Chapter 6: Counting Irreducible Representations

Chapter 6

Counting Irreducible Represent ations Let G be a finite group, let F be an arbitrary field and let cr E Z2(G, F*). Our aim here is t o determine the number of linearly nonequivalent irreducible a-representations of G in terms of natural invariants associated with G, a and F . In case F is algebraically closed, we also determine the corresponding number for projectively nonequivalent representations. As usual, we treat the subject in terms of modules over the twisted group algebra FaG. Hence the former problem can be stated as follows : W h a t is the number of nonisomorphic simple FaG-modules? In the simplest case where a = 1, the answer is provided by the Witt-Berman's theorem which was presented in Vol.1. The general case requires a more sophisticated technique and is proved with the aid of a combinatorial result which plays a role analogous to Brauer's permutation lemma. It should be emphasized that our proof does not use the Wiit-Berman's theorem and relies instead on the special case where F is algebraically closed. This special case was established by Schur (1904) for c h a r F = 0 and by Asano, Osima and Takahasi (1937) if c h a r F # 0. All cohomology groups considered below are defined with respect t o the trivial action of the underlying group on the given abelian coefficient group.

1

Algebraically closed fields

Let F be a n algebraically closed field, let G be a finite group and let a E z 2 ( G , F*). Our aim is to provide the number of linearly nonequivalent

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178

irreducible a-representations of G. In view of Theorem 3.3.2, the required number is the same as the number of nonisomorphic simple FffG-modules. The following classical result is due to Schur (1904) for p = 0 and to Asano, Osima and Takahasi (1937) for p # 0.

Theorem 1.1. Let G be a finite group, let F be a n algebraically closed field of characteristic p 2 0 and let a E Z2(G, F * ) . Denote by r the number of nonisomorphic simple FffG-moudles. Then (i) r equals the number of a-regular conjugacy classes of G if p = 0. (ii) r equals the number of a-regular conjugacy classes of pl-elements of G ifp#O. Proof. (i) Since p = 0, FOG is semisimple by Lemma 2.3.2 (ii). Because F is algebraically closed, we deduce that r = dimFZ(FffG). The desired assertion is therefore a consequence of Theorem 2.6.3. (ii) By Lemma 2.6.2, a is cohomologous t o a normal cocycle. Hence, by Lemma 2.1.1, we may harmlessly assume that a is a normal cocycle. P u t S = [ F f f G ,F a G ] and let

T = (x E F"G(XP"

E S

for some integer

m

.

> 1)

Next choose an F-basis a l , az,. . ,at of T and let the integer m be so large that arm E S , 1 5 i 5 t , and pm is a t least as large as the order of a Sylow p-subgroup of G. Such a choice of m is always possible, since by 1, Proposition 8.3.1 in Vol.1, for any x E [FffG,F f f G ] and any integer n xpn E [ F f f G ,FffG]. Recall also that, by Theorem 17.3.1 in Vol.1, we have

>

Now, by Proposition 8.3.1 in Vol.1, t

(

pm

a i=l

)

t

. ~ ~ ~ ~ a ~ ~ . ~ ( m o c i ~ ) i=l

for all Xi E F. Hence a typical element x = CgEG xgg, xg E F , of F f f G belongs to T if and only if xpm E S or, equivalently, x E T if and only if

X;"X;

m-

gpm E S

(2)

g€G

where for each g E G, A, is a fixed element of F* with gprn = A; also that, by our choice of m, each g ~ mis a pl-element.

m -

gPm. Note

2 Auxiliary results

Let CI, C2,. and put

179

. . ,C, be all a-regular conjugacy classes of p'-elements Di = { g E Glgprn E C;)

(1 5 i

of G

< S)

xgg E T if and only if Applying (2) and Lemma 2.7.2, we see that x = CgEG

for all i E (1, . . . ,s). I t follows that

x

x,Xg = 0

xgB E T if and only if gEDi

g€G

for all i E {I,.. . ,s). Finally, consider the map

$ : FaG + F x F x given by

.x F

( s times)

$(c

~ g g= ) (p17P~7"'7p3)

g€G

where pi = CgEDi xgXg. Then 1C, is a surjective F-homomorphism. Moreover, by (3), Ker$ = T. Thus, by (I), we have

and the result follows. H

2

Auxiliary results

T h e chief aim of this section is t o present a combinatorial result which plays a role analogous t o Brauer's permutation lemma. As an application of this result in the next section, we shall extend Theorem 1.1 t o arbitrary fields. . We begin by recalling the following piece of information. Let F be a n arbitrary field. By a monomial space over F, we understand a triple (V, S, (V,)), where V is a vector space over F, S is a finite set, and ( 6 )is a family of one-dimensional subspaces of V indexed by S such that

Counting Irreducible Representations

180

In particular, we see that dimFV = IS[. Let G be a group and let (V,S , (V,)) be a monomial space over F. By a monomial representation of G on (V,S , (V,)), we understand a homomorphism : G -+ G L ( V )

r

such that for each g E G, r(g)permutes the V,, s E S . Thus I' determines a homomorphism y from G to the permutation group of the set S with

y(g)x = y

if and only if r(g)Vx = Vy

for all g E G and x , y E S . For any given s E S , we denote by G ( s ) the stabilizer of s, i.e. G ( s )= (9 E Glr(g)s = s ) An element s E S is said to be r-regular if for all g E G ( s ) , r(g)is the identity mapping on V,. A G-orbit of S is called I?-regular if each element of this orbit is I?-regular. Finally, by the fixed-point space of I? we mean the set of those v E V for which r(g)v = v for all g E G. Let V * = HomF(V,F ) be the dual space of V and for any linear transformation f of V , let f * be the linear transformation of V * which is dual to j, i.e. ( f * ( $ ) ) ( v )= $ ( f ( v ) ) for all II, E V*, u E V. Recall that if : G -t G L ( V ) is a representation of G , then the contragredient representation r*: G -,GL(V*)is defined by r * ( g ) = I'(g-l)* for all g E G.

r

Lemma 2.1. Let G be a group, let ( V , S , ( V , ) ) be a monomial space over F and let I? : G -+ G L ( V ) be a monomial representation of G on Then (V, (i) The triple ( V * ,S, (V:)) is a monomial space over F where an element f of V * lies in V: if and only if f(Vx) = 0 for all x # s. (ii) The contmgredient representation r* : G -+ G L ( V * ) is a monomial representation of G on ( V * ,S , (T)). (iii) The homomorphisms from G to the permutation group of S determined by r and I?* are identical. In particular, the G-orbits of S determined by r and I'* coincide. (iv) An element s E S is r-regular if and only if s is I"-regular.

s,(vs)).

Proof. (i) Let (v,), s E S , be a basis of V with v, E V, for all s E S . If ( 2 ) : ) ) s E S , is the dual basis of V * , then vf E V,* for all s E S . Hence V * = $,ESV,*, which shows that ( V * ,S,(V;)) is a monomial space over F.

2 Auxiliary results

181

(ii) and (iii) By definition, r ( g ) * permutes the V,: s E S, in such a way that I'(g)V, = Vy if and only if r(g)*V; = Vx*, ( x , y E S ) Hence, for all g E G , F(g)V, = Vy if and only if

I'*(g)V,' = V l

as required. (iv) Given s E S , we have r(g)v, = v, if and only if r*(g)v,' = vb. Hence s E S is I'-regular if and only if s is I'*-regular. H The following result contained in Reynolds (1971a) is an extended version of Berman's lemma (Berman (1958, Lemma 3.1)).

Proposition 2.2. Let (V, S,(V,)) be a monomial space over a n arbitrary field F and let r : G --t GL(V) be a monomial representation of G on (V, s,(K)). (i) If s E S is I?-regular, then so are all the elements in the G-orbit of s. (ii) Let X be a set of all representatives for the I'-regular orbits of S and, for each x E X , let w, be a nonzero element of V,. Put

where T, is a left transversal for G(x) in G. Then the elements v,, x E X, form a n F-basis for the fixed-point space of r. (iii) The dimension of the fixed-point space of r is equal to the number of r-regular G-orbits. (iv) A G-orbit of S is r-regular if and only if it is I?*-regular. In particular, the dimensions of fked-point spaces of I' and r* are the same.

Proof. (i) Assume that s E S is I?-regular and let x be any element in the G-orbit of s . Then there exists g E G such that I'(g)V, = V,, in which case G(x) = gG(s)g-l. Let 0 # v E V, and let t E G(x), say t = gglg-l for some gl E G(s). Because I'(g-l)V, = V, and s is r-regular, we have I'(gl)(I'(g-')v) = I?(g-l)v. Hence

proving that x is r-regular. (ii) Let Y be a set of all representatives for the nonregular orbits of S,

Counting Irreducible Representations

182

let Z = X U Y and, for each z E Z , let U, be the sum of one-dimensional subspaces of V indexed by the elements of the orbit containing z . Then

is a decomposition of V into the direct sum of invariant subspaces. Hence, if W is the fixed-point space of I' and W, = W fl U,, then

Now let v = CsES Asvi, As E F , 0 # v: E K, belong t o W and assume that there is an s E S such that A, # 0. Then, for a given g E G(s), I?(g)vi = p,v: for some pB E F, and so the equality I'(g)v = v implies A, = AspS. It follows that ps = 1, SO s is I'-regular. Hence, by (i), s belongs to a I?-regular orbit and therefore W = @XEXWX Finally, fix x E X and, for each g E T,, put v,, = I'(g)w,. Then the elements { ~ , , ~ lEg T,) form an F-basis of U, and so an arbitrary element v in W, can be uniquely written in the form

Because for all y E G, r ( y ) permutes the v,,,, g E T,, we deduce that v, E W, and that all the coefficients A , , of v are equal, proving (ii). (iii) This is a direct consequence of (ii). (iv) Apply (iii) and the fact that, by Lemma 2.1 (iv), an element s E S IS I'-regular if and only if s is I'*-regular. W We next record some applications of Proposition 2.2. In what follows, H denotes a subgroup of a finite group G and F a G is a twisted group algebra of G over an arbitrary field F. Two elements x, y E G are called H - c o n j u g a t e if y = hxh-' for some h E H . It is clear that the H-conjugacy is an equivalence relation and so G is a union of H-conjugacy classes. For a given g E G, let CH(g) denote the centralizer of g in H . Given a E Z 2 ( G , F * ) , we say that an element g E G is ( a , H ) - r e g u l a r if

= a(g,h)

for all h E CH(g)

Thus g is a-regular if and only if it is (a,G)-regular. I t follows from the definition that if g is ( a , H)-regular, then so is any H-conjugate of g. We

2 Auxiliary results

183

say that a n H-conjugacy class C of G is ( a , H ) - r e g u l a r if a t least one (hence any) element of C is ( a , H)-regular. P r o p o s i t i o n 2.3. Let X be a set of representatives for all ( a ,H ) regular H-conjugacy classes of G, and for each x E X , let

where T, is a left transversal for CH(x) in H . Then

is an F-basis for the algebra CFaG(FaH). Proof. Let V = F"G and, for each g E G, let I/, = (XglX E F ) . Then (V, G, V,)) is a monomial space over F . Moreover, the mapping

defined by

r ( h ) v = fl. v&-'

for all

v E V, h E H

is obviously a monomial representation of H on (V, G, (V,)). The homomorphism y from H t o the permutation group of the set G determined by r is given by y(h)g=hgh-' forall ~ E H , ~ E G Hence H ( g ) = CH(g) for all g E G. By the foregoing, g E G is r-regular if and only if ~ ~ h = - ,j 1

for all

h E CH(g)

or, equivalently, if and only if g is ( a ,H)-regular. Thus a typical r-regular H orbit of G is an (a,H)-regular H-conjugacy class of G. Hence, by Proposition 2.2 (ii), the elements vx, x E X, constitute an F-basis for the fixed-point space of I'. Since the latter is obviously CFaG(FaH), the result follows. Another application of Proposition 2.2 pertaining to an arbitrary field F is given by the following result.

Proposition 2.4. Let a E Z2(G, F * ) , let S be an invariant subset of G and let V = F m S be the subspace of F f f G which is the F-linear span of

Counting Irreducible Representations

184

all S with s E S. For each s E S , put V, = FS and let r : G defined by for all g E G, v E V I?(g)v = jvg-l

-+

G L ( V ) be

Denote by X a set of all representatives for the a-regular conjugacy classes g 3 ij-I where T, of G contained in S and, for each x E X , put v, = CgET, is a left transversal for C G ( x ) in G. Then (i) r is a monomial representation of G on (V, S, (V,)) and the fixedpoint s p c e of I' is FaS n Z(FaG). (ii) The elements v,, x E X constitute an F-basis for FaS n Z ( F a G ) . P r o o f . (i) This is a direct consequence of the definitions. (ii) The hornomorphism y from G t o the permutation group of S determined by I? is given by Y ( s )= ~ gsg-l

forall

~ E G , s E S

Hence G(s) = CG(s) for all s E S . It follows that s E S is I?-regular if and only if for all g E CG(s), g sg-I = S . Thus s E S is r-regular if and only if s is a-regular. This shows that a typical I?-regular G-orbit of S is an a-regular conjugacy class of G contained in S. Hence the desired conclusion follows by virtue of Proposition 2.2 (ii). H Let F be an arbitrary field, let a E Z2(G, F * ) and let p : G --t GL(V) be an a-representation of G over F. Then the map x : G + F given by ~ ( g= ) trp(g) is called the character of p (or an a-character if p is not pertinent t o the discussion). Thus if X is the character of FaG afforded by ) X(g) for all g E G. Since x the corresponding FaG-module V, then ~ ( g = and X determine each other, we say that x is a f f o r d e d by the FaG-module V. If p is irreducible (equivalently, V is a simple FaG-module), then x is called an irreducible a-character . By saying that X I , . . . , x T are all irreducible a-characters, we mean that there is a full set of nonisomorphic simple FaG-modules Vl, . . . ,VT such that X; is afforded by V,, 1 i 5 T .

<

L e m m a 2.5. Let F be an algebraically closed field of characteristic p 2 0, let a E z 2 ( G , F * ) and let Go be the set of all p'-elements of G (i.e. those elements whose order is not divisible by p). Let 21,.. . ,xT be all irreducible a-characters of G and, for each i E (1,. . . ,r ) , let xq = x;[Go. Then x:,. . . ,X: are linearly independent (as functions from Go to F ) .

2 Auxiliary results

185

Proof. By definition x i ( g ) = Ai(g) for all g E G, where X I , . . . , A , are characters of F a G afforded by all nonisomorphic simple FaG-modules. By Proposition 14.1.4 in Vol.1, the characters X I , . . . ,A, are F-linearly independent. Hence, it suffices to verify that if a l , . . . ,a , E F are such that

then the same is true for all g E G. Indeed, in this case, C r = l aiXi vanishes on all g, g E G and hence on all x E FaG. Because A 1 , . . . ,A, are F-linearly independent, it will follow that all ai = 0, as desired. If p = 0, then Go = G and there is nothing to prove. Hence we may assume that p > 0. Now fix g E G and write g = gPgpl, where g, and gp, are the p-part and p'-part of g , respectively. Then there exists an integer m 2 1 m such that g P m = g;l . Let I? be an irreducible representation of F a G with character X and let x be the corresponding irreducible a-character of G. Let wl , . . . , w, be the characteristic roots of I?@). Then wyn', . . . ,wfm are the characteristic roots of r(ij)pm It therefore follows that

.

r(T)

Now there exists a = a ( g ) E F' such that = a r ( g ) p m and hence by the above x ( g P r n >= a ~ ( g ) ~ ~ Since g ~ m= g;,

m

, it follows that

aX(g)~m = bX(gp!)pm

for some b = b(g,r) E F "

Because F is algebraically closed, the latter equality can be written as ~ ( g pIpm t = cPmx ( d p m

for some c E F*

which implies that x ( g P t ) = c ~ ( g ) .Since gpl E Go and since c does not depend on the choice of X , it follows from ( 1 ) that

Hence a l X l ( g ) +

.

+ a T X T ( g =) 0 and the result follows. H

Counting Irreducible Representations

186

>

Let F be an algebraically closed field of characteristic p 0, let cr E Z2(G, F * ) and let Go be the set of all p'-elements of G. Denote by V the F-linear span of all 2 with x E Go. Then (V, Go, (V,)) is a monomial space over F , where V, = Fa: for all x E Go. Let

be the monomial representation of G on (V, Go, (V,)) defined by I'(g)v = g u

p

for all g E G, v E V

Proposition 2.6. Further to the notation and assumptions above, let

be all irreducible characters of F a G and let

be the restriction of Xi to V . Then {AYIl 5 i 5 r ) is an F-basis of the fixed-point space of I?*.

Proof. In the notation of Lemma 2.5, we have A:@) = xY(g) for all g E Go, 1 5 i 5 r . Hence, by Lemma 2.5, A,: . . . ,A: are linearly independent. Now the dimensions of fixed-point spaces of I' and I'* coincide (Proposition 2.2 (iv)) and are equal to the number of a-regular conjugacy classes of p1elements of G (Proposition 2.4 with S = Go). The latter number is equal to r , by virtue of Theorem 1.1. Since each A: belongs to the fixed-point space of r*,the result follows. W We close by recording a simple fact which will be needed in Sec.4.

Proposition 2.7. Let F be an algebraically closedfield, let CY E Z2(G, F*) and let pi : G -, GL(V,) be two irreducible a-representations of G over F , i = 1,2. Denote by X; the character of pi, i = 1,2. Then pl is projectively equivalent to p2 if and only if there exists a homomorphism p : G -t F* such that ~ 2 ( 9= ) P(S)X~(S) for all EG Proof. Assume that pl is projectively equivalent t o pa. Then, by Corollary 3.2.5, there exists a homomorphism p : G + F* and a vector

3 Arbitrary fields

187

space isomorphism f : Vl -t V2 such that p2(g) = p(g) fP1(g) f - l for all g E G. Taking traces of both sides yields x2(g) = p(g)xl(g) for all g E G. Conversely, suppose that there exists a homomorphism p : G -t F* such that ~ ~ ( =9 p(g)x1(g) ) for all g E G. Define the map p : G + GL(V2) by p(g) = p(g)-1p2(g) for all g E G. Then p is an irreducible a-representation of G such that ~1 is the character of p. It follows, from Lemma 2.4 that the FOG-modules affording p and pl &re isomorphic. Thus p is linearly equivalent t o pl. Hence p l is projectively equivalent t o p2 and the result follows.

.

3

Arbitrary fields

Let G be a finite group, let F be an arbitrary field of characteristic p 2 0 and let a E Z2(G,F*). Recall that an element g E G is a pl-element (or g is p r e g u l a r ) if the order of g is not divisible by p. Thus, if p = 0, then all elements of G are pl-elements. In what follows, Go denotes the set of all pl-elements of G. In this section, we return t o the problem of determining the number of nonisomorphic simple FaG-modules in terms of F, a and G. The new feature of our discussion is that the field F need not be algebraically closed. It is not surprising that this will considerably complicate our task. Indeed, even the simplest case where a = 1is by no means trivial. Recall that the corresponding result, due t o Witt and Berman (see Theorem 17.5.3 in Vol.l), asserts that the required number is equal to the number of F-conjugacy classes of pl-elements of G. In the general case, the key to the solution will be provided by the notion of an ( a , F)-regular F-conjugacy class. An introduciton of this notion will be delayed since it requires a number of preliminary observations. Our main result asserts that the number of nonisomorphic simple FaG-modules is equal t o the number of ( a , F)-regular F-conjugacy classes of p'-elements of G. Our first task is t o introduce the notion of ( a , F)-regularity. This will be achieved with the aid of some monomial transformations of E a G , where E is the algebraic closure of F. Here, as usual, we identify a E Z2(G,F * ) with its image in Z 2 ( G ,E*). Recall that, by Lemma 2.3.3, the map

Counting Irreducible Representations

is an isomorphism of E-algebras. We begin by fixing any positive integer n divisible by the order of every element of G . Write n = npnp,,where the factors are the p-part and p'-part of n if p # 0, and where n p = 1, npt = n if p = 0. In what follows, E denotes a fixed primitive npt-th root of 1 in E.

Lemma 3.1. For any given a E G a l ( E / F ) , there exists an integer m ( a ) such that

a ( & )= em(,)

and

m(a)

l(mod n,)

(1)

Moreover, any such m ( a ) is a uniquely determined modulo n.

- -

Proof. Write a ( & )= E @ for some integer p. Since (n,, nPt) = 1, we have tnPt for some integers s , t . Hence m ( a )= p - t n , ~satisfies (1). Now assume that the integers p , X satisfy ( 1 ) . Since e p = e X , we have p X(modn,~). Since p l ( m o d n p ) and X l ( m o d n p ) , we have p X(mod n p ) . Thus p X(mod n ) and the result follows. H p - 1 = sn,

=

+

From now on, the integers rn(a), a E G a l ( E / F ) are chosen so as to satisfy (1). For each g E G, we denote by a, an n-th root of

in E*. Expressed otherwise, ag is an element of E such that

Given a E G a l ( E / F ) , we now introduce an E-linear map

uniquely determined by

Lemma 3.2. With the notation above, the following properties hold : (i) I?, does not depend on the choices of a,, m ( a ) and n . ', to E a H is an (ii) For each cyclic subgroup H of G , the restriction of I algebm automorphism of E a H .

3 Arbitrary Aelds

189

(iii) Let H be a cyclic subgroup of G and let rb be the analog of I', with respect to EaH. Then I'L is the restriction of I?, to E a H . (iv) For all z , g E G, 3I',(g)3-' = F,(3 g 3-l). (v) For each irreducible character x of E a G and for each a E E a G ,

(vi) If p

# 0 and g

is a p-element, then r,(g) = g.

ny.:

Proof. (i) Let ,Bg be any n-th root of a ( g , gi) and let k(a) satisfy (1). Then /3, = agw, where w is an n-th (and hence n,t-th root of I ) , and by Lemma 3.1, k(a-l) = m(a-l) tn for some t E Z.Setting $ = a-l, it follows from ainijn= i that

+

This shows that if n is fixed, then r, does not depend on the choices of a, and m ( a ) . In changing n, we may assume that the new choice of n is a multiple of the old. Then any choice of m(a-l) which satisfies (1) for the new n also satisfies (1) for the old n , and any choice of a, for the old n also works for the new n. Then, since n does not appear explicitly in (2), we see that r, is unchanged. (ii) Let H =< g > for some g E G. It is obvious that the restriction of r, to EaH is a n E-isomorphism of E a H onto E a H . If m is the order of g, then the elements ij, g2,. .. ,gm constitute an E-basis for EaH. Hence i t suffices t o show that I',(gi) = [ra(g)li

for all i 2 1

TOthis end, fix i 2 1 and choose X E F* such that gi = A?. Then

Hence, by (i), we may assume that we have

agi

= X-'a> But then, setting

+ = a-',

Counting Irreducible Representations

as desired. (iii) By hypothesis, H =< g > for some g E G , say of order k. Let w be a primitive kP,-th root of 1 in E and let ,Og be a k-th root of a(g,gi) in E. Because u ( o ) = om(")and m(a)r l(modlc,), it follows from (i) that

nfZl

Because gk = ,f3,k . I and n is a multiple of k, we have gn = proves by (i) that r,,(ij) = +(p,),~;~(*)g~(*)

&' . I, which

Since {?jill5 i 5 k) is an E-basis for E a H , the required assertion follows by virtue of (ii). (iv) Choose X E F*such that xgx-I = XZ g 2-l. Because gn =,a; . I , we have xgx-l = Xnan I= I 9 Hence, by (i), we may assume that axgx-~ = Xa,. It follows that

as required. (v) Since

x is an E-linear map, it suffices to show that

where $ = a-l. Let p be an irreducible representation of EffGwith character and let XI,. . . ,At be the characteristic roots of p(g). Since g n = ~ ( ~ ) i , :ylf ~ ( g , ~we~ have ) , A; = u ( g ) so that X i = a,w;, where wi where u(g) := is an a,,-th root of 1, 1 5 i 5 t . We deduce therefore that

x

3 Arbitrary fields

On the other hand,

L

= $(.,)Cw;

m(*)

7

i= 1

as required. (vi) By (ii) and (iii), we may assume that G is a p-group and n = [GI. Then n,, = 1, so E = 1 and thus we may assume that m(o) = 1 for all a E G a l ( E / F ) . Because n = pk for some t, E F and so o, is purely inseparable over F. Hence a(&,) = og and the required assertion follows. W

at

Lemma 3.3.

Let V = E o G , let Vg= {ASIA E E), g E G and let

be defined by r ( x , o ) ( g ) = zr,(g)z-' Then r is a monomial representation of G space (V7 G, (Vg)).

for all g E G

x

(3) G a l ( E / F ) on the monomial

Proof. It clearly suffices t o verify that for all y , II,E G a l ( E / F ) and all 2, Y,g E G7 W Y , v+)(g) = V ) ~ ( Y$)(9) , Applying Lemma 3.2 (iv), we have

w,

and

Counting Irreducible Representations

192

Hence we need only verify that r,r+ = r,+. To this end, fix g E G and write g = gPgpt. Then ij = a-l(gp, gpt)gpijpt. Invoking Lemma 3.2 (ii), (vi), we obtain

and r,*(g) = a-l(gp, gp')gpr,*(gp') Hence we may assume that g is a pf-element. P u t H =< g > and note that E a H is a direct product of /HI copies of E. Therefore, to show that r,r4(ij) = r,*(g), it suffices to verify that for all irreducible characters x of E a H ,

But, by Lemma 3.2 (iii), (v), we have

as desired. H We are now ready to introduce the notion of ( a , F)-regularity. Let r be the monomial representation of G x G a l ( F ) given by (3). We say that an element g E G is ( a , F ) - r e g u l a r if g is r-regular. Thus, by definition, g is ( a , F)-regular if and only if for any x E G, a E G a l ( E / F ) such that "9 m("-l)x-l = g, we have

Lemma 3.4. (i) If F contains E and each a,, then g E G is ( a , F ) regular if and only if g is a-regular. In particular, g is (a,E)-regular if and only if g is a-regular. (ii) If a is of finite order, then condition (4) reduces to

3 Arbitrary fields

(iii) If g E G is ( a , F)-regular, then g is a-regular. P r o o f . (i) By hypothesis, a ( & )= E and a ( a g ) = a, for all a E G a l ( E / F ) . In particular, each m(o) = 1 and, by (2), condition (4) reduces to 2 g 2-I = ij, as required. (ii) Let e be the exponent of G and let m be the order of a. Then, setting b = em, we see that gk = i for all g E G. Hence, by Lemma 3.2 (i), n can be chosen so that each ag = 1. Thus, in this case, I',(j) = ijm("-'1 for all g E G, a s required. (iii) Assume that xgx-' = g and take a = 1. Then I',(g) = g and so, by (4), 3 g 5-I = g, proving that g is a-regular. To provide some further properties of ( a , F)-regularity, we need t o recall the notion of F-conjugacy. Two elements a , b € Go are said t o be F - c o n j u g a t e if b = xapx-' for some x E G and some integer p such that E H E~ determines an automorphism of F ( E ) over F. I t is clear that Fconjugacy is an equivalence relation and so Go is a union of F-conjugacy classes. By Lemma 2.6.1, if g E Go is a-regular, then so is any F-conjugate of g. We say t h a t an F-conjugacy class C of p'-elements of G is a - r e g u l a r if at least one (hence any) element of C is a-regular.

Lemma 3.5.

Let Vo be the F-linear span of all ij with g E Go and let

be defined by r o ( x 7 o ) ( g )= %I',(g)s-l

for all g E Go

Then the following properties hold : (i) To is a monomial representation of G x G a l ( E / F ) on (Vo, Go,(Vg)). (ii) The F-conjugacy classes of Go are the G x Gak(E/F)-orbits of Go. (iii) An element g E Go is ( a , F)-regular if and only if g is l?o-regular. (iv) If g E Go is ( a , F)-regular, then so is any F-conjugate of g. P r o o f . (i) This is a direct consequence of Lemma 3.3. (ii) By the definition of ro,if a , b E Go are in the same G x G a l ( E / F ) orbit, then a and b are F-conjugate. Conversely, suppose that a , b E Go are F-conjugate. Then b = x a ~ x - I for some x E G and some integer p such

Counting Irreducible Representations

194

that E I+ E~ determines a n automorphism of F ( E ) over F. Let H =< a >, let m be the order of H, and let w be a primitive m-th root of 1 in E. Then there exists a E G a l ( E / F ) such that c r - ' ( ~ ) = E P . I t follows from Lemma 3.2 (iii) that Supp(ra(a)) = ap and hence that

Consequently, a and b belong t o the same G x Gal(E/F)-orbit, as required. (iii) Direct consequence of the definitions of ( a , F ) and ro-regularities. (iv) Apply (ii), (iii) and Proposition 2.2 (i). H Let C be an F-conjugacy class of pl-elements of G. We say that C is ( a , F ) - r e g u l a r if each element of C is ( a , F)-regular. Thus, by Lemma 3.5 (iv), C is ( a , F)-regular if and only if at least one element of C is ( a , F ) regular.

.

C o r o l l a r y 3.6. Let C be an ( a ,F)-regular F-conjugacy class of plelements of G. Then C is a-regular. Proof.

Apply Lemma 3.4 (iii).

We have now come to the demonstration for which this section has been developed. T h e o r e m 3.7. (Reynolds (1971)). Let G be a finite group, let F be a n arbitrary field of characteristic p 2 0, and let a E Z2(G,F * ) . Then the number of nonisomorphic simple FaG-modules is equal to the number of ( a , F)-regular F-conjugacy classes of p'-elements of G. P r o o f . We keep the notation of Lemma 3.5. P u t H = G x G a l ( E / F ) and denote by n ( F a G ) the number of nonisomorphic simple FaG-modules. Owing t o Lemma 3.5, it suffices t o show that n ( F a G ) is equal t o the number, say t , of ro-regular H-orbits of Go. We know, by Theorem 11.1.7 (i) in Vol. 1, that there exists a finite normal field extension L / F such that L is a splitting field for LaG. Hence we may replace E/F by a finite normal field extension L / F such that L is a splitting field for LaG and L contains a primitive npl-th root of 1 as well as a, for all g E G. Since in this case H is finite, it follows from Proposition 2.2 (iii) that, t is equal to the dimension of the fixed-point space of To.

3 Arbitrary fields

195

Let : H -t GL(V:) be the contragredient representation, let t* b e the dimension of the fixed-point space of I?;, and let s be the number of orbits under the action of G a l ( L / F ) on the set of irreducible characters of LffG. Owing to Proposition 2.2 (iv), we have t* = t . On the other hand, by Theorem 14.5.1 (ii) in Vol.1, we have n ( F a G ) = s. Thus we are left t o verify t h a t t* = s. Let X; : E f f G + E, 1 5 i 5 r , be all irreducible characters of E f f G and let be the restriction of xi t o Vo. Let the maps

Xy

r1 : G -+

GL(V;)

and

I'2 : G a l ( E / F )

+ GL(V,')

be defined by

Then rl and tively, and

r2are monomial representations of G and G a l ( E / F ) , respec-

Let W , Wl and W2 be the fixed-point spaces of r;, I'l and r2,respectively. Then, by ( 5 ) , W = Wl n W z . Note also that, by Proposition 2.6, (xPI1 5 i 5 T ) is an E-basis of $Vl, whereas by Lemma 3.2 (v), I ' ~ ( U ) ~=: ax:

for all o E G a l ( E / F ) , i E (1,. . . , r )

Hence W is the fixed-point space of the monomial representation

.

induced by r2. Let s* be the number of orbits of {Xy,. . . ,X:) under the action of G a l ( E / F ) . Then s* = s and, by Proposition 2.2 (iii), s* = t*. Hence t* = s and the proof is complete. Corollary 3.8. Let G be a finite group, let F be a n arbitrary field of characteristic p 0 and let a E Z2(G,F*). Then the number of nonisomorphic simple FOG-modules is less than or equal to the number of a-regular F-conjugacy classes of pl-elements of G .

>

Proof. Apply Theorem 3.7 and Corollary 3.6.

196

Counting Irreducible Representations

An example of strict inequality here is provided by ~ ( 4 ) .Indeed, Q ( a ) S Q a G where G is of the order 2 and a E Z2(G,Q*). Since QaG is commutative, all elements of G are a-regular. Hence the two Q-conjugacy classes of G are a-regular. On the other hand, since QaG is a field, it has exactly one simple QaG-module, up to isomorphism.

4

Counting projectively nonequivalent representat ions

In this closing section, we return to algebraically closed fields and ask : How many projectively nonequivalent irreducible a-representations does a given group G have? A complete solution is provided by the following result.

Theorem 4.1. (Mangold (1966), Tappe (1977)). Let G be a finite group, let F be a n algebraically closed field of characteristic p 0 a n d let a E Z2(G, F*). Then the number of projectively nonequivalent irreducible a-representations of G over F is equal to the number of a-regular conjugacy classes of pl-elements of G contained in GI.

>

Proof, Let S and S' be the sets of all p'-elements of G and GI, respectively, and let V and V' be the F-linear spans of {glg E S) and {gig E S'), respectively. For each s E S , let V, = F S and for each s E S', let V,' = FS. Then (V, S, (V,)) and (V', S', (V,')) are monomial spaces and the maps (g E G , v E V,vl E V') are given by r(g)v = gvg-l, r1(g)v' = gv'g-l monomial representations of G on these spaces. Let W and W' be the fixed-point spaces of I?* and (I")*, respectively, and let n(respectively, m ) b-e the number of projectively nonequivalent irreducible a-representations of G (respectively, the number of a-regular conjugacy classes of pl-elements of G contained in GI). By Propositions 2.2 (iv) and 2.4, m = dimFW1. Furthermore, the maps A: : V + F, 1 5 i 5 r, given by Proposition 2.6, form a n F-basis of W. Since W' = W n ( V 1 ) * it , follows that W' is the subspace of W consisting of all elements of W which take value zero on all g with g E S - S'. Let G* = I£om(G, F * ) and let p : G' + GL(W) be the homomorphism defined by

4 Counting projectively nonequivalent representations

Then p ( 4 ) permutes the A: in such a way that for all $ E G*,

$xi

= Xj if and only if p(+)X: = AS

Here the maps Xi : G + F, defined by x;(g) = X;(g), are all irreducible a-characters of G. Applying Proposition 2.7, we see that n is equal to the number of P - o r b i t s of the set {A:, A;, .. . ,A:). Since all G*-orbits are pregular, it follows from Proposition 2.2 (iii) that n is equal to the dimension of the fixed-point space, say W", of p. We are thus left to verify that W' = W N . Suppose that 0 E W". Then $(g)O(g) = O(j) for all

$ E Hom(G, F * ) , g E S

(1)

If g E S-S', then gG' is a nonidentity p'-element of GIG' and hence $(g) # 1 for some $ E Hom(G, F').Hence, by (I), O(g) = 0 and so 0 E W', proving that W" W'. Conversely, suppose that 6 E W'. Then, for all g E S - S', $ E Hom(G, F*),we have 4(s>O(g) = O(g) = 0 On the other hand, since $(g) = 1 for all $ E Hom(G, F*),g E S', it follows that $(g)O(g) = O(j) for all g E S, $ E Hom(G, F * ) . This shows that 0 E W", hence W' C W" and the result follows. W The following consequence of the above result for the case where F = C and G is abelian is due to Frucht (1931).

Corollary 4.2. Let G be a finite group, let F be a n algebraically closed field and let a E z 2 ( G , F*). Assume that one of the following conditions holds : (a) G is abelian. (b) G' is a p-group and c h a r F = p > 0. Then the following properties hold : (i) All irreducible a-representations of G are projectively equivalent. (zi) If p : G + GL,(F) is a n irreducible a-representation of G, then any irreducible a-representation of G is linearly equivalent to a representation of the form : p, : G

+ GL,(F)

for some p E Hom(G, F*)

198

Coullting Irreducible Representations

where for ,211 g E G , p,(g) = ~ ( g ) p ( g ) .

Proof. (i) By hypothesis, 1 is the only pl-element of G contained in G'. Hence the required assertion follows by Theorem 4.1. (ii) Let p' be any irreducible a-representation of G. Then, by (i), p' is projectively equivalent t o p. Hence, by Corollary 3.2.5, p1 is linearly equivalent to a representation of the form p, for some p E Hom(G,F8). H