Dyck}, quoted}%
\[
S_{2}^{2} = 1,\quad S_{7}^{7} = 1,\quad (S_{7} S_{2})^{3} = 1,\quad (S_{7}^{4} S_{2})^{4} = 1.
\]

\begin{Ex} Shew that the symmetric group of degree five is a group
\index{Defining relations@\Topic{Defining relations} of a group!for the symmetric group of degree~$5$}%
of genus four; and that it is completely defined by the relations
\[
S_{2}^{2} = 1,\quad S_{5}^{5} = 1,\quad (S_{2} S_{5})^{4} = 1,\quad (S_{5}^{-1} S_{2} S_{5} S_{2})^{3} = 1.
\]
\end{Ex}
%% -----File: 322.png---Folio 306-------

\Par{211.} The regular division of a continuous surface into
$2N$~black and white polygons is only one of many methods that
may be conceived for representing a group graphically.

We shall now describe shortly another such mode of
\index{Colour-groups@\Topic{Colour-groups}|(}%
representation, due to Cayley\footnotemark,
\index{Cayley@\Topic{Cayley}, quoted}%
\footnotetext{\textit{American Journal of Mathematics}, Vol.~\VolNo{I}, (1878), pp.~174--176, Vol.~\VolNo{XI},
  (1889), pp.~139--157; \textit{Proceedings of the London Mathematical Society}, Vol.~\VolNo{IX},
  (1878), pp.~126--133.}%
who has called it the method
of \emph{colour-groups}. As given by Cayley, this method is entirely
independent of the one we have been hitherto dealing with;
but there is an intimate relation between them, and the new
method can be most readily presented to the reader by deriving
it from the old one.

Let
\[
1,\ S_{1},\ S_{2},\ \Dots,\ S_{N - 1}
\]
be the operations of a group~$G$ of order~$N$. We may take the
$N - 1$ operations other than identity as a set of generating
operations. Their continued product
\[
S_{1} S_{2} \Dots S_{N - 1}
\]
is some definite operation of the group. If it is the identical
operation, the only modification in the figure, which represents
the group by the regular division of a continuous surface, will
be that the $N$th~corner of the polygon has an angle of two right-angles.

With this set of generating operations, the representation
of the group is given by a regular division of a continuous
surface into $N$~white and $N$~black polygons $A_{1}A_{2}\Dots A_{N}$, the
angle at~$A_{r}$ being~$\dfrac{2\pi}{m_{r}}$, $m_{r}$~being the order of~$S_{r}$. Suppose
now that in each white polygon we mark a definite point.
From the marked point in the polygon~$\Sigma$, draw a line to
the marked point in the polygon derived from it by a positive
rotation round its angle~$A_{r}$. Call this line an $S_{r}$-line, and
denote the direction in which it is drawn by means of an
arrow. Carry out this construction for each polygon~$\Sigma$, and
for each of its angles except~$A_{N}$. We thus form a figure
which, disregarding the original surface, consists of $N$~points
connected by $N(N - 1)$ directed lines, two distinct lines
joining each pair of points. Now if the line drawn from~$a$
%% -----File: 323.png---Folio 307-------
to~$b$, where $a$~and~$b$ are two of the points, is an $S$-line, then
the line drawn from $b$ to~$a$ is from the construction an $S^{-1}$-line.
We may then at once modify our diagram, in the direction
of simplification, by dropping out one of the two lines
between $a$ and~$b$, say the $S^{-1}$-line, on the understanding that
the remaining line, with the arrow-head reversed, will give the
line omitted. If $S$~is an operation of order~$2$, $S$~and $S^{-1}$ are
identical, and two arrow-heads may be drawn on such a line in
opposite directions. The modified figure will now consist of
$N$~points connected by $\frac{1}{2}N(N - 1)$ lines. From the construction
it follows at once that, for every value of~$r$, a single $S_{r}$-line ends
at each point of the figure and a single $S_{r}$-line begins at each
point of the figure; these two lines being identical when the
order of~$S_{r}$ is~$2$.

We may pass from one point of the figure to another along
the lines in various ways; but any path between two points of
the figure will be specified completely by such directions as:
follow first an $S_{r}$-line, then an $S_{s}$-line, then an $S_{t}^{-1}$-line, and
so on. Such a set of directions is said to define a route.
It is an immediate consequence of the construction that, if
starting from some one particular point a given route leads back
to the starting point, then it will lead back to the starting point
from whatever point we begin. In fact, a route will be specified
symbolically by a symbol
\[
S_{u} \Dots S_{t}^{-1}S_{s}S_{r},
\]
and if
\[
S_{u} \Dots S_{t}^{-1}S_{s}S_{r}\Sigma = \Sigma,
\]
then
\[
S_{u} \Dots S_{t}^{-1}S_{s}S_{r} = 1,
\]
and therefore
\[
S_{u} \Dots S_{t}^{-1}S_{s}S_{r}\Sigma' = \Sigma',
\]
whatever operation $\Sigma'$ may be.

\Par{212.} If the diagram of $N$~points connected by $\frac{1}{2}N(N - 1)$
directed lines is to appeal readily to the eye, some method must
be adopted of easily distinguishing an $S_{r}$-line from an $S_{s}$-line.
To effect this purpose, Cayley suggested that all the $S_{r}$-lines
should be of one colour, all the $S_{s}$-lines of another, and so on.
Suppose now that, independently of any previous consideration,
we have a diagram of $N$~points connected by $\frac{1}{2}N(N - 1)$ coloured
directed lines satisfying the following conditions:---
%% -----File: 324.png---Folio 308-------

\Inum{(i)} all the lines of any one colour have either \Inum{($a$)}~a single
arrow-head denoting their directions: or \Inum{($b$)}~two opposed arrow-heads,
in which case each may be regarded as equivalent to two
coincident lines in opposite directions;

\Inum{(ii)} there is a single line of any given colour leading to every
point in the diagram, and a single line of the colour leading from
every point: if the colour is one with double arrow-heads, the
two lines are a pair of coincident lines;

\Inum{(iii)} every route which, starting from some one given point
in the diagram, is closed, \ie\ leads back again to the given point,
is closed whatever the starting point.

Then, under these conditions, the diagram represents in
graphical form a definite group of order~$N$.

It is to be noticed that the first two conditions are necessary
in order that the phrase ``a route'' used in the third shall have
a definite meaning. Suppose that $R$~and~$R'$ are two routes
leading from $a$ to~$b$. Then $RR'^{-1}$~is a closed route and will
lead back to the initial point whatever it may be. Hence if $R$~leads
from $c$ to~$d$, so also must~$R'$; and therefore $R$~and~$R'$ are
equivalent routes in the sense that from any given starting
point they lead to the same final point. There are then, with
identity which displaces no point, just $N$ non-equivalent routes
on the diagram, and the product of any two of these is a definite
route of the set. The $N$~routes may be regarded as operations
performed on the $N$~points; on account of the last property
which has been pointed out, they form a group. Moreover, the
diagram gives in explicit form the complete multiplication table
of the group, for a mere inspection will immediately determine
the one-line route which is equivalent to any given route; \ie\
the operation of the group which is the same as the product of
any set of operations in any given order.

From a slightly different point of view, every route will give
a permutation of the $N$~points, regarded as a set of symbols,
among themselves; no symbol remaining unchanged unless
they all do. To the set of $N$ independent routes, there will
correspond a set of $N$ substitutions performed on $N$~symbols;
and we can therefore immediately from the diagram represent
the group as a transitive substitution group of degree~$N$.
%% -----File: 325.png---Folio 309-------

\Par{213.} It cannot be denied that, even for groups of small
order, the diagram we have been describing would not be easily
grasped by the eye. It may however still be considerably
simplified since, so far as a graphical definition of the group
is concerned, a large number of the lines are always redundant.

If in the diagram consisting of $N$~points and $\frac{1}{2}N(N - 1)$
coloured lines, which satisfies the conditions of §~212, all the
lines of one or more colours are omitted, two cases may occur.
We may still have a figure in which it is possible to pass
along lines from one point to any other; or the points may
break up into sets such that those of any one set are connected
by lines, while there are no lines which enable us
to pass from one set to another.

Suppose, to begin with, that the first is the case. There will
then, as before, be $N$~non-equivalent routes in the figure, which
form a group when they are regarded as operations; it is
obviously the same group as is given by the general figure.
The sole difference is that there will not now be a one-line
route leading from every point to every other point, and therefore
the diagram will no longer give directly the result of the
product of any number of operations of the group.

If on the other hand the points break up into sets, the new
diagram will no longer represent the same group as the original
diagram. Some of the routes of the original diagram will
not be possible on the new one, but every route on the new
one will be a route on the original diagram. Hence the new
diagram will give a sub-group; and since it is still the case that
no route, except identity, can leave any point unchanged, the
number of points in each of the sets must be the same. The
reader may verify that the sub-group thus obtained will be self-conjugate,
only if the omitted colours interchange these sets
bodily among themselves.

\Par{214.} The simplest diagram that will represent the group
will be that which contains the smallest number of colours and
at the same time connects all the points. To each colour
corresponds a definite operation of the group (and its inverse).
Hence the smallest number of colours is the smallest number
of operations that will generate the group. It may be noticed
%% -----File: 326.png---Folio 310-------
that this simplified diagram can be actually constructed from the
previously obtained representation of the group by the regular
division of a surface, the process being exactly the same as
that by which the general diagram was obtained. For if
\[
S_{1},\ S_{2},\ \Dots,\ S_{n}
\]
are a set of independent generating operations, and if
\[
S_{1} S_{2} \Dots S_{n} S_{n + 1} = 1,
\]
we may represent the group by the regular division of a surface
into $2N$ black and white $(n + 1)$-sided polygons. When we
draw on this surface the $S_{1}$-, $S_{2}$-,~\Dots, $S_{n}$-lines, the $N$~points
will be connected by lines in a single set, since from
\[
S_{1},\ S_{2},\ \Dots,\ S_{n}
\]
every operation of the group can be constructed; and the set
of points and directed coloured lines so obtained is clearly the
diagram required.

As an illustration of this form of graphical representation,
\index{Colour-groups@\Topic{Colour-groups}!examples of}%
\index{Graphical representation@\Topic{Graphical representation}!examples of}%
we may consider the octohedral group (§~201), defined by
\begin{gather*}
S_{1}^{2} = 1,\quad S_{2}^{3} = 1,\quad S_{3}^{4} = 1, \\
S_{1} S_{2} S_{3} = 1.
\end{gather*}

On the diagram already given (\PageRef{p.}{290}), we may at once draw
$S_{1}$-,~$S_{2}$- and $S_{3}$-lines. These in the present figure are coloured
respectively red, yellow, and green. On the lines which correspond
to operations of order two (in this case the red lines),
the double arrow-heads may be dispensed with. By omitting
successively the red, the yellow, and the green lines we form
from this the three simplest colour diagrams which will represent
\index{Young@\Topic{Young}, quoted}%
\index{Maschke@\Topic{Maschke}, quoted}%
the group\footnotemark.
\footnotetext{For further illustrations, the reader may refer to Young, \textit{Amer.\ Journal},
  Vol.~\VolNo{XV}, (1893), pp.~164--167; Maschke, \textit{Amer.\ Journal}, Vol.~\VolNo{XVIII}, (1896),
  pp. 156--188.}%
\index{Colour-groups@\Topic{Colour-groups}|)}%
%% -----File: 327.png----------
\index{Burnside@\Topic{Burnside}, quoted}%
\index{Fractional linear group@\Topic{Fractional linear group}!definition of}%
\index{Gierster@\Topic{Gierster}, quoted}%
\index{Jordan@\Topic{Jordan}, quoted}%
\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots, 1)$|(}%
% [Plate 1] \textit{To face page}~310.
\begin{figure}[hp!]
\centering
\Graphic[jpg]{\textwidth}{plate1}
\end{figure}
%% -----File: 328.png----------
% [Blank Page]
%% -----File: 329.png---Folio 311-------


\Chapter{XIV.}{On the Linear Group\protect\footnotemark.}

\footnotetext{The homogeneous linear group and its sub-groups forms the subject of the
  greater part of Jordan's \textit{Traité des Substitutions}. The investigation of its
  composition-series, given in the text, is due to Jordan.

  The complete analysis of the fractional linear group, defined by
  \[
  y \equiv \frac{\alpha x + \beta}{\gamma x + \delta},\ (\mod p),
  \]
  where
  \[
  \alpha\delta - \beta\gamma \equiv 1,
  \]
  is due originally to Gierster, ``Die Untergruppen der Galois'schen Gruppe der
  Modulargleichungen für den Fall eines primzahligen Transformationsgrades,''
  \textit{Math.\ Ann.}\ Vol.~\VolNo{XVIII}, (1881), pp.~319--365. With a few unimportant
  modifications, the investigation in the text follows the lines of Gierster's
  memoir.

  A similar analysis of the simple groups of order $2^{n}(2^{2n} - 1)$, which can be
  expressed as triply transitive groups of degree~$2^{n} + 1$, has been given by the
  author, ``On a class of groups defined by congruences,'' \textit{Proc.\ L.~M.~S.} Vol.~\VolNo{XXV},
  (1894), pp.~132--136.}

\Par{215.} \First{We} shall now, in illustration of the general principles
that have been developed in the preceding chapters, proceed to
discuss and give an analysis of certain special groups. The first
that we choose for this purpose is the group of isomorphisms
of an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots\DPchg{\,}{,} \text{to $n$ units})$.
This group has been defined and its order determined in
§§~171,~172. It is there shewn that the group is simply
isomorphic with the homogeneous linear group defined by all
sets of congruences
\[
\begin{aligned}
y_{1} &\equiv a_{11} x_{1} + a_{12} x_{2} + \Dots + a_{1n} x_{n}, \\
y_{2} &\equiv a_{21} x_{1} + a_{22} x_{2} + \Dots + a_{2n} x_{n}, \\
\multispan{3}{\dotfill} \\
y_{n} &\equiv a_{n1} x_{1} + a_{n2} x_{2} + \Dots + a_{nn} x_{n},
\end{aligned}
\quad(\mod p),
\]
%% -----File: 330.png---Folio 312-------
whose determinants are not congruent to zero; and that its
order is
\[
N = (p^{n} - 1) (p^{n} - p) \Dots (p^{n} - p^{n - 1}).
\]
The operation given by the above set of congruences will be
denoted in future by the symbol
%[** TN: Set on two lines in the original]
\[
(a_{11}x_{1} + \Dots + a_{1n}x_{n},\
a_{21}x_{1} + \Dots + a_{2n}x_{n},\ \Dots,\
a_{n1}x_{1} + \Dots + a_{nn}x_{n}).
\]

\Par{216.} It may be easily verified by direct calculation that, if
$D$~and~$D'$ are the determinants of two operations $S$ and~$S'$ of
the group, then $DD'$~is the determinant of the operation~$SS'$,
all the numbers involved being reduced, $\pmod{p}$. Hence it
immediately follows that those operations of the group, whose
determinant is unity, form a sub-group. If this sub-group is
denoted by~$\Gamma$, the group itself being~$G$, then $\Gamma$~is a self-conjugate
sub-group of~$G$. For if $\Sigma$~is any operation of~$\Gamma$ and
$S$~any operation of~$G$ whose determinant is~$D$, the determinant
of~$S^{-1}\Sigma S$ is $D^{-1}D$ or unity; and therefore $S^{-1}\Sigma S$~belongs to~$\Gamma$.
Suppose now that $S$~is an operation\footnote
  {The operation $(zx_{1}, x_{2}, \dots, x_{n})$ has $z$~for its determinant.}
of~$G$ whose determinant is~$z$,
a primitive root of the congruence
\[
z^{p - 1} \equiv 1\ (\mod p).
\]
Then the determinant of every operation of the set
\[
S^{r}\Gamma
\]
is~$z^{r}$; and therefore, if $r$~and~$s$ are not congruent $(\mod p)$, the
two sets
\[
S^{r}\Gamma\quad \text{and}\quad S^{s}\Gamma
\]
can have no operation in common. Moreover, if $S'$~is any
operation of~$G$ whose determinant is~$z^{r}$, then $S^{-r}S'$~belongs to~$\Gamma$,
and therefore $S'$~belongs to the set~$S^{r}\Gamma$. Hence finally, the
sets
\[
\Gamma,\ S\Gamma,\ S^{2}\Gamma,\ \Dots,\ S^{p - 2}\Gamma
\]
are all distinct, and they include every operation of~$G$; so that
\[
G = \{S, \Gamma\}.
\]

The factor-group $\dfrac{G}{\Gamma}$ is therefore cyclical and of order~$p - 1$.
%% -----File: 331.png---Folio 313-------

\Par{217.} It may be very readily verified that the operations of
the cyclical sub-group generated by
\[
(zx_{1}, zx_{2}, \Dots, zx_{n})
\]
are self-conjugate operations of~$G$. To prove that these are
the only self-conjugate operations of~$G$, we will deal with the
case $n = 3$: it will be seen that the method is perfectly
general. Suppose then that
\[
T = (\alpha   x_{1} + \beta   x_{2} + \gamma   x_{3},\
     \alpha'  x_{1} + \beta'  x_{2} + \gamma'  x_{3},\
     \alpha'' x_{1} + \beta'' x_{2} + \gamma'' x_{3})
\]
is a self-conjugate operation of $G$, while
\[
S = (a  x_{1} + b  x_{2} + c  x_{3},\
     a' x_{1} + b' x_{2} + c' x_{3},\
     a''x_{1} + b''x_{2} + c''x_{3})
\]
is any operation. The relation
\[
ST = TS
\]
involves the nine simultaneous congruences\footnote
  {These and all succeeding congruences are to be taken $\pmod{p}$, unless the
  contrary is stated.}
\begin{align*}
a\alpha + b\alpha' + c\alpha'' &\equiv \alpha a + \beta a' + \gamma a'', \\
a\beta  + b\beta'  + c\beta''  &\equiv \alpha b + \beta b' + \gamma b'', \\
a\gamma + b\gamma' + c\gamma'' &\equiv \alpha c + \beta c' + \gamma c'', \\
\PadTo{a\gamma + b\gamma' + c\gamma''}{\text{etc.},} &
\PadTo{\equiv \alpha c + \beta c' + \gamma c''}{\text{etc.};}
\end{align*}
and these must be satisfied for all possible values of the
coefficients of~$S$. Now
\[
b \equiv c \equiv a' \equiv 0
\]
is a possible relation between the coefficients of~$S$, whether
regarded as an operation of~$G$ or~$\Gamma$; and therefore
\[
\gamma \equiv 0.
\]
In the same way, it may be shewn that
\[
\beta \equiv \alpha' \equiv \gamma' \equiv \alpha'' \equiv \beta'' \equiv 0,
\]
and that
\[
\alpha \equiv \beta' \equiv \gamma'';
\]
so that $T$~is a power of the operation
\[
(zx_{1}, zx_{2}, zx_{3}).
\]
The only self-conjugate operations of~$G$ are therefore the powers
of~$A$, where $A$~denotes
\[
(zx_{1}, zx_{2}, \Dots, zx_{n});
\]
and the only self-conjugate operations of~$\Gamma$ are those operations
%% -----File: 332.png---Folio 314-------
of this cyclical sub-group which are contained in~$\Gamma$. Now the
order of~$A$ is~$p - 1$ and its determinant is~$z^{n}$. Hence the self-conjugate
operations of~$\Gamma$ form a cyclical sub-group~$D$ of order~$d$,
where $d$~is the greatest common factor of $p - 1$ and~$n$; and
this sub-group is generated by~$A^{\efrac{p - 1}{d}}$.

\Par{218.} To determine completely the composition-series of~$G$,
\index{Homogeneous linear group@\Topic{Homogeneous linear group}!composition series of|(}%
it is necessary to find whether $\Gamma$ has a self-conjugate sub-group
greater than and containing~$D$. A simple calculation
will shew that, from
\[
(x_{1} + x_{s}, x_{2}, x_{3}, \Dots, x_{n})
\]
and its conjugate operations, all the operations of~$\Gamma$ may be
generated; and hence no self-conjugate sub-group of~$\Gamma$ which
is different from $\Gamma$ itself can contain an operation of this form.
If then it is shewn that any self-conjugate sub-group of~$\Gamma$,
distinct from~$D$, necessarily contains operations of this form, it
follows that $D$~is a maximum self-conjugate sub-group of~$\Gamma$.

We shall first deal with the case $n = 2$.

If $p = 2$, the orders of $G$,~$\Gamma$ and~$D$ are $6$,~$6$ and~$1$. In this
case, $\Gamma$~is simply isomorphic with the symmetric group of three
symbols, which has a self-conjugate sub-group of order~$3$. The
successive factor-groups of the composition-series of~$G$ are
therefore cyclical groups of orders $2$ and~$3$.

If $p = 3$, the orders of $G$,~$\Gamma$ and~$D$ are $48$,~$24$ and~$2$. The
factor-group $\dfrac{\Gamma}{D}$ has $12$ for its order, and cannot therefore be a
simple group. The reader will have no difficulty in verifying
that, in this case, the successive factor-groups of~$G$ have orders
$2$,~$3$, $2$,~$2$ and~$2$. We may therefore, in dealing with the case
$n = 2$, assume that $p$~is not less than~$5$.

Let us suppose now that $\Gamma$~has a self-conjugate sub-group~$I$
that contains~$D$; and let $S$ or
\[
(ax_{1} + bx_{2},\ a'x_{1} + b'x_{2})
\]
be one of its operations, not contained in~$D$.

If $b$~is different from zero, $\Gamma$~contains~$\Sigma$, where $\Sigma$~denotes
\[
\left(\alpha ax_{1} + \alpha bx_{2}, -\frac{1 + \alpha^{2}a^{2}}{\alpha b} x_{1} - \alpha ax_{2}\right),
\]
%% -----File: 333.png---Folio 315-------
and therefore $I$~contains $\Sigma^{-1} S \Sigma S$, which is
\[
\left(-\alpha^{-2} x_{1}, -\frac{(1 + \alpha^{2})(b' + a\alpha^{2})}{b\alpha^{2}} x_{1} - \alpha^{2}x_{2}\right).
\]

If $b$~is zero, $b'$~is congruent with~$a^{-1}$; therefore, in any case,
$I$~contains an operation~$S'$ of the form
\[
(cx_{1}, dx_{1} + c^{-1}x_{2}).
\]

Again, $\Gamma$~contains the operation~$T$, where $T$~denotes
\[
(x_{1}, x_{1} + x_{2});
\]
and $I$~therefore contains $S' T^{-1} S'^{-1} T$, which is
\[
(x_{1}, (1 - c^{2})x_{1} + x_{2}).
\]

Hence unless $1 - c{^2} \equiv 0$, $I$~must coincide with~$\Gamma$. Now,
when $p > 5$, $c$~can always be chosen so that this congruence
is not satisfied. If $p = 5$, the square of the above operation
$\Sigma^{-1} S\Sigma S$, when unity is written for~$\alpha$, is
\[
\left(x_{1}, \frac{4}{b}(b' + a)x_{1} + x_{2}\right);
\]
unless $b' + a \equiv 0$, this again requires that $I$ coincides with~$\Gamma$.
If finally, the condition $b' + a \equiv 0$ is satisfied in~$S$, it is not
satisfied in~$ST^{-1} S^{-1}T$, another operation belonging to~$I$; and
therefore again, in this case, $I$~coincides with~$\Gamma$.

Hence finally, if $n = 2$, the factor-group $\dfrac{\Gamma}{D}$ is simple, except
when $p$~is $2$ or~$3$.

\Par{219.} When $n$~is greater than~$2$, it will be found that it is
sufficient to deal in detail with the case $n = 3$, as the method
will apply equally well for any greater value of~$n$. Suppose
here again that $\Gamma$~has a self-conjugate sub-group~$I$ which
contains~$D$; and let $S$, denoting
\[
(a  x_{1} + b  x_{2} + c  x_{3},\
 a' x_{1} + b' x_{2} + c' x_{3},\
 a''x_{1} + b''x_{2} + c''x_{3}),
\]
be one of the operations of~$I$ which is not contained in~$D$.
$S$~cannot be permutable with all operations of the form
$(x_{1}, x_{2}, x_{3} + x_{1})$, as it would then be permutable with every
operation of~$\Gamma$. We may therefore suppose without loss of
generality that $S$~and~$T$ are not permutable, $T$~denoting
$(x_{1}, x_{2}, x_{3} + x_{1})$. Then $T^{-1}STS^{-1}$~is an operation, distinct from
%% -----File: 334.png---Folio 316-------
identity, belonging to~$I$. Now a simple calculation shews that
this operation, say~$U$, is of the form
\[
\left(x_{1} - cX, x_{2} - c'X, Ax_{1} + Bx_{2} + Cx_{3}\right),
\]
where $X$~is the symbol with which $S^{-1}$ replaces~$x_{1}$.

If $c$~and~$c'$ are both different from zero, $\Gamma$~will contain an
operation~$V$ of the form
\[
\left(x_{1} - \frac{c}{c'} x_{2}, x_{2}, x_{3}\right);
\]
and $I$~contains $V^{-1}UV$, which is of the form
\[
(x_{1}, \alpha x_{1} + \beta x_{2} + \gamma x_{3}, \alpha' x_{1} + \beta' x_{2} + \gamma' x_{3}).
\]

Moreover, if either $c$~or $c'$ is zero, the operation $U$ itself
leaves one symbol unaltered. Hence $I$~always contains operations
by which one symbol is unaltered.

This process may now be repeated to shew that $I$~necessarily
contains operations of the form
\[
(x_{1}, x_{2}, \alpha'' x_{1} + \beta'' x_{2} + \gamma'' x_{3});
\]
and, since the determinant of the operation is unity, $\gamma''$~is
necessarily congruent to unity. But it has been seen that
the group~$\Gamma$ is generated from the last operation and the
operations conjugate to it. Hence finally, if $n$~is greater than~$2$,
the factor-group~$\dfrac{\Gamma}{D}$ is simple for all values of~$p$.

\Par{220.} The composition-series of~$G$ is now, except as regards
the constitution of the simple group~$\dfrac{\Gamma}{D}$, perfectly definite. It
has, in fact, been seen that $\dfrac{G}{\Gamma}$~and $D$ are cyclical groups of orders
$p - 1$ and~$d$; and therefore if $\alpha$,~$\beta$, $\gamma$,~\Dots\ are distinct primes
whose product is~$p - 1$, and if $\alpha'$,~$\beta'$, $\gamma'$,~\Dots\ are distinct primes
whose product is~$d$; the successive factor-groups of~$G$ are first,
a series of simple groups of prime orders $\alpha$,~$\beta$, $\gamma$,~\Dots: then
a simple group of composite order~$\dfrac{N}{(p - 1)d}$: and lastly, a series
of simple groups of prime orders $\alpha'$,~$\beta'$, $\gamma'$,~\Dots.

The sequence in which the set of simple groups of orders
%% -----File: 335.png---Folio 317-------
$\alpha$,~$\beta$, $\gamma$,~\Dots\ are taken in the composition-series may be clearly
any whatever, and the same is true of the set of factor-groups
of orders $\alpha'$,~$\beta'$, $\gamma'$,~\Dots; but it is to be noticed that, when $d$~is
not equal to~$p - 1$, the composition-series is capable of further
modifications. In this case, $\{A, \Gamma\}$~is a self-conjugate sub-group
of~$G$ of order~$\dfrac{N}{d}$, which has a maximum self-conjugate sub-group~$\{A\}$
of order~$p - 1$. The successive composition-factors of~$G$
may therefore be taken in the sequence
\[
\alpha',\ \beta',\ \gamma',\ \Dots:\ \frac{N}{(p - 1) d}:\ \alpha,\ \beta,\ \gamma,\ \Dots:
\]
and their arrangement may be yet further changed by considering
the self-conjugate sub-group $\{A^{m}, \Gamma\}$, where $m$~is a
factor of $p - 1$ less than~$\dfrac{p - 1}{d}$.
\index{Homogeneous linear group@\Topic{Homogeneous linear group}!composition series of|)}%

\Par{221.} For every value of~$p^{n}$, except $2^{2}$~and~$3^{2}$, it thus
appears that the linear group may be regarded as defining a
simple group of composite order. We shall now proceed
to a discussion of the constitution of the simple groups thus
defined when $n = 2$, $p$~being greater than~$3$\footnotemark.
\footnotetext{For the case $n = 3$ the reader may consult a paper by the author ``On a
  class of groups defined by congruences,'' \textit{Proc.\ L.~M.~S.} Vol.~\VolNo{XXVI}, (1895), pp.~58--106.}
\index{Burnside@\Topic{Burnside}, quoted}%
In this case, the
group~$\Gamma$ is defined by the congruences
\[
\begin{aligned}
y_{1} &\equiv \alpha x_{1} + \beta x_{2}, \\
y_{2} &\equiv \gamma x_{1} + \delta x_{2}, \\
\alpha\delta - \beta\gamma &\equiv 1,
\end{aligned}
\quad (\mod p);
\]
and since $p - 1$ is divisible by~$2$ when $p$~is an odd prime, $d$~is
equal to~$2$. Hence the self-conjugate operations of~$\Gamma$ are
\[
(x_{1}, x_{2}) \quad\text{and}\quad (-x_{1}, -x_{2}).
\]
\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots, 1)$|)}%

The order of~$\Gamma$ is~$p(p^{2} - 1)$, and therefore the order of the
simple group,~$H$, which it defines is~$\frac{1}{2} p(p^{2} - 1)$. Suppose now,
if possible, that $\Gamma$~contains a sub-group~$g$ simply isomorphic
with~$H$. If $S$~is any operation of~$\Gamma$, not contained in~$g$, the
whole of the operations of~$\Gamma$ are contained in the two sets
\[
g,\ Sg.
\]
%% -----File: 336.png---Folio 318-------

Now $(x_{2}, -x_{1})$, whose square $(-x_{1}, -x_{2})$ is a self-conjugate
operation, cannot be contained in the simple group~$g$. Hence
both $(x_{2}, -x_{1})$ and $(-x_{1}, -x_{2})$ are contained in~$Sg$, an obvious
contradiction. Therefore $\Gamma$~contains no sub-group simply
isomorphic with~$H$.

For a discussion of the properties of~$H$, some concrete
representation of the group itself is necessary; this may be
obtained in the following way. Instead of the pair of homogeneous
congruences that define each operation of~$\Gamma$, let us,
as in §~113, consider the single non-homogeneous congruence
\[
y \equiv \frac{\alpha x + \beta} {\gamma x + \delta},\ (\mod p),
\]
where
\[
\alpha\delta - \beta\gamma \equiv 1.
\]
Corresponding to every operation
\[
(\alpha x_{1} + \beta x_{2},\ \gamma x_{1} + \delta x_{2})
\]
of~$\Gamma$, there will be a single operation of this new set; namely
that in which $\alpha$,~$\beta$, $\gamma$,~$\delta$ have respectively the same values. But
since the operations
\[
y \equiv \frac{ \alpha x + \beta}{ \gamma x + \delta} \quad\text{and}\quad
y \equiv \frac{-\alpha x - \beta}{-\gamma x - \delta}
\]
are identical, two operations
\[
( \alpha x_{1} + \beta x_{2},\  \gamma x_{1} + \delta x_{2}) \quad\text{and}\quad
(-\alpha x_{1} - \beta x_{2},\ -\gamma x_{1} - \delta x_{2})
\]
of~$\Gamma$ will correspond to each operation
\[
y \equiv \frac{\alpha x + \beta}{\gamma x + \delta}
\]
of the new set; the two self-conjugate operations
\[
(x_{1}, x_{2}) \quad\text{and}\quad (-x_{1}, -x_{2}),
\]
in particular, corresponding to the identical operation of the
new set. Moreover, direct calculation immediately verifies
that, to the product of any two operations of~$\Gamma$, corresponds the
product of the two corresponding operations of the new set.
Hence the new set of operations forms a group of order~$\frac{1}{2}p(p^{2} - 1)$,
with which $\Gamma$~is multiply isomorphic; the group of order~$2$
formed by the self-conjugate operations of~$\Gamma$ corresponding to
the identical operation of the new group.
%% -----File: 337.png---Folio 319-------

The simple group~$H$, of order~$\frac{1}{2}p(p^{2} - 1)$, which we propose
to discuss, can therefore be represented by the set of operations
\[
y \equiv \frac{\alpha x + \beta}{\gamma x + \delta},\ (\mod p);
\]
where
\[
\alpha\delta - \beta\gamma \equiv 1,
\]
$\alpha$,~$\beta$, $\gamma$,~$\delta$ being integers reduced to modulus~$p$.

\Par{222.} Since the order of~$H$ is divisible by~$p$ and not by~$p^{2}$,
\index{Fractional linear group@\Topic{Fractional linear group}!analysis of|(}%
the group must contain a single conjugate set of sub-groups of
order~$p$. Now the operation
\[
y \equiv x + 1,
\]
or $(x + 1)$ as we will write it in future, is clearly an operation
of order~$p$: for its $n$th~power is~$(x + n)$, and $p$~is the smallest
value of~$n$ for which this is the identical operation. If
$(x + 1)$ and $\left(\dfrac{\alpha x + \beta}{\gamma x + \delta}\right)$ are represented by $P$ and~$S$, then
\[
S^{-1}PS = \left(\frac{(1 - \alpha\gamma)x + \alpha^{2}}{-\gamma^{2}x + 1 + \alpha\gamma}\right).
\]

This is identical with~$P$, only if
\[
\gamma \equiv 0,\quad \alpha^{2} \equiv 1;
\]
and therefore $P$~is permutable with no operations except its
own powers. On the other hand, if
\[
\gamma \equiv 0,
\]
then
\[
S^{-1}PS = P^{a^{2}};
\]
and therefore every operation, for which $\gamma \equiv 0$, transforms the
sub-group~$\{P\}$ into itself. These operations therefore form a
sub-group: a result that may also be easily verified directly.
The order of this sub-group is the number of distinct operations
$\left(\dfrac{\alpha x + \beta}{\delta}\right)$ for which $\alpha\delta \equiv 1$. The ratio~$\dfrac{\alpha}{\delta}$ must be a quadratic
residue, while $\beta$~may have any value whatever. Hence the
order of the sub-group is~$\frac{1}{2}(p - 1)p$; and $H$~therefore contains
$p + 1$ sub-groups of order~$p$. Since $H$~is a simple group, it
follows (§~125) that it can be represented as a transitive
substitution group of degree~$p + 1$.
%% -----File: 338.png---Folio 320-------

This representation of the group can be directly derived, as
in §~113, from the congruences already used to define it. Thus
if, in
\[
y \equiv \frac{\alpha x + \beta}{\gamma x + \delta},
\]
we write for $x$ successively $0$, $1$, $2$,~\Dots, $p - 1$,~$\infty$, the $p + 1$
values obtained for~$y$, when reduced $\pmod{p}$, will be the same
$p + 1$ symbols in some other sequence. For if
\[
\frac{\alpha x_{1} + \beta}{\gamma x_{1} + \delta} \equiv
\frac{\alpha x_{2} + \beta}{\gamma x_{2} + \delta},
\]
then
\[
(\alpha\delta - \beta\gamma)(x_{1} - x_{2}) \equiv 0,
\]
and therefore $x_{1} \equiv x_{2}$.

Each operation of~$H$ gives therefore a distinct substitution
performed on the symbols $0$, $1$,~\Dots, $p - 1$,~$\infty$; and the complete
set of substitutions thus obtained gives the representation of~$H$
as a transitive substitution group of degree~$p + 1$. Since $H$~contains
operations of order~$p$, this substitution group must be
doubly transitive. That this is the case may also be shewn
directly. Thus
\[
\frac{y - a}{y - b} \equiv m \frac{x - a'}{x - b'}
\]
is an operation changing $a'$ into~$a$ and $b'$ into~$b$. This operation
may be written
\[
y \equiv \frac{k(bm - a)x + k(ab' - ma'b)}
              {k(m - 1)x + k(b' - ma')},
\]
and its determinant is
\[
k^{2}m(b - a)(b' - a').
\]

If now $(b - a)(b' - a')$ is a quadratic residue (or non-residue)
$\pmod{p}$, $m$~may be any quadratic residue (or non-residue); and $k$~can
always be chosen so that the determinant is unity. There
are therefore $\frac{1}{2}(p - 1)$ substitutions in the group, changing any
two symbols $a'$,~$b'$ into any other two given symbols $a$,~$b$.
Further, if the operation $\left(\dfrac{\alpha x + \beta}{\gamma x + \delta}\right)$ keeps $x$ unchanged in the
substitution group, $x$~must satisfy the congruence
\[
x \equiv \frac{\alpha x + \beta}
              {\gamma x + \delta},
\]
%% -----File: 339.png---Folio 321-------
that is
\[
\gamma x^{2} + (\delta - a)x - \beta \equiv 0.
\]
Such a congruence cannot have more than two roots; and
therefore every substitution displaces all, all but one, or all but
two, of the $p + 1$ symbols.

\Par{223.} The substitutions, which keep either one or two
symbols fixed, must therefore be regular in the remaining $p$~or
$p - 1$ symbols. Hence the order of every substitution which
keeps just one symbol fixed must be~$p$; and the order of every
substitution that keeps two symbols fixed must be equal to or
be a factor of~$p - 1$. Now it was seen in the last paragraph
that the order of the sub-group that keeps two symbols fixed is
$\frac{1}{2}(p - 1)$. Moreover, if $z$~is a primitive root $\pmod{p}$, the
sub-group that keeps $a$~and~$b$ fixed contains the operation
\[
\frac{y - a}{y - b} = z^{2} \frac{x - a}{x - b},
\]
and the order of this operation is $\frac{1}{2}(p - 1)$. Hence, the
sub-group that keeps any two symbols fixed is a cyclical group
of order~$\frac{1}{2}(p - 1)$; and every operation that keeps two symbols
fixed is some power of an operation of order~$\frac{1}{2}(p - 1)$. Since the
group is a doubly transitive group of degree~$p + 1$, there must
be $\frac{1}{2}(p + 1)p$ sub-groups which keep two symbols fixed; and
these must form a conjugate set. Each is therefore self-conjugate
in a sub-group of order~$p - 1$. To determine the
type of this sub-group, we may consider the sub-group keeping
$0$~and~$\infty$ fixed: this is generated by~$Q$, where $Q$~denotes $\left(\dfrac{zx}{z^{-1}}\right)$.
If $\left(\dfrac{\alpha x + \beta}{\gamma x + \delta}\right)$ is represented by~$S$, then
\[
S^{-1}QS = \left(
  \frac{(z\alpha\delta - z^{-1}\beta\gamma)x - (z - z^{-1})\alpha\beta}
       {(z - z^{-1})\gamma\delta x + z^{-1}\alpha\delta - z\beta\gamma}
\right),
\]
which can be a power of~$Q$ only if
\[
\alpha\beta \equiv 0,\quad \gamma\delta \equiv 0.
\]

Hence either
\[
\beta \equiv \gamma \equiv 0,
\]
in which case $S$~is a power of~$Q$: or
\[
\alpha \equiv \delta \equiv 0,\quad \gamma \equiv -\beta^{-1}.
\]
%% -----File: 340.png---Folio 322-------

In the latter case, we have
\[
S = \left(\frac{-\beta}{\beta^{-1} x}\right),
\]
which is an operation of order~$2$; and then
\[
S^{-1} Q S = \left(\frac{z^{-1} x}{z}\right) = Q^{-1}.
\]

The group of order~$p - 1$, which contains self-conjugately a
cyclical sub-group of order~$\frac{1}{2}(p - 1)$ that keeps two symbols fixed,
is therefore a group of dihedral (§~202) type. Moreover, if $t$~is
any factor of~$p - 1$, this investigation shews that $\{S, Q\}$~is the
greatest sub-group that contains $\{Q^{t}\}$ self-conjugately.

\Par{224.} A substitution that changes all the symbols must
either be regular in the $p + 1$ symbols, or must be such that
one of its powers keeps two symbols fixed. The latter case
however cannot occur; for we have just seen that, if $Q$~is an
operation, of order~$\frac{1}{2}(p - 1)$, which keeps two symbols fixed,
the only operations permutable with~$Q^{r}$ are the powers of~$Q$.
Hence the substitutions that change all the symbols must
be regular in the $p + 1$ symbols, and their orders must be
equal to or be factors of~$p + 1$.

Suppose now that $i$ is a primitive root of the congruence
\[
i^{p^{2} - 1} - 1 \equiv 0\ (\mod p),
\]
so that $i$~and~$i^{p}$ are the roots of a quadratic congruence with
real coefficients; and consider the operation~$K$, denoting
\[
\frac{y - i^{k}}{y - i^{kp}} \equiv i^{2(p - 1)} \frac{x - i^{k}}{x - i^{kp}},
\]
where $k$~is not a multiple of~$p + 1$. On solving with respect to~$y$,
$K$~is expressed in the form
\[
y \equiv \frac{
    \dfrac{i^{\left(\efrac{k}{2} + 1\right)(p - 1)} - i^{-\left(\efrac{k}{2} + 1\right)(p - 1)}}
        {i^{\efrac{k}{2} (p - 1)} - i^{-\efrac{k}{2} (p - 1)}} x
  - \dfrac{i^{(p - 1)} - i^{-(p - 1)}}{i^{\efrac{k}{2} (p - 1)} - i^{-\efrac{k}{2} (p - 1)}} i^{\efrac{k}{2} (p + 1)} %[** End of top-level numerator]
 }{
    \dfrac{i^{(p - 1)} - i^{-(p - 1)}}{i^{\efrac{k}{2} (p - 1)} - i^{-\efrac{k}{2} (p - 1)}} x i^{-\efrac{k}{2} (p + 1)}
  + \dfrac{i^{\left(\efrac{k}{2} - 1\right)(p - 1)} - i^{-\left(\efrac{k}{2} - 1\right)(p - 1)}}{i^{\efrac{k}{2} (p - 1)} - i^{-\efrac{k}{2} (p - 1)}}},
\]
%% -----File: 341.png---Folio 323-------
an operation of determinant unity. It will be found, on writing
$i^{p}$ for~$i$ in the coefficients of this operation, that they remain
unaltered; therefore, since they are symmetric functions of
$i$~and~$i^{p}$, they must be real numbers. The operation therefore
belongs to~$H$. The $n$th~power of this operation is given by
\[
\frac{y - i^{k}}{y - i^{kp}} \equiv i^{2n(p - 1)} \frac{x - i^{k}}{x - i^{kp}},
\]
and therefore, since the first power of~$i$ which is congruent to
unity, $\pmod{p}$, is the $(p^{2} - 1)$th, the order of the operation is
$\frac{1}{2}(p + 1)$. If we write $kp$ for~$k$ in the operation~$K$, the new
operation is~$K^{-1}$; but if $k$~is replaced by any other number~$k'$,
which is not a multiple of $p + 1$, the new operation~$K'$, given by
\[
\frac{y - i^{k'}}{y - i^{k'p}} \equiv i^{2(p - 1)} \frac{x - i^{k'}}{x - i^{k'p}},
\]
generates a new sub-group of order~$\frac{1}{2}(p + 1)$, which has no
operation except identity in common with~$\{K\}$. Now there are
$p^{2} - p$ numbers less than $p^{2} - 1$ which are not multiples of
$p + 1$; therefore $H$~contains $\frac{1}{2}(p^{2} - p)$ cyclical sub-groups of
order~$\frac{1}{2}(p + 1)$, no two of which have a common operation
except identity. The corresponding substitutions displace all
the symbols.

\Par{225.} A simple enumeration shews that the operations of
the cyclical sub-groups of orders $\frac{1}{2}(p - 1)$,~$p$ and~$\frac{1}{2}(p + 1)$,
exhaust all the operations of the group. Thus there are,
omitting identity from each sub-group:

\Inum{(i)} $\frac{1}{2}p(p + 1)$ sub-groups of order~$\frac{1}{2}(p - 1)$, containing
$\frac{1}{4}p(p^{2} - 1) - \frac{1}{2}p^{2} - \frac{1}{2}p$ distinct operations;

\Inum{(ii)} $\frac{1}{2}p(p - 1)$ sub-groups of order~$\frac{1}{2}(p + 1)$, containing
$\frac{1}{4}p(p^{2} - 1) - \frac{1}{2}p^{2} + \frac{1}{2}p$ distinct operations;

\Inum{(iii)} $p + 1$ sub-groups of order~$p$, containing
$p^{2} - 1$ distinct operations; \\
%
and the sum of these numbers, with $1$ for the identical operation,
gives $\frac{1}{2}p(p^{2} - 1)$, which is the order of the group.

Every operation that displaces all the symbols is therefore
the power of an operation of order~$\frac{1}{2}(p + 1)$.
%% -----File: 342.png---Folio 324-------

\Par{226.} We shall now further shew that the $\frac{1}{2}p(p - 1)$ sub-groups
of order~$\frac{1}{2}(p + 1)$ form a single conjugate set, and that
each is contained self-conjugately in a dihedral group of order
$p + 1$. Let $S$~be any operation of~$H$, which is permutable with~$\{K\}$
and replaces $i^{k}$ by some other symbol~$j$. Then $S^{-1}KS$~is
an operation which leaves $j$ unaltered; it may therefore be expressed
in the form
\[
\frac{y - j}{y - j'} = m \frac{x - j}{x - j'}.
\]

This can belong to the sub-group generated by~$K$, only if $j$
and $j'$ are the same pair as $i^{k}$ and~$i^{kp}$. Hence $j$~must be either
$i^{k}$ or~$i^{kp}$; and similarly, if $S$~replaces $i^{kp}$ by~$j'$, the latter must be
either $i^{kp}$ or~$i^{k}$. Hence either $S$ must keep both the symbols
$i^{k}$ and~$i^{kp}$ unchanged or it must interchange them; and conversely,
every operation which either keeps both the symbols
unchanged or interchanges them, must transform $\{K\}$ into
itself. If $S$~keeps both of them unchanged, it is a power of~$K$.
If $S$~interchanges them, it is of the form
\[
\frac{y - i^{k}}{y - i^{kp}} \equiv m \frac{x - i^{kp}}{x - i^{k}};
\]
and a simple calculation shews that
\[
S^{-1}KS = K^{-1}.
\]
If we take $m = 1$, $S$~becomes
\[
x + y \equiv i^{k} + i^{kp},
\]
an operation belonging to~$H$. Hence the cyclical sub-group~$\{K\}$
is contained self-conjugately in the sub-group of~$\{S, K\}$ which is
of dihedral type. If there were any other operation~$S'$, not contained
in~$\{S, K\}$, which transformed~$K$ into its inverse, then
$SS'$~would be an operation permutable with~$K$ and not contained
in~$\{K\}$. It has just been seen that no such operation
exists. Hence $\{S, K\}$, of order~$p + 1$, is the greatest sub-group
that contains $\{K\}$ self-conjugately; and $\{K\}$~must be one of
$\frac{1}{2}p(p - 1)$ conjugate sub-groups.

\Par{227.} The distribution of the operations of~$H$ in conjugate
sets is now known. A sub-group of order~$p$ is contained self-conjugately
in a group of order~$\frac{1}{2}p(p - 1)$, while an operation
%% -----File: 343.png---Folio 325-------
of order~$p$ is permutable only with its own powers. There are
therefore two conjugate sets of operations of order~$p$, each set
containing $\frac{1}{2}(p^{2} - 1)$ operations. Again, each of the operations
of a cyclical sub-group of order $\frac{1}{2}(p - 1)$ or $\frac{1}{2}(p + 1)$ is conjugate
to its own inverse and to no other of its powers. Hence if
$\frac{1}{2}(p + 1)$ is even and therefore $\frac{1}{2}(p - 1)$ odd, there are $\frac{1}{4}(p - 3)$
conjugate sets of operations whose orders are factors of $\frac{1}{2}(p - 1)$,
each set containing $p^{2} + p$ operations; $\frac{1}{4}(p - 3)$~conjugate sets
of operations whose orders are factors of~$\frac{1}{2}(p + 1)$, other than
the factor~$2$, each set containing $p^{2} - p$ operations; and a single
set of operations of order~$2$, containing $\frac{1}{2}(p^{2} - p)$ operations. If
$\frac{1}{2}(p - 1)$ is even and $\frac{1}{2}(p + 1)$~odd, there are $\frac{1}{4}(p - 1)$ conjugate
sets of operations whose orders are factors of $\frac{1}{2}(p + 1)$, each
containing $p^{2} - p$ operations; $\frac{1}{4}(p - 5)$ conjugate sets whose
orders are factors of~$\frac{1}{2}(p - 1)$, other than the factor~$2$, each
set containing $p^{2} + p$ operations; and a single set of $\frac{1}{2}(p^{2} + p)$
conjugate operations of order~$2$. In either case, the group
contains, exclusive of identity, $\frac{1}{2}(p + 3)$ conjugate sets of
operations.

\Par{228.} Since $p - 1$ and $p + 1$ can have no common factor except~$2$,
it follows that, if $q^{m}$~denote\DPnote{** [sic]} the highest power of an odd
prime, other than~$p$, which divides the order of~$H$, $q^{m}$~must be a
factor of~$\frac{1}{2}(p - 1)$ or of~$\frac{1}{2}(p + 1)$; and the sub-groups of order~$q^{m}$
must be cyclical. Moreover, since no two cyclical sub-groups
of order~$\frac{1}{2}(p - 1)$, or~$\frac{1}{2}(p + 1)$, have a common operation except
identity, the same must be true of the sub-groups of order~$q^{m}$.

If $2^{m}$~is the highest power of~$2$ that divides $\frac{1}{2}(p - 1)$ or
$\frac{1}{2} (p + 1)$, $2^{m + 1}$~will be the highest power of~$2$ that divides the
order of~$H$. Moreover, a sub-group of order~$2^{m + 1}$ must contain
a cyclical sub-group of order~$2^{m}$ self-conjugately, and it must
contain an operation of order~$2$ that transforms every operation
of this cyclical sub-group into its own inverse; in other words,
the sub-groups of order~$2^{m + 1}$ are of dihedral type.

Suppose now that two sub-groups of order~$2^{m + 1}$ ($m > 1$) have
a common sub-group of order~$2^{r}$ ($r > 2$). Such a sub-group
must be either cyclical or dihedral: in the latter case, it
contains self-conjugately a single cyclical sub-group of order~$2^{r - 1}$.
%% -----File: 344.png---Folio 326-------
Hence, on the supposition made, a cyclical sub-group of
order~$4$ at least would be contained self-conjugately in two
distinct cyclical sub-groups of order~$2^{m}$. It has been seen that
this is not the case; and therefore the greatest sub-group, that
two sub-groups of order~$2^{m + 1}$ can have in common, must be a
sub-group of order~$4$, whose operations, except identity, are all
of order~$2$. Now every group of order~$2^{m + 1}$ contains one self-conjugate
operation of order~$2$, and $2^{m}$~operations of order~$2$
falling into $2$~conjugate sets of $2^{m - 1}$ each. Moreover, the group
of order~$p ± 1$, which has a cyclical sub-group of order~$2^{m}$ and
contains the operation~$A$ of order~$2$ of this cyclical group
self-conjugately, has $\frac{1}{2}(p ± 1)$ other operations of order~$2$; and
therefore it contains $\dfrac{p ± 1}{2^{m + 1}}$ sub-groups of order~$2^{m + 1}$, each of
which has $A$ for its self-conjugate operation. If now $B$~is
any operation of order~$2$ of this sub-group of order~$p ± 1$, and
if it is distinct from~$A$, then $B$~enters into a sub-group of
order~$2^{m + 1}$ that contains $A$ self-conjugately. But since $A$~is
permutable with~$B$, $A$~must belong to the sub-group of order
$p ± 1$, which contains $B$ self-conjugately; hence $A$~enters into
a sub-group of order~$2^{m + 1}$ which contains $B$ self-conjugately.
The sub-group $\{A, B\}$ is therefore common to two distinct
sub-groups of order~$2^{m + 1}$. Now no group of order~$2^{r}$ ($r > 2$)
can be common to two sub-groups of order~$2^{m + 1}$; and therefore
$\{A, B\}$~must (§~80) be permutable with some operation~$S$ whose
order~$s$ is prime to~$2$. If $s$~is not~$3$, $S$~must be permutable with
$A$ and~$B$: and then $\{A, S\}$ and $\{B, S\}$ would be two distinct
sub-groups of orders~$2s$, whose operations are permutable with
each other. It has been seen that $H$~does not contain such
sub-groups. Hence $s = 3$; and $S$~transforms $A$,~$B$ and~$AB$ cyclically,
or $\{S, A, B\}$ is a sub-group of tetrahedral type (§~202).

The number of quadratic\footnote
  {A non-cyclical group of order~$4$ is called a \emph{quadratic} group.}
sub-groups contained in~$H$ may
\index{Group@\Topic{Group}!quadratic}%
be directly enumerated. A group of order~$2^{m + 1}$ contains $2^{m - 1}$
such sub-groups, which fall into $2$ conjugate sets of $2^{m - 2}$ each;
a single group of order~$8$ containing each quadratic group self-conjugately.
The quadratic groups, contained in the $\dfrac{p ± 1}{2^{m + 1}}$
sub-groups of order~$2^{m + 1}$ of a sub-group of order~$p ± 1$, are
%% -----File: 345.png---Folio 327-------
clearly all distinct, and each quadratic group belongs to just $3$
groups of order~$p ± 1$; thus $\{A, B\}$ belongs to the $3$~groups
which contain $A$,~$B$ and~$AB$ respectively as self-conjugate
operations. Hence the total number of quadratic groups contained
in~$H$ is
\[
\tfrac{1}{2} p (p \mp 1) \frac{p ± 1}{2^{m + 1}} 2^{m - 1} \tfrac{1}{3}
  = \frac{\frac{1}{2}p(p^{2} - 1)}{12}.
\]

\Par{229.} The greatest sub-group of a group of order~$2^{m + 1}$, that
contains a quadratic group self-conjugately, is a group of order~$8$
and dihedral type; and it has been shewn that $3$~is the
only factor, prime to~$2$, that occurs in the order of the sub-group
containing a quadratic group self-conjugately. Hence
finally, the order of the greatest group containing a quadratic
group self-conjugately is~$24$, and the $\dfrac{\frac{1}{2}p(p^{2} - 1)}{12}$ quadratic
groups fall into two conjugate sets of $\dfrac{\frac{1}{2}p(p^{2} - 1)}{24}$ each. The
group of order~$24$, that contains a quadratic group self-conjugately,
contains also a self-conjugate tetrahedral sub-group,
while the sub-groups of order~$8$ are dihedral. Hence (§~84) this
group must be of octohedral type.

Since every tetrahedral sub-group of~$H$ contains a quadratic
sub-group self-conjugately, and every octohedral sub-group
contains a tetrahedral sub-group self-conjugately, there must
also be two conjugate sets of tetrahedral sub-groups and two
conjugate sets of octohedral sub-groups, the number in each set
being $\dfrac{\frac{1}{2}p(p^{2} - 1)}{24}$.

\Par{230.} We have hitherto supposed $m > 1$, or what is the
same thing, $p \equiv ±1\ (\mod 8)$. If now $m = 1$, so that $p \equiv ±3\ (\mod 8)$,
the highest power of~$2$ that divides the order of~$H$ is~$2^{2}$;
and, since $2^{2}$~is not a factor of~$\frac{1}{2}(p ± 1)$, the sub-groups of
order~$2^{2}$ are quadratic. Moreover, since $2^{2}$~is the highest power
of~$2$ dividing the order of~$H$, the quadratic sub-groups form
a single conjugate set. Each sub-group of order~$p ± 1$, which
has a self-conjugate operation of order~$2$, contains $\frac{1}{4}(p ± 1)$
%% -----File: 346.png---Folio 328-------
sub-groups of order~$4$, and each of the latter belongs to $3$ of the
former. The total number is as before
\[
\frac{\frac{1}{2}p(p^{2} - 1)}{12},
\]
and since they form a single conjugate set, each quadratic group
is self-conjugate in a group of order~$12$. Also, for the same
reason as in the previous case, this sub-group is of tetrahedral
type.

Finally, since every sub-group of~$H$ of tetrahedral type
must contain a quadratic sub-group self-conjugately, $H$~must
contain a single conjugate set of $\dfrac{\frac{1}{2}p(p^{2} - 1)}{12}$ tetrahedral sub-groups.
In this case, the order of~$H$ is not divisible by~$24$, and
therefore the question of octohedral sub-groups does not arise.

\Par{231.} The group~$H$ always contains tetrahedral sub-groups;
when its order is divisible by~$24$, it contains also octohedral
sub-groups. Now if $p \equiv ±1\ (\mod 5)$, the order of~$H$ is
divisible by~$60$: it may be shewn as follows that, in these cases,
$H$~contains sub-groups of icosahedral type.

Let us suppose, first, that $p \equiv 1\ (\mod 5)$; and let $j$ be a
primitive root of the congruence
\[
j^{5} \equiv 1\ (\mod p).
\]
Then $\left(\dfrac{jx}{j^{-1}}\right)$, which we will denote by~$A$, is an operation of order~$5$.
The operations of order~$2$ of~$H$ are all of the form~$B$, where
$B$~denotes $\left(\dfrac{\alpha x + \beta}{\gamma x - \alpha}\right)$, since each is its own inverse. Now
\[
AB = \left(\frac{\alpha jx + \beta j^{-1}}{\gamma jx - \alpha j^{-1}}\right);
\]
and (§~203) if $A$~and~$B$ generate an icosahedral group,
\[
(AB)^{3} = 1.
\]
A simple calculation shews that, if this condition is satisfied,
then
\[
\alpha^{2}(j - j^{-1})^{2} \equiv 1.
\]
Also, since the determinant of~$B$ is unity,
\[
\alpha^{2} + \beta\gamma \equiv -1.
\]
%% -----File: 347.png---Folio 329-------
These two congruences have just $p - 1$ distinct solutions,
the solutions $\alpha$,~$\beta$,~$\gamma$ and $-\alpha$,~$-\beta$,~$-\gamma$ being regarded as
identical. There are therefore $p - 1$ operations of order two in~$H$,
namely the operations
\[
\left(\frac{\dfrac{1}{j - j^{-1}} x + \beta}
           {\gamma x - \dfrac{1}{j - j^{-1}}}
\right),
\]
where
\[
\beta\gamma \equiv -1 - \frac{1}{(j - j^{-1})^{2}},
\]
which with $A$~generate an icosahedral sub-group.

The group generated by
\[
\left(\frac{jx}{j^{-1}}\right)
\quad\text{and}\quad
\left(\frac{\dfrac{1}{j - j^{-1}} x + \beta_{0}}
           {\gamma_{0} x - \dfrac{1}{j - j^{-1}}}
\right),
\]
contains $5$ of the $p - 1$ operations of order~$2$ of the form
\[
\left(\frac{\dfrac{1}{j - j^{-1}} x + \beta}
           {\gamma x - \dfrac{1}{j - j^{-1}}}
\right),
\]
viz.\ those for which
\[
\beta  \equiv \beta_{0} j^{n},\quad \gamma \equiv \gamma_{0} j^{-n},\quad (n = 0, 1, 2, 3, 4).
\]
Hence the sub-group~$\{A\}$, of order~$5$, belongs to $\frac{1}{5}(p - 1)$
distinct icosahedral sub-groups. Now each icosahedral sub-group
has $6$~sub-groups of order~$5$; and $H$~contains $\frac{1}{2}p(p + 1)$
sub-groups of order~$5$ forming a single conjugate set. The
number of icosahedral sub-groups in~$H$ is therefore
\[
\tfrac{1}{6} \frac{p - 1}{5}\, \tfrac{1}{2} p(p + 1)
  = \frac{\frac{1}{2}p(p^{2} - 1)}{30}.
\]
The group of isomorphisms of the icosahedral group is the
symmetric group of degree~$5$ (§~173). Now $H$~can contain no
sub-group simply isomorphic with the symmetric group of
degree~$5$. For if it contained such a sub-group, an operation of
%% -----File: 348.png---Folio 330-------
order~$5$ would be conjugate to its own square; and this is not
the case.

Hence (§~165), if an icosahedral sub-group~$K$ of~$H$ is contained
self-conjugately in a greater sub-group~$L$, then $L$~must be
the direct product of~$K$ and some other sub-group. This also is
impossible; for the greatest sub-group of~$H$ in which any cyclical
sub-group, except those of order~$p$, is contained self-conjugately,
is of dihedral type. Hence $L$~must coincide with~$K$, and $K$~must
be one of $\dfrac{\frac{1}{2}p(p^{2} - 1)}{60}$ conjugate sub-groups. The icosahedral
sub-groups of~$H$ therefore fall into two conjugate sets of
$\dfrac{\frac{1}{2}p(p^{2} - 1)}{60}$ each.

In a similar manner, when $p \equiv -1\ (\mod 5)$, we may take as
a typical operation~$A$, of order~$5$,
\[
\frac{y - i}{y - i^{p}} \equiv i^{\efrac{p^{2} - 1}{5}} \frac{x - i}{x - i^{p}};
\]
and it may be shewn, the calculation being rather more
cumbrous than in the previous case, that there are just $p + 1$
operations~$B$, of the form $\left(\dfrac{\alpha x + \beta}{\gamma x - \alpha}\right)$, such that
\[
(AB)^{3} = 1,
\]
and that five of these belong to the icosahedral group generated
by~$A$ and any one of them. It follows, exactly as
in the previous case, that $H$~contains $\dfrac{\frac{1}{2}p(p^{2} - 1)}{30}$ icosahedral
sub-groups, which fall into two conjugate sets, each set containing
$\dfrac{\frac{1}{2}p(p^{2} - 1)}{60}$ groups.

\Par{232.} Finally, we proceed to shew that $H$ has no other
sub-groups than those which have been already determined.
Suppose, first, that a sub-group~$h$ of~$H$ contains two distinct
sub-groups of order~$p$. These must, by Sylow's theorem, form
part of a set of $kp + 1$ sub-groups of order~$p$ conjugate within~$h$.
Now $H$~contains only $p + 1$ sub-groups of order~$p$, and
%% -----File: 349.png---Folio 331-------
therefore $k$~must be unity and $h$~must contain all the sub-groups
of order~$p$; or since $H$~is simple, $h$~must contain and therefore
coincide with~$H$. Hence the only sub-groups of~$H$, whose orders
are divisible by~$p$, are those that contain a sub-group of order~$p$
self-conjugately. They are of known types.

Suppose next that $g$~is a sub-group of~$H$, whose order~$n$
is not divisible by~$p$, and let $S_{1}$ be an operation of~$g$ whose
order~$q_{1}$ is not less than the order of any other operation of~$g$.
In~$H$ the sub-group~$\{S_{1}\}$ is self-conjugate in a dihedral group of
order~$p ± 1$; and the greatest sub-group of this group, which
contains no operation of order greater than~$q_{1}$, is a dihedral
group of order~$2q_{1}$. Hence in~$g$ the sub-group~$\{S_{1}\}$ is self-conjugate
in a group of order $q_{1}$ or~$2q_{1}$, and therefore it forms
one of~$\dfrac{n}{q{_1}}$ or of~$\dfrac{n}{2q{_1}}$ conjugate sub-groups. Moreover, no two of
these sub-groups contain a common operation except identity;
and they therefore contain, excluding identity, $\dfrac{n(q_{1} - 1)}{\epsilon_{1} q_{1}}$ distinct
operations, where $\epsilon_{1}$~is either $1$ or~$2$.

Of the remaining operations of~$g$, let $S_{2}$ be one whose
order~$q_{2}$ is not less than that of any of the others. The
operation~$S_{2}$ cannot be permutable with any of the $\dfrac{n(q_{1} - 1)}{\epsilon_{1} q_{1}}$
operations already accounted for, since $S_{2}$~is not a power of any
one of these operations. Hence, exactly as before, $\{S_{2}\}$~must
form one of~$\dfrac{n}{\epsilon_{2}q_{2}}$ conjugate sub-groups in~$g$, $\epsilon_{2}$~being either $1$ or~$2$;
and these sub-groups contain $\dfrac{n(q_{2} - 1)}{\epsilon_{2} q_{2}}$ operations which are
distinct from identity, from each other, and from those of the
previous set. This process may be continued till the identical
operation only remains. Hence, finally, $n$~being the total
number of operations of~$g$, we must have
\[
n = 1 + \sum_{\nu} \frac{n(q_{\nu} - 1)}{\epsilon_{\nu}q_{\nu}}
\]
or
\[
\frac{1}{n} = 1 - \sum_{\nu}  \frac{q_{\nu} - 1}{\epsilon_{\nu}q_{\nu}}.
\]
%% -----File: 350.png---Folio 332-------

\Par{233.} In this equation, let $r$ of the $\epsilon$'s be~$1$ and $s$~of them be~$2$,
so that $r + s$ is their total number, say~$m$. Then
\begin{align*}
\frac{1}{n}
  &= 1 - \sum \left(1 - \frac{1}{q_{\lambda}}\right)
       - \sum \left(\tfrac{1}{2} - \frac{1}{2 q_{\mu}}\right) \\
  &= 1 - r + \sum \frac{1}{q_{\lambda}} - \tfrac{1}{2} s + \sum \frac{1}{2q_{\mu}} \\
  &\leq 1 - r + \tfrac{1}{2} r - \tfrac{1}{2} s + \tfrac{1}{4} s \\
  &\leq 1 - \tfrac{1}{2} r - \tfrac{1}{4} s \\
  &\leq 1 - \tfrac{1}{4} m.
\end{align*}

Hence, since $n$~is a positive integer, there cannot be more
than three terms under the sign of summation. Moreover, since
\[
\frac{1}{n} \leq 1 - \tfrac{1}{2} r - \tfrac{1}{4} s,
\]
$r$ cannot be greater than~$1$, and therefore not more than one of
the~$\epsilon$'s can be unity. Also, when one of the~$\epsilon$'s is unity, we have
\begin{align*}
\frac{1}{n} &= \frac{1}{q} - \tfrac{1}{2} s + \sum \frac{1}{2 q_{\mu}} \\
  &\leq \tfrac{1}{2} - \tfrac{1}{2} s + \tfrac{1}{4} s \\
  &\leq \tfrac{1}{4} (2 - s),
\end{align*}
so that, in this case, $s$~cannot be greater than unity. The
solutions are now easily obtained by trial.

\Class{\Inum{(i)}}{For one term in the sum, the only possible solution is}
\[
\epsilon_{1} = 1,\quad n = q_{1},
\]
and the corresponding group is cyclical.

\Class{\Inum{(ii)}}{For two terms in the sum, the solutions are}
\begin{align*}
&\Inum{($\alpha$)}\quad \epsilon_{1} = \epsilon_{2} = 2,\quad n = \frac{2q_{1}q_{2}}{q_{1} + q_{2}}; \\
&\Inum{($\beta$)}\quad  \epsilon_{1} = 2,\quad \epsilon_{2} = 1,\quad q_{2} = 2,\quad n = 2q_{1}; \\
&\Inum{($\gamma$)}\quad \epsilon_{1} = 1,\quad \epsilon_{2} = 2,\quad q_{1} = 3,\quad q_{2} = 2,\quad n = 12.
\end{align*}

To the solution~\Inum{($\alpha$)} there corresponds no sub-group; for
$n < 2q_{1}$, and the values $q_{1} = q_{1}$\DPnote{** [sic]}, $\epsilon_{1} = 2$ imply that $g$~has a sub-group
of order~$2q_{1}$.

To the solution~\Inum{($\beta$)} correspond the sub-groups of order~$2q_{1}$
of dihedral type, for which $q_{1}$~is odd, so that the operations of
order~$2$ form a single conjugate set.
%% -----File: 351.png---Folio 333-------

To the solution~\Inum{($\gamma$)} corresponds a sub-group of order~$12$
containing $8$~operations of order~$3$ and $3$~operations of order~$2$,
\ie\ a tetrahedral sub-group.

\Class{\Inum{(iii)}}{For three terms in the sum, the solutions are}
\begin{alignat*}{6}
&\Inum{($\alpha$)}\quad && \epsilon_{1} = \epsilon_{2} = \epsilon_{3} = 2,\quad
  &&& q_{2} &= 2,\quad & q_{3} &= 2,\quad & n &= 2q_{1}; \\
%
&\Inum{($\beta$)} && \Ditto{$\epsilon_{1} = \epsilon_{2}$}\quad\Ditto{$\epsilon_{3} = 2$}
  & q_{1} &= 3,\quad & q_{2} &= 3, & q_{3} &= 2, & n &= 12; \\
&\Inum{($\gamma$)}&& \Ditto{$\epsilon_{1} = \epsilon_{2}$}\quad\Ditto{$\epsilon_{3} = 2$}
  & q_{1} &= 4,      & q_{2} &= 3, & q_{3} &= 2, & n &= 24; \\
&\Inum{($\delta$)}&& \Ditto{$\epsilon_{1} = \epsilon_{2}$}\quad\Ditto{$\epsilon_{3} = 2$}
  & q_{1} &= 5,      & q_{2} &= 3, & q_{3} &= 2, & n &= 60.
\end{alignat*}

To the solution~\Inum{($\alpha$)} correspond the sub-groups of order~$2q_{1}$
of dihedral type, in which $q_{1}$~is even, so that the operations of
order~$2$, which do not belong to the cyclical sub-group of order~$q_{1}$,
fall into two distinct conjugate sets.

To the solution~\Inum{($\beta$)} would correspond a group of order~$12$
containing $3$~operations of order~$2$ and $4$~sub-groups of order~$3$
which fall into two conjugate sets of $2$ each. Sylow's theorem
shews that such a group cannot exist; and therefore there is no
sub-group of~$H$ corresponding to this solution.

Solution~\Inum{($\gamma$)} gives a group of order~$24$, with $3$~conjugate
cyclical sub-groups of order~$4$, $4$~conjugate cyclical sub-groups
of order~$3$, and $6$~other operations of order~$2$ forming a single
conjugate set. No operation of this group is permutable with
each of the $4$ sub-groups of order~$3$; and therefore, if the group
exists, it can be represented as a transitive group of $4$~symbols.
On the other hand, the order of the symmetric group of~$4$~symbols,
which (§~203) is simply isomorphic with the octohedral
group, is~$24$; and its cyclical sub-groups are distributed as
above. Hence to this solution there correspond the octohedral
sub-groups of~$H$.

Solution~\Inum{($\delta$)} gives a group of order~$60$, with $6$~conjugate
sub-groups of order~$5$, $10$~conjugate sub-groups of order~$3$, and
a conjugate set of~$15$ operations of order~$2$. It has been shewn,
in §~85, that there is only one type of group of order~$60$ that
has $6$~sub-groups of order~$5$; viz.\ the alternating group of
degree~$5$: and that, in this group, the distribution of sub-groups
in conjugate sets agrees with that just given. Moreover, the
alternating group of degree~$5$ is simply isomorphic with the
\index{Fractional linear group@\Topic{Fractional linear group}!analysis of|)}%
%% -----File: 352.png---Folio 334-------
icosahedral group. Hence to this solution there correspond the
icosahedral sub-groups of~$H$.

\Par{234.} When $p > 11$, then $\frac{1}{2}p(p - 1) > 60$; and, when $p > 3$,
\index{Fractional linear group@\Topic{Fractional linear group}!generalization of}%
$\frac{1}{2}p(p - 1) > p + 1$. Hence when $p > 11$, the order of the
greatest sub-group of~$H$ is $\frac{1}{2}p(p - 1)$, and the least number of
symbols in which $H$ can be expressed as a transitive substitution
group is $p + 1$.

When $p$~is $5$,~$7$ or~$11$, however, $H$~can be expressed as a
transitive substitution group of $p$~symbols\footnotemark.
\index{Galois@\Topic{Galois}, quoted}%
\footnotetext{This is another of the results stated in the letter of Galois referred to in
  the footnote on \PageRef{p.}{192}.}

For, when $p = 5$, $H$~contains a tetrahedral sub-group of
order~$12$, forming one of $5$ conjugate sub-groups; therefore $H$~can
be expressed as a transitive group of $5$~symbols. It is to be
noticed that in this case $H$~is an icosahedral group.

When $p = 7$, $H$ contains an octohedral sub-group of order~$24$,
which is one of $7$ conjugate sub-groups; and $H$~can therefore
be expressed as a transitive group of $7$~symbols. Similarly,
when $p = 11$, $H$~contains an icosahedral sub-group of order~$60$,
which is one of $11$ conjugate sub-groups; and the group can be
expressed transitively in $11$~symbols.

\Par{235.} The simple groups, of the class we have been discussing
in the foregoing sections, are self-conjugate sub-groups
of the triply transitive groups of degree~$p + 1$, defined by
\[
y \equiv \frac{\alpha x + \beta}{\gamma x + \delta},\ (\mod p),
\]
the existence of which was demonstrated in §~113. In fact,
since $\left(\dfrac{\alpha x + \beta}{\gamma x + \delta}\right)$
and $\left(\dfrac{k\alpha x + k\beta}{k\gamma x + k\delta}\right)$ represent the same transformation,
the determinant, $\alpha\delta - \beta\gamma$, of any transformation may
always be taken as either unity or a given non-residue; and it
follows at once that the transformations of determinant unity
form a self-conjugate sub-group of the whole group of transformations.

If, as in §~113, $\alpha$,~$\beta$, $\gamma$,~$\delta$, are powers of~$i$, where $i$~is a primitive
root of the congruence
\[
i^{p^{n} - 1} \equiv 1,\ (\mod p),
\]
%% -----File: 353.png---Folio 335-------
the triply transitive group~$G$ of degree~$p^{n} + 1$, which is defined
by the transformations, has again, when $p$~is an odd prime, a
self-conjugate sub-group~$H$ of order~$\frac{1}{2}p^{n}(p^{2n} - 1)$, which is
given by the transformations of determinant unity. It follows
from Theorem~IX, §~134, that $G$, being a triply transitive group
of degree~$p^{n} + 1$, must have, as a self-conjugate sub-group, a
doubly transitive simple group; and it is easy to shew that $H$
is this sub-group.

In fact, if a simple group~$h$ is a self-conjugate sub-group of~$G$
it must be contained in~$H$. Also, since $h$~is a doubly
transitive group of degree~$p^{n} + 1$, it must contain every operation
of order~$p$ that occurs in~$G$. Now we may shew that these
operations generate~$H$. Thus $\left(\dfrac{2x + 1}{-x}\right)$ and $(x + 2 - i - i^{-1})$ are
operations of order~$p$ belonging to~$G$. Therefore $\left(\dfrac{(i + i^{-1})x + 1}{-x}\right)$
belongs to~$h$. But this operation is transformed into $\left(\dfrac{ix}{i^{-1}}\right)$ by
$\left(\dfrac{x + i^{-1}}{x + i}\right)$. Hence $\left(\dfrac{ix}{i^{-1}}\right)$ belongs to~$h$; and a sub-group of~$h$
which keeps one symbol unchanged is the group of order
$\frac{1}{2}p^{n}(p^{n} - 1)$ generated by $(x + 1)$ and~$\left(\dfrac{ix}{i^{-1}}\right)$. The order of~$h$
therefore is not less than $\frac{1}{2}p^{n}(p^{2n} - 1)$; in other words $h$~is
identical with~$H$.

When $p = 2$, every power of~$i$ is a quadratic residue, and the
determinant of every transformation is unity. In this case it
may be shewn, by an argument similar to the above, that the
group~$G$ of order $2^{n}(2^{2n} - 1)$ is itself a simple group.

We are thus led to recognize the existence of a doubly-infinite
series of simple groups of orders $2^{n}(2^{2n} - 1)$ and
$\frac{1}{2}p^{n}(p^{2n} - 1)$, which are closely analogous to the groups of order
$\frac{1}{2}p(p^{2} - 1)$ already discussed. For an independent proof of the
existence of these simple groups and for an investigation of
their properties, the reader is referred to the memoirs mentioned
below\footnotemark.
\index{Burnside@\Topic{Burnside}, quoted}%
\index{Fractional linear group@\Topic{Fractional linear group}!generalization of}%
\index{Moore@\Topic{Moore}, quoted}%
\footnotetext{Moore: ``On a doubly-infinite series of simple groups,'' \textit{Chicago Congress
  Mathematical Papers}, (1893); Burnside: ``On a class of groups defined by congruences,''
  \textit{Proc.\ L.~M.~S.} Vol.~\VolNo{XXV}, (1894), pp.~113--139.}%
%% -----File: 354.png---Folio 336-------

\Par{236.} We will now return to the linear homogeneous group~$G$
\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots, 1)$|(}%
\index{Homogeneous linear group@\Topic{Homogeneous linear group}!represented as a transitive substitution group}%
of transformations of $n$~symbols, taken to a prime modulus~$p$;
and consider it more directly as the group of isomorphisms of
an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots\DPchg{\,}{,} \text{to $n$ units})$.
As in §~156, it may be expressed in the form of a substitution
group performed on the $p^{n} - 1$ symbols of the operations, other
than identity, of the Abelian group. In this form it is clearly
transitive, since there are isomorphisms changing any operation
of the Abelian group into any other operation. If $P$~is any
operation of the Abelian group, an isomorphism which changes
any one of the $p - 1$ operations
\[
P,\ P^{2},\ \Dots,\ P^{p - 1},
\]
into any other, will certainly interchange the set among themselves.
Hence, when expressed as a group of degree~$p^{n} - 1$, $G$~is
imprimitive; and the symbols forming an imprimitive system
are those of the operations, other than identity, of any sub-group
of order~$p$ of the Abelian group. If
\[
P_{1},\ P_{2},\ \Dots,\ P_{n}
\]
are a set of generating operations of the Abelian group, an
isomorphism, which changes each of the sub-groups
\[
\{P_{1}\},\ \{P_{2}\},\ \Dots,\ \{P_{n}\}
\]
into itself, must be of the form
\[
\Sub{\PadTo{P_{1}^{\alpha_{1}}}{P_{1}},
     \PadTo{P_{1}^{\alpha_{2}}}{P_{2}}, \Dots,
     \PadTo{P_{1}^{\alpha_{n}}}{P_{n}}}
    {P_{1}^{\alpha_{1}}, P_{2}^{\alpha_{2}}, \Dots, P_{n}^{\alpha_{n}}}.
\]
This isomorphism changes $P_{1}P_{2}$ into $(P_{1} P_{2}^{\efrac{\alpha_{2}}{\alpha_{1}}})^{\alpha_{1}}$; therefore it
will only transform the sub-group $\{P_{1}P_{2}\}$ into itself when
$\alpha_{1} \equiv \alpha_{2},\ (\mod p)$. If then the given isomorphism changes every
sub-group of order~$p$ into itself, we must have
\[
\alpha_{1} \equiv \alpha_{2} \equiv \Dots \equiv \alpha_{n},\ (\mod p).
\]

Hence the only operations of~$G$, which interchange the
symbols of each imprimitive system among themselves, are
those given by the powers of
\[
\Sub{\PadTo{P_{1}^{\alpha}}{P_{1}},
     \PadTo{P_{2}^{\alpha}}{P_{2}}, \Dots,
     \PadTo{P_{n}^{\alpha}}{P_{n}}}
    {P_{1}^{\alpha}, P_{2}^{\alpha}, \Dots, P_{n}^{\alpha}},
\]
where $\alpha$ is a primitive root of~$p$. This operation is the same as
%% -----File: 355.png---Folio 337-------
that denoted by $A$ in §~217. It follows immediately that the
\index{Homogeneous linear group@\Topic{Homogeneous linear group}!represented as a transitive substitution group}%
factor-group $\dfrac{G}{\{A\}}$ can be represented as a transitive group in
$\dfrac{p^{n} - 1}{p - 1}$ symbols. In fact, the operations of~$\{A\}$ are the only
operations of~$G$ which transform each of the $\dfrac{p^{n} - 1}{p - 1}$ sub-groups
of order~$p$ in itself; and these $\dfrac{p^{n} - 1}{p - 1}$ sub-groups must be
permuted among themselves by every operation of~$G$. The
substitution group thus obtained is doubly transitive; for if $P$~and~$P'$
are any two operations of the Abelian group such that
$P'$~is not a power of~$P$, and if $Q$~and~$Q'$ are any other two
operations of the Abelian group subject to the same condition,
there certainly exists an isomorphism of the form
\[
\Sub{P, P', \Dots}{Q, Q', \Dots},
\]
and this isomorphism changes the sub-groups $\{P\}$ and $\{P'\}$ into
the sub-groups $\{Q\}$ and~$\{Q'\}$.

These results will still hold if, instead of considering $G$~the
total group of isomorphisms, we take $\Gamma$~the group of isomorphisms
of determinant unity. Thus the determinant of
\[
\Sub{\PadTo{Q^{\alpha}}{P}, P', \Dots}{Q^{\alpha}, Q', \Dots}
\]
is $\alpha$~times the determinant of
\[
\Sub{P, P', \Dots}{Q, Q', \Dots}.
\]
It is therefore possible always to choose $\alpha$ so that the determinant
of
\[
\Sub{\PadTo{Q^{\alpha}}{P}, P', \Dots}{Q^{\alpha}, Q', \Dots}
\]
shall be unity; and this isomorphism still changes the sub-groups
$\{P\}$ and $\{P'\}$ into $\{Q\}$ and $\{Q'\}$ respectively.

The lowest power of~$A$ contained in~$\Gamma$ is (§~217) $A^{\efrac{p - 1}{d}}$.
Hence the group $\dfrac{\Gamma}{\{A^{\efrac{p - 1}{d}}\}}$ can be represented as a doubly
%% -----File: 356.png---Folio 338-------
transitive group of degree $\dfrac{p^{n} - 1}{p - 1}$. This group is (§~220) simply
isomorphic with the simple group of order $\dfrac{N}{(p - 1)d}$, which is
defined by the composition-series of~$G$.

We may sum up these results as follows:---
\index{Homogeneous linear group@\Topic{Homogeneous linear group}!simple groups defined by}%

\begin{Theorem}
The homogeneous linear group of order
\[
N = (p^{n} - 1) (p^{n} - p) \Dots (p^{n} - p^{n - 1})
\]
when $p^{n}$~is neither $2^{2}$ nor~$3^{2}$, defines, by its composition-series, a
simple group of order~$\dfrac{N}{(p - 1)d}$, where $d$~is the greatest common
factor of $p - 1$ and~$n$. This simple group can be represented as
a doubly transitive group of degree $p^{n - 1} + p^{n - 2} + \Dots + p + 1$.
\end{Theorem}

\Par{237.} The $\dfrac{p^{n} - 1}{p - 1}$ symbols, permuted by one of these doubly
transitive simple groups, may be regarded as the sub-groups of
order~$p$ of an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots\DPchg{\,}{,} \text{to
$n$ units})$. Now every pair of sub-groups of such an Abelian
group enters in one, and only in one, sub-group of order~$p^{2}$;
and every sub-group of order~$p^{2}$ contains $p + 1$ sub-groups of
order~$p$. Hence from the $\dfrac{p^{n} - 1}{p - 1}$ symbols permuted by the
doubly transitive group, $\dfrac{p^{n} - 1 · p^{n-1} - 1}{p - 1 · p^2 - 1}$\DPnote{** TN: No () in original} sets of $p + 1$ symbols
each may be formed, such that every pair of symbols occurs in
one set and no pair in more than one set, while the sets are
permuted transitively by the operations of the group. These
groups therefore belong to the class of groups referred to in
§~148. The sub-group, that leaves two symbols unchanged,
permutes the remaining symbols in two transitive systems of
$p - 1$ and $p^{n - 1} + p^{n - 2} + \Dots + p^{2}$; and the sub-group, that leaves
unchanged each of the symbols of one of the sets of $p + 1$, is
contained self-conjugately in a sub-group whose order is
$(p + 1)p$~times that of a sub-group leaving two symbols unchanged.
This latter sub-group permutes the symbols in two
transitive systems of $p + 1$ and $p^{n - 1} + p^{n - 2} + \Dots + p^{2}$. It may
be pointed out that, when $n$~is~$3$, such a sub-group is simply
\index{Group of isomorphisms@\Topic{Group of isomorphisms}!of an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots, 1)$|)}%
%% -----File: 357.png---Folio 339-------
isomorphic with, but is not conjugate to, the sub-groups that
leave one symbol unchanged: this may be seen at once by
noticing that an Abelian group, of order~$p^{3}$ and type $(1, 1, 1)$, has
the same number of sub-groups of orders $p$ and~$p^{2}$.

\begin{Remark}
\Par{238.} Some special cases may be noticed. First, when $p = 2$,
\index{Homogeneous linear group@\Topic{Homogeneous linear group}!simple groups defined by}%
both $p - 1$ and~$d$ are unity, and the homogeneous linear group is
itself a simple group.

If $n = 3$, then $N = 168$; so that the group of isomorphisms of a
group of order~$8$, whose operations are all of order~$2$, is the simple
group of order~$168$ (§~146).

If $n = 4$, then $N = 2^{6} · 3^{2} · 5 · 7$. This is the order of the
alternating group of $8$~symbols; and it may be shewn that this
group is simply isomorphic with the group of isomorphisms.

The Abelian group of order~$16$ contains $35$~sub-groups of order~$4$;
and it may be shewn that, from these $35$~sub-groups, sets of~$5$ can
be formed in $56$~distinct ways, so that each set of~$5$ contains every
operation of order~$2$ of the Abelian group once, and once only. If
\[
P_{1},\ P_{2},\ P_{3},\ P_{4}
\]
are a set of generating operations of the Abelian group, such a set
of $5$~sub-groups of order~$4$ is given by
\[
\{P_{1}, P_{2}\},\ \{P_{3}, P_{4}\},\ \{P_{1}P_{3}, P_{2}P_{4}\},\
\{P_{1}P_{4}, P_{1}P_{2}P_{3}\},\
\{P_{2}P_{3}, P_{1}P_{3}P_{4}\}.
\]
Now
\[
\Sub{P_{1}, \PadTo{P_{2}P_{3}}{P_{2}}, P_{3}, \PadTo{P_{2}P_{4}}{P_{4}}}
    {P_{1}, P_{2}P_{3}, P_{4}, P_{2}P_{4}}
\]
is an isomorphism of order~$7$ of the Abelian group, and $P_{1}$~is the
only operation of the group, except identity, which is left unchanged
by this isomorphism. It may be directly verified that the $7$~sets of
$5$~groups of order~$4$, into which the given set is transformed by the
powers of this isomorphism, contain every sub-group of order~$4$ of
the Abelian group. The $7$~sets, being interchanged among themselves
by this isomorphism of order~$7$ which leaves only $P_{1}$ unchanged,
must be interchanged among themselves by isomorphisms of order~$7$
which leave any other single operation of the Abelian group
unchanged. There are therefore at least $15$~isomorphisms of order~$7$
which interchange the $7$~sets among themselves. Now the
isomorphisms, which interchange the $7$~sets among themselves, form a
sub-group of the group of isomorphisms, which is isomorphic with a
group of degree~$7$; and the only groups of degree~$7$, which contain at
least $15$ operations of order~$7$, are the symmetric and the alternating
groups. The group of isomorphisms must therefore contain a sub-group
which is isomorphic with the symmetric or with the alternating
group of degree~$7$. Hence at once, since the group of isomorphisms
is simple, it must contain a sub-group which is simply isomorphic
with the alternating group of degree~$7$. Since this must be one of
%% -----File: 358.png---Folio 340-------
$8$~conjugate sub-groups, the group of substitutions itself is simply
isomorphic with the alternating group of degree~$8$.

If $p^{n} = 3^{3}$, then $p^{n - 1} + \Dots + p + 1 = 13$, $d = 1$, and $N = 2^{4} · 3^{3} · 13$.
There is therefore a doubly transitive simple group of degree~$13$ and
order~$2^{4} · 3^{3} · 13$ (§§~145,~149).
\end{Remark}

\Par{239.} The homogeneous linear group may be generalized by
\index{Homogeneous linear group@\Topic{Homogeneous linear group}!generalization of|(}%
taking for the coefficients powers of a primitive root of
\[
i^{p^{\nu} - 1} \equiv 1,\ (\mod p),
\]
instead of powers of a primitive root of
\[
i^{p - 1} \equiv 1,\ (\mod p).
\]
When the coefficients are thus chosen, the order of the group
$G_{p, n, \nu}$, defined by all sets of transformations
\[
\begin{aligned}
  x_{1}^{1} &\equiv a_{11}x_{1} + a_{12}x_{2} + \Dots + a_{1n}x_{n}, \\
  x_{2}^{1} &\equiv a_{21}x_{1} + a_{22}x_{2} + \Dots + a_{2n}x_{n}, \\
  \multispan{3}{\dotfill} \\
  x_{n}^{1} &\equiv a_{n1}x_{1} + a_{n2}x_{2} + \Dots + a_{nn}x_{n},
\end{aligned}
\quad (\mod p),
\]
whose determinant differs from zero, may be shewn, as in §~172,
to be
\[
N = (p^{n\nu} - 1) (p^{n\nu} - p^{\nu}) (p^{n\nu} - p^{2 \nu}) \Dots (p^{n\nu} - p^{(n - 1)\nu});
\]
and the order of the sub-group~$\Gamma$, formed of those transformations
whose determinant is unity, is~$\dfrac{N}{p^{\nu} - 1}$. The only
self-conjugate operations of~$\Gamma$ are the operations of the sub-group
generated by $(ix_{1}, ix_{2}, \Dots, ix_{n})$, which are contained in~$\Gamma$.
If $\delta$~is the greatest common factor of $p^{\nu} - 1$ and~$n$, these
self-conjugate operations of~$\Gamma$ form a cyclical sub-group~$\gamma$ of
order~$\delta$. Finally, the argument of §~219 maybe repeated to
shew that $\dfrac{\Gamma}{\gamma}$~is a simple group.

The homogeneous linear group $G_{p, n, \nu}$, when values of~$\nu$
greater than unity are admitted, thus defines a triply infinite
system of simple groups; it may be proved that these groups
can, for all values of~$\nu$, be expressed as doubly transitive groups
of degree~$\dfrac{p^{n\nu} - 1}{p^{\nu} - 1}$.
%% -----File: 359.png---Folio 341-------

\Par{240.} We may shew, in conclusion, that the group $G_{p, n, \nu}$ is
simply isomorphic with a sub-group of $G_{p, n\nu, 1}$. For this purpose,
we consider the group defined by
\begin{gather*}
x_{1}^{1} \equiv x_{1} + i^{r_{1}},\quad
x_{2}^{1} \equiv x_{2} + i^{r_{2}},\quad \Dots,\quad
x_{n}^{1} \equiv x_{n} + i^{r_{n}}, \\
(r_{1}, r_{2}, \Dots, r_{n} = 0, 1, 2, \Dots, p^{\nu} - 1);
\end{gather*}
the congruences being taken to modulus~$p$. This is an Abelian group
of order~$p^{n\nu}$ and type $(1, 1, \Dots\DPchg{\,}{,} \text{to $n\nu$ units})$. Moreover, the operation
\begin{align*}
  x_{1}^{1} &\equiv a_{11}x_{1} + \Dots + a_{1n}x_{n}, \\
  \multispan{3}{\dotfill} \\
  x_{n}^{1} &\equiv a_{n1}x_{1} + \Dots + a_{nn}x_{n},
\end{align*}
of $G_{p, n, \nu}$ transforms the given operation of the Abelian group into
\[
x_{1}^{1} \equiv x_{1} + i^{s_{1}},\quad
x_{2}^{1} \equiv x_{2} + i^{s_{2}},\quad \Dots,\quad
x_{n}^{1} \equiv x_{n} + i^{s_{n}};
\]
where
\begin{align*}
  i^{s_{1}} &\equiv a_{11}i^{r_{1}} + a_{12}i^{r_{2}} + \Dots + a_{1n}i^{r_{n}}, \\
  \multispan{3}{\dotfill} \\
  i^{s_{n}} &\equiv a_{n1}i^{r_{1}} + a_{n2}i^{r_{2}} + \Dots + a_{nn}i^{r_{n}}.
\end{align*}

Every operation of~$G_{p, n, \nu}$ as defined in §~239, is therefore
permutable with the Abelian group, and gives a distinct isomorphism
of it; or in other words, as stated above, $G_{p, n, \nu}$~is simply isomorphic
with a sub-group of~$G_{p, n\nu, 1}$.

Further, the sub-group
\begin{gather*}
x_{1}^{1} \equiv x_{1} + i^{r},\quad
x_{2}^{1} \equiv x_{2},\quad \Dots,\quad
x_{n}^{1} \equiv x_{n}, \\
(r = 0, 1, 2, \Dots, p^{\nu} - 1),
\end{gather*}
is transformed by the given operation of~$G_{p, n, \nu}$ into the sub-group
\begin{gather*}
x_{1}^{1} \equiv x_{1} + a_{11}i^{r},\quad
x_{2}^{1} \equiv x_{2} + a_{21}i^{r},\quad \Dots,\quad
x_{n}^{1} \equiv x_{n} + a_{n1}i^{r}, \\
(r = 0, 1, 2, \Dots, p^{\nu} - 1).
\end{gather*}
If
\[
a_{21} \equiv a_{31} \equiv \Dots \equiv a_{n1} \equiv 0,
\]
the two sub-groups are identical; but if these conditions are not
satisfied, they have no operation in common except identity.
Moreover,
\[
a_{11},\ a_{21},\ \Dots,\ a_{n1}
\]
may each have any value from $0$ to~$i^{p^{\nu} - 1}$, simultaneous zero values
alone excluded. Hence the sub-group of order~$p^{\nu}$ defined by
\begin{gather*}
x_{1}^{1} \equiv x_{1} + i^{r},\quad
x_{2}^{1} \equiv x_{2},\quad \Dots,\quad
x_{n}^{1} \equiv x_{n}, \\
(r = 0, 1, \Dots, p^{\nu} - 1),
\end{gather*}
is one of $\dfrac{p^{n\nu} - 1}{p^{\nu} - 1}$ conjugate sub-groups in the group formed by combining
the Abelian group with~$G_{p, n, \nu}$; and no two of these conjugate
sub-groups have a common operation except identity.
%% -----File: 360.png---Folio 342-------

\begin{Remark}
The $p^{n\nu} - 1$ operations, other than identity, of an Abelian group
of order~$p^{n\nu}$ and type $(1, 1, \Dots\DPchg{\,}{,} \text{to $n\nu$ units})$, can therefore be
divided into $\dfrac{p^{n\nu} - 1}{p^{\nu} - 1}$ sets of $p^{\nu} - 1$ each, such that each set, with
identity, forms a sub-group of order~$p^{\nu}$; and the group~$G_{p, n, \nu}$ is isomorphic
with a group of isomorphisms of the Abelian group which permutes
among themselves such a set of $\dfrac{p^{n\nu} - 1}{p^{\nu} - 1}$ sub-groups of order~$p^{\nu}$.
\end{Remark}

\begin{Ex}[1.] Shew that the $\dfrac{p^{n\nu} - 1 · p^{n\nu} - p\Dots p^{n\nu} - p^{\nu}}{p - 1 · p^{2} - 1\Dots p^{\nu} - 1}$\DPnote{** TN: No () in original} sub-groups of
order~$p^{\nu}$ of an Abelian group of order~$p^{n\nu}$ and type $(1, 1, \Dots\DPchg{\,}{,} \text{to
$n\nu$ units})$ can be divided into sets of $\dfrac{p^{n\nu} - 1}{p^{\nu} - 1}$ each, such that each set
contains every operation of the group, other than identity, once and
once only; and discuss in how many distinct ways such a division
may be carried out.
\end{Ex}

\begin{Ex}[2.] Shew that the simple group, defined by the group of
isomorphisms of an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots, 1)$,
admits a class of contragredient isomorphisms, which change the
operations of the simple group, that correspond to isomorphisms
leaving a sub-group of order~$p$ of the Abelian group unaltered, into
operations that correspond to isomorphisms leaving a sub-group of
order~$p^{n - 1}$ of the Abelian group unaltered.
\end{Ex}
\index{Homogeneous linear group@\Topic{Homogeneous linear group}!generalization of|)}%
%% -----File: 361.png---Folio 343-------


\Chapter{XV.}{On Soluble and Composite Groups.}

\Par{241.} \First{The} most general problem of pure group-theory, so
far as it is concerned with groups of finite order, is the determination
and analysis of all distinct types of group whose order
is a given integer. The solution of this problem clearly involves
the previous determination of all types of simple groups whose
orders are factors of the given integer.

Now there is no known criterion by means of which we can
say whether, corresponding to an arbitrarily given composite
integer~$N$ as order, there exists a simple group or not. For
certain particular forms of~$N$, this question can be answered in
the affirmative. For instance, when $N = \frac{1}{2}n!$, previous investigation
enables us to state that there is a simple group of
order~$N$; but even in these cases we cannot, in general, say
how many distinct types of simple group of order~$N$ there are.

On the other hand, we have seen that, for certain forms of~$N$,
it is possible to state that there is no simple group of order~$N$.
Thus, when $N$~is the power of a prime, or is the product of two
distinct primes, there is no simple group whose order is~$N$; and
further, every group of order~$N$ is soluble. Again, if $N$~is
divisible by a prime~$p$, and contains no factor of the form
$1 + kp$, a group of order~$N$ cannot be simple; for, by Sylow's
theorem, it must contain a self-conjugate sub-group having a
power of~$p$ for its order.

We propose, in the present Chapter, to prove a series of
theorems which enable us to state, in a considerable variety of
%% -----File: 362.png---Folio 344-------
cases, that a group, which has a number of given form for its
order, is either soluble or composite. If the results appear
fragmentary, it must be remembered that this branch of the
subject has only recently received attention: it should be
regarded rather as a promising field of investigation than as
one which is thoroughly explored.

The symbols $p_{1}$,~$p_{2}$, $p_{3}$,~\Dots\ will be used throughout the
Chapter to denote distinct primes in ascending order of magnitude;
while distinct primes, without regard to their magnitude,
will be denoted by $p$,~$q$, $r$,~\Dots.

\Par{242.} \begin{Theorem}[I.]
If $H$ is one of $n$ conjugate sub-groups
of a group~$G$ of order~$N$, and if $N$~is not a factor of~$n!$, $G$~cannot
be simple.
\end{Theorem}

The group~$G$ is isomorphic with the group of substitutions
given on transforming the set of $n$ conjugate sub-groups by
each of the operations of~$G$. This is a transitive group of
degree~$n$, and its order therefore is equal to or is a factor of~$n!$.
Hence, if $N$~is not a factor of~$n!$, the isomorphism cannot be
simple; $G$~is therefore not simple.

\begin{CorollaryFN}
If a prime~$p$ divides~$N$, and if $m$~is the
\index{Holder@\Topic{Hölder}, quoted}%
greatest factor of~$N$ which is congruent to unity~$(\mod p)$, $G$~will
be composite unless $N$~is a factor of~$m!$.
\end{CorollaryFN}
\footnotetext
  {Hölder, \textit{Math.\ Ann.}\ Vol.~\VolNo{XL}, (1892), p.~57.}

There cannot, in fact, be more than $m$ conjugate sub-groups
whose order is the highest power of~$p$ that divides~$N$; so that
this result follows from the theorem.

\begin{Remark}
In dealing with a given integer as order, it may happen that,
though no single application of this theorem will prove the corresponding
group to be composite, a repeated application of Sylow's
theorem, on which the theorem depends, will lead to that result.
As an example, let us consider groups whose order is $3^{2} · 5 · 11$. Any
such group must contain either $1$ or $45$ sub-groups of order~$11$, and
either $1$ or $11$ sub-groups of order~$5$. Hence, if the group is simple,
it must contain $45$~sub-groups of order~$11$, and $11$~sub-groups of
order~$5$; and each one of the latter must be contained self-conjugately
in a sub-group of order~$45$. Now the $45$~sub-groups of order~$11$
would contain $450$~distinct operations of order~$11$, leaving only
$45$~others; and therefore a sub-group of order~$45$, if the group contain
such a sub-group, must in this case be self-conjugate. This is,
%% -----File: 363.png---Folio 345-------
however, in direct contradiction to the assumption that the group
contains $11$~sub-groups of order~$5$; therefore the group cannot be
simple.
\end{Remark}

\Par{243.} \begin{Theorem}[II.]
Every group whose order is the power
\index{Soluble groups@\Topic{Soluble groups}!special classes of|(}%
of a prime is soluble.
\end{Theorem}

This has already been proved in §~54.

\begin{Theorem}[III\protect\footnotemark.]
A group~$G$ whose order is~$p^{\alpha} q^{\beta}$, where $\alpha$~is
less than~$2m$, $m$~being the index to which $p$ belongs~$(\mod q)$, is
soluble.
\end{Theorem}
\index{Burnside@\Topic{Burnside}, quoted}%
\index{Frobenius@\Topic{Frobenius}, quoted}%
\footnotetext{Frobenius, \textit{Berliner Sitzungsberichte} 1895, p.~190, and Burnside, \textit{Proc.\
  L.~M.~S.} Vol.~\VolNo{XXVI}, (1895), p.~209; for the case where $\alpha \ngtr m$.}

If $\alpha < m$, $G$~contains a self-conjugate sub-group~$H$ of order~$q^{\beta}$.

If $\alpha = m$, $G$~contains either $1$ or $p^{m}$ sub-groups of order~$q^{\beta}$.
In the latter case, if $q^{r}$~is the order of the greatest sub-group
common to two groups of order~$q^{\beta}$, and if $h$~is such a sub-group,
$h$~must (§~80) be contained self-conjugately in a
sub-group~$k$ of order~$p^{x}q^{r + s}$, $x > 0$, $s > 0$; and this sub-group
must contain more than one sub-group of order~$q^{r + s}$. Hence $x$~must
be~$m$; and therefore $h$~must be common to all the $p^{m}$
sub-groups of order~$q^{\beta}$, so that $h$~is self-conjugate, and $s$~is
$\beta - r$. If $r$~is zero, no two sub-groups of order~$q^{\beta}$ have a
common operation except identity, and the $p^{m}$~sub-groups
contain $p^{m}(q^{\beta} - 1)$ distinct operations, so that a sub-group
of order~$p^{m}$ is self-conjugate. Hence when $\alpha$~is equal to~$m$, $G$~contains
either a self-conjugate sub-group of order~$q^{r}$ ($r > 0$) or
else one of order~$p^{m}$.

If $m < \alpha < 2m$, $G$~contains $1$ or $p^{m}$ sub-groups of order~$q^{\beta}$.
If there is only one, it is self-conjugate.

If there are $p^{m}$~sub-groups of order~$q^{\beta}$, and if~$q^{r}$ ($r > 0$) is
the order of the greatest group common to any two of them,
it may be shewn, exactly as in the preceding case, that $G$~contains
a self-conjugate sub-group of order~$q^{r}$.

If $r = 0$, the $q^{\beta} - 1$ operations, other than identity, of each
of the $p^{m}$~sub-groups of order~$q^{\beta}$ are all distinct. Now $G$~is
isomorphic with a group of degree~$p^{m}$. If the isomorphism
is multiple, $G$~must have a self-conjugate sub-group whose order
is a power of~$p$. If the isomorphism is simple, we may regard~$G$
%% -----File: 364.png---Folio 346-------
as a transitive group of degree~$p^{m}$; and every operation,
except identity, of a sub-group of order~$q^{\beta}$ will leave one symbol
only unchanged. Hence $q^{\beta}$~is equal to or is a factor of~$p^{m} - 1$.
Moreover, when $G$~is thus regarded, the sub-groups of order~$p^{\alpha}$
must be transitive; for all the sub-groups of order~$q^{\beta}$ will be
given on transforming any one of them by the operations of a
sub-group of order~$p^{\alpha}$.

Consider now a sub-group of order~$p^{\alpha - m} q^{\beta}$, which keeps one
symbol fixed and contains a sub-group of order $q^{\beta}$ self-conjugately.
It will contain $q^{\gamma}$ ($\gamma \leq \beta$) sub-groups of order~$p^{\alpha - m}$.
Any one of these must keep $p^{x}$~symbols fixed; for it is the
sub-group of a transitive group of order~$p^{\alpha}$ and degree~$p^{m}$, which
keeps one symbol fixed. If $\gamma < \beta$, there are operations of order~$q$
which transform one of these sub-groups into itself, and
therefore interchange among themselves the $p^{x}$~symbols unchanged
by the sub-group. Hence $p^{x} - 1$ must be divisible
by~$q$. This requires that $x = m$, which is impossible. Hence
$\gamma = \beta$, and the sub-group of order~$p^{\alpha - m} q^{\beta}$ contains $q^{\beta}$~sub-groups
of order~$p^{\alpha - m}$. It is obvious that no two of these can enter in
the same sub-group of order~$p^{\alpha}$; for if they did, they would
generate a group of order~$p^{\alpha - m + x}$ ($x > 0$), a sub-group
of this order cannot enter in a group of order~$p^{\alpha - m} q^{\beta}$. Therefore
$G$~contains $q^{\beta}$~sub-groups of order~$p^{\alpha}$.

Since the sub-groups of~$G$, of order~$p^{\alpha}$, are transitive, their
self-conjugate operations must (§~106) displace all the symbols,
and they cannot therefore be permutable with any operation
whose order is a power of~$q$. Suppose now that every
operation of a sub-group~$Q$, of order~$q^{x}$, is permutable with
an operation~$P$ of order~$p^{y}$, and that there is no sub-group
of order~$q^{x + 1}$ of which this is true. If $S$~is a self-conjugate
operation of a sub-group of order~$p^{\alpha}$ to which $P$~belongs,
$P$~is transformed into itself and $Q$~into a new sub-group~$Q'$
by~$S$. Hence the sub-group which contains $P$ self-conjugately
has two, and therefore $p^{m}$, sub-groups of order~$q^{x}$.
No two of these can belong to the same sub-group of order~$q^{\beta}$,
for if they did they would generate a group of order~$q^{x + x'}$
($x' > 0$). Now since $P$~is permutable with~$Q$, it must
leave unchanged the symbol which $Q$~leaves unchanged.
Hence $P$~must leave every symbol unchanged; \ie\ it must
%% -----File: 365.png---Folio 347-------
be the identical operation. The order of every operation of~$G$
is therefore either a power of~$p$ or a power of~$q$.

Suppose now that $p^{r}$~is the order of the greatest sub-group
that is common to any two sub-groups of order~$p^{\alpha}$, and that $h$~is
such a sub-group. Then (§~80) $h$~must be permutable with
an operation of order~$q$. Since no operation of~$h$ is permutable
with any operation of order~$q$, it follows that $p^{r} - 1$ is divisible
by~$q$, and therefore that $r = m$.

Let $k$, of order~$p^{m + t}q^{\gamma}$, be the greatest sub-group that
contains $h$ self-conjugately. If $\gamma = \beta$, $h$~is common to each
of the $q^{\beta}$ sub-groups of order~$p^{\alpha}$, and it is therefore a self-conjugate
sub-group of~$G$.

If $\gamma < \beta$, let $Q$~be a sub-group, of order~$q^{\gamma}$, of~$k$. In~$G$
there must be a group, whose order is divisible by~$q^{\gamma + 1}$,
containing $Q$ self-conjugately. Hence there must be operations
in~$G$, which transform $Q$ into itself and $h$~into a conjugate
sub-group~$h'$. If $h$~and~$h'$ have a common sub-group, it must
be transformed into itself by every operation of~$Q$. This is
impossible, since $p^{s} - 1$ is not divisible by~$q$ when $s < m$.

In the group $\{h', Q\}$, $Q$~is one of $p^{m}$~conjugate sub-groups,
and therefore no operation of~$h'$ can transform $Q$ into itself.
Hence if an operation~$P'$ of~$h'$ transforms $h$ into itself, it must
transform $Q$ into a group~$Q'$, which is conjugate to~$Q$ in~$k$.
Hence, $k'$~being the greatest sub-group that contains $h'$
self-conjugately, $\{Q, Q'\}$~must be common to $k$~and~$k'$. But
$\{Q, Q'\}$ containing two sub-groups of order~$q^{\gamma}$, must contain $p^{m}$
such sub-groups; and therefore the $p^{m}$~sub-groups of order~$q^{\gamma}$
that enter in~$k$ are identical with those that enter in~$k'$.
This is impossible if $H$~and~$H'$ are distinct; for the $p^{m}$~sub-groups
of~$k$, of order~$q^{\gamma}$, generate $\{h, Q\}$. Hence no operation
of~$h'$, except identity, transforms $h$ into itself, and the
number of sub-groups in the conjugate set to which $h$~belongs
must not be less than~$p^{m}$; so that
\[
p^{m} \ngtr p^{\alpha - m - t} q^{\beta - \gamma}.
\]

Now we have seen that
\begin{align*}
p^{m} &\equiv 1\ (\mod q^{\beta}), \\
\lintertext{and}
q^{\beta} &\equiv 1\ (\mod p^{\alpha - m});
\end{align*}
%% -----File: 366.png---Folio 348-------
from these congruences, and the inequality $\alpha - m < m$, it follows
that
\[
p^{m} \nless p^{\alpha - m} q^{\beta} - q^{\beta} + 1.
\]

This inequality is inconsistent with the previous one, which
follows directly from the supposition that $\gamma$~is less than~$\beta$.
Hence $\gamma = \beta$, and $h$~is a self-conjugate sub-group of~$G$.

It follows therefore that, on every possible supposition, $G$~must
have a self-conjugate sub-group. If this sub-group be
represented by~$G_{1}$, the same reasoning may be repeated with
respect to the groups $\dfrac{G}{G_{1}}$ and~$G_{1}$. Hence $G$~is soluble.

\begin{Corollary}
All groups whose orders are $p_{1} p_{2}^{\alpha}$, $p_{1}^{2} p_{2}^{\alpha}$, $p_{1}^{3} p_{2}^{\alpha}$,
$p_{1}^{4} p_{2}^{\alpha}$, $p_{1}^{5} p_{2}^{\alpha}$, $p_{1}^{\alpha} p_{2}$ are soluble.
\end{Corollary}

Since the congruence
\[
p_{1}^{2} \equiv 1\ (\mod p_{2})
\]
is satisfied only by $p_{1} = 2$, $p_{2} = 3$, the results stated follow
immediately from the theorem, except for the cases $2^{4}3^{\alpha}$ and
$2^{5}3^{\alpha}$.

If in these cases there are $16$ sub-groups of order~$3^{\alpha}$ ($\alpha > 1$),
there must (§~78) be sub-groups of order~$3^{\alpha - 1}$ common to two
sub-groups of order~$3^{\alpha}$; and, in the groups of order $2^{4}3^{\alpha}$~or
$2^{5}3^{\alpha}$, such a sub-group, if not self-conjugate, must be one of
either $4$ or~$8$ conjugate sub-groups. The groups must therefore
be isomorphic with groups of degree $4$ or~$8$; from this it follows
immediately that they must be soluble (§~146).

\Par{244.} \begin{Theorem}[IV.]
Groups of order~$p_{1}^{\alpha} p_{2}^{2}$ are soluble.
\end{Theorem}

If a group~$G$, of order~$p_{1}^{\alpha} p_{2}^{2}$, contains only $p_{2}$~sub-groups of
order~$p_{1}^{\alpha}$, it cannot be simple, for a group of order~$p_{2}^{2}$ cannot be
expressed as a substitution-group of $p_{2}$~symbols. Similarly, if $G$~contains
a single group of order~$p_{1}^{\alpha}$, it is not simple.

If $G$~contains $p_{2}^{2}$~sub-groups of order~$p_{1}^{\alpha}$, and if these sub-groups
have no common operations, except identity, there are
$p_{2}^{2}(p_{1}^{\alpha} - 1)$ operations in~$G$ whose orders are powers of~$p_{1}$; and
therefore a group of order~$p_{2}^{2}$ is self-conjugate. If the operations
of the $p_{2}^{2}$~sub-groups of order~$p_{1}^{\alpha}$ are not all distinct, let $p_{1}^{r}$ be
the order of the greatest sub-group common to any two of them;
%% -----File: 367.png---Folio 349-------
and let $h$ be such a sub-group. Then (§~80) $h$~must be self-conjugate
in a group~$k$ of order~$p_{1}^{r + s}p_{2}^{\beta}$ ($s > 0$, $\beta = 1$ or~$2$). If
$\beta = 2$, $h$~is self-conjugate in~$G$; and if $\beta = 1$, $k$~must contain $p_{2}$~sub-groups
of order~$p_{1}^{r + s}$, and therefore
\[
p_{2} \equiv 1\ (\mod p_{1}).
\]

Now (§~78)
\[
p_{2}^{2} \equiv 1\ (\mod p_{1}^{\alpha - r}),
\]
and therefore
\[
p_{2} \equiv 1\ (\mod p_{1}^{\alpha - r}),
\]
unless $p_{1} = 2$; but, if $p_{1} = 2$, then
\[
p_{2} \equiv ± 1\ (\mod p_{1}^{\alpha - r - 1}).
\]

Again, if  $s = \alpha - r$, $h$~is one of $p_{2}$~conjugate sub-groups; as
before, $G$~cannot then be simple.

Suppose now that $H$~is a sub-group of order~$p_{1}^{\alpha}$, and that
the self-conjugate operations of~$H$ form a sub-group~$\Gh$. This
must be one of $1$,~$p_{2}$ or~$p_{2}^{2}$ conjugate sub-groups; in the first
two cases, $G$~cannot be simple. Moreover, if any two sub-groups
of the conjugate set to which $\Gh$~belongs have a common
sub-group, it must be self-conjugate in a group containing
more than one sub-group of order~$p_{1}^{\alpha}$; again, $G$~cannot be
simple.

Let
\[
H,\ H_{1},\ \Dots,\ H_{p_{2} - 1}
\]
be the $p_{2}$~sub-groups (§~80) of order~$p_{1}^{\alpha}$ that contain~$h$; and
suppose that $\Gh$~is not contained in~$h$. Every operation of each
of the $p_{2}$~sub-groups
\[
\Gh,\ \Gh_{1},\ \Dots,\ \Gh_{p_{2} - 1},
\]
is permutable with every operation of~$h$. But these sub-groups,
since they occur in~$k$ and not in~$h$, generate a sub-group of~$k$,
whose order is divisible by~$p_{2}$. Hence there must be an operation~$Q$,
of order~$p_{2}$, which is permutable with every operation of~$h$.
This operation would be permutable with every operation of a
group of order $p_{1}^{r} p_{2}^{2}$ at least, and it would therefore be one of
$p_{1}^{\alpha - r}$ conjugate operations at most. Now, if $p_{1} > 2$, $p_{1}^{\alpha - r} < p_{2}$; and
as a group of order~$p_{2}^{2}$ cannot be represented in terms of less
than $p_{2}$~symbols, $G$~would in this case have a self-conjugate
sub-group. If $p_{1} = 2$, then either $p_{1}^{\alpha - r}$ or $p_{1}^{\alpha - r} = 2(p_{2} + 1)$.
A group of order~$p_{2}^{2}$ cannot be expressed in less than $2p_{2}$
symbols; a cyclical group of order~$p_{2}^{2}$ cannot be expressed in
%% -----File: 368.png---Folio 350-------
$2(p_{2} + 1)$~symbols; and a group of degree~$2(p_{2} + 1)$, which
contains a non-cyclical sub-group of order~$p_{2}^{2}$, must (§~141)
contain the alternating group. Hence again, in this case, $G$~must
have a self-conjugate sub-group.

Finally, suppose that $h$~contains the $p_{2}$~sub-groups
\[
\Gh,\ \Gh_{1},\ \Dots,\ \Gh_{p_{2} - 1}.
\]

If $h$~contains further sub-groups of the conjugate set to
which $\Gh$~belongs, they must be permuted among themselves in
sets of~$p_{2}$ when $h$~is transformed by an operation~$Q$, of order~$p_{2}$,
belonging to~$k$. Hence we may assume that
\[
\begin{array}{*{4}{l}}
\Gh,&         \Gh_{1},& \Dots,& \Gh_{p_{2} - 1}, \\
\Gh_{p_{2}},& \Gh_{p_{2} + 1},& \Dots,& \Gh_{2p_{2} - 1}, \\
\multicolumn{4}{c}{\dotfill} \\
\Gh_{(x-1)p_{2}},&  \Gh_{(x-1)p_{2} + 1},& \Dots,& \Gh_{xp_{2} - 1},
\end{array}
\]
is the complete set of sub-groups of the conjugate set to which
$\Gh$~belongs, contained in~$h$; and that, when transformed by~$Q$,
those in each line are permuted among themselves. If $\{Q, Q'\}$
is a sub-group\footnote
  {If this sub-group is cyclical, then $Q = Q'^{p}$.}
of order~$p_{2}^{2}$, the $p_{2}^{2}$~sub-groups
\[
\Gh,\ \Gh_{1},\ \Dots,\ \Gh_{p_{2}^{2} - 1},
\]
are permuted transitively among themselves, when transformed
by the operations of~$\{Q, Q'\}$. Hence if $Q'$~change\DPnote{** [sic]} $\Gh$ into $\Gh_{yp_{2}}$, it
must change the set
\[
\Gh,\ \Gh_{1},\ \Dots,\ \Gh_{p_{2} - 1},
\]
into the set
\[
\Gh_{yp_{2}},\ \Gh_{yp_{2} + 1},\ \Dots,\ \Gh_{(y + 1)p_{2} - 1};
\]
and therefore
\[
H_{yp_{2}},\ H_{yp_{2} + 1},\ \Dots,\ H_{(y + 1)p_{2} - 1},
\]
is a set of $p_{2}$~sub-groups of order~$p_{2}^{\alpha}$, which have the common
sub-group $Q'^{-1} hQ'$.

Now, since by supposition $h$~is not a self-conjugate sub-group
of~$H$, there must (§~55) be in~$H$ a sub-group~$h'$ conjugate
to and permutable with~$h$. Let this be the sub-group of order~$p_{1}^{r}$,
common to
\[
H,\ H'_{1},\ \Dots,\ H'_{p_{2} - 1}.
\]
No one of the sets
\[
H_{yp_{2}},\ H_{yp_{2} + 1},\ \Dots,\ H_{(y + 1)p_{2} - 1},\quad (y = 0, 1, \Dots\Add{,} x - 1),
\]
%% -----File: 369.png---Folio 351-------
can have more than one group in common with this set, as
otherwise $p_{1}^{r}$ would not be the order of the greatest group
common to two groups of order~$p_{1}^{\alpha}$. Hence, of the set of groups
\[
\Gh,\ \Gh_{1}',\ \Dots,\ \Gh_{p_{2} - 1}',
\]
at least $p_{2} - x$ are distinct from the groups, conjugate to~$\Gh$,
which enter in~$h$. Every group of order~$p_{1}^{r + s}$ occurring in~$k$ will
have a similar set of $p_{2} - x$ sub-groups of the set~$\Gh$, which do
not occur in~$h$. Hence, since the operations of~$h$ are the only
operations common to two sub-groups of order~$p_{1}^{r + s}$ in~$k$, there
must be in~$k$ not less than $p_{2}(p_{2} - x) + p_{2}x$, \ie~$p_{2}^{2}$, sub-groups
of the set to which $\Gh$~belongs.

In other words, $k$~must contain a self-conjugate sub-group
of~$G$.

On every supposition that can be made, we have shewn that
$G$~must have a self-conjugate sub-group. If this sub-group is~$G_{1}$,
and if the order of either $G_{1}$ or $\dfrac{G}{G_{1}}$ contains~$p_{2}^{2}$, we may
repeat the same reasoning; while if the orders of $\dfrac{G}{G_{1}}$ and $G_{1}$ are
both divisible by~$p_{2}$, we have already seen that they must be
soluble. Finally, then, $G$~itself must be soluble.

\Par{245.}
\begin{Theorem}[V.]
A group of order~$p_{1}^{\alpha} p_{2}^{\beta}$, in which the
sub-groups of orders $p_{1}^{\alpha}$ and $p_{2}^{\beta}$ are Abelian, is soluble.
\end{Theorem}

Suppose, first, that there are $p_{2}^{\gamma}$ ($\gamma < \beta$) sub-groups of order~$p_{1}^{\alpha}$.
The group is then isomorphic with a substitution group
of degree~$p_{2}^{\gamma}$. Now the operations of a sub-group of order~$p_{2}^{\beta}$
transform the $p_{2}^{\gamma}$~sub-groups of order~$p_{1}^{\alpha}$ transitively among
themselves; and therefore, in the isomorphism between the
given group and the substitution group of degree~$p_{2}^{\gamma}$, there
corresponds to an Abelian sub-group of order~$p_{2}^{\beta}$ a transitive
substitution group of degree~$p_{2}^{\gamma}$. But (§~124) an Abelian group
can only be represented as a transitive substitution group of a
number of symbols equal to its order. Hence the isomorphism
between the given group and the substitution group is multiple;
and the given group is not simple.

Suppose, next, that there are $p_{2}^{\beta}$~sub-groups of order~$p_{1}^{\alpha}$, and
therefore (§~81) $p_{1}^{\alpha} - 1$ distinct sets of conjugate operations
%% -----File: 370.png---Folio 352-------
whose orders are powers of~$p_{1}$. Let $P$~be an operation whose
order is a power of~$p_{1}$; and let $H$, of order~$p_{1}^{\alpha} p_{2}^{\delta}$, be the greatest
group which contains $P$ self-conjugately. Then $H$~contains at
least $p_{2}^{\delta}$~operations whose orders are powers of~$p_{2}$, and therefore
at least $p_{2}^{\delta}$ distinct operations of the form~$PQ$, where $P$~and~$Q$
are permutable and
\[
Q^{p_{2}^{\delta}} = 1.
\]
Now $P$~is one of $p_{2}^{\beta - \delta}$ conjugate operations; corresponding
to each of these, there is a similar set of $p_{2}^{\delta}$~operations of the
form~$PQ$; while no two such operations can be identical. The
group therefore contains $p_{2}^{\beta}$~operations of the form~$PQ$; and
similarly, it contains an equal number for each set of conjugate
operations whose order is a power of~$p_{1}$. Hence, finally, the
group contains $p_{2}^{\beta} (p_{1}^{\alpha} - 1)$ operations whose orders are divisible
by~$p_{1}$. There is therefore in this case a self-conjugate sub-group
of order~$p_{2}$.

The group therefore always contains a self-conjugate sub-group.
A repetition of the same reasoning shews that it is
soluble.

\Par{246.}
\begin{Theorem}[VI.]
A group of order $p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \Dots\ p_{n}^{\alpha_{n}}$, in
which the sub-groups of order $p_{1}^{\alpha_{1}}$, $p_{2}^{\alpha_{2}}$,~\Dots, $p_{n - 1}^{\alpha_{n - 1}}$ are cyclical, is
soluble\footnotemark.
\end{Theorem}
\index{Burnside@\Topic{Burnside}, quoted}%
\index{Groups whose sub-groups@\Topic{Groups} whose sub-groups of order~$p^{\alpha}$ are all cyclical}%
\footnotetext{\textit{Proc.\ L.~M.~S.} Vol.~\VolNo{XXVI}, (1895), p.~199.}%

If the number of operations of the group, whose orders divide
$p_{r}^{\alpha_{r}} p_{r + 1}^{\alpha_{r + 1}} \Dots\ p_{n}^{\alpha_{n}}$, is exactly equal to this number, it follows
from Theorem~VII, Cor.~I, §~87, that the same is true for the
number of operations of the group whose orders divide
$p_{r + 1}^{\alpha_{r + 1}} \Dots\ p_{n}^{\alpha_{n}}$. Now, when $r = 1$, this relation obviously holds;
and therefore it is true for all values of~$r$. The group therefore
contains just $p_{n}^{\alpha_{n}}$ operations whose orders divide~$p_{n}^{\alpha_{n}}$; hence
a sub-group of order~$p_{n}^{\alpha_{n}}$ is self-conjugate. Any sub-group of
order~$p_{n - 1}^{\alpha_{n - 1}}$ must transform this self-conjugate sub-group of
order~$p_{n}^{\alpha_{n}}$ into itself, so that the group must contain a sub-group
of order~$p_{n - 1}^{\alpha_{n - 1}} p_{n}^{\alpha_{n}}$. If there were more than one sub-group
of this order, the group would contain more than
$p_{n - 1}^{\alpha_{n - 1}} p_{n}^{\alpha_{n}}$ operations whose orders divide this number. Hence
%% -----File: 371.png---Folio 353-------
the group contains a single sub-group of order~$p_{n - 1}^{\alpha_{n - 1}} p_{n}^{\alpha_{n}}$, and
this sub-group must be self-conjugate. By continuing this
reasoning it may be shewn that, for each value of~$r$, the group
contains a single sub-group of order $p_{r}^{\alpha_{r}} p_{r + 1}^{\alpha_{r + 1}} \Dots\ p_{n}^{\alpha_{n}}$, such a
sub-group being necessarily self-conjugate. The group is therefore
soluble; for we have already seen that a group whose
order is the power of a prime is always soluble.

The self-conjugate sub-group of order $p_{r}^{\alpha_{r}} \Dots\ p_{n}^{\alpha_{n}}$ must
contain the complete set of sub-groups of order~$p_{r}^{\alpha_{r}}$. If
then there are $m$ of these sub-groups, $m$~must be a factor of
$p_{r + 1}^{\alpha_{r + 1}} \Dots\ p_{n}^{\alpha_{n}}$; and a sub-group of order~$p_{r}^{\alpha_{r}}$ must be contained
self-conjugately in a sub-group of order
\[
p_{1}^{\alpha_{1}} \Dots p_{r}^{\alpha_{r}}
  \frac{p_{r + 1}^{\alpha_{r + 1}} \Dots p_{n}^{\alpha_{n}}}{m}.
\]
Hence if $i$~and~$j$ are any two indices between $1$ and~$n$, the group
contains sub-groups of order~$p_{i}^{\alpha_{i}} p_{j}^{\alpha_{j}}$. Now the above sub-group,
which contains a sub-group of order~$p_{r}^{\alpha_{r}}$ self-conjugately, is of
the same nature as the original group. Hence, $i$~and~$j$ being
any two indices less than~$r$, it contains a sub-group of order
$p_{i}^{\alpha_{i}} p_{j}^{\alpha_{j}} p_{r}^{\alpha_{r}}$. This process may be continued to shew that, if $\mu$~is
any factor of $p_{1}^{\alpha_{1}} \Dots\ p_{n}^{\alpha_{n}}$ which is relatively prime to
$\dfrac{p_{1}^{\alpha_{1}} \Dots\ p_{n}^{\alpha_{n}}}{\mu}$, then the group contains sub-groups of order~$\mu$.

\begin{Ex}
Shew that, when the groups of order~$p_{n}^{\alpha_{n}}$ are also cyclical,
\index{Groups whose sub-groups@\Topic{Groups} whose sub-groups of order~$p^{\alpha}$ are all cyclical}%
there are sub-groups whose orders are any factors whatever of the
order of the group.
\end{Ex}

\begin{Remark}
\Par{247.} A special case of the class of groups under consideration
is that in which the order contains no repeated prime factor. These
groups have formed the subject of a memoir by Herr O.~Hölder\footnotemark.
\index{Holder@\Topic{Hölder}, quoted}%
\index{Groups whose orders contain@\Topic{Groups} whose orders contain!no squared factor}%
\footnotetext{``Die Gruppen mit quadratfreier Ordnungszahl'' \textit{Göttingen Nachrichten},
  1895: pp.~211--229. Compare also Frobenius, \textit{Berliner Sitzungsberichte}, 1895,
  pp.~1043, 1044.}%
He shews that they are capable of a specially simple form of representation,
which we shall now consider.

Let
\[
N = p_{1} p_{2} \Dots\ p_{n}
\]
be the order of a group~$G$. It has been seen above that a sub-group
of order~$p_{n}$ is self-conjugate. Suppose then that $p_{l}$, $p_{m}$,~\Dots, $p_{n}$ are
%% -----File: 372.png---Folio 354-------
the orders of those cyclical sub-groups of prime order which are
\index{Defining relations@\Topic{Defining relations} of a group!for groups whose orders contain no square factor}%
\index{Groups whose orders contain@\Topic{Groups} whose orders contain!no squared factor}%
contained self-conjugately in~$G$. They generate a self-conjugate
sub-group~$H$ of order $p_{l} p_{m} \Dots p_{n}$, say~$\mu$. Each of the cyclical
sub-groups $\{P_{l}\}$, $\{P_{m}\}$,~\Dots, $\{P_{n}\}$ of~$H$, generated by an operation of
prime order, is permutable with all the rest; and therefore (§~34)
the operations $P_{l}$, $P_{m}$,~\Dots, $P_{n}$ are all permutable. Hence, since
their orders are distinct primes, the sub-group~$H$ is cyclical. Let
now $K$~be a sub-group of~$G$ of order~$\nu$, where $N = \mu\nu$. No self-conjugate
sub-group of~$G$ can be contained in~$K$. For if it contained
such a sub-group of order $p_{\alpha} p_{\beta} \Dots p_{\gamma}$ ($\alpha < \beta < \Dots < \gamma$), this sub-group
would contain a single cyclical sub-group of order~$p_{\gamma}$; and $G$~would
contain a self-conjugate sub-group of order~$p_{\gamma}$, contrary to
supposition. Since then $K$~contains no self-conjugate sub-group of~$G$,
it follows (§~123) that $G$~can be expressed transitively in $\mu$~symbols;
when it is so expressed, an operation~$M$ of order~$\mu$, that
generates~$H$, will be a circular substitution of $\mu$~symbols. The
only substitutions performed on the $\mu$~symbols, which are permutable
with~$M$, are its own powers; and therefore no operation of~$G$
is permutable with~$M$ except its own powers. Every operation
of~$K$ must therefore give a distinct isomorphism of~$\{M\}$. Now
(§~169) the group of isomorphisms of a cyclical group is Abelian.
Hence $K$~must be Abelian; and since its order contains no repeated
prime factor, it must be cyclical. The group~$G$ is therefore completely
defined by the relations
\[
M^{\mu} = 1,\quad N^{\nu} = 1,\quad N^{-1}MN = M^{\alpha},
\]
where $\alpha$~belongs to index $\nu, \pmod{\mu}$.
\end{Remark}

\Par{248.} The theorem of §~246 is a particular case of the
following more general one, due to Herr Frobenius\footnotemark,
\index{Frobenius@\Topic{Frobenius}, quoted}%
\footnotetext{``Über auflösbare Gruppen,~II'': \textit{Berliner Sitzungsberichte}, 1895, p.~1035.}%
from which
a number of interesting and important results may be deduced.

\begin{Theorem}[VII.]
If $G$~is a group, of order $N = mn$, where $m$
and~$n$ are relatively prime, and where
\[
m = p^{\alpha} q^{\beta} \Dots r^{\gamma};
\]
and if sub-groups $P$, $Q$,~\Dots, $R$, of orders $p^{\alpha}$, $q^{\beta}$,~\Dots, $r^{\gamma}$, are
Abelian, while $\theta(P)$ and~$\dfrac{N}{p^{\alpha}}$, $\theta(Q)$ and $\dfrac{N}{p^{\alpha} q^{\beta}}$,~\Dots, $\theta(R)$ and
$\dfrac{N}{m}$, are relatively prime; then $G$~contains \Inum{(i)}~exactly $n$~operations
whose orders divide~$n$, and \Inum{(ii)}~a sub-group of order~$m$ which
has a self-conjugate sub-group of order~$r^{\gamma}$.
\end{Theorem}

The theorem will be proved inductively, by shewing that, if
%% -----File: 373.png---Folio 355-------
it is true for any similar group of order~$m'n'$, where $m'$~is a
factor of~$m$, it is also true for~$G$.

Suppose that the greatest group, which contains $R$ self-conjugately,
is a group~$H'$ of order~$p^{\alpha'} q^{\beta'} \Dots r^{\gamma}n'$, where $n'$~is a
factor of~$n$. Let $N'$~be any operation of this sub-group, whose
order~$\nu'$ is a factor of~$n'$. Then (§~177) since $\theta(R)$ and~$\nu'$ are
relatively prime, $N'$~is permutable with every operation of~$R$.
Hence every operation of~$H'$, whose order divides~$n'$, is permutable
with every operation of~$R$.

Let now $S$~be any operation of~$R$; and suppose that the
order of~$K$, the greatest sub-group which contains $S$ self-conjugately,
is $p^{\alpha_{1}} q^{\beta_{1}} \Dots r^{\gamma}n_{1}$. If the order of~$S$ is~$r^{\delta}$, the
order of~$\dfrac{K}{\{S\}}$ is $p^{\alpha_{1}} q^{\beta_{1}} \Dots r^{\gamma - \delta} n_{1}$, say~$m_{1}n_{1}$. We have seen in
§~176 that, if $P_{1}$~is a sub-group of~$P$, then $\theta(P_{1})$~is equal to or
is a factor of~$\theta(P)$. Hence if $P_{1}$~is a sub-group of order~$p^{\alpha_{1}}$ of~$\dfrac{K}{\{S\}}$,
$\theta(P_{1})$~and $\dfrac{N}{p^{\alpha}}$ are relatively prime; therefore, \DPtypo{à~fortiori}{a~fortiori},
$\theta(P_{1})$ and $\dfrac{m_{1}n_{1}}{p^{\alpha_{1}}}$ are relatively prime. Similarly $\theta(Q_{1})$ and
$\dfrac{m_{1}n_{1}}{p^{\alpha_{1}}q^{\beta_{1}}}$ are relatively prime, and so on. Moreover $m_{1}$~is a factor
of~$m$. We may therefore assume that the theorem holds for~$\dfrac{K}{\{S\}}$.
Hence this group contains exactly $n_{1}$ operations whose
orders divide~$n_{1}$, and it also contains a sub-group of order~$m_{1}$
in which a sub-group of order~$r^{\gamma - \delta}$ is self-conjugate. Therefore
$K$~contains a sub-group of order $p^{\alpha_{1}} q^{\beta_{1}} \Dots r^{\gamma}$ in which a
sub-group of order~$r^{\gamma}$ is self-conjugate, and exactly $n_{1}$ operations
of the form~$SN$, where $S$~and~$N$ are permutable and the order
of~$N$ divides~$n$.

Now $S$~is one of~$p^{\alpha - \alpha_{1}}q^{\beta - \beta_{1}} \Dots \dfrac{n}{n_{1}}$ conjugate operations in~$G$;
corresponding to each of these, there is such a set of $n_{1}$
operations of the form~$SN$, while (§~16) no two of these
operations can be identical. Hence $G$~contains $p^{\alpha - \alpha_{1}} q^{\beta - \beta_{1}} \Dots n$
operations of the form~$SN$, arising from the conjugate set to
which $S$~belongs. If then we sum for the distinct conjugate
sets of operations of~$G$ whose orders are powers of~$r$, the
%% -----File: 374.png---Folio 356-------
number of operations of~$G$, whose orders divide~$r^{\gamma}n$ and do not
divide~$n$, is $n\sum p^{\alpha - \alpha_{1}}q^{\beta - \beta_{1}}\Dots$.

Now since $H'$~is the greatest group that contains $R$ self-conjugately,
the greatest common sub-group of $K$~and~$H'$ is
the greatest sub-group of~$K$ that contains $R$ self-conjugately.
The order of this group certainly contains $p^{\alpha_{1}} q^{\beta_{1}} \Dots r^{\gamma}$ as
a factor; and if the order is $p^{\alpha_{1}} q^{\beta_{1}} \Dots r^{\gamma}\nu$, then $S$~is one of
$p^{\alpha' - \alpha_{1}} q^{\beta' - \beta_{1}} \Dots \dfrac{n'}{\nu}$ conjugate operations in~$H'$. Every operation
of~$H'$ however, whose order divides~$n'$, is permutable with every
operation of~$R$; and therefore $\nu = n'$. Hence the number of
operations of~$H'$, excluding identity, whose orders are powers of~$r$,
is $\sum' p^{\alpha' - \alpha_{1}} q^{\beta' - \beta_{1}} \Dots$, where the summation is extended to all the
distinct conjugate sets of operations of~$H'$ whose orders are
powers of~$r$. It has been shewn (Theorem~IV, Corollary, §~81)
that the number of distinct conjugate sets of operations in~$G$,
whose orders are powers of~$r$, is the same as the number in~$H'$.
Hence the symbols $\sum$ and $\sum'$ represent summations of the same
number of terms.

Finally, in accordance with the induction we are using, we
may assume that the number of operations of~$G$ whose orders
divide~$r^{\gamma}n$ is exactly equal to~$r^{\gamma}n$. For the order of~$G$ may be
separated into the factors
\[
m' = p^{\alpha} q^{\beta} \Dots,
\]
and
\[
n' = r^{\gamma}n,
\]
where $m'$~is a factor of~$m$, while the conditions of the theorem
hold for this separation.

Hence the number of operations of~$G$ whose orders divide~$n$,
is equal to
\[
n\bigl(r^{\gamma} - \sum p^{\alpha - \alpha_{1}} q^{\beta - \beta_{1}} \Dots\bigr),
\]
while at the same time
\[
r^{\gamma} - 1 = \sum p^{\alpha' - \alpha_{1}} q^{\beta' - \beta_{1}} \Dots.
\]
Unless $\alpha' = \alpha$, $\beta' = \beta$,~\Dots, the former of these numbers is
negative, which is impossible; these conditions must therefore
be satisfied. The number of operations, whose orders
divide~$n$, is therefore exactly equal to~$n$; and the order of
the greatest group~$H'$ that contains $R$ self-conjugately is
%% -----File: 375.png---Folio 357-------
$p^{\alpha} q^{\beta} \Dots r^{\gamma}n'$. The factor-group $\dfrac{H'}{R}$ has for its order $p^{\alpha} q^{\beta} \Dots n'$,
and it satisfies the same conditions as~$G$. Hence it contains a
sub-group of order~$p^{\alpha} q^{\beta} \Dots$; and $H'$~therefore contains a sub-group
of order~$p^{\alpha}q^{\beta} \Dots r^{\gamma}$.

\begin{Remark}
The free use of the inductive process that has been made in the
preceding proof may possibly lead the reader to doubt its validity.
He will find it instructive to verify the truth of the theorem
directly in the simpler cases. In fact, the direct verification for
the case in which $m$~is a power of a prime is essential to the formal
proof; it has been omitted for the sake of brevity.
\end{Remark}

\begin{Corollary}[I.]
If $m = m_{1}m_{2}$, where $m_{1}$~and~$m_{2}$ are relatively
prime, $G$~has a single conjugate set of sub-groups of order~$m_{1}$.
\end{Corollary}

That $G$~contains groups of order~$m_{1}$ may be shewn inductively
at once. For since it is clearly true when $m$~contains
only one, or two, distinct prime factors, we may assume it true
when $m_{1}$~does not contain $r^{\gamma}$ as a factor. Now $G$~has a sub-group~$g$
of order~$p^{\alpha} q^{\beta} \Dots r^{\gamma}$, which contains a sub-group~$R$ of
order~$r^{\gamma}$ self-conjugately. The factor-group $\dfrac{g}{R}$ contains then a
sub-group of order~$m_{1}'$ where $m_{1}'$~is any factor of $\dfrac{m}{r^{\gamma}}$ which is
relatively prime to $\dfrac{m}{m_{1}'r^{\gamma}}$; and therefore $g$~contains a sub-group
of order~$m_{1}'r^{\gamma}$.

If now $I$~and~$I'$ are two sub-groups of~$G$ of order~$m_{1}'r^{\gamma}$,
they may be assumed to contain the same sub-group~$R$ of
order~$r^{\gamma}$. For the sub-groups of~$G$ of order~$r^{\gamma}$ form a single
conjugate set; and therefore, if $I'$~contained a sub-group~$R'$ of
this set, $S^{-1}I'S$~would contain $S^{-1}R'S$, which may be taken
to be~$R$. Now $R$~is a self-conjugate sub-group of both
$I$~and~$I'$, and therefore the factor-groups $\dfrac{I}{R}$ and $\dfrac{I'}{R}$ of order~$m_{1}'$
are sub-groups of the factor-group~$\dfrac{H'}{R}$. Hence if the result is
true when $m_{1}$~does not contain the factor~$r^{\gamma}$, it is also true
when $m_{1}$~does contain~$r^{\gamma}$. But when $m_{1}$~is equal to~$p^{\alpha}$, the
result is obviously true; and therefore it is true generally.
%% -----File: 376.png---Folio 358-------

\begin{Corollary}[II.]
If the sub-groups $P$, $Q$,~\Dots, $R$ of~$G$
contain characteristic sub-groups $P_{0}$, $Q_{0}$,~\Dots, $R_{0}$ of orders
$p^{\alpha_{0}}$, $q^{\beta_{0}}$,~\Dots, $r^{\gamma_{0}}$; then $G$~contains a sub-group of order
$m_{0} = p^{\alpha_{0}} q^{\beta_{0}} \Dots r^{\gamma_{0}}$, which has a self-conjugate sub-group of
order~$r^{\gamma_{0}}$.
\end{Corollary}

Assuming the truth of this statement when $m_{0}$~does not
contain the factor~$r$, the factor-group $\dfrac{H'}{R}$ must contain a sub-group
of order~$\dfrac{m_{0}}{r^{\gamma_{0}}}$; and therefore $G$~contains a sub-group~$J$ of
order $p^{\alpha_{0}} q^{\beta_{0}} \Dots r^{\gamma}$ which has $R$ for a self-conjugate sub-group.
Hence, since $R_{0}$~is a characteristic sub-group of~$R$, $J$~contains a
sub-group of order~$p^{\alpha_{0}} q^{\beta_{0}} \Dots$ and a self-conjugate sub-group
of order~$r^{\gamma_{0}}$. It therefore contains a sub-group of order
$p^{\alpha_{0}} q^{\beta_{0}} \Dots r^{\gamma_{0}}$, which has a self-conjugate sub-group of order~$r^{\gamma_{0}}$.

\Par{249.} The reader will have no difficulty in seeing, as has
already been stated, that Theorem~VI is a direct result of the
theorem proved in the last paragraph. We shall proceed at
once to further applications of it.

\begin{Theorem}[VIII.]
A group~$G$ of order
\[
N = p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \Dots p_{n}^{\alpha_{n}},
\]
in which the sub-groups of every order~$p_{r}^{\alpha_{r}}$ \($r = 1, 2, \Dots, n - 1$\)
are Abelian, with either one or two generating operations, is
generally soluble; the special case $p_{1} = 2$, $p_{2} = 3$, may constitute
an exception\footnotemark.
\end{Theorem}
\index{Frobenius@\Topic{Frobenius}, quoted}%
\footnotetext{Frobenius, \textit{loc.~cit.}, p.~1041.}%

If an Abelian sub-group~$P_{r}$ of order~$p_{r}^{\alpha_{r}}$ is generated by a
single operation, it is cyclical and $\theta(P_{r})$~is (§~176) equal to $p_{r} - 1$.
If $P_{r}$~is generated by two independent and permutable operations,
$\theta(P_{r})$~is equal to~$(p_{r} - 1)(p_{r}^{2} - 1)$. Now no prime greater
than~$p_{r}$ can divide $(p_{r} - 1)(p_{r}^{2} - 1)$ unless $p_{r} + 1$ be a prime;
and this is only possible when $p_{r}$~is equal to~$2$. Hence unless
$P_{1} = 2$, $p_{2} = 3$, $\theta(P_{r})$~and $\dfrac{N}{p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \Dots p_{r}^{\alpha_{r}}}$ are relatively prime for
all values of~$r$ from $1$ to~$n - 1$. Omitting for the present this
exceptional case, the conditions of Theorem~VII are satisfied
by~$G$; it therefore contains exactly $p_{n}^{a_{n}}$ operations whose
%% -----File: 377.png---Folio 359-------
orders are divisible by~$p_{n}$. Therefore $G$~has a self-conjugate
sub-group~$P_{n}$ of order~$p_{n}^{\alpha_{n}}$, and hence also one or more sub-groups
of order~$p_{n - 1}^{\alpha_{n - 1}} p_{n}^{\alpha_{n}}$. There are however only $p_{n - 1}^{\alpha_{n - 1}} p_{n}^{\alpha_{n}}$
operations in~$G$ whose orders are divisible by no prime smaller
than~$p_{n - 1}$, and therefore the sub-group of order~$p_{n - 1} ^{\alpha_{n - 1}} p_{n}^{\alpha_{n}}$ must
be self-conjugate. This process clearly may be continued to
shew that $G$~has a self-conjugate sub-group of every order
$p_{r}^{\alpha_{r}} p_{r + 1}^{\alpha_{r + 1}} \Dots p_{n}^{\alpha_{n}}$ ($r = 2, 3, \Dots, n$); from which it follows immediately
that $G$~is soluble.

Returning now to the exceptional case, suppose that a group
of order~$2^{\alpha_{1}}$ is contained self-conjugately in a maximum group
of order~$2^{\alpha_{1}} 3^{\beta} m$. If every operation of the group of order~$2^{\alpha_{1}}$ is
self-conjugate within this sub-group, $G$~contains $2^{\alpha_{1}} - 1$ distinct
conjugate sets of operations whose orders are powers of~$2$;
hence it follows that there are just $\dfrac{N}{2^{\alpha_{1}}}$ operations in~$G$ whose
orders are not divisible by~$2$. In this case, the conditions of
Theorem~VII are satisfied by~$G$; and it is still soluble.

If, lastly, the operations of the sub-groups of order~$2^{\alpha_{1}} 3^{\beta} m$,
whose orders are powers of~$2$, are not all self-conjugate, there
must be an operation~$B$, whose order is a power of~$3$, in this
sub-group which is not permutable with every operation of
the sub-group of order~$2^{\alpha_{1}}$. Let now
\[
A,\ A_{1},\ \Dots,\ A_{r},\ A_{r + 1},\ \dots
\]
be a characteristic series of the group~$A$ of order~$2^{\alpha_{1}}$; and
suppose that $A_{r + 1}$~is the greatest of these groups with every
one of whose operations~$B$ is permutable. Then (§~175) $B$~is
not permutable with every operation of~$\dfrac{A_{r}}{A_{r + 1}}$. Hence $\dfrac{A_{r}}{A_{r + 1}}$~is
a quadratic group; and its three operations of order~$2$ must
be permuted cyclically when transformed by~$B$. It follows
that $\{A_{r}, B\}$ is multiply isomorphic with a tetrahedral group.
Hence finally, under the conditions of the theorem, $G$~is
certainly soluble unless it contains a sub-group which is
isomorphic with a tetrahedral group.

\begin{Corollary}
A group of order $p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \dots p_{n}^{\alpha_{n}}$ in which no
one of the indices $\alpha_{1}$, $\alpha_{2}$,~\dots, $\alpha_{n - 1}$ is greater than~$2$, is soluble
\index{Groups whose orders contain@\Topic{Groups} whose orders contain!no cubed factor}%
\index{Soluble groups@\Topic{Soluble groups}!special classes of|)}%
%% -----File: 378.png---Folio 360-------
unless it has a sub-group which is isomorphic with a tetrahedral
group. For a group of order~$p_{r}^{2}$ is necessarily an Abelian group
which is generated by either one or two independent operations.
\end{Corollary}

\Par{250.} We shall next consider certain groups of even order
\index{Composite groups@\Topic{Composite groups}!of even order|(}%
in which the operations of odd order form a self-conjugate sub-group.
If $G$~is a group of order~$N$, where
\[
N = 2^{\alpha}n,\quad \text{($n$ odd),}
\]
we shall suppose, in this and the following paragraphs, that a
sub-group~$\scrQ$ of order~$2^{\alpha}$ is Abelian, and that $\theta(\scrQ)$ and $n$ are
relatively prime. Let us take first the case in which $\scrQ$~is
cyclical, so that $\theta(\scrQ)$~is unity and the latter condition is satisfied
for all values of~$n$. If such a group is represented in regular
form as a substitution group of $2^{\alpha}n$~letters, the substitution
corresponding to an operation of order~$2^{\alpha}$ will consist of $n$~cycles
of $2^{\alpha}$~symbols each. This is an odd substitution; and therefore
$G$~has a self-conjugate sub-group of order~$2^{\alpha - 1}n$. In this sub-group
there are cyclical operations of order~$2^{\alpha - 1}$. Hence the
same reasoning will apply to it, and it contains a self-conjugate
sub-group of order~$2^{\alpha - 2}n$. Now (Theorem~VII, Cor.~I. §~87) $G$~contains
exactly $2^{\alpha - 2}n$ operations whose orders are not divisible
by~$2^{\alpha - 1}$. Hence the sub-group of order~$2^{\alpha - 2}n$ must be self-conjugate
in~$G$. Proceeding thus, it may be shewn that $G$~contains
self-conjugate (and characteristic) sub-groups of every
order~$2^{\alpha - r}n$ ($r = 1, 2, \Dots, \alpha$)\footnotemark.
\index{Frobenius@\Topic{Frobenius}, quoted}%
\footnotetext{Frobenius, ``Über auflösbare \DPtypo{Grüppen}{Gruppen},~II'' \textit{Berliner Sitzungsberichte},
  1895, p.~1039.}%

\Par{251.} Suppose, secondly, that, under the same conditions, $G$~contains
a self-conjugate sub-group~$\scrQ$ of order~$2^{\alpha}$. Since $\theta(\scrQ)$
and $n$ are relatively prime, every operation of~$\scrQ$ is permutable
with every operation of~$G$ (§~177). If $\scrQ'$~is a sub-group of~$\scrQ$
of order~$2^{\alpha - 1}$, $\scrQ'$~must therefore be self-conjugate in~$G$. Now
the order of the factor group $\dfrac{G}{\scrQ'}$ is~$2n$; hence, by the preceding
result, it has a self-conjugate sub-group of order~$n$. It follows
that $G$~has a self-conjugate sub-group of order~$2^{\alpha - 1}n$. This
group, again, has a self-conjugate sub-group of order~$2^{\alpha - 2}n$,
and so on. Hence $G$~has a sub-group of order~$n$. Now by
Theorem~VII (§~248), $G$~has just $n$~operations whose
%% -----File: 379.png---Folio 361-------
divide~$n$. Hence the sub-group~$H$, of order~$n$, must be self-conjugate;
and $G$~is the direct product of the two groups $\scrQ$
and~$H$.

\Par{252.} Suppose next that $\scrQ$~is not self-conjugate in~$G$; and
let~$I$, of order~$2^{\alpha}n'$, be the greatest group that contains $\scrQ$ self-conjugately.
Every operation of~$I$ is permutable with every
operation of~$\scrQ$; and therefore (§~81) $G$~contains $2^{\alpha - 1}$ distinct
sets of conjugate operations whose orders are powers of~$2$.
The case, in which $\scrQ$~is cyclical, has been already dealt with
and will now be excluded; we may thus assume that $\scrQ$~contains
$2^{\beta} - 1$ ($\beta \nless 2$) operations of order~$2$, and that $G$~contains an equal
number of sets of conjugate operations of order~$2$.

If possible, let no two sub-groups of order~$2^{\alpha}$ have a common
operation except identity. Then if two operations $S$ and~$T$,
of order~$2$, are chosen, belonging to distinct conjugate sets and
to different sub-groups of order~$2^{\alpha}$, they cannot be permutable
with each other. They will therefore generate a dihedral sub-group
of order~$2m$. If $m$~were odd, $S$~and~$T$ would be conjugate
operations. Hence $m$~must be even; and the dihedral sub-group
must contain a self-conjugate operation~$U$ of order~$2$. Since
$U$~is permutable with both $S$~and~$T$, it must occur in at least
two different sub-groups of order~$2^{\alpha}$; and therefore the supposition,
that no two sub-groups of order~$2^{\alpha}$ have a common
operation except identity, is impossible.

Let now $\scrQ'$ of order~$2^{\alpha'}$ be a sub-group common to $\scrQ$ and~$\scrQ_{1}$,
two sub-groups of order~$2^{\alpha}$; and suppose that no two sub-groups
of order~$2^{\alpha}$ have a common sub-group which contains~$\scrQ'$ and is
of greater order. Since every operation of~$\scrQ$ is self-conjugate
in~$I$, $\scrQ'$~must be self-conjugate in~$I$. If then $J$, of order~$2^{\alpha}n'r$,
is the greatest sub-group of~$G$ that contains $\scrQ'$ self-conjugately,
$J$~contains $r$~sub-groups of order~$2^{\alpha}$. Hence the factor-group
$\dfrac{J}{\scrQ'}$ contains $r$~sub-groups of order~$2^{\alpha - \alpha'}$, and no two of these
have a common sub-group; for if they had, some two sub-groups
of order~$2^{\alpha}$ contained in~$J$ would have a common sub-group
greater than and containing~$\scrQ'$, contrary to supposition. The
$r$~sub-groups of order~$2^{\alpha - \alpha'}$ of~$\dfrac{J}{\scrQ'}$ must therefore be cyclical.
%% -----File: 380.png---Folio 362-------

\Par{253.} We will apply the results of the last paragraph to
the case in which all the operations of~$\scrQ$ are of order~$2$. Then
$\scrQ'$~must be of order~$2^{\alpha - 1}$, since $\dfrac{\scrQ}{\scrQ'}$ is cyclical. Suppose now that
$R$~is an operation of~$J$ of odd order, which is permutable with~$\scrQ'$
but not with~$\scrQ$. If $R$~were self-conjugate in a sub-group
whose order is divisible by~$2^{\alpha}$, this sub-group would contain~$\scrQ'$
and therefore one or more sub-groups of order~$2^{\alpha}$ containing~$\scrQ'$.
But $R$~is not permutable with any sub-group of order~$2^{\alpha}$
that contains~$\scrQ'$; and therefore the highest power of~$2$ that
divides the order of the group in which $R$ is self-conjugate is~$2^{\alpha - 1}$,
so that $R$~is one of~$2\mu$ ($\mu$~odd) conjugate operations. If
now $A$~is an operation of order~$2$ that belongs to~$\scrQ$ and not to~$\scrQ'$,
no operation conjugate to~$R$ can be permutable with~$A$.
Hence in the substitution group of degree~$2\mu$, that results on
transforming the set of operations conjugate to~$R$ among themselves
by all the operations of~$G$, the substitution corresponding
to~$A$ is an odd substitution. This substitution group has
therefore a self-conjugate sub-group whose order is half its own,
and therefore $G$~has a self-conjugate sub-group of order~$2^{\alpha - 1}n$.
In the same way it may be shewn that this sub-group has a
self-conjugate sub-group of order~$2^{\alpha - 2}n$; and so on. Hence $G$~has
a sub-group $H$ of order~$n$. But it follows from Theorem~VII,
§~248, that $G$~has exactly $n$ operations whose orders divide~$n$;
and therefore $H$~is a self-conjugate sub-group. Moreover,
since $\dfrac{G}{H}$~is an Abelian group of order~$2^{\alpha}$, it must contain self-conjugate
sub-groups of every order~$2^{r}$ ($r = 1, 2, \Dots, \alpha - 1$);
and $G$~therefore has self-conjugate sub-groups of every order~$2^{r}n$.
In general however these sub-groups are not characteristic,
as is the case when $\scrQ$~is cyclical.

\Par{254.} We will consider next the case where $\scrQ$~is generated
by two operations $A$~and~$B$, of orders $2^{\alpha - 1}$ and~$2$. There are in~$\scrQ$
three operations of order~$2$, namely, $A^{2^{\alpha - 2}}$,~$A^{2^{\alpha - 2}}B$, and~$B$; and
in~$\scrQ$ there are no operations, of order greater than~$2$, of which
the two latter are powers. Suppose that $G$~contains an operation~$B'$,
conjugate to~$B$, with which $A^{2^{\alpha - 2}}$~is not permutable.
Then $\{A^{2^{\alpha - 2}}, B'\}$ is a dihedral group, and if $R$~is an operation
%% -----File: 381.png---Folio 363-------
of odd order of this group, no power of~$A$ is permutable with~$R$.
Since $A^{2^{\alpha - 2}}$ and $B'$ are not conjugate operations in~$G$, there
must be an operation of order~$2$, conjugate to~$A^{2^{\alpha - 2}}B$, and
permutable with~$R$. The sub-group, in which $\{R\}$ is self-conjugate,
therefore contains representatives of each of the
three sets of conjugate operations of order~$2$ that belong to~$G$;
and in this sub-group these representatives form three distinct
conjugate sets. Hence no operation conjugate to~$A^{2^{\alpha - 2}}$ can be
permutable with~$R$; and therefore no power, except identity,
of an operation conjugate to~$A$ is permutable with~$R$. Hence
$R$~is one of~$2^{\alpha - 1}\mu$ ($\mu$~odd) conjugate operations, with none of
which is $A$ permutable. It follows that, in the substitution
group of degree~$2^{\alpha - 1}\mu$, with which $G$~is isomorphic, $A$~is an odd
substitution. Hence $G$~contains a self-conjugate sub-group of
order~$2^{\alpha - 1}n$; and since this is of the same type as~$G$, the
same reasoning may be applied to it. Exactly as before, $G$~will
contain self-conjugate sub-groups of every order~$2^{r}n$
($r = 0, 1, \Dots, \alpha - 1$).

We have assumed that there is an operation~$B'$, conjugate
to~$B$, with which $A^{2^{\alpha - 2}}$ is not permutable. If $A^{2^{\alpha - 2}}$~were permutable
with every operation that is conjugate to~$B$, the
sub-group in which $A^{2^{\alpha - 2}}$ is permutable would contain a self-conjugate
sub-group~$G'$ of~$G$ whose order is divisible by~$2$. In
this case, we may deal with~$\dfrac{G}{G'}$ exactly as we have been
dealing with~$G$.

The results of §§~250--254 may be summed up in the
following form:---

\begin{Theorem}[IX.]
If in a group~$G$ of order~$2^{\alpha}n$, where $n$~is
odd, a sub-group~$\scrQ$ of order~$2^{\alpha}$ is an Abelian group of type
$(\alpha)$, $(\alpha - 1, 1)$, or $(1, 1, \Dots, 1)$, and if $\theta(\scrQ)$ and $n$ are relatively
prime, then $G$~contains self-conjugate sub-groups of each order~$2^{\beta}n$
\($\beta = 0, 1, \Dots, \alpha - 1$\).
\end{Theorem}

\Par{255.} Still representing the order of~$G$ by~$2^{\alpha}n$, where $n$~is
odd, there are two cases, in which the sub-groups of order~$2^{\alpha}$
are not Abelian, where it may be shewn without difficulty that
$G$~contains a self-conjugate sub-group of order~$n$.
%% -----File: 382.png---Folio 364-------

\begin{Theorem}[X.]
If the order of~$G$ is~$2^{\alpha}n$, where $n$~is odd and
not divisible by~$3$, and if a sub-group~$\scrQ$ of order~$2^{\alpha}$ is of the type
\begin{alignat*}{4}
&\Inum{(i)}  & A^{2^{\alpha - 1}} &= 1, & B^{2} &= A^{2^{\alpha - 2}},\quad & B^{-1}AB &= A^{-1}, \\
\lintertext{or}
&\Inum{(ii)}\quad & A^{2^{\alpha - 1}} &= 1,\quad & B^{2} &= 1, & B^{-1}AB &= A^{-1},
\end{alignat*}
{\Loosen $G$~contains a self-conjugate sub-group of each order~$2^{\beta}n$
\($\beta = 0, 1, \dots, \alpha - 1$\).}
\end{Theorem}

For each of these types, $\vartheta(\scrQ)$~is $3$~or~$1$; and if $\scrQ'$~is any
sub-group of~$\scrQ$, $\vartheta(\scrQ')$~is either $3$~or~$1$. Hence if any operation
of~$G$ of odd order is permutable with $\scrQ$~or~$\scrQ'$, it must be
permutable with every operation of $\scrQ$~or~$\scrQ'$. In each type,
the self-conjugate operations form a cyclical sub-group of order~$2$,
namely $\{A^{2^{\alpha - 2}}\}$.

We will consider first the case where $\scrQ$~is of type~\Inum{(i)}. In
this case, $\{A^{2{\alpha - 3}}\}$~and $\{B\}$ are sub-groups of~$G$ of order~$4$, the
former being self-conjugate in~$\scrQ$ while the latter is not. If
they are conjugate sub-groups in~$G$, we have seen (§~82)
that $G$~must contain an operation of odd order~$S$, such that
the sub-groups $S^{-n}\{B\}S^{n}$ ($n = 0, 1, 2, \dots$) are permutable with
each other. This is impossible, since every operation that
is permutable with a sub-group~$\scrQ'$ is permutable with all its
operations. Hence $\{A^{2^{\alpha - 3}}\}$~and $\{B\}$ are not conjugate sub-groups;
and $A^{2^{\alpha - 3}}$~and~$B$ are not conjugate operations.

Now $A^{2^{\alpha - 3}}$ is self-conjugate in~$\{A\}$ and is one of two conjugate
operations in~$\scrQ$; therefore in~$G$ it must be one of $2\mu$ conjugate
operations, where $\mu$~is odd. When these $2\mu$ conjugate
operations are transformed by~$A^{2^{\alpha - 3}}$, two only, namely, $A^{2^{\alpha - 3}}$ and
$A^{-2^{\alpha - 3}}$, remain unchanged. Let us suppose that, in the resulting
substitution, there are $x$~cycles of $2$~symbols and $y$~cycles of $4$~symbols.

When the $2\mu$ conjugate operations are transformed by~$B$,
none can remain unchanged; and we may suppose that the
resulting substitution has $x'$~cycles of $2$~symbols and $y'$~cycles of
$4$~symbols.

Now since
\begin{gather*}
A^{2^{\alpha - 2}} = B^{2}, \\
y = y',
\end{gather*}
and therefore
\[
1 + x = x'.
\]
%% -----File: 383.png---Folio 365-------

Hence one of the two substitutions $A^{2^{\alpha - 3}}$ and $B$ must be odd;
$G$~has therefore a self-conjugate sub-group of order~$2^{\alpha - 1}n$. The
groups of order~$2^{\alpha - 1}$ contained in this self-conjugate sub-group
are either of type~\Inum{(i)}, with $\alpha - 1$ written for~$\alpha$: or they are
cyclical; for, like~$\scrQ$, they can only contain a single operation of
order~$2$. If they are cyclical, the reasoning of §~250 may
be applied; and if they are of type~\Inum{(i)} the same reasoning
will apply to the self-conjugate sub-group of order~$2^{\alpha - 1}n$
that has been used for~$G$. Finally, then, $G$~must contain
sub-groups of orders $n$,~$2n$, $2^{2}n$,~\dots, $2^{\alpha - 1}n$, each of which is
contained self-conjugately in the next; and therefore the sub-group~$H$
of order~$n$ must be self-conjugate.

If $\scrQ$~is of type~\Inum{(ii)}, $A$~is one of $2\mu$ conjugate operations
in~$G$, where $\mu$~is odd. It may be shewn, as in the previous
case, that $B$~and $A^{2^{\alpha - 2}}$ cannot be conjugate in~$G$, so that no one
of the operations conjugate to $A$~has~$B$ for one of its powers.
If now $B$~were permutable with any one of these $2\mu$~conjugate
operations, the group would contain an Abelian sub-group of
order~$2^{\alpha}$, which is not the fact. Hence the substitution, given
on transforming the $2\mu$~operations by~$B$, consists of $\mu$~transpositions
and is an odd substitution. Therefore $G$~contains a self-conjugate
sub-group of order~$2^{\alpha - 1}n$, in which the sub-groups of
order~$2^{\alpha - 1}$ are cyclical. Hence, again, there is a self-conjugate
sub-group of order~$n$.

\Par{256.} The only non-Abelian groups of order~$2^{3}$ are those of
types \Inum{(i)}~and~\Inum{(ii)} of §~255, when $\alpha = 3$. The Abelian groups
of order $2^{2}$~and~$2^{3}$ are all included in the types considered in
Theorem~IX, §~254. For an Abelian group~$\scrQ$, of order~$2^{3}$ and
type~$(2, 1)$, $\theta(\scrQ)$~is~$3$; and for one of order~$2^{3}$ and type $(1, 1, 1)$,
$\theta(\scrQ)$~is~$21$. Hence Theorems IX~and~X shew that, if $2$,~$2^{2}$
or~$2^{3}$ divide the order of a group, but not~$2^{4}$, then the
operations of odd order form a self-conjugate sub-group, with
possible exceptions when $12$~or $56$ is a factor of the order.
Hence:---

\begin{Theorem}[XI.]
A group of even order cannot be simple
unless $12$,~$16$ or~$56$ is a factor of the order\footnotemark.
\end{Theorem}
\index{Burnside@\Topic{Burnside}, quoted}%
\footnotetext{Burnside, \textit{Proc.\ L.~M.~S.} Vol.~\VolNo{XXVI}, (1895), p.~332.}%
%% -----File: 384.png---Folio 366-------
\index{Composite groups@\Topic{Composite groups}!of even order|)}%

\begin{Remark}
\Par{257.} In further illustration of the methods of the preceding
paragraphs, we will deal with another case in which the sub-groups
of order~$p^{\alpha}$ are not Abelian.

{\Loosen Let $G$~be a group of order~$p^{\alpha}m$, where $m$~is relatively prime to
$p(p - 1) \Dots (p^{\alpha} - 1)$; and suppose that a sub-group~$H$ of order~$p^{\alpha}$ is
such that within it every operation is either self-conjugate or one
of $p$ conjugate operations\footnotemark.}
\PageLabel{366}%
\footnotetext{The reader will easily verify that, when this condition is satisfied, $\dfrac{H}{\Gh}$
  is Abelian, $\Gh$~being the sub-group formed by the self-conjugate operations of~$H$.}%
Let $h$ be a sub-group of~$G$ whose order
is a power of~$p$; and let $S$ be an operation of~$G$ whose order is not
divisible by~$p$. Then since $m$~is relatively prime to $p(p - 1) \Dots
(p^{\alpha} - 1)$, if $S$~is permutable with~$h$, it is permutable with every
operation of~$h$. Hence it follows from §~82 that no operation of~$H$,
which is not self-conjugate in~$H$, can be conjugate in~$G$ to
a self-conjugate operation of~$H$. Suppose now that $S_{1}$~and $S_{2}$ are
two operations of~$H$, each of which in~$H$ is one of $p$ conjugate
operations; and that while $S_{1}$~and $S_{2}$ are conjugate in~$G$, they
are not conjugate in~$I$, the greatest group that contains $H$ self-conjugately.
Let $H_{1}$~and $H_{2}$ be the sub-groups of~$H$, of orders\DPnote{** [sic]}~$p^{\alpha - 1}$,
which contain $S_{1}$ and $S_{2}$ self-conjugately. Since $S_{1}$ and $S_{2}$
are conjugate in~$G$, all the sub-groups of order~$p^{\alpha - 1}$ which contain
either of them self-conjugately belong to the same conjugate
set. If then $H_{1}$ and $H_{2}$ are identical, there must be an operation,
of order prime to~$p$, which will transform $S_{1}$ into~$S_{2}$ and $H_{1}$
into itself. This, we have seen, is impossible. If $H_{1}$~and $H_{2}$ are
not identical, there must be an operation which will transform $S_{1}$
into $S_{2}$ and $H_{1}$~into $H_{2}$. Now $H_{1}$ and $H_{2}$ both contain the sub-group~$\Gh$
formed of the self-conjugate operations of~$H$; and since no self-conjugate
operation of~$H$ is conjugate in~$G$ with any operation of~$H$
which is not self-conjugate, the operation in question, which
transforms $S_{1}$ into $S_{2}$ and $H_{1}$~into $H_{2}$, must transform $\Gh$ into itself.
Hence $S_{1}$ and $S_{2}$ are conjugate operations in that sub-group,~$G'$, of~$G$
which contains $\Gh$ self-conjugately. Now in $\dfrac{G'}{\Gh}$ the operations, that
correspond to $S_{1}$~and~$S_{2}$, are self-conjugate operations of~$\dfrac{H}{\Gh}$. Since
then these operations are conjugate in~$\dfrac{G'}{\Gh}$, they must (§~82) be
conjugate in~$\dfrac{I}{\Gh}$. This however is impossible, since every operation
of~$\dfrac{I}{\Gh}$ whose order is a power of~$p$ is self-conjugate. Finally, then, no
two operations of~$H$, which are not conjugate in~$H$, can be conjugate
in~$G$; and the number of distinct sets of conjugate operations in~$G$,
whose orders are powers of~$p$, is equal to the number in~$H$. From this
%% -----File: 385.png---Folio 367-------
it follows, as in previous cases, that the number of operations of~$G$
whose orders divide~$m$ is equal to~$m$.
\end{Remark}

\Par{258.} \begin{Theorem}[XII.]
The only simple groups, whose orders
are the products of four or of five primes, are groups of orders
$60$,~$168$, $660$, and~$1092$: and no group, whose order contains
less than four prime factors, is simple.
\end{Theorem}

Groups, whose orders are $p_{1}$,~$p_{1} p_{2}$, $p_{1}^{2}$, $p_{1}^{3}$, $p_{1}^{2} p_{2}$, $p_{1} p_{2}^{2}$ or~$p_{1} p_{2} p_{3}$
are all proved to be soluble by previous theorems in the present
chapter.

If the order of a group contains four prime factors, it must
be of one of the forms $p_{1}^{4}$,~$p_{1}^{3} p_{2}$, $p_{1}^{2} p_{2}^{2}$, $p_{1}^{2} p_{2} p_{3}$, $p_{1} p_{2}^{3}$, $p_{1} p_{2}^{2} p_{3}$,
$p_{1} p_{2} p_{3}^{2}$ or~$p_{1} p_{2} p_{3} p_{4}$. If $p_{1}$~is an odd prime, groups of any one of
these orders have already been shewn to be soluble; while if $p_{1}$~is~$2$,
the only case which can give a simple group is~$2^{2} p_{2} p_{3}$. If
a group of this order is simple, it follows from Theorem~VIII,
Cor.\ (§~249), that $p_{2}$~must be~$3$; and the order of the group~$12p_{3}$.
A cyclical sub-group of order~$p_{3}$ must be one of $1 + kp_{3}$ conjugate
sub-groups. Hence $1 + kp_{3}$ must be a factor of~$12$, so that $p_{3}$~is
either $5$~or~$11$. If $p_{3}$~were~$11$, the $12$~conjugate sub-groups of
order~$11$ would contain $120$~distinct operations of order~$11$, and
the tetrahedral sub-group, which the group (if simple) must
contain, would be self-conjugate. Hence $p_{3}$~must be~$5$ and the
order of the group is~$60$. We have already seen that a simple
group of order~$60$ actually exists, namely, the icosahedral group;
and that there is only one type for such a group (§~85).

If the order of a group contains five prime factors, it must
be of one of the forms:---
\begin{gather*}
p_{1}^{5},\ p_{1}^{4} p_{2},\ p_{1}^{3} p_{2}^{2},\ p_{1}^{3} p_{2} p_{3},\ p_{1}^{2} p_{2}^{3},\ p_{1}^{2} p_{2}^{2} p_{3},\ p_{1}^{2} p_{2} p_{3}^{2},\ p_{1}^{2} p_{2} p_{3} p_{4},\ p_{1} p_{2}^{4}, \\
p_{1} p_{2}^{3} p_{3},\ p_{1} p_{2}^{2} p_{3}^{2},\ p_{1} p_{2}^{2} p_{3} p_{4},\ p_{1} p_{2} p_{3}^{3},\ p_{1} p_{2} p_{3}^{2} p_{4},\ p_{1} p_{2} p_{3} p_{4}^{2},\ p_{1} p_{2} p_{3} p_{4} p_{5}.
\end{gather*}

If $p_{1}$~is an odd prime, it follows from previous theorems that
none of these forms, except $p_{1}^{3} p_{2} p_{3}$ and $p_{1} p_{2}^{3} p_{3}$, can give simple
groups.

Taking first the order~$p_{1} p_{2}^{3} p_{3}$, let us suppose (without limitation
to the particular case) that a group of order~$p_{1} p_{2}^{m} p_{3}$ is
simple. It must contain $p_{3}$~or $p_{1} p_{3}$ conjugate sub-groups of
order~$p_{2}^{m}$. In either case, the operations of these sub-groups
must be all distinct, or else they must all have a common
%% -----File: 386.png---Folio 368-------
\index{Simple groups@\Topic{Simple groups}!whose orders are the products of not more than $5$~primes|(}%
sub-group, for the order of the group contains no factor
congruent to unity $(\mod p_{2})$, except $p_{3}$~or $p_{1} p_{3}$. Now the group
contains (§~248) just $p_{2}^{m} p_{3}$ operations whose orders are not
divisible by~$p_{1}$; and if there are $p_{3}$~sub-groups of order~$p_{2}^{m}$
whose operations are all distinct, there are $(p_{2}^{m} - 1)p_{3}$ operations
whose orders are powers of~$p_{2}$. Hence, in this case, a sub-group
of order~$p_{3}$ is self-conjugate. On the other hand, if there are
$p_{1} p_{3}$~sub-groups of order~$p_{2}^{m}$, their operations cannot be all
distinct; and the group has a self-conjugate sub-group whose
order is a power of~$p_{2}$. A repetition of this reasoning shews
that a group of order~$p_{1} p_{2}^{m} p_{3}$ is always soluble.

We consider, next, a group whose order is~$p_{1}^{3} p_{2} p_{3}$. If $p_{1}$~is
odd, $p_{1}^{2}$~cannot be congruent to unity, $\pmod{p_{2}}$ or $\pmod{p_{3}}$. The
same is true, when $p_{1} = 2$, if $p_{2}$~is not equal to~$3$. We shall
therefore first deal with a group of order~$p_{1}^{3} p_{2} p_{3}$ on the supposition
that $p_{2}$~is not~$3$.

If neither $p_{2}$~nor $p_{3}$ divides~$p_{1}^{3} - 1$, it follows from §§~248,~257
that there are just $p_{2} p_{3}$ operations in the group whose orders
divide~$p_{2} p_{3}$. When this is the case, the group clearly cannot
be simple.

If $p_{3}$~divides $p_{1}^{3} - 1$, and if there are more than $p_{2} p_{3}$
operations whose orders divide~$p_{2} p_{3}$, a sub-group of order~$p_{1}^{3}$
must be self-conjugate in a sub-group of order~$p_{1}^{3} p_{3}$, so that the
group is isomorphic with a group of degree~$p_{2}$. In this case,
again, the group cannot be simple.

Finally, then, we have only to deal with the case in which a
sub-group of order~$p_{1}^{3}$ is self-conjugate in a sub-group of order~$p_{1}^{3} p_{2}$,
while in this sub-group an operation of order~$p_{2}$ is
\PageLabel{368}%
one of $p_{1}^{3}$ conjugate operations. Now the congruences
\begin{align*}
p_{1}^{3} &\equiv 1\ (\mod p_{2}), \\
p_{2}     &\equiv 1\ (\mod p_{1}),
\end{align*}
are inconsistent; therefore, if a sub-group of order~$p_{2}$ is self-conjugate
in a sub-group of order~$p_{1}^{\alpha} p_{2}$, the latter must be
Abelian. Hence the group contains $p_{1}^{3}(p_{2} - 1)p_{3}$ operations
whose orders are divisible by~$p_{2}$. Now if the sub-groups of
order~$p_{1}^{3}$ have a common sub-group, it must be common to all of
the sub-groups of order~$p_{1}^{3}$, and it is a self-conjugate sub-group.
%% -----File: 387.png---Folio 369-------
If, however, no two have a common sub-group, the group
contains $(p_{1}^{3} - 1)p_{3}$ operations of order~$p_{1}$; and there remain
only $p_{3}$~operations whose orders divide~$p_{3}$. The group is
therefore in any case composite.

If now $p_{2} = 3$, a group of order~$2^{3}3p_{3}$ has, by Sylow's theorem,
a self-conjugate sub-group of order~$p_{3}$ unless $p_{3}$~is $5$,~$7$,~$11$ or~$23$.
If $p_{3} = 23$, the group (if simple) would have just $24$~operations
whose orders divide~$24$; it is thence easily seen to be non-existent.
If $p = 11$, the group (if simple) could be expressed as
a doubly transitive group of degree~$12$. In this form, however,
each operation, which transforms an operation of order~$11$ into
its own inverse, would be a product of $5$~transpositions and
therefore an odd substitution. This group therefore cannot be
simple. If $p_{3} = 5$, the group could be expressed as a doubly
transitive group of degree~$6$. The sub-group of order~$4$ which
transforms a sub-group of order~$5$ into itself would be cyclical;
and the corresponding substitution being odd, the group could
not be simple.

Hence, finally, the only possibility is a group of order~$168$.
That a simple group of this order actually exists is shewn in
§~146; also, there is only one type of such group.

When $p_{1}$~is equal to~$2$, the only case, that requires discussion
in addition to those we have dealt with, is $p_{1}^{2} p_{2} p_{3} p_{4}$.
A group of this order can only be simple (§~249) when it
contains a tetrahedral sub-group. In this case, the operations
of order~$2$ form a single conjugate set, and the group contains
just $3p_{3} p_{4}$ operations whose orders are divisible by~$2$. If a sub-group
of order~$2^{\alpha}3q$ is the greatest that contains a sub-group of
order~$3$ self-conjugately, the group will contain $2^{2 - \alpha}(3 - 1)p_{3}p_{4}$
operations whose orders are divisible by~$3$ and not by~$2$. Hence
the group must contain either $p_{3}p_{4}$,~$5p_{3}p_{4}$, or~$7p_{3}p_{4}$, operations
whose orders divide~$p_{3}p_{4}$. On the other hand, the number of
these operations may be expressed in the form
\[
x(p_{3} - 1)p_{4} + y(p_{4} - 1) + 1,
\]
where $x$~is a factor of~$12$, and $y$~is a factor of~$12p_{3}$ which is
congruent to unity, $\pmod{p_{4}}$.

Hence
\[
x(p_{3} - 1)p_{4} + y(p_{4} - 1) = zp_{3}p_{4} - 1,\quad (z = 1, 5, 7).
\]
%% -----File: 388.png---Folio 370-------

The case $z = 1$ leads to $y = 1$, so that the sub-group of order~$p_{4}$
\index{Simple groups@\Topic{Simple groups}!whose orders do not exceed~$660$|(}%
is self-conjugate.

The case $z = 5$ gives no solution; but when $z = 7$, it will be
found that there are two solutions, namely,
\begin{alignat*}{4}
x &= 6,\quad &y &= 12,\quad &p_{3} &= 5,\quad &p_{4} &= 11, \\
\lintertext{and}
x &= 6,\quad &y &= 14,\quad &p_{3} &= 7,\quad &p_{4} &= 13.
\end{alignat*}

That simple groups actually exist corresponding to the
two orders $660$ and $1092$ thus arrived at, is shewn in §~221.
There is also in each case a single type; the verification
of this statement is left to the reader.

\Par{259.} As has already been stated at the beginning of the
present chapter, the solution of the general problem of pure
group-theory, namely, the determination of all possible types of
group of any given order, depends essentially on the previous
determination of all possible simple groups. A complete solution
of this latter problem is not to be expected; but for orders, which
do not exceed some given limit, the problem may be attacked
directly. The first determination of this kind was due to
Herr O.~Hölder\footnotemark,
\index{Holder@\Topic{Hölder}, quoted}%
\footnotetext{\textit{Math.\ Ann.}\ Vol.~\VolNo{XL}, (1892), pp.~55--88.}%
who examined all possible orders up to~$200$:
he proved that none of them, except $60$~and $168$, correspond to
a simple group. Mr~Cole\footnote
  {\textit{American Journal of Mathematics}, Vol.~\VolNo{XV}, (1893), pp.~303--315.}
continued the investigation, by
\index{Cole@\Topic{Cole}, quoted}%
examining all orders from $201$ to~$660$, with the result of shewing
that in this interval the only orders, which have simple groups
corresponding to them, are $360$,~$504$ and~$660$. The existence
of a simple group of order~$504$ had not been recognised
before Mr~Cole's investigation.

The author\footnote
  {\textit{Proc.\ London Math.\ Soc.}, Vol.~\VolNo{XXVI}, (1895), pp.~333--338.}
has carried on the examination from $661$ to~$1092$,
\index{Burnside@\Topic{Burnside}, quoted}%
with the result of shewing that $1092$~is the only number
in this interval which is the order of a simple group.

As the limit of the order is increased, such investigations
as these rapidly become more laborious, since a continually increasing
number of special cases have to be dealt with. There
is little doubt however but that, with the aid of the theorems
proved in the present chapter, the investigation might be
continued without substantial difficulty up to~$2000$.
%% -----File: 389.png---Folio 371-------

We shall here be content with verifying the result of Herr
Hölder's and Mr~Cole's investigations for orders up to~$660$. The
method used in dealing with particular numbers may suggest
to the reader how the determination might be continued.

\Par{260.} It follows from Theorem~XII that, if an odd number
can be the order of a simple group, it must be the product of at
least $6$~prime factors. Moreover, by previous theorems, we have
seen that $3^{6}$,~$3^{5}p$ and~$3^{4}p^{2}$ cannot be orders of simple groups.
Hence certainly no odd number less than $3^{4} · 5 · 7$ or~$2835$\footnote
  {It is not, of course, here suggested that $2835$~is the order of a simple
  group. It may, in fact, be shewn with little difficulty that no number of the
  form $3^{4}pq$,~$3^{3}p^{2}q$, or $3^{2}p^{3}q$, where $p$~and~$q$ are odd primes, can be the order of a
  simple group. A much higher limit than~$2835$ may therefore be given for the
  order, if odd, of a simple group.}
can be
the order of a simple group. Therefore, by Theorem~XI, we need
only examine numbers less than~$660$ which are divisible by
$12$,~$16$, or~$56$. When each number less than~$660$ which is
divisible by $12$,~$16$, or~$56$ is written in the form $p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}}$\dots, it
will be found that, with eleven exceptions, Theorems II to
XII of the present chapter immediately shew that there are no
simple groups corresponding to them. The exceptions are $60$,~$168$,
$240$, $336$, $360$, $480$, $504$, $528$, $540$, $560$,~$660$. We will first
deal with such of these numbers as do not actually correspond
to simple groups.

\Subsection{$240 = 2^{4} · 3 · 5$.}

%[** TN: No indentation in the original]
A simple group of this order would contain $6$~or $16$ sub-groups
of order~$5$. If it contained $6$~sub-groups of order~$5$, the group
could be expressed as a transitive substitution group of degree~$6$;
and there is no group of degree~$6$ and order~$240$. If there
were $16$~sub-groups of order~$5$, each would be self-conjugate in
a group of order~$15$, which must be cyclical. The group then
would be a doubly transitive group of degree~$16$ and order
$16 · 15$. Such a group (§~105) contains a self-conjugate sub-group
of order~$16$.

\Subsection{$336 = 2^{4} · 3 · 7$.}

%[** TN: No indentation in the original]
A simple group of this order would contain $8$~sub-groups of
order~$7$, each self-conjugate in a group of order~$42$. We have
seen (§~146) that there is no simple group of degree~$8$ and order
$8 · 7 · 6$.
\index{Simple groups@\Topic{Simple groups}!whose orders are the products of not more than $5$~primes|)}%
%% -----File: 390.png---Folio 372-------

\Subsection{$480 = 2^{5} · 3 · 5$.}

Since $2^{5}$~is not a factor of~$5!$, a group of order~$480$ must, if
simple, contain $15$~sub-groups of order~$2^{5}$. Now $2^{5}$~is not congruent
to unity, $\pmod{4}$, and therefore (§~78) some two sub-groups
of order~$2^{5}$ must have a common sub-group of order~$2^{4}$. Such
a sub-group, of order~$2^{4}$, must (§~80) either be self-conjugate, or
it must be contained self-conjugately in a sub-group of order
$2^{5} · 3$~or $2^{5} · 5$. In either case the group is composite.

\Subsection{$528 = 2^{4} · 3 · 11$.}

There must be $12$~sub-groups of order~$11$, each being
self-conjugate in a group of order~$44$. A group of order~$44$
necessarily contains operations of order~$22$, and such an operation
cannot be represented as a substitution of $12$~symbols.
The group is therefore not simple.

\Subsection{$540 = 2^{2} · 3^{3} · 5$.}

There must be $10$~sub-groups of order~$3^{3}$, each self-conjugate
in a group of order~$2 · 3^{3}$. There must also be $36$~sub-groups of
order~$5$, each self-conjugate in a group of order~$15$. Now when
the group is expressed as transitive in $10$~symbols, the substitutions
of order~$5$ consist of $2$~cycles of $5$~symbols each, and
no such substitution can be permutable with a substitution of
order~$3$. Hence the group is not simple.

\Subsection{$560 = 2^{4} · 5 · 7$.}

There must be $8$~sub-groups of order~$7$. A transitive group
of degree~$8$ and order~$8 · 7 · 10$ does not exist (§~146).

That there are actually simple groups of orders $60$,~$168$, $360$,
and~$660$, has already been seen; the verification that there is
only a single type of simple group of order~$360$ may be left
to the reader. It remains then to consider the order~$504$.

\Subsection{$504 = 2^{3} · 3^{2} · 7$.}

A simple group of this order must contain $8$~or $36$ sub-groups
of order~$7$. There is no group of degree~$8$ and order~$504$ (§~146).
Hence there must be $36$~sub-groups of order~$7$.
%% -----File: 391.png---Folio 373-------

Again, there must be $7$~or $28$ sub-groups of order~$3^{2}$; and
since there is no group of degree~$7$ and order~$504$ (\lc), there
are $28$~sub-groups of order~$3^{2}$. If two of these have a common
operation~$P$ of order~$3$, it must (§~80) be self-conjugate in a group
containing more than one sub-group of order~$3^{2}$. If the order
of this sub-group is~$2^{2} · 3^{2}$, it must contain a sub-group of order
$2^{2}$ self-conjugately; and therefore this sub-group of order~$2^{2}$ must
be self-conjugate in a group of order~$2^{3} · 3^{2}$ at least. Such a
group would be one of $7$~conjugate sub-groups, and this is
impossible. Similarly, $P$~cannot be self-conjugate in a group
whose order is greater than~$2^{2} · 3^{2}$. Hence no two sub-groups
of order~$3^{2}$ have a common operation, except identity.

There are therefore in the group $216$~operations of order~$7$,
and $224$~operations whose orders are powers of~$3$, leaving just
$64$ other operations.

Suppose now that $A$~is an operation of order~$2$, self-conjugate
in a sub-group of order~$2^{3}$. If $A$~is self-conjugate in a
group of order~$2^{3}x$ ($x$~being a factor of~$3^{2} · 7$), this group must
contain at least $x$~operations of odd order and therefore at least $x$~operations
of the form~$AS$, where $S$~is an operation of odd order
permutable with~$A$. There are also $x$~operations of this form,
corresponding to each of the $\dfrac{63}{x}$ operations conjugate to~$A$;
so that the whole group contains $63$~operations of the form~$AS$,
where $S$~is of odd order and permutable with~$A$, while $A$~is one
of a set of conjugate operations of order~$2$, each of which is self-conjugate
in a sub-group of order~$2^{3}$. Hence taking the identical
operation with these $63$~operations of even order, all the
operations of the group are accounted for; and there are therefore
no operations whose orders are powers of~$2$ except those of
the conjugate set to which $A$~belongs. The sub-groups of order~$2^{3}$
are therefore Abelian groups whose operations are all of order~$2$;
and since the $7$~operations of order~$2$, in a group of order~$2^{3}$,
are all conjugate in the group of order~$504$, they must (§~81) be
conjugate in the sub-group within which the group of order~$2^{3}$
is self-conjugate. Hence, finally, there must be $9$~sub-groups of
order~$2^{3}$, each self-conjugate in a group of order~$2^{3} · 7$, in which
the $7$~operations of order~$2$ form a single conjugate set.
%% -----File: 392.png---Folio 374-------

A simple group of order~$504$, if it exists, must therefore be
expressible as a triply transitive group of degree~$9$, in which
the sub-group of order~$56$ that keeps one symbol fixed is of
known type. From this, several inferences can be drawn.
Firstly, all the $63$~operations of order~$2$ of the $9$~sub-groups
of order~$2^{3}$ must be distinct, since each operation of order~$2$
keeps just one symbol fixed. There is therefore a conjugate
set of $63$~operations of order~$2$, and there are no other operations
of even order in the group. Secondly, the $224$~operations whose
orders are powers of~$3$ cannot all be operations of order~$3$. For,
if they were, there would be $8$~operations of order~$3$ containing
any three given symbols in a cycle; and therefore the group
would contain operations other than identity which keep $3$~symbols
fixed. This is not the fact; the sub-groups of order~$3^{2}$
are therefore cyclical.

Conversely, we may now shew that such a triply transitive
group of degree~$9$, if it exists, is certainly simple. The distribution
of its operations in conjugate sets has, in fact, been
determined as follows. There are:---
\begin{align*}
&3 \text{ conjugate sets of operations of order } 7 \text{ each containing } 72 \text{ operations} \\
&3 \Ditto{ conjugate sets of } \Ditto{operations of order } 9 \Ditto{ each containing } 56 \Ditto{ operations} \\
&1 \Ditto{ conjugate sets of } \Ditto{operations of order } 3 \Ditto{ each containing } 56 \Ditto{ operations} \\
&1 \Ditto{ conjugate sets of } \Ditto{operations of order } 2 \Ditto{ each containing } 63 \Ditto{ operations}
\end{align*}
A self-conjugate sub-group, that contains a single operation
of order~$7$, must contain all of them; and such a sub-group,
that contains a single operation of order~$9$, must contain all
the operations of orders $3$~and~$9$. Hence if a self-conjugate sub-group
contains operations of order~$9$, its order must be of the form
\[
1 + 224 + 216x + 63y;
\]
and if it contains no operations of order~$9$, its order must be of
the form
\[
1 + 56x' + 216y' + 63z';
\]
each of the symbols $x$,~$y$, $x'$,~$y'$,~$z'$ being either zero or unity.
The order must at the same time be a factor of~$504$. A very
brief consideration will shew that all these conditions cannot be
satisfied, and therefore that no self-conjugate sub-group exists.

That there is a triply transitive group of degree~$9$ and order
$9 · 8 · 7$, has already been seen in §~113; and the actual formation
%% -----File: 393.png---Folio 375-------
of the substitutions generating this group will verify that it
satisfies the conditions just obtained.

We may however very simply construct the group by the
method of §~109; and this process has the advantage of shewing
at the same time that there is only one type.

There is no difficulty in constructing the sub-group of order~$56$
as a doubly transitive group of $8$ symbols. If we denote by
$s$~and~$t$ the two substitutions
\[
(1254673)\quad \text{and}\quad (12)(34)(56)(78),
\]
$\{s, t\}$ is, in fact, a group of the desired type. If now $A$~is any
substitution of order~$2$ in the symbols $1$, $2$,~\Dots, $8$,~$9$ which
contains the transposition~$(89)$, and satisfies the conditions
\[
AsA = S_{1},\quad AtA = S_{2}AS_{3},
\]
where $S_{1}$,~$S_{2}$,~$S_{3}$ belong to~$\{s, t\}$: then (§~109) it follows that
$\{A, s, t\}$ is a triply transitive group of degree~$9$ and order~$504$.

Now every operation of order~$2$ in the group, which interchanges
$8$~and~$9$, must transform~$s$ into its inverse. Also from
the $9$~symbols only $7$~operations of order~$2$, satisfying these
conditions, can be formed; and if one of them belongs to the
group, they must all belong to it. Hence there cannot be more
than one type of group satisfying the conditions.

Finally, if
\[
A = (89)(23)(46)(57),
\]
then
\[
AsA = s^{-1},
\]
and $AtA$~can be expressed in the form
\[
(1635842) A (1685437),
\]
where the two operations $(1635842)$ and $(1685437)$ belong
to $\{s, t\}$. It follows, then, that $\{A, s, t\}$ is a triply transitive
group of degree~$9$ and order~$504$, which satisfies all the conditions,
and that there is only a single type of such group.
\index{Simple groups@\Topic{Simple groups}!whose orders do not exceed~$660$|)}%

\Par{261.} The order of a non-soluble group must be divisible by
the order of some non-cyclical simple group. Moreover, if $N$~is
the order of a non-cyclical simple group and if $n$~is any other
number, there is always at least one non-soluble group of order~$Nn$;
for, $G$~being a simple group of order~$N$ and $H$~any group
of order~$n$, the direct product of $G$~and~$H$ is a non-soluble group
%% -----File: 394.png---Folio 376-------
of order~$Nn$. Herr Hölder\footnote
  {``Bildung zusammengesetzter Gruppen,'' \textit{Math.\ Ann.}\ Vol.~\VolNo{XLVI}., (1895),
  pp. 321--422; in particular, p.~420.}
\index{Holder@\Topic{Hölder}, quoted}%
\index{Composite groups@\Topic{Composite groups}!non-soluble|(}%
has determined all distinct types of
non-soluble groups whose orders are less than~$480$. He shews
that, besides the three simple groups of orders $60$,~$168$, and
$360$, there are $22$~non-soluble groups whose orders are given by
the following table.
\[
\begin{array}{|*{7}{c|}}
\hline
\Strut
120 &  180  & 240  & 300  & 336 & 360 & 420 \\
\hline
\Strut
 3  &   1   &  8   &  1   &  3  &  5  &  1 \\
\hline
\end{array}
\]

For the proof of all these results except the first, and for
the very interesting and suggestive methods that lead to them,
the reader is referred to Herr Hölder's memoir. The case
of a non-soluble group of order~$120$ is susceptible of simple
treatment; and we will consider it here as exemplifying how
composite groups may be constructed when their factor-groups
and the isomorphisms of the latter are completely known.

\begin{Remark}
\Par{262.} The composition-factors of a non-soluble group~$G$ of order~$120$
are $2$~and~$60$. If these may be taken in either order, then $G$~contains
a self-conjugate sub-group of order~$2$ and a self-conjugate
sub-group of order~$60$. The latter, being simple, must be an
icosahedral group and cannot contain the former. Hence, in this
case, $G$~must (§~34) be the direct product of an icosahedral group
and a group of order~$2$.

Next, suppose that the composition-factors can only be taken in
the order $2$,~$60$. Then $G$~contains a self-conjugate sub-group~$H$ of
icosahedral type and no self-conjugate sub-group of order~$2$. Hence
if $S$~is any operation of~$G$ which is not contained in~$H$, the
isomorphism of~$H$ which arises on transforming its operations by~$S$
must (§~165) be contragredient. It follows that $G$~is simply isomorphic
with a group of isomorphisms of~$H$. Now $H$~is isomorphic
with the alternating group of degree five, and its group of isomorphisms
is therefore (§~173) simply isomorphic with the symmetric
group of degree five. Hence, in this case, $G$~is simply isomorphic
with the symmetric group of degree five.

Lastly, suppose that the composition-factors can only be taken
in the order $60$,~$2$. Then $G$~has a self-conjugate sub-group of order~$2$
and no self-conjugate sub-group of order~$60$. Hence (§~35) $G$~has
%% -----File: 395.png---Folio 377-------
no sub-group of order~$60$. Suppose that $A$~is the self-conjugate
operation of order~$2$, and arrange the operations of~$G$ in the sets
\[
1,\ A;\quad S_{1},\ S_{1}A;\quad S_{2},\ S_{2}A;\ \Dots;\quad S_{59},\ S_{59}A.
\]
Since $\dfrac{G}{\{A\}}$ is simply isomorphic with an icosahedral group, it must
(§~203) be possible to choose three sets
\[
S',\ S'A;\quad S'',\ S''A;\quad S''',\ S'''A;
\]
such that either operation of the first, multiplied by either operation
of the second, belongs to the third set; while at the same time the
cube of either operation of the first set, the fifth power of either
operation of the second set, and the square of either operation of the
third set, belong to the set $1$,~$A$. Now if
\[
S'^{3} = 1,
\]
then
\[
(S'A)^{3} = A.
\]
Hence we may assume, without loss of generality, that
\[
S'^{3} = 1,\quad S''^{5} = 1,\quad S'S'' = S''' \quad\text{or}\quad S'''A.
\]

If $S'''$~is of order~$2$, so also is~$S'''A$, and $G$~would then contain an
icosahedral sub-group. Hence
\[
S'''^{4} = 1,
\]
and
\[
S'''^{2} = A.
\]

Now
\[
S_{1}^{3} = 1,\quad S_{2}^{5} = 1,\quad (S_{1}S_{2})^{2} = 1,
\]
is a complete set of defining relations for the icosahedral group; so
that, from the symbols $S_{1}$~and~$S_{2}$, exactly $60$ distinct products can be
formed. Hence from $S'$,~$S''$, and~$A$, where
\[
S'^{3} = 1,\quad S''^{5} = 1,\quad (S'S'')^{2} = A,\quad A^{2} = 1,
\]
$A$~being permutable both with $S'$ and~$S''$, exactly $120$ distinct
products can be formed. It follows that the relations just given are
the defining relations of a non-soluble group of order~$120$, which has
a self-conjugate sub-group of order~$2$ and no sub-group of order~$60$;
and that there is only one type of such group. We have already
seen (§~221) that there must be at least one such type of group; viz.\
the group defined by
\[
\begin{aligned}
x_{1}' &\equiv \alpha x_{1} + \beta x_{2}, \\
x_{2}' &\equiv \gamma x_{1} + \delta x_{2}, \\
\alpha \delta - \beta \gamma &\equiv 1,
\end{aligned}
\quad (\mod 5).
\]
\end{Remark}

\begin{Ex} If $p$~is an odd prime, shew that a group, whose composition-factors
are $60$~and~$p$, must be the direct product of an icosahedral
group and a group of order~$p$. \Attrib{Hölder.}
\end{Ex}

\Par{263.} Let $G$~be a composite group; and suppose that the
factor-groups of the composition-series of~$G$ are the two non-cyclical
simple groups $H'$~and~$H$ in the order given. Then
%% -----File: 396.png---Folio 378-------
(§~165) if $H_{1}$~is the group formed of all the operations of~$G$
which are permutable with every operation of~$H$, the direct
product $\{H, H_{1}\}$ of $H$~and $H_{1}$ is a self-conjugate sub-group of~$G$.
Now the composition-factors of $\{H, H_{1}\}$ are composition-factors
of~$G$. Hence $H_{1}$~must either be isomorphic with~$H'$, or it must
reduce to the identical operation. The latter alternative can
only occur if $\dfrac{L}{H}$~has a sub-group simply isomorphic with~$H'$,
$L$~being the group of isomorphisms of~$H$. In the cases of the
simple groups whose groups of isomorphisms have been investigated
in Chapter~XI, \ie\ the alternating groups, the doubly
transitive groups of degree~$p^{n} + 1$ and order $\frac{1}{2}p^{n}(p^{2n} - 1)$, and the
triply transitive groups of degree~$2^{n} + 1$ and order $2^{n}(2^{2n} - 1)$,
$\dfrac{L}{H}$~was found to be an Abelian group. (These groups include
all the simple groups whose orders do not exceed~$660$.) If $H$~and
$H'$ belong to these classes of groups, then $G$~must be the
direct product of simple groups simply isomorphic with $H$~and~$H'$.
For instance, a group of order~$3600$, whose composition-factors
are $60$~and~$60$, must be the direct product of two icosahedral
groups.

This result may clearly be extended to the case of a group,
the factor-groups of whose composition-series are a number of
non-cyclical simple groups $H_{1}$, $H_{2}$,~\Dots, $H_{n}$, such that no two
of them are of the same type, while for each of them $\dfrac{L}{H}$~is
soluble. Such a group must be the direct product of $n$~simple
groups which are isomorphic with $H_{1}$, $H_{2}$,~\Dots, $H_{n}$.

If, however, several of these groups are of the same type, the
inference is no longer necessarily true. Thus if $G$~is the direct
product of five icosahedral groups, and if $L$~is the group of
isomorphisms of~$G$, the order of~$\dfrac{L}{G}$ is~$2^{5} · 5!$, and this group
contains icosahedral sub-groups. A group of order~$(60)^{6}$, the
factor-groups of whose composition-series are all of order~$60$,
is not therefore necessarily the direct product of $6$~icosahedral
groups.
\PageLabel[endpar]{263}
\index{Composite groups@\Topic{Composite groups}!non-soluble|)}%
%% -----File: 397.png---Folio 379-------

\Note{\textit{Note to} §~257.}

\begin{Remark}
The property stated in the footnote on \PageRef{p.}{366} may be proved as
follows. Let $S_{1}$~and~$S_{2}$ be two operations of the group which are
not permutable with each other; and suppose that
\[
S_{2}^{-1} S_{1}S_{2} = S_{1}\Sigma.
\]
The sub-group of order~$p^{n - 1}$, which contains $S_{1}$ self-conjugately, also
contains $S_{1}\Sigma$ self-conjugately (§~55); therefore it contains $\Sigma$ self-conjugately.
Similarly the sub-group, which contains $S_{2}$ self-conjugately,
contains $S_{2}\Sigma^{-1}$ and therefore $\Sigma$ self-conjugately. Hence $\Sigma$,
being contained self-conjugately in two distinct sub-groups of order~$p^{n - 1}$,
is a self-conjugate operation. From this it follows at once that
$\dfrac{H}{\Gh}$~is Abelian.
\end{Remark}


\Note{\textit{Note to} §~258.}

\begin{Remark}
The statement, on \PageRef{p.}{368}, that the congruences $p_{1}^{3} \equiv 1\ (\mod p_{2})$
and $p_{2} \equiv 1\ (\mod p_{1})$ are inconsistent, is subject to an exception
if $p_{1}^{2} + p_{1} + 1$ is a prime. When this is the case and when
$p_{2} = p_{1}^{2} + p_{1} + 1$, a group of order~$p_{1}^{2} p_{2} p_{3}$ can only be simple, under
the conditions assumed in the text, if it contains $(p_{1}^{3} - 1)p_{3}$ operations
of order~$p_{1}$, $p_{1}^{2}(p_{2} - 1)p_{3}$ operations of order~$p_{2}$, $p_{1}^{2} p_{2}(p_{3} - 1)$
operations of order~$p_{3}$, and no other operation except identity. The
relation
\[
p_{1}^{3} p_{2} p_{3}
  = (p_{1}^{3} - 1)p_{3} + p_{1}^{2}(p_{2} - 1)p_{3} + p_{1}^{2} p_{2}(p_{3} - 1) + 1
\]
cannot however be satisfied; the group is therefore composite.
\end{Remark}


\Note{\textit{Note to} §~260.}

\begin{Remark}
No simple group of odd order is at present known to exist. An
investigation as to the existence or non-existence of such groups
would undoubtedly lead, whatever the conclusion might be, to
results of importance; it may be recommended to the reader as
well worth his attention. Also, there is no known simple group
whose order contains fewer than three different primes. This
suggests that Theorems III~and~IV, §§~243,~244, may be capable of
generalisation. Investigation in this direction is also likely to lead
to results of interest and importance.
\end{Remark}
%% -----File: 398.png---Folio 380-------
\BackMatter


\Section{Appendix.}

\begin{Remark}
\First{The} technical phraseology that has been used in this book is
borrowed almost entirely from French or German; and far the
greater number of important memoirs on the subject of finite groups
are written in one or the other of those languages. To enable the
reader to refer, with as little trouble as possible, to the writings of
foreign mathematicians, a table is here given of the French and the
German equivalents of the more important technical terms. Even
abroad, the phraseology of the subject has not yet arrived at that
settled state in which every writer uses a technical term in the
same sense; but the variations of usage are not very serious. In
his recently published \textit{Lehrbuch der Algebra}, however, Herr Weber
has introduced or adopted several deviations from ordinary usage;
the chief of these are noted, by the addition of his name after the
term, in the subjoined table.

\begin{longtable}{*{2}{>{\hangindent1em}p{0.45\textwidth}}}
Group & Groupe \\
      & Gruppe \\
Abelian group & Groupe des opérations échangeables \\
      & Abel'sche Gruppe \\
Alternating group & Groupe alterné \\
      & Alternierende Gruppe \\
Complete group & Vollkommene Gruppe \\
Composite group & Groupe composé \\
      & Zusammengesetzte Gruppe \\
Factor-group & Groupe facteur \\
      & Factorgruppe \\
Primitive or imprimitive group & Groupe primitif ou non-primitif \\
      & Primitive oder imprimitive Gruppe \\
Transitive or intransitive group & Groupe transitif ou intransitif \\
      & Transitive oder intransitive Gruppe \\
Simple group & Groupe simple \\
      & Einfache Gruppe \\
Soluble group & Groupe résoluble \\
      & Auflösbare Gruppe \\
      & Metacyclische Gruppe\footnotemark\ (\textit{Weber})
\footnotetext{This term is used by other German writers to denote the holomorph of a
group of prime order.} \\
Substitution group & Groupe des substitutions \textit{or} système
des substitutions conjuguées \\
      & Substitutionengruppe \\
Symmetric group & Groupe symétrique \\
      & Symmetrische Gruppe \\
Group of isomorphisms & Gruppe der Isomorphismen \\
%% -----File: 399.png---Folio 381-------
Sub-group & Sousgroupe \\
      & Untergruppe \\
      & Theiler (\textit{Weber}) \\
Conjugate sub-groups & Sousgroupes conjugués \\
      & Gleichberechtigte Untergruppen \\
      & Conjugirte Theiler (\textit{Weber}) \\
Self-conjugate sub-group & Sousgroupe invariant \\
      & Ausgezeichnete Untergruppe \textit{or} invariante Untergruppe \\
      & Normaltheiler (\textit{Weber}) \\
Maximum self-conjugate sub-group & Sousgroupe invariant maximum \\
      & Ausgezeichnete \textit{or} invariante Maximaluntergruppe \\
      & Grösster Normaltheiler (\textit{Weber}) \\
Characteristic sub-group & Charakteristische Untergruppe \\
\hline
\Strut
Composition-series & Suite de composition \\
      & Reihe der Zusammensetzung \\
      & Compositionsreihe (\textit{Weber}) \\
Composition factors & Facteurs de composition \\
      & Factoren der Zusammensetzung \\
Chief series & Hauptreihe \\
Characteristic series & Lückenlose Reihe charakteristischer
  Untergruppen \\
\hline
\Strut
Simple isomorphism & Isomorphisme holoédrique \\
      & Holoëdrischer \textit{or} einstufiger Isomorphismus \\
Multiple isomorphism\footnotemark & Isomorphisme meriédrique
\footnotetext{It is to be noticed that, if $G$~is multiply isomorphic with~$H$, then $H$~is
  ``meriédriquement isomorphe'' with~$G$.} \\
      & Meriëdrischer \textit{or} mehrstufiger Isomorphismus \\
Isomorphism of a group with itself & Isomorphismus der Gruppe in sich \\
Cogredient or contragredient isomorphism & Cogredient oder contragredient Isomorphismus \\
\hline
\Strut
Degree (of a group) & Degré \\
      & Grad \\
Order \Ditto{(of a} \Ditto{group)} & Ordre \\
      & Ordnung \\
      & Grad (\textit{Weber}) \\
Genus \Ditto{(of a} \Ditto{group)} & Geschlecht \\
\hline
\Strut
%% -----File: 400.png---Folio 382-------
Substitution & Substitution \\
      & Substitution \textit{or} Buchstabenvertauschung \\
Circular substitution & Substitution circulaire \\
      & Cirkularsubstitution \\
Even and odd substitutions & Substitutions positives et négatives \\
      & Gerade und ungerade Substitutionen \\
Regular substitution & Substitution régulière \\
      & Reguläre Substitution \\
Similar substitutions & Substitutions semblables \\
      & Ähnliche Substitutionen \\
Transposition & Transposition \\
      & Transposition \\
\hline
\end{longtable}

One other term, introduced by Herr Weber, for which no simple
equivalent is used in this book, must be mentioned. The greatest
sub-group which is common to two groups he calls the ``Durchschnitt''
of the two groups.

The ratio of the order of a sub-group~$H$, to the order of the group~$G$
containing it, is called by French writers the ``indice,'' by German
the ``Index,'' of~$H$ in~$G$. The phrase is most commonly used of a
substitution group in relation to the symmetric group of the same
degree.

The smallest number of symbols displaced by any substitution,
except identity, of a substitution group is called by French writers
the ``classe,'' by Germans the ``Klasse,'' of the group.
\end{Remark}
%% -----File: 401.png---Folio 383-------


% [** TN: \printindex macro prints "(The numbers refer to pages.)"]
\printindex

\iffalse
Abel@\Topic{Abel}, quoted#Abel 46


Abelian group@\Topic{Abelian group}
  definition of 46
  sub-groups of 47, 48, 55-59
  existence of independent generating operations of 52
  invariance of the orders of a set of independent generating operations of 54
  symbol for, of given type 55


Abelian group@\Topic{Abelian group}
  of order~$p^{m}$ and type $(1, 1, \dots, 1)$#Abelian group 58
  number of sub-groups of, whose order is given 60
  number of distinct ways of choosing a set of independent generating operations of 59
  group of isomorphisms of 243, 244
    (|seealso \Topic{Homogeneous linear group})
  holomorph of 245


Alternating group@\Topic{Alternating group}
  definition of 139
  group of isomorphisms of 246
  is simple, except for degree~$4$#simple 154


Bochert@\Topic{Bochert}, quoted#Bochert 153


Bolza@\Topic{Bolza}, quoted#Bolza 162


Burnside@\Topic{Burnside}, quoted#Burnside 65, 97, 157, 311, 317, 335, 345, 352, 365, 370


Cauchy@\Topic{Cauchy}, quoted#Cauchy 90


Cayley@\Topic{Cayley}, quoted#Cayley 306


Characteristic series@\Topic{Characteristic series}
  definition of 232
  invariance of 233
  of a group whose order is a power of a prime 233-235


Characteristic sub-group@\Topic{Characteristic sub-group}
  definition of 232
  groups with no, are either simple or the direct product of simply isomorphic simple groups 232


Chief composition series@\Topic{Chief composition series}, or \Topic{Chief series}
  definition of 123
  invariance of 123


Cole@\Topic{Cole}, quoted#Cole 196, 370


Cole and Glover@\Topic{Cole} and \Topic{Glover}, quoted#Cole and Glover 137


Colour-groups@\Topic{Colour-groups} 306-310
  examples of 310


Complete group@\Topic{Complete group}
  definition of 236
  group of isomorphisms of a simple group of composite order is a 238
  holomorph of an Abelian group of order~$p^{m}$ and type $(1, 1, \dots, 1)$ is a#holomorph 239
  holomorph of a cyclical group of odd order is a 241
  symmetric group is a, except for degree~$6$#symmetric 246
  groups which contain a, self-conjugately are direct products 236
%% -----File: 402.png---Folio 384-------


Composite groups@\Topic{Composite groups}
  definition of 29
  non-soluble 376-378
  of even order 360-365


Composition factors@\Topic{Composition factors}
  definition of 118


Composition-series@\Topic{Composition-series}
  definition of 118
  invariance of 122
  examples of 128, 129


Conjugate operations@\Topic{Conjugate operations}
  definition of 27
  complete set of 31


Conjugate sub-groups@\Topic{Conjugate sub-groups}
  definition of 29
  complete set of 32
  operations common to or permutable with a complete set of, form a self-conjugate sub-group 33


Dedekind@\Topic{Dedekind}, quoted#Dedekind 89


Defining relations@\Topic{Defining relations} of a group
  definition of 21
  limitation on the number of, when the genus is given 283
  for groups of genus zero 291
  for groups of genus unity 301, 302
  for groups of orders $p^{2}$, $p^{3}$, $p^{4}$#groups 87, 88, 89
  for groups of order~$pq$#groups 100
  for groups of order~$p^{2}q$#groups 133-137
  for groups whose orders contain no square factor#groups 354
  for the holomorph of a cyclical group#holomorph 240, 241
  for the symmetric group of degree~$5$#symmetric 305
  for the simple group of order~$168$#simple 305


Degree@\Topic{Degree} of a substitution group
  definition of 138
  is a factor of the order, if the group is transitive 140


Direct product@\Topic{Direct product} of two groups
  definition of 40
  represented as a transitive group 190


Direct product@\Topic{Direct product} of two simply isomorphic groups of order~$n$ represented as a transitive group of degree~$n$#product 146, 147


Doubly transitive groups@\Topic{Doubly transitive groups}
  generally contain simple self-conjugate sub-groups 192
  the sub-groups of, which keep two symbols fixed 212
  with a complete set of triplets 214
  of degree~$n$ and order~$n(n - 1)$#degree 155-157


Dyck@\Topic{Dyck}, quoted#Dyck 22, 44, 172, 195, 255, 292, 301, 305


Dyck's theorem@\Topic{Dyck's theorem} that a group of order~$n$ can be represented as a regular substitution group of degree~$n$#Dyck's 22-24


Factor groups@\Topic{Factor groups}
  definition of 38
  set of, for a given group 118
  invariance of 118


Forsyth@\Topic{Forsyth}, quoted#Forsyth 280, 283


Fractional linear group@\Topic{Fractional linear group}
  definition of 311
  analysis of 319-333
  generalization of 334, 335


Frobenius@\Topic{Frobenius}, quoted#Frobenius 39, 45, 65, 71, 90, 97, 108, 110, 114, 166, 232, 250, 251, 345
  354, 358, 360


Frobenius and Stickelberger@\Topic{Frobenius and Stickelberger}, quoted#Frobenius and Stickelberger 46


Frobenius's theorem@\Topic{Frobenius's theorem} that, if $n$~is a factor of the order of a group, the number of operations of the group whose orders divide~$n$ is a multiple of~$n$#Frobenius 110-112


Galois@\Topic{Galois}, quoted#Galois 192, 334


General discontinuous group@\Topic{General discontinuous group} with a finite number of generating operations 256
  relation of special groups to 257, 258


Genus@\Topic{Genus} of a group, definition of 280


Gierster@\Topic{Gierster}, quoted#Gierster 311


Graphical representation@\Topic{Graphical representation}
  of a cyclical group 260, 261, 288
  of a general group 262-266
  of a special group 266-269
%% -----File: 403.png---Folio 385-------
  of a group of finite order 273-278
  examples of 269-272, 279, 288-290, 294, 296, 299, 300, 303, 310


Group@\Topic{Group}
  definition of 11, 12
  continuous, discontinuous, or mixed 13, 14
  order of 14
  multiplication table of 20, 49
  fundamental or generating operations of 21
  defining relations of 21
  cyclical 25
  simple and composite 29
  Abelian 46
  soluble 130
  symmetric 139
  alternating 139
  complete 236
  group of isomorphisms of a 223
  general 256
  special 257
  quadratic 326
  dihedral 287
  tetrahedral 289
  octohedral 289
  icosahedral 289
  holomorph of a 228
  genus of a 280
  symbol for a 27
  (|seealso \Topic{Substitution group}).


Group of isomorphisms@\Topic{Group of isomorphisms}
  contains the group of cogredient isomorphisms self-conjugately 226
  of a cyclical group 239-242
  of an Abelian group of order~$p^{n}$ and type $(1, 1, \Dots, 1)$, 243, 244, 311-317#Abelian 336-338
  of the alternating group 246
  of doubly transitive groups of degree~$p^{n} + 1$ and order $\frac{1}{2}p^{n}(p^{2n} - 1)$#doubly 246-248


Groups@\Topic{Groups}
  of genus zero 286-292
  of genus one 295-302
  of genus two 302


Groups whose order@\Topic{Groups whose order is~$p^{m}$}, where $p$~is a prime 61
  always contain self-conjugate operations 62
  every sub-group of, is contained self-conjugately in a greater sub-group 65
  number of sub-groups of given order is congruent to unity, $\pmod{p}$ 71
  case in which there is only one sub-group of a given order 72-75
  number of types which contain self-conjugate cyclical sub-groups of order~$p^{m - 2}$#self-conjugate 76-81
  determination of distinct types of orders $p^{2}$, $p^{3}$, $p^{4}$, where $p$~is an odd prime#types 81-88
  table of distinct types of orders $8$ and~$16$#table 88, 89


Groups whose orders contain@\Topic{Groups} whose orders contain
  no squared factor 353, 354
  no cubed factor 359


Groups whose sub-groups@\Topic{Groups} whose sub-groups of order~$p^{\alpha}$ are all cyclical#cyclical 352, 353


Heffter@\Topic{Heffter}, quoted#Heffter 215


Holder@\Topic{Hölder}, quoted#Hölder 38, 87, 119, 137, 224, 236, 246, 249, 344, 353, 370, 376


Holomorph@\Topic{Holomorph}
  definition of 228
  of a cyclical group 240-242
  of an Abelian group of order~$p^{n}$ and type $(1, 1, \dots, 1)$#Abelian 245


Homogeneous linear group@\Topic{Homogeneous linear group}
  definition of 244
  composition series of 314-317
  represented as a transitive substitution group 336, 337
%% -----File: 404.png---Folio 386-------


Homogeneous linear group@\Topic{Homogeneous linear group}
  simple groups defined by 338, 339
  generalization of 340-342


Hurwitz@\Topic{Hurwitz}, quoted#Hurwitz 24, 280, 282


Identical operation@\Topic{Identical operation}
  definition of 12


Imprimitive groups@\Topic{Imprimitive groups}
  definition of 171
  of degree~$6$#degree 181, 182


Imprimitive systems@\Topic{Imprimitive systems}
  definition of 172
  of a regular group 173
  of any transitive group 176
  properties of 185, 186


Imprimitive self-conjugate sub-group@\Topic{Imprimitive self-conjugate sub-group} of a doubly transitive group 193, 194


Intransitive groups@\Topic{Intransitive groups}
  definition of 140
  properties of 159-162
  transitive sets of symbols in 159, 166
  of degree~$7$ with transitive sets of $3$ and~$4$#degree 163, 164


Isomorphism of two groups@\Topic{Isomorphism of two groups}
  simple, definition of 32
  multiple, definition of 36
  general, definition of 41


Isomorphisms@\Topic{Isomorphisms} of a group with itself
  definition of 222
  cogredient and contragredient, definition of 224
  class of, definition of 224
  limitation on the order of 252


Isomorphism@\Topic{Isomorphism} of general and special groups 257, 258


Jordan@\Topic{Jordan}, quoted#Jordan 46, 119, 143, 146, 149, 150, 153, 189, 196, 198, 204, 311


Klein@\Topic{Klein}, quoted#Klein 224, 289, 292


Klein and Fricke@\Topic{Klein} and \Topic{Fricke}, quoted#Klein and Fricke 162


Limitation@\Topic{Limitation}
  on the order of a group of given genus 282
  on the number of defining relations of a group of given genus 283


Maillet@\Topic{Maillet}, quoted#Maillet 108


Marggraff@\Topic{Marggraff}, quoted#Marggraff 198


Maschke@\Topic{Maschke}, quoted#Maschke 310


Mathieu@\Topic{Mathieu}, quoted#Mathieu 155


Maximum sub-group@\Topic{Maximum sub-group}, definition of 35


Maximum self-conjugate sub-group@\Topic{Maximum self-conjugate sub-group}
  definition of 35


Miller@\Topic{Miller}, quoted#Miller 105, 196, 259


Minimum self-conjugate sub-group@\Topic{Minimum self-conjugate sub-group}
  definition of 124
  is a simple group or the direct product of simply isomorphic simple groups 127


Moore@\Topic{Moore}, quoted#Moore 215, 335


Multiplication table@\Topic{Multiplication table} of a group 20, 49


Multiply isomorphic groups@\Topic{Multiply isomorphic groups}
  definition of 36


Netto@\Topic{Netto}, quoted#Netto 141, 198, 215


Number of symbols@\Topic{Number of symbols} unchanged by all the substitutions of a group 165, 166


Operations@\Topic{Operations}
  common to two groups form a group 27
  of a group, which are permutable with a given operation or sub-group, form a group 29, 31
  common to or permutable with each of a complete set of conjugate sub-groups form a self-conjugate sub-group 33


Order@\Topic{Order}
  of a group, definition of 14
  of an operation, definition of 14


Permutable operations@\Topic{Permutable operations}
  definition of 12


Permutable groups@\Topic{Permutable groups}
  definition of 41


Permutability@\Topic{Permutability} of an operation with a group
  definition of 29


Primitive groups@\Topic{Primitive groups}
  definition of 171
  when soluble, have a power of a prime for degree 192
  limit to the order of, for a given degree 199
  with a transitive sub-group of smaller degree 197, 198
  of prime degree 201
  of degrees $3$,~$4$ and~$5$#degrees 205
  of degree~$6$#degree 205, 206
%% -----File: 405.png---Folio 387-------
  of degree~$7$#degree 206-208
  of degree~$8$#degree 209-211


Primitivity@\Topic{Primitivity}, test of 184


Regular division@\Topic{Regular division} of a surface, representation of a group by means of 278


Representation of a group@\Topic{Representation of a group}
  in transitive form; \ie\ as a transitive substitution group 22, 173-179
  in primitive form 177


Representation@\Topic{Representation}, graphical (|see \Topic{Graphical representation}).


Self-conjugate operation@\Topic{Self-conjugate operation}
  definition of 28
  of a group whose order is the power of a prime 62
  of a transitive substitution group must be a regular substitution 144


Self-conjugate sub-group@\Topic{Self-conjugate sub-group}
  definition of 29
  generated by a complete set of conjugate operations 34
  of a primitive group must be transitive 187
  of an imprimitive group 187, 188
  of a $k$-ply transitive group is, in general, $(k - 1)$-ply transitive#transitive 189


Simple groups@\Topic{Simple groups}
  definition of 29
  whose orders are the products of not more than $5$~primes#products 367-370
  whose orders do not exceed~$660$#exceed 370-375


Simply isomorphic groups@\Topic{Simply isomorphic groups}
  definition of 22
  said to be of the same type 22


Soluble groups@\Topic{Soluble groups}
  definition of 130
  properties of 131, 132
  special classes of 345-359


Sub-group@\Topic{Sub-group}
  definition of 25
  order of a, divides order of group 26
  (|seealso, \Topic{conjugate}, \Topic{self-conjugate} and \Topic{characteristic sub-group}).


Substitution@\Topic{Substitution}
  definition of 1
  cycles of 2
  identical 4
  inverse 4
  order of 6
  circular 7
  regular 7
  similar 7
  permutable 8
  even and odd 10
  symbol for the product of two or more 4


Substitutions@\Topic{Substitutions} which are permutable
  with a given substitution 215, 216
  with every substitution of a given group 145, 146, 217
  with every substitution of a group, whose degree is equal to its order, form a simply isomorphic group 146


Substitution group@\Topic{Substitution group}
  degree of, definition of 138
  regular, definition of 24
  transitive and intransitive, definition of 140
  primitive and imprimitive, definition of 171
  multiply transitive, definition of 148
  representation of any group as a regular 22
  degree of transitive, is a factor of its order 140
  order of a $k$-ply transitive, whose degree is~$n$, is a multiple of $n(n - 1)\Dots (n - k + 1)$#order 148
  construction of multiply transitive 150
  limit to the degree of transitivity of 152
  transitive, whose substitutions, except identity, displace all or all but one of the symbols 141-144
  doubly transitive, of degree~$n$ and order~$n(n - 1)$#doubly 155-157
  triply transitive, of degree~$n$ and order~$n(n - 1)(n - 2)$#triply 158
  quintuply transitive, of degree~$12$#quintuply 170
  conjugate substitutions of, are similar 144


Substitution groups@\Topic{Substitution groups} whose orders are powers of primes 218, 219
%% -----File: 406.png---Folio 388-------


Sylow@\Topic{Sylow}, quoted#Sylow 62, 90


Sylow's theorem@\Topic{Sylow's theorem} 92-95
  some direct consequences of 97-100
  extension of 110


Symbol@\Topic{Symbol}
  for the product of two or more substitutions 4
  for the product of two or more operations 12
  for a group generated by given operations 27
  theta@$\vartheta(P)$ and $\theta(P)$, definition of#definition 250, 251


Symmetric group@\Topic{Symmetric group}
  definition of 139
  of degree~$n$ has a single set of conjugate sub-groups of order~$(n - 1)!$ except when $n$ is~$6$#degree 200
  of degree~$6$ has $12$~simply isomorphic sub-groups of order~$5!$ which form two distinct conjugate sets#degree 200
  is a complete group, except for degree~$6$#complete 246


Transforming an operation@\Topic{Transforming an operation}
  definition of 27


Transitive group@\Topic{Transitive group}
  definition of 140
  representation of any group as a 175-179
  number of distinct modes of representing the alternating group of degree~$5$ as a#modes 179, 180


Transpositions@\Topic{Transpositions}
  definition of 9
  representation of a substitution by means of 9
  number of, which enter in the representation of a substitution is either always even or always odd 10


Triply transitive@\Topic{Triply transitive} groups of degree~$n$ and order $n(n - 1)(n - 2)$#triply 158


Type of a group@\Topic{Type of a group} (|see \Topic{Simply isomorphic groups}).


Types of group@\Topic{Types of group}, distinct, whose order is
  $p^{2}$, $p^{3}$, or $p^{4}$, where $p$~is an odd prime#types 87, 88
  $8$ or~$16$#types 88, 89
  $pq$, where $p$~and $q$ are different primes#types 100, 101
  $p^{2}q$, where $p$~and $q$ are different primes#types 133-137
  $24$#types 101-104
  $60$#types 105-108


Weber@\Topic{Weber}, quoted#Weber 46


Young@\Topic{Young}, quoted#Young 63, 68, 87, 310
\fi % ** End of index text
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\begin{PGtext}
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William Burnside

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[7] [8] (./40395-t.aux)

 *File List*
    book.cls    2007/10/19 v1.4h Standard LaTeX document class
    bk12.clo    2007/10/19 v1.4h Standard LaTeX file (size option)
inputenc.sty    2008/03/30 v1.1d Input encoding file
  latin1.def    2008/03/30 v1.1d Input encoding file
 fontenc.sty
   t1enc.def    2005/09/27 v1.99g Standard LaTeX file
  ifthen.sty    2001/05/26 v1.1c Standard LaTeX ifthen package (DPC)
 amsmath.sty    2000/07/18 v2.13 AMS math features
 amstext.sty    2000/06/29 v2.01
  amsgen.sty    1999/11/30 v2.0
  amsbsy.sty    1999/11/29 v1.2d
  amsopn.sty    1999/12/14 v2.01 operator names
 amssymb.sty    2009/06/22 v3.00
amsfonts.sty    2009/06/22 v3.00 Basic AMSFonts support
mathrsfs.sty    1996/01/01 Math RSFS package v1.0 (jk)
   alltt.sty    1997/06/16 v2.0g defines alltt environment
footmisc.sty    2009/09/15 v5.5a a miscellany of footnote facilities
multicol.sty    2008/12/05 v1.6h multicolumn formatting (FMi)
 makeidx.sty    2000/03/29 v1.0m Standard LaTeX package
indentfirst.sty    1995/11/23 v1.03 Indent first paragraph (DPC)
    calc.sty    2007/08/22 v4.3 Infix arithmetic (KKT,FJ)
fancyhdr.sty    
longtable.sty    2004/02/01 v4.11 Multi-page Table package (DPC)
   array.sty    2008/09/09 v2.4c Tabular extension package (FMi)
graphicx.sty    1999/02/16 v1.0f Enhanced LaTeX Graphics (DPC,SPQR)
  keyval.sty    1999/03/16 v1.13 key=value parser (DPC)
graphics.sty    2009/02/05 v1.0o Standard LaTeX Graphics (DPC,SPQR)
    trig.sty    1999/03/16 v1.09 sin cos tan (DPC)
graphics.cfg    2009/08/28 v1.8 graphics configuration of TeX Live
  pdftex.def    2009/08/25 v0.04m Graphics/color for pdfTeX
rotating.sty    2009/03/28 v2.16a rotated objects in LaTeX
geometry.sty    2008/12/21 v4.2 Page Geometry
   ifpdf.sty    2009/04/10 v2.0 Provides the ifpdf switch (HO)
  ifvtex.sty    2008/11/04 v1.4 Switches for detecting VTeX and its modes (HO)
geometry.cfg
hyperref.sty    2009/10/09 v6.79a Hypertext links for LaTeX
 ifxetex.sty    2009/01/23 v0.5 Provides ifxetex conditional
 hycolor.sty    2009/10/02 v1.5 Code for color options of hyperref/bookmark (HO
)
xcolor-patch.sty    2009/10/02 xcolor patch
  pd1enc.def    2009/10/09 v6.79a Hyperref: PDFDocEncoding definition (HO)
etexcmds.sty    2007/12/12 v1.2 Prefix for e-TeX command names (HO)
infwarerr.sty    2007/09/09 v1.2 Providing info/warning/message (HO)
hyperref.cfg    2002/06/06 v1.2 hyperref configuration of TeXLive
kvoptions.sty    2009/08/13 v3.4 Keyval support for LaTeX options (HO)
kvsetkeys.sty    2009/07/30 v1.5 Key value parser with default handler support 
(HO)
     url.sty    2006/04/12  ver 3.3  Verb mode for urls, etc.
  bitset.sty    2007/09/28 v1.0 Data type bit set (HO)
 intcalc.sty    2007/09/27 v1.1 Expandable integer calculations (HO)
bigintcalc.sty    2007/11/11 v1.1 Expandable big integer calculations (HO)
pdftexcmds.sty    2009/09/23 v0.6 LuaTeX support for pdfTeX utility functions (
HO)
ifluatex.sty    2009/04/17 v1.2 Provides the ifluatex switch (HO)
 ltxcmds.sty    2009/08/05 v1.0 Some LaTeX kernel commands for general use (HO)

atbegshi.sty    2008/07/31 v1.9 At begin shipout hook (HO)
 hpdftex.def    2009/10/09 v6.79a Hyperref driver for pdfTeX
supp-pdf.mkii
   color.sty    2005/11/14 v1.0j Standard LaTeX Color (DPC)
   color.cfg    2007/01/18 v1.5 color configuration of teTeX/TeXLive
 nameref.sty    2007/05/29 v2.31 Cross-referencing by name of section
refcount.sty    2008/08/11 v3.1 Data extraction from references (HO)
 40395-t.out
 40395-t.out
  t1cmtt.fd    1999/05/25 v2.5h Standard LaTeX font definitions
    umsa.fd    2009/06/22 v3.00 AMS symbols A
    umsb.fd    2009/06/22 v3.00 AMS symbols B
   ursfs.fd    1998/03/24 rsfs font definition file (jk)
  omscmr.fd    1999/05/25 v2.5h Standard LaTeX font definitions
./images/fig01.png
./images/fig02.png
./images/fig03.png
./images/fig04.png
./images/fig05.png
./images/fig06.png
./images/fig07.png
./images/fig08.png
./images/fig09.png
./images/fig10.png
./images/fig11.png
./images/fig12.png
./images/fig13.png
./images/fig14.png
./images/fig15.png
./images/fig16.png
./images/fig17.png
./images/fig18.png
./images/plate1.jpg
 40395-t.ind
 ***********

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