Chapter 23 Zeros of Characters

Chapter 23

Zeros of Characters This chapter is devoted entirely to the study of zeros of C -characters x of a finite group G, i.e. to the study of those elements g E G for which x(g) = 0. Our first result, due to Gallagher (1966), asserts that if H is a subgroup of G and x is an irreducible character of G with x(g) # 0 for all g E G - H , then XH is irreducible and, for any g E G - H,x(g) is a root of unity. As an immediate consequence, we derive a classical theorem of Burnside (1903), which shows that any nonlinear irreducible character of G has at least one zero. Let H be a subgroup of G and let x be an irreducible character of G such that x ( g ) = 0 for all g E G not conjugate to elements of H . What information can be obtained about such a character ? One of the results presented (Ferguson and Isaacs (1989)) shows that X H is reducible and that there exists a positive integer n and a generalized character X of H such that X~ = nx. A separate section is devoted to counting the number n(x) of zeros of a given irreducible character x. For example, it is shown that n(x) 2 ( ~ ( 1 -) ~l)lZ(x)l, and n(x) 2 x ( l ) z ~ Z ( x )unless l 1x1 takes only values 0, 1 and ~ ( 1 ) .We also present a powerful condition for a character value to be zero. Namely, we establish a classical result, due to Brauer and Nesbitt (1941), which asserts that if the degree of an irreducible character x of G is divisible by the order of a Sylow p-subgroup of G, then x(g) = 0 for each g E G whose order is divisible by p . A separate section is devoted to some recent results due to Zmud. In particular, we prove an elegant theorem, due to Zmud (1977, 1990), which asserts that if G is a nonabelian group, then G is generated by the zeros of 837

Zeros of Characters

838

a suitable irreducible

(I: -character

of G.

1 Burnside’s theorem Throughout this section, G denotes a finite group and all characters are assumed to be (I:-characters. Here we demonstrate that if x is an irreducible nonlinear character of G, then x ( g ) = 0 for some g E G. This will be derived as a consequence of a more general result which will be needed later.

Lemma 1.1. Let a1,. . . ,an be the complex roots of a monic polynomial f ( X ) E Z [XIof degree n 2 1 and let Ia;I = 1 for each i E (1,. ,n } . Then each cri is a root of unity.

..

Proof. For each k 2 1, put fh(X)= nbl(X- at). Then b ( X ) E Z [XIsince, up to the sign, the coefficients of fk(X)are the elementary symmetric functions of at,. .. ,a! and each of these functions is a polynomial

.

of the elementary symmetric functions of a1 ,.. ,on. Since ~ ( Y=~1I for each i E {l,...,n } , we see that the coefficients of various fk(X) are uniformly bounded. Hence there are only finitely many such polynomials. Therefore there are only finitely many roots of these polynomials. Hence some two distinct positive powers of a; must coincide, as desired.

Lemma 1.2. Let (Y # 0 be an algebraic integer with all conjugates a1, . . . ,a, (each repeated the same number of times). Then T

i=l

with equality if and only if

(Y

is a root of unity.

Proof. We know that the arithmetic mean of positive real numbers is at least the geometric mem and that the equality occurs if and only if these numbers are equal. In particular, l-

In:=’=,

with equality if and only if all lail are equal. But a;[is a positive rational integer, so (1) holds with equality if and only if all lail are equal

1 Burnside’s theorem

839

nr==,

and I ail = 1. Since the latter is equivalent to all [ail= 1, the result follows by virtue of Lemma 1.1.

Lemma 1.3. Let g E G be an element of order rn, let

c = {s”I(P,m) = 1) and let

x be a character of G with x(g) # 0.

c

x€C

Then

lx2(x)l 2 ICl

with equality if and only if x(g) is a root of unity.

Proof. Let E be a primitive rn-th root of unity. Then for any p 2 1 with ( p , r n ) = 1, we have

x2(sP> = q.t(x2(s>> where up E G a l ( Q ( E ) / Q ) is defined by up(&)= E P . Hence {x2(x)1x E C} are all conjugates of the aIgebraic integer x2(g) # 0 (each repeated the same number of times). Since x2(g) is a root of unity if and only if x(g) is a root of unity, the result follows from Lemma 1.2. 4

Theorem 1.4. (Gallagher (1966)). Let H be Q subgroup of G and let x be a n irreducible character of G such that x(g) # 0 for all g E G - H . Then X H is irreducible and, for any g E G - H , x(g) is a root of unity. Proof. We have < x , x >= 1 and < X H , X H >2 1 with equality if and only if XH is irreducible. The equivalence relation x-y,=,y>

on G

partitions G-H into disjoint subsets, say G-H = UfZ1Cj. Then, by Lemma 1.3, IGI =

c 1x2(s)l c

g€G

=

hEH

2 IHI

cc t

Ix2(h>> i-

i = l gEC,

Ix”g>l

+ IG - HI = /GI

Hence C h E H 1x2(h)l = IHI and &C, 1x2(g)1= ICiI,1 I i I t. Thus XH is irreducible and, by Lemma 1.3, x(g) is a root of unity for all g E G - H . H

Zeros of Characters

840

Corollary 1.5.

(Burnside (1903)). Let

x

be an irreducible character

of G with x(1) # 1. Then x(g) = 0 for some g E G .

Proof. Apply Theorem 1.4 for H = 1. W

2

Characters vanishing off subgroups

Let H be a subgroup of G and let x be an irreducible character of G such that x ( g ) = 0 for all g E G not conjugate to elements of H. What conclusion can be drawn about such a character x? In this section we show that XH is reducible and that there exists a positive integer n and a generalized character X of H such that XG = nx. The proof of the following result is due t o D. Gluck (see Isaacs (1983)).

Theorem 2.1. Let H be a subgroup of G and let x be an irreducible character ofG. Assume that x(g) = 0 for all g E G not conjugate to elements of H . Then XH is reducible.

Proof. By Proposition 19.1.7 (ii), (xH)' = ( 1 ~ x. ) ~Furthermore, 2 1 whenever x(g) # 0 (by hypothesis) and

(lH)'(g)

( 1 ~ )1)~=( (G : H ) > 1 Applying Frobenius reciprocity, we then have

as asserted.

Our next aim is to demonstrate that if x is an arbitrary character of G satisfying the hypothesis of Theorem 2.1, then there exists a generalized character X of H and a positive integer n such that nx = XG. The following lemma due to Ferguson and Isaacs (1989) will clear our path.

2 Characters vanishing off subgroups

841

Lemma 2.2. Let a be a generalized character of G such that a(g) # 0 for all g E G and let s = o ( g ) , where g runs over a set of representatives for the conjugacy classes of G. Then s E Z and s/a,defined by ( s / a ) ( g )= s a ( g ) - ' , g E G , is a generalized character of G.

n,

Proof. Our hypothesis on a ensures that Q has an inverse in the ( c space c f ( G ) of class functions of G. Hence the map r : c f ( G ) + c f ( G ) defined by I'(cp) = ya is a nonsingular linear transformation of cf(G). The matrix A of r with respect to the basis Irr(G),consisting of all irreducible characters of G , has entries in Z . On the other hand, the characteristic functions of the conjugacy classes of G constitute another basis for c f ( G ) . With respect to this basis, the matrix of I? is diagonal, with entries equal to the various a(g). Thus s = det(I') = d e t ( A ) E Z . Let A* denote the adjoint of A so that A-' = s-'A*. Then A* has integer entries since so does A . Hence sA-' has integer entries and therefore S r - ' ( l G ) is a Z -linear combination of I r r ( G ) . Because S r - ' ( l G > = s / o , the result follows. H

Theorem 2.3. (Ferguson and Isaacs (1989)). Let x be a character of G and let H be Q subgroup of G such that x(g) = 0 for all g E G not conjugate to elements of H . Then there exists Q positive integer n and a generalized character X of H such that XG = nx. ) ~see, that a ( h ) # 0 for all h E H . Now Proof. Setting a = ( l ~ we define the class function P on G by

Then both Pxa and x are zero on elements g E G which are not conjugate to elements of H and P ( g ) a ( g ) = 1 for the remaining g E G. Thus

x

= =

Pxa = ( P x ) ( W G ( ( P x ) H ~ H ) ~ (by Lemma 19.3.2

(i))

= (PHXH)~ On the other hand, by Lemma 2.2, n P H is a generalized character of H for some positive integer n. Thus = n & x H is a generalized character of H such that XG = nx, as desired.

Zeros of Characters

842

3

Gallagher's theorems

The results of this section will require a considerable amount of notation. Therefore it will be especially useful to assemble most of it in one place. In what follows G denotes a finite group and all characters are assumed to be (I:-characters. Moreover the following notation will be used : Irr(G) is the set of all irreducible characters of G. n(x) is the number of elements g E G such that x(g) = 0, x E Irr(G). n(g) is the number of x E Irr(G) such that x(g) = 0, g E G. Z(X) = (9 E GI Ix(s>l= X ( W K(g) is the smallest normal subgroup of G containing {[g,z]Iz E G} r ( G ) is the number of conjugacy classes of G.

Theorem 3.1. (Gallagher (1962 a)). With the notation above, we have (i) For each x E Irr(G), n(x) L ( ~ ( 1 )-' 1)lZ(x)l. (ii) For each x E Irr(G),n(x) 2 x(l)'lZ(x)l unless 1x1 takes only values 0, 1 andx(1). (iii) For each g E G, n(g) 2 r(G) - ICG(g)l (iv) For each g E G, n(g) = r(G)- IcG(g)l if and only af K(g) = G' and Ix(g)l = 0 or 1 for all x E Irr(G).

Proof. (i) and (ii). Given x E Irr(G), it follows from Lemma 1.3 that

with equality only if Ix(g)l takes only the values 0, 1 and ~ ( 1 )Because . n(x) is a multiple of IZ(x)l, inequality implies n(x) 2 x(l)'~Z(x)l, as desired. (iii) and (iv) For each g E G, we have

where the first sum is over those x with Ix(g)l = x( 1) and the second sum is over the others. Note that g E Z(x) if and only if x is a character of G/K(g). Hence the first sum is IG/IC(g)l and the number of characters in the first

3 Gallagher's theorems

sum is r(G/K(g)).Thus, if X I , . . then t = r ( G )- r ( G / K ( g ) )and

843

. ,xt are the characters in the second sum,

Let E be a primitive m-th root of unity, where m is the exponent of G. Then the group Gal(Q ( e ) / Q ) acts on the set { X I , . . . ,x t } via

If S1,.

..

)

Sk are the orbits of { X I , . . . , x t } , then by Lemma 1.2,

with equality only if Ix;(g)l takes only the values 0 and 1. Hence, by (l),

with equality only if Ix(g)l takes only the values 0, 1 and x ( 1 ) for each x E I r r ( G ) . Because IG/K(S)I 2 r ( G / K ( g ) ) with equality if and only if K ( g ) = G', it follows that

with equality if and only if K ( g ) = G' and Ix(g)l = 0 , l or x ( 1 ) for each x E I r r ( G ) . But 1x(g)1 = x(1) only for the characters of G / K ( g ) ,which is only for x(1)= 1 in this case. This completes the proof of the theorem. H The following assertion is a part of Clifford theory which is a subject in itself and will be examined at a later stage. However, since the result is of a very elementary nature, its verification can be achieved by an ad hoc argument which we present below.

Lemma 3.2. Let N be. a normal subgroup of G such that GIN is cyclic of order n. If x is a G-invariant irreducible character of N , then there exist precisely n irreducible characters of G extending x and xG is their sum.

Zeros of Characters

844

Proof. By Frobenius reciprocity, the number of irreducible characters of G extending x cannot exceed n . On the other hand, if X is a character of G extending x , then A is irreducible; hence so are Ap1,. . .,Apn (where p 1 , . . . ,pn are all linear characters of GIN) since each Api extends x . Furthermore, each Ap, is a constituent of xG by Frobenius reciprocity and ~ ~ (= 1n ( )X p j ) ( l ) ,so xG = C7=lXpj. Thus it suffices to show that x can be extended to a character of G. Let 'l : N + GL,((L:) be an irreducible matrix representation of N which affords x and let g E G be such that G I N =< gN >. Since g x = x and g x is afforded by the matrix representation gI' given by ( T ) ( n )= I?(g-lng) for all n E N , there is a matrix p ( g ) E GL,((C ) such that p ( g ) - l r ( n ) p ( g )= r ( g - l n g ) for all n E N . Then

.

(for all n E N, i E (1,. . ,n)) P(g)-ir(n)P(s)' = Replacing p ( g ) by its scalar multiple, if necessary, we may assume that p ( g ) n = r(n0) where g n = no E N . Defining I?* : G + GL,(C) by r * ( g j , ) = p(g)jr(n)

it follows that

(0

5 j 5 n - 1, n E N )

I'* is a representation of G which extends I?, as desired.

Theorem 3.3. (Gallagher (1966)). Let N be a normal subgroup o f G with N # G and let G j i x each conjugacy class of N. Then, f o r any nonlinear irreducible character x of G, there exists g E G - N such that x ( g ) = 0 .

Proof. Assume by way of contradiction that x is a nonlinear irreducible character of G with x ( g ) # 0 for all g E G - N . Put H =< g l N > where g1 E G - N . Then H / N is cyclic, H fixes each conjugacy class of N and x ( g ) # 0 for all g E G - H . Hence, by Theorem 1.4, X H is irreducible and so we may assume that GIN is cyclic of order n. Let XI,. . . ,xr be all irreducible characters of N . By Corollary 17.4.3, each xi is G-invariant since G fixes all conjugacy classes of G. Hence, by Lemma 3.2, each xi extends in exactly n ways to characters x j j , 1 5 j 5 n, of G and each irreducible character of G is precisely one of the xij, 1 5 i 5 r , 15jSn. By Theorem 1.4, x ( g ) is a root of unity for all g E G - N . Hence the character x X - 1 vanishes off N (here x ( g ) = x ( g ) for all g E G). Thus < X X - 1, x;j > depends only on i and so

4 Applications and related results

045

for suitable integers ai. By Corollary 1.5, x ( g ) = 0 for some g E G. Hence g E N and evaluating (2) at g, we get

naiX;(g)= 0 (mod n),

-1 = i

a contradiction.

A strengthened version of the result below is given by Theorem 5.11 to be proved in Sec. 5 . Corollary 3.4. Each nonabelian group G is generated b y the set of all zeros of all of its irreducible characters.

Proof. Let N be the subgroup of G generated by all of the zeros of all irreducible characters of G. By Theorem 1.4, the restriction of each irreducible character of G to N is irreducible. It follows that G fixes the characters of N and hence G fixes the conjugacy classes of N . Because no irreducible character of G has zeros off N ,it follows from Theorem 3.3 that N = G, as required.

4

Applications and related results

We first present the following remarkable consequence of Corollary 1.5 discovered by Brauer (1954) and Wielandt.

Theorem 4.1. Let C1, ...,CT be all conjugacy classes of G and let

CT = Cscc, g , 1 5 j 5 r . Then G = G' if and only if r

t

Proof. Let x1 = l ~~ 2, , .. . ,xT be all irreducible characters of G and let w1,.. . ,wT be the central characters of C G given by Theorem 21.1.1 (i). Then, by Theorem 21.1.1 (iv) and the orthogonality relation

j=1

j=1

Zeros of Characters

846

Let m E Q be defined by 21.1.1 (iv),

'&lCjl

= m n j Z 1ICjl. Then, by Theorem

n

w i ( C c?)= m j=1 lCjl= wl(mjn c:) =1 r

r

t

j=1

(3)

Applying (2) and (3), we see that (1) holds if and only if ui(& CT) = 0 for i # 1, or, equivalently, for any i # 1 there exists zi E G with xi(zi) = 0. But xj(zi) = 0 implies xi(1) # 1; hence by Corollary 1.5, (1) is equivalent to the fact that x1 is the only linear character of G, or equivalently G = G'. This completes the proof of the theorem. Our next result is extracted from a classical paper of Feit and Thompson (1963, Lemma 4.3).

Proposition 4.2. Let N be a normal subgroup of G and let x be an irreducible character ofG with KerX 2 N . If g E G is such that CG(g)nN = 1, then x(g) = 0.

Proof. Let XI, X2,. . . ,At be all irreducible characters of GIN and let XI,~ 2 , . .. ,xs be the remaining irreducible characters of G. If C G ( g ) n N = 1, then the natural homomorphism G GIN induces an injective homomorphism cG(fJ) --+ C,,N(gN). Thus, by Theorem 19.2.3 (iii), --f

t

i=l

.

j=1

Hence xj(g) = 0 for all j E { 1,.. ,s}, as required.

As a preliminary to the next result, let us record the following elementary fact. Lemma 4.3. Let R be a commutative ring of characteristic 0 and let n,m be coprime positive integers. If r E R is such that n divides mr, then n divides r.

Proof. By hypothesis, mr E Rn and 1 = an Hence T = Tan rpm E Rn, as desired.

+

+ pm for some a ,p E 23 .

5

Zmud’s theorems

847

We close by providing the following powerful condition for a character value to be zero. Theorem 4.4. (Brauer and Nesbitt (1941)). Let x be a n irreducible character of G and suppose that p is a prime such that x(1) is divisible by the order of a Sylow p-subgroup of G. Then x(g) = 0 for each g E G whose order is divisible by p .

Proof. (Gallagher (1966)). Denote by a the class function on G given by

{

4 7 ) =

;(g)

if p divides the order of g otherwise

We must show that a = x. Because 0 < < a,a >I< x,x >= 1, with equality if and only if a = x, it suffices to show that a is a generalized character of G. By Theorem 20.2.1 (ii), it therefore suffices to prove that < ~ E , X > E Z for each elementary subgroup E of G and each irreducible character X of E . For each such E , we have E = P x Q, where P is a p-group and Q is a group of order prime to p . Hence a(.) = 0 for all 5 E E - Q and QQ = XQ. Therefore, for each irreducible character X of E ,

< a E , x >= I P I - ’ <

xQ,XQ >

(4)

For z E Q, we have Cc(.) 2 P, so (G : CG(Z))divides (G : P ) . On the (in the other hand, by Theorem 21.1.1 (iv), x(1) divides x(s)(G : CG(S)) ring of algebraic integers of Q (x)). Hence x( 1) divides x(z)(G : P) and so (P(x(1)divides IGlx(z). From the hypothesis that x(1) is divisible by the order of a Sylow psubgroup of G it now follows from Lemma 4.3 that ]PI divides ~ ( s for ) all x E Q. Hence IPI divides IQI < XQ,XQ > and so IPI divides < XQ, XQ >. Thus, by (4), < a ~A , > is an integer as required. W

5

Zmud’s theorems

Throughout, G denotes a finite group. Our aim is to obtain sharpened versions of Theorem 1.4 and Corollary 3.4. One of the main results (Theorem 5.11) asserts that if G is nonabelian, then G is generated by the zeros of a suitable irreducible (c -character of G. If G is nonabelian and nilpotent, then it is shown that G is generated by the zeros of each irreducible nonlinear (c character of G.

Zeros of Characters

a48

Lemma 5.1. Let p be a prime and let g E G be of order pn with n > 0. Then, for any C -character a of G and any x E C G ( g ) ,

a(gz)P" where R = Z [ E ] and

E

=a(q"

(modpR)

is a primitive complex pn-th root of unity.

Proof. Put H = C G ( g ) and write CIH = a1 t - - t . aT,where each ai is an irreducible (I: -character of H . Since g E Z ( H ) , we have a i ( g ) = E ; ( Y ; ( ~ ) where E i is a p"-th root of unity. Since x E H , it follows that q ( g x ) = E ~ o ~ ( z )(1 5 i 5 r ) We deduce therefore that t

a(gz)P"

t

= ( c a i ( g z ) ) p " = (CE;ai(z))P" i=l

i=l

as required.

In what foIlows, the subgroup generated by an empty set is understood

to be 1.

Lemma 5.2. Let p be a prime, let x be an irreducible (I: -character of G and let N be the (necessarily normal) subgroup ofG generated by the zeros of x. (i) If G has a p-element g with Ix(g)l = 1, then x(1)= f l ( m o d p ) . (ii) If G # N and p divides [ G I N / ,then x ( 1 ) 1 f l ( m o d p ) . (iii) ( ~ ( l) )G,/ N ) = ) 1.

Proof, (i) We may assume that the order of g is pn with n a = xx and 5 = 1 it follows from Lemma 5.1 that a(g)P"

> 0.

Setting

= a(1)P" = a( 1 ) ( m o d p R )

where the last congruence holds since for any integer a, up" ~ ( m o d p ) . Now a ( g ) = lx(g)I2 = 1, a(1) = ~ ( 1and ) ~so ~ ( 1 E) 1 ~( m o d p R ) . Hence

5 Zmud's theorems

849

~ ( 1E)1~ ( m o d p ) and therefore x(1) = f l ( m o d p ) , proving (i). (ii) Let P be a Sylow p-subgroup of G. Since p divides \GIN\, we have P N. Hence we may choose g E P - N . Then, by Theorem 1.4 (applied to H = N ) , Ix(g)l = 1. The desired assertion is now a consequence of (i). then x(1) (iii) We may assume that G # N . If p divides [GIN[, f l (rnodp) by virtue of (ii). But then p [ ~ ( l as ) ,required. H

=

Lemma 5.3. Let H be a proper subgroup of G and let x be a n irreducible (1: -character of G such that x = +G for some character of H and such that the zeros of x generate a proper subgroup N of G. Then (i) G = N H and H n g-lHg c N n H for all g E G - H . (ii) nseGg-lHg E N n H .

+

Proof. (i) Since x(1) = (G : H)+(l), it follows from Lemma 5.2 (iii) that ((G: H),(G : N ) ) = 1 Since (G: N H )divides both (G: H )and (G : N ) , we deduce that G = N H . Hence GIN E H / H o , where HO = H n N , and therefore, since G # N , we have Ho # H . Now put m = (G: H ) so that

mlHol= IN1

(1)

Next put S = G - N and let g E S. Then, since x ( g ) = ]HI-'

+(tgt-')

(+(z) = 0

for z E G - H )

tfG

and since x(g) # 0, we have g E t - * H t for some t E G. Hence, setting H t = t - l H t , we have

s c utecHt

Since S is a G-invariant subset of G,we have

s = S n(

u ~ ~ H = ugl(s ~ ' ) nP

i )

= u"a=1 (S n H ) ~ ;

where t l , . . . ,t , is a right transversal for H in G.Applying (1) and the fact that S n H = H - Ho, we deduce that

Zeros of Characters

850

= m(lHI - IHol) = IGl -IN1 = IS1

whence I UEl (S n = Czl I(S n H)ttl. Hence the sets (S n H ) t l , 1 5 i 5 m, are mutually disjoint. We may, of course, assume that t l = 1. Then, setting X = S n H , we have X n = 8 for all i > 1. Finally, fix g E G - H . Then g E H t i for some i > 1. Since X is H-invariant, we have X n Xg = X n X t l = 8. Taking into account that Ho n X9 = X n Hog = 8, we deduce that H)tlI

Xtl

H n H g = (Houx)n(H;uXg) = (Ho n H i ) u ( H n~ xg)u ( X n H;) u (Xn x = H ~ ~ H : c_ Ho,

g )

as required.

(ii) This is a direct consequence of (i). H

Lemma 5.4. Let x be an irreducible (I: -character of G such that the zeros of x generate a proper subgroup N of G and let M be a normal subgroup of G which is not a proper subgroup of N. Then X M is irreducible and, under N, the zeros of X M generate a proper the stronger assumption that A4 subgroup of M .

Proof. If M = G, then the assertion is trivial and if M = N then the assertion is a consequence of Theorem 1.4. We may therefore assume that M N and M # G. By Clifford's theorem (Proposition 19.1.12), X M = e(A1+ * .+A,) where XI = X is irreducible, A 1 , . . . ,A, are all distinct conjugates of A, e is the ramification index and n = (G : H ) where H is the inertia group of A. Assume that n > 1. Then H # G and by Clifford's theorem, x = t,hG for some character $ of H . Hence, by Lemma 5.3 (ii), a

a contradiction. Thus n = 1 and so X M = eX. P u t S = G - N . S i n c e M g N , w e h a v e M n S + @ . F i x g E M n S . Since x(g) = e$(g) and, by Theorem 1.4, x(g) is a root of unity, then e divides 1

5 Zmud's theorems

851

in the ring of algebraic integers. Hence e = 1 and so X M = A is irreducible. Finally, put T, = ( 9 E GIX(g) = 0 } , Tx{g E GIX(g) = 0) and let Ir' be the subgroup of M generated by Tx. Because Tx = M n T,, we have

Thus K

#M

and the result follows.

We are now ready to prove the following theorem which is a strengthened version of the first assertion of Theorem 1.4. (Zrnud (1990)). Let x be an irreducible a! -character of x genemte a proper subgroup N of G and let M be a subnormal subgroup of G which is not a proper subgroup of N . Then X M is an irreducible character of M and, under the stronger assumption that M N , the zeros of X M generate a proper subgroup of M .

Theorem 5.5.

G such that the zeros of

-

Proof. Let M = Gk a - - 4 GI a G be a part of a composition series for G passing through M . As in the proof of Lemma 5.4,we may assume that M # G a n d M g N. S i n c e M g N , w e h a v e G j g N f o r a l l i E ( 1 , . . . , I c } . In particular, since G1 d G and G1 g N , it follows from Lemma 5.4 that X G is ~ an irreducible

character of G1 whose zeros generate a proper subgroup of G I . Since G2 d G1 and G2 K where Ir' is generated by the zeros of X G ~ it, follows similarly that xc2 is an irreducible character of G2 whose zeros generate a proper subgroup of G2. Continuing in this way, we finally conclude that X M is an irreducible character of G whose zeros generate a proper subgroup of M. This completes the proof of the theorem.

Corollary 5.6. Let x be an irreducible (c -character of G such that the zeros of x generate a proper subgroup N of G and let M be a subnormal subgroup of G with M g N . Then X M ~ Nis an irreducible character of MnN. Proof. By Theorem 5.5, X = X M is an irreducible character of M whose zeros generate a proper subgroup K of M . Since M n N is a subnormal subgroup of M which is not a proper subgroup of K , it follows from Theorem 5.5 that X M ~ N is an irreducible character of M n N. Finally, since

Zeros of Characters

852

the desired assertion follows.

Let x be an irreducible U2 -character of G whose zeros generate a proper subgroup of G. If M is a maximal normal subgroup of G , then X M is irreducible. Corollary 5.7.

Proof. This is a direct consequence of Theorem 5.5. H Corollary 5.8. Let x be an irreducible nonlinear (c -character of G whose zeros generate a proper subgroup N of G and let M be a normal subgroup of G which does not contain zeros of x. Then M C N .

Proof. Assume by way of contradiction that that M is not a proper subgroup of N . Then, by Lemma 5.4, X M is irreducible. Since x(1) # 1, x ( g ) = 0 for some g E M (by Corollary 1.5). Since this is contrary to the assumption that A4 does not contain zeros of x, the result follows.

Let x be an irreducible nonlinear CI: -character of a nilpotent group G. Then G is generated by the zeros of x, Corollary 5.9.

Proof. Since G is an M-group (Corollary 18.12.4), x is induced from a linear character of a proper subgroup. Hence x is induced from a character of a maximal subgroup M of G which must be normal since G is nilpotent. Then X M is reducible and therefore the desired assertion follows from Corollary 5.7. W

As a preliminary to our next theorem, we record the following result. Lemma 5.10. Let x be an irreducible C -character of G and let x(x) = 0 for some x E G . Then CG(X) c N where N is the subgrozsp of G generated by the zeros of x. Proof. We first show that CG(X) C N. Assume by way of contradiction that &(x) N. Then N does not contain at least one Sylow subgroup of CG(Z).Hence there exists a prime p and a Sylow p-subgroup P of CG(Z) such that P N . Let g E P - N . Then sg fZ N and so, by Theorem 1.4, x ( z g ) is a root of unity. Let o(g) = p". Setting cr = xx, it follows from

5

Zmud's theorems

a53

Lemma 5.1 that a(gz)P"

= cr(5)P"

(modpR)

Since x ( z g ) = x ( g s ) is a root of unity, a ( g s ) = lx(gz)12 = 1. On the other hand, a(%)= lx(z)12 = 0 and so 1 = O(modpR), a contradiciton. Thus CG(Z) N . Assume by way of contradiction that C G ( Z )= N . Then s E Z ( N ) . Since, by Theorem 1.4, X N is irreducible, we have z E Z ( x ) . But then x(z) # 0, a desired contradiction. H We have done most of the work to prove our second theorem.

Theorem 5.11. @mud (1977,1990)). Let G be a nonabelian group. Then G is generated by the zeros of a suitable irreducible (c -character of G. Proof. Assume by way of contradiction that the zeros of each irreducible (c -character of G generate a proper subgroup of G. Let M be any maximal normal subgroup of G. Then, by Corollary 5.7, X M is irreducible for any irreducible CC -character x of G. Hence each irreducible C -character of M is extendible to G which implies, by Theorem 18.7.3 (ii), that each M-conjugacy class is a G-conjugacy class. Hence

Now choose x to be a nonlinear irreducible (c -character of G. Let N be the subgroup of G generated by the zeros of x and let M be a maximal normal subgroup of G containing N . By Corollary 5.7, X M is irreducible and, since x ~ ( 1 #) 1, there exists g E M with x(g) = 0 (see Corollary 1.5). Hence, by Lemma 5.10, c G ( g ) C N C M . But then c G ( g ) = c M ( g ) , contrary to ICG(g)I > I C M ( g ) l . This completes the proof of the theorem. H