GROUPS AND CHARACTERS
PURE AND APPLIED MATHEMATICS
A Wiley-Interscience Series of Texts, Monographs, and Tracts
Founded by RICHARD COURANT
Editor Emeritus: PETER HILTON
Editors: MYRON B. ALLEN III, DAVID A. COX,
HARRY HOCHSTADT, PETER LAX, JOHN TOLAND
A complete list of the titles in this series appears at the end of this volume.
GROUPS AND CHARACTERS
LARRY C. GROVE
Department of Mathematics
University of Arizona
Tucson, Arizona
A Wiley-Interscience Publication
JOHN WILEY & SONS, INC.
New York • Chichester • Weinheim • Brisbane • Singapore • Toronto
This text is printed on acid-free paper.
Copyright © 1997 by John Wiley & Sons, Inc.
All rights reserved. Published simultaneously in Canada.
Reproduction or translation of any part of this work beyond
that permitted by Section 107 or 108 of the 1976 United
States Copyright Act without the permission of the copyright
owner is unlawful. Requests for permission or further
information should be addressed to the Permissions Department,
John Wiley & Sons, Inc., 605 Third Avenue, New York, NY
10158-0012
Library of Congress Cataloging in Publication Data:
Grove, Larry C.
Groups and characters / Larry C. Grove.
p. cm. — (Pure and applied mathematics)
"A Wiley-Interscience publication."
Includes bibliographical references and index.
ISBN 0-471-16340-6 (cloth : alk. paper)
1. Group theory. I. Title. II. Series: Pure and applied
mathematics (John Wiley & Sons : Unnumbered)
QA174.2.G77 1997
512'.2—dc21
96-29776
CIP
10 9 8 7 6 5 4 3 2 1
Contents
Preface
1
2
3
4
vii
Preliminaries
1.1 Some Notation and Generalities
1.2 Permutation Actions
1.3 Coset Enumeration
1.4 Semidirect Products
1.5 Wreath Products
1.6 Transitivity and Primitivity
1.7 Some Linear Algebra
1
3
9
13
14
16
20
S o m e Groups
2.1 Aut(Z n )
2.2 Metacyclic Groups
2.3 Sylow Subgroups of Symmetric Groups
2.4 Affine Groups of Fields
2.5 Finite Groups in 2 and 3 Dimensions
2.6 Some Linear Groups
2.7 Mathieu Groups
2.8 Symplectic Groups
27
29
30
32
33
39
46
52
Counting with Groups
3.1 Fixed Points and Orbits
3.2 The Cycle Index
3.3 Enumeration
3.4 Generating Functions
3.5 The Petersen Graph
61
64
68
73
75
Transfer and Splitting
4.1 Transfer and Normal Complements
4.2 Hall Subgroups
4.3 Mostly p-groups
77
84
89
v
vi
5
6
7
8
9
CONTENTS
Representations and Characters
5.1 Representations
5.2 Characters
5.3 Contragredients and Products
97
105
121
Induction and Restriction
6.1 Modules
6.2 Induction
6.3 Normal Subgroups and Clifford Theory
6.4 Mackey Theorems
6.5 Brauer Theorems
127
128
136
142
147
Computing Character Tables
7.1 Burnside
7.2 Dixon
7.3 Schneider
151
154
158
Characters of Sn and An
8.1 Symmetric Groups
8.2 Alternating Groups
161
168
Frobenius Groups
9.1 Frobenius Groups and Their Characters
9.2 Structure of Frobenius Groups
171
176
10 Splitting Fields
10.1 Splitting
10.2 The Schur Index
10.3 R versus C
183
184
190
Bibliography
203
Index
209
Preface
The present volume is intended to be a graduate-level text. It covers some
aspects of group theory, concentrating mainly, but not exclusively, on finite
groups.
The presentation has been strongly, and positively, influenced by a number
of earlier texts and monographs. Particular mention should be made of books
by Curtis and Reiner [18], Dornhoff [22], Feit [23], Gorenstein [29], Isaacs [39],
and Passman [50].
Chapters 1, 2, and 4 could serve as the text for a basic one-semester
course on group theory. Chapter 2 consists entirely of examples, so it could
in principal be omitted, but not without radically altering the flavor of the
undertaking. Chapter 3 is more easily omitted; on the other hand Chapter 3
is easy and fun, and it provides important applications to combinatorics.
Chapters 5 and 6 contain a basic introduction to ordinary character theory
— they do not depend heavily on the preceding four chapters. Chapters 7
through 10 can be read independently of each other in any order following
Chapter 6. In fact, Chapter 7 only requires Chapter 5. It should be noted,
though, that Chapter 9, on Frobenius groups, makes fairly heavy use of some
of the group theory from Chapter 4.
Some attention has been paid to computational aspects of the subject.
For example, the Schreier-Sims algorithm, Todd-Coxeter coset enumeration,
and various algorithms for calculating character tables are discussed, typically
in the context of the very powerful (and free!) computational group theory
package GAP.
It is assumed throughout that the reader has assimilated most of the
material from a standard first-year graduate abstract algebra course in a
U.S. university, such as in [31]. This includes elementary group theory, such
as Sylow theorems (although a proof is included in Chapter 1), presentations,
solvability and nilpotence, etc.; as well as basic facts about rings, modules,
and field extensions. It is important for the reader to have a reasonable
facility with linear algebra. Nevertheless, some basic linear algebra is included
in the text, on the grounds that it may not always be covered in standard
undergraduate courses.
vn
vin
PREFACE
I wish to thank the faculty, students, and staff of Lehrstuhl D für Mathematik at the RWTH in Aachen for support and stimulation during two sabbatical leaves. Particular acknowledgment is in order for Professor Joachim
Neubiiser, the founding father of GAP; it was my privilege to attend many
of his wonderfully clear lectures on groups and representations.
My special thanks to Robert Beals and Olga Yiparaki for careful readings
of parts of the manuscript.
Larry C. Grove
The University of Arizona
August 1996
GROUPS AND CHARACTERS
Groups and Characters
by Larry C. Grove
Copyright © 1997 John Wiley & Sons, Inc.
Chapter 1
Preliminaries
In this chapter we present a variety of concepts and ideas, in part to establish
terminology, usage, and notation. Everything presented will appear in later
chapters.
It is assumed that the reader is familiar with the elementary grouptheoretical material normally covered in a standard first graduate-level algebra course (in the United States), and also the material from a junior/seniorlevel linear algebra course. Any gaps can be filled by browsing in one or more
of the many texts available for such courses.
1.1
Some Notation and Generalities
If G is a group and x , y € G we shall write xv to denote y~1xy and vx to
denote yxy'1. Note that xvz = (xv)z and yzx = y(zx) for all x,y ,z € G.
If G is a group and H Ç G is a subgroup we write H < G, or G £ H,
as usual. If H < G then a set T of (right) coset representatives for H in G
will be called a (right) transversal; we will always assume that T C\ H = 1,
i.e. that the representative of H itself is the identity element. Note that
|T| = [G:H], the index of H in G.
If T is a transversal for H in G we define a "transversal function" from G
to T, denoted x ►
►
- 3?, via x = t, where t G T and Hx = Ht, or equivalently
Hx C\T = {x}. Note that the notation depends completely on the choice of
a fixed transversal T; if S is any transversal then {s: s 6 S} = T.
The first proposition lists three trivial but important facts about the
transversal function.
Proposition 1.1.1 Suppose that T is a transversal for H <G.
1. x = x,
2. x x " 1 6 H, and
1
Then
CHAPTER
2
1.
PRELIMINARIES
3. xy = xy
for all x,y G G.
Proof Parts 1 and 2 are obvious. For part 3 note that Hxy = Hxy =
(Hx)y = Hxy, so xy = xy.
A
Corollary 1.1.2 IfteT
and x E G then txx~l
= t.
Proof txx~l = tex-1 = t = t.
A
Suppose now that H <G and G = {X), i.e. that X is a set of generators
for G. Write X - 1 = { x - 1 : x € X} and X± = X U X~x. Suppose that T is a
transversal for H in G, and set
A = {txtx~1:t
ET, xE X * } Ç H,
B = {txtx~1:teT,xeX}
Ç A.
Proposition 1.1.3 In the setting above A Ç BliB_1
L){1}.
Proof Take t E T and x E X. It will be sufficient to show that te_1te_1 _ 1 €
B _ 1 . Set h = tx~l E T. Then tixhx~x
E B, and ( f i X i T x - 1 ) - 1 =
1 1
x
l
1
-1
-1
1 _1
hxx- ^
= tx- xx- tx= te te"
€
fl_1.
A
Corollary 1.1.4 (A) = (B).
Bear in mind that if T is a transversal for H in G then it is always assumed
that T n H = 1.
Theorem 1.1.5 (Schreier) Suppose that G is a group, G = (X), H < G,
and T is a transversal for H in G. Then H is generated by the set
B =
{txtx~l:tET,xEX}.
Proof By Corollary 1.1.4 it will suffice to show that H = (A), with A =
{tarte - 1 :* E T, x E Xa1}. For any y E H write y = xix 2 • • x*, with each
Xj E X±. Set Í! = 1; then t 2 = tjxi, t 3 = t 2 x 2 , ..., tk+\ = tkxk e T. Next
set
1 < i < k, and observe that
aia2--ak
=
( t i X i t ^ X t ^ t J 1 ) •• • ( t * ! * * ^ )
=
í i x i x 2 • • ■ xjfef^ = 1 • yfj^j G # .
But then tk+i E # , since y € # , and also tk+1
y = o 1 o 2 ---a f c G (i4).
E T, so tk+i
= 1, i.e.
A
1.2. PERMUTATION
ACTIONS
3
Corollary 1.1.6 If G is a finitely generated group, H < G, and [G: H] is
finite, then H is finitely generated.
P r o o f Say G = (X), with X finite, and let T be a transversal for H in G.
Then H = (B), as in the theorem, and | B | < |T||A"| = [G: H]\X\.
A
By way of contrast if G is the free group on two generators and H = G',
the commutator subgroup, then H is known to be free on countably infinitely
many generators, so [G: H] must be infinite.
The set B = {txtx _ 1 : t 6 T, x 6 X} is called the set of Schreier generators
for H <G, relative to the generating set X and transversal T.
For an example take G to be the symmetric group Sn and H to be the
alternating subgroup A„. Let a be the transposition (12) and r the ncycle (12 ••■ n). Then 5„ is generated by X = {a, r } , and T = {1, a}
is a transversal for A„ in 5„. Assume that n is odd, so r € An. Schreier
generators then are ICTICT -1 = o~o~l = 1, l r l r - 1 = r ,CTCTCTCT-1= 1, and
error _ 1 = ara, so A n = (r,CTTCT).Note that ara = (213 • • • n). What are
the Schreier generators for An if n is even?
1.2
Permutation Actions
If S is a set write Perm(S) to denote the set of all permutations of S,
i.e. all bijections a: S -> 5 . It will usually be convenient to use (right)
"exponential" notation for permutation actions: if s € 5 and a € Perm(S)
then s" is the image of s under the action of a.
lia, T e Perm(S) we define their composition product trr as usual: s"T =
T
(s") , all s € S. Note that with exponential notation a acts first, followed by
T, in or, i.e. multiplication is left-to-right. With composition as the product
Perm(S) is a group.
Very often S will be the set { 1 , 2 , . . . , n}, in which case we will as usual call
Perm(S) the symmetric group and denote it by Sn or Sym(n). Furthermore
we will use the usual cycle notation, etc., bearing in mind that multiplication
proceeds from left to right.
If 5 is a set and G is any group we say that G acts on S if there is a
homomorphism 9: G -> Perm(5), in which case for s € 5 and x 6 G we have
s >-► s6x. If Ö is one-to-one the action is called faithful. In most situations the
6 will be suppressed, and we will simply write s ^ sx. With that convention
the defining properties of a group action are s 1 = s and sxy = (sx)v, all s G S
and all i , y 6 G.
The cardinality \S\ is often called the degree of the action.
In some situations the exponential notation for group actions can be confused with, or even conflict with, other established notations. To avoid such
problems we will usually write instead s i - m (or in some cases s •-* xs, or
s ►>
- xs), with obvious corresponding changes in the governing properties.
CHAPTER
4
1.
PRELIMINARIES
If G acts on 5 and s S 5 recall that the orbitoî sis Orbc(s) = {sx:x € G},
and that the stabilizer of s is Stabo(s) = {x € G: sx = s} < G. Alternate notations that will sometimes be useful are sG for Orbc(s) and Gê for Stabo(s).
Exercise
Suppose that G acts on S, x € G, and s e S. Show that G g . = GJ, i.e.
stabilizers of elements in the same orbit are conjugate.
Recall also that G is said to act transitively on S if Orbo(s) = S for some
(hence all) s € S.
If Gi acts on S\ and G2 acts on £2 we say that the actions are equivalent
if there is an isomophism tp:Gi -> G2, together with a bijection / . S i -> 52
such that the diagram
Si - ^
Si
<l ^ ['
Si
x
is commutative; i.e. f(s )
v x)
= f(s) ^ '
—-r
S2
for all x € Gi, a € Si.
Exercises
1. If H < G show that G acts transitively on the set {Hy} of all right cosets
of H in G, with the action of x € G being given by x: iîj/ •-► Hj/x.
2. Show that the kernel of the action in exercise 1 is Ci{Hx: x 6 G}, which
is called the core of H in G and denoted core(H).
3. Show that core(i/) is the largest subgroup of H that is normal in G.
4. If G acts transitively on S and s € S show that the action of G on S
is equivalent with the action (above) of G on the set of cosets of the
stabilizer H = Stabe (s).
The next proposition is a slight variation on the familiar (and very basic)
fact that the size of an orbit is the index of the stabilizer.
P r o p o s i t i o n 1.2.1 Suppose that G acts on S, s 6 S, and T is a transversal
for Stabe (s) "» G. Then the map <p: t >-► sl is a bijection from T to Orbe(s).
P r o o f Check that ip is 1-1 and onto.
A
Let us illustrate the proposition by means of an example. Set a =
( 1 2 3 4 5 ) , T = (2354), and G = (a, T) < S 5 . Then Ia = 2, VT = 2 T = 3,
l'T" = 3" = 4, and VT* = 3 r = 5, so G is transitive, O r b ö ( l ) = S =
{1,2,3,4,5}. Thus Gi = Stab G (l) has index 5 in G.
"Graphically" we have
1.2. PERMUTATION
1
O
5
ACTIONS
2
T
O
" 3
" 4 ,
^
5
and furthermore we have a transversal T = {l,o,OT,OTO,or2}
for Gi in
G. This is in fact an example of a Schreier transversal, characterized by
the fact that each element of T is obtained from an earlier element of T via
multiplication (on the right) by one of the given generators for G.
Exercise
Carry out the same procedure for the subgroup G of 57 generated by
o = (2 4) (5 6) and T = (12 5 7) (3 4), and find a transversal for Gi in G.
We shall pursue the ideas above somewhat further in order to sketch an
algorithm for determining the order of a (finite) permutation group.
If G acts on S and s i , s 2 S S write C?«i,»2 for GSl D G,3, the 2-point
stabilizer, and similarly for 3-point stabilizers, etc. In general, if T Ç S write
GT to denote (~){Gt:t € T } , the subgroup that simultaneously stabilizes all
points of T.
A sequence (¿>i, 6 2 , . . . , 6*) of distinct points from S is called a base for G
if (J6i,6a,...,6fc = 1-
For example, if G = S4 then B = (1,2,3) is abase; if G = ((13), (123 4)),
a dihedral subgroup of 54, then B = (1,2) is a base.
Proposition 1.2.2 Suppose that B = (¿>i,... ,6/t) is a 6ose /or G, x, y € G,
and 6f = of, 1 < i < k. Then x — y. Thus elements of G are determined by
their action on a base.
Proof We have b'y
= &,, all i, so xy~l € G^
6fc = 1 and x = y.
A
The usefulness of the proposition derives mainly from the fact that surprisingly often even large groups of high degree have bases that are rather
short.
Now for some slightly elaborate notation. Suppose G acts on S, and that
B = ( 6 X , . . . , bk) is a base for G. Write G^ = G, G<2> = G^ = Gbl, G™ =
G
?J = G h,!», etc., then d = bfW, 02 = b§W, etc. Let T¿ be a transversal
for G<i+1) in G « , i = l,...,k.
Note that G(fc+1) = GhlM
b„ = 1. Note
also that |T¡| = \Oi\, all i, by Proposition 1.2.1.
CHAPTER
6
1.
PRELIMINARIES
Proposition 1.2.3 With the notation and setting above each x € G can be
written uniquely as x = tk • • • ¿2*1» with U € T¿, all i.
Proof Let ii = x G ï \ (relative to G (2) < G (1) ) and write x = x2t\
(uniquely), with x2 € G*2'. Then let t2 = x j € T2 and write X2 = X3Í2
(uniquely), with x 3 e G< 3 \ hence x = x 3 í 2 íi, etc. At stage k we have
x = xk+itk- - - i i , a n d x * + i e G ( k + 1 ' = 1.
A
Corollary 1.2.4 |G| = Il*=i |T,| = ü t i l<U
Two questions arise naturally. How can we find a base B for G? Once we
have B, how can we determine the subgroups in the stabilizer chain G^ >
G(3) > . . . > c ? ( * ) ?
One more definition. If B is a base for the group G acting on a set 5, then a
strong generating setis a set X of generators for G such that (Xi)G^)
= G^
for 1 < i < k.
We return to the example on page 4 to illustrate how to obtain a base B
and strong generating set X. The basic idea is to add elements to B one at
a time, get a transversal via Proposition 1.2.1, calculate Schreier generators
íxíx _ 1 for the stabilizer, and continue until a stabilizer is trivial.
Recall that G = (<T,T), with a = ( 1 2 3 4 5 ) and r = (2354). It was
determined that 0\ = S — {1,2,3,4,5}, and we have a transversal T\ =
{1,0,<?T,O-T<T,(TT2}
for Gi = G<2> in G. We begin with B = (1) and X =
{<r, r } .
The table below shows the Schreier generators for G\.
t
1
a
CT
ara
ar2
1
a
ar
ara
ar2
X
a
a
a
a
a
T
T
T
T
T
tx
a
ar
ara
ar2
1
1
ar
ar2
a
ara
txtx *
1
r-1
1
r-1
r2
T
1
1
T2
T2
How are the calculations done? For example, in the second line of the table
we have tx = a2, which is not a transversal element. But a2 = ( 1 3 5 2 4 ) ,
which carries 1 to 3, as does the transversal element ar; hence a2 = ar. The
Schreier generator is thus a2r~1a~1, which is r _ 1 .
We see then that G<2) = (T). Note that X n G<2' = {r}, which generates
2
G^ \ so there is no need to enlarge X.
1.2. PERMUTATION
7
ACTIONS
We proceed by adding another point to B, say B :— (1,2). For O2 and
T-i we may use
2
T
► 3
T
5
T
4,
so 02 = {2,3,5,4} and T2 = {1, r, r 2 , r 3 } , a transversal for G^ = Gi, 2 in
G<2>.
Easy calculations show that the Schreier generators for G ^ are all trivial.
Thus G ^ = 1, and it is clear that B = (1,2) is a base for G with strong
generating set X = {a, r } . Furthermore we see that \G\ = \Ti\\T2¡ = 5 • 4 =
20.
The procedure sketched above is called the Schreier-Sims algorithm.
The algorithm as sketched has the potential for producing large numbers
of generators, with adverse effects on efficiency. There is a simple observation,
however, that allows determination of a reasonable bound on \X\. Suppose
e.g. that G < Sn has base B = (bi,&2!--)- Suppose that o\ and 02 are
generators that move 61, and that b\l = b\a. Then <j\ and a-¡. can be replaced
by o\ and 020^, and oio^x stabilizes 61. When all such replacements are
made there can be at most n - 1 generators that actually move b\. Similar
replacements assure at most n - 2 generators that stabilize 61 and move hi,
etc. Thus as we move down the stabilizer chain we see that we may keep
2
|X| < (n - 1) + (n - 2) + • ■ • + 2 + 1 =
^—^.
Exercises
1. Apply the Schreier-Sims algorithm to the subgroup of Se generated
by (13)(46) and (125). You should find a base of length 3, and the
original generating set will have to be enlarged to include generators for
stabilizers. Find |G|.
2. Apply the algorithm to the subgroup G of 57 in the exercise on page 5.
As the exercises above show it soon becomes rather tedious to carry out
the algorithm by hand. Fortunately a sophisticated version of it has been implemented in the group theoretical software package GAP [57]. The symbolic
algebra package Maple [12] also has rudimentary permutation group operations available, including calculations of orders, which presumably also relies
on an implementation of the Schreier-Sims algorithm.
Here is a brief sample of a GAP session, during which the Schreier-Sims
algorithm is carried out in the background.
gap> G :=
GToyip(.(.í,2,3,A,5,&,7,S,9,ÍO,U),
(3,7,11,8X4,10,5,6));
GroupCd,2,3,4,5,6,7,8,9,10,11), (3,7,11,8) (4,10,5,6))
CHAPTER
8
1.
PRELIMINARIES
gap> Size(G);
7920
The group in the example will reappear in Chapter 2.
GAP is implemented for most commonly used computer platforms, such
as Unix, DOS, and Macintosh. It is available at no charge from a number of
sources in Europe and the United States. Readers are urged to search for it
on the World Wide Web, to download it, and to use it.
Exercise
Use GAP to find the order of the subgroup of S^ generated by the two
permutations in the example above together with the permutation
(1,12)(2,11)(3,6)(4,8)(5,9)(7,10).
The Sylow theorems for finite groups can be proved simultaneously by
means of appropriate group actions. Recall that if p is a prime and pk is the
exact power of p dividing the order of a group G, then a Sylow p-subgroup
of G is any subgroup of order p*. Write np for the number of distinct Sylow
p-subgroups in G.
Theorem 1.2.5 (L. Sylow) Suppose that G is a finite group, p G N is a
prime, and \G\ =pkm, withpjfm.
Then
SI: Sylow p-subgroups exist,
S2: any 2 Sylow p-subgroups are conjugate in G, and
S3: np = 1 (mod p).
Proof We may assume k > 1. Denote by V the set of all p-subgroups of
G that are maximal relative to inclusion. Then G acts on V by conjugation,
as does any subgroup of G. Note that if P ^ Q in V then Qp ^ Q (i.e.
Orbp(Q) 5e {Q}), or else P < PQ, a p-subgroup since PQ/Q ^ P/P n Q,
contradicting the maximality of P. Fix P e V and let O denote Orbß(P).
Then P is the only fixed point in the action of P on O, so \ö\ = 1 (mod p).
If Q € V \ Ö then there are no fixed points in the action of Q on O, so \0\
= 0 (mod p), a contradiction, and hence O = V. Thus S2 and S3 are true if
SI is true.
As for SI, if \P\ < pk then p | [NG(P) : P) (since [G : NG{P)] =\V\
= 1 (mod p)), and so NQ(P)/P
has a subgroup L/P of order p by Cauchy's
Theorem. But then \L\ — p\P\, a final contradiction.
A
Since every p-subgroup of G is contained in a maximal p-subgroup the
following corollary follows immediately.
Corollary 1.2.6 If H is a p-subgroup of G then H is contained in a Sylow
p-subgroup ofG.
1.3. COSET
1.3
ENUMERATION
9
Coset Enumeration
Recall that a group G is finitely presented if it is determined by a finite set
of generators which are subject to a finite set of defining relations. Typical
notation is G = (xi,..., xm | TI = 1 , . . . , rn = 1). The relatora r i , . . . , r m are
words of the form yiyí-yt,
where each yi is either one of the generators Xj
or its inverse. The actual meaning of the notation is that G = Fx/N, where
Fx is the free group based on the set X and N is the normal closure in Fx
of the subgroup generated by the set R of relators (e.g. see [31]).
In 1936 J. A. Todd and H. S. M. Coxeter published a paper ([62]) giving
a procedure for enumerating the cosets of a subgroup in a finitely presented
group. The subgroup must be given by a finite set of generators, which are
words from X * = X U X~l, X being the set of generators for the group.
In order to describe and illustrate the procedure let us suppose that G
has generators X = { x i , . . . ,xm} and relators R = { n , . . . , r „ } , and that H
has generators S = { s i , . . . , s*}. Suppose further that Tj = x¿iXj2 ■ • • Xj*4 and
that Si = j/iiy»2 • • Viti, all i, each x¡j and yy being in X±.
The cosets of H in G are labeled by positive integers, beginning with
1 := H. The generators s< (as words) are each put at the top of a subgroup
table, which will have just one row in it. At the beginning and ending of
that row are the label 1 for the coset H, reflecting the fact that lsj = 1 (i.e.
H Si = H, since s¿ 6 H). Each relator r¿ (also as a word) is put at the top of a
relator table, which may have many rows (one for each coset, as they become
labeled). At the outset the label 1 is placed at the beginning and end of the
first row of each relator table, reflecting the fact that each r,- = 1, and hence
lr¿ = 1 (i.e., Hn = H).
Two further tables are set up, basically for bookkeeping purposes - a
coset table, in which are stored the relations between cosets., and a. deftnitions/bonuses for D/B^ table, which will be explained shortly.
The procedure is started by defining a second (presumably different) coset
labeled 2, and placing it in a blank position in one of the subgroup or relator
tables, normally next to one of the l's. For example, since ri begins with i n
the definition could be 2 := l x n , meaning that the 2nd coset is Hx\\. Once
coset 2 has been defined 2's are placed at the beginning and end of the second
rows of each relator table, the definition is written in the definition part of the
D/B table, and the fact that l x n = 2 (as well as that 2 x n 1 = 1) is recorded
in the coset table. Then the subgroup and relator tables are scanned to see
whether the information available is sufficient to fill in any further blanks.
After the scanning another coset is defined, and the process continues.
When a row fills up in a subgroup or relator table it is because some coset
k fills the last blank space just to the left of another coset £, and at the top of
the table between k and I is a generator (or inverse), say y, forcing on us the
information that ky = I. Three possible situations then arise: (1) It may be
already known that ky = I, in which case we simply continue. (2) It may be
already known that ky = j , where j and £ had been presumed different from
CHAPTER
10
1.
PRELIMINARIES
each other. In this case they must be acknowledged to be the same (this is
usually called a collapse), and it is necessary to replace the larger of the two
by the smaller in all tables, then proceed. (3) The fact that ky = I is simply
new information (called a bonus, which is entered in all tables, including the
bonus part of the D/B table.
When (and if) all rows of all tables fill up consistently, then the number
of rows in each relator table is the index [G: H] (for a careful proof of this
fact see page 11 of [48].
Let us illustrate the procedure first with a very simple example, viz. G =
(x x 3 = 1), and we take H = 1, so no subgroup tables will be required and
the procedure will simply determine the order of G. The relator table, coset
table, and D/B table begin as follows.
D
B
Define 2 := lx and enter the information in the tables.
X
x- 1
1 2
2
1
D
B
2 = lx
The 1 in the second row of the relator table resulted from scanning. Next
define 3 := 2x and enter the information. Before scanning we have the following.
x
1
2
3
3
1
->
1
2
3
2
3
3
x
1
2
D
2 = lx
3 = 2x
Notice that the first row of the relator table has filled up, and we have
obtained a bonus, viz. 3x = 1 (marked for convenience in the table by an
arrow). When that information is entered we see that all blanks are filled in
the coset table, ensuring that all rows in the relator table can be completed,
and the procedure finishes.
x- 1
1 2 3
2 3 1
3 1 2
X
3
1
2
->
1
2
3
D
2 = lx
3 = 2x
B
3x = l
We see that \G\ — 3 (as expected, of course), but also that the procedure
has provided a permutation representation of G, viz. the multiplication action
of the generator x on the set of 3 cosets of the trivial subgroup. Directly from
1.3. COSET
11
ENUMERATION
the coset table we see that x i-> (123), determining a homomorphism from
G into S3 (in fact an isomorphism from G to .A3).
For another example, take G = (x,y | x 3 = y 2 = (xy)2 — 1), this time
with H = (x2y). Thus we begin with a subgroup table for the generator of
H (recall that it will have only one row), 3 relator tables, and as always the
coset table and D/B table. After the first definition 2 := lx has been made,
and after scanning, the situation is as follows.
X
1 2
2
x-1
y
D
B
lx
1
Note that there is no column for y _ 1 in the coset table, since it is clear from
the presentation that ¡/ _1 = y. Defining 3 := 2x we obtain the bonuses 3y = 1
in the subgroup table, then 3x = 1 in the first relator table. Further scanning
in the third relator table yields 2y — 2, and all tables can be completed.
x
X
2
1
1
2
3
y
3
y
1
2
3
1
x
1
2
3
1
y x"
1 2 3
3
2 3 2
1
3 1 1
2
X
y
-►
1
2
3
2
2
y
->
D
2 = lx
3 = 2x
3 ->
1
x
í»
3
y
1
2
3
1
2
3
B
3ff = 1
3x = 1
2y = 2
Thus [G: H] = 3, and the permutation action of G on the cosets is given by
xi-> (12 3), y ►-> (13), providing a homomorphism from G onto S3. Observe,
though, that (x 2 y) 2 = 1 (why?), so \H\ < 2 and \G\ = [G: H]\H\ < 6, and in
fact G s S 3 .
CHAPTER
12
1.
PRELIMINARIES
Note that we can determine a (Schreier) transversal T for H in G directly
from the definition table. We begin with 1 € 1 (= H), then observe that
2 = lx, so we may take x as a representative for 2. Next 3 = 2x = l x 2 , so
we take x 2 as a representative in 3, and we have T = { l , x , x 2 } .
For a final example take G = {x,y \ xy2 = 1, x2y3 = 1) and H = 1.
Begin with the definitions 2 := lx, 3 := 2y, and 4 := 2x. The first bonus is
3y = 1, then, from the first row of the 2nd relator table, 4y = 2. When that
information is inserted in the second row of the first table a further bonus of
2y = 2 is obtained, and the tables are as follows.
x
1
2
3
4
x
1
2
3
4
x
1 2
2 4
3
4
y
y
3 -► 1
2 -► 2
2
3
4
2
4
x
2
4
4
y
->
y
2
y
3
2
y
x-1
3
1
2
1
y-1
3
4
2
2
1
2
3
4
£)
B
2 = l x 3y = l
3 = 2y 4y = 2
4 = 2x 2y = 2
Observe now, though, that we have 2y = 3 by definition and 2y = 2 as a
bonus, so we must conclude that 3 = 2, a collapse. Normally this would call
for replacing each 3 by a 2, copying all the information from 3's row into 2's
row, removing 3's row, and continuing. In this case, however, we see from the
coset table that 2y = 3, which becomes 2y = 2, and 3y = 1, which becomes
2y = 1, so also 2 = 1, another collapse. Finally, l x = 2 and 2x = 4 allows
us to conclude as well that 4 = 1 , and we have total collapse. All tables fill
(with l's), and we conclude that \G\ = 1.
It is not difficult to draw the same conclusion from the relations: 1 =
x 2 y 3 = x{xy2)y = xy, then 1 = xy2 = (xy)y = y; hence also x = y'1 = 1.
The last example can be generalized by setting
G = (x,y | xnyn+l
= x n + 1 y n + 2 = 1),
1.4. SEMIDIRECT
PRODUCTS
13
where n is any positive integer. Just as above it is easy to check that G is
trivial, but if the cosets of H = 1 are enumerated it is clear that 2n cosets
must be defined before any row can fill up, thus before there is any hope of a
collapse occurring. This simple example shows that there can be no general
bound in terms of the index of the subgroup for the number of possibly
superfluous cosets that may have to be defined to carry out the enumeration.
Exercises
1. Let G = (x, y, z, u, v | xy = z, yz = u, zu = v,uv = x, vx = y). Enumerate the cosets of H = (x) in G, and conclude that G is cyclic (in fact it
is cyclic of order 11).
2. Do a coset enumeration to determine the order of
G = (x, y I xyx = y, yxy = x).
Do you recognize G?
It is easy to tell from a completed coset table for H in G whether or not
H <G. Say H has generators { « i , . . . , s*} and [G: H] = k. Then H < G if
and only if xs¿x - 1 € H for 1 < i < k and all x e G, which is if and only if
(HX)SÍ = Hx for all i and x. But the cosets Hx are now labeled as 1, . . . , k,
so that condition becomes simply £SÍ = £ for every coset I and every generator
st of H.
In the second example above, with H = (x2y), we observe e.g. that
2
2x y = 3xy = \y = 3 ^ 2, so H is not normal in G.
Coset enumeration is at the heart of several important procedures in computational group theory, including the sophisticated version of the SchreierSims algorithm mentioned in the previous section, a method for finding a
presentation for a subgroup of a finitely presented group, and a method for
determining all conjugacy classes of subgroups of low index in a finitely presented group. All of these procedures, and more, are available in GAP [57].
For a more thorough discussion of coset enumeration and some of its ramifications see [48], [43], or [44].
1.4
Semidirect Products
We review briefly the definition and construction of semidirect products of
groups, as they will prove useful in later chapters.
Recall that a group G is an extension of K by H if K < G and G/K 2 H.
As a special case G is called a split extension, or (internal) semidirect product,
of K by H if K < G, H < G, K D H = 1, and G = KH. It will be denoted
by G = K x H. For example if G = Sn, K = An, and H = ((12)), then
G = K x H.
CHAPTER
14
1.
PRELIMINARIES
Note that if G = K x H then every x G G can be written uniquely as
x = kh, with A; G if, h G H, and the multiplication proceeds as follows:
{kih1)(k2h2)
= kih^h^hthi
= (h
■hlk2)(h1h2).
To construct a semidirect product (external) we require the groups K and
H, together with a homomorphism 9:H-¥ Aut(if). For present purposes it
will be useful and suggestive to write the action of any <p G Aut(if ) using
what might be called left exponential notation; i.e. if A; G if then vk is the
image of k under the action of ip.
Now the construction: define G as a set to be the cartesian product if x if,
and define the product via
(kuh1)(k2,h2)
=
(k1-e^k2,h1h2).
It will be denoted G = K ~A% H, or simply by G = K x H if 6 is understood.
Exercise
Show that G = i f x f l i / i s a group and that it has subgroups K\ S K
(normal) and H\ Si H such that G = K\ x H\ (internal).
It is common practice to identify K with K\ and H with Hi (in the exercise), thereby blurring the distinction between internal and external semidirect products.
For a familiar class of examples take K = C„ = (a), cyclic of order n,
H = C 2 = (6), cyclic of order 2, and define 6: H -> Aut(iT) via eba = a - 1 .
Then Dn = H x K is the dihedral group of order 2n.
For another class of examples let K be any group, set H = Aut(if), and
take 6 to be the identity map. Then the semidirect product G = K x H is
called the holomorph of K, denoted Hol(/Q.
Exercises
1. If if is Klein's 4-group then Hol(if ) is a familiar group; determine which
one.
2. If G is a group and H <G recall that H is characteristic in G (notation:
H chai G) ïffheH
for all ^ G Aut(G) and all h G H. For any group
K and L < K show that L char if if and only if L < Hol(if ).
1.5
Wreath Products
Suppose that G is a group and that 5 is a set. Write Gs = {f:S -> G},
the set of all functions from S to G. Then G s is a group, the operation
1.5. WREATH
15
PRODUCTS
being point-wise multiplication: (fg)(s) = f{s)g{s). In fact Gs is simply the
cartesian product of \S\ copies of G.
If G, H are groups, and if H acts on a set S, then also H acts on Gs by
means of xf(s) — f(sx), all x 6 H, f G G 5 , and a G S. It is easy to check
that <*»>/ = x(yf).
The action of H on Gs determines a homomorphism 9 from H to Aut(G 5 ),
given by 6{x) = 9x:f >-> */> all a: G i/, / G G s . To see that observe first
that *{fg){8) = (fg)(sx) = f(ax)g(ax) = xf(s) ■ xg(s) = (xf ■ xg)(a), so 9X is
an endomorphism of Gs. Next, ker 9X consists of all those / € Gs for which
*f(a) = f(ax) = 1 € G for all s € 5, hence for all ax G S, i.e. f = 1 € Gs.
Eachflj is onto, since 9X(X~1 f) = / f o r any / G G s , soo* G Aut(G 5 ). Finally,
check that 9xy = 9x9y, all x, y G H .
We may now define the wreath product of G and H, relative to S, denoted
GlH,to be the semidirect product Gs x>e H. Note that \G\H\ = | G | | 5 | | i / | .
The multiplication is given by (f,x)(g,y) = (/ ■ xg,xy).
Exercise
Show that Gl H is abelian if and only if both G and H are abelian and
the action of H on S is trivial.
For an easy example take G = H = Sym(2), with the natural action of H
on S = {1,2}. Then G l H has a normal subgroup isomorphic with G x G, a
Klein 4-group, and \G I H\ = 8, so GI H is dihedral.
Suppose that G, H, and 5 are as above, and suppose further that G acts
on a set T. We may then define a group action oî G Ï H on the cartesian
product T x S via (i,s)^- x ) = (t^*\sx).
It is straightforward to check that
the action really is a group action.
Proposition 1.5.1 / / G acts on T with kernel K and H acts on S with
kernel L then the action of G l H onT x S has kernel K\L.
Proof We have (/, x) in the kernel of the action if and only if (t, s)^,x^ =
{tf('\sx)
= (t,s) for all t G T, s G 5, or equivalently sx = s and «/<•> = t, all
s and t, i.e. x 6 L and f(s) G K, all s, or / G ATS, as required.
A
Corollary 1.5.2 The wreath product G I H acts faithfully on T x S if and
only if G and H act faithfully on T and S, respectively.
More general notions of wreath product exist; e.g. see [60]. In general,
wreath products are often used for constructing pathological examples. For
example a wreath product construction can be used to construct an infinite pgroup having trivial center and a proper subgroup that is its own normalizer.
For present purposes the construction will mainly be used to describe Sylow
subgroups of symmetric groups.
Exercise
If (f,x) eGlH
calculate ( / , x ) _ 1 explicitly.
CHAPTER
16
1.6
1.
PRELIMINARIES
Transitivity and Primitivity
Suppose that a group G acts on a set S, with \S\ = n, finite. If k < n say that
G acts k-transitively if, for each pair of fc-tuples ( a i , . . . , a*) and ( 6 i , . . . , fejt)
of disii'nci elements of S, there is some x 6 G for which af = b¡, 1 < i < k.
Clearly fc-transitivity implies ¿-transitivity for 1 < £ < k. For an obvious
example note that the symmetric group S„ acts n-transitively on { 1 , . . . , n } .
Exercise
Show that a transitive group G of degree n is fc-transitive, for 2 < fc < n,
if and only if the stabilizer G„ is (k — l)-transitive on S \ {s} for each
seS.
Proposition 1.6.1 If n > 3 then the alternating group An acts (n — 2)transitively, but not (n — I)-transitively, on { 1 , . . . , n } .
Proof We use induction on n, beginning with the fact that A3 is transitive,
but not 2-transitive, on {1,2,3}. For n > 3 we have Stabyi n (n) = A„-i, and
A„-i is (n - 3)-transitive on { 1 , . . . ,n - 1} by the induction hypothesis, so
An is (n - 2)-transitive by the exercise above. If it were (n — l)-transitive
there would be some a € An fixing each of 1, 2, . . . , n — 2 and carrying n — 1
to n. But the only a € Sn that does this is the transposition (n — 1 n) ^ A„.
A
It is customary to view Sn and An as trivial examples of highly transitive
groups. Every group has a transitive action (on itself - that is Cayley's
theorem). We shall see later many examples of doubly and triply transitive
groups, and even some 4- and 5-transitive groups. It is a very deep fact that
there are no nontrivial fc-transitive groups for k > 6.
If G is transitive on S define the rank of G to be the number of G,-orbits
in S (for any s € S). Clearly the rank is at least 2 (if \S\ > 2), since {s} is a
G„-orbit. Note that if G is fc-transitive on S and fc > 2 then rank(G) = 2.
Proposition 1.6.2 The rank of a transitive permutation group G on a set
S is equal to the number of distinct double cosets GsxG, of the stabilizer G,
for any s £ S.
Proof Fix s e S. For x,y € G, s x and sv lie in the same G,-orbit if and
only if sxG- = syG- = sG-xG' = sG^G>. The latter equality holds if and only
if G,xG, = G,yGB, for if sG'xG- = sG^G- then there are o, b e G, such that
saxb _ s ! / ) h e n c e O I t y - i e Q^ axo e Qfy g G,yG„ and so G,xG, Ç G,yGs.
The reverse inclusion is similar, and they are equal.
A
Corollary 1.6.3 If G is k-transitive on S, k> 2, then
G = G s U GsxGe
for any s g S and any x €
G\Ge.
1.6. TRANSITMTY
AND
PRIMITIVITY
17
Proposition 1.6.4 If G is k-transitive on S and \S\ = n then
In fact, ifTÇS
and \T\ = k then [G:GT\ = n\/(n - fc)!.
Proof Write T = {s1:. ..,sk}.
Then \G\ = n ■ | G „ | by Proposition 1.2.1,
and G 4l is (fc - l)-transitive on S \ {si} by. the exercise on page 16. Similarly
|G 4 l | = (n - l ) | G S l , , J , so \G\ = n(n - l)|G S l , i 2 |, etc.
A
Say that G is sharply fc-transitive on S if it is fc-transitive and there is
a subset T Ç S, \T\ = fc, with GT = 1. If G is finite and \S\ = n, finite,
then the definition is equivalent, by Proposition 1.6.4, with requiring that
\G\ = n!/(n — fc)!. Thus, for example, S„ is sharply n-transitive, and An is
sharply (n — 2)-transitive, since n\/(n — (n - 2))! = n!/2 = \An\.
We shall see examples of sharply 2-transitive and sharply 3-transitive
groups in Chapter 2.
For examples of sharply 1-transitive actions think back again to Cayley's
theorem. The right regular action of any group G on itself is determined by
right multiplication: x: y »-► yx; it is transitive and the stabilizer of any point
is trivial.
In general, if G is sharply 1-transitive on S we say that G is regular on 5, in
which case | 5 | = \G\. Note, in fact, that if we fix s £ S then S = {sx:x
€G},
and if x, y € G then (sx)y = sxy. It follows easily that the action of G on 5
is equivalent with the right regular action of G on itself.
We shall return to regular actions after introducing the notion of primitivity.
If G acts on 5 then a block is a subset B Ç S such that B ^ S, \B\ > 1,
and if a; 6 G then either Bx = B or B n Bx = 0 . If G is transitive on S and
has no blocks we say that G is primitive on S.
For a negative example let G be the dihedral group D4 acting on the
vertices {1,2,3,4} of a square. If 1 and 3 are opposite vertices then clearly
{1,3} is a block, so D4 is not primitive.
Exercise
If G is transitive but imprimitive on S and B is a block for G show that
\B\ I \S\. Conclude that if \S\ = p, a prime, then G, if transitive, must be
primitive.
Proposition 1.6.5 Suppose that G acts transitively on S. Then G is primitive if and only if each stabilizer G,, s € S, is a maximal (proper) subgroup
ofG.
Proof Exercise.
A
CHAPTER 1.
18
PRELIMINARIES
Proposition 1.6.6 If G is doubly transitive on S then G is primitive.
Proof Suppose that B Ç S, with \B\ > 1 and B ¿ S. Choose a, b G B,
a ^ 6, and c G S \B. Then choose x £ G with ax = a and bx = c. Then
a G Bx n B so Bx n B ¿ 0 , and c G B* \ B , so Bx ¿ B. Thus B is not a
block for G.
A
Proposition 1.6.7 Suppose that G is primitive on S and K<G is the kernel
of the action. Suppose that N <G and N £ K. Then N is transitive on S.
Proof Since N £ K there is an TV-orbit in 5 having more than one element,
say B = Orb,v(a) and \B\ > 2. If x G G then Bx = aNx = axN (since N<G).
Thus Bx = OTONÍO,"). Since 5 is partioned into its TV-orbits we have either
Bx = B or Bx n B = 0 , and consequently B = S, or else B would be a block
for G.
A
Proposition 1.6.8 Suppose that G acts on S, N <G,
onS. If se S then G = Stab G (s) • N.
Proof If i € G then sx = sv for some y € N. Thus sxy
Stab G (s) and x G Stab G (s) • y Ç Stab G («) • N.
and N is transitive
= s, so x y _ 1 €
A
The next theorem will be applied in Chapter 2.
Theorem 1.6.9 (Iwasawa) Suppose that G is faithful and primitive on S
and G' — G. Fix s € S and set H = Stab G (s)- Suppose there is a solvable
subgroup K <H such that G = (U{KX: x 6 G}). Then G is simple.
Proof Suppose that 1 ^ N <G. By Proposition 1.6.7 N is transitive on 5 ,
and by Proposition 1.6.8 G = HN. Thus KN < HN = G. If x e G then
Kx < (KN)X = KN, so Li{Kx:x €G}CKN
and hence KN = G. Since K
is solvable K^m^ = 1 for some K^m^ in the derived series. Check inductively
that (KN)W < K&N for all /. Thus G = G^ = (KN)W < K^N
= N
and N = G.
A
Theorem 1.6.10 Suppose that G is primitive and faithful on S, and that Gs
is simple, s € S. Then either
1. G is simple, or
2. G has a normal subgroup N that is regular on S.
Proof If G is not simple choose N<G, 1¿ N ¿G. If s 6 S then NnGs <GS,
which is simple, so N D G, = Gs or 1. If N n G8 = Gs then G„ < N, but
G, ^ N since N is transitive by Proposition 1.6.7. But then N = G by
Proposition 1.6.5, a contradiction. Thus N DG, = 1, i.e. N, — 1, all s e 5 ,
so N is regular on 5 .
A
1.6. TRANSITIVITY
AND
PRIMITIMTY
19
Theorem 1.6.10 could be a useful simplicity criterion if regular normal
subgroups were relatively rare, so let us consider them.
Note first that if N < G then G acts by conjugation on N* = N \ {1}. If
G also acts on a set 5 , then each G„ s € S, acts on N* as well.
Proposition 1.6.11 Suppose that G acts on S, N <G, and N is regular on
S. If s € S then the action ofG, on N* by conjugation is equivalent with the
action of G, on S\ {s}.
Proof Since AT is regular on S we have \N\ = \S\ and S = {sy:y € N}.
Thus f(y) = sv defines a bijection from N* to S \ {s}. If x € G, then
A
f{yx) = sx yx = syx = f(y)x, so the actions are equivalent.
Recall that a finite group is an elementary abelian p-group (p a prime) if
it is isomorphic with a direct product of cyclic groups of order p.
Theorem 1.6.12 Suppose G acts on S, N <G, and N is regular on S.
1. If G is doubly transitive on S then N is elementary abelian and \S\ = pk
for some prime p and some k € N.
2. If G is 3-transitive on S then either N is an elementary abelian 2-group
or N is cyclic of order 3, so \S\ = 2* or 3.
3. If G is 4-transitive on S then N is a Klein four-group and \S\ = 4.
4- G cannot be 5-transitive.
Proof (1) Since G is doubly transitive G, is transitive on S \ {s}, and
therefore also on N* by Proposition 1.6.11. Choose a prime p such that p | |AT|
and choose x € N with |x| = p. By transitivity |y| = p for all y € N*, so N
is a p-group. The action (by conjugation) of each x € Gs on N determines
an automorphism of N, so Z{N)X = Z(N), and Z(N) ¿ 1, hence Z(N) = N
by transitivity; i.e. N is abelian. Thus N 5i Z* for some k by the structure
theorem for finite abelian groups.
(2) Now G, is doubly transitive on N*. By part (1) AT is an elementary
abelian p-group. If p > 2 choose x G N* and note that x / x - 1 . The
stabilizer in G, of x is transitive on N* \ {x}, but also any z € S t a b e (x)
effects an automorphism of N by conjugation, so ( x _ 1 ) z = x - 1 . It follows
that N* \ {x} = { x - 1 } , hence JV = { l , x , x - 1 } S Z 3 .
(3) Now G, is 3-transitive on N*, so |JV*| > 3 and \N\ > 4. By parts (1)
and (2) AT is an elementary abelian 2-group. If |AT| > 8 then N contains a
Klein four-group A" as a subgroup, say with K* = {a,b,c}, and also some
d e N\K.
By 3-transitivity there is some x £ G, with ax = a, bx = b, and
c" — d. But then d = c* = (ab)x — axbx = ab = c, a contradiction, so in fact
|JV| = 4.
(4) If G, were 4-transitive then we would have \N*\ > 4, but also, by part
(3), \N\ = 4, a contradiction.
A
20
CHAPTER
1.
PRELIMINARIES
As an application we may (re-)prove the simplicity of the alternating
groups An for n > 5.
To begin, recall that A$ has conjugacy classes of sizes 1,12,12, 15, and 20,
so A$ is simple by Lagrange's theorem. This starts an induction. Assume that
n > 6 and that An-\ is simple. Write G = A„ and Gn = Stabo(n) = An-\.
Then G is 4-transitive but of degree greater than 4, so it cannot have a regular
normal subgroup by Theorem 1.6.12; hence it is simple by Theorem 1.6.10.
1.7
Some Linear Algebra
The material developed in this section will be used in the next chapter in the
discussion of symplectic groups.
Suppose that F is a field and V is a vector space over F. Then its dual
space, V* = Homf(V), is the set of all linear functionals from V to F; it is
also an F-vector space. If {vi,..., vn} is a basis for V define { u j , . . . , u*} Ç
V* via v*(vj) = ôij, all i and j (and of course extend by linearity). This
= a
defines the dual basis for V*. If £)t a « u ? = 0 t n e n 0 = 51iaivï(vj)
i->
all j ; if / € V* set f(vi) = &¡ and check that / = ^biV*.
In particular
dimp(V*) = dimp{V) (when finite).
A bilinear form on V is a function B: V x V -> F that becomes a linear
functional in either variable if the other variable is fixed, i.e.
B(u + v,w)
B(u,v + w)
=
=
B(u,w) + B(v,w)
B(u,v) + B(u,w)
and B(au,w) = aB(u,w);
and B(u,aw) — aB(u,w)
for all u, v, w € V, all a € F.
If B is a bilinear form on V and {v\,... ,vn} is a basis for V set bij =
B(vi,Vj), all i, j . Then B = [b^] is called the matrix of B relative to {vi}.
If it becomes necessary to exhibit the dependence on the basis we will write
B{Vi\. If u,w € V write u = ^ O j U j and w = Ylic*vii s o u anc ^ w a r e
represented by column vectors (coordinate vectors) û = (ai,... ,a„Y and
w = ( c i , . . . ,0«)'. Then
ß(u,tü) = B^aiVi^CjVj)
»
= ^2aiB{vi,Vj)cj
i
=
tfBw.
i,j
Conversely, if B — [bij] is any n x n matrix over F we may define a
bilinear form (depending on a choice {i>j,... ,vn} of basis) by decreeing that
B(vi,Vj) = bij and extending by linearity.
If { u n , . . . ,wn) is another basis write uij = ^djj-Wj, dy G F , for all
j . Then B(u>i,Wj) = £*,i d *« ß ( u *»*>/H,.; = E*,/ d *tí'*/d/,j, the ¿¿-entry
of D'BD, where D = [dy] is the (invertible) change-of-basis matrix. In
particular, any two representing matrices have the same rank, which we define
to be the rank of the form B.
1.7. SOME LINEAR
21
ALGEBRA
Classically, square matrices M and N are called congruent if M = DlND
for some invertible matrix D, so two representing matrices for a bilinear form
B are congruent. Note that then det M = (det D)2 det N.
Write F x 2 = {a2: a 6 F*}, the subgroup of squares in the multiplicative
group F*. If Z? is a bilinear form on V and B is a representing matrix define
the discriminant of 2? to be
discrü?) = í °
~
{
'
\(detB)Fx2eF*/F*2
if det £ = 0,
otherwise.
Note that discr(B) is independent of the choice of basis. Say that B is
nondegenerate if discr(B) ^ 0.
If B is a bilinear form on V define two maps L and R from V to V* as
follows: L:v y+ Lv, R:v i-+ A,,, with L„(u;) = B(u,u>) and Rv(w) = B(w,v),
all iu € V. It is easy to check that L and R are linear transformations.
Define the left and rt<//i£ radicals of F (relative to ¿?) to be
rad¿(V) = ker L = {v € V: £(«,«;) = 0, all w € V},
r a d ñ ( F ) = ker fl = {v € V: B(w, v) = 0, all tu € V}.
Proposition 1.7.1 A bilinear form BonV
radL{V) = 0 if and only if radR(V) = 0.
is nondegenerate if and only if
Proof Choose a basis {vi,... ,v„} and write B = [6¿j] for the matrix representing B. Note that v e radfl(V) if and only if B(vi,v) = 0, all i. Write
v = YLjaivj^ ai G F. Then B(VÍ,V) = J2j^ijai = 0, all i, if and only
if the column vector {a%,... , a n ) 4 is a solution to the homogeneous system
BX = 0. Thus radfi(V) = 0 if and only if det ß # 0, and the equivalence
with radi,(V) = 0 is similar.
A
Corollary 1.7.2 A bilinear form B on V is nondegenerate if and only if
V* = Im L = Im R, i.e. if and only if given any f e V* there exist u, v € V
such that f(w) = B(w, u) = B{v,w), all w € V.
The corollary generalizes.
Corollary 1.7.3 Suppose that B is a nondegenerate bilinear form on V and
that W is a subspace of V. If f € W* then there exist u, v € V such that
f = Ru\w = Lv\wProof We may choose a basis {wi,... ,uik} for W and enlarge it to a basis
{tüi,... ,«;„} for V. Extend / to A g V* via fx\w = f, / x ( V i ) = 0 for
i > k. By the first corollary 3u,v € V with fi = Ru = Lv, and hence
/ = Ru\w = Lv\wA
CHAPTER
22
1.
PRELIMINARIES
For any subset S Ç V define
B(v, w) = 0, all w € 5 } ,
±L (S) = {veV:
-L« (5) = { c € V:ß(«;,u) = 0, all w € 5 } .
Thus v e l i (5) if and only if Lv\s = 0,V€±R
(S) if and only if fi^s = 0.
Note that 1L (V) = rad L (V) and ±R {V) = raAR{V).
Exercise
Show that
1. ±L (S) and A-R (S) are subspaces of V,
2. _U
(_LH
(5)) D S and ± f i ( ± ¿ (5)) D S, and
3. if 5 C T then ±L (T) c ± ¿ (5) and ±R (T) C±R (5).
Proposition 1.7.4 If B is a nondegenerate bilinear form on V and W is a
subspace of V then
dim LL (W) = dim ±R (W) = dim V - dim W.
Proof The map 0:v i-> L„|w is a linear transformation from V onto W*
(Corollary 1.7.3), and ker 0 =±L (W) by definition. Thus dim V = dim W*+
dim ±£ (W), and similarly dim V = dim W* + dim ±R (W).
A
Corollary 1.7.5 If B is a nondegenerate bilinear form on V and W is a
subspace ofV then ±L {±R (W)) =±R (±L (W)) = W.
Proof By the exercise above J_L (±R (W)) D W, and
dim J_£ (±R (W))
=
=
dim V - dim ±R (W)
dim V - (dim V - dim W) = dim W.
A
Exercise
If B is nondegenerate on V and S is any subset of V show that
±L (±R (S)) =±R (±L (S))
is the subspace spanned by S.
Suppose that B is a bilinear form on V. Say that B is symmetric if
B(v,w) = B{w,v) for all v,w G V. Say that B is alternate if fî(u,u) = 0 for
all v e V. Note that ß is symmetric if and only if any representing matrix
B is symmetric, i.e. Bl = B.
1.7. SOME LINEAR
23
ALGEBRA
If B is an alternate form and v,w e V then 0 = B(v + w,v + w) =
B(v, v) + B(v, w) + B(w, v) + B{w, u>), so B(w, v) = -B(v, w). If char F = 2
this shows that B is symmetric; in general it shows that B is skew symmetric
(also called antisymmetric). If char F ± 2 then the concepts of alternate
form and skew symmetric form are equivalent, for then skew symmetry entails
B(v,v) = -B(v, v), or B(v,v) = 0. In general, a representing matrix B =
[bij] for an alternate form is an alternate, or skew symmetric, matrix, meaning
that bu = 0, all i, and Bl = -B.
Proposition 1.7.6 Suppose that B is a bilinear form on V satisfying
B(u, v)B(w, u) = B(v, u)B(u, w)
(*)
for all u,v,w € V. Then B is symmetric or alternate.
Proof Take u = v and conclude that
B{v, v)[B(w, v) - B(v, w)\ = 0
(**)
for all v,w £ V. We wish to show that either B(v,v) = 0 for all v € V, or
B(w, v) = B(v, w) for all v, w € V. If that is not the case then 3a;, y, z € V
with B(y,y) ^ 0 and B(x,z) ^ B(z,x). Apply (**) and obtain (i) B(x,x) =
B{z,z) = 0, (it) B(x,y) = B(y,x), and (Hi) B(y,z) — B(z,y). Use (*), with
u = x, v = y, and w = z, to conclude from (ii) that B(x,y) = B(y,x) = 0.
Interchange u and w in (*) and get, again with u = x, v — y, and w = z, that
B(z,y) = B(y,z) = 0 from (tü). Thus B(x,y + z) = B(x,z) ¿ B(z,x) =
B(y + z,x), and, by (**),
0 = B(y + z, y + z) = B(y, y) + B(y, z) + B(z, y) + B(z, w) = B(y, y) ¿ 0,
a contradiction.
A
For any bilinear form B on V, and v, w e V, say that v is orthogonal to
w, and write v ± w, if B(v, w) — 0. If orthogonality is a reflexive relation on
V (i.e. v -L w =*• w A. v) then the form B is called reflexive. Note that if
B is a reflexive form then _L¿ (5) = ± A (5) for all subsets S ÇV; -we agree
to write S± = ± ¿ (5) = l f i (5) in that case. If W is a subspace then W is
called the orthogonal complement of W. Note that W O WL = 0 if and only
if B\w*w is a nondegenerate form on W. If VF n W1- = 0 we say that W is
a nondegenerate subspace (relative to B).
In general, if B is a reflexive bilinear form on V and W is a subspace we
define the radical of W to be rad W = W n i y x .
Exercise
Suppose that B is a reflexive bilinear form on V, and V = U © W, with
Î/ -L W. Show that rad V = rad U © rad W.
CHAPTER
24
1.
PRELIMINARIES
Proposition 1.7.7 A bilinear form B on V is reflexive if and only if it is
either symmetric or alternate.
Proof «=: A symmetric form is clearly reflexive. If B is alternate it is skew
symmetric, so v Lw means 0 = B(v,w) = — B(w,v), and therefore w ± v.
=>: Take any u, v, w 6 V and set x = B(u,v)w-B(u,w)v.
Then B(u,x) =
B(u, v)B(u, w) - B(u, w)B(u, v) = 0, so u ± x. Thus also x ± u, which says
0 = B{x,u) = B{u, v)B{w,u) - B(v,u)B(u, w). Apply Proposition 1.7.6. A
Suppose that B\, B-i are bilinear forms on spaces Vi, V-j, respectively. An
isometry relative to B\ and B^ is an F-isomorphism a: V\ -* V2 satisfying
B2{ov,ow) — B\{v,w) for all v,w 6 V\. If an isometry exists the forms are
called equivalent.
Proposition 1.7.8 Bilinear forms B%, B2 on spaces Vi^V^ are equivalent if
and only if there are bases for V\, Vi relative to which B\ = B2 ■
Proof =>: Let a: V¡, -> V2 be an isometry. Choose a basis {vi,... ,vn} for
Vx, and set Wi = av¡, all i, to get a basis for V¡. Then B\ = [Bi(v¡,Vj)] =
\B2{ovi,aVj)\ = [B2{wi,Wj)\ = B2.
<=: Suppose B\ — B2 relative to bases { u j and {w¿} for Vi and V2.
Define a:V\ —> V2 via a(vi) = u>i, all 1, an isomorphism. If u,w Ç. V\ write
u = J \ ÜÍVÍ and w = J ^ b,Vi. Then
Bi (u, w)
=
^2 aibjB\
(vi, vj) = 5 Z
=
¿ J aibjB2{o-Vi,avj) = B2{^2 Oitrvi, ^
aibjB2(wi,Wj)
bjVVj) = ¿^(fu, aw).
A
Proposition 1.7.9 Suppose that B is a reflexive bilinear form on V, and
that W is a nondegenerate subspace of V. Then V = W © W x .
Proof We may choose a basis {vi,... , w* } for W and adjoin further vectors
{ufc + i,... ,vn} to extend it to a basis for V. Represent B by its matrix
B = [bij] relative to {u¿}, and represent vectors in V correspondingly as
column vectors in F". If v = £V a ^ € V, so v t-> ( 0 1 , . . . , a«)', then v € W x
if and only if 0 = B{VÍ,V) = £ \ B(vi,Vj)aj = £_, 6^0,- for 1 < i < fc, i.e. if
and only if ( a i , . . . ,an)1 is in the solution space of the homogeneous system
AX = 0, whose coefficient matrix A is the k x n matrix over F whose rows
are the first A: rows of B. Thus dim W1 = n - rank A > n - k. But W
nondegenerate means that rad W = W D Wx = 0 , s o W + WL = W 0 WX,
and d\m{W®W*-) = dim W+dim WL > k+(n-k) = n; hence W®W*- = V.
A
1.7. SOME LINEAR
25
ALGEBRA
Assume for the remainder of this section that B is an alternate bilinear
form on V.
0 1
For a simple example take V of dimension 2 and B =
Thus,
-1 0
if u =
Xi
yi
XT.
and v =
, then B(u,v)
= xyyi - X2yi- We'll see that
this example is surprisingly general.
If u,v 6 V and B{u,v) ^ 0 then {u,v} is linearly independent, for
B{u,au) = aB{u,u) = 0. If B(u,v) = b ^ 0 set ui = 6 _ 1 u and vi = v.
Then B(ui,vi) = 1. The subspace W spanned by u\ and vx is called a hyperbolic plane, with hyperbolic pair («i,t>i) as basis. Note that the restriction
0 1"
of B to W has representing matrix - 1 0 relative to a hyperbolic pair
as basis.
A bit of notation: If U and W are subspaces of V, with U D W = 0 and
also U -L W, we will denote their sum (which is a direct sum) by U@V.
T h e o r e m 1.7.10 If B is an alternating form on V then
V =
W^WT®
■ ■ ■ ®Wr® rad V,
a direct sum of mutually orthogonal subspaces, with each Wi a hyperbolic
plane. Consequently V has a basis
{ « l . V l , . . . ,Ur,t>r,U;i,...
,W„-Tr}
relative to which the representing matrix B has block diagonal form
M
0
M
0
where M =
0
-1
1
0
0„- 2r
Proof If B = 0 then rad V = V and there is nothing to prove. Otherwise,
choose a hyperbolic pair (ui, vi ), as above, and let W\ be the hyperbolic plane
that they span. Then W\ is a nondegenerate subspace (the determinant of
the restriction of B to Wi is 1), so V = WQWf by Proposition 1.7.9. Also
rad V = VL = ( W i © W i x ) x = Wf n W x x = r a d ( W x ) . The result follows
by induction on the dimension.
A
Corollary 1.7.11 B has even rank; if B is nondegenerate then dim V is
even.
Corollary 1.7.12 Alternate bilinear forms B\ and B 2 on spaces Vy and Vi
are equivalent if and only if dim Vx = dim Vi and rank B\ = rank B2-
26
Proof Proposition 1.7.8.
CHAPTER 1. PRELIMINARIES
A
Corollary 1.7.13 The determinant of any representing matrix for an alternating form is a square in F.
Proof If B is as in the theorem then its determinant is 0 or 1; relative to another basis the matrix takes the form DlBD, with determinant
(detD) 2 detB.
A
Groups and Characters
by Larry C. Grove
Copyright © 1997 John Wiley & Sons, Inc.
Chapter 2
Some Groups
The point of this chapter is simply the introduction of several examples, and
classes of examples, of groups, in fairly complete detail. They will serve as a
useful source of illustrative examples in later chapters.
2.1
Aut(Z n )
We write Z„ for the ring Z / n Z of integers mod n. Throughout this section
we will abuse the notation as is customary and let context determine whether
an integer k is being viewed as an integer or is taken mod n and hence is
being viewed as an element of Z n .
If 6 e End(Z„) say that 0(1) = m. Then for any k € Z„ we have 6(k) =
0(1 + ■ • • + 1 ) = k9(l) = mk, and it follows easily (verify!) that End(Z„) S
Z„. Also, 6 € End(Z„) is an automorphism if and only if 0(1) = m is another
generator of Z„, i.e. if and only if (m, n) = 1, so Aut(Z„) = £/(Z n ), the group
of units in Z„.
If n = n»=i PV 1S t n e prime factorization of n then by the Chinese Remainder Theorem Z„ = 0 { Z p «i, so
Aut(Z„) Si C/(Zn) S C/(Zp«0 x • • • x U(Zpl„).
Thus we may concentrate on Aut(Z p «), p a prime. Recall from elementary
algebra that
|Aut(Z p .)| = V » ( p e ) = p e - 1 ( p - l ) ,
where ip denotes Euler's totient function.
Proposition 2.1.1 If p is an odd prime then Aut(Z p «) is cyclic.
Proof If e = 1 then Z p is a field and it is a standard fact that t/(Z p ) = Z*
is cyclic. Suppose, then, that e > 1. Choose a generator x for C/(ZP), and
note that y = x + p is also a generator for t/(Z p ). If xp~l = 1 in Zp3
27
28
CHAPTER 2. SOME
GROUPS
then y"- 1 = (x + p ) p _ 1 = xp~l + (p - l ) p x p _ 2 = 1 - px p ~ 2 ¿ 1 in Zpa
since p\ x. Thus we may assume that x p _ 1 = 1 in Z p but that x p _ 1 ^ 1
in Zp3. Let us show that the multiplicative order of x in Zp« is <p(pe) =
p e - 1 ( p — 1). As above we may write x p _ 1 = 1 + pa, with p j a. Thus
x*«»"> = (x"-1)"*"1 = (1 +pa)P'-1 = 1 + E C CDù»)* = 1 in Zp «>
so
IxP"1! \jf~l. However, ( x " - 1 ) " ' - 2 = (1 + p a ) p ' ~ 2 = 1 + a p « " 1 / 1 in Z p . ,
since p^ a. Thus | x p _ 1 | = p e _ 1 . Suppose now that x m = 1 in Z p . . Then also
^p-i^m _ 1 ; s o p e - i I m W r i t e m _ pe-i£ Then 1 = x m = ( x " ' " 1 ) ' = xl in
Z p (since x p = x in Z p ), so p - 1 11. Thus p e - 1 ( p - 1) | m, and we conclude
A
that |x| = |i/(Z p «)|.
It should be remarked that the proof above is not constructive. Even in
the case of Z p there seems to be no efficient algorithm for finding a generator
- called in number theory a primitive root.
Note that t/(Z 2 ) = 1 and f/(Z 4 ) = {1,3} = (3). However, i/(Z 8 ) =
{1,3,5,7} = ( - l ) x (5).
P r o p o s i t i o n 2.1.2 If e > 3 then Aut(Z 2 «) Si f/(Z 2 .) = (-1) x (5).
Proof Use induction on e to see that 5 2 ' = 1 + 2 e _ 1 in
e = 3, so assume the result for some e > 3 and consider
(1 + 2 e _ 1 ) 2 = 1 + 2 e + 2 2 e - 2 in Z, and that 2e - 2 > e + 1.
that 5 2 ' - 3 = 1 + 2 e _ 1 in Z 2 «, i.e. that 5 2 ' - 3 + b ■ 2 e = 1 +
b € Z. Squaring, we have
Z 2 «. It is true for
e + 1. Note that
We have assumed
2 e " 1 in Z for some
5 2 '" 2 + b ■ 5 2 e _ 3 2 e + 1 + 6 2 2 2e = 1 + 2 e + 2 2 e - 2 .
2
2
( +,) 3
(*)
e
1
Since 2e - 2 > e + 1 this says that 5 '" = 52 ' - = i + 2( + )" 1 in
Z 2 .+i. Thus |5| > 2 e _ 3 in Z2«. But also (*) shows that 5 2 ' 2 = 1 in Z 2 «, so
|5| = 2 e - 2 .
To finish the proof it will be sufficient to show that the elements ( — l ) ' ^ ,
for 0 < i < 1 and 0 < j < 2e~2, are all distinct. Say ( - 1 ) ^ = (-1)*5* in
Z2«, with the exponents in the indicated ranges. Then (—1)' = (—1)* in Z4,
and it follows that i = k. But then also j = £, since |5| = 2 e - 2 .
A
We summarize in a theorem.
Theorem 2.1.3 Suppose that n € N has prime factorization n = 2e n*=i PV ■
Then
Aut(Z„) Si t/(Z n ) ^ i/(Z 2 «) x t/(Zp.,) x • • • x U(Zp:,);
i/(Z 2 ), i/(Z 4 ), and [/(Zp.*) are all cyclic; ife>3
then t/(Z 2 «) = (-1) x (5).
Corollary 2.1.4 Aut(Z„) is abelian; it is cyclic precisely when n = 2, 4, p*,
or 2pl, p an odd prime.
Proof Exercise.
A
2.2. METACYCLIC
2.2
29
GROUPS
Metacyclic Groups
A group G is metacyclic if it is an extension of a cyclic group by a cyclic group,
i.e. if G has a normal cyclic subgroup A = (a) with G/A = {Ab) also cyclic.
If the extension splits over A, so there is a cyclic subgroup B = (b) < G with
G = A x B, then G is called split metacyclic.
Suppose that G is metacyclic, as above, and finite, say with \A\ = \a\ = m
and [G: A] = s, so \G\ = ms. Then b" € i4, so there is some t 6 Z, 0 < r < m,
with 6* = a'. Note that if £ = 0 then G splits. Conversely, if G splits over A
then it is possible to choose b so that t = 0 (however, see Exercise 2 on page
30).
Since A < G there is also r € Z, 1 < r < m, with 6 _ 1 a6 = o r ; note
that (r,m) = 1, since |a r | = \a\. For all fc > 1 we have b~kabk = ar ,
and in particular a = b~"abs = a r ' , so r* = 1 (mod m). Note also that
a ^ - D = ( o O ' a - ' = (6-*o6)'a-* = fc-Vha"* = fc-^kr* = a ' a " ' = 1, so
tr — t (mod m).
For convenience we gather all this arithmetic information in the next
proposition.
Proposition 2.2.1 If G — (a,b) is metacyclic with parameters m, s, t, and
r as above, then
(r,m) = 1,
r* = 1 (mod m),
and
tr = t
(mod m).
Since ab = baT all elements of G can be written in the form ft'o-7, and in
fact
G = { 6 V : 0 < i < s - 1, 0 < j < m - 1}.
Proposition 2.2.2 Suppose that G is split metacyclic, with t = 0, and assume that (m,s) = 1. Then all Sylow subgroups of G are cyclic.
Proof Since \G\ = ms and (m,s) = 1 it is clear that Sylow subgroups of
A = (a) and of B = (b) are Sylow in G, and all Sylow subgroups of G (up to
conjugacy) appear in either A or B.
A
The converse of Proposition 2.2.2 is true but not quite so simple; it will
be proved in Chapter 4 (Theorem 4.1.17).
The final proposition of this section provides a presentation for a finite
metacyclic group.
Proposition 2.2.3 Suppose that m,s,t,r
(r,m) = 1, rs = 1 (mod m), and tr = t
€ N satisfy 1 < r < m, 0 <t < s,
(mod m), and set
G = (a, b | am = 1, b* = a1, b~lab = aT).
Then G is metacyclic of order ms, with A = (a) <G and G/A = (Ab).
CHAPTER 2. SOME
30
GROUPS
Exercises
1. Prove Proposition 2.2.3. (Suggestion: construct such a group from
cyclic groups A and B of orders m and s by defining an appropriate
multiplication on the set A x B.)
2. If G = (a,c | a 8 = l,c 2 = a 6 , c - 1 a c = a 5 ) show that G is split metacyclic, even though t ^ 0. (/finí Try replacing c by b = ca' for some
appropriate choice of i.)
3. If G = {a, b I a 4 = 63 = 1,6-1ai> = a 2 ) show that G is cyclic of order 6.
4. If G = (a, b | a 4 = 1, b4 = a 3 ,6 _ 1 o6 = a 3 ) show that G is cyclic of order
2.
5. If G is metacyclic, as in Proposition 2.2.1, show that its derived group
is G' = ( a r _ 1 ) , of order m/(m,r — 1). Determine the center Z(G).
Exercises 3 and 4 are intended to point out the necessity of the arithmetic
conditions in Proposition 2.2.3.
Probably the best known specific examples of groups as in Proposition
2.2.3 are the dihedral groups (which are split) of order 2m,
Dm = (a, 6 | a m = b2 = 1,6_1afc = a" 1 " 1 ),
and the generalized quaternion groups (which are not split) of order 4m,
Qm = (c,d\ c 2 m = l,cP = cm,d-xcd
2.3
= c2"-1).
Sylow Subgroups of Symmetric Groups
If p € Z is a prime and 0 / m 6 Z we write vp(m) to denote the exact power
of p that divides m, i.e. p"»>(m) | m but p"p( Tn ) +1 j m. Also, if a € E we write
\a\ to denote the greatest integer that is less than or equal to a. Examples:
i/5(1000) = 3 = [it\.
If p € N is a prime set J p = ((12 • • • p)), a cyclic subgroup of order p
in the symmetric group Sp. Set 5 = { 1 , 2 , . . . , p } . By Corollary 1.5.2 the
wreath product Jp I Jp is a faithful permutation group on S x 5; it has order
pP ■ p = pP+1 and degree p 2 . Thus J p I Jp is isomorphic with a subgroup of
Sp*.
Since
l-V I = (P2)! = (P-P) • • • ( ( ? - 1)P)---((P-2)P)
2
(2p)-.-p---l
we have ^ p ((p )!) = p + 1, so in fact Jp I Jp is isomorphic with a Sylow
p-subgroup of Spi.
Our goal is to generalize from p 2 to arbitrary n > 0 in N.
2.3. SYLOW SUBGROUPS
OF SYMMETRIC
31
GROUPS
Proposition 2.3.1 Suppose that a, b, c, € Z, with b, c > 0. Then
[[a/b\/c\
= [a/bc¡.
Proof Write a = bq + r, with 0 < r < 6, and q = cq\ + n , with 0 < J*I < c.
Then [o/6J = g and [q/c\ — LLa/*J/cJ = 9i- B u t a ^ 0 a = HcQi +ri) + r =
bcqi +bri+r, and 0 < bri + r < b(c- 1) + b = be, so a/6c = qi + (bri + b)/bc,
hence also [a/bc\ = qiA
Proposition 2.3.2 If p is a prime and 0 < n € Z then
i/p(n!) = [n/pj + [n/p2¡ +■■■.
(L n /pJp) " " '"> a ^
Proof Note that n! = 1 • • p- ■ ■ (2p)
L"/PJ
n(M=Pl"/pJ • \n/p\\.
Thus ^ p (n!) = [n/pj + ^ p ([n/pJ!). The result now follows by induction and
Proposition 2.3.1.
A
For an example take p = 7 and n = 1000. Then [1000/7J = 142,
[1000/72J = L142/7J = 20, and |.1000/73J = [20/7J = 2, so i/7(1000!) = 164.
Given arbitrary p and n, as above, write n = ao + aip +
1- akpk, with
ak # 0, Oj 6 Z, and 0 < a¿ < p, all i. Then
akpk~l,
[n/pj
=
ai + a 2 p + • • • +
|n/p 2 J
=
a2 + a3p + ■ ■■ + akpk~2,
etc., and so
i/p(n!)
=
ai +a2(l+p)
k
+ p2) + --- +
+ a3{l+p
j-i
k
,
ak(l+p+---+pk-1)
,
= £"»Zy-5>(£r)
¿=1
i=0
j=l
y
is the order of a Sylow p-subgroup of Sn.
Back to wreath products.
Recall that Jp = ((12 • • • p)) < 5 P . Define Jjfp(0) = 1, J?r{l) = JP,
Jpr{2) = Jp I Jp, and inductively J™r(r) = Jp l J p J ' r ( r - 1 ) for r > 3. We shall
call Jpr'
the rth wreath power of Jp.
Proposition 2.3.3 / / 1 < r € Z then Jpr{r) has order
degree pT. It is isomorphic to a subgroup of Spr.
P
+P+-+P"~1
1
and
32
CHAPTER 2. SOME GROUPS
Proof By Corollary 1.5.2, and induction, Jpr(r) is faithful on Sr = S x S x
■ ■ ■ x S, so its degree is pr. Assume inductively that \Jp
\ = pl+p+-+pr
Then
|j? p ( r ) l = \JPUpr(r~l)\
=ppr_1 •p1+p+-+pr"2 =p1+^-+"'"1.
A
If G and if are groups acting on sets S and T, respectively, then G x H
acts naturally on S x T via (s,i) (ff, ' ,) = (s9,th). In particular G2 = G x G
acts on 5 2 = 5 x 5, and likewise G m acts on Sm if 1 < m € N; if G has
degree d then G m has degree dm.
Theorem 2.3.4 / / 1 < n € N and p G N is prime write n = ao + aip H
h
J
li) ai
ajtp*, with a* # 0, a¿ € Z, and 0 < a4 < p, all i. Then P = l\ï=o( p"' )
is isomorphic with a Sylow p-subgroup of the symmetric group Sn ■
Proof Note that the group P has order n£=i \Jpr{i)\a' = p"»(,,!) and degree
5Zi=o °»P' = n - ^ n e a c t i ° n is faithful so P is isomorphic with a Sylow psubgroup of S„.
A
For example, if n = 1000 and p = 7 we have seen that 1/7 (1000!) = 164,
and 1000 = 6 + 2-7 + 6 - 7 2 + 2-7 3 . Thus a Sylow 7-subgroup of S1000 is
isomorphic with
(J?r{0))6 x (J7""-(1))2 x (J7"""(2))6 x (JJ" p(3, ) a .
2.4
Affine Groups of Fields
If F is a field we write F* for its multiplicative group F \ {0}. Denote by
Aff(F) the group (under composition) of all functions Tb,a'F -> F, where
a € F*, b € F, and r^ 0 (x) = ax + 6, all x 6 F. Thus if F is finite, say
|F| = 9 , t h e n | A f r ( F ) | = 9 ( 9 - l ) .
Note that T(,I0T<ÍIC = T(,+0(iiac, so Aff(F) 5i Fx-gF*, where 0 maps the multiplicative group F* to automorphisms of the additive group F via 6(a): d i->
ad. Internally Aff(F) has the normal subgroup T = {n^ib G F} of translations, and the subgroup H = {T0ia:a e F*}, with Aff(F) =T x H.
Clearly Aff(F) < Perm(F), so we may naturally view Aff(F) as a permutation group acting on F.
Proposition 2.4.1 If F is afield then the affine group G = Aff(F) is sharply
2-íranstítve on F.
Proof Clearly the translation subgroup T is transitive on F, so G is transitive. Furthermore, the subgroup H = {ro,a:o € F*} is Stabe (0), and H
is transitive on F*, so G is doubly transitive on F by the exercise on page
16. Finally, if r = n,o 6 G stabilizes both 0 and 1 then 0 = r(0) = b and
1 = r(l) = a, so T — 1 and G is sharply 2-transitive.
A
33
2.5. FINITE GROUPS IN 2 AND 3 DIMENSIONS
Corollary 2.4.2 If p is a prime and 0 < n € N then there exists a sharply
2-transitive group of degree pn.
Exercises
Let G = Aff(F) =TxH
as above.
1. If r € G \ H show that H n HT = 1. Conclude that H
NG(H).
2. If 1 ¿ T G T show that CG{T) = T.
The results in the two exercises above reflect the fact that Aff(F) is an
example of a Frobenius group, of which we shall hear more later.
2.5
Finite Groups in 2 and 3 Dimensions
If V is a real vector space with an inner product then the orthogonal group
0(V) is the group of all linear transformations r on V that preserve the inner
product, i.e. (TU,TV) = (u,v) for all u, v G V. Equivalently r G 0{V) if and
only if T preserves lengths of vectors.
Recall from linear algebra that if r G 0(V) then det r = ± 1 , and that
if A G C is an eigenvalue of r then |A| = 1. If A is the matrix representing
T relative to an orthonormal basis for V then A is an orthogonal matrix,
meaning that its transpose is its inverse.
In this section we shall take V to be either R2 or R 3 , in each case with
the standard inner product, commonly called the dot product. We will write
e i , e 2 , . . . for the standard orthonormal basis vectors in each case. Vectors
will be column vectors, e.g. v = a , but for typographical convenience
b
they will often be written as row vectors, v = (a,b).
Suppose first that r G 0 ( R 2 ) . Say that r(ei) = (a, b). Then a? + tí2 = 1.
Choose 9, 0 < 9 < 2n, so that a = cosö, b = sinö. We have re-i X. re\, so
re-i = ±(6, - a ) .
Case 1. If re-x — (—6, a) then r is represented by the matrix
cos 9
sin 9
— sin 6
cos 9
det A = 1, and r is a (counterclockwise) rotation through angle 9 about the
origin as center.
Case 2. If re-i = (6, —a) then
A =
cos 6
sin 9
sin 9
— cos 9
CHAPTER
34
2. SOME
GROUPS
and det A — — 1. If we set
, 0 . 0 .
. . 9
$
vi = ( c o s - , s i n - ) , v2 = ( - s i n - , c o s - )
it is easy to check that v\ ± v2, TVI = v\, and TV2 = —v2. It follows that T
is geometrically a reflection, with mirror the line spanned by vi.
Note that products and inverses of rotations are rotations, and that reflections have order 2. If G is any subgroup of 0(R 2 ) then the set H of rotations
in G is a normal subgroup of G.
Exercise
Show that the product of two reflections in 0(R 2 ) is a rotation. What is
the angle of the rotation?
Suppose that G is a finite subgroup of 0 ( R 2 ) , and let H be its rotation
subgroup. If H ^ 1 choose p ^ 1 in H whose angle 9 = 0(p) is minimal.
For any r € H choose m G N so that m0(p) < 0(T) < (m + l)9(p). Then
0 < 0(T) - m6{p) < 6(p). But 9{T) - m9{p) = 0 ( p - m r ) , and 0(p) is minimal,
so 9(p~mr) = 0, i.e. p~mr = 1, or r = pm and H = (p) is cyclic. Its order is
n = 2TT/0(P); we write H = c£]'.
If G / H choose a reflection a € G\H. Since det(p*<r) = — 1 for each k
the coset Ha contains n distinct reflections. If r 6 G is any reflection then
det(rff) = 1, so TO G H and r € Ho. Thus if has index 2 in G, and
G = (p, o) = {1, p, p2,...,
pn~l, o, po,...,
pn~xo}
of order 2n. Since <rp is a reflection we have {op)2 = 1, or op = p~lo, so G
is dihedral. It will be denoted by T>„ ■
We have proved the next theorem, which has been attributed by Hermann
Weyl [65] to Leonardo da Vinci.
Theorem 2.5.1 If G is a finite subgroup of 0(R2)
group C„ or a dihedral group Vn , n — 1,2,3,
then G is either a cyclic
Next consider three dimensions: V = R 3 . If p £ 0 ( R 3 ) and det p = 1,
then p is called a rotation.
Theorem 2.5.2 (Euler) / / p e 0(R 3 ) is a rotation then it is in fact a
rotation about a fixed line as its axis, in the sense that p has an eigenvector
v having eigenvalue 1 such that the restriction of p to the plane V = v1 is a
rotation ofV.
Proof If Ai, A2, and A3 are the eigenvalues of p then one of them, say
Ai, must be real, as they are the roots of a cubic with real coefficients. If
A2 £ R then its conjugate A2 is also a root, so A2 = A3 in that case. Since
det p = A1A2A3 = 1 the only possibilities are (relabeling if necessary)
2.5. FINITE GROUPS IN 2 AND 3 DIMENSIONS
35
1. Ai = 1, A2 = A3 = ± 1 , and
2. Ai = 1, A2 = Al" g R.
In either case 1 is an eigenvalue. Choose a corresponding eigenvector v and
note that also v = p~lpv = p~lv. If u L v then (pu,v) = (u,p~1v) = (u,v) =
0, so V = vx is p-invariant. Since the determinant of the restriction p\p is
the product of the other two eigenvalues A2 and A3, we see that det(p\-p) = 1
and P\T> is a rotation of V.
A
A reflection in 0 ( R 3 ) is a transformation a whose effect is to map every
point of R3 to its mirror image with respect to some plane V through the
origin. Thus it is characterized by the fact that au = u for all u 6 V and
av = — v if v Ç. V1-. If r is chosen to be a unit vector in Vs- (which is a line),
then a is given by the formula av = v - 2(v, r)r, all v € R 3 . It is not difficult
to show, by arguments similar to those in the proof of Euler's Theorem above,
that if T € 0 ( R 3 ) and det r = - 1 then the effect of r is that of a reflection
through a plane V followed by a rotation about an axis perpendicular to V.
If V is a plane (through the origin) in R3 then each rotation p in Ö(V)
extends naturally to a rotation in 0 ( R 3 ) ; it suffices to set p(v) = v for v e P x
and extend by linearity.
By extending each rotation in a cyclic subgroup Cn in that way we obtain
an isomorphic group of rotations in 0 ( R 3 ) , which will be denoted Cn •
On the other hand, if a € 0(V) is a reflection then a can also be extended
to a rotation in 0 ( R 3 ) ; in this case set a(v) = —v for v € Vx. The result is
a rotation through angle -n with the original reflecting line in V as its axis.
Thus each r in a dihedral group P „ can be extended to a rotation in
0 ( R 3 ) , the result being an isomorphic subgroup denoted by T>„ .
Note that C2 and T>[ each consist of the identity and a rotation through
angle n, so they are indistinguishable.
Consider next the rotational symmetry groups of regular solids in R 3 .
There are (up to similarity) just five regular solids: the tetrahedron, cube,
octahedron, dodecahedron, and icosahedron. Only three groups result, however, for simple geometric reasons. If midpoints of adjacent faces of a cube
are joined by line segments then those line segments are the edges of an octahedron having the same rotational symmetry group as the cube. Similar
remarks apply to the dodecahedron and icosahedron.
Denote by T the rotation group of a regular tetrahedron. Clearly T is
transitive on the set of 4 faces, and the stabilizer of any face is cyclic of
order 3, so \T\ = 12 by Proposition 1.2.1. Being isomorphic to a subgroup of
Sym(4), it is clear that T must be isomorphic with Alt(4), the only subgroup
of order 12 in Sym(4).
Denote by Ö the rotation group of a regular octahedron (or cube); O
is transitive on the set of 6 faces of the cube and the stabilizer of a face is
36
CHAPTER 2. SOME
GROUPS
cyclic of order 4, so \0\ = 24. A moment's thought convinces one that Ö acts
faithfully on the set of 4 diagonals of the cube, so Ö 2í Sym(4).
Note that a regular tetrahedron can be inscribed in a cube, its 6 edges
being appropriately chosen diagonals of the faces of the cube. It follows that
we may view T a s a subgroup of O.
Finally, denote by I the rotation group of a regular icosahedron (or dodecahedron); it is transitive on the 20 faces of the icosahedron and the stabilizer
of a face has order 3, so \T\ = 60. It is possible to inscribe a cube in a
dodecahedron; each of the 12 edges of the cube is an appropriately chosen
"diagonal" of one of the 12 faces of the dodecahedron. In fact, since each
pentagonal face has 5 different diagonals, it follows that there are exactly 5
such cubes inscribed in the dodecahedron. The group I acts faithfully on the
set of 5 inscribed cubes, and hence I = Alt(5).
The unit sphere {v € M3: ||u|| = 1} is invariant under 0 ( R 3 ) . If 1 ¿ p G
0(K 3 ) is a rotation then by Euler's Theorem (2.5.2) there are exactly two
points on the unit sphere satisfying pv = v, viz. the two points where the
axis of p pierce the sphere. Those two points will be called the poles of p.
If G < 0(K 3 ) write S = SG for the set of poles of nonidentity rotations
inG.
Proposition 2.5.3 If G < 0(M 3 ) then G acts on its set S of poles.
Proof If v € S then v is a pole for some rotation p £ G- If r e G then
(rpr~l)TV = rpv = TV, so TV is a pole for the rotation rpr~l and TV G S.
A
Each cyclic group Cn , n > 1, has just two poles. No rotation in Cn
carries one to the other, so S has two 1-point orbits, and each stabilizer is all
ofd3).
The dihedral group P „ has n rotation axes in the plane V on which T>n
acts, and one more axis VA-, so V(n ' has 2n -I- 2 poles. The two poles on VAconstitute one orbit. If n is odd the n poles that are vertices of a regular
n-gon in V make up another orbit, and the set of their negatives make up
another. If n is even the vertices of the regular n-gon again make up an orbit,
and the poles that are on axes joining midpoints of opposite sides of the n-gon
make up a third. In both cases there are three orbits in 5 , and stabilizers
have orders n, 2, and 2, respectively.
It is easy to analyze the actions of T , O, and X on their sets of poles; all
the data are tabulated below.
G
r>(3)
T
O
1
\G\
n
2n
12
24
60
Orbits
2
3
3
3
3
\s\
2
2n + 2
14
26
62
Orders
n
2
2
2
2
of
n
2
3
3
3
stabilizers
n
3
4
5
2.5. FINITE GROUPS IN 2 AND 3 DIMENSIONS
37
Theorem 2.5.4 IfG< 0(K 3 ) is a finite group of rotations then G is one of
C(n\ n>\;V{n\n>
2; T; O; or X.
Proof We may assume G ^ C[ . Let <S be the set of poles of G and let
U denote the set of all ordered pairs {p,v), where 1 / / > £ G and v G S
is a pole of p. Set |G| = n, and for each v G S set m„ = |Orbc(t>)| and
nv = | Stabo(v)|. Thus n = mvnv, all v G «S.
Since each p ^ 1 in G has two poles we have \li\ = 2(n - 1). On the
other hand, we may count the elements of U by counting the number of
group elements corresponding to each pole. Suppose that v i , . . . , v* is a set
of poles, one from each of the G-orbits in S, and write mVi = mi, nVi = TIJ.
Since each v G Orb(uí) has n„ = n¡ we have
M = $3í n »- 1 : v € 5 >
-
Jk
k
^^¿(TIÍ-1) = ^(n-mj).
Thus 2n - 2 = ^ ( n — m¿), and, dividing by n, we have 2 - 2/n =
E Î U i 1 - ! / « • ) • But 1 < 2 - 2 / n < 2 , andeachrii > 2, so 1/2 < l - l / n ¿ < 1,
and k must be either 2 or 3.
If k = 2 then
2 - £n = (l--J-)
+ (l--J-),
ni
ri2
or 2 = n / n i + n/n2 = mi + 7712, so mi = n»2 = 1, ni = 712 = n. Thus G has
just one axis of rotation and is a cyclic group C„ .
If k = 3 we may assume that n\ < n 2 < »13. If «! were 3 or greater we
would have £)t(l - l/n¿) > ^,¡{1 — 1/3) = 2, a contradiction. Thus ni = 2
and
*
2
1 .,
1.
.,
1.
n
¿
7i2
"3
or
1 2
1 1 1
- + - = — + — >-.
2 n
712 ri3
2
If 712 were 4 or greater we would have I/712 + l/ri3 < 1/2, a contradiction, so
7i2 = 2 or 3.
If 7i2 = 2 then 713 = n / 2 , and mi = 7712 = n / 2 , m3 = 2. Setting m = n / 2
we have the data in the table above for the dihedral group T>m •
If 712 = 3 we have 1/6 + 2/n = l/ri3, and the only possibilities are (1)
713 = 3, n = 12, (2) 713 = 4,n = 24, and (3) n3 = 5, n = 60; for n3 > 6 would
require 2/n < 0. Thus in the cases 1, 2, 3 we have precisely the data for T,
O, and I .
In each case the group G not only shares the data in the table with one of
the groups in the table, but in fact is the group. For example, in case 1 the
poles in either of the 4-element orbits are the vertices of a regular tetrahedron
CHAPTER
38
2. SOME
GROUPS
centered at the origin. The tetrahedron is invariant under G so G < T, but
also \G\ = \T\, so G = T. Further discussion of this point for the other groups
may be found e.g. in [69].
A
The group Ö* < 0(R 3 ) of all symmetries of a cube is larger than O, e.g.
since - 1 € O* \ O. Observe, however, that if r G O* \ O then - r G O, since
- r = - 1 • T G O* and d e t ( - r ) = 1. Thus O* = Oil O ( - l ) and in fact
Ö* =
Ox{±l}.
Proposition 2.5.5 IfG< 0 ( R 3 ) has rotation subgroup H then either H =
G, or H has index 2 in G; in particular H <G.
Proof Fix T G G\H and take any a G G\H.
so err'1 G H, a G HT, and G = H U HT.
Then d e t ^ T " 1 ) = ( - 1 ) 2 = 1,
A
If G < 0(R 3 ) with rotation subgroup H jt G let us distinguish between
two cases, viz. - 1 G G and - l & G.
If - 1 G G then just as for O* above we have G = H x {±1}. On the
other hand, if H is any group of rotations in 0 ( R 3 ) , we may form G =
H* = H x {±1} and obtain a subgroup G of 0(R 3 ) having H as its rotation
subgroup.
Suppose then that — 1 £ G, and say that HT is the coset different from
H in G. Then it is easy to check (do so!) that the set K = H U H(-T) is a
group of rotations in 0(R 3 ) having H as a. subgroup of index 2.
To turn the tables, suppose that K is a group of rotations in 0 ( R 3 ) and
that K has a subgroup H of index 2. Set G = H ö {-T:T € K \ H} and
check that G is a subgroup of 0 ( R 3 ) having H as its rotation subgroup. We
will denote this group by G = K]H; as an example 0]T is the full symmetry
group of a regular tetrahedron (why?).
Combining Theorem 2.5.4 with the discussion above we have proved a
theorem due originally to the crystallographer J. F. C. Hessel.
Theorem 2.5.6 (Hessel, 1830) If G is a finite subgroup ofO(R3)
is one of
then G
1. C£\ n > 1; V™, n > 2; T; O; 1;
2. d 3 ) *,
n > I; P ! 3 ) * , n > 2; T*;
3. C^)C^\
n > 1; V^)C^,
O*; 1*;
or
n > 2; V^)V^,
n > 2; Ö\T.
Note that there are three groups of order 2 in the theorem, viz. C^ ,
{C[ ')*, and Q ]C\ '. As abstract groups they are all, of course, isomorphic.
But they are geometrically different — the technical meaning of that is that
they are not conjugate as subgroups of 0 ( R 3 ) . Of course the entire spirit of
Hessel's Theorem is to list the finite subgroups of 0(R 3 ) up to conjugacy. For
the subgroups of order 2 the nonidentity elements are, respectively, a rotation
2.6. SOME LINEAR
39
GROUPS
through angle 7r, an inversion through the origin, and a reflection. From
another viewpoint they have —1 as an eigenvalue with differing multiplicities
2, 3, and 1.
It is a mildly tedious exercise (not assigned) to check that all the groups
listed in Hessel's Theorem are geometrically distinct.
Although it is not the usual definition, it is true that a group G in Hessel's
Theorem is a point group for a crystal if and only if the orders of elements in
G are 1, 2, 3, 4, or 6.
Exercise
There are 32 crystallographic point groups in 0(K 3 ). Find them.
A final remark: the results of this section are in a sense more general than
they may appear to be. If V is a finite dimensional real vector space and G is
any finite group of invertible linear transformations on V then G is conjugate
to a subgroup of the orthogonal group 0(V). For a proof see e.g. page 97
of [32]; or, for a complex analogue (whose proof is similar), see Proposition
10.3.13 below.
2.6
Some Linear Groups
If V is an n-dimensional vector space over a field F then GL(K) denotes the
general linear group of V, i.e. the group of all invertible linear transformations
on V. Choosing a basis for V provides an isomorphism of GL(V) with the
group GL(n, F) of all invertible n x n matrices over F.
The determinant is a homomorphism from GL(V) (or GL(n, F)) onto the
multiplicative group F*; its kernel is the subgroup SL(V) (or SL(n, F)), the
special linear group.
If F is finite, with q elements, then the matrix groups are often denoted
by GL(n,q) and SL(n,q).
Exercises
1. Show that
Z(GL(V))
= {al:aeF*}
and Z(SL(V))
= {aha G F*, an = 1}.
2. Let G be the (additive) elementary abelian group Z " of order p " , where
p is a prime. Show that the holomorph of G is isomorphic with the
subgroup of G L ( n + l , p) consisting of all matrices having the partitioned
form
A v
0 1 '
with A € GL(n,p) and v eG.
CHAPTER
40
2. SOME
GROUPS
If 0 ^ v € V write [v] for the line Ft; through the origin spanned by v,
and call it a projective point. The set of all distinct projective points [v] is
called the projective space of dimension n — 1 based on V, and is denoted
by Pn-i(V),
or simply by P(V). There is a natural permutation action of
GL(V) on P(V), given by T[V] = [TV] for all r € GL(V), [v] € P(V). It
is easy to check that the kernel of the action is Z(GL(V)), and likewise that
the kernel of the action of SL(V) on P(V) is Z(SL(V)).
Exercise
If dim V = n and | F | = q, finite, show that |P(V)| = (qn - l)/(q - 1).
Define the projective general linear group of V to be
PGL(y) = GL(V)/Z(GL(V))
and the projective special linear group to be
PSL(V) = SL(V)/Z(SL(V));
each of them acts faithfully on P{V). There are isomorphic copies based on
the corresponding matrix groups, with obvious corresponding notation.
Proposition 2.6.1 / / | F | = q, finite, then
1. | GL(n, 4)| = g"«"- 1 )/ 2 \[{qk - 1:1 < fc < n } ,
2. |SL(n,ç)| = |PGL(n,<z)| = | GL(n,q)\/(q
- 1), and
3. |PSL(n,ç)| = | SL(n,<7)|/(n,g-1).
Proof The first row of a matrix in GL(n, q) can be any of the qn — 1 nonzero
row vectors over F\ the second must be independent of the first, so there are
qn — q choices, etc. Thus
|GL(n,ç)|
=
=
n(9B-í*)=í1+-+("-1>n(9*-l)
i
g «(«-i)/
2
i
Y[{q" - 1:1 < jfc < n } .
Part 2follows, since |F*| = q-1, Z(GL(n,q)) S F*, and GL(n,q)/SL(n,q)
S
F*. Since F* is cyclic it follows from the exercise on page 39 that a e
Z(SL(n, q)) if and only if a n = a«" 1 = 1, which is if and only if a^-«- 1 ) = 1.
In other words, a must be in the unique subgroup of order (n, q — 1) in F*.
A
Although SL and PGL have the same order they are usually not isomorphic.
2.6. SOME LINEAR
41
GROUPS
Exercise
If F is any field denote by M2{F) the group of all /»near fractional transformations (sometimes called Möbius transformations) of F, i.e. all rational functions f{x) 6 F(x) of the form
f{x)
with ad-bc¿
ax + b
= ^Td'
0. Show that M2(F)
S PGL(2,F).
Let C2(F) be the subgroup of M2(F) consisting of those f(x) for which
ad - be is a square in F*. Show that C2(F) * PSL(2, F).
A hyperplane in F is any subspace of codimension 1. If 1 ^ r € GL(V)
then r is called a transvection if there is a hyperplane W such that T\W = lw
and TV — v € W for all v € V; W is called the /ired hyperplane of r.
If r is a transvection with fixed hyperplane W choose a basis for V consisting of first some v\ € V \ W and then a basis {v2, •.. ,vn} for W. It is
clear from the matrix representing r relative to this basis that d e t r = 1, so
in fact r € SL(V).
Exercises
1. Show that the inverse of a transvection is a transvection.
2. Suppose that V is a subspace of a space V\, that v £ V\ \ V, and that
T is a transvection on V with fixed hyperplane W. Show that r can be
extended to a transvection T\ on V\ whose fixed hyperplane W\ contains
v.
Proposition 2.6.2 Ifu and v are linearly independent in V then there is a
transvection T with TU = v.
Proof Choose a hyperplane W in V with u — v € W but u £ W, and define
r by means of T\W = l w , ru = v. If x € V write x = au + w, where a & F
and w G W. Then TX — x = av + w — au — w = a(v — u) G W, so r is a
transvection.
A
Proposition 2.6.3 Suppose that W\ and W2 are two distinct hyperplanes in
V and that v e V \ (W\ U W2). Then there is a transvection r with TW\ = W2
and TV = v.
Proof Note that W^ + W2 = V, so dim Wl n W2 = n - 2 and W = Wi n
W2 + Fv is another hyperplane. Write v = x + y, with x € Wi and j/ € W^Then x £ W 2 , so Wi =WiDW2
+ Fx, and likewise W2 = WinW2+
Fy.
Thus y = Wi n W2 + Fx + Fy. It follows that x & W, or else y = v - x is
also in W and hence V Ç W\\ Define r via T\W = lw and TX = y. Then
T is a transvection just as in the proof above, TV = v since t; € W, and
rWi =T{Wlr\W2
+ Fx) = Wxr\W2 + Fy = W2.
A
CHAPTER 2. SOME
42
GROUPS
Theorem 2.6.4 The set of transvections generates SL(V).
Proof Fix p 6 SL(V), then choose a hyperplane W in V and choose v €
V\W. If v and pv are linearly independent then by Proposition 2.6.2 there is
a transvection T\ with T\pv = v. liv and pv are linearly dependent then first
choose a transvection r 0 so that v and r0pv are linearly independent, then a
transvection T{ SO that T[TQPV = v, and set T\ = T[TO. Thus in either case
we have T\pv = v and T\ is a product of transvections. Note that v & TipW.
If T\pW = W set T2 = lv- If TipW / W apply Proposition 2.6.3 to get a
transvection T2 with r-¡,r\pW = W and X2V = v. Set a = T2Tip. Since av = u
it follows that cr|w G SL(W). Now use induction on n = dim V. If n = 2
then (TIW = l w , so cr = 1 and p = TÍ" 1 T 2 ~ I . If n > 2 then by induction er|w
is a product of transvections on W, each of which extends to a transvection
fixing v on V by Exercise 2 on page 41. Thus a is the product of the extended
transvections, and p = T 1 _1 T 2 _1 <T, a product of transvections.
A
Proposition 2.6.5 If T\ andTi are transvections on V then they are conjugate in GL(V). Ifn>2
they are conjugate in SL(V).
Proof For ¿ = 1,2 write W, for the fixed hyperplane of n, choose n € V\Wi,
and set w¿ = r¿Xj — x¿ € H-V Choose bases {wi,U3,... tt„} for Wi and
{^2,^3,. • • ,vn} for W%. For each a e F* define aa € GL(V) by setting
<Tax\ = ^2, CTQWI = wi, aaUi = t)j for 3 < i < n — 1, and, if n > 2,
^o u n = ÛV„. Then a straightforward calculation verifies that aar\a~^ and T-I
agree on X2, W2, «3,. •., un, and hence (TaTio-"1 = T?.
If n > 2 we may set 6 = (det<7i) _1 and obtain o\, € SL(V).
A
Proposition 2.6.6 Suppose that dim V = 2, and Zei {«1,^2} &e any east's
/or V. Every transvection is conjugate in SL(V) to one whose matrix relative
to {vi,V2} is of the form
1
°
1
,aeF*.
Proof If T is a transvection with fixed hyperplane (line) W choose v 6 V \ W
and set w = TV — v 6 W. Relative to the basis {v, w} the matrix representing
T IS
1
1
. If M is the matrix that represents r relative to {t>i,U2} then
there is a matrix B in GL(2,F) with B~lMB
we may set A =
a
0
=
1
1
0
. If det B = a - 1
1
0
. T h e n B 4 e S L ( 2 , F ) , and
1
{BA)~lM{BA)
= A- 1
1
1
0
1
A =
1
a
0
1
A
2.6. SOME LINEAR
43
GROUPS
Theorem 2.6.7 7/n > 3 and G = SL(V) then G' = G.
Proof By Theorem 2.6.4 and Proposition 2.6.5 it will suffice to exhibit a
transvection in G'. Choose a basis {v\,...,vn}
for V and define T\, T^ via
T\: v\ »-> v\ — V2, Vi t-t Vi if 2 < i < n,
T2'.Vi l-¥ V\, V2 >-* V2 - V3, Vi I-* Vi if 3 < t < fl.
Then
so
TIT2T^1T2~1
T\T2T^lT2~l: V\ •-> V\ - V3, Vi 1-4 U¿ if 2 < Î < n,
is a transvection in G'.
Corollary 2.6.8 Ifn>3
A
then PSL(V) is eguo/ to its derived group.
Theorem 2.6.9 Ifn = 2, \F\ > 3, and G = SL(V), iften G' = G.
Proof Choose a basis {vi,«2} for V and choose a € F*, a ^ ± 1 . Define
er 6 SL(V) via <T(«I) = a~1vi, afa) = av2, and for each 6 Ç. F* define
TÍ € SL(V) via n(vi) = vi +bv2, nfa) = «2- Then erT(,er-1T6-1 is represented
by the matrix
' a~l
0
0 '
a
1 0 '
6 1
a
0
0
a" 1
1
-6
0
1
=
1
6a(a — a - 1 )
0
1
The theorem follows from Theorem 2.6.4 and Proposition 2.6.6, since b € F*
is arbitrary.
A
Corollary 2.6.10 If n = 2 and \F\ > 3 i/»en PSL(V) is equal to its derived
group.
Exercise
If n > 3 or | F | > 3 show that the derived group of GL(V) is SL(V).
Determine the derived groups of SL(2,2), SL(2,3), and GL(2,3).
Proposition 2.6.11 If n > 2 then PSL(V) acts faithfully and doubly transitively on the protective space P(V). In particular it is primitive.
Proof It was observed earlier that PSL(V) is faithful on P(V). Take [vi] ¿
[v2] and [wi] ^ [1U2] i n P ( V ) , so {vi, v2} and {wi,ui2} are linearly independent
sets in V. If n = 2 they are bases. If n > 3 set V\ = Fv\ + Fu 2 and V2 —
Fwi + Fw2. Then either Vi = V2 ¿ V or else V\ ¿V2, in which case Vi U V2
is not a subspace - in either case Vi U V2 ^ V. If t>3 G V \ (Vi U V"2) then both
{ui,v2,i>3} and {iüi,«;2,t;3} are linearly independent. The argument may be
repeated to obtain two bases for V, {v 1, v2, V3,... vn} and {wi ,w2,V3,...,
vn}.
CHAPTER
44
2. SOME
GROUPS
For any b G F* define r& G GL(V) via ui i-4 bwi, v-i H4 tu2, u¿ i-> u¿ for
3 < t < TI. If tUj; = J ^ o y ü i , j = 1,2, then det T¡, = b{a\\a-i2 — 012021)Choose b so that det 7& = 1; then r> G SL(V0> and it carries [vi] to [wi], [«2]
to [1^2].
A
Proposition 2.6.12 If 0 jí v e V and A = Stabs¿(v)(M) then A has an
abelian normal subgroup B whose conjugates in SL( V) generate SL(V).
Proof Choose a hyperplane W with v £ W, so V = W + Fv. If a G A and
w £ W write aw = a' w + awv, with <x'tu G W and aw G F . It is easy to
check that a' G GL(W) and that ip: a t-t a' is a homomorphism from A into
GL(W). Thus B = ker <p is a normal subgroup of A.
for W. For any b G F* define ri, G -A
Choose a basis {wi,...,wn-i}
via lui ►-»• t«i + bv, Wi i-t Wi for 2 < i < n — 1, and v ■-> u. Then 74 is a
transvection (with hyperplane spanned by {tz;2,.. .,wn-i,v}),
and if «; G W
then Tt «; = w + fry, so r¿ = \\v and T& G B. If n = 2 the matrix representing
" 1 0 "
7j, relative to the basis {wi, v} is
, and b G F* is arbitrary. It follows
b 1
from Theorem 2.6.4 and Propositions 2.6.5 and 2.6.6 that the conjugates of
B generate SL(V) for all n > 2. If a G B then a' = lw, so <r u;¿ = Wi + a¿u,
all i, Oj G F . Thus, if o\, ai G B , their representing matrices relative to the
basis {wi,... ,wn-i,v}
have the partitioned form
/
0
Ui
1
and
/
0
1*2
1
Since
I
Ui
I
u2
I
0
/
0
J
0
Ul + U 2
1
U2
1
Ui
1
we see that B is abelian.
Theorem 2.6.13 Except for PSL(2,2) and PSL(2,3), every PSL(V) is a
simple group.
Proof Choose v / 0 in V, and take A = StabSL(V)([t>]) and B<A as in Proposition 2.6.12. Write Z = Z(SL(V)) and set H = A/Z = Stab P S L (v)(H),
K = BZ/Z < H. Then all the conditions of Iwasawa's Theorem (1.6.9) are
met, and PSL(V) is simple.
A
Exercise
Use the action on the projective line P to show that PSL(2,2) Si Sym(3),
PSL(2,3) Si Alt(4), and PSL(2,4) Si Alt(5). Note also that PSL(2,5) Si
Alt(5) (why?).
2.6. SOME LINEAR
45
GROUPS
Note that |PSL(3,4)| = 20160 = |Alt(8)|, and that both groups are
simple. Let us show that they are not isomorphic. For any a € SL(3,4) write
ä for its image in PSL.
If W is any element of order 2 in PSL(3,4) then <r2 € Z(SL(3,4)), so
6
a = 1. Replacing a by r = a3 we have r = a and r of order 2 in SL as well
as in PSL. Let W = ker(l + r) Ç V, so dim V = 3 = dim W + dim (1 + T)V.
Note that v € W if and only if TV = v (characteristic 2), so W / V. Also
T ( 1 + T > = (T + 1)U, allv e F;i.e. ( 1 + T ) U is fixed by r . Thus ( 1 + T ) V Ç W.
Now dim (l-t-r)V < dim W, and it follows that dim (1+T)V = 1, dim W = 2.
Consequently W is a hyperplane and TV — v 6 W for all u € V; i.e. r is a
transvection. By Proposition 2.6.5 we see that all elements of order 2 are
conjugate in PSL(3,4).
Since Alt(8) has nonconjugate elements of order 2, e.g. (12)(34) and
(12)(34)(56)(78), it is immediate that PSL(3,4) 9» Alt(8).
Exercise
Note also that PSL(4,2) and PSL(3,4) are of the same order. Show that
PSL(4,2) j£ PSL(3,4) by considering the matrices
' 1 0
0 0
0 1 0 0
0 0
1 0
0 0
1 1
and
1 0 0 0
1 1 0
0
0 0
1 0
0 0
1 1
inSL(4,2).
It is in fact true that PSL(4,2) Si Alt(8). One proof proceeds by finding
a set of generators for PSL(4,2) that satisfy the relations in a presentation
for Alt(8); see Huppert [37], page 157, for details.
It can be shown that any simple group of order 168 is isomorphic to
PSL(2,7) (see [37], page 184). That applies in particular to PSL(3,2), the
collineation group of the world's smallest projective plane P, with 7 points
and 7 lines.
Finally, |PSL(2,9)| = 360 = | Alt(6)|. In this case PSL(2,9) has a subgroup isomorphic with PSL(2,5), hence of index 6, so the action on its cosets
gives an isomorphism into Sym(6), and PSL(2,9) Si Alt(6) since it is the
unique subgroup of index 2 of Sym(6). Again, see [37] for details.
The argument used in the proof of Proposition 2.6.11 shows that P G L ^ )
acts faithfully and doubly transitively on the projective space P(V). Note
that | PGL(2,g)| = (q + l)q(q - 1), a product of three consecutive integers,
and that the projective line P = p2(q) has q + 1 points. A glance at page 17
suggests the possibility of sharp 3-transitivity.
Theorem 2.6.14 If dim V = 2 then PGL(V^) is sharply triply transitive on
the projective line P = P(V).
46
CHAPTER 2. SOME GROUPS
Proof Write G for PGL(F), and if T G GL(V) write r for its image in G.
We know that G is doubly transitive on P. For 3-transitivity it will suffice
to choose [u] ^ [v] in P and to show that the 2-point stabilizer (?[„],[„] is
transitive on P' = P\{[u], [v]}. Since {u, v} is a basis for V every point in P'
can be written as [au + bv], where a,b G F and ab ^ 0. Take any such point
[w] G P ' , w = au + bv. Define r € GL(V) by setting TU = au and TV = bv.
Then r 6 G and r([u + u]) = [au + bv] = [w], and the G-orbit of [u + v] is all
ofP'.
For sharpness take w = u + v, so [tv] # [u] or [v], and suppose that
f € G[u],[«],[«;]• Say that TU = au, TV = bv, and TW = cw. Then
cu + cv = cw =
T(U + V)
=
TU
+
TV
so a = c = b and r = 1 G G.
2.7
= au + bv,
A
Mathieu Groups
We begin this section with some ideas about permutation groups that could
well have gone into Chapter 1.
Suppose that G acts transitively on a set S and a & S. Set 5* = S U {a}
and extend the action of G from S to 5* by simply agreeing that ax = a, all
x e G. The hope is to embed G within a larger group G* acting transitively
on S* and having G = S t a b e (<*)• If tn &t can be done then G* is called a
one-point extension of G. Note that if G is (sharply) fc-transitive on 5 , and
such a G* exists, then G* is (sharply) (k + l)-transitive on S*.
For a rather obvious example take G = 5„, S = {1,2,..., n } , a = n + 1,
and G* = Sn+i- Unfortunately, interesting one-point extensions are rather
rare. The next theorem, while somewhat technical, provides the possibility
of some interesting constructions.
Theorem 2.7.1 (E. W i t t , 1938) Suppose that G is faithful and k-transitive on S, k > 2, and S* = S U {a}, with a <¿ S. Set ax = a, all x eG.
Suppose that there are elements x € G, h € Perm(S*), and s,t G S, satisfying
1. sx — t and tx = s;
2. sh = a,ah
= s, and th = t;
3. {xhf e G and h2 G G; and
4- hG,h = G,.
Then G* = (G, h) is a one-point extension of G.
Proof Note that x $ G„ so G = G, U G„xG, by Corollary 1.6.3. Let
us show that hGh Ç G U GhG. Since sh* = ah = s and h2 G G we have
2.7. MATHIEU
47
GROUPS
h2 G Ga, and by part 4 hG„ = G,h. Also (xh)3 = (xhx)(hxh)
= x ^ / i G . Thus
hxh G x~lh~xG = x~lh~lh2G
/iG/i
=
=
G G, so
h{G, U G,xG,)h = hG,h U hG.xG.h
G.L>GshxhG,ÇGllG,x-lhGG,ÇGoGhG.
It follows that GöGhG is a group, for GhGGhG = GhGhG Ç GllGhG and
(GhG)-1 = Gh~lG = Gh~lh2G = GhG, and in fact GuG/iG = (G, h) = G*.
But then G* is clearly G, since no element of GhG stabilizes a.
A
If F is a field and V = F 2 then the points of the projective line P = P(V)
are of two types: first there is the set of points [(a, 1)], a G F, which can
be identified with F itself via [(a, 1)] <-► a, and there is one additional point
[(1,0)], commonly called the point at infinity and denoted oo. With that
identification P = F U {oo}.
Recall now the group C^(F) of linear fractional transformations in the
exercise on page 41. It is isomorphic with PSL(2,F), which acts doubly
transitively on the projective line. If /(x) = (ax + b)/(cx + d) G C2{F)
then /(x) determines a function from F U {oo} to itself via substitution, with
/(oo) = a/b ( = oo if 6 = 0), etc. As a result £2 acts on P = F U {00},
and it is easy to check that the isomorphism with PSL(2, F ) provides an
equivalence of actions. In particular, £2 is faithful and doubly transitive on
P by Proposition 2.6.11.
If F is finite and of odd characteristic then F* is cyclic of even order, and
the set F x 2 of squares in F* is just half of F* (each a2 G F * 2 has two square
roots, viz. ±a).
Suppose now that F = Fg. Recall that the map x 1-4 x 3 is an automorphism of Fg. We enlarge £2(9) by adjoining the transformations g(x) —
(ax3 + b)/(cx3 + d), where a,b,c,d G F and ad - be G F * \ F x 2 , and call the
resulting set Mi0.
Exercise
Show that M 1 0 is a group having £2(9) as a subgroup of index 2.
The action of £2(9) on the 10-point projective line P = F 9 U {00} extends
naturally to an action of Mw on P (also by substitution), it is at least doubly transitive since £2 is doubly transitive. The 2-point stabilizer (Mi0)o,oo
consists of the transformations of the form /(x) = ax e , a G F*, with e = 1
if a G F x 2 , e = 3 otherwise. Clearly the (Mio)o.oo-orbit of 1 is all of F*, so
AÍ10 is 3-transitive on P. If f(x) = axc is in the 3-point stabilizer (Mio)o,i,oo
then 1 = / ( l ) = a, and e = 1 since 1 G F x 2 ; i.e. f(x) is the identity element
of Mio, and Mw is sharply triply transitive. Its order is 10 ■ 9 • 8 = 720.
The field F 9 is a degree 2 extension of F3, so it is generated over F 3 by a
root of an irreducible quadratic. Since x 2 + x - 1 is irreducible over F 3 we
may take F 9 = F 3 (6), with Ó2 = 1 - Ó. Thus
F 9 = {a + bS:a,
6GF3}.
CHAPTER
48
2. SOME
GROUPS
Exercise
Define 51 and 52 in M\o via gi(x) = 62x and 52(1) = Sx3.
(<7i) 02) is a quaternion group of order 8.
Show that
We are now in a position to carry out the procedure of Witt's Theorem
(2.7.1). Choose some point a £ P = F 9 U {00} and set P* = PL) { a } . Define
h 6 Perm(P*) by setting
h(a) = 00, h(oo) = a, and h(a + bó) — a — bô
for a,b € ¥3. Take / e Mio defined by f(x) = 1/x; it will play the role of x
in Witt's Theorem. The roles of s and f are played, then, by 00 and 0, and
the first two conditions of the theorem are met. Clearly h? = 1, and fh has
orbits {1}, {-1}, {0,oo, a } , {<5, - 1 -6,1 - 6}, and {-6,1 + ¿ , - 1 + 6}, so also
(fh)3 = 1.
The stabilizer (Mi0)oo has order 72; it contains the translation subgroup T
consisting of all fa, a € F 9 , where fa{x) = x + a, and the two transformations
<?i and 02, with gi{x) = S2x and 52(2;) = ¿x 3 . It follows from the exercise
above that (Mio)oo = (51,52, fa-o. G F 9 ), so to verify condition 4 in Witt's
Theorem it will suffice to conjugate those generators by h. Straightforward
calculations show that hgih = 5251, hg^h = g^1, and if a = b + cS € F9 then
hfah = fb-cSSet M n = (Mio,h), the first Mathieu group.
We may repeat the procedure and extend M n . Choose a point ß & P*
and set P** = P* U {/3} = F 9 U {00, a, 0 } . Define k 6 Perm(P**) via
k(a) = ß, k(ß) = a, fc(oo) = 00,
and fc(x) = x 3 for x € F 9 .
It is clear that the first two conditions of Witt's Theorem are met, with
x,h,s,t, and a replaced, respectively, by h,k,a, 00, and ß. Also fc2 = 1, and
the orbits of hk are
{0}, {1}, {-1}, {oo,a,/J}, {6,-1 + 6,1 + 6},
{-6,1-6,-1-6},
so (hk)3 = 1. The stabilizer (Mu)a is M i 0 . For / € Mio, say with / ( x ) =
{axf+b)l{cx(+d),
we have kfk(x) = (a3x(+b3)/(c3xc+d3),
anda3d3-63c3 =
3
x2
2
(ad - be) € F
if and only if ad - be 6 F * , so kfk e M 1 0 .
Set M12 = ( M n , fc), the second Mathieu group. We have proved
Theorem 2.7.2 The first Mathieu group M\\ is a one-point extension of
Mio; it is sharply A-transitive of degree 11, and has order 11 ■ 10 • 9 • 8 =
7,920 = 2 4 - 3 2 - 5 1 1 . The second Mathieu group M12 is a one-point extension
°f Mu; it is sharply 5-transitive of degree 12, and has order 12• 11 ■ 10■ 9• 8 =
95,040 = 2 6 - 3 3 - 5 1 1 .
2.7. MATHIEU
49
GROUPS
To begin another round of one-point extensions take F = F4 and G =
PSL(3,4). If S is a primitive element for F then <53 = 1, 62 = 1 + 6, and
F = F2(6) = {0,1,6,1 + 6}.
As representative vectors in V = F3 for the points of the projective plane
P — PÏ{V) w e may take
va,b = {a,6, 1), va = (a, 1,0),
and
v^ = (1,0,0),
with a,beF,
so \P\ = 21. Recall that G is doubly transitive on P (Proposition 2.6.11). Choose a point a^P and set P* = P U {a}, as usual.
Since a •-> o 2 is an automorphism of F the map f\ defined by /1 (a, b, c) =
2
(a + bc,b2,c?) is a permutation of F 3 , and it induces a permutation gi of P
via <h[t>] = [fiv]. Set «i = [i>oo] and ti = [uo] in P. Define /ii 6 Perm(P*)
via h(a) = si, h(s\) = a, and /ii = 91 on P\ {s\}; note that /ii(ti) = ti and
ft2 = l.
Next define i i G G via x\: [(a,b,c)] >-> [(6,o,c)]. Check that
(Xi/n) 3 : [(a, 6,c)] ^ [{ac + b2(l + c 3 ), 6c + a 2 (l + c 3 ), c 2 )].
If c ?£ 0 the image is [(aCjbc,^)] = [(a,6,c)]; if c = 0 but ab ^ 0 we may
take 6 = 1 and o = 1, 6, 62, and check that each point is fixed by (x\hi)3.
Similarly s\, h, and a are fixed, and we conclude that ( i i / i i ) 3 = 1.
If y € GSl then y = Y, with
1 u
0 v
0 w
so det Y = vz — yw = 1. Thus fiYfi
i
y
z
GSL(3,4),
maps (a, 6,c) to
(a -I- 6(u2 -I- vw) + c(x2 + yz), bv2 + cy2,bw2 + cz2),
which is the result of multiplication by the matrix
Z =
1
0
0
u2 + vw
x2 + yz
y'
€SL(3,4).
w
Also hiyhi fixes a, so it follows that hiGtlhi = G,x.
The conditions of Witt's Theorem are thus met, and we define the third
Mathieu group M22 to be the one-point extension (G, hi). It has order
22 • I PSL(3,4)| = 443,520.
Keeping the notation above, choose a point ß & P* and set P** = P* U
{/?}. Define f2 on F3 via / 2 (a,6,c) = (a 2 ,6 2 ,c 2 i), then g2 € Perm(P) via
92H = [Î2v\. Set s 2 = a and í 2 = [UQO] in P*, and define h2 G Perm(P**)
by setting h2{ß) = s2, h2(s2) = ß, h2 = g2 on P; note that h2(t2) = t2 and
CHAPTER 2. SOME
50
GROUPS
h2, = 1. Set x-i = hi € M 2 2 . It is straightforward (using f\fi) to calculate
that (x2fe2)3 = 1.
Note that (M 22 )„ 2 = (M 2 2 ) a = G = PSL(3,4), and that G = G,, U
GtlxiGtl
by Corollary 1.6.3. Thus to show that fe2(M22),2fe2 = (M22)»2 it
will suffice to show that /i2G,1/i2 < G and that /i2xife2 € G. For the first
of these we may use y = Y G G»,, as above, and calculate that the effect of
f2Y$2 on F3 is the same as multiplication by
1 u2 xH2
y262
0 V2
2
0 w 6 z2
€SL(3,4),
so A2Î//12 6 G. Finally /12X1/12 = i i Define next the fourth Mathieu group M23 to be the one-point extension
<M22, fea) of order 23 • |M 2 2 | = 10,200,960.
One more time. Choose 7 g P** and set P*** = P** U {7}. Define / 3
on F 3 by setting f3(a,b,c)
= (a2,b2,c2), then g3 on P via </3[v] = [/ju].
Define fe3 e Perm(P***) via ^3(7) = /3, h3(ß) = 7, fe3(o;) = a, and fe3 = 5 3
on P . Set i 3 = /12 € M 2 3 , a3 = ß, í 3 = a, and observe that (M2 3 ) i s =
M22 = (G,hi) = {GSl,xi,hi).
It is easy now to check that h\ = (i 3 fe 3 ) 3 = 1,
hzG^hz = Gtl, fe3iife3 = i i , and fe3feife3 = fei. Thus we may define the
fifth Mathieu group M24 to be the one-point extension (M 23 ,fe 3 ), of order
24 |M 2 3 | = 244,823,040.
We summarize in a theorem.
T h e o r e m 2.7.3 The third Mathieu group M22 is a one-point extension of
PSL(3,4); tr is 3-transitive of degree 22 and has order 443,520 = 2 7 • 3* ■ 5 •
7 11. The fourth Mathieu group M 2 3 is a one-point extension of M22,' it is
^-transitive of degree 23 and has order 10,200,960 = 2 7 ■ 3 2 • 5 • 7 • 11 • 23. The
fifth Mathieu group M24 is a one-point extension of M23; it is 5-transitive of
degree 24 and has order 244,823,040 = 2 10 • 3 3 • 5 • 7 • 11 • 23.
The groups Mi2 and M24 do not admit further transitive extensions. A
proof can be found in [38].
Since M u has degree 11 (and is faithful) we may view it as a subgroup
of Sym(ll), acting on S = { 1 , 2 , . . . , 11}. Similar remarks apply to the other
Mathieu groups as well. In fact, a bit later we will exhibit generators for
them in the appropriate symmetric groups.
Theorem 2.7.4 The five Mathieu groups are all simple.
Proof We begin with G = M u . Choose E G Syl u (G); it is cyclic of
order 11, so a generator for it must be an 11-cycle and E is transitive. Set
C = CG(E) > E. Take r € Stabc(l). For any i e S = { 1 , . . . , 11} choose
a e E with 1" = i and observe that iT = \aT = V = V = i, so r = 1. Thus
C is regular on S, so \C\ = 11 and C = E. Next set N = NG(E); N acts by
2.7. MATHIEU
GROUPS
51
conjugation on E as a group of automorphisms. The kernel of the action is
C = E, and | Aut(£)| = 10, so if k = [N: E) then k | 10 and |JV| = llfc. The
number of distinct Sylow ll-subgroups of G is £ = [G: N] = \ (mod 11), so
\G\ = \\kl = 11 • 10 • 9 • 8. This forces k = 5 and I = 144.
Suppose now that 1 ^ K <G. Then K is transitive on S by Proposition
1.6.7, so 11 | \K\ and we may assume E < K. In fact, since K <G, K
must contain all of the 144 Sylow ll-subgroups of G, and \K\ is a multiple
of 11 • 144 = 11 • 2 • 9 • 8; thus [G: K] is 1 or 5. If K ¿ G then 5 j \K\, so
N n K = NK(E) = E. By a theorem of Burnside, which we borrow from
a later chapter (Theorem 4.1.9), there is a characteristic subgroup H of K
(hence H <G) such that K = H x E. But now 11 J \H\ is in conflict with
H < G because of Proposition 1.6.7, so in fact K = G and G = Mu is simple.
The others are now easy. Since the stabilizer of a point in M12 is M u ,
which is simple, then if M12 were not simple it would have a regular normal
subgroup by Theorem 1.6.10. But that contradicts Theorem 1.6.12, since
M12 is 5-transitive. The stabilizer of a point in M22 is PSL(3,4), also simple.
Thus if M22 were not simple it would have a regular normal subgroup, by
Theorem 1.6.10, and would have degree 2* or 3 by Theorem 1.6.12, whereas
the degree is 22. The same reasoning applies to M23 and M24.
A
There is much more to be said about the Mathieu groups. They were
discovered in 1861 by E. Mathieu [46], although his arguments were not totally
convincing. In fact C. Jordan and G. A. Miller tried to argue in subsequent
papers that M24 could not exist. In 1938 E. Witt (see [66]) gave essentially
the constructions that appear above, and he also discussed the groups as
automorphism groups of combinatorial objects called Steiner triple systems
(see [67], also [4]). The groups Mi 2 and M24 are also automorphism groups
of certain Golay codes (see [2]).
For over a hundred years the Mathieu groups were the only known simple
groups that did not fit into infinite families such as the alternating groups and
projective special linear groups; as such Burnside called them sporadic groups.
Further sporadic simple groups were discovered beginning in the 1960s - the
list is now complete, there are all told 26 of them. It is a consequence of the
classification of finite simple groups (see [30]) that M12 and M24 are the only
nontrivial examples of 5-fold transitive groups.
For purposes of calculation it is useful to have (permutation) generators
for the groups. The ones given below are a slight variation on generators
given by Mathieu himself in [47]. Set
A = (3,7,ll,8)(4,10,5,6),
ß = (1,2,3,4,5,6,7,8,9,10,11),
i/ = (l,12)(2,ll)(3,6)(4,8)(5,9)(7,10),
n = (1,21,10,7,8,12,9)(2,19,15,6,11,18,20)(3,16,17,13,4,5,14),
CHAPTER
52
2. SOME
GROUPS
p = (1,6,20,12,10,21,11)(2,8,13,15,5,16,22)(3,4,17,19,9,14,18),
a = (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23),
n = (1,24)(2,23)(3,12)(4,16)(5,18)(6,10),
T2 = (7,20)(8,14)(9,21)(11,17)(13,22)(15,19),
and r =
TIT 2 .
Then Mu
(71-,a), and M 2 4 =
= (A,/x), Mí2 = (A,¿t,i/), M 2 2 = (7r,p), M 2 3 =
(^,T).
Exercise
If F is a finite field show that Aff(F), acting on F, has M2(F) as a
one-point extension, the new point being oo in P = F U {oo}. Use this
information to reprove Theorem 2.6.14, at least for finite fields.
2.8
Symplectic Groups
Suppose that V is a vector space of dimension n = 2m over a field F, and let B
be a nondegenerate alternate form on V. An ordered basis {iti, u i , . . . , um, vm}
for V built up of hyperbolic pairs {ui,Vi}, as in Theorem 1.7.10, is called a
symplectic basis for V, V is called a symplectic space, and somewhat loosely
speaking the study of (V, B) is called symplectic geometry.
An invertible linear transformation r of V is said to be symplectic if
B(TV,TW)
= B(v,w) for all v, w G V, i.e. if r is an isometry of V. The
set of all symplectic transformations of V is evidently a subgroup of GL(V);
it is called the symplectic group on V and denoted Sp(V). The dependence
of Sp(V) on the form B is not significant. Since any other nondegenerate
alternate form Bi on V is equivalent with B (Corollary 1.7.12), the resulting
symplectic group (relative to B\) is conjugate to Sp(V) in GL(V). In other
words, relative to appropriately chosen bases the two groups are represented
by the same matrices in GL(n, F).
Suppose, then, that a symplectic basis has been chosen for V. If r €
GL(V) say that r is represented, relative to that basis, by the matrix T.
An easy calculation (as on page 20) shows that r 6 Sp(V) if and only if
TlBT = B. The resulting group of matrices is denoted Sp(n, F), or also by
Sp(n, q) if | F | = q, finite.
Proposition 2.8.1 / / d i m V = 2 then Sp(V) = SL(V).
Proof We may calculate with Sp(2, F) via a symplectic basis, hence
B =M =
0
-1
1
0
2.8. SYMPLECTIC
By the remarks above T =
l
T MT
53
GROUPS
a b
c d
£ GL(2, F) is in Sp(n, F) if and only if
= M, i.e.
0
—ad +bc
0 1
-1 0
ad— be
0
Exercise
be a symplectic basis for V, but re-order the
Let {ui,t>i,... ,um,vm}
vectors as
{ t i i , u 2 , . . . ,um,vi,v2,.
■ ■ ,vm}.
Determine the resulting representing matrix for B. Show that every
A
0
0
(A1)-1
i€GL(m,F),
represents a symplectic transformation relative to this basis, and conclude
that Sp(2m,F) has a subgroup isomorphic to GL(m,F).
Recall that if 1 ^ T e SL(V), then r is called a transvection, with fixed
hyperplane W, if T\W = W and r t; — v € W for all v € V.
Let T be a transvection with fixed hyperplane W, and suppose that r €
Sp(V). By Proposition 1.7.4 dim Wx = 1, so say that Wx = Span(w). By
Corollary 1.7.5 we have u x = W, but also u e u x = (W-1)1- = W. Choose
= z, say, in W. For any
x 6 V \ W, so V = Span(i) 0 W. Then O^TX-X
v £ V write v = bx + w, b € F and u; € W, and define f € V* via / ( f ) = b;
thus ker / = W. By Corollary 1.7.2 3y e V such that /(u) = B(v,y), all t; 6
V. Note that TV = r(6a;+u;) = brx+w — b(x + z)+w = v + 6z = i;-l-/(u)2: =
v + B(v, y)z, 3L\\V Ç.V. In particular if w e W then «; — T W = w + B(w, y)z,
so B(w, y) = 0, all w, and y G W1- = Span(u). Say that y = cu, c G F*. Also,
since T € 5p(V), B(w,x) = B(TW,TX)
= B(w,x + z) = ß ( w , x ) + B{w,z),
so ß(ti;,z) = 0 for all w € IV, and z € W x , say z = du, d € F*. Thus
TU = V + ß(u, cu) ■ du = v + cdB(v, u)u. Set a = cd and we have
rt; = v +
aB(v,u)u,
all u e V, i.e. T is determined by the scalar a and the vector u. Write
T = T U I „, and call r a symplectic transvection with direction u.
Conversely, given any nonzero u £ V and any a € F* we may define
T = Tu,o via TV = v + aB(v,u)u, all v. Then clearly r is a transvection with
fixed hyperplane W = u±, and it is easy to check directly that B(TV,TW)
=
B(v,w), all v,w € V, so T UJ0 € Sp(V).
It will be convenient to extend the notation slightly and define rUiQ by
the formula above even when a = 0. Of course rUio is simply the identity
transformation.
CHAPTER 2. SOME GROUPS
54
Exercise
Fix u ,¿ O in V.
S h o w t h a t Tu<aTu<b = T„,<,+(,, T6u,o = T"u,a62 >
an(
i T~a = T U ) _ a .
2. Conclude that {rUia:a G F } is a group isomorphic with the additive
group F.
3. If a G Sp(F) then °Yu,a = Tau<a.
If V is a symplectic space write T = (r U i a :0 ^ u G V, a G F*), the
subgroup of Sp(V) generated by symplectic transvections. It will be shown
eventually that T = Sp(V), so Sp(V) is generated by symplectic transvections.
Proposition 2.8.2 IfV is symplectic then the group T < Sp(V) is transitive
onV\{0}.
Proof Take v ^ w in V\{0}. If B(v,w) ^ 0 set a — l/B(v,w)
Then
T
. .
u,a(V)
_.
.
= V + aB(V,U)U
= VH
B(v,v — w) .
—.
r (V
B(v,w)
.
and u = v—w.
.
.
— W) — V — (V — W) = W.
Suppose then that B(v,w) = 0. Choose f 6 V* with f(v) ^ 0 and
f(w) ,¿ 0. By Corollary 1.7.2 3u G V such that f{v) = B(v,u) ¿ 0 and
f(w) = B{w, u) ^ 0. By the first part of the proof 3T\, T?. G T with T\ V = u
and T2 u = w, hence T<¡,T\ V = W.
A
Recall that a hyperbolic pair in V is an ordered pair (u, v) of vectors with
B{u,v) = 1.
Proposition 2.8.3 IfV is symplectic then the group T < Sp(V) is transitive
on the set S of all hyperbolic pairs in V.
Proof Take (ui,vi), (1*2,^2) G «S. First use Proposition 2.8.2 to choose
T G T with TU\ = U2, and hence T:(UI,VI)
t-t (02,^3) G <S, U3 = TV\.
Now we need a G T with <r:(u2,i>3) *-¥ (ii2,t;2)) i e . cr G Stabr(u2) and
<TU3 = U2- If B(v3,V2) ^ 0 u s e °~ = Tu,a, with u = U3 — t>2, as in the proof
of Proposition 2.8.2. Then avz = t>2 and also B(it2,i>3) = 1 = B(u2,V2), so
¿?(ii2,tt) = ß(u 2 ,w3 - V2) = 0, and therefore <rit2 = U2 + aJ3(u2,u)u = U2.
Suppose then that B(v3,vi) = 0. Note that (u2,«2 + "3) G <S, and that
B{v3,U2 + "3) ^ 0) s o by what was just proved 3<TI G T with cri u 2 = 1x2 and
<7i t>3 = "2 + W3. But also B(u2 + v3,V2) = B{u2,V2) ^ 0, so 3(72 € T with
cr2 U2 = U2 and «72(^2 + U3) = v2. Setting a = V20~\ G T we have <ru2 = t¿2
and (Tt)3 = <T2(it2 + v3) = u2- Finally, o r G T and or: (ui,ui) 1-+ (u 2 ,U2). A
T h e o r e m 2.8.4 The symplectic group Sp(V) is generated by its symplectic
transvections.
2.8. SYMPLECTIC
GROUPS
55
Proof Use induction on m, where 2m = n = dim V. The case m = 1
is covered by Proposition 2.8.1 and Theorem 2.6.4. Choose a hyperbolic
pair (u,v) in V and set W = Span(u,v), a hyperbolic plane. Then V =
W © ^ - 1 - by Proposition 1.7.9. Take any a e Sp(V). Then (<ru,<7v) is a
hyperbolic pair, so by Proposition 2.8.3 3r € T with rau = u and TOV = v,
so TO\W = l\v- But also TO~\WL 6 Sp(W x ). By induction TO-\W± is a
product of symplectic transvections on W x . Any such transvection, say Tw<a,
determines, by the same formula, viz. Tw<a(v) = v + aB(v, w)w, a symplectic
transvection on V whose fixed hyperplane contains W. Since also TO~\W = l w
we conclude that ra € T, and hence that a 6 T.
A
Corollary 2.8.5 Sp(V) < SL(V).
Proposition 2.8.6 The center o/Sp(V) is {±1}.
Proof Note that ru<a v = v if and only if B(v, u) = 0, so u x is the space of
fixed points for r u>0 . If a € Z(Sp(V)) then in particular a commutes with
every r Ui0 . If v e u x then ru<a(av) = crru>av = av, so cru € u x , i.e.
ffui=«i
x
for all u. But Span(u) = rad u and a is symplectic, so a carries Span(u) to
Span(u); in particular, cru = cuu for some cu 6 F , all u G V. Choose u and
v linearly independent in V. Then o~{u + v) = cu+v(u + v) = c u u + c„u, so
c« = cu+v = cv = c, say, independent of u and u, and a = civ- But then
B(u, v) = B((7 u, a v) = <?B{u, v) for all u, v, so c = ± 1 .
A
Proposition 2.8.7 / / | F | > 4 then the derived group of Sp(V) is Sp(F).
Proof Fix u ¿ 0 in V and
c = a / ( l - 6 2 ), ci = — b2c.
exercise on page 54. Choose a
Then OT~*O~1 = OTU,-CO~1 =
Tu.cCfT'lcr'1 = Tu,cTUtd = TUta
a € F*. Choose 6 G F \ {0,±1} and set
Then c + d — a, so Tu.cTu.d = r u , a by the
e Sp(V) with au = bu (Proposition 2.8.2).
r a u ,_ c = rt Ui _ c = TUI_&2C = ru>d, and hence
€ Sp(V)'. Apply Theorem 2.8.4.
A
Proposition 2.8.8 / / \F\ = 3 and dim V > 4 i/ien ifte derived group of
Sp(V) Ù Sp(V).
Proof Choose a symplectic basis {ui, u i , . . . , um, vm} and define linear transformations a and r as follows:
cr:ui i-+ ui + W2, ui •-> t>2, U2 i-> ui, u2 >-► «i — «2, and
Ui ►
►
- Uj, Uj t-> Uj if i > 2;
r:i¿i i-¥ ui - v\ +V2, «i »-> «1, u 2 H-> u 2 + Ki, v2 »-► u 2 , and
Ui >-> Ui, Vi >-> Vi if i > 2.
It is an exercise to show that a and r are in Sp(V) and to show that
ara~lT~l = TVI<1. Thus TVUI G Sp(V)'. Given any u ¿ 0 in V, choose
56
CHAPTER 2. SOME
GROUPS
7 G Sp(V) with jvi = u; then "fTVuif 1 = r 7 t I l i l = r„,i € Sp(V)' and also
T~\ = T„,_I € Sp(V)'. Apply Theorem 2.8.4.
A
The case \F\ = 3 and dim V = 2 is an exception, since Sp(2,3) = SL(2,3)
of order 24.
P r o p o s i t i o n 2.8.9 / / \F\ = 2 and dim V > 6 t/ven the derived group of
Sp(V) is Sp(V).
Proof The proof is exactly as for Proposition 2.8.8, except that
a:u\ >-> u\ +U3, V\ >-¥ V3, U2 >-> Ui, «2 "-► «i + f3,
U3 i-> U2, f3 >-► ^2, and
Ui >-► u¿, u» t-4 fi if i > 3;
r:tii !-»• Ui + 1*2, Vi •-> V\, U2 •-► t>l + 1*2 + t>2 + V3, ^2 >-> «2,
«3 !-> f2 + "3 + t»3, V3 •-► u 3 , and
Ui ■-» Ui, Ui i-> fj if ¿ > 3.
A
Since Sp(2,2) = SL(2,2) a 5 3 it is an exception to Proposition 2.8.9. It
will be shown later that Sp(4,2) a S 6 , so it is also an exception.
Since Sp(V) < GL(V) there is an action of Sp(V) on the projective space
P = P„_i(V). The kernel of the action is £(Sp(V)) = {±1}, and we define the projective symplectic group PSp(V) to be Sp(V)/Z(Sp(V)); it acts
faithfully and transitively (see Proposition 2.8.2) on P.
Proposition 2.8.10 Sp(V) is primitive on P =
Pn-i(V).
Proof We may assume that n > 4 by Proposition 2.8.1, since SL(2, F)
is doubly transitive. Suppose that S Ç P, \S\ > 1, and either a S = S or
er SC\S = 0 for each a € Sp(V). We show first that there are [u], [v] G S with
B(u, v) ^ 0. Suppose to the contrary that B(u, v) = 0 for all [«], [v] G S.
Choose [u] ¿ [v] in 5 and choose f e V* with /(u) = 1, f(v) = 0. By
Corollary 1.7.2 3x G V with B(u,x) = 1, B(v,x) = 0. Set W = Span(w,i),
a hyperbolic plane, and set H = {a G Sp(F):or|w = 1^} < Sp(V). Since
V = WQIV-1- it is clear that every a G S p ^ - 1 ) extends trivially to a' G
Sp(V) with o'\w = \\v 1 so in fact Sp(W^x) = {T\WL:T
G H}. Choose w ¿ 0
in W1. Since u G WL there is some T € H with r u = w, by Proposition 2.8.2.
Since r u — u we have [u] G 5 n r S, so T 5 = 5 . Also [w] = r[u] G 5 , and «;
was an arbitrary nonzero vector in Wx. Since Wx ^ 0 there is a hyperbolic
pair (y,z) in Wx; hence [y], [z] G 5, but B(y,z) = 1, a contradiction.
Thus we may in fact choose [u], [v] G 5 with B(u,v) ^ 0, so we may
assume that B(u,v) = 1 and {u,v) is a hyperbolic pair. Now take any [tu] G
P. If B(u, w) ^ 0 we may assume that (u, w) is a hyperbolic pair and by
2.8. SYMPLECTIC
57
GROUPS
Proposition 2.8.3 find a € Sp(V) with au = u and av = w. Thus u 6 SHcr S,
so <TS = S and [w] e S. On the other hand, if B(u,w) = 0 choose / € V*
with / ( u ) = /(u>) = 1 and thereby obtain x € V with Z?(tt, x) = 2?(w,z) = 1.
As above, [x] € 5 , and also 3T € Sp(V) with TU = w and T I = I . Again
TS = S, since [x] G S D r 5 , and [tu] = T[U] € S. Thus S = P.
A
Theorem 2.8.11 Except for PSp(2,2), PSp(2,3), and PSp(4,2) every projective symplectic group PSp(V) is simpie.
Proof We know that PSp(V) acts faithfully on P = Pn-i(V), and the action
is primitive by Proposition 2.8.10. With the exceptions noted, PSp(V) is its
own derived group by Propositions 2.8.7, 2.8.8, and 2.8.9. Fix [u] € P and set
H = Stab S p (v)(M), H = H/{±1} = Stab P S p(v)(M). Set K = {Tu,a:ae F}.
By the exercise on page 54 K < H and K ^ F, so K is abelian. If a e Sp(V)
then "K D {Tau<a'.o- 6 F}, so \j{aK:a € Sp(V)} contains all symplectic
convections by Proposition 2.8.2, and hence generates Sp(V) by Theorem
2.8.4. Thus ÇŒ-.ô e PSp(V)) = PSp(V), and with the exceptions noted
PSp(V) is simple by Iwasawa's Theorem (1.6.9).
A
Theorem 2.8.12 / / | F | = q, finite, then
m
m
\Sp(n,q)\=q *Y[(q2i-l),
¿=1
where n = 2m, and
Proof Any one of the qn — 1 nonzero vectors can serve as the first vector
u in a hyperbolic pair. Since \q-L\ = qn_1 there are qn — q n - 1 vectors v
with B(u,v) ji 0. Thus there are (qn — q"~1)/(q — 1) = <7n_1 choices of v
for which B(u, v) = 1, and consequently (qn — l ) g n _ 1 distinct hyperbolic
pairs. If (u, v) is a hyperbolic pair and W is the hyperbolic plane with {u, v}
as basis, then a 6 Stabg p (v)((u,v)) if and only if a\w = l w , and we saw
in the proof of Proposition 2.8.10 that Stab S p ( v ) ((u,j;)) 2* SptW-1-). Thus
|Sp(n,g)| = (qn - l ) g n _ 1 | S p ( n - 2,q)\ by Proposition 1.2.1, and we may
argue by induction. When n = 2 then |Sp(2,g)| = |SL(2,g)| = q(q2 - 1).
Assume inductively that | Sp(n - 2 , q)\ = q(m-1)2 fl^iq2*
- 1), then
\Sp(n,q)\=q2m-1(q2m
m-l
m
i=l
1
- l)^" 1 ) 2 f [ (92i - I) = qm'Y[(q2i " ^
The relation to | PSp(n,g)| is immediate from Proposition 2.8.6.
A
Note, for example, that |Sp(4,2)| = 720 = | 5 6 | . Let us prepare to show
that Sp(4,2) S 5 6 .
CHAPTER
58
2. SOME
GROUPS
If V has dimension 4 over F = F 2 then | V| = 16. Choose vi ^ 0 in V, and
then choose V2 G V \ vx. Then i>2 € t ^ \ vj-, so u/- + v2 = V, and therefore
d i m ^ n t ) ^ - ) = 3 + 3 - 4 = 2and|t; 1 J -nu^| = 4. Thus IUJ'-UUJH = 8 + 8 - 4 =
12, and we may choose v3 € V \ {v¡- I) v2}- Similar inclusion-exclusion
arguments show that \v¡- U v2 U VJ¡-\ = 14, so 3u 4 € V \ {vx U v2 L) v^}.
Then {v^ U v2 U v3 Ö v^\ = 15, and the last remaining element u5 of V is in
V \ {vf U t>2" U V3- U U4-}. The resulting set { « i , . . . , «5} satisfies ¿?(ui, Vj) = 1
if i ^ j , and is a maximal nonorthogonal set in V.
Now we approach maximal nonorthogonal sets from another direction.
Choose a hyperbolic pair («1,^2) in V and set W = Span(ui,t;2). Then
V = WQW1- and we may choose a hyperbolic pair (xi,Xi) in Wx. If we set
«3 = ui + «2 + i i , V4 = t>i + «2 + x*i a n c l u 5 = wi + u2 + £1 + ^2, then it is
a routine matter to check that {i>i,... ,1*5} is a maximal nonorthogonal set
(m.n.s.).
It is in fact the unique m.n.s. containing {vi,i>2}. The reason: given
(^ii ^2), suppose that we try to adjoin w so that {v\, v2, w} is a nonorthogonal
set. Write w = a\Vi + a2v2 + bix\ + 62^2- Then 1 = B(vi,w) = 02 and
1 = B(v2,w) = ai, and tu = vi+V2+hx\+b2X2.
If we try taking ¿>i = 62 = 0,
giving, say, w' = v\ + V2, then w' ± w for any other choice of values of 61
and b2, so {vi,V2,w'} is not contained in a larger nonorthogonal set. There
remain only the other three possible assignments of values to 61 and b2.
By the proof of Theorem 2.8.12 there are (2 4 - 1) • 2 4 - 1 = 120 hyperbolic
pairs in V. Each m.n.s. contains 5 • 4 = 20 hyperbolic pairs, so there are
exactly 120/20 = 6 distinct m.n.s.'s in V.
Proposition 2.8.13 Sp(4,2) = S6.
Proof Let S be the set of 6 m.n.s.'s in V, as above. It follows from Proposition 2.8.3 that Sp(4,2) acts transitively on S, so there is a homomorphism
<¿>:Sp(4,2) -> Se- Let K = ker <ç. Choose a hyperbolic pair (v\,V2), set
W = Span(vi, V2), and choose a hyperbolic pair (xi,i2) in W±. As above we
obtain two m.n.s.'s {VÍ} and {XJ}, with
V3 = «I + V2 + Xi,
Vi — V\ + V2 + X2,
V5 = V\ + t>2 + Xi + X2',
X3 — Xi + X2 + Vi, X4 = Xi + X 2 + V2, X5 = Xi + X2 + V\ + U2-
Set z = v\ + V2 + xi + x 2 and note that {u¿} D {x¿} = {z}. Every a 6 K fixes
{VÍ} and {xj}, so a fixes { u j D {x¿}; i.e. az = z, all a G K. But Sp(V) is
transitive on P{V) = V \ {0}, and for each r € Sp(F) we have Ta(r z) = T z;
i.e. Tz is fixed by all Ta e # , which is all of K. Since Sp(V) = PSp(V) acts
faithfully on P(V) we conclude that K = 1. Thus <¿> is 1-1, and hence is an
isomorphism since |Sp(V)| = |Se|A
Thus Sp(4,2) is not simple, but its derived group (= A6) is simple and of
index 2.
2.8. SYMPLECTIC
59
GROUPS
Exercises
1. Suppose that V is a space with an alternate bilinear form B and that
W is a subspace. Set U = W/ rad W. Show that U is a symplectic
space if we define
B(wi + rad W, w2 + rad W) = B(wi, wz).
2. (This is an alternate approach to Proposition 2.8.13). Take V of dimension 6 over F = F2 with basis {vi,...,ve}.
Define B via B(VÍ,VJ) =
1 + Sij, and verify that (V, B) is a symplectic space. Observe that 56
is embedded isomorphically in Sp(V) by permuting {v\,...,ve}.
Set
W = Span(t>i + ■•• + VQ); W and W1- are Se-invariant, so Se acts on
U = W^/W, a 4-dimensional symplectic space by Exercise 1 above.
The transposition (12) € Sß embeds as the transvection r,, 1+ „ 2i i in
Sp(V) and maps to a transvection in Sp(i/). Its conjugates in Se yield
all 15 transvections in Sp(U), and therefore S6 — Sp(f/). Provide the
details.
Note that | PSp(4,3)| = 3 4 • (3 2 - 1)(3 4 - l ) / 2 = 25,920. A. Cayley and
G. Salmon proved in the 1840s that each sufficiently general cubic surface
has 27 straight lines lying on it. C. Jordan (who first defined the symplectic
groups) determined the group of incidence-preserving permutations of the 27
lines, and found it to have order 51,840. The group is in fact the Weyl group
of type E% (e.g. see [32] or [36]), and it has a copy of PSp(4,3) as a subgroup
of index 2.
In classical plane geometry it was known that each sufficiently general
quartic curve has 28 "bitangents," i.e. lines that are tangent to the curve at
two distinct points. This seems to have first been proved rigorously by K.
Jacobi in 1850. Jordan showed that the symmetry group of the 28 bitangents
is a simple group of order 1,451,520 - it is isomorphic with PSp(6,2).
For further discussion of the history of these configurations and relations
between them, and for further references, see [63]. Also see the Atlas [14],
where the groups are denoted S4(3) and Se (2).
Groups and Characters
by Larry C. Grove
Copyright © 1997 John Wiley & Sons, Inc.
Chapter 3
Counting with Groups
This chapter is concerned with enumeration techniques for orbits under group
actions. The techniques afford a wide variety of applications to combinatorics,
and thereby to many other disciplines. For example, one of the early sources
[51] was concerned with enumerating chemical polymers.
Except for the Orbit Formula (Proposition 3.1.2), which is used later, the
chapter is by and large independent of the other chapters.
3.1
Fixed Points and Orbits
Suppose the group G acts on a set S (assume that both are finite). If x G G
and s G S then s is a fixed point for x if sx = s. Write Fix(x) for the set of all
fixed points in S of x. Define the character 6 of the action via 8(x) = | Fix(x) |,
all x G G. Thus 8 is a function from G to N.
Proposition 3.1.1 The character 6 of a group action is a class function on
G, i.e. 6 is constant on conjugacy classes of G.
Proof If x, y e G verify that Fix(x w ) = Fix(x) y , and so
9{xv) = | Fix(i»)| = | Fix(s)»| = | Fix(x)| = 0(x).
A
Proposition 3.1.2 (The Orbit Formula) If G acts on S with character 9
then the number of distinct G-orbits in S is
Jfc = | G r 1 ^ { Ö ( x ) : x G G } .
Proof Set S = {(s, x) G S x G: sx = s}. For fixed s G S there are | Stab G (s)|
such ordered pairs; for fixed x e G there are 9{x) of them. Thus
| 5 | = £ { | S t a b G ( s ) | : S G S) = J > ( x ) : x G G}.
61
62
CHAPTER 3. COUNTING
WITH
GROUPS
To evaluate the first sum let 0\, ..., Ok be the distinct G-orbits in S, and
choose Si 6 0¿, 1 < i < k. Then
k
£{|Stab G (s)|:s e 5} = £ ICMIStabcfo)! = k\G\
t=i
by Proposition 1.2.1, so k\G\ = £
x
6{x).
A
Corollary 3.1.3 If G is transitive on S, and \S\ > 1, then G has an element
x having no fixed points.
Proof We have 1 = | G | _ 1 YH&{X):X G G}, which can be interpreted as
saying that the average number of fixed points for elements of G is 1, so not
all elements can have more than 1 fixed point. In fact, there is an element
(x = 1) that has more than 1 fixed point, so to compensate there must be at
least one element having fewer than 1 (therefore 0) fixed points.
A
The Orbit Formula is often referred to as Burnside 's Orbit Formula, e.g.
in [31], probably because it appeared in Burnside's book [10]. However it was
proved earlier by Frobenius, and a special case of it even earlier by Cauchy.
See [49] and [68] for a discussion of the history.
For an easy example of the use of the Orbit Formula imagine coloring the
faces of a regular tetrahedron using 2 colors, say red (r) and blue (£>), each
face being a solid color. Since each of the 4 faces can be one of 2 colors there
are 2 4 = 16 distinct ways the coloring can be done. Distinct does not mean
distinguishable, however. For example any one of the 4 ways of coloring with
1 r and 3 6's can be made to look exactly like any other one by simply rotating
the tetrahedron appropriately. Clearly any 2 colorings are indistinguishable
precisely when one can be rotated to the other.
Something slightly subtle has happened. We are no longer talking about
the action of the rotation group T on the set of 4 faces, but on the set of 16
colorings of the faces. To count distinguishable colorings we need to count
the orbits under that action. To that end we count first fixed points of group
elements. The identity 1 has of course 16 fixed points. There are just 2 other
types of rotation in T; there are 3 rotations of order 2 about axes joining
midpoints of opposing edges, and there are 8 rotations of order 3 about axes
through a vertex and the center of the opposing face. Since faces sharing
an edge through which an axis passes must be the same in a fixed coloring,
the order 2 rotations each have 4 fixed colorings. Similarly each of the order
3 rotations has 4 fixed colorings. By the Orbit Formula, then, there are
k = (1/12)(1 - 1 6 + 3-4 + 8-4) = 5 orbits, hence 5 distinguishable colorings.
They are of course rrrr, rrrb, rrbb, rbbb, and bbbb.
Exercises
1. How many distinguishable ways are there to color a regular tetrahedron
if each face is red, blue, or green? List them all.
3.1. FIXED POINTS AND
ORBITS
63
2. How many distinguishable ways are there to color the edges of a square
with 2 colors? With 3 colors?
For a rather different sort of application, we may use the Orbit Formula
to calculate class numbers of certain groups - the class number c(G) is simply
the number of distinct conjugacy classes in G.
Suppose that H <G and that [G: H] = p, a prime. Assume that c(H) = d
and that the conjugacy classes of H are labeled as Ci, C2, . . . , Cd- Note that
G acts by conjugation on the set S = {Ci, Ci, ■ ■ ■, Cd) - to see why suppose
that Ci = {yz: z G H), with y e H. Take x G G, then Cf = {yzx: z G H) =
{{yx)z' : z G H), another Jí-conjugacy class since yx € H and zx ranges over
H as z does. Also H is in the kernel of that action, so in fact G/H acts
on S. Since G/H is cyclic of order p every x G G \ P determines the same
orbits in S. We may assume the CVs labeled so that Cf = C¿, 1 < i < k for
all x G G \ H, and the remaining C / s lie in G-orbits each of size p. Clearly
C\,..., Ck are each also G-conjugacy classes, and the unions of each of the
size p orbits are also G-conjugacy classes. Thus we see that the total number
of G-conjugacy classes lying in H is k + (d - k)/p.
P r o p o s i t i o n 3.1.4 ( B u r n s i d e , [10]) Suppose that G and H are as above.
Then
, c(H) - k
c(G) = pk + -i—i
,
P
where k is the number of H-classes that are also G-classes.
P r o o f It has already been observed that there are k + (d — k)/p distinct
G-classes in H, d = c(H); we proceed by considering classes not in H.
If x € G\H and h, z 6 H, then {hx)z = hz ■ [z,x~x] ■ x e Hx since
G' < H. Thus H acts by conjugation on the coset Hx. In fact, each iï-orbit
in Hx is a G-conjugacy class. To see that take a particular element hx in
some i/-orbit. Since G/H is cyclic of prime order it will suffice to show that
the conjugate of hx by any one element of G \ H lies in the same i/-orbit as
hx, so let us conjugate by x~l. But (hx)x~ = (hx)h, in the same i/-orbit as
hx.
Thus to complete the proof we need to count the Jï-orbits in each of the
p - 1 cosets of the form Hx, x # H. To that end we need to calculate numbers
of fixed points of elements of H. Consider first an element a lying in one of
the first k conjugacy classes C\,... ,Ck of H, which are also classes of G.
Thus xa = ab for some b G H, and (hx)a = hx, a fixed point, if and only if
h = a^h ■ xa = a~lhab, if and only if ah = ab. That holds if and only if
hb-1 € CH{a), or he CH(a)b, and so 6(a) = \CG(a)\.
Next take a G Ç,, k + 1 < j < d, and recall that Cf ¿ C,. Let us
show that a has no fixed points in Hx. If (hx)a = hx, then a I _ 1 = a*, a
contradiction since ah is in Cj but ax~ is not. Thus 9(a) = 0.
CHAPTER 3. COUNTING
64
WITH
GROUPS
We may now apply the Orbit Formula. Choose ai £ Ci, 1 < i < k. The
number of G-conjugacy classes in Hx is
\H\~l J > ( a ) : a£H} = \HI"1 ¿ |C*|JC^JET(o*)| = |#I"1 £ M = *■
i=l
Thus c(G) = k + (d- k)/p + (p-l)k
i=l
= pk + (d- k)/p.
A
Exercises
1. Suppose that G is metacyclic as in Proposition 2.2.1, with s = p, a
prime. Determine c(G).
2. Use (1) to determine c(Dm) and
c(Qm).
3. If p is an odd prime there are two nonabelian groups of order p3, viz.
(x, y, z I xp = yp = zp = 1, xy = yxz, i z = zx, yz = zy)
and
(x, y | xp = yp = 1, x - 1 j/x = î/ p + 1 ).
Show that both have class number p2 +p— 1.
The main result of this chapter, the Redfield-Pólya Theorem, is in essence
a refinement of the Orbit Formula that makes it possible to count orbits in a
wide variety of situations. One further bit of technical apparatus, the cycle
index, will be needed.
3.2
The Cycle Index
If G acts on S and | 5 | = n then each x € G corresponds to an element
a £ Sym(n), which is uniquely a product of disjoint cycles. Recall that the
cycle type (ji, J2, • • •,in) of a is simply a listing of numbers of cycles; a has
jk fc-cycles for each k. We say as well that x e G has that same cycle type.
Note that conjugate elements have the same cycle type. Define the monomial
of x to be
Mon(x) = ííV 2 2 ---í¿»,
where ¿i, £2, •••) tn are distinct (commuting) indeterminâtes. Then define
the cycle index of the action of G on 5 to be the polynomial (say over Q)
Z = ZG = ZG,s = |C?|_1 £ { M o n ( x ) : x € G}.
3.2. THE CYCLE
65
INDEX
Note that if G has conjugacy classes Ki, ..., Km, with representative elements Xi € Ki, then
m
m
»=i
i=l
Let us compute several examples of cycle indices by tabulating the relevant
information about conjugacy classes and cycle types.
1. Let G = S3 and S = {1,2,3}.
|cl(<r)|
type
1
(3,0,0)
(12)
3
(1,1,0)
(123)
2
(0,0,1)
Thus 2 = | ( í ? + 3<ií2 + 2í 3 )2. Let G = A4 and 5 = {1,2,3,4}.
|cl(a)|
type
1
(12)(34)
1
3
(4,0,0,0) (0,2,0,0)
(123)
4
(1,0,1,0)
(132)
4
(1,0,1,0)
Thus Z = j$(t\+ 3<2 + 8*1*3). Note that this is the same as the cycle
index of the rotation group T of a regular tetrahedron acting either on
the faces or the vertices of the tetrahedron.
Exercise
Calculate the cycle index for the action of T on the set of 6 edges
of the tetrahedron.
Let G = Ö, the rotation group of the cube. We shall calculate the cycle
indices for three different actions of O: (1) on the set F of 6 faces, (2)
on the set V of 8 vertices, and (3) on the set E of 12 edges of the cube.
Recall that O =■ 54, so we may list elements of S4 as representatives of
the conjugacy classes. Note that (1234) corresponds to a 90° rotation
and (13) (2 4) to its square, whereas (12) corresponds to the other kind
of 180° rotation, with axis joining the midpoints of opposite edges.
a
|C1(<7)|
F- type
V- type
E-type
1
1
(6,0,...)
(8,0,...)
(12,0,...)
(12)
6
(0,3,0,...)
(0,4,0,...)
(2,5,0,...)
(123)
8
(0,0,2,0,...)
(2,0,2,0,...)
(0,0,4,0,...)
CHAPTER 3. COUNTING WITH GROUPS
66
a
(1234)
|cl(<r)|
6
F- type (2,0,0,1,...)
V-type (0,0,0,2,...)
E- type (0,0,0,3,...)
(13)(24)
3
(2,2,0,...)
(0,4,0,...)
(0,6,0,...)
Thus we have
1. ZC¡F = ¿(ff + 6t| + 8*| + 6í2t4 + it\t\),
2. Z0,v = ¿ ( t ? + 9í£ + 8t?i?, + 6t\), and
3. 2 0 ) £ = ¿ ( í } 2 + 6í?í5 + 8*S + 6t\ + 3i§).
Exercise
Calculate the cycle index for the action of the rotation group I on
the set of 12 vertices of the icosahedron (equivalently on the set of
12 faces of the dodecahedron).
4. Let G = Cn = {a\&n = 1), acting as a plane rotation group on the set
S of n vertices of a regular n-gon. If we label the vertices consecutively
as 1, 2, . . . , n then we may take a to be the n-cycle (12 • • • n). There
is a unique (cyclic) subgroup of order d for each positive divisor d of n,
and a cyclic group of order d has (p(d) generators. Furthermore each
ak G Cn has order n/(n,k) and is a product of (n,k) disjoint n/(n,k)cycles. Thus if 1 < d < n and d | n there are ip(d) elements of order d,
each with cycle type (0,..., n/d,..., 0). As a result
2=-
^{y{d)tnd,d:
1 < d < n, d | n}.
For example, if n = 18 then Z = ±(t\s + 1 \ + 2t% + 2t% + 6$ + 6t 18 ).
5. Let G = Dn, dihedral of order 2n, acting on the set of vertices of a
regular n-gon. In the presentation
Dn = {o,T\an=T2
= {oT)2 = \)
we may assume that a is a generating rotation as in the example above,
so the cycle structure of each ah is already determined. Write A = (a).
Let T be a reflection whose mirror passes through one of the vertices
of the n-gon. Since a~krak = a~2kT we have cl(r) = {a2kr:k 6 Z}.
If n is odd then (a 2 ) = (a), so C1(T) is the entire coset AT in that
case. Each reflection fixes the vertex on its mirror and exchanges the
remaining vertices in pairs, so the cycle type is (1, (n - l)/2,0,...).
3.2. THE CYCLE
67
INDEX
If n is even there are two classes of reflections, viz. cl(r) and C1(OT),
each of size n / 2 . Reflections in cl(r) have mirrors through 2 opposite vertices and cycle type (2, (n - 2 ) / 2 , 0 , . . . ) . Those in cl(trr) have
mirrors passing through midpoints of opposite edges and cycle type
(0,71/2,0,...).
If n is odd then
Z = ¿ E M d ) ^ : 1 < d < n, d | n} + niit<B-1)/a];
if n is even then
Z
=¿ 5 > W
:
l<d<n,d\n} + ¡(iî4"-2)/2 + £")].
Exercise
If a € Sn has cycle type ( j i , . . . , jn) show that
|cl(a)|
n!
nr=i (*•*!)■
(This is perhaps best done by "overcounting" ; i.e. fill in the slots in
all n! ways and then determine how many times each element has
been counted and re-counted.)
6. Let G = Sn, the symmetric group. Use the abbreviation (j) for the
cycle type (ji,- ■ ■ ,jn), and write V for the set of all possible cycle
types (equivalently the set of all partitions of n). It is immediate from
the exercise above that the cycle index for 5„ is
As particular examples we have
Zs< = —(([+ U\t2 + Zt\ + 8ht3 + 6t4)
and
ZSt = Y2ö(f? + 10í?í2 + 15íif! + 20í 2 í 3 + 20í2Í3 + 30íií 4 + 24f 5 ).
68
CHAPTER 3. COUNTING WITH
3.3
GROUPS
Enumeration
In the discussion above of colorings of a tetrahedron we passed from a permutation action on the set of 4 faces to the set of 2 4 = 16 colorings of the
faces. If the faces are labeled 1, 2, 3, and 4 write D = {1,2,3,4} for the set
of faces. Let R = {r,b}, the set of colors. Then the set of colorings can be
identified with the set of all functions f:D —► R, since a coloring is simply an
assignment of a color to each face.
In general, suppose that D (the domain) and R (the range) are 2 finite sets,
and write RD for the set of all functions from D to R. Note that \RD\ = |fl| | D | .
If G acts on D, then G also acts on RD if we define " f{d) = f(d") for all
f £RD,aeG,
and de D.
The G-orbits in the set RD are commonly called patterns.
Suppose that 5 is a commutative ring with 1 containing the rational field
Q as a subring, with I s = 1Q (S will often be a polynomial ring, for example).
A function w: R -> S is called a weight assignment on R. For each r G R
the ring element w(r) is called the weight of r. A weight assignment w on R
induces a weight assignment W on RD via
W(f) =
Y[{w(f(d)):dtD}
for all / 6 RD.
Suppose that w is a weight assignment on R and that / i , / 2 € RD are in
the same pattern, say with / 2 = "fi, a eG. Then
W(f2)
=
nM/2(d)):de£>} = n{wr/1(d)):d€L»}
= n M / i ( d f f ) ) : d 6 ö } = lV(/1)
since a permutes the elements of D. Thus the induced weight assignment is
constant on the patterns in RD. If F is a pattern define its weight W(F) to
be W{f) for any / G F.
If U is any subset of R define the inventory of U to be
Inv([/) = £ { w ( r ) : r € ! / } ,
with the convention that lnv(0) = 0. Similarly if T Ç RD define its inventory
to be Inv(T) = J2{W{f): f € T } .
In the example above (coloring the tetrahedron) we had domain D =
{1,2,3,4} and range R = {r, b}. Take 5 = Q[<i, i 2 ], polynomials in 2 indeterminates, and define a weight assignment w: R -ï S via w(r) = t\, w(6) = i 2 .
As we saw, there are 5 patterns F\, ..., F&, represented by / i = rrrr G F\,
f2 = rrrb € F2, h = rrbb € F3, f4 = rbbb € F 4 , and fs = bbbb G F5. It
is easy to check that |Fi| = |F 5 | = 1, | F 2 | = |F 4 | = 4, and \F3\ = 6, and it
3.3.
69
ENUMERATION
is clear that W(Fi) = t\, W(F¡) = t\t2, W(F3) = t\t22, W(F4) = txt\, and
W(F5) = 4. Consequently
lnv(RD)
=
t\ + 4í?í 2 + 6ÍÍÍ2 + 4*i*2 + 4
=
(ti+t2)4
= \hiv(R)]M.
The example illustrates a special case of the next result.
Proposition 3.3.1 Suppose that R and D are finite sets and that
w.R-tS
is a weight assignment on R. Suppose that D is partitioned as a union of
disjoint subsets D\, ..., Dm, and denote by T the set of functions f € RD
that are constant on each of the subsets D{. Then
m
Inv(T) = n £ { w ( r ) | D l 1 : » - e #}■
¿=i
Proof Define rp: D -+ { 1 , 2 , . . . , m} by agreeing that ip(d) = j if and only
if d € Dj. For each / e T define <j> = <¡>f. { 1 , . . . ,m} -► R via <t>{j) = / ( d )
if and only if d € Dj. Clearly then each / € T factors as / = <f>fip. If
R = { r i , . . . , r*} then the right hand side of the equation to be proved is
m
n { w ( r 1 ) l D ' l + --- + w ( r * ) | D i | } .
¿=i
To multiply we choose one term from each factor in all possible ways, multiply
the terms together, and add the resulting products. To choose a term from
each factor is to choose a function <f>: { 1 , . . . , m} -> R, then multiplying yields
IliLi w (0(*)) | £ , i '! after which we add the results for all such <¡>. But
w((A(¿))|D<l = l[{w(<t>(i>(d))):d e A } = n W / ( d ) ) : d € A } ,
where / = <fnp € T. The choice of all possible <>
/ results in all / Ç. T, so the
effect of adding is
z n n w (/w)=E ww=inv(T)m
/eTi=ld€D¡
/€T
A
Corollary 3.3.2 lnv(RD)
Proof If D = {du...
= [Inv(fl)]lDl.
,dn} take D{ = {d¿}, 1 < i < n.
A
Suppose now that D and R are finite sets as usual, w: R -> S is a weight
assignment, and G is a finite group acting on D, hence also on RD. Recall
CHAPTER 3. COUNTING WITH GROUPS
70
that if / € RD then the weight of / is \\{w{f{d)):d € £>}, and if F is a
pattern (i.e. A G-orbit in RD) then its weight is W(F) = W(f) for any
/ G F. Define the pattern inventory of G to be
PI = J2{W(F): all patterns F Ç RD}.
Suppose, for example, that we define w(r) = 1 e Q for all r € R. Then
W{f) = W{F) = 1 for all / e RD and all patterns F, and so the pattern
inventory PI is just the number of patterns. Thus PI, if it can be computed,
contains at least as much information as is afforded by the Orbit Formula
(Proposition 3.1.2) applied to the action of G on RF.
Theorem 3.3.3 (Redfield-Pólya) Suppose that G is a finite group acting
on a finite set D, \D\ = k, with cycle index Z, and that R is a finite set with
Then the pattern inventory for the action of
a weight assignment w.R-¥S.
G on RD is
PI = Z ( $ > ( r ) , 5 > ( r ) 2 , . . . , £ w ( r ) * ) .
refl
refi
r€fi
Proof Let W\, W2, ... be the distinct weights of patterns in RD, and set
Ti = {f E RD : W(f)
= Wi},
i — 1,2,
Clearly Ti is a union of patterns; let m¿ denote the number
of distinct patterns in Ti. Thus PI = ¿» m tW»- Note that G acts on each
Ti, and that the G-orbits in Ti are just the m¿ patterns whose union is Ti.
If we write 0< for the character of the action of G on T% then, by the Orbit
Formula, m¿ = |G| _ 1 ^{0i{a):a e G}. Consequently
= ICI"1 ^ { p ¡ ( * : ^ G}.
PI = ^rmWi
t
t
Fix a e G. Then Y.iei(aWi is the inventory of {/ € RD:"f
0i(a) is just the number of / 6 T fixed by a, and so
= / } , since
PI = IGT1 ¿2 (iW(fy- f € RD and ' / = / } ) .
<r€G
If "f = f and d € D then
f(d) = °f(d) = f{d") = -fid") = f(d°3) = -..,
and / is constant on the cr-cycles in D. Conversely, if / 6 RD is constant on
the <r-cycles in D then "f = f (why?). Thus, if the <r-cycles partition D into
disjoint sets DUD2,...
then {/: "f = / } is just the set of all / G RD that
3.3.
71
ENUMERATION
are constant on each of the sets Z>¿, i.e. the set T of Proposition 3.3.1. By
that proposition
^{Wify.'f
= /} = n (£í w ( r ) | D i l : r e R})
i
for each fixed a € G. If a has cycle type {j\,. ..,jk) then among the sets £>¿
there are j \ of them for which \Di\ — 1, 22 f° r which \D¡\ = 2, etc. Thus
n(EWr)
| D i |
:r6fi} =
»
(5;w(r)f...(^w(r)f
r€fl
rgfi
for each a, and consequently
pi = ii?r l E(E w W)' ,(ff) (E w w 2 ) i2(a) ---(E w ( r ) fc r w
<r€G
=
rg«
rgfi
r€ft
Z(]Tw(r),£w(r)2,...,;[>(r)*).
r6fi
rgiî
r€fl
Corollary 3.3.4 The total number of different patterns in RD is
Z(\R\,\R\,...,\R\)P r o o f Set w(r) = 1, all r e R.
A
The theorem was first proved by Redfield [52] in 1927, but it promptly
fell into obscurity. It was rediscovered by Pólya ([51]) ten years later. N. de
Bruijn wrote an excellent survey article [19], still very much worth reading,
in 1964. The theorem has been used extensively for enumerating graphs with
various conditions imposed on them; see [35].
For an example take D = F, the set of 6 faces of a cube, and R =
{red,blue,yellow}. Let S = Q[r, b,y], where r, b, and y are distinct indeterminates, and assign weights by means of w(red) = r, w(blue) = b, and
w(yellow) = y. Recall from page 65 that the cycle index ZQ,F is
Z = —{t\ + 6tl + 8f| + 6ífí 4 +
3t\tl).
By the Redfield-Pólya Theorem
PI
=
¿ [ ( r + 6 + j/) 6 + 6(r 2 + 62 + j/ 2 ) 3 + 8 ( r 3 + 6 3 + j/ 3 ) 2
+6(r + b + y)2{rA + 64 + y4) + 3(r + 6 + y)2(r2 +b2 + y 2 ) 2 ]
=
r 6 + b6 + y6 + r5b + r5y + rb5 + ry5 + b5y + by5
CHAPTER 3. COUNTING
72
WITH
GROUPS
+2(r3b3 + r V + b3y3 + r V + r*y2 + b*y2
+b2y* + r2b4 + r2yA + r4by + rb4y + rby*)
+3(r 3 6 2 y + r3by2 + r2b3y + r2by3 + rb3y2 + rb2y3) + 6r2b2y2.
(Verify.) The total number of distinguishable colorings is obtained by setting
r = b = y = 1; there are 57 of them. Note also, though, that the monomial
summands of the polynomial PI give the numbers of patterns of various types.
For example the monomial r 6 represents the unique pattern with all faces red;
the monomial 3r3b2y represents 3 distinct patterns each having 3 red faces,
2 blue faces, and 1 yellow face. The 2 blue faces can be either adjacent or
opposite, and when they are adjacent the yellow face can be adjacent either
to both of them or to only one of them.
For a second example let us determine the number of distinguishable necklaces that can be made by using 9 red or blue beads. We assume that beads of
the same color are indistinguishable, and that 2 necklaces are indistinguishable if one can be made to look like the other by rotating it and/or turning
it over. In effect, then, we imagine each necklace laid out so the beads are at
the vertices of a regular 9-gon, which is acted on by the dihedral group Dg.
Take D to be the set of 9 beads, R = {red, blue}, 5 = Q[r, b] as above,
and w(red) = r, w(blue) = b. The cycle index for Dg is
2 = ¿ ( t ? + 2 t a + 6 t » + 9M2),
so the pattern inventory is
PI
6 ) 9 + 2 ( r 3 + 6 3 ) 3 + 6 ( r 9 + 69) + 9(r + 6 ) ( r 2 + 6 2 ) 4 ]
=
l[(r
=
r9+ r»b + 4r7b2 + 7r6b3 + 10r5ft4
lo
+
+10r 4 6 5 + 7r 3 6 6 + 4r 2 6 5 + rb8 + b9.
Set r = 6 = 1 to see that there are 46 distinguishable necklaces, and
observe, for example, that there are 7 patterns each using 3 red beads and 6
blue beads.
Exercises
1. Let p € N be a prime. Discuss the possible necklaces with p beads (a)
if red and blue beads are used, and (b) if red, blue, and green beads are
used. In particular, determine the number of distinguishable necklaces.
2. In how many distinguishable ways can the faces of a regular tetrahedron
be colored if there are n colors available?
3.4. GENERATING
3.4
73
FUNCTIONS
Generating Functions
A different sort of weight assignment is often useful in combinatorics. Suppose
that D, R, and G are as usual, let x be an indeterminate, and set S = Q[x].
Let k: R -¥ N be a function, and call k(r) the content of r e R. For / G RD
define the content of / to be K{f) = 52{k(f(d)):d G £>}. Assign weights on
R by means of w(r) = x* ( r ) , all r e fi. Note that then, for / € RD,
W(f) = n W ^ M e D} = n{xfc<'(d»:d6D} = ^-»(/eo) = x*<'>.
For each m € N write aTO = |{r € R: k(r) = m}\, and define the generating
function for R (and A;) to be
o(x) = ]>^{a m x m :m e N},
a polynomial whose coefficients count the numbers of elements in R having
various contents.
If / i and / 2 are in the same pattern F in RD it is easy to verify that
K(h) = K{f2), so we may define K(F) = K(f) for any / € F. If m € N set
An = |{F: F is a pattern and AT(F) = m } | ,
and define the pattern counting function to be
A{x) = Y^A™xm dZ[x\.
m
Theorem 3.4.1 Suppose that R and D are finite sets and G is a finite group
acting on D, with cycle index Z, generating function a(x), and pattern counting function A(x). Then
A(x)=Z(a(x),a(x2),...,a(xk)).
Proof Since PI = J^{W(F):aü
patterns F } , and W(F) = x * ( F ) , we have
PI = A0 + Aix + ■ ■ • = A(x).
But by the Redfield-Pólya Theorem
PI
=
2(Erw(r),Erw(r2),...)
=
Z(a0 + dix + a 2 x 2 -\
=
¿?(a(x),a(x 2 ),...,a(x*)).
, ao + a\x2 + a 2 x 4 H
, ...)
A
(m)
If \D\ = k define £> , for 0 < m < k, to be the collection of all m-sets
in D, i.e. subsets of size m. Thus |D (m >| = ( £ ) . If a group G acts on D
then it also acts naturally on each D^:
if a G G then {di,d2,... ,dm}a —
CHAPTER 3. COUNTING WITH
74
GROUPS
Theorem 3.4.2 Suppose that G acts on D with cycle index Z, and \D\ — k.
If x is an indeterminate and 0 < m < k then the number ofG-orbits in £)( m )
is the coefficient of xm in 2 ( 1 + x, 1 + x 2 , . . . , 1 + x fc ).
Proof Take R = {0,1}, define a content k:R->Z
via fc(0) = 0, fc(l) = 1,
and hence assign weights via w(0) = 1, w(l) = x. If / € RD then, as
observed,
W{f) = *KU) = * E * /(d) .
Thus W(f) = xm if and only if precisely m elements of D are mapped by /
to 1, i.e. if and only if / - 1 ( 1 ) € D^m\ There is thus a 1-1 correspondence
between G-orbits of m-sets in D<m) and patterns of content m in RD. But the
number of patterns of content m is the coefficient Am in the pattern counting
function A{x). Since the generating function is a(x) = 1 + x the result follows
from Theorem 3.4.1.
A
Corollary 3.4.3 The number of G-orbits in D is the coefficient of x in
2 ( l + x , l + x 2 , . . . , l + x*).
Proof Identify D with D^\
A
Corollary 3.4.4 The total number of orbits of G acting simultaneously on
all £)("»), for 0 < m < k, is
Ao + A1 + --- + Ak = Z(l + l,l + l2,...)
= Z{2,2,...,
2).
For an example take G = O, acting on the set V of 8 vertices of a cube.
Recall that the cycle index is Z = ¿ ( i ? + 94 + 8í?í| + 6tf). Thus
A(x)
=
Z ( l + x , l + x 2 , . . . , l + x8)
=
¿ 1 ( 1 + * ) 8 + 9(1 + x 2 ) 4 + 8(1 + x ) 2 ( l + x 3 ) 2 + 6(1 + x 4 ) 2 ]
=
l + x + 3x 2 + 3x 3 + 7x4 + 3x 5 + 3x 6 + x 7 + x 8 .
(Verify.) Thus, for example, there are 3 orbits of 2-sets: one orbit, of size
12, consists of pairs of vertices that share an edge; another, also of size 12,
consists of pairs of vertices at opposite corners of faces; and the third orbit,
of size 4, consists of pairs of vertices at opposite corners of the cube.
Exercises
1. Describe the 3 orbits of 3-sets and the 7 orbits of 4-sets in the example
above.
2. If O acts on the set E of edges of a cube compute the numbers of orbits
of m-sets, 0 < m < 12, and describe the orbits of 2-sets explicitly.
3.5. THE PETERSEN
75
GRAPH
3. Suppose that a hollow regular tetrahedron has thin walls, and both the
inside and outside of each face is to be colored red or blue. How many
distinguishable colorings are there, and how many of them have 3 or
fewer of the faces blue?
3.5
The Petersen Graph
This final section can be viewed as an extended list of exercises.
We begin by constructing a well-known graph T, the Petersen graph. For
the set V of vertices of T take the set of 10 transpositions in the symmetric
group 55; we agree that two vertices are joined by an edge in T if and only if
they are disjoint as permutations. See the figure below.
An automorphism of a graph is any permutation a of its vertex set V such
that u and v are joined by an edge in T if and only if u" and v" are joined
by an edge. Denote by G the group Aut(r) of all graph automorphisms of T.
(12)
1
The Petersen Graph
1. Show that 5 5 acts faithfully on V by conjugation, and that each a e 5 5 is
a graph automorphism. Conclude that S5 can be viewed as a subgroup
ofG.
2. Show that G is transitive on V.
3. If H denotes the stabilizer in G of the vertex (12) show that \H\ = 12.
What is HI [Hint H includes automorphisms that restrict to give all
possible permutations of the three vertices (34), (35), and (45) that
are adjacent to (12), and there is a unique (nonidentity) a € H that
fixes all of (34), (35), and (45).]
4. Conclude that \G\ = 120, and hence that G = 5 5 .
CHAPTER 3. COUNTING WITH GROUPS
76
5. Relabel the vertices as 1 through 10 (see the figure again), so that
G = Ss is now represented as a subgroup of Sio- Show that
a = (l,2,3,4,5)(6,7,8,9,10)andi> = (3,7)(8,9)(4,10)
generate G6. Calculate the cycle index Z of G < Sio- (Suggestion. You may wish to
use the GAP command ConjugacyClaßses(G) to determine representatives and sizes of the conjugacy classes, cycle structure, etc.)
7. We now wish to color the vertices of T each red or blue, and two colorings
will not be distinguished if there is a graph automorphism carrying one
to the other. How many distinguishable colorings are there, and how
many of them have half the vertices red, half blue? (Suggestion. You
may wish to use a computer algebra package, such as Maple [12], to
substitute the powers r* + 6* into the cycle index.)
8. Try to determine the distinguishable colorings having 3 red and 7 blue
vertices.
9. Answer questions similar to the previous two questions if three (or more)
colors are available.
Groups and Characters
by Larry C. Grove
Copyright © 1997 John Wiley & Sons, Inc.
Chapter 4
Transfer and Splitting
This is the most serious "purely group-theoretical" chapter in the book. We
develop some of the techniques that are central to the further study of the
structure of finite groups. Substantial applications of some of the results will
appear in Chapter 9, where we discuss the structure of Frobenius groups.
4.1
Transfer and Normal Complements
Suppose that G is a group and that H < G has finite index n. Let T =
{ti,...,tn}
Ç G be a (right) transversal for H, and write x >-ï x for the
transversal function, i.e. HxdT = {x} (see Chapter 1).
If A is an abelian group and 8: H -► A is a homomorphism define a
function V: G -> A via V{x) = H"=1 9{UxUx _ 1 ) . Note that V is well defined
since A is abelian; it is called the transfer (or Verlagerung) from G to A via
0. The transfer was first introduced by I. Schur in 1902.
Proposition 4.1.1 The transfer V:G -* A is a homomorphism.
Proof If i , y € G then
V(xy)
= Y[e(tixyt¡x^-1)
i
=
Y[9(tixüí-1t¡EyWy~1)
=
i
ï[e(tixûï-l)-Y[6(uïyïïïy~-l)
i
= V(x)V(y)
i
(we have used Proposition 1.1.1 and the fact that {t\x,...
transversal).
,tnx}
is also a
A
Proposition 4.1.2 The transfer V:G -* A is independent of the choice of
transversal.
77
CHAPTER 4. TRANSFER
78
AND
SPLITTING
Proof Write V = VT for the transfer as defined above. Let S = {s\,...,
sn)
be another tranversal, denote its transversal function by x *-* x, and write
Vs for the transfer defined in terms of S. We may label the elements of
S so that HSÍ = Hti for all i, so s¿ = U and t¡ = s¿. Write s¡ = hit¡,
with hi € H, all i. Now fix x e G. Then {s¡x : 1 < i < n} - S, so
there is a permutation a G Sym(n) such that s¡x — s^, all z. Note that
HtiX = HsiX — Hs¡x = H Si" = Ht^, so íjX = íj» as well. Now we
calculate:
Vs(x) = n e<<sixii~x~i)=n ^(^íi^v 1 )=n ^xuxt-^h-})
i
i
i
= Y[o(hi) ■ IpCv 1 ) -Yieitixûï-1) = i • vT(x).
i
i
i
A
Exercise
Take G = Sym(4) and let H = ((1234), (13)>, a dihedral subgroup. Take
K = ((13)(24)) < H, set A = H/K, and let 6 : H -> A be the quotient
map.
Determine ker V in this case.
Let us construct a very particular transversal for H in G. Fix x 6 (? and
consider the action of x as a permutation on the set G/H of (right) cosets
of H in G. Write the permutation as a product of disjoint cycles (including
1-cycles) as follows:
x !-+ (Hyi,Hyix,...,
ifj/ix* 1 - 1 ) • • • (Hym, Hymx,...,
Hy^*™-1).
Then T = {tij = j/<x^: I < i < m, 0 < j < kt — 1} is a. transversal for H
in G. Furthermore the sum of the lengths of the cycles is n = \G/H\, i.e.
J2T=i ki = nIn view of Proposition 4.1.2 we may use the transversal T to evaluate the
transfer V at x. We first investigate the effect of the transversal function.
If j < k{ - 1 then UjX = j/jX-'4"1 = í¿(¿+i) € T, so UjX = tijX and
tjjxtjjX ~l = 1. However, if j = fc¿ - 1, then i / i ^ x = HyiXki = Hyi, so
i y z = y« and UjXtiji - 1 = j/jX^y," 1 . Note also that yiXkiy^1 € í í for each i.
We have proved the next proposition, which will be used later in an application of the transfer.
Proposition 4.1.3 Fix x 6 G. Then there are elements ylt...
positive integers fci,..., km such that
1- Z?=1ki = [G:H),
2- yiXkiy~l
G H, alii, and
3- V{x) = X\™=ie{yixk<y-1).
,ym € G and
4.1. TRANSFER
AND NORMAL
COMPLEMENTS
79
Exercise
If H < Z{G) and n = [G: H] show that V(x) = 6{xn), all x € G.
Recall that the derived subgroup G' of G is the smallest normal subgroup of G for which the quotient group is abelian. Thus in particular
G' = D{K: K < G and G/K is abelian}.
Let us "localize" the above observation. For any prime p define
G'p = n{K: K <G and G/K is an abelian p-group} < G.
Note that each K in the definition contains G'. Thus G'p > G', and G/G'p is
abelian. The subgroup G'p is called the p- commutator subgroup of G.
Proposition 4.1.4 For each prime p the p-commutator subgroup G'p is normal in G, and G/G'p is an abelian p-group.
Proof The only part of the proposition requiring proof is that G/G'p is
a p-group. To establish that it will suffice to show that if H, K < G, with
G/H and G/K both p-groups, then G/(H fl K) is also a p-group. Define
e-.G/H x G/K via 6(x) = (Hx,Kx), and note that ker 6 = H n K. Thus
G/(H D K) is isomorphic with a subgroup of the p-group G/H x G/K.
A
Exercise
Calculate G'p for all p if (a) G = Alt(4), (b) G = Sym(n), and (c) G is
the split metacyclic group {a,b\a10 = 66 = 1, b-1ab = a - 1 ) .
Proposition 4.1.5 If G is a finite group then G' = <~)pG'p and G/G'
UPG/G'P.
^
Proof Since G/G' is abelian it is the product of its distinct primary subgroups, say G/G' = Hi/G' x • ■ • x Hk/G', with Hi/G' pj-primary. Each Hi
is normal in G. Fix i and define Ki via Ki/G' = f ] ^ Hj/G', so K¡ <G and
G/G' = Hi/G' x Ki/G'. Then G/Ki S G/G'/Ki/G'
S Hi/G', the abelian
Pi-group of maximal order in G/G'. Since all abelian images of G are images
of G/G' it follows that K{ = G'p.. Now Hi/G' S G/G'p., so G/G' S ft G/Gp,
and also n^Ki/G') = G' = 1G'/G> = (npG'p)/G', i.e. ñpG'p = G'.
A
Proposition 4.1.6 If G is a finite group and P € SyL(G) then PG'p = G
andG/G'p~P/(PnG'p).
Proof The index [G: G'p] is a power of p and a divisor of |G|, so if q ^ p is a
prime and qe \ \G\ then qe \ \G'p\. Thus \PG'p\ \ \G\ and G = PG'p. Furthermore
G/G'p = PG'p/G'p S P/(P n G'p) by the standard isomorphism theorem. A
CHAPTER
80
4. TRANSFER
AND
SPLITTING
Recall that the commutator of two group elements x and y is [x, y] =
x~xy-lxy.
If H, K < G then [H,K] = {[x,y]:x € H, y 6 K); in particular
G' = [G,G}.
If G is a group and H <G define the focal subgroup of H in G to be
FocG(H) = {[y,x]:yeH,
x e G, and
[y,x]eH).
Clearly if' < F o c G ( # ) < H n G', and in particular H/FocG(H)
(why is Foc G (#) « # ? ) •
is abelian
Theorem 4.1.7 Suppose that P e Syl p (G). Then
FocG{P) = PnG'p
and P/FocG{P)
~ G/G'p.
Proof It is clear from the definitions that Foc G (P) < P n G' < P n G p .
For the reverse inclusion set A = P/FocG(P), let 0 : P -+ A be the quotient
map, and apply the transfer V from G to P via 0. If x € P then, by
for appropriate j/¿ G G,
Proposition 4.1.3, V(x) = Tl'¡^iFocG(P)yiXkiy¡~1
e P , and so
ki 6 N. But [xfci,j/i] € Foc G (P) since xfci € P , yiXkiyrl
í/ti*'!/, 7 " 11- *' = [ i * S y t ] - 1 € P- As a result FocoCPÍJ/ia;*'!/,-1 = Foco(P)x fci ,
and consequently
V(x)
Y[FocG(P)xki
=
i
FOCGÍP)!^*'=FocG(P)x[G:P].
=
Since x € P and p\ [G:P], x and xl G:P ) have the same order, so in fact
FOCG(P)X is also in the image of V. Thus V maps P (and o fortiori G) onto
P/FOCG(P).
We now have G/ ker V 5í P / F O C Q ( P ) , an abelian p-group, and so ker V >
Gp. Apply Proposition 4.1.6:
[G:kerV]
=
=
[P:Foc G (P)] > [ P : P n G ; ]
[PG;:G p ] = [ G : G ; ] > [ G : k e r n
and all are equal. The last inequality holds since G/ ker V is an abelian
A
p-group, hence ker V > G'p. For the isomorphism see Proposition 4.1.6.
Proposition 4.1.8 Suppose that P € Syl p (G) and set N = NG(P).
abelian then F O C G ( P ) = Focjv(P).
If P is
Proof Inclusion right-to-left is clear. Let [y,x] be a generator for F O C G ( P ) ,
i.e. y € P , x e G, and [y,x] = y~lyx G P . Thus also z = yx € P , and z € Px,
so P and Px are both Sylow p-subgroups of CG(z) ( P is abelian). By the
Sylow Theorem there is some w € CG(z) with P = Pxw, so xw € N. But
= y^z = y~lyx = [y,x], and Foc G (P) <
now [y,xw] = y~xyxw = y-xzw
Foc;v(P).
A
4.1. TRANSFER
AND NORMAL
81
COMPLEMENTS
Now an important definition. If G is a group and H <G, then a normal
complement for H in G is a normal subgroup KofG satisfying G = KH and
K n H = 1, i.e. G = Kx H, a split extension. If G is finite then a normal
complement for a Sylow p-subgroup P is called a normal p- complement.
For an easy class of examples suppose that G is metacyclic, as in Proposition 2.2.3, with s = p e , p a prime, and p^ m. Then A is a normal p-complement.
Exercise
If a group G has a normal p-complement K show that K is unique, and
hence Ä"char G. Show that K = {x € C7:p| | i | } .
Theorem 4.1.9 (Burnside) If P e Sylp(G) and P < Z(NG(P))
has a normal p-complement K, which is characteristic in G.
then G
Proof Set N = NG(P). Note that P is abelian, so Foc G (F) = FOCAT(P)
by Proposition 4.1.8. Also FOCN(P) = ([y,x]:y G P,x 6 N) = 1 since
P < Z{N). Thus P D G'p = 1 and G = PG'p by Proposition 4.1.6 and
Theorem 4.1.7, so K = G'p is a normal p-complement. For Jfchar G see the
exercise preceding the theorem.
A
Note that the hypothesis P < Z(NQ(P))
in Burnside's Theorem is equivalent with NG(P) = CG(P)I the theorem is sometimes stated accordingly.
There are numerous applications of Burnside's Theorem; we shall see just
a few of them.
Proposition 4.1.10 If H < G then CG(H)<NG(H)
isomorphic with a subgroup of Aut(H).
and NG(H)/CG(H)
is
Proof Define a homomorphism 6 from NG{H) to Aut H via x >-¥ 0X, where
ex(h) = xh,he
H, and check that ker 6 = CG{H).
A
Theorem 4.1.11 Suppose that p is the smallest prime divisor of \G\, and
that P e Syl p (G) is cyclic. Then G has a normal p-complement, and in
particular G is not simple.
1) since P is cyclic, and
Proof If \P] = p* then | Aut P\ = <p(pk) = pk~l{pP < CG(P), so p\ \NG(P)/CG(P)\,
and [NG(P):CG(P)} divides p - 1 by
Proposition 4.1.10. But also [NG(P): CG(P)] | \G\, whose smallest prime divisor is p, so [NG(P):CG{P)] = 1, NG{P) = CG{P), and Burnside's Theorem
(4.1.9) applies.
A
Corollary 4.1.12 If G is simple, p is the smallest prime divisor of \G\, and
P € Sylp(G), then P is not cyclic.
CHAPTER 4. TRANSFER
82
AND
SPLITTING
Note that the corollary rules out many possible orders for simple groups,
viz. all pm, where pj^ m and all prime divisors of m are larger than p. The
first candidate for a Sylow 2-subgroup of a simple group is a Klein 4-group it occurs in Alt(5).
If n G Z and p is a prime let us write pk |) n to indicate that pk exactly
divides n, i.e. p* | n but pk+1 j n.
Proposition 4.1.13 Suppose that G is simple, p is a prime, and p || |G|. If
P e Syl p (G) then 1 # [NG(P): CG(P)] | p - 1.
Proof Since | Aut P\ = p - 1 the divisibility follows from Proposition 4.1.10.
If Na{P) = CG(P) then G would not be simple, by Burnside's Theorem. A
As a sample application suppose that G is simple and that \G\ = 1900 =
2 2 • 5 2 • 19. Then | Syl 19 (G)| = 20. If P € Syl 19 (G) then [G: NG(P)] = 20, so
|ATG(P)| = 5 - 1 9 . By the proposition [NG(P):CG{P)] ¿ 1, so \CG(P)\ = 19.
But then the proposition also tells us that 5 19 — 1 = 18, a contradiction.
Exercise
If p and q are primes, say with p < q, and if \G\ = p2q2, show that G is
not simple.
One more (direct) application of Burnside's Theorem.
Theorem 4.1.14 Suppose that G is a nonabelian simple group and p is the
smallest prime divisor of \G\. If p is odd then p 3 \G\; if p — 2 then either
8 | | G | or 12 | \G\.
Proof Suppose that p 3 j \G\, and choose P € Syl p (G); note that P is abelian.
In fact, by Corollary 4.1.12 we see that P S Z p © Z p , so Aut P 5! GL(2,p)
and | Aut P\ = (p - l) 2 p(p + 1) by Proposition 2.6.1. Write N = NG{P) and
C = CG(P). By Burnside's Theorem (4.1.9) [N:C] ¿ 1, and by Proposition
4.1.10 [N:C] | | Aut P\. But also [N:C] | \G\, and p is the smallest prime
divisor of \G\, so [N:C] is prime to p - 1. Furthermore P < C, so [iV:C]
is also prime to p, hence [TV: C] | p + 1. But then [N: C] must eçua/ p + 1,
and p + 1 must be prime, for otherwise [N: C] and p + 1 would share a prime
factor smaller than p. That is possible only if p = 2 and p + 1 = 3; thus 3
and 4 are both divisors of \G\.
A
Burnside conjectured ([10], page 503) that simple groups must have even
order. That was proved in 1963 in the monumental paper [24] of Feit and
Thompson.
Burnside also noticed that all simple groups known at the time his book
was written had orders divisible by 12 ([10], page 330), but he seems not to
4.1. TRANSFER
AND NORMAL
COMPLEMENTS
83
have made a corresponding conjecture. The first (nonabelian) simple groups
with orders prime to 3 (hence divisible by 8) were discovered by Suzuki [61]
in 1960.
To conclude this section let us aim at a converse for Proposition 2.2.2, as
was promised in Chapter 2.
Proposition 4.1.15 If all Sylow subgroups of G are cyclic then G is solvable.
Proof Induction on |G|; the result clearly holds for |G| sufficiently small. If
p is the smallest prime divisor of \G\ and P G Sylp(C?) then G has a normal pcomplement K by Theorem 4.1.11. By induction K is solvable, and G/K ^ P
is solvable, so G is solvable.
A
Proposition 4.1.16 Suppose that G is a finite group and that for some k >
2 the terms G^*-1) > G^ > G(fc+1) in the derived series have quotients
+1
+1
G (fc-i)/ G (*) a n áG(*)/G(* > that are both cyclic. Then G<* > = G ^ .
H™ =
Proof Set H = G<fc-2>/G(*+1>, so that H' = G^-^/G^+l\
ç* £(*-!)/<;(*), so W/HM and
G{k)/G(k+i)^
&nd H(3) = j
A l s o H'/H(2)
# ( 2 ) are both cyclic; say H^
= (x). Map H into Aut H^ by restricting
inner automorphisms; the kernel of the map is CH(H^)
= C//(x). Since
is abelian
H^ is cyclic Aut H^ is abelian (Theorem 2.1.3), so H/CH(X)
and hence H' < CH{x) = CH(H^),
i.e. ¿7<2> < Z(H'). But then H'/Z(H')
is cyclic, being a homomorphic image of H'/H&K Thus H' is abelian, so
i/( 2 ) = l , i . e . G ( ' + 1 ' = G W .
A
The parameters m, s, r, and t in the next theorem are as in Proposition
2.2.1.
Theorem 4.1.17 Suppose that all Sylow subgroups of G are cyclic. Then
G is split metacyclic, with A = G' of order m, s = [G:G'], t = 0, and
(m,s(r — 1)) = 1.
Proof If G is abelian it is cyclic and the conclusion holds trivially, so we may
assume it is not abelian. Note that all subgroups and homomorphic images of
G have all Sylow subgroups cyclic. In particular G'/G^ and G*2) /G^3\ being
abelian, are cyclic. Thus G (3) = G (2) by Proposition 4.1.16, and G (2) = 1
by Proposition 4.1.15. Consequently G is metacyclic, with A — G' = (a),
say, of order m, and G/A = (Ab) of order s, with b~lab = ar and bs = a1
as in Proposition 2.2.1. Recall (Exercise 4, page 30) that G' = ( a r _ 1 ) , so
(m,r - 1) = 1. Also m\t(r - 1) and 0 < i < m , s o i = 0 and G is split. If
p is prime and p \ (m,s), then the subgroup (a m / p ,6*/ p ) has order p 2 and is
not cyclic, a contradiction, and the proof is complete.
A
Corollary 4.1.18 If G is a group having square-free order then G is split
metacyclic with (m,s(r — 1)) = 1.
CHAPTER 4. TRANSFER
84
AND
SPLITTING
Exercise
Let G = (a,6|a 14 = 63 = 1, b~lab = a9), split metacyclic of order 42.
Since (m, s) = (14,3) = 1 all Sylow subgroups are cyclic by Proposition
2.2.2. Conclude from Theorem 4.1.17 that ( m , r - l ) = 1. But ( m , r - l ) =
(14,8) = 2, so 2 = 1??? What went wrong?
4.2
Hall Subgroups
A subgroup H of a finite group G is called a Hall subgroup if \H\ and [G: H]
are relatively prime. For example, if G has all Sylow subgroups cyclic then
its derived group G' is a Hall subgroup by Theorem 4.1.17.
For any finite group G write ir(G) to denote the set of prime divisors of
\G\.
If n is any set of prime numbers write n' for the complementary set of
primes. An integer n is called a n-number if all its prime divisors are in n.
Say that a group G is a it-group if \G\ is a 7r-number. If H < G then H
is a Hall 7r-subgroup if H is both a Hall subgroup and is a 7r-group, and in
addition [G:H] is a 7r'-number. Write Hall»(G) for the set (possibly empty)
of all Hall 7r-subgroups of G.
Subgroups H and K of a group G are called complements of each other if
HK = G and H D K = 1 (no assumptions about normality).
Exercises
1. If H and K are complements in G and if is a Hall subgroup show that
K is a Hall 7r(ii)'-subgroup. Conversely, if H and K are Hall subgroups
with n(H) D ir(K) = 0 and ir(H) U n(K) = n(G), show that they are
complements to one another.
2. Suppose that H is a normal Hall subgroup of G and \H\ = n. Show
that H = {xeG:xn
= 1}.
Proposition 4.2.1 (The Frattini Argument) Suppose thatH<G andP €
Syl p (tf)- ThenG = NG(P)H.
Proof If x € G then Px < Hx = H, so Px G Syïp(H).
Take h G H with
Px = Ph, so Pxh~' = P, xh~l G NG(P),
Ç Na(P)H.
and * G NG(P)h
A
Theorem 4.2.2 (I. Schur) A normal abelian Hall subgroup A of G has a
complement.
4.2. HALL
85
SUBGROUPS
Proof Write G/A = {a, 0 , 7 , . . . } and choose a transversal {tQ:a € G/A}
for A in G, with, of course, ta 6 a. Then Atatß = aß = Ataß, and so
tatff = f{a,ß)taß for some f{a,ß) € A, all a, ß. It follows (verify) from the
associative law that
f{a,ß)f{aßn)
t
=
°f{ß,1)f{a,ß1),
all a, ß, 7 (the function / : G/A x G/A -> .4 is called a factor set, or a factor
system). Set n = |^4| and m = [G: A], so (n,m) = 1. For each a € G/A set
g(a) = Y\{f(a,ö)'-S e G/A}, and observe that
]\t°nß,l)f{c*,ßl)l
]}[/(<*,/3)/(oA 7)] =
7
7
[n/(Q^)][n^a^^]=tt,[n^^)][n/(a'^)]>
7
7
and so
7
7
/(a,/3) m 5 (a/3) = t » 5 ( ^ ( Q )
for all ß, 7 e G / A Choose fc € Z so that fcm = \{mod n), set /i(<5) = g{S)~k,
all ¿ 6 G/A, and take — fcth powers to see that
f(a,ß)-1h(aß)
= t°h(ß)h(a)
(*)
for all a, ß 6 G / A
Now set s a = h(a)ta, all a e G/A, to obtain a new transversal for A in
G, and set B = {s a : a Ç. G/A}. If ß is a subgroup of G then it is clearly a
complement for A.
If sa, Sß € B then
sas0
=
=
h(a)tah(ß)t0
= h(a) ■
f{a,ß)~'ih{aß)tat0
=
by equation (*) above, and the proof is complete.
ta
h(ß)tat0
h{aß)ta3=saß
A
Theorem 4.2.3 (Schur-Zassenhaus) A normal Hall subgroup H of G has
a complement.
Proof We use induction on |G|, the result being trivially true if G is small
enough.
Suppose first that there is a prime p dividing \H\ and a Sylow p-subgroup
P of H that is not normal in G. Then G ¿ N = NG(P)By the Frattini
Argument (Proposition 4.2.1) G = NH, and G/H = {NH)/H S
N/{NnH),
so [N: NDH} = [G: H). Thus N D H is a normal Hall subgroup of N, and
by induction there is a complement K to N n H in N. But then \K\ =
[N: Nf)H] = [G: H], so K is also a complement to H in G.
CHAPTER 4. TRANSFER AND SPLITTING
86
Thus we may suppose that all Sylow subgroups of H are normal in G. In
particular H is nilpotent, and Z = Z(H) ^ 1. Since Zchar H <G we have
Z<G. Now H/Z is a normal Hall subgroup of G/Z, and by induction there is
a complement K/Z to H/Z in G/Z. Note that \K/Z\ = [G/Z: H/Z] = [G: H).
UK^G
then Z is a Hall subgroup of K (since Z < H and [K: Z] = [G: H]),
and by induction there is a complement L to Z in ÜT. Then \L\ = [K: Z] =
[G: H], so L is also a complement to H in G.
Finally, suppose that K = G. Then [G: i/] = [K: Z] = [G: Z], so H = Z
is abelian, and H has a complement by Theorem 4.2.2.
A
Recall that a finite abelian group A is called elementary abelian if all
nonidentity elements have order p for some prime p. Thus A S ©¿LjZp for
some k, and A (written additively) is a Z p -vector space.
Proposition 4.2.4 Suppose that G is a finite solvable group and that N ^ 1
is a minimal normal subgroup ofG. Then N is an elementary abelian p-group
for some prime p.
Proof Suppose that K ^ 1 is a minimal normal subgroup of N. Then
( U {Kx : x € G } ) is contained in N and is normal in G, so it is equal to
N. Each Kx is minimal normal in N and two of them, Kx and Kv, are either
equal or intersect only in 1. Thus for some subcollection {KXl,...,Kx'}
we
have N = AT*1 x • • • x Kx>. lîL<KXi
then L<N. Thus L = 1 or KXi, since
KXi is minimal normal in N; i.e. KX< is simple. Since it is also solvable we
have KXi 5í Z p for some prime p.
A
Theorem 4.2.5 (H. Zassenhaus) Suppose that H is a normal Hall subgroup of G, and suppose that either H or G/H is solvable. Then any two
complements to H in G are conjugate.
Proof Let K\ and K2 be complements for H; note that they are both
transversals for H in G. We consider three cases.
Case 1. Assume that H is abelian. For each a € K-¿ there are a unique
x Ç. K\ and a(x) € H so that a = a{x)x, determining a function a: Ki -+ H.
If also 6 e K2, with b = a(y)y, then Hab = HaHb = HxHy = Hxy, so
ab = a(xy)xy. On the other hand ab = a(x)xa(y)y = a(x) • xa(y)xy, so
a(xy) = a(x) ■ xa(y)
for all x,y € K\ (the function a is called a cocycle).
Say that \Ki\ = [G:H] = n. Set u = U{a(y)-y
equation (*) we have
u = l[a(xy)
V
(*)
e Kx) 6 H. By the
= JJ[o(«) • x «(y)] = <*(x)n ■ xu,
V
all a; € K\. Since H is abelian and {\H\,n) = 1 the map h i-+ hn is an
automorphism of # . Choose v e H so that t>" = u; then u = a(x) • xv for
4.2. HALL
SUBGROUPS
87
all x € K\. But now "x = vxv~lx xx = v(xv) lx = a(x)x = a € K2, all
a; e Ku and "tfi =/iT 2 .
Case 2. Assume that i / is solvable and use induction on \G\. Suppose
first that there exists N<G with 1 / N < H. Then KiN/N and K2N/N are
conjugate in G/N by induction, since they are complements to H/N. Thus
3a: 6 G such that XK\ < K2N. But then XKX and K2 are complements of N
in Ä"2iV < G, and N is solvable. By induction 3j/ € /f2./V with yxKi = K"2,
as desired.
Suppose then that no such N exists. Then H is solvable and has no
nontrivial characteristic subgroups; in particular H' = 1, so H is abelian and
we are back in Case 1.
Case 3. Assume that G/H is solvable.
Some preparation will be useful. If a; € G then x = hk\, uniquely, with
h e H, hi € K\, and the map W\:x *-* ki is a homomorphism from G onto
Ki with kernel H. It induces an isomorphism 0\\G/H -+ K\. Similarly there
is an isomorphism 0 2 : G/H -> K2, and so 6 — 920^1 is an isomorphism from
K\ to K2. Note that 0 provides a 1-1 correspondence between subgroups Lx
of K\ and L2 of K2 as follows: 0: L\ ■H L2 = L\H D /¡T2, and L\ < Ä"i if and
only if L2<K2.
Now the proof. Once again use induction on \G\. Since K\ a G/H it
is solvable, and a minimal normal subgroup of K\ is an elementary abelian
p-group Pi for some prime p, by Proposition 4.2.4. Set P2 = P\HC\K2 Si Pi,
minimal normal in K2 by the discussion above. Note that both Pi and P 2 are
in Sylp(Pitf), so 3xe PxH with Pf = Pi. Set N = NG(Pi). Since Pi <ATi
and Pi =P2X<KX we have Kx, Kx <N. Thus N = G D N = (HKi) n N =
(H n AOtfi (since Ki < N), and similarly N = {H n JV)fff, i.e. Ki and /¡Tf
are complements to HC\N in AT. Now Ki/Pi and K2/P\ are complements to
{HnN)P1/Pl
in ./V/Pi, and by induction 3y e N so that ^ / P i =
K^/Pi,
which says that tfj" < KiPi = K i , and finally Kxy = Kx.
A
It must be noted that at least one of \H\ and \G/H\ is an odd number in
Theorem 4.2.5, since if is a Hall subgroup, and consequently at least one of
H and G/H must be solvable by the Feit-Thompson Theorem mentioned on
page 82. Thus the conclusion to the theorem holds under all circumstances.
It should perhaps be mentioned that the proofs of Theorems 4.2.2 and
4.2.5, in utilizing factor sets and cocycles, represented brief excursions into
the subject of cohomology of groups, a subject of considerable importance in
its own right, especially for the general study of group extensions. See e.g.
[33] or [37] for an introductory account.
As an application of the Schur-Zassenhaus Theorem (4.2.3) we prove next
a remarkable generalization, for solvable groups, of the Sylow Theorems, due
to Philip Hall.
Theorem 4.2.6 (P. Hall, 1928) Suppose that G is a finite solvable group
and that IT is a set of primes. Then G has a Hall v-subgroup, and any two
88
CHAPTER 4. TRANSFER
AND
SPLITTING
of them are conjugate. Furthermore every it-subgroup of G is contained in a
Hall ^-subgroup.
Proof If K is a 7r-subgroup of G let us use induction on \G\ to show that
K < H for some H G Hall*(G). Take N minimal normal in G; N is an
elementary abelian p-group for some prime p by Proposition 4.2.4. Then
KN/N is a 7T-subgroup of G/N; by induction there is a Hall 7r-subgroup
M/N of G/N with K < M. Now [M:N] is a 7r-number and [G: M] is a
ir'-number, so if p G IT then M 6 Hall*(G), and the proof is complete with
H = M. Suppose then that p G n', so N is a normal Hall subgroup of M ,
and N has a complement L in M by Theorem 4.2.3 (or 4.2.2 for that matter).
Now \L\ = [M:N] is a 7r-number, and [G:L] = [G:M][M:L] = [G:M]\N\ is
a 7r'-number, so L 6 Hall* (G). Note that K and N are complements in KN,
since p i 7T. Also \Lf\KN\ = \L\\KN\/\LKN\
= |L||A:||iV|/|M| = \K\, so
L n KN is another complement to N in KN, which is solvable. Hence, by
Theorem 4.2.5, 3x G KN with K = (Lf) KN)X. Thus K < Lx e Hall^G).
The conjugacy conclusion remains to be proved. Let us again use induction on \G\. Suppose that H, K € Hall„(G), and choose N minimal normal
in G, a p-group as above. Then HN/N and KN/N are in Hall* (G/N) so
they are conjugate by induction and 3a; € G with Hx < KN. If p G 7r then
KN is a 7r-group, so KN = K (since K G Hall^G)), and Hx = K in that
case. Suppose, finally, that p £ n. Then N is a normal Hall subgroup of KN,
and it has K and Hx as complements. Apply Theorem 4.2.5 again to see that
K and Hx are conjugate, and hence so are K and H.
A
The first group for which the conclusions of Theorem 4.2.6 could fail is
Alt(5), and indeed Alt(5) has neither a Hall {2,5}-subgroup nor a Hall {3,5}subgroup. (Why?) That is no accident, as in fact Hall also proved a rather
strong converse to the theorem, providing a characterization of finite solvable
groups.
The proof (below) requires another borrowing of a theorem of Burnside
(the paqb Theorem, 5.2.39) from a later chapter. This practice is perhaps less
than satisfying from a pedagogical point of view, but it presents no logical
difficulties, as Hall's Theorem will not be used in the proof of Burnside's
Theorem.
Hall's converse to Theorem 4.2.6 does not require the existence of Hall
7r-subgroups for all sets ix of primes, only of p-complements (not necessarily
normal) for all p G n(G), which are of course just Hall {p}'-subgroups.
Exercise
Suppose that p ^ q in 7r(G), and suppose that H and K are ap-complement
and a q-complement, respectively, in G. Show that HOK is ap-complement
in K and a q-complement in H, and show that H n K G Hall{ Pi ,}.(G).
(Hint See Exercise 1.2.2 of [31].)
89
4.3 MOSTLY p-GROUPS
Theorem 4.2.7 (P. Hall) / / G has a p-complement for every p in n(G)
then G is solvable.
Proof Suppose that the theorem is false, and choose a nonsolvable group G
of minimal order having p-complements for all p. The Burnside paqb Theorem
(5.2.39) asserts that groups whose orders have only 2 prime divisors are solvable, so 7T(G) > 3. Fix p € 7r(G), and let H be a p-complement in G. Then H
has a g-complement for all g € n(H) = n(G)\{p} by the exercise preceding
the theorem, so H is solvable, since \G\ is minimal. Choose a minimal normal
subgroup N of H, an r-group for some prime r by Proposition 4.2.4. Choose
q / p, r in n(G) and let K be a q-complement in G; K is solvable just as H
is. If R € Sylr(Ä") then 3 i £ G so that N < Rx < Kz, also a g-complement.
Thus we may replace K by Kx and assume that N < K.
Set L = {U{Nx:x € G}) <G. Since G = HK we have
L = (ö{Nhk: he H, k€K})
= (U{Nk: k € K}) < K,
so L is solvable. But now G/L satisfies the hypotheses of the theorem, so it
is solvable (since \G\ is minimal), forcing G to be solvable, a contradiction.
A
4.3
Mostly p-groups
If G is any group then its Prattini subgroup $(G) is defined to be the intersection of all maximal (proper) subgroups of G (if G has no maximal subgroups
then $(G) = G). If G is finite (which will be assumed throughout this section), and G / 1, then clearly $(G) < G. Furthermore, since any ip G Aut G
permutes the maximal subgroups of G, it is clear that $(G) char G.
Proposition 4.3.1 Suppose that H <G. Then G = HK for some K < G if
and only if H £ *(G).
Proof We have H ¿ $(G) if and only if H ¿ K for some maximal K
in which case G = HK.
<G,
A
A subgroup K < G with G = HK, as in the proposition, is sometimes
called a partial complement for H in G.
An element y in a group G is called a nongenerator if for any subset
X Ç G with G = (X U {y}) it is also true that G = (X). For example
y = 1 is a nongenerator in any group; if G = Sym(3) then 1 is the only
nongenerator. In the quaternion group of order 8 the unique element of order
2 is a nongenerator.
Proposition 4.3.2 The Prattini subgroup $(G) is the set of nongenerators
inG.
CHAPTER
90
4. TRANSFER
AND
SPLITTING
Proof Take y € $(G), so y e M for all maximal subgroups M. If G ^ (X)
but G = (X U {y}) choose a maximal subgroup M < G with M > (X).
Then y € M, so G < M, a contradiction. Conversely, suppose that y is a
nongenerator and that M < G is maximal. If y £ M then (M U {y}) = G
but (M) / G, also a contradiction.
A
Exercise
If G/$(G) is cyclic show that G is cyclic. What is $(G) if G is cyclic?
Proposition 4.3.3 The Prattini subgroup $(G) is nilpotent.
Proof It is sufficient to show that all Sylow subgroups of $(G) are normal.
Take P € Syl p ($(G)). Then G = NG(P)${G) by the Frattini Argument
(Proposition 4.2.1), so G = NG(P) by Proposition 4.3.1, i.e. P < G and so
P<$(G).
A
Proposition 4.3.4 Suppose that G is a p-group for some prime p. Then
1. G' < $(G),
2. G/$(G)
is elementary abelian, and
3. $(G) = 1 if and only if G is elementary abelian.
Proof Each maximal M < G is normal and of index p, so G/M is abelian
and G' < M; hence G' < $(G). Furthermore x" € M for all x 6 G and all
maximal subgroups M, so xp € $(G). Now G/$(G) is abelian and the pth
power of each of its elements is the identity, so it is elementary abelian. It is
immediate then that G is elementary abelian if $(G) = 1. On the other hand,
if G is elementary abelian it is clear that its maximal subgroups intersect in
1.
A
Corollary 4.3.5 If G is a p-group then $(G/$(G)) is trivial.
Exercise
If G is a p-group, N <G, and G/N is elementary abelian show that N >
$(G); conclude that $(G) is the smallest normal subgroup yielding an
elementary abelian quotient.
If G is a p-group then by Proposition 4.3.4 it is clear that V - G/$(G)
can be viewed as a vector space over the field F = F p with p elements. That
is the point of view of the next theorem.
4.3 MOSTLY
91
p-GROUPS
Theorem 4.3.6 (Burnside's Basis Theorem) Suppose that G is a finite
p-group, write $ for $(G), and set V = G / $ , viewed as an F = F p vector
space. Suppose that [G: $] = pn. Then n is the minimal size for sets of
Ç G and vt — $£¿ e V, 1 < i < n,
generators for G; if X = {xi,...,xn}
then G — (X) if and only if {vi,..., vn} is a basis for V.
Proof Clearly diniF(V) = n. Thus V has a basis consisting of n vectors
and has no smaller generating sets. The set {v\,...,vn}
is a basis for V if
and only if it generates V, which is if and only if G = (X U $(G)) = (X), by
Proposition 4.3.2. Everything follows.
A
Corollary 4.3.7 If G is a p-group with a unique subgroup H of index p then
G is cyclic.
Proof This is an immediate consequence of the Basis Theorem, since $(G) =
H; hence dimf-(G/$) = 1, and G has 1 generator. However, a simple alternative proof is also available. If G is not cyclic then every (x), x e G, is
contained in a maximal subgroup, of which H is the only one, so G < H, a
contradiction.
A
Exercises
1. Let G = Qm = (a,b \ a2m = 1,62 = o m ,6- 1 afc = a 2 " 1 - 1 ), the (generalized) quaternion group of order 4m, as on page 30. Set A = (a). Show
that if x € G\^4 then \x\ = 4; conclude that G has a unique element of
order 2.
2. Show that Qm has the presentation (a, b \ am = b2 = (ab)2).
The next few results go back to Frobenius and Burnside (see Chapter VIII
of [10]); the exposition below is based on that in [70], with further inspiration
from [54].
We begin with some slightly technical commutator identities.
Proposition 4.3.8 Suppose that G is a group with G' < Z(G), x, y € G,
and m, n € Z. Then
1. [x,y]m = [xm,y] =
m
n
2. {x1y ][x,y }
3. (xy)m =
[x,ym},
m+n
= [x,y ), and
xmym[x,y-(-m-^m/2].
Proof Note that
[x,y][x -1 ,y] = x-l\x-l,y]y-lxy
= x^xy^x^yy^xy
= 1.
Thus [x, y ] - 1 = [x _ 1 , y], anycd it will suffice to assume that m > 0 and argue
by induction.
CHAPTER 4. TRANSFER AND SPLITTING
92
For Part 1 we have
[x,y}m+1
=
[x,y}[x,y}m =
=
x-\xm,y\y-lxy
[x,y}[xm,y]
= x-lx-my-1xmyy-lxy
=
[xm+\y],
and also
[x,y]m = [y,x]-m = [ym,x]-1 = [x,ym].
For Part 2
[x,ym][x,yn]
=
=
x-1y-mx[x,yn]ym
x-1y-mxx-iy-nxynym
=
[x,ym+n}.
Finally, for Part 3,
(xy)m+1
=
=
=
=
by Part 2.
(xy)mxy =
xmymxy[x,y^m-^m/2]
xm+l(x-1ymxy-m)ym+1[x,y-^m'l)m/2]
xm+1î/m+1[x,y-m][x,î/-(m-1)m/2]
xm+12/m+1[x,2/-m(m+1)/2],
A
Theorem 4.3.9 Suppose that G is a p-group having two different cyclic subgroups of index p but only one subgroup of order p. Then G is the quaternion
group Q of order 8.
Proof Note that G is not abelian. Say that \G\ = pm, with [G: A] = [G: B] =
p, A = (a) 5¿ B = (b). Then A < G, B < G, and G = (a,b). IÎC = AnB
then [G:C] = p2, C < Z(G), and C > G' since G/C is abelian. In fact,
Z(G) = C, since Z(G) cannot have index p. Also G/C = (Ca, Cb) is not
cyclic, so x p € C for all x € G.
If p is odd and x,y e G then [ x , ^ ^ - 1 ) / 2 ] = 1, since yp G Z{G). Then
by Proposition 4.3.8 we have (xy)p = xpyp. Thus the map f i H i p is a
homomorphism; its kernel is {x G G:xp = 1}. But [G:ker ip] = |Im <p| <
p m _ 2 , so | ker <p\ >p2, and ker tp has more than one subgroup of order p.
Thus p = 2. Say that C = (c); since both a2 and 62 generate C we may
choose a, i», and c so that a 2 = c and b2 = c - 1 . Thus (ab) 4 = a4b4[a, b~6] = 1,
but (ab)2 = a 2 6 2 [a,6 - 1 ] = [a,6 - 1 ] ^ 1, since G is not abelian. So \ab\ = 4,
and clearly ab & C. If \C\ > 2 choose d e C having order 4. Then (ai>, d)
is abelian and not cyclic, so it has 2 distinct elements of order 2. Conclude
that \C\ = 2, so \G\ = 8, and G 5í Q since D 4 has 5 elements of order 2.
Alternatively, observe that (o6)2 = c, so a 2 = ft2 = (ai>)2, and apply the
exercise on page 91.
A
4.3 MOSTLY
93
p-GROUPS
Exercises
1. If AT « G show that CG(N) < G.
2. If G is finite and nilpotent, and N < G, show that N n Z{G) ¿ 1.
Proposition 4.3.10 If G is finite and nilpotent, but not abelian, and A is a
maximal abelian normal subgroup of G, then CG(A) = A.
Proof Suppose not; write C = CG(A) and note that C <G. Since G/A is
nilpotent there is an x G G such that A ¿ Ax G (C/A) D Z(G/A), by the
exercise above. Thus (A U {x})/A < G/A, so (A U {x}) « G, and ( 4 U {x}) is
abelian, since x G C, so the maximality of A has been contradicted.
A
Theorem 4.3.11 Suppose that \G\ = pm for some prime p and that G has a
unique subgroup of order p. Then G is either cyclic or generalized quaternion.
Proof Suppose first that p is odd and apply induction on m. The result is
clear for small m. Suppose then that m > 1, and assume the result for lower
powers. Thus every subgroup of index p is cyclic, so G is cyclic by Corollary
4.3.7 and Theorem 4.3.9.
Suppose next that p = 2. The result is clear if G is abelian, so suppose
also that it is not abelian, and let A be a maximal abelian normal subgroup.
Thus A is cyclic, and CG(A) = A by Proposition 4.3.10. Choose b G G\A
with b2 G A; note that A ¿ (b2) since b £ CG(A). Choose C = (c) < A
with [C: (b2)] = 2, and set H = (b,c), B = (b). Now BnC = (b2), so \H\ =
|5||C|/|(b 2 )| = 2\C\, or [H:C] = 2. Also [H:(b2)] = [H:C][C: (b2)] = 4, so
[H: B] = 2. Thus H has a unique subgroup of order 2 but 2 subgroups, B and
C, of index 2, so H is quaternion of order 8 by Theorem 4.3.9; in particular
|6| = | c | = 4 .
Say that |a| = 2 n + 1 ; then a2" = b2, since both have order 2. The argument
can be repeated with b replaced by ab, since ab & A but (ab)2 = a2b2[a, ¿>-1] €
A. Thus also a2 — (ab)2, and (a,b) is a homomorphic image of Q 2 n by the
exercise on page 91. Since \{a,b)\ > 2n+2 we have (a,b) S Q2n.
It remains to be shown that G = (a,b), and it will suffice to show that
[G: A] = 2. If that is not the case choose x € G with x 2 £ A but x 4 £ A.
By the argument above (a,x2) Si <52", and in particular x2ax~2 = a - 1 .
Let 6 denote the homomorphism (Proposition 4.1.10) from G to Aut A that
restricts inner automorphisms. Say that 9x(a) = ak, and note that k is odd
since (ak) = A. Thus fc2 = l(mod 4). But also 92x(a) = a*2 = 6xi(a) = a" 1 ,
so fc2 = - l ( m o d 2 n + 1 ) ; hence k2 = - l ( m o d 4), a contradiction. We conclude
that x2 e A for all x G G\A and also that 9x(a) - a~l. Thus the image of 8
in Aut .4 has order 2, and ker 9 - CG(A) = 4 , so [G: .4] = 2.
A
94
CHAPTER 4. TRANSFER
AND
SPLITTING
Corollary 4.3.12 Suppose that \G\ = pm, 1 < k < m, and that G has a
unique subgroup H of order pk. Then G is cyclic.
Proof Choose K < G with [K: H] = p, then K is cyclic by Corollary 4.3.7,
hence H is also cyclic. Any L < G of order p is contained in a subgroup of
order pk, therefore in H, so G has only one subgroup of order p, and is either
cyclic or generalized quaternion by the theorem. Similarly any M < G of
order p 2 is contained in a subgroup of order pk, therefore in H, so G has only
one subgroup of order p 2 . But (?2*> has more than one subgroup of order 4,
so G must be cyclic.
A
Exercise
If G is a finite p-group show that G has exactly one subgroup of order p
if and only if every subgroup of order p 2 is cyclic.
Proposition 4.3.13 Suppose that G is a group, x, y G G, x2 = y2 = 1, and
that 1 ^ H = (x, y) is finite. Then H is a dihedral group.
Proof I f z = l o r j / = l o r x = j / then H =í D\, so suppose that 1 ^
i / y / 1. Set z — xy and say that \z\ = m; note that H = (z,x). Since
x ^ y at least one or the other of x and y is not in {z), so \H\ > 2m. Also
xzx = x(xy)x = yx = z~l, so z and x satisfy defining relations for Dm; hence
H S D m.
A
Corollary 4.3.14 A nontrivial homomorphic image of a dihedral group is
dihedral.
Corollary 4.3.15 A nontrivial homomorphic image of the generalized quaternion 2-group #2» is either isomorphic with Q2" or is dihedral.
Proof Take Q2» = (a,b\ a 2 " +1 = 1, b2 = a 2 ", b~xab = a - 1 ) , as usual.
Suppose that 9 is a homomorphism from Q2* onto H, and assume that
1 ^ K = ker 6. Then the unique element a 2 " = b2 of order 2 is in K,
so 6(b)2 = 0(b2) = 1. We have H = (0(a),0(6)>, with 0(a) 2 " = 6(b)2 = 1
and 6(b)~16(a)0(b) = 0 ( a ) - 1 , so H is a homomorphic image of D 2 »- Apply
Corollary 4.3.14.
A
Some of the results above will be applied in a later chapter in the study
of Probenius groups. For the present let us give some applications to Sylow
subgroups.
If p € N is a prime set q = pk and let F = ¥q. Recall that | SL(2,q)\ =
(9 - 1)9(9 + !)• The subgroup
■{
1
a
?1 :«i'l
4.3
95
MOSTLYp-GROUPS
has order q, so P is a Sylow p-subgroup of SL(2,g). Clearly P is isomorphic
with the additive group F, which, being a vector space over its prime field,
is an elementary abelian p-group.
Suppose now that p is odd. An easy calculation shows that if r 6 SL(2, q)
and T 2 = 1 then in fact r = ± 1 , so SL(2,<f) has a unique element of order
2. By Theorem 4.3.11 a Sylow 2-subgroup T of SL(2,g) is either cyclic or
generalized quaternion.
If q = 3 recall that PSL(2,3) Si Alt(4), whose Sylow 2-subgroup is a Klein
4-group. Thus the Sylow 2-subgroup of SL(2,3) is (isomorphic to) Q = Q2- If
q > 5 then PSL(2,ç) is simple (Theorem 2.6.13), and its Sylow 2-subgroups,
which are homomorphic images of T, cannot be cyclic by Corollary 4.1.12.
Thus T = Q 2 m for some m in all cases.
If T is a Sylow 2-subgroup of SL(2,q) then T / { ± 1 } is a Sylow 2-subgroup
of PSL(2,ç) - it is dihedral by Corollary 4.3.15.
Proposition 4.3.16 Suppose that G has more than one Sylow 2-subgroup,
and that if 7\ ¿ T2 in Syl 2 (G) then Ti n T2 = 1. Then G has exactly one
conjugacy class of involutions.
Proof Choose involutions x and y from two different Sylow 2-subgroups.
Then H = (x, y) 2 Dm for some m by Proposition 4.3.13. If m is odd then x
and y are conjugate in H. Could m be even? If so then z = (xy)m/2 e Z(H),
and \z\ = 2. But then Hi = (x,z) and H2 = (z,y) are both Klein 4-groups,
and each is contained in a Sylow 2-subgroup, say Hi < 71. Then Xi ^ T2 by
the initial choice of x and y, but 1 ^ z € T\ D T 2 , a contradiction.
A
Suppose now that q = 2k. As noted above
-{[ÜM
is a Sylow 2-subgroup of G = SL(2,q), and it is elementary abelian of order
q. An easy calculation shows that the normalizer of T is
" = {[? .î.]=.^.»eJ-},
which has order q(q - 1). Thus there are [G:N] = q + 1 different Sylow 2
subgroups in G. The elements of order 2 are all matrices
of which there are q — 1, and all
2
a
b
c
a
, a € F*,
a 1
where a £ F and b € F* are
arbitrary, but c = (1 + a )/b (verify this), so there are q(q - 1) such matrices.
All told, then, there are q2 - 1 = (q - l)(q + 1) involutions in G. Since they
are distributed among the q + 1 Sylow 2-subgroups, q — 1 in each of them,
we see that if Tx ¿ T2 in Syl 2 (G) then 7\ n T2 = 1. By Proposition 4.3.16 we
conclude that all involutions in SL(2,2fc) are conjugate.
Groups and Characters
by Larry C. Grove
Copyright © 1997 John Wiley & Sons, Inc.
Chapter 5
Representations and
Characters
In this chapter we introduce the basic ideas of the (ordinary) representations
and characters of finite groups, concentrating on characters. The applications
are legion.
Group theoretical applications of character theory appear prominently in
this chapter as well as later in the book, especially in Chapter 9. Some of
the many other examples of applications of character theory appear in the
books by Sternberg [59] (physics), James and Liebeck [41] (chemistry), and
Diaconis [20] (probability and statistics).
5.1
Representations
If F is a field and V is a finite dimensional vector space over F write GL(V)
for the general linear group of all invertible linear transformations of V. If
dimf (V) = n, and a basis is chosen for V then we obtain an isomorphism (via
representing matrices) from GL(V) to GL(n, F ) , the group o f n x n invertible
matrices over F.
An F-representation of a group G is a homomorphism T from G to GL( V)
for some V. If a basis is chosen then the composite map
G^GL(V)-»GL(n,F)
is called an F-matrix representation; it will be denoted by T.
If T is a representation of G and T is 1-1 then it is called faithful. As
a more-or-less general rule, any adjective that applies to a representation T
applies as well to any corresponding matrix representation T.
If 5 and T are F-representations of G on spaces V and W then we say
that S and T are equivalent, and write 5 ~ T, if there is an F-isomorphism
97
98
CHAPTER
5. REPRESENTATIONS
AND
CHARACTERS
9: V -> W for which the diagram
V
S(z]
•1
w
T(^
V
1»
w
is commutative, i.e. 0S(x) = T(x)0, or T(x) = eS{x)ß-1, for all x 6 G.
If T is a representation of G on V, then a subspace VT of V is called
T-tTiDortoni if T(x)W Ç W for all x 6 G. Note that then each T(x) can be
restricted to a linear transformation of W, resulting in a representation of G
on W. Denote it by T\wIf W is T-invariant, and if a basis is chosen for W and extended to a basis
for V then the resulting matrix representation T has the form
T(x) =
A(x)
0
C(x)
B(x)
where A = T\wRemark B is also a matrix representation of G, corresponding to the action
of G (via T) on the quotient space V/W.
If V has a proper nonzero T-invariant subspace W then T is called reducible, otherwise T is irreducible, or simple.
W a r n i n g ! Just as for polynomials, the question of reducibility versus irreducibility of representations depends (heavily) on the field F of scalars.
Exercise
Take G = C4 = (x \ x 4 = 1), cyclic of order 4, and define a (matrix)
0 -1
representation via T(x) =
. If F = TSL the reals, show that T
1 0
is irreducible. If F = C, the complexes, show that T is reducible.
Suppose that S and T are F-representations of G on V and W, respectively. Define S © T on V © W via (S 0 T)(x): (v,w) ^ (5(x)u,T(x)u;) for
all x e G, t; € V, u; € W. Then S © T is an F-representation of G, called
the direct sum of 5 and T. In like manner we may of course define the direct
sum T\ © T-i © • • • © Tk of any finite number of F-representations.
If T is an F-representation of G and T is equivalent with a direct sum
of some number of irreducible F-representations then T is called completely
reducible (or semisimple). Note that as a special case an irreducible representation is completely reducible.
If T is an F-representation of G on V and dimf(V) = n we say that
T has degree n, and write deg(T) = n. If deg(T) = 1 then T is called a
5.1.
99
REPRESENTATIONS
linear representation. If T is linear then T is a homomorphism from G to the
multiplicative group F* — F\ {0}.
Examples
1. For any G, any F, and any V define T(x) = l v , the identity transformation on V, hence T = I, an identity matrix, for all x e G. Then T
is called a trivial representation. If T is trivial and deg(T) = 1 then T
is called the principal F-representation of G and is usually denoted I G .
2. Let G = Cn = (x | x n = 1), take F = C, and let Ç e C be a primitive
nth root of unity, e.g. C = e2niln. Define f (x) = C- Then T is a faithful
irreducible linear representation of G.
3. Let G = Sn, the symmetric group of degree n, or more generally any
subgroup of Sn, and let F be any field. Define
<p/ \ _ Í 1
J
W - \ _ 1
if x is even,
if x is odd.
Then T is called the alternating representation of G.
4. Suppose that G acts as a permutation group on a set X = {xi , . . . , x„}.
For any field F let V be a vector space over F of dimension n, and let
{ « ! , . . . , « „ } be a basis. If g € G and x? = x¡ define T(g)vi — Vj,
determining T(g) £ GL(V) and hence an F-representation T. It is
called the permutation representation corresponding to the action of G
on X. Note that each T(g) is a permutation matrix, i.e. each row and
each column has exactly one nonzero entry, which is a 1.
5. In 4 above take X = G, and let G act on X by right multiplication
(as in Cayley's Theorem). The resulting permutation representation
of G is called the right regular representation of G; denote it by R.
Similarly the permutation action of G on X by left multiplication results
in a permutation representation called the left regular representation,
denoted L.
Exercise
Write out the right regular (matrix) representation for generators for the
groups d, C2 x Ci, S3, and D4 (dihedral of order 8).
Theorem 5.1.1 (Maschke's Theorem) Suppose that G is a finite group,
F is a field, and char(F)j' |G|. Then every F-representation T of G is completely reducible.
100
CHAPTER 5. REPRESENTATIONS AND CHARACTERS
Proof We use induction on n = deg(T). If n = 1 then T is irreducible,
hence completely reducible. Assume then that n > 1, and assume the result
for all F-representations of degree less than n. Assume that T is reducible
and choose a basis so that
T(x) =
A{x)
0
C(x)
B(x)
all x G G. If x,y € G then, since T is a representation, we have T{xy) =
T{x)T(y), which says that
A(xy)
0
A(x)A(y)
0
C(xy)
B{xy)
A(x)C(y) + C{x)B(y)
B(x)B(y)
Thus C{xy) = A{x)C{y) + C(x)B(y), and
C(xy)B((xy)-1)B{x)
= C(xy)B(y-1)
l
Define D = \G\~ T.{C{z)B(z
over all y in G to obtain
l
= A(x)C(y)B(y-1)
+ C(x).
(*)
):z € G}, and sum the equation (*) above
\G\DB(x) = \G\A(x)D + \G\C(x).
Hence
DB(x) = A(x)D + C(x),
all
x£G,
since \G\ ^ 0 in F. Observe now that
/ D
0 /
A(x)
0
0
B(x)
A(x)
0
A{x)
0
A(x)
0
DB(x)
B(x)
A{x)D + C(x)
B{x)
C{x)
I D
B{x)
0 /
It follows that T ~ A ® B, and the induction hypothesis can be applied to A
and B to complete the proof.
A
Proposition 5.1.2 (Schur's Lemma) Suppose that S and T are irreducible F-representations of G onV and W, respectively, and that A: V -> W is
a linear transformation such that AS(x) = T(x)A for all x € G. Then either
A = 0 or else A is an isomorphism (and hence S ~ T).
Proof Suppose that A ¿ 0. Set Vx = ker(4) and Wx = Im{A). If v € Vi
then 0 = T(x)Av = AS(x)v, all x, so S(x)v € ker(4) and Vi is 5-invariant.
But Vi ,¿ V, so Vi = 0 since S is irreducible, and hence A is 1-1.
5.1.
101
REPRESENTATIONS
If w e Wl write w = Au, u € V. Then T{x)w = T(x)Au = A(S(x)u) C
Im(A) = Wi, all x. Thus W\ is T-invariant and nonzero, so W\ = W since
T is irreducible, and A is onto.
A
If T is an ^-representation of G on V define the centralizer of T to be the
algebra of all linear transformations A: V -¥ V for which AT(x) = T(x)A for
all x e G. We write C(T) or C F ( T ) for the centralizer of T. The centralizer
C(T) of a corresponding matrix representation T is the algebra of all matrices
over F that commute with all T(x), x € G.
Proposition 5.1.3 (1) If T is an irreducible F-representation of G on V
then C(T) is a division algebra. (2) If ch&r{F)]( \G\ and C(T) is a division
algebra then T is irreducible.
Proof (1) If A ^ 0 in C(T) then A is an automorphism of V by Schur's
=
Lemma (5.1.2), so A~l exists. If x 6 G then A~lT{x) = (Tix'^A)-1
(ATix'1))-1
= T(x)A-\
so A~l € C(T) and C(T) is a division algebra.
(2) Suppose that T is reducible. By Maschke's Theorem (5.1.1) V =
V\ © V2, with both Vi nonzero and T-invariant, so we may assume that T =
Ti © T 2 , with Ti = T\v¡- Let P be the projection of V onto V\ along V-i, i.e.
if v = vi + v2 € V, with Vi € Vi, then Pv = v\. Note that P is not invertible,
since ker(P) = V2 £ 0. However,
FT(ï)t) = PT(x)(wi + t>2) = P(T(x)vi
+ T(x)v2) = T{x)vi =
T{x)Pv,
so P € C(T), contradicting the assumption that C(T) is a division algebra.
A
Exercise
0 -1
. Compute C(T). If
1 -1
F Ç E show that T is irreducible. If F = C find a matrix M such that
u) 0
Af-xr(i)M
, where or + u + 1 = 0 in C.
0 u>2
Let G = C3 - (x \x3 = 1) and set f(x)
=
If V is a vector space over F and K is an extension field of F we may
set VK = Ä" <8>F V to extend the field of scalars from F to K. Recall that if
{ « i , . . . ,t>„} is an F-basis for V then { l ® v i , . . . , l ® u „ } is a A"-basis for VK.
If T is an F-representation of G on V define T * on VK via T K (o;): 1 ® u< >-»
1 (81 T(X)DJ, 1 < ¿ < n, all x € G. Note that, relative to the bases indicated,
TK = T. This process is referred to as extending the ground field.
If T is an irreducible F-representation of G and if TK remains irreducible
for every extension field K of F , then T is said to be absolutely irreducible.
If F is a field for which every irreducible F-representation is absolutely irreducible then F is called a splitting field for G.
102
CHAPTER 5. REPRESENTATIONS
AND
CHARACTERS
Proposition 5.1.4 (1) IfT is an absolutely irreducible F-representation of
G thenC(T) = {XI: X G F} 2 F. (2) Ifdai{F)jf
|G|, andifC(T) = {A1:A G
F}, then T is absolutely irreducible.
P r o o f (1) Let F be an algebraic closure of F. Then TF is irreducible and
CF(T) Q Cjr(f). Take A G CF(f) and let A € F be an eigenvalue of A. Thus
A - XI is singular. Clearly A - XI 6 C-p(T), which is a division algebra by
Proposition 5.1.3. Thus A - XI = 0, so A = XI, and in particular A € F.
(2) We show first that CK(TK) = {XI: X G K} for every extension field
K DF. Take A G CK(TK)
and choose a basis (for VK) so that A is represented by a matrix M. Write
M = axMi +
\-akMk,
where each Mi is an F-matrix, each a¿ G K, and the set {a¡i,... ,ak} is
linearly independent over F (why is this possible?). Then MT(x) = T(x)M,
all x G G, or E*=i a ¿ M ¿ ^ ( x ) = E*=i ajT(x)M¿. Read that equality entry
by entry to conclude (since {a¿} is linearly independent) that MiT(x) =
f{x)Mi, 1 < i < k. Thus Mi G CF(T), so M¿ = XJ for some A¿ G F , and
so M = 53 ¿ ajMj = (EjOiiAi)/ G {AI: A G i f } . In particular we see that
CK(TK)
is a division algebra for all K D F , and so TK is irreducible by
Proposition 5.1.3. Thus T is absolutely irreducible.
A
Although we will not need the more general statement it should be noted
that the conclusion of part 2 of Proposition 5.1.4 holds without the assumption that char(F) j |G|. For a proof see Isaacs [39], page 145.
Corollary 5.1.5 (Of the proof of Proposition 5.1.4.) / / c h a r ( F ) | |G| and if
F is algebraically closed, then F is a splitting field for G.
Exercise
Suppose that G has a faithful irreducible C-representation T. Show that
the center Z = Z{G) is cyclic. (Hint If z G Z then T{z) G C(T).)
Proposition 5.1.6 Suppose that S and T are F-representations of G, F is
an infinite field, and K is an extension field of F. If SK ~ TK then S ~ T,
i.e. inequivalent representations cannot be made equivalent by extending the
ground field.
P r o o f Since^S* and TK are equivalent there is an invertible if-matrix M
such that MS(x) = f(x)M, all x eG. As in the proof of Proposition 5.1.4
write M = £ . = 1 ciiMi, where the Mi are F-matrices and {cti,.. .,ak] Ç K
5.1.
103
REPRESENTATIONS
is F-linearly independent. Since MS(x) = T{x)M we have, as above, that
MjS(x) = T(x)Mi, 1 < i < k and all x G G. Define a polynomial
P(tu.
..,tk)=
d e t ^ i M i + • • ■ + tkMk) G F[tu...,
tfc].
Note that P(ai,... ,ak) = det(M) ^ 0, so P is not the zero polynomial.
Since F is infinite there are ßi,... ,ßk € F so that
P ( Ä , . . . , & ) = d e t a i l * ! + • ■ • + AM*) 5¿ 0.
(Why?) Set N = 5Z» A M , an invertible F-matrix, and observe that NS(x) =
f (x)AT, all i G G, so 5 ~ T.
A
Remark The result above is also true for finite fields F , but the proof is more
complicated (e.g. see Theorem 29.7 of [18]).
Proposition 5.1.7 Suppose that G is abelian and that T is an absolutely
irreducible F-representation of G onV. Then deg(T) = 1.
P r o o f Since G is abelian T(x) € C(T) for all x e G. Thus T(x) = \xl for
some Ax € F , all x € G. But then every subspace of V must be T-invariant
and so deg(T) = dim(V) = 1 by irreducibility.
A
Theorem 5.1.8 (Schur) (1) Suppose that S and T are inequivalent irreducible F-representations. Choose bases so that S(x) = [s¿¿ (x)] and T(x) =
[tij(x)], all x eG. Then
^KWiH(i"')^6G}=0
for all i,j,k, and I.
(2) IfT is absolutely irreducible, and n — deg(T), then
nY,{Uj{x)tki{x-l):x£
for all i,j,k,
G) = 6uSjk\G\,
and I.
P r o o f (1) For any F-matrix M (of appropriate size) define
L = 5 3 { 5 ( x ) M T ( x - x ) : x G G).
If y G G then
S(y)L
=
=
^{SMSfiJMftW-'lfMugG}
[YíS(yx)MT({yx)-1)]f{y)
= LT(y).
104
CHAPTER 5. REPRESENTATIONS
AND
CHARACTERS
By Schur's Lemma (5.1.2) L = 0 for every choice of M. Choose M = Ejk,
the matrix with 1 in position jk and O's elsewhere. Then the ¿/-entry of L is
(2) Set L = £ { T ( x ) M T ( x - 1 ) : x e G} for any M and check, as above,
that if y € G then f (y)L = LT(y), so L e C(f). Again take M = Ejk. By
Proposition 5.1.4 L = Xjkl for some \jk € F . Thus the ¿/-entry of L is
Xjk&u
=
^2tij(x)tki{x~1)
^tij(x~1)tki(x)
=
X
i
X
= ^r tki(x)tij(x~ ) = Xuôkj.
X
Choose ¿ = / and j ^ kto see that Xjk = 0 when j ^ k. Next choose i = I
and j = k to see that AÜ = A¿, = A (say). So, A = ^x tij(x)tji(x~l)
for all i
and j . Now sum over j and get
nA
= £5>j(*)*¿i(*-1) = £ i = iGi,
since J V iij(i)tjj(a; - 1 ) is the ¿¿-entry of f (x)f ( i _ 1 ) = f (1) = J„. Thus
n ^2 tij{x)tki(x~l)
= nXjkSu = (nX)ójkSu = \G\6jk6u.
x€G
Theorem 5.1.9 (Frobenius and Schur) Suppose that char(F) jf \G\, and
letTi,...,Tk
be mutually inequivalent absolutely irreducible F-representations
of G. Choose bases so that Ts{x) = [t\' (x)] for each s and all x. Then the
set {t\f:l < s < k, all i,j}
from G to F.
is an F-linearly independent set of functions
Proof Suppose that £ M i j a j ' ^ = 0, a\f
Choose arbitrary r, /, and m, and note that
»EG
s,i,j
€ F . Say that deg(r„) = n..
= E^VE^*)^*-1)
»,i,j
so a|¿, = 0 for all r, /, and m.
x
5.2.
105
CHARACTERS
Corollary 5.1.10 Ifn,
= deg(T,), all s, then £ * = 1 n2, < \G\.
Proof There are n\-\
\-nj. matrix entry functions, and the dimension of
the space FG of all functions from G to F is \G\.
A
Corollary 5.1.11 There are at most \G\ inequivalent absolutely irreducible
F-representations o}G.
Recall (Corollary 5.1.5) that algebraically closed fields are splitting fields.
It is usually not necessary to extend all the way to an algebraic closure,
however, in order to obtain a splitting field for a group G.
P r o p o s i t i o n 5.1.12 Ifchax(F))(
that is a splitting field for G.
\G\ then there is a finite extension K of F
Proof Let F be an algebraic closure for F , a splitting field for G by Corollary 5.1.5. Let {Ti,...,Tfc} be a full set of inequivalent irreducible (hence
absolutely irreducible) F-representations of G. Choose bases to obtain corresponding matrix representations T,, say with Ta(x) = [í¿* (x)], all x £ G.
The matrix entries t\*'(x) are all algebraic over F (they are in F), and there
are only finitely many of them. Adjoin them all to F to obtain a field K
which is affinité extension of F , and observe that K is a splitting field for G
since all T,, 1 < s < k, are K-representations.
A
Corollary 5.1.13 For any G there is an algebraic number field (i.e., a finite
extension of Q) that is a splitting field for G.
5.2
Characters
Recall that the trace of an n x n matrix A = [a¿j] is tr(^4) = YLl=iaaIt is easy to check that tr(AB) = tr(BA) for all A, B In particular if M
is invertible then tr(M~1AM)
= tr(^). Consequently, if T:V -> V is a
linear transformation, we may unambiguously define tr(T) = tr(A) for any
representing matrix A of T.
If T is an F-representation of G define the character x — XT of T to be
the function from G to F determined by x(z) = tr(T(x)), all x € G. The
character of an F-representation will often be called an F-character.
Remark If if is an extensionjfield of F then the character of TK is the same
as the character of T, since TK = T for appropriate bases.
As a general rule any adjective that can be applied to a representation
will be applied as well to its character. Thus x c a n D e faithful, reducible,
irreducible, linear, etc.
Note that x r ( l ) = tr(7) = deg(T) • 1 G F . Thus if char(F) = 0 then x(l)
(as an integer) is the degree of T, and in that case we call it the degree of x
and write deg(x) = x(l)-
106
CHAPTER 5. REPRESENTATIONS
AND
CHARACTERS
Proposition 5.2.1 If x — XT is a character of G then x w o class function;
i.e. x *s constant on conjugacy classes of G.
Proof If x, y € G then
X(î/ _1 zy)
T{y~xxy)
=
tr
=
tv[T{y-l)T{x)T{y)\
= tr T(x) = *(x).
Proposition 5.2.2 If S and T are F-represeníaítons of G, and S ~T,
then
Xs = XT-
Proof For appropriately chosen bases we have S =T.
A
Proposition 5.2.3 If S andT are F-representations of G then
XS&T = XS + XT-
S
Proof This is clear, since S © T <
0
0 f
Corollary 5.2.4 Suppose that c h a r ( F ) / |£7| and that x w an F-character
of G. Then there are irreducible F-characters Xi>-->X* o / G so that x =
Xi+'-'
+ Xk-
Proof Say that x = XT- By Maschke's Theorem (5.1.1) there are irreducible
F-representations 7 \ , . . . , 7* so that T ~ T\ ©• • • ©Tk. Let Xt be the character
of Ti for each i and apply Proposition 5.2.3.
A
Proposition 5.2.5 Suppose that F C C and that x is an F-character of G.
Then x ( x - 1 ) = x(x) for all xeG.
Proof Say that X = XT- Set H = {x) < G and restrict T to H, giving a
representation of H. Extend the field to C (x is unchanged). By Maschke's
Theorem (5.1.1), Corollary 5.1.7, and Proposition 5.1.7 (applied to T\H) we
may choose a basis so that
Ci
0
0
a
T(x) =
5.2.
107
CHARACTERS
with each Q € C. Say that |x| = m. Then
'cr
0 "
2 :* ) = / =
m
cr.
0
j
so C™ = 1- all i. But tl en Ç"1 = Ci) and so
1
fix- ) =
"cr1
0
cr 1 .
0
Hence xi*'1)
'
"CT
0 '
. 0
Ci.
= G + C2 + • • ■ + Cfc = Ci + ■ • • + Cfc = *(*)■
Remark It is not necessary to assume in Proposition 5.2.5 that F is closed
under complex conjugation.
Exercise
Suppose that F Ç C, \ is an F-character of G, and z e G is an involution,
i.e. an element of order 2. Show that xiz) € Z, and that x(z) = x(l)
(mod 2).
We will a s s u m e until further notice that F C C .
If if and 6 are functions from G to F define
(¥>,«) - M " 1 ^ M x ) Ö ( x - x ) : x 6 G}.
It is easy to verify that this defines a symmetric bilinear form on the vector
space of all such functions.
Remarks If x IS an F-character of G then
\G\-I^2x(x)xjx)>0
(X,X) =
X
(by Proposition 5.2.5). Furthermore, if ip is another F-character then (x, t/>) G
R, for
(x,t/>)
=
IGr^xisM*-1)
z
=
l
\G\- Y,x(x-l)rp(x)
=
(^x)=(x^).
X
Theorem 5.2.6 (1) / / x and ip are distinct irreducible F-characters
(X>V0 = 0. (2) If x is absolutely irreducible then (x,x) = 1-
then
108
CHAPTER 5. REPRESENTATIONS
AND
CHARACTERS
Proof (1) Say that \ — XT and ip = V>s- Choose bases so that T = [ty] and
S = [sij]. Then
u,*) = iGr 1 E^w x " 1 ) = iGi"1EEi"(x)s»(x_1)
*
I
Í
X
t,J
X 1 =0
= IGI- 5;^ «( )^( " )
«,J
«
by Theorem 5.1.8.
(2) Just as in the proof of part 1 we have
(x,x) = IGI-'EE*«^)*«^" 1 )
Sa\G\
= IGI- 171
E deg(T)
«.J
1
deg(T)
v
deg(T)
deg(T)
=
again using Theorem 5.1.8.
Corollary 5.2.7 Any set {\i,.
G is F-linearly independent.
A
■ ■ ,Xk} of distinct irreducible F-characters of
Proof Suppose that 5Z¡ a«X« = 0, ai 6 F. Then
0 = (0,Xj) = (XI a «'Xt,Xj) = ^atiiXuXj)
i
= <Xj(Xj,Xi),
i
so atj = 0, all j , since (Xj>Xj) > 0-
A
The next theorem is simply another corollary of Theorem 5.2.6; it is singled out as a separate theorem because of the important role it plays in
character theory.
Theorem 5.2.8 (The First Orthogonality Relation) / / F Ç C is a
splitting field for G, and if xi, ■ ■ ■, X* are a^ tne (absolutely) irreducible Fcharacters ofG, then
(xi,Xj) = Sij
for all i, j .
The following two corollaries to Theorem 5.2.8 are often referred to as the
"Fourier analysis of finite groups." Continue to assume, as in the theorem,
that F Ç C is a splitting field for G.
5.2.
109
CHARACTERS
Corollary 5.2.9 //Xi> • • • >X* are all ^ne absolutely irreducible characters of
G, and x is any F-character, then
k
X = ^2(x,Xi)Xi»=i
Proof It follows from Corollary 5.2.4 that x — E¿=i ntXi> where 0 < n¿ € Z,
A
all i. Thus (x,Xj) = (E< ™ÍXÚXj) = E i nt(x»>Xj) = "¿> a11 3Those absolutely irreducible Xi for which n^ = (x, Xi) > 0 a r e called the
constituents of x, and the integers n¿ = (x, X») axe t n e multiplicities of the
constituents.
Corollary 5.2.10 J / x i > - > X * a r e a " ^ e oôso/ute/y irreducible characters
ofG, and ifx = E¿=i m¿X¿ a n ¿ i> — E t = i n»X« a r e anV ^wo F-characters of
G, then (x,V0 = E i m ¿ " t > 0 in Z. In particular, (x,x) = E i m i Proof (x, tp) = ( E i m»Xi> E j n iXi) = E¿,j "»jiijixi, Xj) = E i m « n ¿-
A
Theorem 5.2.11 Ifx is an F-character of G then x is absolutely irreducible
if and only i / ( x , x ) = 1Proof (=>) Theorem 5.2.6.
(«=) Let K D F be a splitting field for G, and say that x i , • • •, X* are a u
the (absolutely) irreducible Af-characters of G. Write x = E. n »X¿- Then
1 = (x> x) == E « n ? by Corollary 5.2.10, and so one n¿ = 1 and all others are
0. Thus x = Xi is absolutely irreducible.
A
Theorem 5.2.12 Suppose that S and T are F-representations
XS = XTThenS~T.
of G, with
Proof Let K D F be a splitting field for G and let T i , . . . , Tk be a full set of
inequivalent irreducible ii-representations of G, with characters xi > • • •, X* •
For any m > 0 in Z write mTi to denote the direct sum of m copies of T{.
By Maschke's Theorem (5.1.1) TK ~ m\T\ © • ■ • @mkTk, and likewise SK ~
niTi © ■ • • © nkTk, 0 < m^m € Z. Thus xs = XT = E . m . X i = E< "¿Xi,
and m< = n¿ for all i by Corollary 5.2.7, and SK ~ TK. But then 5 ~ T by
Proposition 5.1.6.
A
Remark The assumption that char(F) = 0 is essential in Theorem 5.2.12.
For example, if char(F) = 2, G is cyclic of order 3, and if 5 and T are both
trivial representations, but of different odd degrees, then they have the same
character but are obviously inequivalent.
Suppose that G acts as a permutation group on a set X = { x i , . . . , x „ } .
Let T be the corresponding permutation representation (see Example 4 on
110
CHAPTER 5. REPRESENTATIONS
AND
CHARACTERS
page 99). If 9 is the character of T and g & G note that 6(g) is just the
number of l's on the diagonal of T(g), which is the number of fixed points
of g in X. As a special case (Example 5, page 99) we have the right regular
representation R - denote its character by p = po, the regular character of
G. Observe that
if ff = l,
= /|G|
otherwise.
*>-{?
Proposition 5.2.13 If p is the regular character, and Xi, • • • >Xfc are a^ ^ne
absolutely irreducible characters of G, then p = E»=i X»(l)Xi/ i e - every
absolutely irreducible character of G is a constituent of p, with multiplicity
equal to its degree.
Proof Apply Corollary 5.2.9 to see that the multiplicity of x¿ is
(p,Xi) = IGT 1 ^ M a O x i í x - 1 ) : * € G} = ¡G]"1 p(l)Xi(l)
= *(!)•
A
Theorem 5.2.14 / / x i , X 2 > - >X*
tersofGthenY!¡=lXiO.?
= \G\.
Proof \G\=p{\)
are
° " the absolutely irreducible charac-
= £ i *(!)*(!)•
A
Notation Until further notice F Ç C will denote a splitting field for G,
Ti,T2, . ■ .,Tk a full set of inequivalent absolutely irreducible F-representations, with characters xi = 1 G , X 2 , - •• , Xfc- Write Irr(G) to denote the set
{xi>X2,-• • ,X*} of (absolutely) irreducible characters. Let m = c(G) be
the class number of G (the number of conjugacy classes), and let K\ =
{1}, Ki,..., Km be the conjugacy classes.
The character table of G is the kxm matrix whose rows and columns are
indexed by the absolutely irreducible characters and the conjugacy classes of
G, respectively, and whose entries are the character values at the classes.
For an easy example take G to be the symmetric group 53, with Ki =
{1}, Ki = cl(123), and K¡ = cl(12). There are two linear characters, xi = I G
and X2 the alternating character (Example 3, page 99). The permutation
character 6 of 5 3 has values ô(Ki) = 3, #(#2) = 0, and 6(K3) = 1 (count
fixed points). It is not absolutely irreducible - since (9,0) = 2 it must have
two constituents. Since (9,\i) = 1, Xi is a constituent of 9 with multiplicity
1, and hence X3 = 9 - Xi is the other constituent. The list Xi,X2,X3 is
complete by Theorem 5.2.14, because the sum of the squares of the degrees
is 6. The character table is as follows.
Ki
Xi
X2
X3
1
1
2
K2 K3
1
1
1 -1
-1
0
5.2.
111
CHARACTERS
Exercise
Find a (matrix) representation of 53 whose character is X3 above. (One
possible approach is to recall that 53 can be viewed as the symmetry
group of an equilateral triangle.)
Remark In calculating (xi,Xj) fr°m the character table it is necessary to
take account of the sizes of the conjugacy classes. Observe in fact that
\G\-l^\K.\xi(Kt)Xj(Kt).
(Xi,Xi) =
This can be useful when applying the First Orthogonality Relation (5.2.8) in
cases where some of the values of one of the characters are not known yet.
Proposition 5.2.15 For each Xi and Kj define
My = £{!■,(*):*€ Kt) and u,y = 1 A Í M .
XiK1)
Then My G C(T¿), and in fact My = wyl.
= ^{TiiV1 xy): x G Kj) = My,
Proof If y G G then T^y)'1 M^y)
so My G C(Ti). Thus My = Ayl for some Ay G F by Proposition 5.1.4.
Now tr(My) = Ayxj(l) = £ { X i ( x ) : * G Kj} = \Kj\Xi(Kj),
and so Ay =
\Ki\xi(Ki)/xi{l)=u>ii.
A
Proposition 5.2.16 Ifx€K„
n
define
ijs = \{(y,z)
€ Ki x Ky.yz = x}\
for 1 < i, j , s < m. Then
1. Hij, depends only on i, j , and s, and not on the choice of x G K„;
2. utiu>tj = 52,nij,uju; and
3. \Ki\\Kj\xt(Ki)xt(Kj)
forl<t<k.
=
xt(l)Zsnij*\K,\Xt(K,),
Proof (1) This is easy; just conjugate x,y and z.
(2) Multiply:
MtiMtj
=
^{Ttiyy.yeKi^iTtizy.zeKj}
=
£ { T t ( x ) : z = yz, (y,z) G Kt x Kj}
112
CHAPTER 5. REPRESENTATIONS
=
^2nijs
AND
]T¡{7í(z):a: 6 K.) =
CHARACTERS
^2nijsMu.
Apply Proposition 5.2.15 to conclude that
a
(3) This is now an easy consequence of part 2. Multiply both sides by
Xt(l) 2 and recall the definition of the u's.
A
Exercise
Show that riijk = Tijik for all i,j, k.
Theorem 5.2.17 (The Second Orthogonality Relation) The columns
of the character table of G are orthogonal with respect to the usual inner
product of complex column vectors. Explicitly
¿Xt(ffiOT^) =
1 < i>j < m> where xi,...,Xk
classes.
SjJ
r£r- = Ä«|<7o(s4)|,
are representative elements in the conjugacy
Proof Say that K~l = Kt. Then, by Proposition 5.2.16,
IKiWKelxtiKMKt)
= Xt(l)5>H.|*.|Xt(Jr.),
8
all t. Sum over t and apply Proposition 5.2.13:
\Ki\\Ki\'£,Xt(Ki)xt(Kï1)
t
=
^n^lüT.l^x^l)^^)
a
t
= 52nu.\KMKi)
= mti\G\,
a
where p is the regular character. But
nm = \{(y,z) e Kt x Kf.yz = 1}| = \Kj\Sij
since Kt = -f^"1- The theorem follows upon division by |Äj||Äj|.
A
Remark Note in particular that £ t Xt(Ki)xt(Ki) = |G|/|üfj| = \CG(x)\ for
any x £ Ki. Thus orders of centralizers and sizes of conjugacy classes are
easily determined from the character table (as is |G|).
5.2.
113
CHARACTERS
For example, the following is the character table of a group G.
Ki K2 K3 K4 K6
1
1
1
1
1
1
1
1 -1 -1
2
2 -1
0
0
3 -1
0
1 -1
3 -1
0 -1
1
Xi
X2
Xz
Xi
X5
Thus we may read from the table that |G| = 24, and that the sizes of the
conjugacy classes are, respectively, 1, 3, 8, 6, and 6. Does that sound familiar?
Theorem 5.2.18 The number k of absolutely irreducible characters of G is
equal to the class number m, so the character table is a square matrix.
Proof Let G act on X = G by conjugation, with permutation character 6.
Thus 8(g) — | C G ( Ö ) | , all g 6 G, and there are m = c(G) orbits. Apply the
Orbit Formula (3.1.2) and the Second Orthogonality Relation (5.2.17) to see
that
m = (9,lc) = |Gr153{|CG(ff)|:p6G}
=
\G\~l'EUx'MX'b-1)
k
g€Gt=l
= ^2(xt,Xt) = k-l = k.
t=i
A
Corollary 5.2.19 The set Irr(C?) = {Xi>--,X*} « a & asts for
cfj?(G) of F-valued class functions on G.
^e space
Proof Irr(G) is a linearly independent set by Corollary 5.2.7, and it is clear
that the dimension of cf(G) is k = c(G).
A
Corollary 5.2.20 G is abelian if and only if every absolutely irreducible
character Xi is linear; i.e. x¿(l) = 1Proof (=>) Proposition 5.1.7.
(<=) If all xi(l) are equal to 1 then £ * = 1 * i ( l ) 2 = k = \G\ = c(G), and G
is abelian.
A
Thus it is a trivial matter to tell from its character table whether or not
a group is abelian.
The next two corollaries simply observe that the "Fourier analysis" (see
page 109) of characters extends to arbitrary class functions. The proofs are
virtually unchanged.
114
CHAPTERS.
REPRESENTATIONS
Corollary 5.2.21 J/Irr(G) = {xu---,Xk}
AND
CHARACTERS
cf(G) then
andipe
k
ip = '%2(ip,Xi)Xi1=1
Corollary 5.2.22 / / Irr(G) = {xi, •••,**} and ip, 6 G cf(G), and if tj> =
E*=i aiXi and 0 = £ * = 1 hxi, ait bi € F, then {rp,6) = £V afo. In particular
(M)=E 4 o?-
Recall that a G C is an algebraic integer if it is algebraic over Q and its
(monic) minimal polynomial m(x) over Q is in Z[x]. Thus, for example, y/2
is an algebraic integer, but v^2/2 is not.
Proposition 5.2.23 If a G C then the following are equivalent:
1. a is an algebraic integer,
2. Z[a] is a finitely generated abelian group,
3. a is a root of a monic polynomial f(x) G Z[x].
Proof (1) => (2) If the minimal polynomial of a is
m(x) =b0 + bix + --- + xk,
bi£Z,
then
ak = -bo-1-bi-a
¿>fc_i • a * - 1 G Z • 1 + • • • + Z • a * - 1 ,
and it follows easily that {1, a , . . . , a * - 1 } generates Z[a] as an abelian group.
(2) => (3) Let {/i(a), / 2 ( a ) , . . . , /m(a)} be a generating set for Z[a], with
each fi(x) G Z[x]. Set
n — max{l + deg fi(x): 1 < i < m}.
Since a " € Z[a], we may write a " = YJïLi bifi{a),
bi G Z. Set
m
/(x) = x" - 5 > / ¿ ( x ) G Z[x].
i=l
Then / ( x ) is monic and / ( a ) = 0.
(3) =>• (1) Let m(x) G Q[x] be the minimal polynomial of a.
/(x) = m(x)g(x) for some g(x) G Q[x]. Write
Then
m(x) = (^)/i(x) and </(x) = (-)fc(x),
with a,b,c,d G Z and h{x),k(x) primitive polynomials in Z[x]. We may
assume that both have positive leading coefficients. By Gauss's Lemma
h(x)k(x) is also primitive. But then, since bd ■ f(x) = ac ■ h(x)k(x), we
have bd = ac, and hence /(x) = h(x)k(x) in Z[x]. As a result h(x) and k(x)
are monic, and hence m(x) = h(x) G Z[x].
A
5.2.
115
CHARACTERS
Corollary 5.2.24 If a,ß Ç.C are algebraic integers then a ± ß and aß are
algebraic integers. In particular, the set of all algebraic integers is a subring
ofC.
Proof Let {a\,..., a m } and {ß\ ,...,ßm}
be generating sets for Z[a] and
Z[/3], respectively, as abelian groups. Then {ai,ßj} is a finite generating set
for Z[a,/?], and Z[a ± ß] and Z[a/?] are subgroups of Z[a,ß], so they are
finitely generated.
A
Corollary 5.2.25 If a € Q, and a is an algebraic integer, then a € Z.
Proof The minimal polynomial over Q of a is m ( i ) = x—a, and m(x) € Z[x],
so a G Z.
A
Corollary 5.2.26 / / x w an F-character of G, and x € G, then x(x) is an
algebraic integer.
Proof Say that | i | = m and x(l) = n. See the proof of Proposition 5.2.5 to
see that \(x) = Ci H
•" Cm where each C» G C is a root of xm — 1, monic in
Z[x\. Thus x{z) is an algebraic integer by Proposition 5.2.23 and Corollary
5.2.24.
A
Recall from Proposition 5.2.15 that we defined
_
\Kj\xijKj)
where Xi € Irr(G) and Kj is a conjugacy class.
Proposition 5.2.27 Each u)ij is an algebraic integer.
Proof Fix i and j ; let t; be the column vector (w ¿ i,... ,uJik)1, and let A^ be
the A; x A; matrix
Ujii
■■■
Ujkl
■ ■ ■
Ujik
Ujkk
Then Nv = UijV by Proposition 5.2.16, so wy is an eigenvalue of N. Thus
Uij is a root of the characteristic polynomial of A^, and hence is an algebraic
integer by Proposition 5.2.23, since N has entries in Z.
A
Theorem 5.2.28 The degrees X¿(1) of the absolutely irreducible characters
of G are all divisors of\G\.
116
CHAPTER 5. REPRESENTATIONS
AND
CHARACTERS
Proof For each i we have
*(i)
-
xi(iyx,,x,)
= E i ^f ii »(ifj- i )=x;^(^- 1 ).
which is an algebraic integer by Corollary 5.2.26 and Proposition 5.2.27. Thus
|G|/x«(l) € Z by Corollary 5.2.25.
A
Remark
FCC.
For the next few paragraphs it is sufficient to assume only that
If X = XT define the kernel of x to be ker(x) = ker(T); it is well defined
by Theorem 5.2.12.
Suppose that H < G and T is an F-representation of G/H on V with
character x- Write 77: G -> G/H for the usual quotient map. Then the
composite map G A G/H -> GL(V) determines a representation T' of G,
viz. T'(x) = T(xH) for all x € G. Clearly H < ker(T'), and if x' is the
character of V then x'fa) = XO^ÉOI all x € G. We say that T and x have
been ft/ted from G/H to G.
Conversely, if T" is an F-representation of G, with character x' and with
H < ker(T'), then we determine a representation T of G/H via T(xH) =
T'(x), well defined since H < ker(T'), and again x'(x) = X(XH), all x 6 G.
It is clear in both cases that x ' s irreducible if and only if x' is irreducible.
Thus there is a 1-1 correspondence between the set of F-characters of
G having H in their kernels and the set of all F-characters of G/H. It is
customary to write x' = X a n d let the context determine whether it is a
character of G or of G/H.
Proposition 5.2.29 / / x is an F-character of G then
ker(x) = { x € G : x ( x ) = x(l)}Proof Say that x = XT- If x € ker(x) then T(x) = 1, so x(z) = deg(T) =
X(l). Suppose conversely that x(^) = x(l)- As in the proof of Corollary
5.2.26 we have x(a;) = Ci +
r Cn, n = x ( l ) , where each Ç« G C is a root of
unity. Thus
x(x) = x(i) = ICi + ■ • ■ + Cnl = ICil + • • • + ICnl;
i.e. equality holds in the triangle inequality in C. Thus each £» is a positive
multiple of Ci, and hence Çt = Ci since all have absolute value 1. But then
n = nCi, so Ci = 1, f(x) = I, and so x e ker(T) = ker(x).
A
Remark It follows from the proof above that |x(a;)| < x(l) f° r all x e G.
117
5.2. CHARACTERS
Exercises
1. Calculate the character tables of the alternating group A4 and the dihedral group £>4 of order 8. (Hint A4 has a normal subgroup K of
order 4. Obtain three linear characters for A\¡K as in Example 2, page
99, and lift them to linear characters of A4 by composition with the
quotient map A4 -* A4/K. Similar remarks apply to D4, except that
the quotient is Klein's 4-group.)
2. Find the character table of Q2, the quaternion group of order 8, and
compare it with the table of D4.
3. Find the character table of the dihedral group
A , = (a, b I o 5 = b2 = l, ab = 6a _ 1 ).
(Hint You will need the quadratic irrational a = cos 2n/5, which is
equal to ( - 1 + v / 5)/4. To see this use the fact that Ç = e2ni/5 is a root
°f
1
1
x4 + x3 +x2 + x + l = x2(x2 + x + l + - + -~)1 = 0.
X
X
Make the substitution y = x + 1/x, solve for y, then for x, and hence
find a explicitly.)
4. If p e Z is a prime and G is a nonabelian group of order p 3 show that
G must have p2 linear characters and p — 1 that are nonlinear (and
irreducible), each of degree p. Conclude that c(G) — p2 + p — 1 (see
page 64).
Resume at this point the standing assumption that F C C
is a s p l i t t i n g field for G, w i t h c o r r e s p o n d i n g n o t a t i o n for
Irr(G), conjugacy classes, etc.
For each Xi G Irr(G) set Ni — ker(x¿), so Ni<G. Note that by Proposition
5.2.29 the subgroups Ni can be obtained (as unions of conjugacy classes) from
the character table.
P r o p o s i t i o n 5.2.30 If x is any F-character of G write x = Ylin'XiThen
ker(x) = n {Nf. n¡ > O}, the intersection of the kernels of the constituents
ofxP r o o f Say that x = XT, and write J = {i: 1 < i < k and n¿ > 0}. Since
T ~ ®{mTi:i € J} it is clear that T(x) = 1 if and only if Ti(x) - 1 for all
i e J; hence that ker(x) = n { N i : m > 0}.
A
118
CHAPTER 5. REPRESENTATIONS AND
CHARACTERS
Corollary 5.2.31 n{7Vj: 1 < i < k} = 1.
Proof The right regular representation R is faithful, so ker(ß) = ker(p) = 1.
But also p = EiX»(l)Xi by Proposition 5.2.13, so n< = x>(l) > 0, all t, and
hence 1 = ker(/9) = n{N¿: 1 < i < k}.
A
Corollary 5.2.32 If H <G then H = n{JV¿: H < Ni}.
Proof IIT(G/H) consists of those x» € Irr(G) for which H < Nt = ker(xi).
By the preceding corollary the intersection of their kernels is the identity
element of G/H, i.e. is H.
A
Corollary 5.2.33 G is simple if and only if Ni = 1 for 2 < i < k.
Note that by the corollaries above all normal subgroups of G can be determined from the character table, and in particular it can be determined
whether or not G is simple.
Proposition 5.2.34 The set of linear characters of G is just lrr(G/G'),
where G' is the derived group, or commutator subgroup, of G. Furthermore
G' = n{Ar¿:x«(l) = 1}, so G' is determined by the character table.
Proof Since G/G' is abelian each x» € Irr(G/G') is linear by Corollary
5.2.20. On the other hand, if x € Irr(G) is linear then x= G -> F* is a
homomorphism, and G' < ker(x) since F* is abelian, so x 6 Irr(G/G')- In
particular, TVj > G' if and only if x« is linear, so G' = níiV^Xiíl) = 1} by
Corollary 5.2.32.
A
To illustrate some of the ideas above, we consider the group G whose
character table appears on page 113. It has a normal subgroup H = Nz =
ker(x3) = K\ U K?, of order 4. The absolutely irreducible characters of G/H
are just Xii X2, and X3- When X4 and Xs are omitted from the table it appears
as follows.
Kx K2 K3 KA K5
1
1
1
1
1
Xi
1
1 -1 -1
1
X2
2 -1
0
2
0
X3
Observe that columns 1 and 2 are identical, as are columns 4 and 5. That
simply reflects the fact that classes K\ and K-¡, have merged to form a single
class at the level of G/H (in fact the trivial class of G/H), and similarly for
classes K\ and K5. Thus we may simply omit the repeated columns (and
relabel slightly) to obtain the table of G/H.
Xi
X2
X3
Kx K2 K3
1
1
1
1
1 -1
2 -1
0
5.2.
119
CHARACTERS
Note that it is the same as the character table of 53 derived on page 110.
The ideas in the proof of Proposition 5.2.29 suggest another useful subgroup of G associated with an F-character \- Define
Z(x) =
{x€G:\x(x)\=x(l)}-
Clearly N(x) Ç Z(x). A slight variation on the proof of Proposition 5.2.29
shows easily that Z(x) consists precisely of those elements x of G that are
represented by scalar multiples of the identity transformation for any representation T whose character is x- Two immediate consequences of that
description are recorded in the next proposition.
Proposition 5.2.35 / / x is an F-character of G then Z(x) <G.
faithful then Z(x) < Z(G).
If x *s
For each Xi € Irr(G) write Z{ = Z(xi).
Exercise
For each Xi € Irr(G) show that
conclude that Z¿ = Z(G).
ZÍ/NÍ
= Z{G/Ni).
If Xi is faithful
Proposition 5.2.36 Z(G) = n{Zj: 1 < i < k}.
Proof Since Z(G)Ni/Ni < Z(G/Ni) = Zi/Ni (by the exercise above) we
have Z(G) < Z¿, all i, and so Z(G) < CljZj. Conversely, if x 6 Zu all
i, then xNi Ç. Z(G/Ni) (also by the exercise), so for any y G G we have
[xNi,yNt] = [x,y]Ni = N{ (commutator in G ¡Ni). Thus [x,y] € n ¿ ^ = 1
(Corollary 5.2.31), and hence x € Z(G).
A
Thus we see that Z{G) can be determined from the character table, and
consequently the ascending central series can be determined, using first the
character table of G, then that of G/Z{G), and so on. In particular, it can
be determined whether or not G is nilpotent.
Note that there is another easy way to determine Z(G) from the character
table, since x e Z(G) if and only if |cl(z)| = 1, and the size of each class
is determined by the inner product of its column with itself (see the remark
following the Second Orthogonality Relation (5.2.17)).
Proposition 5.2.37 Suppose that x«(l) a " d \Kj\ are relatively prime for
some i,j. Then either Xi(Kj) = 0 or else \xi(Kj)\ = Xi(l)> , n which case
Kj Ç Z{.
Proof Choose a,b € Z for which axi(l) + b\K¡\ = 1. Multiply through by
Xi(*0)/X»(l) to obtain
aXi{Kj) +
-xlirT
- -5SÖT'
120
CHAPTER 5. REPRESENTATIONS
AND
CHARACTERS
or
aXi(Kj)
+ lK,ij =
^ - .
Thus Xt(.Kj)/Xi(i) is an algebraic integer by Corollary 5.2.26 and Proposition
5.2.27, and \XÍ(KJ)/XÍ(1)\
< 1, as has been observed earlier (page 116). If the
elements of Kj are of order n let C € C be a primitive nth root of unity, and
let G be the Galois group of Q(Ç) over Q. Since Xt(Kj) is a sum of powers of
C the same is true of a(xi(Kj)) for all a G G- Thus also <r(x«(-Kj)/Xt(l)) is
an algebraic integer and \o{Xi(Kj)IX\W)\
^ 1- Set
■-n{-(»-«4
Then a is an algebraic integer, | Q | < 1, and, since it is clearly fixed by all
a € G, a is in Q. But then a € Z by Corollary 5.2.25, so a = 0 or ± 1 . If
a = 0 then Xi{Kj) — 0> whereas if a = ± 1 then every factor of a must have
absolute value 1. In particular | x ¿ ( ^ ) | = X»(l), a n ^ hence Kj Ç Z{.
A
Theorem 5.2.38 (Burnside) If G is a nonabelian simple group then the
only conjugacy class of G that has cardinality a prime power is K\ = {1}.
Proof Suppose that p is a prime, \K¡\ = pm, and Kj jt Kx. Note that
m > 1, since Z(G) = 1. Relabel the nonprincipal irreducible characters so
that p j x»(l) if i < fco, but p | Xi(l) if i > ^o- Since G is simple Zi = 1 for all
i > 1. By Proposition 5.2.37 it follows that XiiKj) = 0 for 2 < i < fc0. Say
that x«(l) = P^i f° r e a c h ' > ^0- Apply the Second Orthogonality Relation
(5.2.17) to columns j and 1 of the character table to obtain
0 = 1 +pJ2{mXi(Kj):ko
<i<k)
(in particular fco < fc, for otherwise 0 = l!). But then
Q
= H mXi(Kj) = -i>k0
F
is an algebraic integer, contradicting Corollary 5.2.25.
A
The final theorem in this section is our first powerful application of character theoretical methods to finite group theory. It was proved by Burnside in
1904, and the first "group theoretical" (i.e. not character theoretical) proof
(for odd primes) was given by D. Goldschmidt ([26]) in 1970 (see also [3]).
Theorem 5.2.39 (Burnside's p a q 6 -Theorem) Ifp and q are primes in N
and G is a group of order paqb then G is solvable.
5.3. CONTRAGREDIENTS
AND
PRODUCTS
121
Proof We may assume p ^ q since p-groups are solvable. If G is a minimal
counterexample then it must be simple, for otherwise it would have either
a smaller normal subgroup or quotient group with the same property. Let
P be a Sylow p-subgroup of G, choose x jt 1 in Z(P), and write K for the
conjugacy class (in G) of x. Then P < CG(X), SO \K\ = [G: CG(X)] is a divisor
of [G: P] = qb, and \K\ is a prime power, contradicting Theorem 5.2.38
A
Corollary 5.2.40 The order of a finite nonabelian simple group has at least
three distinct prime divisors.
5.3
Contragredients and Products
In this section we may assume once again that F is an arbitrary field and
impose further conditions as they are needed.
If V is an F-vector space recall (e.g. see [34]) that the dual space is
defined to be V* = Homf(V, F ) , the space of linear functionals from V to
F. If {v\,... ,v„} is a basis for V then V* has a corresponding dual basis
{v*,..., u*}, characterized by V*(VJ) = Sy for all i and j .
If T: V -¥ V is a linear transformation define its adjoint transformation
T*: V* -► V* via T*(<p) = tp o T (composition of functions) for all <p 6 V*.
If 5 : V -> V is another linear transformation then (TS)*tp = ¡p o (TS) =
(tp o T) o S = (T*ip) oS = S*{T*<p) = {S*T*)<p, so (TS)* = S*T*.
Suppose that T is represented by a matrix A = [ay] relative to a given
basis {vi,..., v„}, i.e. TVJ = £ ^ aijVi, all j . Then
T X : W j •-* vf(Tvj)
= vfiY^aijVi)
= ^ ° y w * ( t ' < ) = °'i
i
i
for each I and all j . If we define ¡pi = 52m im m e ^ * > t n e n w( w j) =
E m °im«m(wj) = u í j , e a c n ' a x i á a » Í- I<; follows that T*uf = <¿>t = J^ m o/ m i; m ,
and hence the matrix that represents T* relative to the dual basis is A*, the
transpose of A. In particular tr(T*) = tr(T).
Now suppose that T is an F-representation of G on V. Define the contragredient representation T* on V* via T*(x) = T(x~1)* for all x eG. Observe
that T*(xy) = Tiy-1!-1)*
= [ r f a " 1 ) ^ * " 1 ) ] * = T*(x)T*(y), so T* really is
a representation of G. Note also that deg(T*) = deg(T).
a
v
We summarize the above information in the next proposition.
Proposition 5.3.1 IfT is an F-represention of G onV then its contragredient T* is an F-representation ofGjon the dual space V*. Relative to a basis
and its dual basis we have T*(x) = T ^ x - 1 ) ' , all x € G. If x is the character
ofT then the character x* ofT* is given by x*(x) = x ( ^ - 1 ) Corollary 5.3.2 / / F C C and x is an F-character then the complex conjugate x is also an F-character.
122
CHAPTER 5. REPRESENTATIONS
Proof By Proposition 5.2.5 x
IS
AND
CHARACTERS
the character contragredient to x-
&
Proposition 5.3.3 / / F C C and if x, i> are F-characters of G then
(x*,V»*) = (x,i&).
Proof This is a simple calculation.
A
Corollary 5.3.4 / / F C C then x* »* absolutely irreducible if and only if x
is absolutely irreducible. In particular, the complex conjugate of any row in
the character table of G must also be a row in the table.
Suppose that <p is a linear F-character of G and that T is an F-representation, with character x- Then the pointwise product ipT, defined by
(ipT)(x) = tp(x)T(x), is also an F-representation, since tp(x) € F for all x £
G. The character of ipT is clearly just <pXi the pointwise product of functions
from G to F . If F Ç C an easy calculation shows that (fX,fX) — (x>x)> s o
<PX is absolutely irreducible if and only if x is.
In particular, if ip, 8 are two linear F-characters of G then ip9 is also a
linear F-character. If ip* is the contragredient of ip then tp*ip = \Q, the
principal character.
The next proposition should now be clear.
Proposition 5.3.5 The set of all linear F-characters of G is an abelian
group under pointwise multiplication; it acts via multiplication as a permutation group on the set of all F-characters.
Suppose now that S and T are F-representations of G on V and W, with
respective characters x and ip. Then S(x) <g> T(x) is a linear transformation
on V ® W for each x € G, with
(S{x) (8) T(x))(%2 Vi ® Wi) = ^2 S(x)vi ®
i
T(x)Wi.
i
It is easy to check that x i-> S(x) <8> T(x) is an F-representation of G on
V ® W; we denote it by S ® T.
If {vi}, {wj} are bases for V and W then {u< g> Wj} is a basis for V <8> W,
so deg(5 ® T) = deg(S) ■ deg(T). Relative to the bases mentioned (with a
natural ordering of {u< ® uij }) the matrix representations are related by
S®T(x)
=
[Sij(x)f(x)}
in block form, where S(x) = [sij(x)]. Consequently the character of S ® T is
Xip, pointwise product of functions.
5.3. CONTRAGREDIENTS
AND
PRODUCTS
123
Again we summarize.
Proposition 5.3.6 If x andip are F-characters of G then the product x^ **
also an F-character.
Remark If x and ip are nonlinear irreducible characters then xV> is very often
reducible. For example, if F Ç C, x is absolutely irreducible but not linear,
and t/> = x*, then (xx*. I G ) = (x,X • I G ) = (x,x) = 1, so 1 G is a constituent
of XX* a n d XX* cannot be irreducible.
Example
We now have sufficient techniques available to derive quite easily the character table that first appeared on page 113. It is in fact the table of the symmetric group 5 4 . The classes are Kx = {1}, K2 = cl(12)(34), K3 = cl(123),
K4 = cl(1234), and K$ = cl(12). Since S4' = A4 there are just two linear
characters, the principal character xi = I G and the alternating character X2
(see page 99). Thus 22 = £ 3 X¿(1)2> forcing the remaining degrees to be 2,
3, and 3. The permutation character 6 of the action on {1,2,3,4} has xi
as a constituent and gives X4 = 0 — Xi> t n e n Xs = X2 • X4- Finally X4 has
degree 9, and we see by taking inner products that it has Xi> X4> and xs as
constituents, each with multiplicity 1. Subtracting them from xi yields X3Although it won't reappear until late in the next chapter, this seems to
be the appropriate section in which to introduce the next concept, that of the
character ring of G.
If F Ç C is a splitting field for G define Char(G) to be the set of all
Z-linear combinations of the characters in Irr(G):
k
Char(G) = { ^ n ¿ x ¿ : n ¿ € Z } .
i=l
It follows from Proposition 5.3.6 (and from Maschke's Theorem (5.1.1)) that
Char(G) is a ring (with pointwise multiplication and addition), a subring of
the ring of all functions from G to F. Elements of Char(G) are sometimes
called generalized characters, or virtual characters. We call Char(G) the character ring of G; it is variously referred to in the literature as the ring of generalized characters, the ring of virtual characters, or even the Grothendieck
ring of G.
Suppose that tp = YliniXi G Char(G), and set ip = 5Z{n»X«:"i > 0}
and 9 = £ ) { - n « X i : n t < 0}. Note that ip and 8 are both characters of G
(provided that they are nonzero), and <p = ip -6. Thus Char(G) can also be
characterized as the set of differences of F-characters of G.
Suppose that G is the (internal) direct product of its subgroups H and
K, that 5 is an F-representation of H with character (p, and that T is an
124
CHAPTER 5. REPRESENTATIONS
AND
CHARACTERS
F-representation of K with character %¡>. We may extend S and T to representations S and f of all of G by setting
S{hk) = S{h)
and
f(hk)
= T(k)
for all h € if, fc € K. The resulting characters of G are given by
tp(hk) = tp(h)
and
V>(hk) = *P(k)-
By Proposition 5.3.6 the product ¡pip is an F-character of G. Denote it by
<p x xj;; thus for each hk € G we have (i¿> x ip)(hk) = (p(h)ijj(k).
Proposition 5.3.7 Suppose that G is the direct product of its subgroups H
and K, F Ç C is a splitting field for both H and K, Irr(H) = {tpi,..., ¥?»•},
and Irr(AT) = {i¡>i,...,ip$}.
Then F is also a splitting field for G, and
Irr(G) = fat x ipy. 1 < i < r, 1 < j < s}.
Proof An easy calculation shows that
(if i X V'i.Vm X *l>n)G = (<Pi, VmJi/ÍV'j, Í>n)l( =
Sim6jn,
so each <¿>< x ipj is absolutely irreducible and they are all distinct from one
another. Since the sum of the squares of their degrees is $ 3 ¿ } <¿>¿(1) V j ( l ) 2 =
\H\\K\ = \G\, they constitute all of Irr(G).
'
A
Corollary 5.3.8 If G is a finite abelian group and F Ç C is a splitting field
for G then the group Irr(G) of linear F-characters of G is isomorphic with
G.
Proof It is sufficient to prove this in the case that G = (x) is cyclic, say
of order m. In that case there must be a primitive mth root C °f unity in
F, and if we define x(x) — Ci * n e n X *s a linear character. Its powers x'i
0 < i < m - 1, are distinct linear characters, so they are all of Irr(G).
A
Corollary 5.3.9 If F ÇC is a splitting field for G then the group of linear
F-characters of G is isomorphic with G/G'.
Proof Proposition 5.2.34 and the corollary above.
A
We close this section, and the chapter, with a theorem of Burnside about
powers of a faithful character.
Theorem 5.3.10 Suppose that F Ç C is a splitting field for G and ip is
a faithful (but not necessarily irreducible) F-character of G that takes on
exactly m distinct values oi = ip(l),a2,... ,am € F. Then every x € Irr(G)
1Q,ip,ip2,...,^)m_1.
is a constituent of one of the powers ip° =
5.3. CONTRAGREDIENTS
AND
125
PRODUCTS
Proof (Brauer) Set Aj = {x 6 G:x¡){x) = a¿}, 1 < j < m, and observe that
Ai = {1} since ip is faithful. Let bj = Yl{x(x):x
€ Aj} for each j . Note that
m
m
W>\x) = \G\-1 ^ ^ { ^ ( x î i Ô ô i a : 6 A,-} = Id" 1 J > & .
If the theorem were false for x we would have (ipl, x) = 0 for 0 < i < m — 1,
and therefore the following system of equations.
&i
a\b\
a^h
+
+
62
0262
+
a?-%
+
+
•■• +
••• +
6m
o,mbm
+ ■■■ + cC~xbm
=
=
0
0
= 0
The coefficient matrix is a Vandermonde matrix [aj] whose determinant is
rij<i (°i — aj) Í 0- Thus all bi = 0, but that contradicts the fact that
61 = x(l) > 0!
A
The theorem is not generally very helpful for finding the character table, since it is usually not easy to decompose the characters ip* into their
irreducible constituents. The regular character p might be viewed as an extreme example; every x 6 Irr(G) is a constituent, but p can almost never be
decomposed into its constituents without substantial further information.
Exercise
Let G be the generalized quaternion group
Qm = (a, b | a2m = 1, b2 = am, ab = 6a _ 1 ).
Write the elements of G as b'aj, 0 < i < 1, 0 < j < 2m - 1.
1. Show that G' = (a2), and that G/G' is Klein's 4-group if m is even,
cyclic of order 4 if m is odd.
2. Show that c{G) = m + 3 by finding the classes explicitly - the class
sizes are 1, 1, 2, . . . , 2, m, m. (Also see the exercise on page 64.)
3. Let C = e'i/m
Tj(a) =
€ C Define
0
0
and
Tj(b) =
0
(-iy ■
1
0
J'
1 < j < m - 1. Show that Tj is a representation of G, and if Xj is i t s
character show that Xj G Irr (G), all j .
4. Write out the character table of Q.
Groups and Characters
by Larry C. Grove
Copyright © 1997 John Wiley & Sons, Inc.
Chapter 6
Induction and Restriction
In this chapter we explore the relations between characters of a group and the
characters of its subgroups. It is perhaps fair to say that most of the power
of character theory results from exploiting these relationships.
6.1
Modules
Suppose that F is a field and that G is a finite group. Denote by FG the
set of all functions from G to F. With the usual addition of functions and
multiplication by scalars it is clear that FG is an F-vector space. By a mild
abuse of notation we may view each a; € G as an element of FG - we identify
x with the function whose value is 1 at x and 0 elsewhere. Clearly then every
o € FG can be written uniquely as $ 3 I 6 G °* ' x-> w i t n a* = a(x) e F- I n
particular the elements of G are a basis for FG, and its dimension is |G|.
The basis G is closed under (group) multiplication, and that multiplication
extends naturally to a multiplication on all of FG via
( ^2 axxj ( ^2 bvV)
=
^{axbvxy:x,yeG}
=
£{(£ax&y)u:u€G}.
xy=u
Exercise
Verify that FG is a ring, in fact an F-algebra, with the multiplication
defined above. Show that the multiplication is given by
ab(x) = ^{aixz-^biz):
z G G}
for all a, b € FG, all x € G (it is sometimes called the convolution product).
127
128
CHAPTER 6. INDUCTION AND RESTRICTION
The F-algebra FG is called the group algebra of G over F.
If T is an F-representation of G on V we may extend the definition of T
to elements of FG linearly, i.e. T{J^axx) = ^axT{x).
The resulting map
T-.FG -t Homp(V, V) is easily checked to be an F-algebra homomorphism,
i.e. an F-representation of the algebra FG. The extended representation can
now be used to define a (left) FG-module action on V via
(^2axx)v =
^axT{x)v
for all t; € V.
Thus each representation yields a unitary FG-module. Conversely, a unitary FG-module V is automatically an F-vector space, and if the module
action is restricted to elements of G C FG an F-representation T is determined by T(x)v = xv, all x G G, v 6 V.
Exercises
1. Suppose that 5 and T are F-representations of G on V and W, respectively.
a. Show that S is irreducible if and only if V is a simple FG-module.
b. Show that 5 ~ T if and only if V and W are isomorphic as FGmodules.
2. View FG as a (left) FG-module via multiplication. Show that the
resulting F-representation of G is the left regular representation L.
6.2
Induction
If H < G we may view F H as a subalgebra of FG in a natural way - if
a = ^2 avV e FH
we agree that ax = 0 for all x € G \ H, hence
o = ^2 a*x € FG.
Suppose now that H < G, and that T is an F-representation of H on V,
so that V is a left Fif-module. View FG as an (FG-F#)-bimodule and set
VG = FG ®FH V.
Then VG is naturally a left FG-module, called the module induced from V.
The FG-module Va determines the induced representation TG on the vector
6.2.
129
INDUCTION
space VG, and if x is the character of T then the induced character xG is the
character of TG.
Example
Take H = 1 and T = 1H on V = F, so FH S F. Then
VG * FG
®F
F S* FG,
and the induced representation is (equivalent with) the left regular representation of G by Exercise 2 above.
Choose a (left) transversal (i.e. left coset representatives) x 1 , X 2 , . . . , x m
for H in G. If H = {huh2,...}
then
G = {xihj: 1 < i < m, 1 < j < \H\).
For any a e F G set ay = a(xjhj), so that
o = 2_,aijXihj = ¿ ^ ^ ( ¿ J a y / i y ) = )
i,j
i
j
Xjbj,
i
where 6< = J^ • a^/ij G F / Í . Thus
FG = xiFH + x2FH + ■■■ + xmFH,
a sum of right Fü-submodules. It is in fact a direct sum, for if ^li x¿6j = O
then every a^ = 0, and hence every 6¿ = 0. Thus, at least as F-vector spaces,
we have
Va
=
FG
=
(xiF/i©--©a;TOF7/)(8)F//^
®FH
V
Proposition 6.2.1 If H <G andT is an F-representation of H onV then
deg(T G ) = [G:i/]deg(T).
Proof Each XiFH above is isomorphic as a right FJ7-module with FH via
the map x¡b >-► b, all b e FH, and so
XiFH
<S>FH
V *FH
®FH
V ^ V
a
as vector spaces. Thus V is isomorphic with the direct sum of m = [G: H]
copies of V and the proposition follows.
A
Retain the notation above for the next few propositions.
Proposition 6.2.2 If {v\,...
,vn) is a basis for V then
{xi ® vy. 1 < i < m, 1 < j < n)
is a basis for
Va.
CHAPTER
130
6. INDUCTION AND
RESTRICTION
Proof There are mn conceivably different x< ® i>/s, and mn is the dimension
of VG by Proposition 6.2.1, so it will suffice to show that they span Va. If
u e Va write u = 5 3 r o r <8> u r , ar € FG, uT e V. Write o r = 5Zi x »^« 5
6 rj 6 FH, s o u = £r,¿a;«&r» ig) ur = 52P,»X» ® &r»Ur- But briur G V, so
write briUr = Yljarijvj> arij € F, and substitute to see finally that u =
Now let T be the matrix representation of H relative to the given basis
{vi,...,vn},
say T(x) = [ty(x)], x € G. Order the basis elements for V G
"along the rows," i.e. as follows:
xi ® u i , . . . , x i ®vn;x2
® « ! , . . . , i 2 ®w„; . . . ; x m ® t>i,... , x m ® vn.
Then
T G ( X ) ( X J ® u¡) = x{xj
® vj) = xxj
<g> v¡.
We may write XXJ = Xjj/ for some i and some y € H, and so
TG(X)(XJ
<8>i>i) =
=
Xjj/ ® i>( = x¿ ® yvi = Xi CS) T(y)i;i
Xj®^iri(y)ur = ^irj(y)(xi®ur)r
Note, though, that y =
X^'XXJ,
TG(x)(Xj
r
and therefore
®Vt) = ^trtix^1
XXj)(Xi ® Vr).
r
Thus there are (possibly) nonzero coefficients only for the basis elements in
"row i" ; viz. x¿ ® vi,..., x¿ <g> v n .
It j s convenient to extend the definition of T to all of G by agreeing
that f (x) = 0 if x € G \ H. (Warning The extended f is usually not a
representation of G!) With this convention the discussion above has proved
the next proposition.
Proposition 6.2.3 In block form the matrix representation of TG is given
b
y
TG(x)
x
with T(xj xxj)
=
[fix^xxj)},
the ij'th block, and with T(z) = 0 for any z&H.
We remark that there is only one nonzero block in each (block) row of
TG(x), since for given i there is only one j for which X^XXJ € H. Likewise
there is only one nonzero block in each column.
Example
Take H = 1 and T = \H. Then TG(x) has f{Xrlxxj)
= 1 in entry
ij if and only if xx¿ = x¿, and we see again that TG is the left regular
representation.
6.2.
131
INDUCTION
Assume now that F C C . Just as for T it is useful, in order to calculate
the induced character xG, to extend x t o a function \ defined on all of G to
F by agreeing that X\H = X and X\G\H = 0- Of course x tends not to be a
character of G.
Proposition 6.2.4 If F Ç C, H < G, x is an F-character of H, and
{x\,..., x m } is a left transversal for H in G, then
m
X (x) = £ x{*Tl**i) = \HI"1 EM" 1 *«): « € G}
G
«=i
/or a/f x € G.
Proof We have
XG{x)
=
tr(fô(x))=tr[f(x-Ixxj)]
=
^trfix^xxi)
m
¿=i
m
=
^x(xrlxxi).
t=i
If 2/ € i / a n d u G G, thenj/ _ 1 uj/ € ¿/ if and only if u 6 # , so x{y~l xj1 xxiy) =
x(£, 7 l x a ; «) f° r all y € i / . Every t € G can be written uniquely as í = x¿y, j / e
i i , so
^2 x ( í _ 1 x í )
=
^ { ^ " V ^ i î / ) :
=
\H\J2x(xT1xxi)
yeH,l<i<m}
t€G
=
\H\XG(x).
Exercise
Let G = S 3 and H = A3. Calculate the values of xG at (12) and (12 3)
for each x € lrr(H). Which x G are irreducible? Write out TG for one of
them.
It will prove useful to generalize the formula in Proposition 6.2.4. Recall
that the space of all class functions from G to F is denoted cf(G), or cff(G)
if we wish to emphasize the field. It is easy to check that cf(G) is in fact
a ring (hence an F-algebra), multiplication being the pointwise product of
functions.
Suppose that H <G. If ¡p 6 cf(ii) extend <p to a function ¡p:G -* F via
G
<P\G\H = 0. Then define the induced class function ip via
<pa(x) =
\H\-lY,{<P(t-1*t)--t€G}
CHAPTER
132
6. INDUCTION AND
RESTRICTION
for all x € G. Clearly <pG <E cf(G).
Just as in the proof of Proposition 6.2.4 it is straightforward to verify that
if { i i , . . . ,xm} is a left transversal for H in G then
m
i=l
a slight variation on the proof shows that if {y\,...,
then
fG(x)
ym} is a right transversal
m
= ^v'(i/« a ; y. _ 1 )»=i
Proposition 6.2.5 If H <G, <p,0 € d(H),
((p + 0)G = <pG + 6°
and
and a£ F, then
{a<p)G = a(pG.
Thus <p t-> <pG is a linear transformation from ci(H) to cf(G). Also <p°(l) =
[G:H]<p(l).
Proof This is clear.
A
Proposition 6.2.6 (Transitivity of Induction) Suppose that K < H <
G and ip e cf(ff). Then (<£>")G = <pG.
Proof IfxeG
then
(<¿>")G(*) = \H\-1
1
¿2(<p»y(t-lxt)
t€G
= i^r ¿ IÄT 1 52 vi«-1*-1**«)
teG
»€H
= i^r'iHi-^Êv«*«)-1*^))
s€Ht€G
»6Hu€G
=
\K\-
1
£
u6G
^(«-»xii) = <pG(x).
A
Remark If <¿> happens to be a character of G then Proposition 6.2.6 is a
consequence of the fact that
(VH)G
= FG ®FH (FH ®FK V) S FG ®FK V =
Va.
133
6.2. INDUCTION
Proposition 6.2.7 Suppose that H <G, <p£ d(H),
<pG-9 =
and 6 e cf(G). Then
(<p-e\Hf.
Thus the set of class functions induced from H form an ideal in the ring
cf(G).
Proof If x e G then
(<pd\Hf(x) = izrr 1 ^(r^tWIirXt- 1 *«)
t€G
=
1
l i / p ^2 <p{t-lxt)e{t~lxt)
(since <j> = 0 off H)
teG
=
| H | - x 5 2 £(t" x xt)0{x)
tec
(since 0 € cf(G) )
= ^G(x)0(x) = &Ge)(x).
A
Theorem 6.2.8 (Frobenius Reciprocity) If H < G, ¡p €
0 € cf(G), then
{<pG,0)a = (<p,0\„)H.
d(H),
and
Proof We calculate:
(<pG,0)
=
=
IGl-^^iiWx"1)
z6G
1
IGI- !//!-1^^-1^)^-1)
t
z
= IGI-MÄI-^E^»)^»"1)
t
=
¡HI'1 £
y£G
piyWiy'1)
(since <¿ = 0 off H)
y€H
=
(<P,0\H)H.
A
Corollary 6.2.9 Suppose that H<G,FCCisa
splitting field for both
H and G, ip e Irr(H), and x 6 Irr(G). Then the multiplicity of \ as a
constituent ofipG is the same as the multiplicity oftp as a constituent ofx\fí.
CHAPTER 6. INDUCTION AND RESTRICTION
134
Exercise
If A < G, A is abelian, and x 6 Irr(G), show that x(l) < [G: A].
The next proposition ([39], page 64) looks rather technical, but it can be
very useful for calculations of induced characters.
Proposition 6.2.10 Suppose that H < G, x € G, K = C1G(X), and that
<p e d(H). IfHC\K = 0 then <pa(x) = 0. // H n K ¿ 0 write H n K =
Ù^clnixi).
Then
vP(x) = \CG(x)\ ¿
1 7 ^gL
= í ^ l ¿ | d* («,)M*<).
Proof K # n K = 0 then t _ 1 xí g # for all í € G, and so v?G(x) = 0.
Suppose then that H(~\K ¿0. Fix x¿ and choose s € G so that s~1xs = Xj.
Then {/ G G: í - 1 xí = x¿} is just the coset CG(X)S (verify). Thus
<PG(X)
= in r 1 5 ; ^(*-i«t)
m
= IHI-^ICoWbíxOlclirííOI
m
= ICoí^l^^O/ICffíxOI;
¿=i
the second equality follows immediately.
A
Example
By way of illustration let us calculate the character table of the alternating
group A&. Label the classes as
* i = {l},
K2 = cl(12)(34),
K4 = cl(12345),
Jfs = cl(123),
K5 = cl(13524),
of respective sizes 1, 15, 20, 12, and 12. The only linear character is the
principal character I Q , since A'h — A$. The permutation character 6 on
{1,2,3,4,5} has 2 constituents, one of them IG, SO the other is (say) X4, with
values 4, 0, 1, - 1 , - 1 . The degrees xi(l) = 1 and *4(1) = 4 are already
sufficient to force the remaining degrees to be 3, 3, and 5 by Theorem 5.2.14.
Now let H = At < A5, of index 5. Let (p be the character of H (see Exercise
1 on page 117) with values
tpQ) = ¥>(12)(34) = 1,
ip(123)=u,
¥>(132) = u;2,
6.2.
135
INDUCTION
where w 2 + w + l = O i n C Then <¿>G(1) = 5. If x = (12) (34) € K2 then
K2nH
= c\H(x), |cl t f (z)| = 3, and, by Proposition 6.2.10, <pG(K2) = 1.
Next set x = (123) € K3. Then
HnK3
= el*(123) u cl*(132),
with | el/,(123)| = | cl H (132)| = 4, so
p 0 (ür 3 ) = (5/20) (4u;2 + 4w) = - 1 .
Since AT4 n H = # 5 n H = 0 we have « ^ ( i ^ ) = <PG{K5) = 0. Check that
(<pG,<pG) = 1 and set Xs = <PGNext set H = (CT) of index 12, where a is the 5-cycle (12345). Take
C = e2*i/5 € C and define ip(a) = Ç. Note that
H D K2 = H D K3 = 0 ,
iín/C^K^
- 1
},
and H n A r 5 = {ír 2 ,í7 3 }.
By Proposition 6.2.10
TPG{K4)
—
2ir
= Ç + Ç = 2 cos —
5
=
2a,
V>G(AT5) = C2 + C3 = 4a 2 - 2 = - 2 a - 1,
where a = ( - 1 + %/5)/4 (see Exercise 3 on page 117). Thus rpG has values 12,
Check that (V>G,*4) = W>G,Xs) = 1, set * 2 = VG - * 4 -Xs,
0, 0, 2a, -2a-1.
and check that (X2,X2) = 1- The final character X3 is now easily obtained by
the Second Orthogonality Relation, and the table is as follows.
Ki
1
Xi
3
X2
X3 3
4
Xi
X5 5
K2 K3
Ki
Kt
1
1
1
1
-1
0 2a + 1
-2a
-1
2a+ 1
-2a
0
0
1
-1
-1
1 -1
0
0
- l + >/5
" = — 4 —
Note that it is apparent from the table that A5 is a simple group.
Proposition 6.2.11 Suppose that G acts transitively on a set 5 , s € 5 , and
H = Gs = Stabö(s). Then (1//) G is the permutation character 6 of G on S.
Proof If x 6 G recall that Fix(i) denotes the set of fixed points of x in 5.
Now {lH)G(x) = \H\~l Y,t€G Ífí(txt-1).
But txt-1 € H = G, if and only
if s i x t
= s if and only if (s') 1 = s', which is if and only if s1 G Fix(x).
CHAPTER 6. INDUCTION AND RESTRICTION
136
However, if ii, ij 6 G, then stl = sh if and only if t\t2 1 G G, = H, or
ii € Ht2, so each fixed point s', and hence each txt~x G H, occurs with
multiplicity \H\. Thus (l ff ) G (x) = | Fix(x)| = 6(x).
A
Recall (page 16) If G acts transitively on a set S, with |5| > 1, and s € S,
then G, = Stabo(s) also acts on 5. The number r (> 2) of G«-orbits in S is
called the rank of G on 5. Note, for example, that r = 2 if and only if G is
doubly transitive on S.
Theorem 6.2.12 (Burnside) Suppose that G acts transitively on S, with
permutation character 6. Then the rank of G on S is r = (6,6).
Proof Set H = G, and note that 6\H is the permutation character of H
on S. Apply the Orbit Formula (Proposition 3.1.2), Frobenius Reciprocity
(Theorem 6.2.8), and Proposition 6.2.11 to see that
r = (6\H,1H)H
= (6,(1H)G)G
= (6,6).
A
Corollary 6.2.13 If G is transitive on S then it is doubly transitive if and
only if6 = la + x for some \ 6 Irr(G).
Proof As observed above G is doubly transitive if and only if it has rank 2,
hence if and only if (6,6) = 2, which means it has exactly 2 constituents, one
of which is IG by the Orbit Formula (3.1.2).
A
Corollary 6.2.14 If n > 2 then the symmetric group Sn has an absolutely
irreducible character of degree n — 1, as does the alternating group An ifn> 4.
6.3
Normal Subgroups and Clifford Theory
Suppose that H < G and that T is an F-representation of H. If x G G define
Tx via Tx(y) = T(xy) = T(xyx~1), all y e H. It is easy to check that Tx
is also an F-representation of H - it is called the conjugate of T by x. If \
is the character of T then the character of Tx is the conjugate character \ x ,
with x*(y) = X(xy), all y 6 H. Note that if x,z £ G then (xx)z = x"More generally, given any ip G d(H) and x € G, define the conjugate ipx
via <px(y) = (p(xy), and check that tp" G d(H), (<px)z = ipxz, and ip* = ip if
xeH.
UFÇC,ip,6e
cîF(H), and x G G then
(ÍPX,6X)
=
=
\H\-lJ2fx(y)6x(y-1)
veH
1
Itfl" Y,
rtxyx-'Wxy-'x-1)
= (ip,6).
6.3. NORMAL SUBGROUPS AND CLIFFORD
THEORY
137
In particular, if x € Irr(if) then also xx £ Irr(if).
We summarize.
Proposition 6.3.1 If H <G then G acts by conjugation as a permutation
group on ci(H) and on the set of F-characters of H. If F Ç C is a splitting
field for H then G acts by conjugation on ITT(H) .
Note that H is in the kernel of the group actions mentioned in the proposition; whenever it is convenient we may view G/H as the group that acts.
If H <G and ip is an F-character of H set IG(IP) = Stabe (V0 and call
it the inertia group of ip in G. Thus ipx = ip if and only if x e IGW- If
t = [G: IG{IP)} then | Orbc;(t/0| = t and ip has t distinct G-conjugates. In fact,
if {x\,..., xt} is a (right) transversal for IG (ip) in G then tp = ipXi,..., xpXt
are the distinct conjugates of tp.
Exercise
Suppose that H <G and tp € cf(G), with %p\¡j = tp G Irr(i/). Show that
<pG\H = [G:H)v, <pG\G\H = 0, and (vG,<pG) = [G:H].
Proposition 6.3.2 Suppose that H < G, ip is an irreducible F-character of
H, and { x i , . . .,xt} is a transversal for IQ {ip) in G. Then
(ipG)\H = [lG(ip):H]J2ri.
t=l
Proof Take y 6 H. Then
iPG(y) = \H\-iY,M*yx-1) = \n\-1J2^x(y)
XÇ.G
(
i€<3
= i/fn/owoiE^'iv).
A
Theorem 6.3.3 (Clifford) Suppose that H <G, F Ç C is a splitting field
for both G and H, x € Irr(G), and ip € lrr(H) is a constituent of X\H- Then
t
Xi
x\H=e^ip ,
i=l
where {x\,...,xt}
is a transversal for IG(IP)
multiplicity of ip as a constituent of X\H-
in G and e = (X\H,IP),
the
CHAPTER
138
6. INDUCTION
AND
RESTRICTION
Proof Set e = (x|/f,V0> and n o t e t n a t e = (Xii>G) by Frobenius Reciprocity (6.2.8). By Proposition 6.3.2 {I¡)G)\H has the characters ipXi as its
constituents, each with multiplicity [IG(II>):H]. If </? is any other irreducible
F-character of H then ( E i ^ S v ) = 0, hence 0 = ((II>G)\H,<P) = W»G,¥>G)
by Probenius Reciprocity (6.2.8). But x w a constituent ofipG, therefore not
of <¿>G, so ( X J V 0 ) = (x|tf>¥>) = 0- Thus the only possible constituents of X\H
are the various ipXi, and we have X\H = ^»(xlffiV'*')^' 1 '- B u t ( x ^ V ^ ) =
(xXi\H,ipXi) = ( x * 1 ! * . ^ * ) = ((Xlff)*',^*') = (x|//,V>) = e for all i, and the
proof is complete.
A
Corollary 6.3.4 / / 1 # is a constituent of X\H then H < ker(x).
Proof Clearly {lH)x = 1 * for all i E G, so / c ( l t f ) = G, t = 1, and
X\H = elH. But then e = x ( l ) , so xlff = x(l)ltf and x(î/) = x ( l ) for all
y € H, hence if < ker(x) by Proposition 5.2.29.
A
Remark The common multiplicity e of the conjugates of ip in Clifford's theorem is called the ramification index of X\HExercise
Suppose that H<G, \ e Irr(G), and 0 €
IIT(H),
with (x|ff,0) # 0. Show
that 0(1) I x(l).
Proposition 6.3.5 Suppose that \ € Irr(G) and Z = Z{G). Then \\z
X(l)<p, where <p G Irr(Z) »s linear, and if z E Z, x € G, then xizx)
<P(z)x(x)-
=
=
Proof Let ip be an absolutely irreducible constituent of x\z- Then tpx = tp
for all i € G, so 7G(V) = G, í = 1, and xlz = e<iP- But <¿?(1) = 1, so e = x(l)If T is a representation of G with character x then
T(z) =
0
<p(z)
all * € Z. Thus if a; G G then
Xizx) = tx(f(z)f(x))
= tv(V(z)f(x))
= <p(z)x(x).
6.3. NORMAL SUBGROUPS AND CLIFFORD THEORY
139
Exercise
Suppose that p € Z is a prime, G is a p-group with center Z, [G: Z] = p 3 ,
and x € Irr(G).
1. Show that \\z = e\, where e = x ( l ) , A e Irr(Z), and A(l) = 1.
2. Show that (x,A G ) = e.
3. Conclude that x ( l ) 2 < P34. Conclude that x ( l ) = 1 or p.
For some examples of groups as in the exercise take G to have four generators a, b, c and d, subject to defining relations
ar = V = <f = iF = (a, 6) = (a,c) = (a,d) = (b,c) = l,(fc,d) = a,(c,d) = b,
where p is an odd prime.
Recall If Xi € Irr(G) then
Zi = Z{xi) = {x € G: |x¿(x)| =
Xi(l)}
> Nt = ker(xi).
Theorem 6.3.6 If Xi e Irr(G) <Aen x«(l) | [G:Z¿].
Proof (Isaacs) We may view x« as a character of G/Ni, and hence we may
assume that Xt is faithful. In that case Zi = Z(G) = Z (see the exercise on
page 119). Define an equivalence relation R on G via xRy if and only if there
is some z g Z such that x ~ zy (~ denoting conjugacy in G). By Proposition
6.3.5 Xi(zv) = vMXtiy) f° r some <p € Irr(Z) (note that <p is faithful since
Xi is faithful). If xRy then |x»(x)| = |Xi(*î/)l = M*)Xt(î/)| = \Xi(y)\ since
|<£(2)| = 1, i.e. |Xt| is constant on the equivalence classes in G induced by
R. Let C\,...,CT be the iî-equivalence classes on which Xi ¥" 0> and choose
representatives x¿ € Cj. Note that x € Cj if and only if x ~ ZXJ for some
2 € Z, i.e. 2 - 1 x = j / € CIG(XJ), or x € 2C1G(X¿) Ç Z C I G ( X J ) . TO repeat x € Cj if and only if x = zy, with 2 G Z and y € C1G(X¿). If zx, z2 € Z and
î/i, y2 e CIG(XJ), and if zxyi = z2y2, then <p(.zi)xi(¡/i) = v( 2 2)x¿(í/2), and
Xi(yi) = Xi(í/2) ¿ 0, since t/i ~ y2 ~ Xj, so <¿>(2i) = ^(22). But then zx = z2
since <p is faithful; hence also j/i = î/2, and the representation oí x as zy is
unique. Thus \Cj\ - \Z\\ C1G(XJ)|. Label the classes in G so that Kj = CIG(XJ)
for all j . Then
|G| = |G|(Xi,Xi) = E l ^ ( * ) l 2 = ¿I^IIXi(*¿)| 2
xeG
j=i
r
r
i=\
¿=1
= \Z\ £ \Kj\xi{Kj)x~&j) = \Z\Xi{l) £"ijXiixJ 1 )
140
CHAPTER 6. INDUCTION AND
RESTRICTION
(see Proposition 5.2.15 for the definition of ufy). It follows from Corollary
5.2.26 and Proposition 5.2.27 that [G:Z]/x<(l) = T,jUJnXi(xJ1)
is both an
algebraic integer and a rational number, and hence is a rational integer by
Corollary 5.2.25.
A
Corollary 6.3.7 (Schur's T h e o r e m ) If\i
of[G:Z{G)}.
€ Irr(G) then Xi(l) »'* a divisor
Proof This is clear from the theorem, since Z(G) < Z{.
A
Theorem 6.3.8 Suppose that H<G, 9 6 Irr(H), and L = IG{9).
1. Ifrpe Irr(L) is a constituent of 8L then tpG € Irr(G) and {ip°,0G)
=
2. The map V i-> ipG is a bijection between the set of absolutely irreducible
constituents of 8L and the set of absolutely irreducible constituents of 0G.
Proof (1) Choose x 6 Irr(G), a constituent of ipG. Then 8 is a constituent
of ¡P\H and V is a constituent of xl¿, so 0 is a constituent of X|JÏ- Say
that [G: L] = t and let öi = 8,..., 6t be the distinct G-conjugates of 6, so
X\H = e ¿ - = 1 OÍ for some e > 0 in Z by Clifford's Theorem (6.3.3). Also,
since L = IG(&) = IL(&), we have X¡>\H = e'6 for some e' € Z, and since ip is
a constituent of X\L we have e' < e. So
X(l) = et6(l) < ipG(l) = iV(l) = e't9(l) < et0(l),
and all are equal. Thus x = i>G, and furthermore
WG,0G) = (X,e°) = (X\H,6) = e = e' = (rP\H,d).
(2) Let x G Irr(G) be a constituent of 8G, and let V G Irr(L) be a
constituent of X\L- Since xltf = X\L\H has 0 as a constituent we may
choose %p so that 8 is a constituent of ip\¡j. Thus ^ is a constituent of
8L, and x = ipG by part 1. To prove the 1-1-ness suppose that Vi> i>i
in Irr(L) are distinct constituents of 9L but that ipG = ipG = x- Then
(X\H,8) = (X\L\H,8)
> ((tpi+tp2)\H,8) > (ipi\H,8), which by part 1 is equal
to (4>G,8G) = (x,9°) = {X\H,0), a contradiction.
A
Theorem 6.3.9 (Ito, 1951) If A <G, A is abelian, and x € Irr(G), then
X(1)\[G:A].
Proof Choose 8 6 Irr(.4) to be a constituent of x U , and set L = IG{8). By
Theorem 6.3.8 x = ^G for some ip 6 Irr(L), and VU = e ö . w i t h e = V»(l)
since 8 is linear. If a € A then \ip(a)\ = e\8(a)\ =e = V(l), so A < Z(ip). Also
V(l) | [L:Z{t¡>)] by Theorem 6.3.6, so V(l) | [L:A]. But then # ( 1 ) | t[L:A],
and t = [G:L], *V(1) = V G (1) = x ( l ) , so x(l) | [G:L][L:A] = [G:A].
A
6.3. NORMAL SUBGROUPS AND CLIFFORD
THEORY
141
Note in particular that the dihedral groups Dn, n > 3, and the quaternion
groups Qm, m > 2, have nonlinear absolutely irreducible characters only of
degree 2, as they have abelian subgroups of index 2.
Theorem 6.3.10 (Blichfeldt) Suppose that G has a normal abelian subgroup A and that G/A is a p-group for some prime p, F Ç C is a splitting
field for G, and \ € Irr(G). Then \ = AG for some linear C-character X of a
subgroup H < G.
Proof There is no harm to the statement of the theorem if we assume for
the proof that F has been enlarged to C, in order to have ready access to
absolutely irreducible characters of subgroups of G. Proceed by induction
on x(l)- If x(l) = 1 there is nothing to prove, so assume that x(l) > 1
(hence x(l) is a power of p by Ito's Theorem (6.3.9)). Denote by $ the set
of all linear characters <p of G for which ip\ = Xi clearly a subgroup of the
group of all linear characters of G (in fact the stabilizer of x)- If V € * then
(XX>V) = (x,X¥>) = (X-X) = 1, so ip has multiplicity 1 in XX- Write
XX = Y1&- ¥ > € $ } + ^ V>¿,
where the V« are the other constituents (if any) of XX- If any rpi were linear
we would have 0 < (xxW't) = (XiX^)* forcing (XiXV'O = 1> consequently
X = x^i, and V>¿ € #! Thus V¿(1) > 1 for all i. Since x ( l ) x ( l ) and all V<(1)
are powers of p it follows that p | $ | , and hence there is an element tpo € $
having order p. Then tpo is a homomorphism from G to the subgroup of order
p in C*, and if K = ker(y>0) then [G: K\ = p.
Since both l o and <¿>o restrict to IK on K, the principal character 1*
has multiplicity at least 2 in x x k , so ( x x k - l j c ) * = (X\K,X\K)K
> 2 and
X|K is reducible. Let 6 be an absolutely irreducible constituent of X|K, SO
6(1) < x(l)- Both are powers of p, sopö(l) < x(l)- By Frobenius Reciprocity
(6.2.8), (X,0G)G = (X\K,6)K
> 1, and therefore x(l) < Ö G (l) =pÖ(l). Thus
G
X = 0 . By the induction hypothesis there is a subgroup H < K and a linear
character A on H such that XK = 8. But then, by Proposition 6.2.6,
AG
= (xKf = e° = x.
A
Corollary 6.3.11 If G is nilpotent and x € ^(G)
linear C-character X of a subgroup H <G.
then x = AG for some
Exercises
1. Prove Corollary 6.3.11. Recall that a finite group is nilpotent if and
only if it is the direct product of its Sylow subgroups.
Assume for the remaining problems that H <G, and that F Ç C is a
splitting field for G and H.
CHAPTER
142
6. INDUCTION AND
2. Show that (1H)G = T,{<fi(l)<P:tP e
{Hint
ITT(G/H)}.
RESTRICTION
Compare with
PG/H-)
3. If x is any F-character of G show that
4. If x> ^
are
(IH)GX
=
(X\H)°-
F-characters of G show that
(XM\H)
= E M 1 ) ( W , # * ' € Irr(G//f)}.
Assume further for the remaining problems that G / J Ï is abelian and
write G / H for the group Irr(G/i/).
5. If X) V' € Irr(G) show that the restrictions
onlyifxeOrbg^VO6. If x € Irr(G) show that
(X\H,X\H)
X\H
and
I¡>\H
= I Stabg^(x)|.
7. If X € Irr(G) show that the restriction \\H is in kr(H)
S t a b ^ C x ) = 1.
6.4
are equal if and
if and only if
Mackey T h e o r e m s
Suppose that H <G and that T is an F-representation of H with character ip.
UxeG define Tx on tf1 via Tx{yx) = T(y) for all y€H. Then T* is clearly
an F-representation of Hx, with character Vx given by i>x(yx) = ip(y), and
if also z € G then (T 1 )* = Txz and (V'I)Z = V " on H". As usual we may
broaden the definition of conjugate character so that it applies to arbitrary
class functions: if <p 6 cf(ii) and x € G then ipx 6 ci(Hx) is defined by
Vx(ux) = V(y) for y £ H, or equivalently fx(z) = <£(Xz) f° r z G # x (and
hence xz £ H).
The next two theorems, and their consequences, are due to G. Mackey
[45]. They analyze quite generally the interactions between induced class
functions.
It will be convenient to establish notation before stating the theorems.
Suppose that K, H < G and let x\,... ,x, be a set of (ÜT-if)-double coset
representatives in G, with x\ = 1. For each i set K^ = KXi and Hi =
KW nH.Ifipe
ci(K) define ^ 6 cf(K^) via V¿ = V*'T h e o r e m 6.4.1 If ^ € d(K)
then (r¡;G)\H = E U d f c k ) " -
P r o o f The subgroup H of G acts (by right multiplication) on the set {Kx}
of right cosets of K in G; for a fixed Kxi the union of the cosets in the //-orbit
of Kxi is the double coset KxiH, and it is easy to check that Stabtf(ifa:¿) is
KXiC\H = Hi. Let ha,..., hUi be a (right) transversal for Hi in H, so Kx{H
6.4. MACKEY
143
THEOREMS
is the disjoint union KxihnÙ■ --UKxihi,.. Thus {x¿/iy: 1 < t < «, 1 < j <
Si} is a transversal for K in G.
Make the trivial observation that if z G H then z G A' (t) if and only if
z G Hi, and hence (V'ÍIJ/J'C*) = ^t(^) for all i. So, if y € H,
»
j
i
A
Theorem 6.4.2 If F C C, v? G cf(tf), and V G cf(A") then
Proof We apply Frobenius Reciprocity (6.2.8) twice and the theorem above:
(^G,VG) = {VMG)\H) = ¿foMI»«)") = ¿ M ^ . ^ I A ) .
t=l
A
t=l
For the next theorem we specialize to the case K = H. If x G G define
/f( x ) to be Hf)Hx; thus in the notation introduced above we have Hi = H(Xi).
Theorem 6.4.3 Suppose that H < G and that F Ç C is a splitting field for
G and its subgroups. If i¡> G lrr(H) then i/)G G Irr(G) if and only if t/f\n(m)
and IPX\H(I) have no irreducible constituents in common for all x G G\H.
Proof This follows directly from Theorem 6.4.2, with tp = ip, and the fact
that any x G G \ H can be taken to be one of the double coset representatives
Xi, i > 2.
A
Corollary 6.4.4 If H <G and xjj G Irr(H) then V>G G Irr (G) if and only if
xpx ¿ i¡> for
allxeG\H.
Proof Since H < G, H(x) = H, all x&G.
A
Exercise
Suppose that H < G, i¡>, 9 G Irr(tf), and ipG, 9° G Irr(G). Show that
ipG = 0G if and only if ip = 9X for some x e G, i.e. if and only if ip and 9
are in the same G-orbit in ITT(H).
For a very special case of Clifford's Theorem (6.3.3) suppose that [G: H] =
2. Then G/H Sí 52 has just two linear characters, viz. the principal character
IG/H and the alternating character CG/H, which lift, respectively, to I G and
e = ce which is 1 on H and - 1 on G \ H. Call e the alternating character of
G relative to H.
CHAPTER
144
6. INDUCTION AND
RESTRICTION
Proposition 6.4.5 Suppose that [G: H] = 2, and let F Ç C be a splitting
field for G and H. Let e be the alternating character of G relative to H, and
take xeG\H.
If x € M G ) set 9 = X\H- Then either
1- X ^ ¿X> *n which case 9 6 lir(H), 9 = 6X, and 9° = x + ^X! or
%■ X = ¿X> , n which case 9 = ip + ipx for some ip € Irr(iï), ip ^ ipx, and
4>G = x-
All absolutely irreducible characters of H occur either as 9 in part 1 or as if)
in part 2.
Proof Let ip e Irr(/f) be a constituent of 6. We have
i = (x,x) = |G|-1(ElxWI2 + E W M 2 )
= J(M) +IGT 1 E M M I 2 ,
or
(e,9) + \H\-1J2\x(hx)\2=2.
Thus either (1) (9,9) = 1, or (2) (9,9) = 2. In case 1 9 = ip € \rr(H), and
in the context of Clifford's Theorem e = t = 1, IG(9) = G, so 9X = 9. Also
S h g / / IxC 11 )! 2 f^ 0, so x doesn't vanish on G \ H, i.e. x ^ eX- Furthermore
9G\H
= 29, 9G\G\H = 0,SO9G
= X + *X-
In case 2, (9,9) = 2, 9 has two constituents, so in Clifford's Theorem
IG(ip) = H, t = 2, e = 1, 9 = tp +1/»1, and ij)x ¿ ip. Also i/>G = x by Corollary
6.4.4 and Frobenius Reciprocity.
Why do the two cases describe all possibilities? Given any ip e Irr(li) set
T) = ipG = ip + ipx, so T)\H = ip + ipx- Then n 6 Irr(G) if and only if ip ^ ipx,
which is case 2 above, again by Corollary 6.4.4. If n splits it can only split as
r\ = xi + X2. Xi € Irr(G), by Frobenius Reciprocity, since n(l) = 2ip(l), and
both xi and X2 restrict to ip on H. Also n\G\H = 0, so X2 = «Xi / Xi> case
1 above.
A
Proposition 6.4.5 will be useful later for calculating character tables of
alternating groups.
Recall that a group G is the semidirect product for split extension) of a
subgroup A by a subgroup B if
A<G,
B<G,
G = AB,
and AnB = l.
In that case we write G = A x B (or, less often, G = B x A).
For example, the symmetric group S„, n > 3, is the semidirect product
An » B, where A„ is the alternating group and B = ((12)). If G is the dihedral
group Dn = (a,b\an = b2 = (ab)2 = 1), then G = A x B, with A = (a) and
B = (b).
6.4. MACKEY
145
THEOREMS
Assume for the remainder of this section that G = A x B,
with A abelian, and that F Ç C is a splitting field for G
and its subgroups.
Then B acts by conjugation on lii(A), and if y € Irr(.A) we'll write Bv
to denote Stabe (v). If T is an F-representation of Bv on V define y x T o n
A x By via
{<p x T)(ab) = <p(o)T(6).
The result is an F-representation of A x Bv, for
( V xT)((a 1 & 1 )(a 2 6 2 ))
=
^(a 1 ( 6 l a 2 ))r(6 1 6 2 )
=
¥ p(oi)^
=
ip{ai)T{bx)<p{a2)T{b2)
=
(<p x T ) ^ ) ^ x T)(a 2 6 2 )
1
(a2)T(6 1 )T(6 2 )
(we used the fact that (p is linear). If ^ is the character of T, then the
character of ¡p x T is ip x ip, where (<p x ip)(ab) = <p(a)rp(b).
Proposition 6.4.6 If tp € Irr(A), ^ e I r r ^ ) «hen <pxrp e
Ir^AB^).
Proof Calculate: (<p x rp,tp x x¡>) = (ip,tp)(ip,ip) = 1.
Proposition 6.4.7 If <p G
\TT{A)
and i¡) 6 lrr(Bv)
then (<p x xp)G e Irr(G).
Proof Set H = AB^, so <pxtp € Irr(ff). Then by Theorem 6.4.3 it will suffice
to choose x = ab 6 G \ H (i.e. be B\B^,) and show that ipxip and (tp x tp)x
have no common constituents when restricted to HC\HX = A{Bvr\Bb0). Even
more so it will suffice to show that the restrictions to A have no common
constituents. But an easy calculation shows that (¡p x VOU = VK^Vi whereas
{tp x tp)ab\A = VKlOv6! and the constituents are different since b^Bv.
Alternatively, it is easy to check that Ici'P x ip) = AB^,, so the conclusion
follows from part 1 of Theorem 6.3.8.
A
Proposition 6.4.8 Suppose thatip, 6 € Irr(A), i/> € I r r ^ ) , and77 G Irr(ß e ).
1. Ifip$ Orbß(ö) then (<p x i/>)° ¿ {6 x n)G.
2. If<p = 6butxp¿n
then (ip x \¡))G ^ {ip x TJ)G .
A
CHAPTER
146
6. INDUCTION AND
RESTRICTION
P r o o f Let us set the stage to apply Theorem 6.4.2. Set K = ABV and
of (Bv-B$)-dovb\e coset representatives in B is
H = ABg. A set {b\,...,ba}
also a set of (K-H)-doub\e coset representatives in G. Set K^ = Khi = ABfy
and Hi = K(i) D H = ABU where B{ = B$ n Be. Then by Theorem 6.4.2,
8
((<pxrl>)G,(0*T,)a)
=
£(^x^iU
Ö x r
?)
6 i
UÜ
¿=i
= ¿(V,^)(^lfl i ,»i 6< |fl < )Thus in case 1 the inner product is 0, since 6bi ^ tp for all i.
In case 2 the inner product becomes
((ip x v)G,(*> x nf) =
¿(^VOMB^IBJ,
i=l
but now (<p,tpbi) = 0 unless bi G ß v , i.e. unless i — 1. Furthermore the
A
summand for ¿ = 1 is (y,<p)($,r)) = 0, since ip ^ n.
T h e o r e m 6.4.9 Suppose that G = AxB, with A abelian. Let ip\,..., ipm 6e
representatives of the B-orbits in Irr(yl). For each i let Bi = B^;, and write
Irr(ßj) = {tpij: 1 < j ' < c(B¿)}. For eac/i i and j set \ij — Pi x V'tj- rAen
Irr(G) = { X g: 1 < ¿ < m, 1 < j < c(B,)}
P r o o f By Propositions 6.4.7 and 6.4.8 the characters \fj are all irreducible,
and they are distinct, so it will suffice, by Theorem 5.2.14, to show that the
sum of the squares of their degrees is \G\. Note that [G: ABi] = [B: Bi], and
that each tpi is linear, so Xy(l) = [B'-Bi]rpij(l). Thus
£xS(D2 = D ß : ß ^ i ) 2 = Bß:ß<]2£<M1)2
i,j
i,j
=
:
=
i
j
¿ [ B : Bif\Bi\ = \B\ Ç [ B : B{] = \B\ £ | Orb B (^)l
i
i
|B||Irr(A)| = |B||A| = |G|.
A
The procedure indicated in Theorem 6.4.9 for finding all the characters of
G = A x B is sometimes called the little group method of Mackey and Wigner.
For an easy example let us apply the method to determine the characters
of the dihedral group Dm = {a,b | am — b2 = (ba)2 = 1), with m — 2n being
even (the easier case of m odd is an exercise below).
The conjugacy classes are K0 = {1}, Ki = {a*,a'*} for 1 < i < n - 1,
Kn = {an}, Kn+i = {baj:j even}, and Kn+2 = {baj: j odd}. Set A = (a) and
6.5. BRAUER
147
THEOREMS
B = (b). If C = e2ni/m € C and if ip(a) = < then Irr(>l) = (<p); we write <¿>j for
^ ' , 0 < i < m - 1 . The B-orbits in ÎIT(A) are {<¿>o}> {<£>«, Vm-¿}, 1 < i < n - 1 ,
and {v?n}, so we may take as orbit representatives {w.Q < i < n}. The
stabilizers in B are Bo = B„ = B, Bi — 1 for 1 < i < n - 1. Write
Irr(B) = {^o,^i}, with Vo = I ß and ^i(ft) = — 1 - Then there are four linear
characters on G, viz. ¡po x ip0, ip0 x i^ii <Pn x V'o, and f ) n x ^ , which we will
relabel as xi through X4, respectively. The remainder of Irr(G) consists of
the degree 2 characters Xj+i = V ^ » l ^ i ^ n _ l > whose values are easily
calculated as in Proposition 6.2.4. The character table follows.
Ko
Xi
X2
X3
X4
Xj+4
1
1
1
1
2
Ki
1
1
(-I)'
(-I)1
^n
1
1
(-1)"
(-1)"
cj + c-0' i-iy-2
A'n+l Kn+2
1
1
-1
-1
-1
1
1
"I
0
o
Compare the table of D^m (order 4m) with that of the generalized quaternion
group Qm (exercise, page 125).
Exercises
1. Calculate the character table for the dihedral group Dm if m is odd.
2. Calculate the character table of the group
G = (a,b\a7
6.5
=b3 = l,6o = a2ft).
Brauer Theorems
This section is devoted to some powerful theorems of Richard Brauer.
We begin with a very general fact ([39], page 128) about rings of Z-valued
functions on finite sets.
Proposition 6.5.1 (Isaacs-Banaschewski) Let X be a finite set and suppose that R is a ring of functions from X to Z, with pointwise operations.
Write lx for the constant function with value 1 on X. Suppose that for each
x G X and each prime p there is a function / I i P 6 R such that p)f
fx,P(x).
Then lx € R.
CHAPTER 6. INDUCTION AND
148
RESTRICTION
Proof For each x € X set Ix = {f(x): f € R}, evidently an ideal in Z. If
Ix were not all of Z it would be contained in some ideal (p), p a prime. But
then p I / ( x ) , all / G R, contradicting p^ / Z i P (x). Thus Ix — Z, all x, and in
particular there exists gx € R with gx{x) = 1. Then Y\x&x(^x-9x)
= 0 and,
when the product is expanded, l x can be expressed as a sum of elements of
R.
A
Suppose that H is a family of subgroups H of G so that if H, K G V. and
x G G then i/nA" 1 G ft. Define ß G CH), the Burnside ring of G relative to U,
to be the additive abelian group generated by all the permutation characters
(iHf,
Hen.
Proposition 6.5.2 The Burnside ring BG(H) is a subring of the character
ring Char(G).
Proof It will suffice to show that the product of any two of its generators
is in BGCH). Take H, K G H and apply Propositions 6.2.6 and 6.2.7 and
Theorem 6.4.1 to see that
(IH)G(IK)G
=
(1H(1K)G\H)G
((1K)G\H)°
=
= (E(1«i)//)G = E(1^)G^G(ft),
z
i
since each H{ = H f~l KXi 6 U.
A
A group H is called elementary if H = C x P, where P is a p-group for
some prime p and C = (c) is cyclic of order prime to p. More generally, H is
called quasi-elementary if H = C x P , with C and P as above. If it is necessary
to specify the prime p we say that H is p-elementary or p-quasi-elementary.
Note that subgroups of elementary groups are elementary, and likewise
subgroups of quasi-elementary groups are quasi-elementary. Thus both the
family S of elementary subgroups of G and the family Q of quasi-elementary
subgroups of G satisfy the requirement imposed above in the definition of a
Burnside ring.
Proposition 6.5.3 The principal character
IG
is in the ring
BG(Q)-
Proof We will apply Proposition 6.5.1 to R = BG{Q)- Given a prime p
and i € G write |x| = pen, with {n,p) = 1, set c = xp' and C = (c), cyclic
of order prime to p. Let P be a Sylow p-subgroup of N = NG(C),
and
set H = CP € Q. Then let / I ) P be the permutation character (IH)GWe
need to show that p j {lH)G(x). By Proposition 6.2.11 we have (lH)G{x) =
|{yii:j/ € G and x y / i = y i i } | , the number of fixed points in the action of
G on cosets of H. But xyH = yH if and only if xv G H, in which case
Cv < H. But C contains all the Sylow ç-subgroups of H for g ^ p so it
follows that C» = G and y G AT. Thus the cosets fixed by x are all cosets of
6.5. BRAUER
149
THEOREMS
H in JV, so (l//) G (x) = (lH)Nt{x). lîyeN
then cyH = y&H = yH (since
C < N and C < H)\ that is xp'yH = yH. Thus the permutation effected by
x on {yH:y eJV} has order dividing p e , and any nontrivial orbits have size
divisible by p. But the total number of cosets is [N: H], which is prime to p,
so the number of cosets fixed by x (i.e. 1-point orbits) must also be prime to
p,i.e. p\(lH)N(x)
A
= fx<p(x).
The next two theorems will be proved simultaneously; the corollary follows
immediately.
Theorem 6.5.4 (Brauer's Induction Theorem) If ip £ Char(G) there
of G, linear characters X, G IIT(HÍ),
are elementary subgroups H\,--,Hm
and integers at so that i¡) = J^íILi OiAG.
Theorem 6.5.5 (Brauer's Characterization of Characters) / / rp is a
class function on G then ip G Char(G) if and only if %1>\H G Char(JÏ) for
every elementary subgroup H of G.
Corollary 6.5.6 / / x¡> G cf(G) then tp G Irr(G) if and only if
1. Í¡)\H € Char(ií) for every elementary subgroup H of G,
2. (ip,ip) = 1, and
3. rp(l) > 0.
Proof Let TZ be the ring
{V e d{G):<p\H € Char(H), all H € 8}
(where £ is the set of elementary subgroups of G), and let 2 = To be the
additive group generated by all AG, where A is a linear character of some
H G £. Evidently I Ç Char(G) Ç TZ, and to prove both theorems it will
suffice to show that I = 71, since I = Char(G) proves 6.5.4 and Char(G) = TZ
proves 6.5.5.
We show first that 1 is an ideal in TZ. If tp € TZ and r\ G I write n =
^ ¿ a j A p , with a¿ G Z and A< a linear character of Hi G £, all i. Then
ffj = ¿i a »(<P|tf,A«) G by Proposition 6.2.7. Write y>|//,. G Char(H¿) as (p\H¡ =
Hj bijZij, with all bij G Z and each &j G Irr(i/¿). Since elementary subgroups
are clearly nilpotent, we have £y = /x"' for some linear character ßij of
a subgroup ÄV, (also automatically elementary) of Hi by Corollary 6.3.11.
Substituting and applying Propositions 6.2.6 and 6.2.7, we see that
tpr) = ^2aibij(Xi\KijHij)G
G 2.
Now it will suffice to show that I G G X. We may assume inductively
that the result of Theorem 6.5.4 holds for all proper subgroups of G. Thus
it will suffice to show that I G can be written as a Z-linear combination of
CHAPTER 6. INDUCTION AND
150
RESTRICTION
characters induced from proper subgroups of G, since then by induction each
of those characters is a Z-linear combination of characters induced from linear
characters of elementary subgroups, hence so is \Q by transitivity of induction
(Proposition 6.2.6).
If we knew that 1^ £ 1 « for every quasi-elementary subgroup H oî G
we could conclude that I G 6 IG by Proposition 6.5.3 (and the transitivity
of induction), so we may assume now that G = C x P is quasi-elementary.
Set N = NG(P) and note that N = (N (~l C) x P 6 £. If N = G there
is nothing to prove, so we assume that N is a proper subgroup. Note that
({IN)G,IG)
= (1N,1N)
(lNf
= 1, so
= 1 G + J2iaiXi-ai
€ Z, Xi e Irr(G), and i > 2}.
Let us show that if a* ^ 0 then Xi must be nonlinear. If to the contrary
Xi were linear, then XÍ\N has ljv as a constituent by Frobenius Reciprocity
(6.2.8), and then N <K = ker(x<) by Corollary 6.3.4. But then, since P is a
Sylow p-subgroup of K, the Frattini Argument (Proposition 4.2.1) entails that
G = NK < K, contradicting the fact that K is a proper normal subgroup.
Thus Xi(l) > 1- By Blichfeldt's Theorem (6.3.10) each Xi with a{ ¿ 0 can be
written as Xf for some linear character Aj of a (necessarily proper) subgroup,
and finally
1G = (1JV)G - 2 > A ? ,
i
completing the proof.
A
R. Brauer first proved the theorems above in 1947, in a paper [7] on Lseries. The proof given here, which appeared in 1975 and is due to D. Goldschmidt and I. M. Isaacs [28], is considerably shorter than Brauer's original
proof.
Applications of Brauer's theorems will appear in later chapters.
Groups and Characters
by Larry C. Grove
Copyright © 1997 John Wiley & Sons, Inc.
Chapter 7
Computing Character
Tables
In this chapter we discuss some algorithms for calculating the character table
of a finite group G. We assume throughout that G has conjugacy classes
Ki, K2, ..., Kk, that F Ç C is a splitting field for G, and that Irr(G) =
{Xi> ■ • ■ > X*}- We take K\ = {1} and Xi = IG as usual.
7.1
Burnside
Recall from Chapter 5 (Proposition 5.2.15) that we defined
^
»(i)
-
'
and that if z G Kt then
nr,t = \{(x,y) € Kr x Ka:xy = z}|
is independent of the choice of z. Prom Proposition 5.2.16 we have
k
UjrWj, = }] TlrstVit
(*)
for i,r, s = 1,..., k.
As in the proof of Proposition 5.2.27 we may interpret the equation (*)
as follows: for 1 < i < k define column vectors Vi = (uu,... ,Wik)1, and for
1 < r < k define a k x k matrix Nr = [nrst], often called a class matrix. Then
the Uj's are all eigenvectors of each NT, and the eigenvalue corresponding to
Vi is Uir.
Proposition 7.1.1 The eigenvectors {i>i,... ,«*} are linearly independent.
151
CHAPTER 7. COMPUTING
152
CHARACTER
TABLES
Proof Suppose that J ^ c&i = 0, with all c¿ Ç. F. Entry by entry we see that
Yli^^ij = 0) all 3, hence from the definition of wy that 5Zi( c »/Xt(l))Xt = 0,
and so all Cj = 0 by Corollary 5.2.7.
A
Notation Determine a permutation j i-> j ' of { 1 , . . . , A;} by means of Ky =
K~l for each conjugacy class K¡.
Proposition 7.1.2 / / 1 < i,j < k then
¿-
\Kr\
- Xi(l)Xj{l)
"
Proof Calculate:
\Kr\xi{Kr)\KT\Xj{Krl)
v
4- I*-I
V
=
Xi(l)Xi(l)
M^MKMK^)
\G\(Xi,Xj) _
¿ii\G\
Xi(i)xj(i)
XiWxiiD
'
A
Corollary 7.1.3 //w¿ r is known for allr, then x¿(l) *s determined.
Proof Take j = i in the Proposition.
A
In [10], §223, Burnside presented a procedure for calculating character
tables which, he wrote, "requires only algebraical processes" provided that
the integers nr,t are known. Based on his description of the procedure, we
sketch next an algorithm for determining the character table.
The Burnside Algorithm
1. Determine the conjugacy classes of G.
2. Calculate all nrat, hence all matrices Nr.
3. Find k linearly independent eigenvectors for each Nr, normalized to
have first entry 1 (since un = \Ki\ = 1). The ijth entry of Vi is o ^ .
4. Use Corollary 7.1.3 to calculate Xt(l), all i.
5. Calculate Xi(Kj) = UÍJXÍ(1)/\KJ\,
all i and j .
153
7.1. BURNSIDE
For a very simple example take G = 53, with K\ = {1}, Ki = cl(l 2 3),
and K3 = cl(l 2). Easy calculations yield
AT X =
1 0
0'
0 1 0 ,
0 0 1
N2 =
' 0 1 0 '
2 1 0
0 0 2
,
and
N3 =
' 0 0 1
0 0 2
3 3 0
For example, n 332 = 3 since (123) = (12)(13) = (13)(23) = (23)(12). The
respective characteristic polynomials are
/!(A) = (1-A) 3 , /2(A) = - ( 1 + A)(2-A)2, and /3(A) = -A(3-A)(3 + A).
Since N3 has distinct eigenvalues it has linearly independent eigenvectors.
When normalized to have first entry 1 they are
vi
= (u>ii,wi2,wi3)'= (1,2,3)'
V2
=
(w 2 l,W22,W2 3 ) t = ( 1 , 2 , - 3 ) '
(A = - 3 ) ,
V3
=
(W31,W32,W33)t = ( 1 , - 1 , 0 ) '
(A = 0).
(A = 3),
They are of course automatically eigenvectors for N\ and N2 as well.
Next Corollary 7.1.3 yields the degrees Xi(l) = X2U) = 1 and X3(l) = 2.
Finally, we calculate the remaining values, X3(K2) = ^32X3(1)/\K21 = — 1,
etc. The completed character table, as we have seen earlier, is
Xi
X2
X3
Ki K2 Ki
1
1
1
1
1 -1
2 -1
0
Exercise
Apply Burnside's algorithm to calculate the character tables of the dihedral groups D4 and D5.
One possible approach to calculating the integers nrat is to work within
the group algebra FG. For a given conjugacy class K, we may define C =
£ { x : x € K} Ç FG, i.e. C is the function from G to F defined by
i{xeK
C(x) = {1
>
y. 0 otherwise.
If the conjugacy classes of G are Ki,..., Ä*, as usual, we obtain corresponding C i , . . . , Ck € FG. Note, for example, that C\ = \G\~lpG.
Clearly C i , . . . ,C* € cf(G). ïn fact it is not difficult to see that the set
{Ci,..., Ck) is a basis for the center of FG.
CHAPTER
154
7. COMPUTING
CHARACTER
TABLES
In the present context, however, the important point is that
k
CrCê = / ^ nr3tCt
t=i
for all r and s. Thus the calculation of the nr,t can be carried out via
multiplications in the group algebra.
7.2
Dixon
Even for small examples such as those in the exercise above it soon becomes
clear that Burnside's algorithm requires a considerable amount of calculation,
particularly for step 3.
There are of course readily available computer routines for finding eigenvalues and eigenvectors. They tend, however, to do the calculations numerically rather than symbolically. As J. Dixon pointed out in [21], most
theoretical uses of character tables require knowing the entries in exact form
as algebraic integers. Consequently Dixon produced a revised form of the algorithm that overcame that difficulty, yet could be programmed for machine
computation.
Some preparation is required.
Let m = exp(G), the least common multiple of the orders of elements of
G. If x € G and x« € Irr(G) then Xi( x ) ls a s u m °f complex mth roots of
unity. If C € C is a primitive mth root of unity then all the character values
Xi(z)lieinZ[C]={/(C):/(*)€Z[x]}.
Choose a prime p € Z such that p > 2y/\G\ and m | p - 1. Thus p is
in the arithmetic progression {km + l : i g N } , and the existence of such a
prime is the result of a deep theorem of Dirichlet (e.g. see [11], page 120). It
is interesting to note that the usual proof of Dirichlet's theorem utilizes the
theory of characters of abelian groups.
Since all prime divisors of |G| are divisors of m it is clear that pi \G\.
The requirement that p > 2y/\G~\ ensures that p > x¿(l) for all x¿ € I rr (<?)
by Theorem 5.2.14.
Next choose z € Z having multiplicative order m when viewed as an
element of Z*. Then z is a root mod p of the cyclotomic polynomial $ m ( x ) €
Z[x]. Since $ m ( x ) is the minimal polynomial for Ç over Q it follows that
$m(x) I f(x) for any / ( x ) 6 Z[x] satisfying /(Ç) = 0. As a consequence there
is a well-defined map 6: Z[C] -► Z p given by 6: /(C) »-> f(z) (mod p). Clearly
8 is a ring homomorphism; it maps Z[C] onto Z p since 0(1) = 1 (mod p) is
in its image. Extend 0 in the obvious fashion so that it maps vectors with
entries from Z[Ç] to vectors with entries from Z p .
We continue using the notation of the previous section.
7.2.
155
DIXON
Proposition 7.2.1 The image under the map 9 of {v\,... ,Vk} is a set of
k linearly independent eigenvectors of the matrices N\,..., Nk (mod p); the
eigenvalue corresponding to 9{vî) for Nr is 9{u)ir).
Proof Only the linear independence needs proving - the rest is a consequence
of the fact that 6 is a ring homomorphism.
Define a weighted Hermitian inner product on the space of ¿-dimensional
column vectors over C as follows: if u = ( « i , . . . , Uk)1 and w = (u>¿, ...,Wk)1
set
Then we may interpret Proposition 7.1.2 as saying
and {v\,... ,Vk} is an orthogonal set relative to the new inner product.
Let W — Z*, the Z p -space of column vectors, and use 9 to define a bilinear
form on W, also denoted (*,*), via (9u,9v) = 6(u,v) for vectors u,v € Z[£]fc
(this is well-defined since p j |G| and p | |A¿|, all i, the denominators of the
weights).
Observe now that
(recall that p does not divide the character degrees). Thus 9v\,... ,9vk are
orthogonal relative to the bilinear form and hence are linearly independent
by the standard argument from linear algebra.
A
Dixon's idea was to use the homomorphism 9 to transfer the calculations to
Z p . One further technicality is required to enable us to recover the character
values in F after the calculations mod p.
Suppose that T is an F-representation of G, with character \ € Irr(G)
and d = x(l)- Suppose that x € G has order k (so k | m). Let f = £ m / fc ,
a primitive fcth root of unity. Then xix) is the sum of the d eigenvalues of
T(x), each a kth root of unity in C, so x(z) — £*' + £"2 H
1- fSd, say. Note
Sd
also that x(x<) = £*>< + £»»' + • • • + £ ' for any i e N . Retain this notation
in the statement of the next proposition.
Proposition 7.2.2 For each s € N, 0 < s < k - 1, denote by /x(s) the
multiplicity o / í " as an eigenvalue ofT(x).
Then
{Y,x(xl)Cal.
n(s) =
1=0
CHAPTER 7. COMPUTING CHARACTER TABLES
156
Proof Recall that if a G C is an rth root of unity then
ao + ai +
... + a r-i
=
/ r ifa = l,
10 otherwise.
We calculate:
rËxoOr" = ^E«*1'+ ••■+€"')*-'
1=0
t
= I y^£(»i-«)* + ... + 1 VM»*-»)*.
k
k
i
i
The ith summation adds to 0 if Sj ^ s, to k if s< = s, and the proposition
follows.
A
Corollary 7.2.3 The multiplicity fi(s) is uniquely determined by
1 *_1
<=0
and the fact that p > /i(s) € N.
Proof Since /x(s) < d this is a consequence of p > 2y/\G\.
A
Corollary 7.2.4 *(*) = E ^ 1 p(a)?.
We may now state Dixon's variation of the Burnside algorithm.
The Burnside-Dixon Algorithm
1. Determine the conjugacy classes of G.
2. Calculate all n r » t , hence all matrices Nr.
3. Find k linearly independent eigenvectors 8v\,...,8vk for the matrices
Ni,...,Nk (mod p), normalized to have first entry 1. The jth entry of
$Vi is Oujij.
4. Calculate x»(l)> all i, via (6VÍ,6VÍ) = l/x<(l) 2 (mod p). There is a
unique solution for x¿(l) G N satisfying Xi(l) < p/2, as must be the
case since p > 2^/¡G[.
5. Calculate 9{Xi{Kj)) = tf(w4i)x<(l)/|Ä'j|, all i and j .
6. Calculate all x»(-Kj) from
Proposition 7.2.2.
6{XÍ{KJ))
by means of the corollaries to
7.2.
157
DIXON
To carry out step 3 for a given Nj the elements of Z p can be tried in
succession to see whether Nj — XI, X G Z p , has a nontrivial nullspace - if so
a basis for the resulting eigenspace is determined. Thus V = Z* is split as
a direct sum of eigenspaces, say V = Vi © • • • © VT. The next N, say Nj+i,
then acts by restriction on each of the V¿'s of dimension more than 1 to see
whether it will split them into sums of lower-dimensional subspaces. That
continues until all the eigenspaces are of dimension 1.
For an example let us take G = Aff(F), for F = F5, so G is Frobenius of
order 5 • 4 = 20. It has the presentation
G = (a,b\a5
= b* = 1, ba = a2b).
It is easy to check that the conjugacy classes are
# i = {l}> K2 = c\(a) = {ai:l<i<4],
K3 = c\{b) = {0*6:0 < i < 4},
K4 = cl(62) = {aW.O < i < 4}, K5 = cl(63) = {aW-.O < i < 4}.
The matrices Ni are easily computed, e.g. via calculating in the group
algebra. Since not all Ni will be needed for the algorithm we list only the
two that will be used:
N2 =
0
4
0
0
0
1
3
0
0
0
0
0
4
0
0
0 0
0 0
0 0
4 0
0 4
and
ÍV3 =
0
0
0
0
5
0
0
0
0
5
1
4
0
0
0
0 0
0 0
5 0
0 5
0 0
We have 2y/\G\ = 4-y/5 < 12 and exp(G) = 20, so we need a prime that is
at least 12 and is = 1 (mod 20). The first one is p = 41, so we calculate mod
41.
The only eigenvalues of N2 in Z 4 1 are 4 and 40 = — 1. The eigenspace
for - 1 is 1-dimensional; we label it V5 and take as its normalized generator
Ovs = (1, —1,0,0,0)'. The eigenspace for 4 is 4-dimensional - the basis that
results from Gaussian elimination is
{(1,4,0,0,0), (0,0,1,0,0), (0,0,0,1,0), (0,0,0,0,1)}.
If we denote the eigenspace by \\ then we have V = 7L\X = V\ ® V5. Next N3
acts on Vi ; its action there is represented by the 4 x 4 matrix
A =
0
0
0
5
1
0
0
0
0 0
5 0
0 5
0 0
The matrix A has 4 distinct eigenvalues mod 41, viz. A = 0, 8, 9, and
40 = - 1 . For example, A = 9 has as an eigenvector (1,5,5,5)' and, using its
CHAPTER 7. COMPUTING
158
CHARACTER
TABLES
entries as coefficients of the basis vectors for Vi, we obtain Ovi = (1,4,5,5,5)'.
In like manner we obtain Qvi = (1,4, - 4 , - 5 , 4 ) ' , 6v3 = (1,4,4, - 5 , - 4 ) ' , and
Ovi = ( 1 , 4 , - 5 , 5 , - 5 ) ' , and Vi is split by iV3 into the direct sum of four
1-dimensional subspaces with the indicated normalized basis vectors.
As a sample of the determination of degrees note that
(0v5,6v5) =
i-i
20 1
, (-i)-(-i)l
4
16
1
(mod 41),
X5(l) 2
so X5(l) = 4. We see similarly that Xi> • • ■ >X4 are all linear.
Take Ç = e2*i/2°, a primitive 20th root of unity in C , and set z = 2,
which has multiplicative order 20 in Z41. The final step is to calculate the
other character values, using the fact that 6xi{Kj) — (Xi{l)/\Kj\)0vijFor
example, let us evaluate X s ( ^ ) = Xs(o). Since \a\ = 5 we use £ = C4 and
correspondingly use z4 = 2 4 in Z41.
We need the multiplicities /x(s):
"<°> = 5
,(D = ¡
4 - - — 1 — 1 — 1= 0,
4
(-)-2-"-l-2-8-l-2-12-l-2
-16
(calculating in Z41), and similarly p(2) = /x(3) = /x(4) = 1. Thus
xs(K2) = t+e+e+c4
= -i,
since £ = £4 is a primitive 5th root of unity. The complete character table
follows.
Xi
X2
X3
Xi
X5
7.3
Kx K2
1
1
1
1
4 -1
K3 K4 Ks
1
1
1
-1
1 -1
—i - 1
i
i - 1 —i
0
0
0
Schneider
Computer implementations of the Burnside-Dixon algorithm were written by
several people, beginning with Dixon himself. In 1990 G. Schneider discovered
a variation on the algorithm that in many cases speeds it up considerably.
The variation will be described briefly below, after some further preliminary
discussion.
We retain the notation of the previous two sections. Recall, in particular, that Kji = KJ1 for each conjugacy class Kj. We require first some
elementary information concerning the entries nrst of the class matrix Nr.
7.3.
159
SCHNEIDER
Proposition 7.3.1 For all values ofr, s, and t we have
1- nr,t = nS'T't',
2. nr,t = n,rt, and
3. nr,t\Kt\ = nt,'r\Kr\.
Proof For parts 1 and 2 note that xy = z if and only if y~lx~l = z~l if and
and K, = { y i , . . . , y » } ,
only if yx = zx. For part 3 write KT = {x\,...,xu}
say. Note that \xiK, n Kt\ = \XJK, n Kt\ for all i, j , and the sets are
disjoint if i ^ j . Since (x1Kt H Kt) U ■•■ U (xuK, D Kt) is the set of all
elements of Kt exhibited as products ijj/j in all possible ways we see, writing
xi = x, that IÄVHXÄ, D Kt\ = nrst\Kt\. We may assume the yjs labeled
so that xK, D Kt = {xy\,... ,xyw}, and set xy¡ — Zi, 1 < i < i«. Thus
x = ziyf 1 = • • • = ^u/î/^1 are precisely the distinct ways that x € Kr can be
written as a product of elements from Kt and K,>, i.e. w = n t i < r , and the
result follows.
A
Exercise
Show that \KT\\K,\ =
EtnT>t\Kt\.
Proposition 7.3.2 Ifl<r,s<k
then X)*=i Xi{Kt)natr
=
vwXiiKr)-
Proof By Proposition 5.2.16 we have
UirWigi = y
ynTa'tWjt,
t
which can be written as
,,
W
»»' "
\Kr\Xi(Kr)
77\
=
Xi(l)
,r/
V
^nrt't ■
l*tlX»(*t)
XiTTT
^>
•
Cancel the common factor of l/x¿(l), and apply part 3 of Proposition 7.3.1.
The equation becomes
Y,nt,r\KT\Xi(Kt).
^.'l-KrlXi(tfr) =
t
Apply part 2 of Proposition 7.3.1 and cancel the common \Kr\ to complete
the proof.
A
Proposition 7.3.2 admits the following interpretation: the row vector v =
(x»(-^i),--,X»(^t))> which is just row i of the character table, is a left
eigenvector of the class matrix N, with eigenvalue <*;<,<, i.e.XS vN, = UiS> v.
The chief difference in Schneider's version of the algorithm is to use Proposition 7.3.2, hence to calculate left eigenspaces rather than right eigenspaces,
CHAPTER
160
7. COMPUTING
CHARACTER
TABLES
although it should be pointed out that he introduced further refinements
as well. In particular, he first calculates all the linear characters as characters of G/G'. He incorporates tests that allow for calculating only certain
columns of some of the N,, and for avoiding calculation of N, if it will not contribute to splitting of eigenspaces. Furthermore he has developed an independent method that sometimes works to complete the splitting of 2-dimensional
eigenspaces. See [56] for details.
The calculations are done in Z p for an appropriate prime p, and the actual
values in C are recovered just as in Dixon's version.
The Dixon-Schneider version of the algorithm is incorporated into the
system GAP (see [57]). For example, GAP produced the following character
table of the Mathieu group M\\ (see Chapter 2).
Kx K2
Xi
X2
X3
X4
X5
X6
X7
X8
X9
XlO
1
10
10
10
11
16
16
44
45
55
K3
1
1
2
2
0 -2
0 -2
-1
3
0
0
0
0
0
4
1 -3
-1 -1
K4
1
0
K5 K6 K7 Ks
1
1
1
1
0
0 -1
1
Q
0
1
1
—Q
—a
a
0
1
1
-1 -1
1
2
0
0
0
1
0 -2
0
0
1
0 -2
0
0 -1
1 -1
-1 -1
0
0
0
1
1
1
0 -1
K9 ■ft"io
1
1
-1
-1
-1
-1
-1
-1
0
0
7
ß
7
ß
0
0
1
1
0
0
Character Table of M\\
H e r e a = C + C3, /? = £2 + £ 6 + £ 7 + £8 + £ 10 , and 7 = £ + £ 3 + £ 4 + £ 5 + É9,
with ( e C a primitive 8th root of unity and Ç e C a primitive 11th root of
unity.
The actual GAP commands used to create the table are as follows.
gap> mil := G r o u p ( ( l , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 1 0 , l l ) ,
(3,7,li,8)(4,10,5,6));
gap> mil.name := ' ' m i l " ;
gap> CharTable(mll);
Exercise
Use GAP to find the character table of the Mathieu group Mi2-
Groups and Characters
by Larry C. Grove
Copyright © 1997 John Wiley & Sons, Inc.
Chapter 8
Characters of Sn and An
There is little doubt that the world's best-loved finite (nonabelian) groups are
the symmetric groups Sn and the alternating groups An. In this chapter we
present reasonably practical algorithms for calculating their character tables.
8.1
Symmetric Groups
If 0 < n € N recall that a partition of n is a sequence A = ( A i , . . . , Ar) of
positive integers (the parts of A), satisfying
Ai > A2 > • • • > Ar
and
Ai -f- A2 + • • ■ + Ar = n.
It will be convenient to use another form of exponential notation for partitions - one example should make it clear. The partition A = (4,4,3,2,2,2,1)
of n = 18 will be written as A = (4 2 ,3,2 3 ,1).
The partitions of n determine the conjugacy classes in Sn - if K is a
conjugacy class and a G K write a = o\o~i ■• ■ ar, a product of disjoint cycles
(including 1-cycles), with |tr¿| > |CT¿| for i > j . Setting A¿ = \o~i\, all i, we
have a partition A = ( A i , . . . , Ar) of n. It will be convenient to refer to a as
an element of type A (note that this is related to, but not the same as, the
usual notion of cycle type of a).
It is often convenient to agree that each partition A of n has n parts, which
can be attained by simply adjoining sufficiently many 0 parts at the end.
Each partition A of n has a Young diagram, which can be thought of as
a (left-aligned) stack of empty boxes, the number of boxes in each row being
the parts of A. For example, the partition A = (3,2 2 ,1 2 ) of n = 9 has the
161
162
CHAPTER 8. CHARACTERS OF S„ AND An
Young diagram
Each Young diagram has an opposite diagram (sometimes called the conjugate
diagram), obtained by transposing it. If the diagram is of type A we will
denote the type of the opposite diagram by A'. For example if A = (3,2 2 , l 2 ),
as above, then A' = (5,3,1), with Young diagram
If A = (Ai, A2,...) and A' = (A'j,X'2,-- •) then it is easy to see that
Aj = |{A i :A j ><}|.
Let us write [ñ] for the set {1,2,.. .,n}. Thus, for example, Sn —
Perm([n]). A set partition A of [if] is as usual a partitioning of [ñ] into
a disjoint union of nonempty subsets, say [ñ] = AiÙ---ÙAr, labeled so
that |A<| > |Aj| if i > j . If we set Aj = |A»|, all », it is immediate that
A = (Ai,..., Ar) is a partition of n, which we will call the type of A; we write
X = A.
The symmetric group Sn acts naturally on the family of all set partitions
of [ñ} the orbits are determined by types. If A is a set partition of type
A then the stabilizer of A in S n will be denoted SA and is called a Young
subgroup of type A (or type A). If the sets in the partition are A i , . . . , Ar
then SA = Perm(Ai) x • • ■ x Perm(Ar).
If the elements of [ñ] are distributed among the boxes of a Young diagram
the resulting object is called a Young tableau. For A = (3,2 2 ,1 2 ), as above,
one possible Young tableau is
3
7
2
5
8
1
9
4
6
8.1. SYMMETRIC
GROUPS
163
The rows of a Young tableau determine a set partition of [ ñ ] , as do the
columns. For the example above the row partition is
({3,1,6}, {7,9}, {2,4}, {5}, {8}),
and the column partition is
({3,7,2,5,8}, {1,9,4}, {6}).
The Young subgroup of the row partition is
Perm({3,l,6}) x Perm({7,9}) x Perm({2,4}) x Perm({5}) x Perm({8})
Si S 3 x S 2 x S 2 x Si x Si
and that of the column partition is
Perm({3,7,2,5,8}) x Perm({l,9,4}) x Perm({6})
~S5 xS3xSi.
Note that the row partition and column partition are of opposite types. It is
clear that the intersection of any set in the row partition with any set in the
column partition has at most 1 element in it. As a consequence their Young
subgroups intersect trivially.
In general, two set partitions A = ( A i , - - , A r ) and M = (Mi, • •■,M s )
of [ñ] are said to be disjoint if \A¡ n Mj\ < 1 for all i and j . They are
called opposite if they occur as the row partition and column partition for
some Young tableau. Thus opposite set partitions are disjoint. Note that a
given set partition typically has several opposites; e.g. A = ({1,2,3}, {4})
is opposite to each of Mi = ({1,4}, {2}, {3}), M 2 = ({2,4}, {1}, {3}), and
M 3 = ({3,4},{2},{1}).
Proposition 8.1.1 If A is a set partition of\n^ then the Young subgroup SA
acts transitively on the collection of all partitions that are opposite to A.
Proof If T\ and T2 are two Young tableaux both having A as row partition
then Ti can be transformed to T2 by permuting the entries of its rows, i.e.
by an element of SA.
A
Proposition 8.1.2 Set partitions A and M o/[]ñ] are opposite if and only if
they are disjoint and of opposite types.
Proof =>■: This is clear.
•^: There is a simple algorithm for producing a Young tableau T whose row
partition is A and column partition is M: the ij-entry of T (when it exists)
is the unique element of A¿ n M,. The proof that this works becomes clear
(and is easily formalized) after seeing an example. See the following exercise.
A
CHAPTER 8. CHARACTERS
164
OF Sn AND An
Exercise
Apply the algorithm suggested in the proof above to the set partitions
A = ({2,6,9}, {4,7}, {3,8}, {1}, {5})
and
M = ({1,3,5,7,9}, {4,6,8}, {2}).
There is a useful partial ordering on the set of partitions of n. If A =
( A i , . . . , An) and p. = (pi, ... ,/¿„) say that A is dominated by /x, and write
A < n (or /i » A), if
i=l
i=l
for 1 < fc < n. Thus, for example, (2,2,2) -C (3,2,1), but (2,2,2) and
(3,1,1,1) are not comparable.
Define a lexicographic total ordering on the set of partitions of n as follows.
If A = ( A i , . . . , An) and ¡i = (pi,..., /i„) are partitions, write A < ß if A = fi
or if for some fc we have A¿ = /i¿ for i < k but A* < /i*.
Exercises
1. Show that dominance and the lexicographic ordering are the same if
n<5.
2. Show in general that if A «; /x then A < fjt.
Proposition 8.1.3 If A and M are disjoint set partitions o/jjT] having types
A and \i, then A C ( i ' .
Proof Let T\ and Tß be Young tableaux whose row partitions are A and
M, respectively. Imagine distributing the entries of T\, row by row, into the
boxes of the diagram of Tu. To begin, the Ai entries of row 1 of T\ must
go into different rows of Tu, of which there are y.\, so Ai < n\. Assume
inductively that the entries of the first A; — 1 rows of T\ have been distributed
and that £ i = 1 ^« ^ X)i=i /*i> 1 < ^ < * _ 1- Since row entries in Tu can be
rearranged without impairing disjointness we may assume that the entries of
the first Ä; - 1 rows of T\ have been distributed among the first k — 1 columns
of Tß. Thus the number of still vacant boxes in the first k columns of Tu is
£ i = 1 n\ — Y^iZi ^»> snd there are at most that many rows of Tu with vacant
boxes. Since it is possible to distribute the entries of rowfcof T\ into different
rows of Tu, it must be the case that Xk < £ * = i fi[ - YliZi •*«> as required.
A
The converse of the proposition above is also true (but more difficult
to prove); the two together are a version of the Gale-Ryser Theorem, an
important result in combinatorics. For an amusing proof see [27].
8.1. SYMMETRIC
165
GROUPS
Corollary 8.1.4 Suppose that A and p. are partitions of n. If there exist
Young subgroups of types A and ¡x having trivial intersection then A <C /i'.
Suppose now that A is a partition of n. Choose a set partition A of type A
and get thereby a Young subgroup 5A- Set r¡>\ = lf^, the principal character
of 5 A induced to 5„. Since ip\ is the permutation character of the action of
S„ on the set of all set partitions of type A it is independent of the particular
choice of A.
Denote by e\ the alternating character of 5A and set ¡p\ = e^ n , also
independent of the choice of A.
Proposition 8.1.5 If X and /x are partitions of n and (il)\,<pß) ^ 0, then
A«/x'.
Proof Choose partitions A and M of QT] of types A and /i, and denote the
respective Young subgroups by K and H. By the Mackey Theorem 6.4.2 we
have
»
i=l
where Hi = KXi f\H, the Xj's being (K-H)-double coset representatives, with
i i = 1. Thus each J/¿ is an intersection of Young subgroups of types A and /x.
Note that e#. = e ^ , since parity is preserved by conjugation. If any Hi ^ 1
then it is a nontrivial Young subgroup; hence it contains an odd permutation,
so IH{ ^ £H¡ and (l#¡,fj/j) = 0. Thus for {^\,ipß) to be nonzero there must
exist disjoint partitions of types A and ¡i, in which case A C n' by Proposition
8.1.3.
A
T h e o r e m 8.1.6 / / A is a partition of n then {ip\,^>y) = 1, so ip\ and <¿>v
have a unique common irreducible constituent, occurring with multiplicity 1
in each of them; it will be denoted x\Proof Choose opposite partitions A and A' of [n] of types A and A', and
denote the respective Young subgroups by K and H, so KC\H = 1. As in the
proof of Proposition 8.1.5 we have contributions of 1 to (ip\,<p\') from each
Hi = 1, and Hi = K D H = 1, so it must be shown that Hi ^ 1 for i > 1.
Suppose then that Hi = KXi n H = 1. Thus A*' and A' are disjoint, hence
opposite by Proposition 8.1.2. By Proposition 8.1.1 we have \Xih = A for
some h e H, so xth 6 K and Xi € KH, the (Ä"-ii)-double coset containing
x\ — 1; hence n = x\ and % = 1.
A
For each partition A of n denote by K\ the conjugacy class in 5„ of
elements of type A. Order the classes according to the lexicographic ordering
of the partitions. Thus Ä^i«) is first and K(n) is last. Order the collection
of characters \ \ according to reverse (descending) lexicographical ordering of
partitions, so X(n) is first and X(i») is last. Then form the square matrix X =
[X\(Kß)]- With the same indexing of rows and columns define Y = [rp\(Kß)].
166
CHAPTER 8. CHARACTERS
OF Sn AND An
Proposition 8.1.7 If Xv w a constituent ofip\ then A < v.
Proof Since Xv is also a constituent of <¿v we have (VA> w ) ^ 0, so A <S i/
by Proposition 8.1.3. Thus A < v by Exercise 2 on page 164.
A
Theorem 8.1.8 The characters xx are all distinct, so X is the character
table of Sn.
Proof Suppose that xx = Xu f° r partitions A and /*. Then xx is a constituent
of tpp and Xii is a constituent of ip\, so A < p. < A by Proposition 8.1.7, and
A = /x.
A
Corollary 8.1.9 VA = E{Wx,X»)Xv-v
> A}.
If A and \i are partitions of n then a fi-refinement of A is an expression of
each of the parts of A as a sum of parts of /z, the parts of ¿x being exhausted
in the process. An example should make this clear. Take A = (5,4) and
\i = (3,2 2 , l 2 ) , which we write as (3,2i,2 2 , l i , I2). Then there are 5 different
/¿-refinements of A:
(3 + 2 1 ,2 2 + l 1 + l 2 ) , (3 + 2 2 ,2i + li + l 2 ) , (3 + l 1 + l 2 , 2 1 + 2 2 ),
( 2 i + 2 2 + l i , 3 + l2), and (2i + 2 2 + 1 2 ,3 + l i ) .
Proposition 8.1.10 If X and n are partitions of n then the character value
ip\{Kß) is the number of distinct fi-refinements of A.
Proof Recall that VA is the permutation character of Sn on the collection
of all set partitions of [ñT] of type A. Choose a € K^ - we need to count the
set partitions of type A that are fixed by a. Write a = 0\ ■• • a„ with ai a
/¿¿-cycle, all i, and take M = ( M i , . . . , M,), the corresponding set partition,
i.e. Mi is the orbit ofCTJ.If A = ( A i , . . . , A r ) is of type A, then a fixes A if
and only if each Mj is contained in some A<, which is clearly if and only if M
determines a /x-refinement of A.
A
Thus calculation of the matrix y is a relatively straightforward combinatorial problem.
Proposition 8.1.11 If X is a partition ofn then xx1 = « X A , where as usual
e denotes the alternating character of Sn.
Proof We have <px = e|^ and e • VA = e • i f ; = (e|sA • l s A ) s " = efA by
Proposition 6.2.7, so <p\ = e ■ t¡>\. Now XA is the unique common constituent
of VA and <p\>, so exx is the unique common constituent of CVA = VA and
e w = Vv, which is x v - Thus xx' = exxA
The results above make possible an efficient recursive algorithm for calculation of the character table X, assuming that Y has been calculated. In fact
8.1. SYMMETRIC
167
GROUPS
because of Proposition 8.1.11 it is only necessary to calculate approximately
half of the rows of X, and as a consequence only those rows of Y are needed;
denote the truncated versions by X' and Y'.
Since X(n) = ls„ t n e first r o w OI" X' has all entries 1. If Xv has been
calculated for all v > A then by Corollary 8.1.9 we see that xx = *J>\ ~
5Z{(V'AiXi')x«':i/ > ^ } - Thus we need only take inner products of row A of
Y' with the already constructed rows of A", subtract those multiples of the
rows of X' from row A of Y', and insert the result as row A of X'.
To illustrate, let us calculate the character table of S^. We will simply
index the rows and columns of both X and Y' by the partitions of 5, ordered
as indicated above. The sizes of the classes have been inserted in Y' to
facilitate taking inner products.
First Y'.
\Kx\
(5)
(4,1)
(3,2)
(3,1 2 )
2
(I 5 ) (2, I 3 ) (2 ,1) (3, I 2 ) (3,2) (4,1) (5)
10
20
30 24
1
15
20
1
1
1
1
1
1
1
1
3
1
2
0
0
5
4
2
1
1
0
0
10
2
0
0
20
6
0
0
Then we begin with X{5) = ls 5 - Since (^(4,1),X(5)) = 1 we have
X(4,l) =^(4,1) ~X(5)Next (^(3,2), X(5)) = W>(3,2),X(4,1)) = 1, SO
X(3,2) = ^(3,2) - X(5) - X(4,l)2
Finally (V>(3,i )>X(5)) = 1, W(3,ii).X(4,i)) = 2 , and (^(3,i2),X(3,2)) = 1. so
X(3,l2) = ^(3,12) - X(5) - 2X(4,1)) - X(3,2)The remaining rows of X are then obtained via Proposition 8.1.11. Note that
(5)
(4,1)
(3,2)
(3,1 2 )
(2M)
a
(2,1 )
(I 5 )
(I 5 ) (2, I 3 )
1
1
2
4
1
5
6
0
-1
5
-2
4
-1
1
(2M)
1
0
1
-2
1
0
1
(3, I 2 ) (3,2) (4,1) (5)
1
1
1
1
-1
1
0 -1
-1
1
-1
0
0
0
1
0
-1
-1
1
0
1
1
0 -1
1
-1
-1
1
Characters of 5s
Exercise
Determine the character tables of Sg and 57.
168
8.2
CHAPTER 8. CHARACTERS
OF Sn AND An
Alternating Groups
Let us assume throughout this section that n > 2, so [Sn- An] = 2.
Since An is a normal subgroup of Sn it is a union of S n -conjugacy classes.
If K is an 5„-conjugacy class and K C An there are two possibilities: K
may also be an An-conjugacy class, or it may split up as a union of smaller
A n -classes.
Choose a € K. Observe that CA„ (O~) = An D Csn (<?)> so that CAn (<?) has
index either 1 or 2 in CSn{<r). If the index is 2 then |Ä"| = \Sn\/\CSn(cr)\ =
(2\An\)/(2\CAn(a)\)
= | c U » | , so K = c U » . If the index is 1 (i.e.
Csja) < An) then \K\ = [Sn:CsAa)} = [Sn:CAn(<r)} = 2[An:CAn(tr)) =
2| cU n (<r)\. Thus K splits in that case into two .¿„-classes, each of size \K\/2.
We see then that the key issue is whether or not there is an odd element
of Sn that commutes with a.
Proposition 8.2.1 If K is a conjugacy class in Sn, K Q An, and a 6 K,
then K is a conjugacy class in An if and only if some odd element of Sn
commutes with a; if that is not the case then K splits as the union of two
An-classes, each of size \K\/2. If X is the (partition) type of a then K splits
if and only if the parts of A are all odd and all different from each other.
Proof The first statement is clear from the discussion above.
If A has an even part then a has a cycle of that length, which is odd and
commutes with a. If A has 2 odd parts of the same length it is also possible to
construct an odd element that commutes with a - an example will make this
clear. If a has (abc)(def) in its cycle decomposition then r = (ad)(be)(cf)
is odd, and r commutes with a since conjugation of a by r simply permutes
the cycles (abc) and (de/). On the other hand, if the parts of A are odd and
distinct, and if some r € Sn commutes with o, then inspection of aT = a
shows that the cycles of r are either 1-cycles or have the same (odd) lengths
as cycles of a; hence T must be even.
A
In the case of splitting we may clearly take any o € K as a representative
of one i4,,-class, then o~T as a representative of the other, where r is any odd
element in Sn, e-g- T = (12).
For an example, note that the even classes in S% are those of types ( l 6 ) ,
2 2
(2 , l ) , (3, l 3 ), (3 2 ), (4,2), and (5,1). Of these only the class of 144 5-cycles,
of type (5,1), splits, and as representatives of the 2 i4„-classes we may take
a = (12345) and oT = (12345)<12) = (21345).
Now the characters.
If A ^ A' then it follows from Proposition 8.1.11 that \x and Xx agree on
the classes that lie in An. By Proposition 6.4.5 those characters restrict to
irreducible characters of An.
If A = A' then x\ splits as a sum of 2 conjugate characters, also by
Proposition 6.4.5, say X\\A„ = X\} + x{2)', each of degree X A ( 1 ) / 2 .
8.2 ALTERNATING
169
GROUPS
Since A = (Ai,...,A r ) = A' the Young diagram of A is symmetric; it
can be thought of as made up of "hooks", the first hook being the first row
together with the first column. Let us write s for the total number of hooks,
and Si for the length of the ¿th hook, i.e. the number of boxes in the hook.
Thus ¿i = 2Ai - 1, S2 - 2(A2 - 1) - 1 = 2A2 - 3, and so on.
For example, A = (4,3,2,1) has diagram
with two hooks of lengths ¿i = 7 and ¿2 = 3.
The hook lengths of A determine a partition S = ( ¿ i , . . . ,6,); it is clear
from Proposition 8.2.1 that K¿ splits in An into two classes, call them Kg
and K,(2)
an
For each class Kß, with \i ^ ¿, we have \ \«)/\Kß) = XÁK^/Z,
d it
remains only to determine the entries in the 2 x 2 box where the two split
characters \ \ take their values on the two split classes Kg.
S e t A = n¿=i <5 ¿To complete the story, note first that xx(Ks) — (—1)("-*>/2; abbreviate
it as t. Then we may write
{1)
Xx (K^)
=
P)ttrW)
X?(K¡
=
i + \A A
2
(1)
Xx
(*1 2) ) = l
-
NA •A
2
and
i — y/i ■
A
Xx2)(K 1
2)
)=
L + y/i
A
The proof of this is surprisingly (?) long and involved and will not be repeated
here. See [5] or [40].
For an example we may determine the character table of A5 from the table
for i s in the previous section. The even classes are (l 5 ), (2 2 ,1), (3, l 2 ) , and
(5), and only (5) splits into two j45-classes, which we will denote by (5i) and
(02). The first three characters of S5 restrict to irreducible characters of A5;
denote them simply by Xi. X2, and X3- Only X(3,i*) splits, and A = (3, l 2 )
has hook length partition ó = (5), since A has only one hook. For that pair
A, Ô we have t = ( - l ) ^ - 1 ) / 2 — l and A = 5. Denote the two conjugate
characters that are constituents of X(3,i2)U5 by X* a n d X5> and we have the
character table as follows.
CHAPTER 8. CHARACTERS
170
Xi
X2
X3
Xi
X5
(I 5 )
1
4
5
3
3
(2M)
(3, I 2 )
1
1
0
1
1
-1
-1
0
-1
0
(5i)
1
-1
0
OF Sn AND An
(52)
1
-1
0
2
l-\/5
2
2
2
1+V5
Characters of A$
Compare the table with the one on page 134.
Exercise
Determine the character tables of AQ and Aj.
The character tables of the alternating groups An, for 5 < n < 13, appear
exphcitly in the Atlas of Finite Groups [14], and a scheme appears in the
Atlas for constructing the character tables of the corresponding symmetric
groups.
There is a formidable literature on the representation theory of S„ (and
to a lesser extent that of An), including a great variety of combinatorial algorithms and the general theory of symmetric functions.Two recent references
that have become standard are the books of James and Kerber [40] and of
Sagan [55]; [40], in particular, has a very extensive list of further references.
Groups and Characters
by Larry C. Grove
Copyright © 1997 John Wiley & Sons, Inc.
Chapter 9
Frobenius Groups
Results from earlier chapters join forces in this chapter to establish many of
the important properties of a class of groups called Frobenius groups, which
are originally defined as permutation groups but have numerous equivalent
descriptions.
Frobenius groups occur naturally in a wide variety of group-theoretical
settings. For example, they played a prominent role in the proof by Feit and
Thompson [24] that groups of odd order are solvable.
9.1
Frobenius Groups and Their Characters
A transitive permutation group G on a set S (with \S\ > 1) is called a
Frobenius group if Stabe (s) ^ 1 for each s e S, but each g ^ 1 in G has at
most one fixed point, or equivalently Stabe («i) H Stabe («2) = G»i,«2 = 1 for
«i ,¿ 32 in S.
For a class of examples let F be a finite field with more than two elements
and let G = Aff(F), the (one-dimensional) affine group on F, i.e.
G =
{Tb<a:aeF*,b6F},
where TJ,I0: F -> F is defined by TÍ, I0 (X) = ax + 6, and the group operation is
composition of functions (see Chapter 2). Then G acts transitively on S = F,
and it is easy to check that each T^J , 6 ^ 0 , has no fixed points, whereas if a ^
1 then Tft)0 has the unique fixed point 6/(1 - o ) . Thus G is a Frobenius group.
The translation subgroup M = { T ^ : b 6 F} is normal in G; it is isomorphic
with the additive group F. The subgroup H = Stabe(0) = {ro,a-a € F*} is
isomorphic with the multiplicative group F*, and G is the semidirect product
Mxff.
The next proposition is the first of several purely group-theoretical characterizations of Frobenius groups.
171
172
CHAPTER 9. FROBENIUS GROUPS
Proposition 9.1.1 A group G is a Frobenius group if and only if it has a
proper subgroup H ^ 1 such that H n Hx = 1 for all x £G\H.
Proof (=>) Say that G acts on 5, take s G S, and let H = StabG(s). If
x G G \ if then sx ¿ s, and if 1 ^ y G H then a*" ^ s* (since y fixes only s),
and hence xyx~l$H; i.e. y £ ZP, so H n i / 1 = 1.
(<=) Set 5 = {Hx:x G G}. Then G acts transitively on 5 by right
multiplication. If we take s = H £ X then Stabc(s) = H, and if x £ G then
StabG(sz) = if*. In particular, if x G G \ H then Stabo(s) D Stabo(ax) =
A
H n Hx = 1, and it follows that G is a FVobenius group.
Corollary 9.1.2 If G is a Frobenius group and H < G is the stabilizer of a
point then NG(H) = H.
If G is a FVobenius group on S and H = Stabe (s) for some a € 5 then H
is called a Frobenius complement in G. Denote by M* the set of all a; € G
having no fixed points in S, and set M = M* U {1}. Thus
M = (G\U{/P:xeG})U{l};
M is called the Frobenius kernel of G. Note that M x = M for all ar G G.
Proposition 9.1.3 Suppose that G is a Frobenius group with complement H
and kernel M. Then
1. \M\ = [G:H] > 1 , and
2. ifK<G withKrtH = l then K CM.
Proof (1) Since H = NG(H) there are [G:H] distinct conjugates of H, so
| U {Hx:x € G}\ = [G:H](\H\ - 1) + 1 = |G| - [G:H] + 1. Thus \M\ =
| G | - ( | G | - [ G : t f ] + l) + l = [G:tf].
(2) Since K D if = 1 and K < G we have if D i / 1 = 1 for all x € G and
hence K CM.
A
Proposition 9.1.4 Suppose that G is a Frobenius group with complement
H, and suppose 6 6 d(H), with 6(1) = 0. Then (9G)\H = 9.
Proof Note first that 0G(1) = [G:Jf]0(l) = 0 = 0(1). If 1 ¿ y 6 H then
9G(y) = IH]-1 E t e o ^ * " 1 ^ ) . and r t y G # if and only if y G H n 4 # if
A
and only if í G H, in which case o(í -1 yí) = 9(y), so 0G(y) = 9(y).
Recall from page 123 that the character ring of a group G is
Char(G) = {Y^mXi-.Xi G Irr(G), m G z } .
9.1. FROBENIUS
GROUPS AND THEIR CHARACTERS
173
Theorem 9.1.5 (Frobenius) If G is a Frobenius group with complement H
and kernel M then M is a normal subgroup of G.
Proof Note that the issue here is not normality but rather the fact that
M is a subgroup. We shall show that M is the intersection of the kernels of
certain absolutely irreducible characters of G. Take (p ^ ljj in Irr(H), and set
9 = <P-V(1)1H 6 Char(tf), so0(1) = 0. Then (9G',9a) = (9, (9°)\H) = (6,9),
by Proposition 9.1.4, and
(9,9) = (<p- <P(1)1H,V-
¥>(1)1») = 1 + V(l) 2 -
Also (9°,1G) = (0,1//) = - V ( l ) Thus, if we set <p' = 9G + <p(l)lG, then
tp* € Char(G) and ((p*,lG) = 0. Furthermore
(¥>*, </) = 1 + <p(l)2 - 2y>(l)2 + <p(l)2 = 1,
so either (p* or — <p* is in Irr(G). Apply Proposition 9.1.4 again to see that if
y € H then
¥>*(!/) = * G (y) + V(l) = 6(y) + <p{l) = <p(y),
and hence tp'ln = ¡p. In particular, </?*(l) = <p(l) > 0, and <¿>* G Irr(G).
Set AT = n{ker(y>*): lH / <p € Irr(#)} « G. If y 6 K D # then y?(y) =
V*(y) = V*(l) = v ( l ) for all v? ^ Iff in Irr(if), and hence y = 1 by Corollary
5.2.31. Consequently Ä" Ç M by Proposition 9.1.3. On the other hand, if
1 ^ x € M then x ^ / F for any z G G, so
< / ( * ) = 0 G (x) + <¿>(1) = (¿(I) = v*(l),
and x € ker(y>"). Thus K = M <G.
A
Corollary 9.1.6 G »s í/ie semidirect product M x H.
Remark Frobenius proved the theorem above in 1901. Attempts have been
made to prove the theorem without using character-theoretic methods, but to
date no one has succeeded. Group-theoretical proofs have been given under
the extra assumption that H is solvable, by techniques involving the transfer
and cohomology theory.
Exercise
If p and q are primes and G is a nonabelian group of order pq, show that
G is a Frobenius group; find its kernel and a complement.
Proposition 9.1.7 Suppose that G is a Frobenius group with complement H
and kernel M. Ifl¿x£M
then CG(x) < M.
174
CHAPTER 9. FROBENIUS
GROUPS
Proof First suppose that h G H D C G (x). Then hx = h Ç. H D Hx = I, so
h — 1. In general, if y G CG(X) but y & M, then I jt y £ Hz for some z € G ,
and so *y € if n CG{ZX) = 1, and y = 1, a contradiction.
A
Proposition 9.1.8 Suppose that G is a Frobenius group with complement H
and kernel M. Then \H\ \ \M\ - 1.
Proof The complement H acts by conjugation on M, and if 1 £ x G M
then Stabjj(x) = C//(x) = 1 by Proposition 9.1.7. Thus M \ {1} is a union
of i/-orbits, each of size \H\.
A
Recall A subgroup K of G is called a Hall subgroup \î \K\ and [G: K] are
relatively prime.
Corollary 9.1.9 The complement H is a Hall subgroup ofG, and the kernel
M is a normal Hall subgroup.
Proposition 9.1.10 If G is Frobenius with complement H of even order then
the kernel M is abelian.
Proof Choose h G H with \h\ = 2. If 1 ¿ x € M then x ¿ xh G M by
Proposition 9.1.7. Take x,y G M and suppose that xh ■ x~l = yh ■ y~l. Then
(y h ) _ 1 x' 1 = (j/ - 1 !)' 1 = y~lx, so y = x, again by Proposition 9.1.7. Thus
the map x *-¥ xh ■ i - 1 is 1-1 on M to M, and M = {xA • x - 1 } . Note that
(x h • x~l)h = x ( x h ) - 1 = (xA • x - 1 ) - 1 . Thus the automorphism z >-> zh of M
coincides with the map Í H Z " 1 , and so M must be abelian.
A
Proposition 9.1.11 Suppose that G = M » H is a Frobenius group with
kernel M and complement H, and let z be an involution in H. Then zx = x _ 1
for all x € M.
Proof Since \H\ is even M is abelian by Proposition 9.1.10. If x € M
then z(xz)2 = (zxz)xz = x(zxz)z = {xz)2z, i.e. z centralizes (xz) 2 . But
(xz) 2 € M, so (xz) 2 = 1, or equivalently xx = x - 1 , by Proposition 9.1.7. A
Set S = { 1 , 2 , . . . , n} and X — S x S. If a is a permutation of X and
A = [ay] is an n x n matrix over a field F, define A" = [by], where bij = au,
with (fc, I) = (i,j)a. Check that if r is another permutation of X then ^4ITT =
{A°)T, so any permutation action on X determines a permutation action on
the set of n x n matrices.
Proposition 9.1.12 (Brauer's Lemma) Suppose that G is a permutation
group on X = 5 x S as above, that FCC,
and that A is an invertible
n x n matrix over F. Suppose further that for each a G G the matrix A*
can be obtained from A either by permuting the rows of A or by permuting
the columns of A, so G can be viewed either as a permutation group Gr on
the set of rows of A or Gc on the set of columns of A. Then the permutation
characters 0r and 6C of GT and Gc are equal.
9.1. FROBENIUS GROUPS AND THEIR CHARACTERS
175
Proof If a G G then there are permutation matrices R(a) and C{o~) for
which R(a)A = A" = AC{a); in fact, <7 t-» ^(cr)' and <r t-> C(a) are the
permutation representations of Gr and G c , respectively. Thus A~lR{a)A =
C{a), so tr R{aY = tr fí(<r) = tr C(<x), all <r G G, and 0 r = 6C.
A
Corollary 9.1.13 7h the setting of Brauer's Lemma the numbers of orbits
of Gr and Gc are equal.
Proof Since 0r = 9C we have (0 r , I G ) = (6C, I G ) , and these are the numbers
of orbits by the Orbit Formula (Proposition 3.1.2).
A
A fairly typical application of Brauer's Lemma appears in the proof of
the next proposition. It should be remarked that any character table is an
invertible matrix by the Second Orthogonality Relation (5.2.17).
Proposition 9.1.14 If G is a Probenius group with kernel M and if\p ^ I M
in Irr(M) then ip has inertia group IG(<P) — M■
Proof If A is the character table of M then G acts on rows of A by conjugating characters and on columns of A by conjugating M-conjugacy classes.
If x G G, x € Irr(M), and L is an M-class, then the actions are related by
XX{L) = XÍ 1 ^), s o Brauer's Lemma (9.1.12) applies.
Choose x G G \ M and suppose that XL = L for some M-conjugacy
class L, say L = C1A/(J/). Thus xy G L, so xy = my for some m € M,
m
X
V — V-, so m~xx G CG{V) and x G TUCGÍV)- But then x G M by
Proposition 9.1.7, a contradiction unless y = 1. Thus 6c{x) = 1 for all
x G G \ M, and so also 9T(x) = 1 by Brauer's Lemma. But that says that
<px # if if 1 M ^ <P G Irr(M).
A
If G is Frobenius with kernel M and complement H then H 5í G/M, so
any character of H can be viewed as a character of G/M, hence also as a
character of G by lifting, as discussed in Chapter 5. In particular, ITT(H) can
be viewed as a subset of Irr(G) in a natural way.
Theorem 9.1.15 Suppose that G is a Frobenius group with complement H
and kernel M.
1. If 1 M # <P G Irr(M) then yG G Irr(G).
2. Irr(G) = lir{H)Ù{<pG: lM ± <p G Irr(M)}.
3. If x — VG>
character of H.
as
in part 1, then
X\H
=
<P{1)PH,
where pu is the regular
Proof (l) Proposition 9.1.14 and Corollary 6.4.4.
(2) Take x £ M ^ O and let ip G Irr(M) be a constituent of X\MIÍ¡P¿1M
then (tp°,x) - (*P,X\M) > 1, so x = VG since both are irreducible. If ip = 1 M
then M < ker(x) by Corollary 6.3.4, so we may view x as a character of G/M,
hence of H.
CHAPTER 9. FROBENIUS
176
(3) Take h ¿ 1 in H. If x € G then hx & M, since M<G.
X(h)
GROUPS
Thus
= v>G(fc) = IMP 1 5 3 ¿ ( z _ 1 M = 0,
x€G
whereas *(1) = ¥>G(1) = [G: M] V (1) = <p(l)\H\, so X | „ =
<P(1)PH.
A
Theorem 9.1.16 Suppose that G is Frobenius with complement H and kerThen tp° = 6G if and only if
nel M, and that ip, 8 e Irr(M) \ {1M}.
9 € Orbj/Op). Furthermore, | Orb/f (VJ)| = |ff|, so G has [c(M) - l]/\H\
distinct irreducible characters of the form <pG', cp € Irr(M).
Proof For the first statement see the exercise on page 143. For the second
use Proposition 9.1.14 and the fact that [G: M] = \H\.
A
Corollary 9.1.17 If G is Frobenius with complement H and kernel M then
c(G)=c(H) +
[c(M)-l]/\H\.
Exercise
Use the theorems above to compute the character table of the affine group
Aff(Ä") if if is a field with q > 2 elements.
9.2
Structure of Frobenius Groups
We begin with two more group-theoretical characterizations of Frobenius
groups.
T h e o r e m 9.2.1 A finite group G is Frobenius if and only if it has a nontrivial proper normal subgroup M such that i / 1 / i í M then Co(x) < M.
Proof (=►) Proposition 9.1.7.
(■^) We show first that M is a Hall subgroup of G. If not there is a
prime p, a Sylow p-subgroup P of M, and a Sylow p-subgroup Q of G with
1 7E P < Q but P ¿ Q. Choose x e Z(Q), with | i | = p, and note that
x $ P, for otherwise CQ(X) > Q. Choose y ¿ 1 in P. Then x € G G ( J / ) , but
x $ M since P = M ("IQ. That is a contradiction to Co(y) < M, so M is a
normal Hall subgroup of G. By the Schur-Zassenhaus Theorem (4.2.3) there
is a complement H to M in G.
If x G G and H D Hx ¿ 1 write x = hz, with h € H and z € M. Then
x
H = Hhz = Hz and H n Hz ¿ 1. Choose y ¿ 1 in H with zy € H. Then
z
yy~1 = z{yz-ly-1)
€ i / n M = l , s o y € G G (¿), contradicting CG(z) < M
unless 2 = 1. Thus x = h e H and G is Frobenius by Theorem 9.1.1.
A
9.2. STRUCTURE
OF FROBENIUS
GROUPS
177
T h e o r e m 9.2.2 Suppose that \G\ = mn, with (m,n) = 1, that either xm = 1
or xn = 1 for all x € G, and that M = {x € G:xm = 1} < G. Then G is
Frobenius with kernel M. Conversely, if G is Frobenius with kernel M and
complement H, and if \M\ = m, \H\ = n, then either xm = 1 or xn = 1 for
allxzG
and M = {x € G: xm = 1}.
Proof Note first that (|M|,n) = 1, since exp(M) is at least a divisor of m.
Since \M\ | \G\ = mn we may conclude that \M\ | m. But for each prime p
dividing m there is a Sylow p-subgroup of G in M, so \M\ = m and M is a
normal Hall subgroup of G. By the Schur-Zassenhaus Theorem (4.2.3) there
is a complement H to M in G. If H n Hx ^ 1 but x $ H we may assume that
x e M . Choose M l i n i / n i P . Then ^/i/i" 1 = i ( / i i - 1 / i - 1 ) e H n M = l,
so hx = x/i. Also |x| | m and \h\ | n, so |x/i| = |x||/i|, contradicting the fact
that \xh\ must be a divisor of either m o r n . Thus in fact H must be a
Probenius complement by Theorem 9.1.1.
The converse is clear from the definition of the Frobenius kernel.
A
Exercise
If G is Frobenius with kernel M show that M char G; show in fact that
a(M) < M for every endomorphism a of G, so M is fully characteristic
inG.
Proposition 9.2.3 Suppose that G is Frobenius with kernel M and complement H and that 1 ¿ M x < M, 1 / Hi < H, with Hx < NG(Mr).
Then
Gi = M\Hi is Frobenius with kernel Mi and complement Hi.
Proof If 1 JÉ x e Mi then C G l (x) < (Mii7i) n M = Mi. Apply Theorem
9.2.1.
A
Proposition 9.2.4 Suppose that G is Frobenius with kernel M and complement H, and that K < M, K ¿ M, and K < G. Then G/K is Frobenius,
with kernel M/K.
Proof Clearly M/K is a normal Hall subgroup of G/K. If \M/K\ = m and
[G: M] = n, then M/K = {xK € G/K: (xK)m = 1} by the exercise on page
84, and if xK $ M/K then x n = 1, so (xK)n = 1 in G/K. Apply Theorem
9.2.2.
A
Exercises
1. If G is Frobenius with kernel M and K < G show that either K < M or
M <K.
2. If G is Frobenius with kernel M and K <G show that (1) K < M, or
(2) K <1 M = 1, or (3) K is Frobenius with kernel KClM.
178
CHAPTER 9. FROBENIUS
GROUPS
Proposition 9.2.5 If G is abelian, and the only characteristic subgroups in
G are 1 and G, then G is elementary abelian.
Proof Choose a prime divisor p of \G\. The subgroup
H={xeG:xp
= 1}
is characteristic, so H = G and G is elementary abelian.
A
A Frobenius group G is said to be minimal if no proper subgroup of G is
a Frobenius group.
Theorem 9.2.6 If G is a minimal Frobenius group with kernel M and complement H then M is elementary abelian and H has prime order.
Proof By Proposition 9.2.3 \H\ is prime, say \H\ = q. If P is a Sylow
p-subgroup of M and N = NG(P) then G = NM by the Frattini Argument
(Proposition 4.2.1), and hence q | \N\ since |G| = q\M\. Since if is a Sylow
g-subgroup of G we may assume that H < N. Thus HP is Frobenius by
Proposition 9.2.3, so HP = G by minimality of G, and we may conclude as
well that M = P. If K ^ 1 is a characteristic subgroup of M then K <G,
and HK is Frobenius, again by 9.2.3, so K = M by minimality of G. In
particular, Z(M) = M and M is abelian. Thus M (= P) is an elementary
abelian p-group by Proposition 9.2.5.
A
Theorem 9.2.7 If G is Frobenius with complement H then no subgroup of
H is Frobenius.
Proof If the result is false we may assume, by Proposition 9.2.3 and Theorem
9.2.6, that H itself is Frobenius and minimal, hence that its kernel K is
elementary abelian and its complement Q is cyclic of prime order q. If r is a
prime, r | \M\, R is a Sylow r-subgroup of M, and N = NG(R), then G = NM
by the Frattini argument, so \H\ | \N\. Since Ni)M is a normal Hall subgroup
of N it has a complement L by the Schur-Zassenhaus Theorem (4.2.3). Note
that G/M = NM/M a N/(N n M), so \L\ = [G: M] = \H\. Thus G = ML,
L a G/M ^ H, and L is minimal Frobenius. Since L < N = NG{R), RL is
a subgroup of G, and if 1 ^ x € R, then CRL(X) < (RL) n M = R, so RL is
Frobenius with kernel R.
In other words, we could have assumed at the outset that M is an r-group,
so we proceed now to do so. Furthermore we may assume by Proposition
9.2.4 that no nontrivial normal subgroups of G are properly contained in M.
In particular, M has no nontrivial proper characteristic subgroups, so M is
an elementary abelian r-group by Proposition 9.2.5. We may now view M
(written additively) as a vector space over Z r . The action of if on M by
conjugation is then a Z ^representation T of H on M. If ft e H and T(h) = 1
then ft € CG(M), so ft = 1 by Proposition 9.1.7, and T is faithful. The only
9.2. STRUCTURE
OF FROBENIUS
GROUPS
179
T-invariant subspaces are 0 and M, since M has no other subgroups normal
in G; i.e. T is irreducible. Choose a finite extension F of Z r that is a splitting
field for both H and K (5.1.12).
Say that TF ~ Si © • • ■ © Si, with each Si absolutely irreducible, and say
that S = Si acts on the F-subspace V of MF. Write V = Vi © • • • © V*, with
each V¡t an irreducible if-invariant subspace. But K is abelian, so each V<
is one-dimensional, and if x G K then S(x) is a diagonal matrix. Combine
the subspaces Vi so that V = Wi © • • • © W u , where S(x) restricts to a scalar
matrix on each W<, all x €_ Ä", with different scalars for some x if i ^ j .
Observe that if « e V and 5(x)u = Xxv, Xx G F , for all x £ K, then « e W j
for some i.
Say that Q = (y) and choose i; G Wi. For each x G Ä" we have S(x)S(y)v =
S(y)S(y~lxy)v
= S(y)XxvV — XxvS(y)v, so 5(j/)u G W} for some j . Thus
for each i there is some j = j(i) so that S(y)Wi = Wj^). In other words,
Q acts as a permutation group on the set {Wi}, and it has no fixed points
since a fixed point would be both Q-invariant and (see above) if-invariant
as a subspace, hence if-invariant, whereas 5 is irreducible on V. Since Q
is cyclic, of prime order q, S(y) permutes each Q-orbit in {Wi} cyclically as
a g-cycle. Relabel if necessary so that S(y)Wi = W2, S(y)W2 = W3, . . . ,
S(y)Wq-i = Wq. Choose w ¿ 0 in Wi and set v = w + S(y)w + S(y2)w H .
Then clearly S(y)v = v ^ 0, so 1 is an eigenvalue of S(y), hence of T(y).
That means that for some z € M, with z / 1 (i.e. z nonzero in the vector
space M) we have T(y)z = vz = z. But then y € CQ(Z), contradicting
Proposition 9.1.7 and proving the theorem.
A
Theorem 9.2.8 The Frobenius kernel of a Frobenius group is unique; i.e.
there do not exist distinct nontrivial proper normal subgroups each of which
contains the centralizer of each of its nonidentity elements.
Proof Suppose that M\ and M-¿ are both Frobenius kernels in G, say with
\Mi\ = nu and [G:M<] = n¿. Set N = Mx n M 2 . Then {MXM2)IN =
(M\/N) x (M2/N) (internal direct product). If M i ^ M 2 then there is a prime
p with p | mi, p | T7i2, and if M2^M\ there is a prime q with qf m\, q ni2, so
q I n i . In that case choose X./V e M\/N of order p and yN G M2/N of order
q. Then {xyN)mi =ymiN
¿ N and {xyN)ni = x ni JV ¿ N, contradicting
Theorem 9.2.2, according to which either (xy)mi = 1 or (xy)" 1 = 1. We
may assume, then, that M\ < M2. Let H\ be a complement for M\, and set
K = Hi n M2 < HL If 1 # x G K then CHl (*) < Hi n M 2 = K, so # i is
Frobenius with kernel K by Theorem 9.2.1. This contradicts Theorem 9.2.7,
so K = Hi n M 2 = 1. But then \M2\ < [G: Hi] = \MX\, and hence Mj = M 2 .
A
Remark Since M is a normal Hall subgroup it follows from the remark following the proof of the Schur-Zassenhaus Theorem (4.2.3) that the Frobenius
180
CHAPTER 9. FROBENIUS
GROUPS
complement H is uniquely determined up to conjugacy in G. This does not
depend on the Feit-Thompson Theorem ([24]), since the kernel M must in
fact be solvable (see below).
Proposition 9.2.9 (Burnside) Suppose that G is Probenius with kernel M
and complement H, and that p, q are primes in N. If K < H and \K\ = pq
then K is cyclic.
Proof If p 5e q this follows from Theorem 9.2.7 and Exercise 3 on page
173. Suppose then that p = q. Thus if K of order p 2 is not cyclic then
it is elementary abelian. By the reductive arguments used in the proof of
Theorem 9.2.7 we may assume that H = K and that M is an elementary
abelian r-group for some prime r ^ p. As in that proof the action of H on
M by conjugation determines a faithful irreducible Z ^representation T oí H
on M. Choose a finite extension F of Z r that is a splitting field for H, so
TF ~ Ti® • • • © Tjfe, with deg T¿ = 1, all i. Thus for an appropriate choice
of basis T(x) is diagonal, all a; € H, and each diagonal entry T¿(x) is a pth
root of unity in F. There are at most p distinct pth roots of unity in F, and
\H\ = p 2 , so there are x, y € H, x ¿ y, with 7\(x) = Ti(y), or Ti(a;y _1 ) = 1.
If z — xy~l this says that 1 is an eigenvalue of T(z), so there is an eigenvector
u € M, u ^ 0 (i.e. multiplicatively u ^ 1), and T(z)u = u, or zu = u. Thus
1 T¿ z € CG(U) n H, contradicting Proposition 9.1.1.
A
Theorem 9.2.10 Suppose that G is Probenius with complement H, thatp \\H\
is prime, and that P is a Sylow p-subgroup of H. Ifp is odd then P is cyclic;
if p = 2 then P is either cyclic or a generalized quaternion group Q2* > °f
order 8 or greater.
Proof Take K < Z(P) with \K\ = p. If there is a subgroup L of P
with \L\ — p and L ^ K then KL is noncyclic of order p 2 , contradicting
Proposition 9.2.9. Thus P has only one subgroup of order p, and the theorem
follows from Theorem 4.3.11.
A
If a Sylow 2-subgroup of a Frobenius complement H is cyclic then it follows
from the theorem above and Theorem 4.1.17 that H is split metacyclic, i.e.
a split extension of a cyclic group by another cyclic group. In particular, H
is solvable. Zassenhaus has shown more generally (see [50], page 196) that
if H is a solvable Frobenius complement then H has a normal subgroup H0
which is split metacyclic and such that the quotient H/H0 is isomorphic with
a subgroup of the symmetric group 54.
There are nonsolvable Frobenius complements, however. An example with
H = SL(2,5) will follow shortly. In this case Zassenhaus has shown (see [50],
page 204) that it is approximately the only such example, in that if H is a
nonsolvable Frobenius complement then H has a subgroup Ho of index 1 or
2 with H0 = SL(2,5) x K, where K is split metacyclic of order prime to 30.
9.2. STRUCTURE
OF FROBENIUS
181
GROUPS
For the example we observe first that H = SL(2,5) has the presentation
(x,y,z\x3
via
X i~¥
= y5 =z2 = 1, xz = x,yz =y, {xy)2 = z),
0 1
4 4
, y>->
1 0
i i
,
Z >-¥
-1
0
0
-1
The matrices satisfy the relations in the presentation and they generate
SX(2,5), so there is a homomorphism onto SL(2,5). A Todd-Coxeter coset
enumeration (see Chapter 1) of the subgroup (y, z) shows that it has index
12, so the group presented has order 120 or less and hence is isomorphic with
H.
Let F = Z29 and note that 122 = - 1 , l l 2 = 5 (any finite field of characteristic not 2, 3, or 5 in which - 1 and 5 are squares would do as well).
Define
X =
-1 1
-1 0
Y =
0
-12
-12
-6
Z =
-1
0
0
-1
in SL(2,29). Verify that the matrices X, Y, Z satisfy the relations in the
presentation for H. It follows that H maps isomorphically into SL(2,29) via
i 4 X , y i4 y , z 4 Z. With only a slight abuse of notation we take the
point of view that H is that subgroup of SL(2,29).
Now set M = F © F, so H < Aut(M), and finally let G be the resulting
semidirect product M x H.
Exercise
If v € M , but v ¿
1A/, show that CG{v) = M. Thus G
is
Frobenius, with complement H = SL(2,5). (Hint If {w,g) € CG{V) \ M
then v is an eigenvector of g for eigenvalue 1. But then both eigenvalues
of g are 1 (since det g = 1), and the order of g is 29.)
If G is any group and if a € Aut(G) fixes only 1, then a is called a fixedpoint-free (abbreviated f.p.f.) automorphism. It was known to Frobenius
that any (finite) group with an f.p.f. automorphism of order 2 or 3 must be
nilpotent, and he conjectured that any group with an f.p.f. automorphism of
prime order must be nilpotent.
If G is Frobenius then every x ^ 1 in H provides an f.p.f. automorphism
of M by conjugation, since x would be in the centralizer of any fixed point.
In particular, M has f.p.f. automorphisms of prime order, which is the basis
for the conjecture by Frobenius that M must be nilpotent. J. Thompson, in
his Ph.D. dissertation in 1959, proved the more general Frobenius conjecture:
a group with an f.p.f. automorphism of prime order is nilpotent. For a proof
see [29], page 337.
CHAPTER 9. FROBENIUS
182
GROUPS
It seems in fact that in Frobenius's time no examples were known of
nonabelian FVobenius kernels. L. Weisner thought in 1939 that he had proved
all FVobenius kernels are abelian, but in 1940 O. J. Schmidt produced an
example of a nonabelian kernel. We will give a later example due to N. Ito.
Take p, n € N, with p prime, n odd, and n > 1, and let F be the field
with pn elements. For each o, b € F define a matrix
H(a, b) =
1 a
0 1
0 0
1
Since the map a *-* a? is an automorphism of F it is easy to check that
n(a,b)fj,(c,d) = fi(a 4- c,adp + b + d),
and it follows that M = {/x(a, b):a, b,€ F} is a nonabelian group of order
p2n.
Choose r € F of (multiplicative) order (p" - l ) / ( p - 1), and define a
map a: M —► M via ap(a, b) = fi(ra,rp+lb).
Again it is easy to check that
a is an automorphism of M, so H = (a) < Aut(M). Define G to be the
corresponding semidirect product M » H.
Exercises
1. Show that if 1 ^ x = /x(a, b) € M (above) then CG(X) < M, and
conclude that G is a FVobenius group with nonabelian kernel M.
Let F = Z 5 and H = (a, b) < SL(2,5), where
a=
3 4
3 1
and
b=
3 0
0 2
Set M = F © F, so H < Aut M, and set G = M x H.
(a) Show that H 5i Q 2 » C3, of order 24. (¿fini
(b) Show that ti 1 ¿ x e M (i.e. x ^
Consider (6,6 o ).)
jj ) then CG(x) < M. (Hint
It is sufficient to show that if 1 ^ c 6 i î then c does not have 1 as
an eigenvalue.)
(c) Conclude that G is FVobenius with complement H and that H is
solvable and nonabelian.
Groups and Characters
by Larry C. Grove
Copyright © 1997 John Wiley & Sons, Inc.
Chapter 10
Splitting Fields
This chapter explores various aspects of the effects on characters and representations of a finite group when the ground field is extended. The particular
case of K extended to C receives special attention, with interesting consequences relating to elements of order 2.
10.1
Splitting
Recall that the exponent of a finite group G is the least common multiple of
the orders of all the elements of G; it is the minimal integer m > 1 for which
xm = 1 for all x € G. Denote the exponent of G by exp(G).
It seems to have been conjectured by H. Maschke in about 1900 that if
exp(G) = m and C 6 C is a primitive mth root of unity then F = Q(Ç) is a
splitting field for G. The conjecture was proved by R. Brauer [6] in 1945.
The next theorem should probably be called Blichfeldt's Splitting Field
Theorem. It will be subsumed immediately by the more general theorem of
Brauer; it is included because its proof requires much less preparation but
also for historical interest - it provided an early proof of a special case of
Maschke's conjecture..
T h e o r e m 10.1.1 Suppose that G is nilpotent of exponent m. LetC> = Cm € C
be a primitive mth root of unity, and set F = Q(C)- Then F is a splitting
field for G.
P r o o f Recall from Corollary 6.3.11 that if x € Irr(G) then x = * G for
some linear character A of a subgroup H <G. It follows that if S is a (onedimensional) matrix representation of H with character A then T = SG is a
matrix representation of G with character x- But S(x) = X(x) 6 F for all
x e G, and consequently all matrix entries of T(x) are in F by Proposition
6.2.3, i.e. T is an F-representation. Thus a full set of inequivalent absolutely
183
CHAPTER
184
10. SPLITTING
FIELDS
irreducible matrix representations of G can be written over F and F is a
splitting
field.
A
Theorem 10.1.2 (Brauer's Splitting Field Theorem) Suppose that G is
of exponent m, let Ç = £ m G C be a primitive mth root of unity, and set
F = 0(C)- Then F is a splitting field for G.
Proof Let T be an irreducible matrix representation of G over C, with
character x £ Irr(G). It will suffice to show that T is equivalent with an F representation. Suppose by way of contradiction that it is not. By Brauer's
Induction Theorem (6.5.4) we may write x = #i — #2, where each 0< is a
character that is a Z-linear combination of characters induced from linear
characters of subgroups of G. As in the proof of Theorem 10.1.1 there are
F-matrix representations Si and 52 of G having characters 0i and 62, respectively. Since x+02 = #ii T(BS2 is equivalent, over C, to S\ by Theorem 5.2.12.
Thus, even though T is not equivalent with an F-representation, T (B S2 is
for some F-representation 52- We may assume that 52 is of lowest possible degree with that property. By Maschke's Theorem (5.1.1) we may write
\-r)k and 62 = £1 H
1- £1, where each t]i is the character of an
0i = 771 H
irreducible F-matrix representation Ui, and likewise each & is the character
of an irreducible F-matrix representation V¿. Thus
x + ft + • • • + íi = m + ■ ■ ■ + VkSince (x + (>2,£t) > 0 we have iß\,£t) > 0, and hence (r]i,£i) > 0 for some
i; we may assume that {r¡k,Ce) > 0. By Theorem 5.2.6 (i) it follows that
7)k = £e, and so
X + Éi + • ■ • + £<_i = 171 + • • ■ + Tjfc.
i-
But then T© V\ © • • • © V/_i is equivalent with U\ © • • • © Uk-i, contradicting
the minimality of the degree of 52 and completing the proof.
A
10.2
The Schur Index
The dihedral group Z)4 = (a, b | a 4 = b2 = (ab)2 = 1) has a Q-matrix representation T with
T(a) =
0
1
-1
0
and
T(b) =
1
0
0
-1
with character XT whose values at 1, a 2 , a, b, ab are 2, - 2 , 0, 0, 0.
The quaternion group Q2 = {c,d | c4 = l.c 2 = cP,cd = dc~x) has a
character x that "equals" XT, in the sense that it takes on the same values
on the five classes of Q2 (suitably ordered; see Exercise 2 on page 117). If
185
10.2. THE SCHUR INDEX
S is a matrix representation with character \ t n e n S(c) must have order
4 and must be diagonalizable. If S(c) is to have real entries it must have
i
0
eigenvalues ±i and be similar (over C) to
, so all possible S(c) are
0
-i
similar (over R; e.g. see [31], page 143), and without loss of generality we
0 - 1
may take S(c) =
0
x y
with rational, or at least real,
u v
-y
x , and therefore
entries. Since cd = d e - 1 we have —u —v
x
y
—v u
u = y and v = —x. Since d2 is of order 2 in the center we must have
-1
0 "
S(d*) =
o - i , so
Let us try next to find S(d) =
x
y
x
y
y
-x
x2+y2
0
y
-x
0
x2 + y2
-1
0
0
-1
which is of course impossible over R.
Thus, even though the values of x lie in Q, x 1S not a Q-character, i.e. it
is not the character of a Q-representation of Qi. If we extend Q to K — Q(i)
(i 2 = —1 as usual), then it is easy to check that with S(c) as above and
S(d) =
[ ¿
o 1
. w e obtain a .^-representation with character x-
We shall return to this example after a brief digression.
If F is a field and A" is a finite extension of F it is customary (and
useful) to view K as an F-vector space. Each a € K determines an F-linear
transformation of K (call it a) by multiplication; i.e. ä{b) = ab, all bÇ. K. If a
basis {b\,..., 6„} for K over F is chosen then each ä is represented by an n x n
F-matrix à. The mapping a >-+ à is called the regular (matrix) representation
of K over F - it is an isomorphism from K to a field of matrices over F .
For each a € K define the trace from AT to F of a to be TrK/F(a) =
tr(â) = tr(à), which is of course independent of the basis chosen for K over
F . The function TX^/F' K -> F is an F-linear transformation, and if b 6 F
then TrK/F{b) = [K: F] ■ b (since b = bl).
Returning to the example above, take {1, ¿} as a basis for K over Q and
observe that
0
0 -1
1 =
i =
1
1
0
so for each a — u + vi € K we have
ó =
u
V
-.—V
u
186
CHAPTER
10. SPLITTING
FIELDS
Replace each matrix entry a in the representation S by its regular representation ó to determine a Q-representation 5 ' , with
ro o
0 0
1 0
0 1
-1
0
0
0
0]
"0
1
0
0
-1
0
0
0
-1
0
0
0
0
0 -1
0"
0
1
0
Note that the character of S' is x' — 2*. A routine calculation with block
matrices shows that the centralizer CQ(S') consists of all rational matrices of
the form
u
V
a;
y
—V
X
u
-v
— V —u
X
y
X
V
—u
-y
which is Hamilton's division algebra HQ of rational quaternions (e.g. see [31],
page 48), so S' is an irreducible Q-representation by Proposition 5.1.3.
To summarize, x is an absolutely irreducible character of G = Q2 having
values in Q; it is not a Q-character, but 2\ is the character of an irreducible
Q-representation.
Proposition 10.2.1 Suppose that F and K are fields, with F Ç K Ç C and
[K: F] finite, and that T is a K-(matrix) representation of G with character
X- Then there is an F-representation T" of G with character x' = TXK/F °XProof Choose a basis for K over F. If T(x) = [Uj(x)], x G G, let T'(x)
be the F-matrix whose partitioned form is [iy(a;)], with entries of [iy(a;)]
replaced by their regular representation matrices for K over F. Then T" is
an F-representation and
X'(x)
=
trT'(x)=Çtr[t«(x)] = ÇTrK/J,[t«(x)]
i
= T r K / F ( £ '«(*)) =
i
^K/FÍXÍX)).
i
A
For any field F C C and any C-character t}> of G define F{ip) to be the field
obtained by adjoining to F all the values of ip, i.e. F(V>) = F({ip(x): x € G}),
a finite extension of F .
Proposition 10.2.2 / / F C C and x € Irr(G) then there is some m > 0 in
N such that mx is an F(x) -character.
Proof Choose a finite extension K of F(x) for which x i s a /^-character
(e.g. K could be a splitting field for G). Apply Proposition 10.2.1 to x,
ox is the character of an F(x)-representation of G. Set
so that TrK/FM
187
10.2. THE SCHUR INDEX
m = [K:F(x)].
a; eG.
Since x(x) € F(x) we have TrK/F{x)(x(x))
= mx(x) for all
A
The proposition above calls for a definition. If x G Irr(G) and F C C then
the Schur index mF(x) of X relative to F is the least positive integer m for
which mx is an F(x)-character, i.e. the character of an F(x)-representation
ofG.
Clearly trif (x)x is the character of an irreducible F(x)-representation, by
the minimality of mp(x).
Proposition 10.2.3 Suppose that x € Irr(G), F C C , and 0 < n G N.
is an F'(x)-character then mF(x) \ n-
Ifnx
Proof Say that T is an F(x)-representation of G with character nx- By
Maschke's Theorem (5.1.1), T ~ 7\ 0 • • • © 7*, with each Tt an irreducible
F(x)-representation. If V« is the character of T¡ then tpi = n<x, with 0 <
rij G N for each i and J2in» = n - Now n¿x and mF(x)x ^ e both irreducible
F(x)-characters, and (niX,mF{x)x)
= nimF{x) > 0, so n¿x = mF(x)x bY
Theorem 5.2.6. Thus n = kmF(x).
A
Proposition 10.2.4 Suppose that x € Irr(G), FÇK Ç C
ond X *s a K-character of G. Then mF(x) \ [K: F(x)].
IÜÍÍA
[if: F] ^m'te,
Proof Set n = [K: F(x)] and apply Proposition 10.2.1. There is an F(x)representation T whose character is TiK/F^x) ox. Since x(x) € F ( x ) , all
a; G G, we have TrKy F{x )(x(:r)) = [K:F(x)]x(x)
= nx(x), and n x is an
F(x)-character. Thus mF(x) \ n by Proposition 10.2.3.
A
Remark Given F Ç C and x € Irr(G) there is a field L, with F(x) Ç L Ç C ,
such that x is an L-character and [L: F(x)] = mF(x). Thus mF(x) also could
have been defined as the minimal degree of an extension field K of F(x) for
which x is a A"-character. For a proof (which involves the theory of central
simple algebras) see [18], page 469, or [22], page 128.
Suppose that F is any subfield of C. If exp(G) = n and we set K = F(Q,
where Ç G C is a primitive nth root of unity, then K is a splitting field for G
by Brauer's Splitting Field Theorem (10.1.2). The Galois group G(K:F) is
abelian (e.g. see [31], page 97).
If 0 is any C-character of G then F(9) Ç K, and F{9) is a normal extension
of F since G{K: F) is abelian. Let G - G(F(6): F ) , viewed as a subgroup (via
restriction) of G{K: F). If T is a A"-matrix representation with character 9,
say with T(x) = (iy(i)], all x, then for each a G G we may define T" via
T"(x) = [a(tij (x))]. Then T" is also a /^-representation of G, whose character
is clearly 9", with 9"{x) = o{9{x)), all x G G. Note that if 1 ¿ a G G then
9a / 9, by elementary Galois theory. Furthermore {9°,9") = (0,9), so 9" is
absolutely irreducible if and only if 9 is.
CHAPTER
188
10. SPLITTING
FIELDS
The characters 9", a € G = G(F(6):F), are called the Galois conjugates
of 6 over F. Note that £ { ö < 7 ; a € </} has its values in F.
The fact that Irr(G) contains the Galois conjugates of each of its characters can be very useful in calculating character tables. For example, the
final character X3 of AB obtained on page 134 via the Second Orthogonality
Relation could have been written immediately as a Galois conjugate of \2The next proposition is a bit of a digression; it illustrates an easy application of the Galois conjugates discussed above.
Proposition 10.2.5 Suppose that exp(G) = n, x G G, and that 6 is a Ccharacter of G. Suppose that if m € N is relatively prime to n then x ~ xm
(conjugate in G). Then 6(x) € Q.
Proof Let G = £(Q(C):Q)> as above, consisting of the automorphisms am
determined by o-m'C, *-> Cm> w^ith (m,n) = 1 and 0 < m < n. As in the
proof of Proposition 5.2.5 we have 0(x) = J ^ C 6 ' - Since x ~ xm we have
6{x) = e(xm) = Zi Cim = Ei *m(Cei) = <7m(0(x)). Thus 8(x) is fixed by all
crm € G and hence must be in Q.
A
Remark In fact, of course, 6(x) € Z in the proposition above, since 6(x) is
also an algebraic integer.
Corollary 10.2.6 Characters of the symmetric group Sn are integer-valued.
Proof The exponent of S„ is clearly LCM{k € N : 1 < f c < n } , s o m € N i s
relatively prime to exp(5„) if and only if it is relatively prime to n\. In that
case x and xm have the same cycle structure, so x ~ xm.
A
It is in fact true that all of the absolutely irreducible representations of
Sn can be realized over the field Q of rational numbers (e.g. see [40] or [55])
i.e. all have Schur index 1 relative to Q.
Exercise
Suppose that 6 is a C-character of G, with 9(x) £ Q for all x € G. If
y, z eG and (y) = (z) show that 8(y) = 6(z). (If \y\ = n think about the
Galois group G = £ ( Q « n ) : Q).
Back to the Schur index.
Proposition 10.2.7 Suppose that F is a subfield of C and that xp is an
irreducible F-character of G. Then there is some x € Irr(G) such that
tp = Tn F (x)E{x":<7 6 G}, where G = G(F{x)-F).
Furthermore ip is the
unique irreducible F-character for which (tp, x) ^ 0.
189
10.2. THE SCHUR INDEX
Proof Choose x € Irr(G) with (ip,x) = m ^ 0. Then {ip,x°) = (V'.x) = m
for ail er G £, since V has its values in F . Thus ip = m ^ { x " : ( T £ 5} +
V>o, where ipo is either 0 or is a character for which (ipo,x") = 0 for all
a € Q. By the remark following the proof of Proposition 10.2.4 there is a
field L D F(x) such that x is an L-character and mp(x) = [L:F(x)j- Set
<p = mF(x)^2{x<T:°~ € G}, and observe that then ip = mF(x)TrF(xyF
ox
(e.g. see [31], page 117), so
<p = [L: F(x)] T r F ( x ) / F oX = TrL/F(x)
o T r F ( x ) / F oX = T r £ / F oX,
and ip is an F-character by Proposition 10.2.1. Since ip is an irreducible F character and (ip, (p) ^ 0 we have <p = ip + ipi, with V>i either 0 or a character.
Thus
V = mF(x) ^{x"-
o-eÇ}=m^{x"-
o € G} + Vo + ^ i -
Comparing summands, we see that m < m/r(x) and ipo = 0.
Observe next that ip and mF(x)x are both F(x)-characters and that
mF(x)x is irreducible as an F(x)-character. Thus
m ^ { x " : ^ 6 G) = ip = m F ( x ) x + M>2,
where ip2 is either 0 or is an F(x)-character. Comparing summands again,
we see that mF(x) < m, and so
V> = m F ( x ) ^ { x < T : < r e ö }
as desired.
For the uniqueness - if n is an irreducible F-character with (n, x) ^ 0 then
(Wi^P) i1 0 (since x is a constituent of both), so r\ — ip by Theorem 5.2.6. A
Corollary 10.2.8 / / F C C , x ^ Irr(G), and 0 < m € N, iAen m ¿s í/ie
Schur index of x if and only if m »'s íAe multiplicity of x as a constituent of
some irreducible F-character.
Proof It is sufficient to observe that there is some irreducible F-character
ip having x as a constituent.
A
Theorem 10.2.9 (Schur) Suppose that F Ç C, x € ^r(G), and n is an
F-character of G having x as a constituent. Then mF(x)
(v>x)Proof Let ip be an F-irreducible summand of n having x as a constituent,
so {ip, x) = rnF(x) as in the proof of Proposition 10.2.7. Also ip is unique by
Proposition 10.2.7, so if n = kip+6, where 0 is an F-character with (ip, d) = 0,
then (x,0) = 0 as well; hence (n, x) = kmF(x).
A
CHAPTER
190
Corollary 10.2.10 If FCC
10. SPLITTING
FIELDS
and x€ Irr(G) then m F ( x ) | x(l)-
Proof Take n to be the regular character p, which is an F-character, and
A
recall from Proposition 5.2.13 that x(l) = (p, x)It follows from the corollary above and Ito's Theorem (6.3.9) that if G is
a dihedral group Dn, or a quaternion group Qn, then every x € Irr(C?) has
Schur index 1 or 2.
Exercise
If F Ç C show that mF{x) = 1 for all x € Irr(£>„) (see page 147). {Hint
If x is nonlinear what is its multiplicity in if,?)
10.3
R versus C
An element i G G i s called a real element if it is conjugate to its own inverse,
x ~ i _ 1 . A conjugacy class K of G is called a real class if all elements of K
are real, i.e. if K = K~l.
Proposition 10.3.1 If x € G, then x is real if and only if Xi{x) € K for all
Xi € Irr(G).
Proof (=») If x is real and Xi £ I r r (G) then x¡(x) = Xi(£ _ 1 ) = Xi(x) by
Propositions 5.2.1 and 5.2.5.
(<=) If all Xi(x) ar e r e a l t n e n Xi(x) = Xi(x) = Xtfa - 1 ), all i, again by
Proposition 5.2.5. Thus x ~ a; -1 , since the set Irr(G) is a basis for cf(C?)
(Corollary 5.2.19).
A
Proposition 10.3.2 The number of real-valued absolutely irreducible characters of G is equal to the number of real conjugacy classes in G.
Proof Let A denote the character table of G, and denote by H = (a) a
group of order 2. Then H permutes the entries of A either by acting on rows
via xf = Xi, or by acting on columns via "Kj — KJ1. The two actions
agree, since Xi(Kj) = XtC^T*1) ^ y Proposition 5.2.5. By Brauer's Lemma
(9.1.12) the two permutation characters 0r and 9C of H on the rows and
columns, respectively, are equal. But 0r(<r) counts fixed rows, i.e. real-valued
A
characters, and 6c(a) counts fixed columns, i.e. real classes.
Theorem 10.3.3 (Burnside) / / G has odd order then 1 G is the only realvalued absolutely irreducible character of G.
Proof By Proposition 10.3.2 it will suffice to show that K = {1} is the only
real conjugacy class in G. Suppose that y~lxy = x~l. Then y~2xy2 = x, so
y2 e CG{X). Since \G\ is odd (y) = (y2), and y 6 C G (x). But then z = x~\
so x2 = 1, and x = 1.
A
191
10.3. R VERSUS C
Corollary 10.3.4 (Burnside) / / \G] is odd then \G\ = c(G) (mod 16;.
Proof By the theorem we may label the characters in Irr(G) as xo = I G ,
then Xi,---,X*; X»+i>---.Xa«, where x»+i = XT, 1 < i < s. Say that Xi(l) =
2m¿ + 1, all i. Then, by Theorem 5.2.14,
8
\G\
=
l + 2 ^ ( 2 m ¿ + l) 2 = l + 2 ^ ( 4 m ? + 4m i + l)
¿=1
=
:
l + 2s + 8 ^ m i ( m i - l - 1 ) = c(G) + 8m,
where m = £ ¿ mi(mi + 1) is even.
A
Exercise
If |G| is odd, and if x(l) = 1 (mod 2*) for all x € Irr(G), show that
\G\ = c(G) (mod 2fc+2).
Recall If x is the character of an F-representation 5 on a vector space U
then x 2 is the character of T = S <g> S on V = U ®F U.
The map f:U x U -> V defined by f(u,v) = v <S> u is easily checked
to be "balanced," so there is a uniquely determined and well-defined linear
transformation * on V with (u®ti)' = v <8> u, all u, v 6 U (e.g. see [31], page
158). The square of * is the identity, so its minimal polynomial divides x2 — 1.
Let Va and Va denote the respective eigenspaces for 1 and — 1, respectively,
of *, the so-called subspaces of symmetric and antisymmetric tensors.
Thus Va = {v € V:v* = v} and Va = {v G V:v* = —v}, and clearly
V,nVa = 0 . Ifv € ^ t h e n v + v * € V,,v-v' € V a ,andu = i ( u + v * ) + i ( u - u * ) ,
so V = V, © V«.
Since T(x)(v*) = (T(x)u)* for all x € G, it follows that T preserves
the eigenspaces of *, i.e. that V, and Va are T-invariant subspaces of V.
Thus, if T, and T a are the restrictions of T to V, and VQ, respectively, then
Î 1 = Ts ®Ta, and we may write x 2 = X» + Xai the corresponding sum of
characters; x« a n d Xa are sometimes called the symmetric square and the
antisymmetric, or alternating, square, respectively, of xIf { u i , . . . , u „ } is a basis for U then {t>y = u¿<g>Uj:l < i',j < n} is a basis
for V. Then Vs and Va have bases as follows:
V„:
{vij + Vji: 1 < i < j < n},
Va:
{vij - Vji\ I <i < j
<n},
2
so their respective dimensions are (n + n)/2 and (n 2 - n)/2. In order to
compute Xo say that S = [sy]. Then
T a (x)(t;jj - Uji) = 5(i)u¿ ® S ( I ) U J - 5 ( i ) u j (8 5(x)u ¿
CHAPTER
192
k
I
10. SPLITTING
k
FIELDS
I
=
^2(Ski{x)stj(x)
k,l
- Skj(x)sli{x)){uk
0 Ul)
=
^2(ski(x)Stj(x)
- Skj(x)sii(x))(Vkl
- Ujfc),
all x € G. The "diagonal entries" are Sü(x)«^-(a;) - sy(x)sji(x), so
X a ( x ) = Y^(Sii(X)aij(X)
-
S¡j(x)sii(x))-
Doubling Xo> we see that
2Xo(i)
=
^(s»(aOs>¿(aO-St¿(z)s¿¿(x))
»
j
*.j
(the summands with i = j cancel out). Thus
2 Xo (x) = x 2 (s) - tr(S(x) 2 ) = xHx) - x(* 2 )We have proved the next proposition.
Proposition 10.3.5 If \ is an F-character ofG, with symmetric and antisymmetric squares \ , and Xa, then
X.(x) = \(X2(X) + x(x2))
and
X o (x)
= i ( X 2 ( x ) - *(x 2 )),
all x 6 G. In particular, the function x i-> x ( i 2 ) *s * n Char(G), since it is a
difference 2x« - X2 = X2 ~ 2Xo of characters.
If x € G set 02 (x) = | {3/ € G: y2 = x}|, the number of "square roots" of x
in G. Clearly 02 € cf(G), and Irr(G) is a basis for cf(G) by Corollary 5.2.19,
so we may write
02 = 5 > ( x k : * e I r r ( G ) }
for some coefficients u(x) € C.
Proposition 10.3.6 If x € Irr(G) then
v(x) = \G\-lY,{x(*2Y-xeG}.
10.3.
193
R VERSUS C
Proof We have u(x) = (x,h) = l ^ l - 1 E v 6 GX(y)0 2 (î/)- Fix y and observe
that x(î/)02(y) = Y,{x(x2yx
€ G and x2 = y}. Then let y vary over G. A
The function v. Irr(G) —► C is called the Frobenius-Schur indicator.
Theorem 10.3.7 (Frobenius and Schur) If x £ hr(G) then v(x) = 1,
— 1, or 0, with u{x) =0 if and only if x is not real-valued.
Proof By Proposition 10.3.5
Hx) = \G\-lY,xtf)
X
\G\-l^X2(x)-2\G\-l^2xa(x)
=
X
=
X
2
(X ,lG)-2(Xa,lG).
Suppose that X ^ X- Then 0 = (x,x) = ( X 2 I ! G ) - But then also
(Xo> I G ) = 0, since Xa is a summand of x 2 , and consequently u(x) = 0.
Suppose next that x = X- Then 1 = (x,x) = ( X 2 , 1 G ) - Thus (XO,1G)
must be 0 or 1, and consequently u(x) = 1 — 2(Xa, I G ) = ±1A
Corollary 10.3.8 Suppose that t is the number of involutions in G, i.e.
d2(l) = t+l.
Then
Í + 1 = $ > ( X ) X ( 1 ) : x 6 Irr(G)} < £ { x ( l ) : X e Irr(G)}.
Proof By definition 02 = J2X "(x)x-
A
For example, recall that the dihedral group D\ and the quaternion group
(?2 have identical character tables. If G = D\ then the number t of involutions
is 5, so 6 = £ \ v{xt)XiiX) = 4 + 2t/(xs) and u{Xb) = 1. If G = Q 2 then t = 1,
so 2 = ¿2i "(Xi)Xi(l) = 4 + 2J/(X5), and i/(xs) = - 1 . In particular the
indicator is not in general determined just by the character table, although
it is fairly easily calculated from the table provided that we also know the
effect of the squaring map x >-¥ x2 on the conjugacy classes.
Exercise
Show that
E(^xcMSJ)>l|J(2"fc)Ä.
Remark The inequality is actually equality, since i/(x) = 1 for all x G
Irr(5 n ).
CHAPTER
194
10. SPLITTING
FIELDS
Proposition 10.3.9 Suppose that G has even order and that t is the number
of involutions in G. Then there is a real element i e C , i j ¿ l , with
[G:C G (x)]<(i^i) 2 .
Proof Label the characters in Irr(G) so that xo = IG I Xi> • • • > Xr are the
nonprincipal real-valued characters; and Xi ^ Xi if * > r - Likewise label the
conjugacy classes so that KQ = {1}; Ä2, • • •, Kr are the real classes; and the
remaining classes are not real (see Proposition 10.3.2).
Then, as a slight refinement of Corollary 10.3.8, we have
r
0<t<5>,(l).
i=l
Let u = ( 1 , 1 , . . . , 1) and v = ( x i ( l ) , . . . ,*>(!)) in l r . Applying the CauchySchwarz inequality to the standard inner product on Kr we have
t2 < ( ¿ X i ( l ) ) 2 = (u,v)2 < \\u\\2\\v\\2 = r 5 > ( 1 ) 2 < r(|G| - 1).
t=l
:
Multiply by \G\ - 1 and divide by t2 to obtain
| G |-l<r(£L=i) 2 .
If \Kt\ > ((|G| - l ) / i ) 2 for 1 < i < r then
¿|^|>r(fci) 2 >|G|-l,
i=i
l
a contradiction, so there must be a real conjugacy class K¡ with \Kj\ <
((\G\ - l ) / i ) 2 . Since \Kj\ = [G: CG(x)] for any x G Kh the result follows. A
Theorem 10.3.10 (Brauer and Fowler, 1955) / / 0 < n € N then there
are only finitely many nonisomorphic simple groups G having an involution
z with \CG(Z)\ = n.
Proof Given such a G, say that G has t involutions. Then t > [G: CG(Z)] =
\G\/n, so n > (\G\ - l ) / i . By Proposition 10.3.9 there is an element x € G
with 1 < [G: CG(X)] < n2. The action of G on cosets of CG{X) thus gives an
isomorphism (G is simple) from G into Sym(n 2 — 1), which of course has only
finitely many subgroups.
A
This theorem of Brauer and Fowler, while not terribly deep, played an
important role in the classification of finite simple groups (e.g. see [30]).
When the Feit-Thompson Theorem [24] was proved it became clear that every
195
10.3. R VERSUS C
(nonabelian) simple group has an involution, and the idea arose of classifying
the groups in terms of centralizers of involutions - the basic idea being as
follows. Take a simple group, or possibly some class of them, determine the
centralizer of one of its involutions, usually taken in the center of a Sylow
2 - subgroup, and try to determine all of the finitely many simple groups
having an involution with that same centralizer.
In many cases the simple group is determined up to isomorphism by the
centralizer. That is the case, for example, for each PSL(4,g) with q = 3
(mod 4).
The classification also requires negative results - for example, Brauer and
Suzuki proved in 1959 that no simple group has a (generalized) quaternion
group as Sylow 2-subgroup.
In like manner various groups had to be ruled out as possible centralizers
of involutions. However, in studying the case of a simple group G with T €
Syl 2 (G), 2 an involution in Z{T), and CG(z) Si (z) x Alt(5), Z. Janko in 1965
was unable to prove nonexistence. Instead he proceeded to construct the first
of the sporadic groups following the Mathieu groups. It is commonly called
the first Janko group J\ (he found more as well); it has order 175,560 and
was constructed by Janko as a subgroup of GL(7,11) (see [42]). Some of the
other sporadic groups arose in similar fashion.
See [39], pages 52-55, for some further interesting group theoretical applications of Corollary 10.3.8.
Exercise
If X is a linear character of G, show that v{\) = 1 if and only if x — X\
v{x) = 0 if and only if \ ¥" XWe resume the discussion of \ 2 as the character of T = S <8> S on V =
U ® F U that preceded Proposition 10.3.5.
The vector space V is isomorphic with the space of all n x n matrices over
F via v i-> A{v), where v = ^2tj aijVij and A(v) = [ay]. If x 6 G then
T(x)v
=
T(x) "^2 «.ijVij = ^2aij(S(x)ui
*,/
<g> S(X)UJ)
i,j
and so
A(T{x)v)
= S(x)A{v)S{x)*.
Exercise
Show that A(v") = A(v)\
all v G V.
(*)
CHAPTER
196
10. SPLITTING
FIELDS
Proposition 10.3.11 Suppose that x € Irr(G) is the character of representation S on U. If x , a real-valued then there is a matrix A ^ 0 such that
S(x)AS(x)t
= A for all x G G. Any such matrix A is either symmetric or
antisymmetric, and in fact A1 = v(x)A.
Proof Since x is real-valued, (x 2 , I G ) = (XiX) = 1- Thus I G has multiplicity
1 in x 2 , and it follows from Maschke's Theorem that V = U®U has a unique
one-dimensional subspace Vo for which T(x)v = v for all v € Vo, all x G G.
Thus we see by equation (*) above that S(x)AS(x)t
= A if and only if
A = A(v) for some v £ VoSince x 2 = X» + Xa and (x 2 , I G ) = 1, we must have either (x«, I G ) = 1
(if Vo Ç V.) and v(x) = L ° r (x«, I G ) = 1 (if V0 Ç Va) and u(X) = " 1 (see
the proof of Theorem 10.3.7). Thus for each v € Vo we have v* = f(x) u Finally, apply the exercise preceding the proposition to see that if v G Vo
then A(y)* = A(v') = A(v(x)v) = u(x)A{v).
A
Corollary 10.3.12 If x 6 Irr(G) is an R-character then i/(x) = 1.
Proof We may assume that S is an R-representation with x as character.
Set B = £ V € G S(y)S{y)1, and note that B* = B and S(x)BS(xY
= B for all
x £ G. Furthermore each S(y)S(yY has positive diagonal entries, so B ^ 0,
and B = Bl = u(x)B; hence u{x) = 1.
A
The converse of Corollary 10.3.12 is true as well. The proof will require
more work, however.
Recall that the adjoint of a complex (square) matrix A is the conjugate
transpose of A; denote it by A*, so A* = A1. Note that, with the standard
inner product for complex column vectors, (Au, v) = (u, A*v) for all vectors
u and v.
If A* = A then A is called Hermitian, and if A* — A~l then A is called
unitary. If AA* = A* A then i4 is called normal - clearly Hermitian and
unitary matrices are normal. It is a well-known fact from linear algebra (e.g.
see [16] or [34]) that A is normal if and only if it is unitarily diagonalizable,
i.e. there is a unitary matrix U so that U*AU is diagonal. A Hermitian
matrix A is called positive definite if (Av, v) > 0 for all nonzero vectors v; if
A is positive definite then all its eigenvalues are positive real numbers.
A matrix representation U (over C) of G is called a unitary representation
if U(x) is a unitary matrix for all x G G.
Proposition 10.3.13 Every C-matrix representation T of G is equivalent
with a unitary representation.
Proof Set A = £ l € G T(x)*T(x) and note that A is Hermitian and positive
definite. Since A can be unitarily diagonalized, it is easy to see that there is
a positive definite Hermitian matrix B with B2 = A. Observe that
{AT(x)A-1)*
= A~lT{x)*A
= A-lT{x)*
^T(j/)*T(y)
v
197
10.3. R VERSUS C
A'1 ^(r(x)*T(y)*T(y)T(a ; ))T(x- 1 )
=
y
=
A-1AT(x-1)=T(x~1),
all x e G. Set í/(ar) = B T ( x ) ß - \ all x. Then
i/(z)*
(BT(x)B-1)*
=
=
{B-lAT{x)A-lB)*
=
B(AT(x)A-1)*B-1
=
BTix-^B'1
=
U{x)-\
and U is unitary.
A
Exercise
Use Proposition 10.3.13 to reprove Maschke's Theorem (5.1.1) in the case
that F = C. (If W is a T-invariant subspace show that its orthogonal
complement W1- is also T-invariant.)
Proposition 10.3.14 If D is a diagonal matrix (over C) then there is a
diagonal matrix E with E2 = D and such that any complex matrix commutes
with D if and only if it commutes with E.
ßm € C for the distinct diagonal entries
Proof Choose square roots ß\,...,
c*i,... ,am of D. By Lagrange Interpolation (e.g. [31], page 76) there is a
polynomial f(z) 6 C[z] with /(a¿) = ßu all i. Set E = f(D).
A
Proposition 10.3.15 Suppose that A is both unitary and symmetric (so A =
A"1). Then A = BB~l for some matrix B.
Proof Choose U unitary so that U*AU = D, diagonal, and write D = E2
as in Proposition 10.3.14. Then
D = Dt = {U*AU)1 = IPAÜ =
tfUDirV
=
V~lUDU*V,
so U*IJD = DU*TJ, hence also U*Ü~E = EU*Ü, and consequently VÉU* =
UEU*. Since DD = D*D = I the diagonal entries of D are of absolute value
1, and likewise for those of E, so ~EE = I as well. Set B = UE~U*, and
observe that
BB~X = VÉÏFUEU*
= {UEU*)2 = UDU* = A.
A
CHAPTER
198
10. SPLITTING
FIELDS
Theorem 10.3.16 / / x 6 Irr(G) and v(x) = 1 then x is a" R-character.
Proof By Proposition 10.3.13, x is t n e character of a unitary representation U. Since x is real-valued U ~ U by Theorem 5.2.12, so there is
a matrix A with A~lU(x)A
= U(x), all x € G. Thus U{x)AU{xf
=
AU{x)U{xf
= A(U(x)U(x)*)1 = A, and consequently At = u{x)A = A
by Proposition 10.3.11. From U(x)A = AU(x) we see that A*U(x)* =
U(x) A*, or A*U{x~l) = U(x~l)A*, all x, and replacing x by x - 1 we have
A*U(x) = U(x)A*, all x. But then AA*U{x) = AU(x)A* = U(x)AA*,
all x, so AA* € C(I/) and .A.A* = al for some a G C by Proposition
5.1.4. Note that if v is any nonzero complex column vector of length 1 then
a = a(v,v) = (AA*v,v) = (A*v, A*v) > 0 in R. Thus we may assume
that A has been replaced by ßA, where 0 < ß € R and ß2 = 1/a, i.e. we
may assume that AA* = I, so A is unitary as well as symmetric. Then
by Proposition 10.3.15 we may write A as BB~1 for some B, and so from
U{x)BB-1 = B~B-lU{x) we see that
B~lU(x)B
=B
1
t/(x)ß = ß
l
U(x)B,
all x 6 G, and Î7 is equivalent with a real representation.
A
Probenius and Schur introduced the following terminology: x in Irr(G) is
of the first kind if it is an R-character, x is of the second kind if it is real-valued
but is not an R-character, and x is of the third kind if it is not real-valued. If
X is of the second kind and T is a C-matrix representation with x as character
then, as noted in the proof of Theorem 10.3.16, T must be equivalent with
T, but T is not equivalent with a real representation.
The next theorem summarizes the results above.
Theorem 10.3.17 (Frobenius and Schur) If x € Irr(G) then
1. x is of the first kind if and only if i>(x) = 1,
2. x is of the second kind if and only if v(x) = — 1, and
3. x is of the third kind if and only if i/(x) = 0.
Corollary 10.3.18 If x e Irr(G) then x has Schur index 2 relative to R if
and only if x is of the second kind, hence if and only if v{x) = — 1Exercise
Calculate the Frobenius-Schur indicator v(x) for each x € Irr(Q m ) (see
page 125). What are the Schur indices relative to R?
Proposition 10.3.19 If x € Irr(G) and x ( l ) is odd then x is not of the
second kind, and hence x has Schur index 1 relative to R.
199
10.3. R VERSUS C
Proof Suppose, to the contrary, that u(\) = — 1. By Proposition 10.3.13
X is the character of a unitary representation U, and by Proposition 10.3.11
there is a nonzero matrix A, with A = —A* and U{x)AU(x)t = A for all
x eG. As in the proof of Theorem 10.3.16, AA* = ai # 0, so A is invertible.
Thus 0 ^ det A = det A1 = det(—.4) = - det A, since the degree is odd, a
contradiction.
A
In a survey article [8] in 1963 R. Brauer cited a "rather old problem which
is still open," namely that of determining by group theoretical properties of
G the number of characters x € Irr(C?) of the first kind. Although answers
have been obtained for a few special classes of groups, the problem seems to
be still open in general.
Let us write u+(G), v~{G), and v°{G) to denote the number of x € Irr(G)
of the first, second, and third kinds, respectively, so
i/ + (G) + //-(G) + ^ ( G ) = c(G).
Thus Brauer 's question asks for a determination of v+(G). Note that, by
Proposition 10.3.2, v+{G) + v~(G) is the number of real conjugacy classes of
G, so of course v°(G) is the number of non-real classes. Burnside's Theorem
(10.3.3) says that v+{G) = 1 if \G\ is odd.
To conclude, we consider Brauer's question in the context of Frobenius
groups. Recall that the characters of a Frobenius group have been described
in Theorem 9.1.15.
Proposition 10.3.20 Suppose that G = M x H is a Frobenius group and
that 1M ¿ (pe Irr(M).
1. If \H\ is even then v(tpG) = 1.
2. If \H\ is odd then i/ G (v G ) = vM(<p)Proof (1) Choose an involution z e H and set L = M x (z), so [L: M] = 2.
Then <PL\M = <p + <pz (e.g. by Theorem 6.4.1, or by an easy calculation) and
\M\ is odd, so VM(<P) = *>MWZ) = 0 by Theorem 10.3.3. Thus
vL{<pL) = ILr^Mx^
=
+ v'ix^
+ ipttxzft +
ip'dxzfy.xeM}
2["Aí(<¿>) + « W ) + l + l ] = l ,
since (xz)2 = 1 for all x e M by Proposition 9.1.11. Thus ipL is an Rcharacter, and hence ipG = (<pL)G is an R-character.
(2) Since \H\ is odd all elements of G\ M have odd order. Hence if x $ M
then x 2 i M. Also <pa\M = "Z{fh:he
H} (6.4.1 again) and <pG\G\M = 0.
Thus
i/(<pG) = ICI"1 J ^ V ) : / » € H, x € M}
CHAPTER
200
10. SPLITTING
= l i / l " 1 £>*#(¥>*):/» € H} =
FIELDS
vu{}p),
since UM(iph) = VM{<P) for all h € H.
A
Proposition 10.3.21 Suppose that G = M » H is Frobenius and that x G
Irr(G), with X\H G Irr(ff). Then uG{X) = VH(X\H)Proof Calculate:
"(x) = \G\-1Y,{x{(xh)2):xeMtheH}
\G\-l'£{x(x-hx-ti,):x€M,heH}
=
=
m-^H]-1
^ { x ( / i 2 ) : i G M, /i G H} =
(recall that if a; G M and h G H then xixh)
= xCO
smce
UH(X\H)
M < ker(x)).
A
We may use the propositions above to reduce Brauer's problem for a
Frobenius group G to the corresponding problems for M and H.
Theorem 10.3.22 Suppose that G is a Frobenius group with kernel M and
complement H.
1. If \H\ is even then
v+{G)
{
\H\
u-(G)
=
u°(G)
= "°(#).
h
»-(H),
2. If \H\ is odd then
v+{G)
""(G)
u°(G)
=
1+
i/+(M) - 1
w\
v-{M)
\H\ '
v° {M)
+ c(H) - 1.
m
Proof (1) Each character in Irr(if) extends to a character of the same
kind in Irr(G) by Proposition 10.3.21. The kernel M is abelian, the \M\ - 1
nonprincipal characters in Irr(M) lie in H-orbits of size \H\, and each of them
induces to a character of the first kind on G by Proposition 10.3.20.
10.3. R VERSUS C
201
(2) All nonprincipal characters in Irr(/f) are of the third kind by Theorem
10.3.3, and they remain so when extended to G. Of course Ijj extends to IG,
of the first kind. The nonprincipal characters in Irr(M), lying in //-orbits
of size | H |, induce to characters of the same kinds in Irr(G) by Proposition
10.3.20.
A
In many cases it is possible to answer Brauer's question for M and H
fairly explicitly, since fairly complete information is available concerning the
structure of Frobenius groups (see the previous chapter). For example, since
M is nilpotent we have M = K x T, with \K\ odd and T a 2-group, so
u+(M) = i/+(T).
Exercise
Let G = Aff(F), where F is a finite field with q > 2 elements. Determine
f + (G), v~(G), and u°(G) (see the exercise on page 176).
Groups and Characters
by Larry C. Grove
Copyright © 1997 John Wiley & Sons, Inc.
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Groups and Characters
by Larry C. Grove
Copyright © 1997 John Wiley & Sons, Inc.
Index
characteristic subgroup, 14
class
matrix, 151
number, 63
Clifford's Theorem, 137
cocycle, 86
collapse, 10
complement, 84
partial, 89
completely reducible, 98
congruent matrices, 21
constituent, 109
content, 73
core, 4
coset
enumeration, 9
table, 9
cycle
index, 64
type, 64
absolutely irreducible, 101
adjoint, 121, 196
affine group, 32
algebraic integer, 114
alternate form, 22
alternating
character, 143
square, 191
antisymmetric, 23
square, 191
tensor, 191
base, 5
bilinear form, 20
Blichfeldt's Theorem, 141
block, 17
bonus, 10
Brauer
characterization of characters, 149
Induction Theorem, 149
Lemma, 174
Splitting Field Theorem, 1
Burnside
algorithm, 152
ring, 148
Burnside's
Basis Theorem, 91
Theorem, 81, 120, 136
Burnside-Dixon algorithm, 156
definitions/bonuses table, 9
degree, 3, 98, 105
dihedral group, 14, 30
direct sum, 98
discriminant, 21
dual
basis, 20, 121
space, 20, 121
elementary
abelian group, 19, 86
group, 148
equivalent
actions, 4
forms, 24
centralizer, 101
character, 61, 105
kernel, 116
ring, 123
table, 110
209
INDEX
210
equivalent (cont.)
representations, 97
Euler's Theorem, 34
exponent, 183
extension, 13
factor set, 85
faithful
action, 3
character, 105
representation, 97
finitely presented group, 9
first kind, 198
fixed point, 61
fixed-point-free, 181
focal subgroup, 80
Frattini
Argument, 84
subgroup, 89
Probenius
complement, 172
group, 171
kernel, 172
reciprocity, 133
Frobenius-Schur
indicator, 193
Theorem, 104, 193, 198
fully characteristic, 177
Galois conjugate, 188
GAP, 7, 13, 76, 160
general linear group, 39
generalized
character, 123
quaternion group, 30
generating function, 73
group
action, 3
algebra, 128
Hall
7T-subgroup, 84
subgroup, 84
Hall's Theorem, 87, 89
Hermitian, 196
Hessel's Theorem, 38
holomorph, 14, 39
hook, 169
hook length, 169
hyperbolic
pair, 25
plane, 25
hyperplane, 41
induced
character, 129
class function, 131
module, 128
representation, 128
inertia group, 137
invariant subspace, 98
inventory, 68
irreducible
character, 105
representation, 98
isometry, 24
Ito's Theorem, 140
fc-transitive, 16
kernel, 116
left radical, 21
lift, 116
linear
character, 105
fractional transformation, 41
representation, 99
little group method, 146
Möbius transformation, 41
Mackey's Theorem, 142, 143
Maple, 76
Maschke's Theorem, 99
Mathieu groups
Mn,48
M 1 2 , 48
M 2 2 , 49
M 2 3 , 50
M 2 4 , 50
metacyclic group, 29
monomial, 64
multiplicity, 109
211
INDEX
nondegenerate
form, 21
subspace, 23
nongenerator, 89
normal
p-complement, 81
complement, 81
matrix, 196
one-point extension, 46
opposite diagram, 162
orbit, 4
Orbit Formula, 61
orthogonal, 23
complement, 23
group, 33
matrix, 33
Orthogonality Relation
First, 108
Second, 112
p-commutator subgroup, 79
partition, 161
parts, 161
pattern, 68
counting function, 73
inventory, 70
Petersen graph, 75
point
at infinity, 47
group, 39
pole, 36
positive definite, 196
primitive
action, 17
root, 28
projective
general linear group, 40
point, 40
space, 40
special linear group, 40
symplectic group, 56
quasi-elementary group, 148
radical, 23
ramification index, 138
rank, 16
real
class, 190
element, 190
Redfield-Pólya Theorem, 70
reducible
character, 105
representation, 98
refinement, 166
reflection, 34, 35
reflexive form, 23
regular
action, 17
character, 110
representation, 99, 185
relator, 9
table, 9
representation, 97
alternating, 99
contragredient, 121
matrix, 97
permutation, 99
principal, 99
regular, 99
right
radical, 21
regular action, 17
rotation, 33, 34
scan, 9
Schneider, G., 158
Schreier
generators, 3
transversal, 5
Schreier-Sims algorithm, 7
Schur index, 187
Schur's
Lemma, 100
Theorem, 84, 103, 140, 189
Schur-Zassenhaus Theorem, 85
second kind, 198
semidirect product, 13
set partition, 162
sharply transitive, 17
INDEX
212
skew symmetric, 23
special linear group, 39
split
extension, 13
metacyclic group, 29
splitting field, 101
sporadic groups, 51
square
alternating, 191
antisymmetric, 191
symmetric, 191
stabilizer, 4
strong generating set, 6
subgroup table, 9
Sylow Theorems, 8
symmetric
form, 22
group, 3
square, 191
tensor, 191
symplectic
basis, 52
group, 52
space, 52
transvection, 53
tensor
antisymmetric, 191
symmetric, 191
third kind, 198
trace, 105, 185
transfer, 77
transitive, 4
transvection, 41, 53
transversal, 1
function, 1
unitary, 196
representation, 196
Verlagerung, 77
weight, 68
assignment, 68
wreath product, 15
Young
diagram, 161
subgroup, 162
tableau, 162
Zassenhaus's Theorem, 86
Groups and Characters
by Larry C. Grove
Copyright © 1997 John Wiley & Sons, Inc.
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