Irreducible characters which are zero on only
Institute of Advance Studies in Basic Sciences, Zanjan, Iran
Suppose that G is a …nite solvable group which has an irreducible
which vanishes on exactly one conjugacy class.
show that G has a homomorphic image which is a nontrivial 2-transitivepermutation group. The latter groups have been classi…ed by Huppert.
We can also say more about the structure of G depending on whetheris primitive or not.
Mathematics Subject Classi…cation 2000: 20C15 20D10 20B20
be an irreducible character of a …nite group G. A well-known theorem
on at least one conjugacy class of G. Groups having an irreducible characterthat vanishes on exactly one class were studied by Zhmud’in  (see also ).
Chillag [2, Lemma 2.4] has proved that if the restriction of
vanishes on exactly one class of G, then G is a
Frobenius group with a complement of order 2 and an abelian odd-order kernel.
Our purpose in this paper is to show that, if an irreducible character
…nite solvable group G vanishes on exactly one conjugacy class, then G has ahomomorphic image which is a nontrivial 2-transitive permutation group. Thelatter groups have been classi…ed by Huppert: they have degree pd where p isprime, and are subgroups of the extended a¢ ne group A L(1; pd) except for sixexceptional degrees (see Remark 8 below).
We shall initially assume that our character is faithful, and make the followingassumptions:
(*) G is a …nite group with a faithful irreducible character
only one class which we denote by C. Furthermore, G has a chief factorK=L which is an elementary abelian p-group of order pd such that therestriction
must be nonlinear, the latter condition clearly holds whenever G is
solvable, but for the present we shall not assume solvability.
Proposition 1 Suppose (*) holds. Then C = K n L, K = hCi and L is equalto L0 := fu 2 G j uC = Cg. In particular, C consists of p-elements (since L doesnot contain a Sylow p-subgroup of K). Moreover, either:
pd is the sum of pd distinct G-conjugate irreducible
Proof. Since K is irreducible, the theorem of Burnside quoted above shows
Now since K=L is an abelian chief factor, and
from [9, (6.18)] that either (i) d is even,
pd is the sum of pd distinct G-conjugate irreducible characters
We shall consider these two cases separately.
In case (i) we note that, since C \ L = ;, the irreducible character
is assumed to be faithful, L is contained in the centre Z(G) of G. On the otherhand, for each z 2 Z(G), (z) is a scalar of the form 1: Thus for each x 2 C wehave
K for all z 2 Z(G). This shows that Z(G) is a normal
is a nonlinear irreducible character of K, we conclude that Z(G) = L. Finally,since K=Z(G) is abelian, [9, (2.30)] shows that K n L = C.
K is an irreducible constituent of ( 1)K and so comparison
normal subgroup L, and so K n L = C in this case as well.
Finally since jC [ f1gj > 1 jKj, therefore K = hCi. Finally, it is easily seen
that L0 is a normal subgroup of G, and that L0
C = K n L is a union of cosets of L, we see that L
C * L0 since C is not a subgroup. Therefore L0 C G and L
Corollary 2 Under the hypothesis (*) every normal subgroup N of G eithercontains K (when N is irreducible) or is contained in L (when N is reducible).
In particular, K=L is the unique chief factor such that
is reducible and K=L is the socle of G=L. Since K has a nonlinear irreduciblecharacter, K is not abelian and so L 6= 1.
Remark 3 Both cases (i) and (ii) in Proposition 1 can actually occur. Thegroup SL(2; 3) has three primitive characters of degree 2 which satisfy (*) (case(i) with jKj = 8 and jLj = 2 for each character), and S4 has an imprimitivecharacter of degree 3 which satis…es (*) (case (ii) with jKj = 12 and jLj = 4).
Proposition 4 Suppose that the hypothesis (*) and case (i) of Proposition 1hold. Then L = Z(G) has order p, K is an extraspecial p-group and
Proof. Let z 2 L. Then for any x 2 C we have zx 2 C and so zx = y 1xy
for some y 2 G. Since K=L is an elementary abelian p-group, zpxp = (zx)p =y 1xpy = xp, and so zp = 1.
represented faithfully as a group of scalar matrices by a representation a¤ording
, it follows that L is cyclic and hence jLj = p.
(K) = L = Z(K) and so K is an extraspecial p-group.
is primitive. Indeed, otherwise there is a maximal
induced character shows that G is 0 on each conjugacy class disjoint fromH. As is well-known every proper subgroup of a …nite group is disjoint fromsome conjugacy class, and so we conclude that C is the unique class such thatC \ H = ;. By Proposition 1 this implies that H \ K
(1) = pd=2, we obtain a contradiction. Thus
Proposition 5 Suppose that the hypothesis (*) and case (ii) of Proposition 1hold (so
is imprimitive). Then there exists a subgroup M of index pd in G
= G for some 2 Irr(M), G = MK and M \K = L = coreG(M).
Proof. As noted in the proof of Proposition 1
irreducible constituents i. Because K is irreducible, these constituents are K-conjugates (as well as G-conjugates): Let M := IG( 1) be the inertial subgroup…xing the constituent
Then jG : Mj = pd and G = MK because K acts
jK : M \ Kj, we conclude that M \ K = L: On the other hand, since G is0 on any class which does not intersect M , the hypothesis on
that ux does not lie in any y 1M y, and hence ux 2 C. Thus with the notationof Proposition 1, coreG(M )
L0 = L. Since L is a normal subgroup contained
in M , the reverse inequality is also true and so coreG(M ) = L.
The proof of the next result requires a theorem of Isaacs [10, Theorem 2]
Let H be a …nite group with centre Z and K be a normal subgroup of H
with Z = Z(K). Suppose that H centralizes K=Z and jHom(K=Z; Z)jjK=Zj. Then H=Z = K=Z
Proposition 6 Under the hypothesis (*) the centralizer CG(K=L) equals K.
is primitive, then Proposition 4 shows that the hypotheses of
Isaacs’theorem are satis…ed for H := CG(K=L) (the condition jHom(K=Z; Z)jjK=Zj is trivial since the irreducibility of H implies that Z is cyclic). Also,since
K is irreducible, CG(K ) = Z (G) = L, and so Isaacs’theorem shows that
is imprimitive, then using the notation of Proposition 5 we can show
that M \ H = L where H := CG(K=L): Indeed, it is clear from Proposition5 that L
M \ H. To prove the reverse inequality suppose that u 2 M \ H.
Then for each x 2 K we have xu = yux for some y 2 L. Choose i such that
in x 1M x. Since this is true for all x 2 K, it follows from Proposition 5 thatu 2 coreG(M) = L. Thus M \ H = L as claimed. Finally H = H \ MK =(H \ M)K = LK = K as required.
Corollary 7 Under the hypothesis (*) G acts transitively by conjugation on thenontrivial elements of the vector space K=L and the kernel of this action is K.
Thus G=K is isomorphic to a subgroup of GL(d; p) which is transitive on thenonzero elements of the underlying vector space.
Remark 8 Huppert [8, Chapter XII Theorem 7.3] has classi…ed all solvable sub-groups S of GL(d; p) which are transitive on the nonzero vectors of the underly-ing vector space. Apart from six exceptional cases (where pd = 32; 52; 72; 112; 232
or 34), the underlying vector space can be identi…ed with the Galois …eld GF (pd)in such a way that S is a subgroup of the group
t is an automorphism of the …eld. The group
1)d. A classi…cation for nonsolvable groups has been carried out by Her-
ing , . It is considerably more complicated to state and prove, but amongother things it shows that such groups have only a single nonsolvable compositionfactor (a summary is given in [8, page 386]).
Since the latter half of hypothesis (*) is certainly satis…ed in a solvable group,
we can specialize to solvable groups and drop the condition that
Theorem 9 Let G be a …nite solvable group which has an irreducible character
which takes the value 0 on only one conjugacy class C. Let K := hCi : Then:
(b) There is a unique normal subgroup L of G such that K=L is a chief
factor of G and K n L = C (we set jK : Lj = pd).
(c) G=K acts transitively on the set (K=L)# of nontrivial elements of the
vector space K=L and so is one of the groups classi…ed by Huppert.
is imprimitive, then G=L is a 2-transitive Frobenius group of degree
Remark 10 We also note that (c) and Huppert’s classi…cation show that theinteger k in (a) is bounded. Indeed, since
except in the six exceptional cases. Computations using GAP  show that inthe remaining cases k
Proof. (a) Let k be the largest integer such that K
we know that the restriction G(k+1) is reducible, and so G(k+1)
(b), (c) and (d) follow from Proposition 1, Corollary 7 and Proposition 4.
(e) Let M be the subgroup de…ned in Proposition 5. Since
jMj : Since G = MK and G=K acts transitively on (K=L)# we
1 by Proposition 1, so equality must hold throughout.
1. Hence M=L acts regularly on (K=L)# and so G=L =
(M=L)(K=L) is a 2-transitive Frobenius group.
Remark 11 Not all groups having an irreducible character which takes 0 ona single conjugacy class satisfy the second half of hypothesis (*).
ple, the Atlas  shows that A5 has three characters with this property and itscentral cover 2 A5 also has three.
the required property and each of the groups L2(2k) (k = 3; 4; :::) appears tohave one such character (of degree 2k). It would be interesting to know if thesewere the only simple groups with this property, or whether a group with such acharacter can have more than one nonabelian composition factor (see Remark8).
Another question which can be asked is what can be said about the kernel
of such a character; evidently this kernel is contained in the normal subgroupL0 := fu 2 G j uC = Cg.
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(2), Vol. 181, Mathematical Monographs, Amer. Math. Soc., Rhode Island,1999.
 D. Chillag, On zeros of characters of …nite groups, Proc. Amer. Math. Soc.
 J.H. Conway et al, Atlas of Finite Simple Groups, Clarendon Press, Oxford,
 The GAP Group, GAP–Groups, Algorithms and Programming, Version
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 Ch. Hering, Transitive linear groups and linear groups which contain irre-
ducible subgroups of prime order, Geometriae Dedicata 2 (1974), 425–460.
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