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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 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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