arXiv:1609.00262v1 [math.RT] 1 Sep 2016

CERTAIN IRREDUCIBLE CHARACTERS OVER A NORMAL
SUBGROUP
GABRIEL NAVARRO AND NOELIA RIZO
Abstract. We extend the Howlett-Isaacs theorem on the solvability of groups of
central type taking into account actions by automorphisms. Then we study certain
induced characters whose constituents have all the same degree.

1. Introduction
The celebrated Howlett-Isaacs [HI] theorem on groups of central type solved a
conjecture proposed by Iwahori and Matsumoto in 1964: if Z is a normal subgroup
of a finite group G, λ ∈ Irr(Z) is a G-invariant complex irreducible character of Z, and
the induced character λG is a multiple of a single χ ∈ Irr(G), then G/Z is solvable.
(In this case, it is said that λ is fully ramified in G/Z and that G is of central
type if furthermore Z = Z(G).) This theorem, proved in 1982, is one of the first
applications of the Classification of Finite Simple Groups to Representation Theory.
Fully ramified characters are essential in both Ordinary and Modular Representation
Theory.
Our first main result in this note is the following generalization.
Theorem A. Suppose that Z ⊳ G, and let λ ∈ Irr(Z). Assume that if χ, ψ ∈ Irr(G)
are irreducible constituents of the induced character λG , then there exists a ∈ Aut(G)
stabilizing Z, such that χa = ψ. If T is the stabilizer of λ in G, then T /Z is solvable.
In a different language of projective representations, Theorem A was obtained
by R. J. Higgs under some solvability conditions [H]. His proof is mostly sketched,
among other reasons because he uses some of the arguments in [HI] or [LY] (where,
as a matter of fact, the Iwahori-Matsumoto conjecture was wrongly proven) or in
some other papers by the author. Here, we choose to present a complete proof of
Theorem A, in the language of character theory, and by doing so we shall adapt
several arguments in all these papers. We would like to acknowledge this now.
2010 Mathematics Subject Classification. Primary 20D; Secondary 20C15.
Key words and phrases. Howlett-Isaacs theorem.
The research of the first author is supported by the Prometeo/Generalitat Valenciana, and
Proyecto MTM2013-40464-P. The second author is supported by a Fellowship FPU of Ministerio de
Educación, Cultura y Deporte.
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GABRIEL NAVARRO AND NOELIA RIZO

Theorem A is only one case of a more general problem, which seems intractable by
now: if all the irreducible characters of G over some G-invariant λ ∈ Irr(Z) have the
same degree, then G/Z is solvable. (See Conjecture 11.1 of [N].)
In the second main result of this note, we study this latter situation under some
special hypothesis.
Theorem B. Suppose G is π-separable and let N = Oπ (G). Let θ ∈ Irr(N) be
G-invariant. Then all members of Irr(G|θ) have equal degrees if and only if G/N is
an abelian π ′ -group.
As the reader will see, the proof of Theorem B uses a lot of deep machinery. The
proof that we present here is an improvement by M. Isaacs of an earlier version which
we reproduce it here with his kind permission.
2. Transitive Actions
In general, we follow the notation in [Is]. If G is a finite group, then Irr(G) is
the set of the irreducible complex characters of G. If Z ⊳ G and λ ∈ Irr(Z), then
Irr(G|λ) is the set of the irreducible constituents of the induced character λG . By
Frobenius reciprocity, this is the set of characters χ ∈ Irr(G) such that the restriction
χZ contains λ.
Lemma 2.1. Suppose that Z ⊳ G, and let λ ∈ Irr(Z) be G-invariant. Assume that
all characters in Irr(G|λ) have the same degree dλ(1). Let P/Z ∈ Sylp (G/Z). Then
dp λ(1) is the minimum of {δ(1) | δ ∈ Irr(P |λ)} and |Irr(P |λ)| ≤ |Irr(G|λ)|p .
Proof. By Character Triple Isomorphisms (see Chapter 11 of [Is]), we may assume
that λ(1) = 1. Write Irr(G | λ) = {χj | 1 ≤ j ≤ s}, and observe that the multiplicity
of χj in λG is χj (1).
P Since by hypothesis, all of the degrees χj (1) are equal, we
can write λG = d j χj , where d = χj (1) for all j. Also, we have sd2 = |G : Z|.
Write P
Irr(P | λ) = {δi | 1 ≤ i ≤ t},
observe that because λ(1) = 1, we have
P and
2
P
λ = i di δi , where di = δi (1) and
di = |P : Z|. We can write
δiG

=

s
X

dij χj ,

j=1

and it follows that δiG (1) is a multiple of the common degree d of the χj . Then d
divides |G : P |di , and hence the p-part dp of d divides di for all i. We conclude that
dp divides the greatest common divisor e of the di .
We also have that
t
X
(χj )P =
dij δi
i=1

CERTAIN IRREDUCIBLE CHARACTERS OVER A NORMAL SUBGROUP

3

by Frobenius reciprocity, and thus e divides χj (1) = d. Since e is a p-power, we see
that e divides dp , and thus e = dp . Then we have that
|P : Z| =

t
X

di 2 ≥ e2 t = (dp )2 t .

i=1
2

Taking p-parts in sd = |G : Z|, we obtain that sp ≥ t.
The following is a character-theoretical version of Theorem 1.2 of [H].
Theorem 2.2. Suppose that Z ⊳ G, λ ∈ Irr(Z) is G-invariant, p is a prime and
P/Z ∈ Sylp (G/Z). Let A = Irr(G|λ) and B = Irr(P |λ). Suppose that A is a finite
group acting on A and B in such a way that
[(χa )P , δ a ] = [χP , δ]
for all χ ∈ A, δ ∈ B and a ∈ A. Assume further that χa (1) = χ(1) for χ ∈ A and
a ∈ A. Let B ∈ Sylp (A). If A acts transitively on A, then B acts transitively on B
and |A|p = |B|.
Proof. Write A = {χ1 , . . . , χs } and B = {δ1 , . . . , δt }. By hypothesis, we have that all
the characters in A have the same degree dλ(1). Hence
|G : Z| = d2 s .
By Lemma 2.1, we have that dp λ(1) is the minimum of the degrees in B and that
t ≤ sp . Write
t
X
(χi )P =
dij δj
j=1

so that
(δj )G =

s
X

dij χi

i=1

by Frobenius reciprocity. Let B be a Sylow p-subgroup of A. Let δj be such that
δj (1) = dp λ(1).
Now, let S = Bδj be the stabilizer of δj in B. We have that S acts on the set
Irr(G|δj ) of irreducible constituents of δjG . Let O1 , . . . , Or be the set of S-orbits. Let
ψi ∈ Oi . We may write
r
X
X
G
ξ) .
(δj ) =
bk (
k=1

Hence
|G : P |dp λ(1) =

r
X
k=1

ξ∈Ok

bk |Ok |ψk (1) = dλ(1)

r
X
k=1

bk |Ok |

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GABRIEL NAVARRO AND NOELIA RIZO

and therefore p does not divide

r
X

bk |Ok | .

k=1

Therefore there is k such that bk |Ok | is not divisible by p. In particular, we see that
there is an irreducible constituent ψk of (δj )G that is S-fixed. Hence S = Bδj ⊆
Bψk ⊆ B. Also Bψk ⊆ R ∈ Sylp (Aψk ) for some Sylow p-subgroup R of Aψk . Since A
acts transitively on Irr(G|λ), we have that s = |A : Aψk |. Thus
sp = |A|p /|Aψk |p = |A|p /|R| = |B|/|R| ≤ |B|/|Bψk | ≤ |B|/|Bδj | = |B : Bδj | ≤ t ≤ sp .
Thus t = sp , |B : Bδj | = t and everything follows.

3. Auxiliary results
Of course, if A acts by automorphisms on G, then A also acts on Irr(G). If
χ ∈ Irr(G) and a ∈ A, then χa ∈ Irr(G) is the unique character satisfying that
χa (g a ) = χ(g) for g ∈ G.
Hypotheses 3.1. Suppose that Z ⊆ N ⊳ G, where Z ⊳ G. Let λ ∈ Irr(Z). Suppose
that if τi ∈ Irr(N|λ) for i = 1, 2, then there exists g ∈ G such that τ1g = τ2 .
We say in this case that (G, N, λ) satisfies Hypothesis 3.1.
If N ⊳ G and τ ∈ Irr(N), then we denote by IG (τ ), or by Gτ , the stabilizer of τ in
G.
Recall that induction defines a bijection
Irr(IG (θ)|θ) → Irr(G|θ)
by the Clifford correspondence (Theorem (6.11) of [Is]).
Lemma 3.2. Suppose that (G, N, λ) satisfies Hypotheses 3.1. Let Z ⊆ K ⊆ N,
where K ⊳ G. Then the following hold.
(a) Let τi ∈ Irr(K|λ) for i = 1, 2. Then there exists g ∈ G such that τ1g = τ2 .
(b) Suppose that L ⊳ G is contained in K. Let ǫ ∈ Irr(L). Suppose that γi ∈
Irr(IK (ǫ)|ǫ) are such that (γi )K lie over λ for i = 1, 2. Then there is g ∈ IG (ǫ)
such that γ1g = γ2 .
(c) Let τ ∈ Irr(K|λ). Let γi ∈ Irr(IN (τ )|τ ) for i = 1, 2. Then there exists g ∈ IG (τ )
such that γ1g = γ2 .

CERTAIN IRREDUCIBLE CHARACTERS OVER A NORMAL SUBGROUP

5

Proof. (a) Let γi ∈ Irr(N) over τi . By hypothesis, we have that γ1x = γ2 for some
x ∈ G. We have that τ1x and τ2 are under γ2 , so by Clifford’s theorem there is n ∈ N
such that τ1xn = τ2 . Set g = xn.
(b) By part (a), there is g ∈ G such that ((γ1 )K )g = (γ2)K . Now, ǫg and ǫ are
under (γ2 )K , so by replacing g by gk for some k ∈ K, we may assume also that ǫg = ǫ.
Then g ∈ IG (ǫ). By the uniqueness in the Clifford correspondence, we deduce that
(γ1 )g = γ2 .
(c) By the Clifford correspondence, we have that (γi )N ∈ Irr(N) lie over λ. By
hypotheses, there is g ∈ G such that ((γ1 )N )g = (γ2 )N . Now, τ g and τ are Nconjugate by Clifford’s theorem, so by replacing g by gn, for some n ∈ N, we may
assume that τ g = τ . Notice now that g ∈ IG (τ ). Also, γ1g = γ2 , by the uniqueness in
the Clifford correspondence.
Theorem 3.3. Assume Hypotheses 3.1, with Z ⊆ Z(N). Let U ⊆ N, with U ⊳ G.
Suppose that q is a prime dividing |U|. Then q divides |Z ∩ U|.
Proof. Let K = UZ ⊳ G. If q does not divide |K : Z| = |U : U ∩ Z| then we are
done. Let 1 6= Q/Z ∈ Sylq (K/Z). By Lemma 3.2(a), we know that Gλ = IG (λ) acts
transitively on Irr(K|λ). By the Frattini argument, we have that Gλ = KNGλ (Q).
Notice then that A = NGλ (Q) acts transitively on Irr(K|λ). Also A acts on Irr(Q|λ)
and [(χa )Q , δ a ] = [χQ , δ] for a ∈ A, χ ∈ Irr(K|λ) and δ ∈ Irr(Q|λ). By Theorem 2.2,
we have that A acts transitively on Irr(Q|λ).
Suppose now that q does not divide |Z ∩ U|. Let ν = λZ∩U . Then ν has a
canonical extension ν̂ ∈ Irr(Q ∩ U) of q ′ -order. By Corollary (4.2) of [Is2], we know
that restriction defines a natural bijection
Irr(Q|λ) → Irr(Q ∩ U|ν) .
Let ρ ∈ Irr(Q|λ) be such that ρQ∩U = ν̂. In particular, ρ is linear. Also ρZ = λ. Let
a ∈ A. Then a fixes λ, and therefore ν. Now, a normalizes Q and U, so a normalizes
U ∩ Q. By uniqueness, we have that (ν̂)a = ν̂. Thus ρa = ρ by uniqueness. Since
A acts transitively on Irr(Q|λ), it follows that Irr(Q|λ) = {ρ}. Since ρZ = λ, by
Gallagher Corollary (6.17) of [Is], we know that |Irr(Q|λ)| = |Irr(Q/Z)|. We conclude
that Q = Z, and this is the final contradiction.
As in [HI], we shall only use the Classification of Finite Simple Groups in the
following result. If X is a finite group, recall that M(X) is the Schur multiplier of
X.
Theorem 3.4. Let X be a non-abelian simple group. Then there exists a prime p
such that p divides |X|, p does not divide |M(X)|, and there is no solvable subgroup
of X having p-power index.
Proof. This is Theorem (2.1) of [HI].

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GABRIEL NAVARRO AND NOELIA RIZO

4. The Glauberman Correspondence
The idea to use the Glauberman correspondence in the Iwahori-Matsumoto conjecture appears in [HI]. As we shall see in the proof of our main theorem, we need to
do the same here, in a more sophisticated way. For the definition and properties of
the Glauberman correspondence, we refer the reader to Chapter 13 of [Is].
We remark now the following. If Q is a p-group that acts by automorphisms on a
p -group L, then the Glauberman correspondence is a bijection ∗ : IrrQ (L) → Irr(C),
where IrrQ (L) is the set of Q-invariant irreducible characters of L and C = CL (Q) is
the fixed point subgroup. Furthermore, for χ ∈ IrrQ (L), we have that
′

χC = eχ∗ + p∆ ,
where p does not divide the integer e and ∆ is a character of C (or zero). In particular,
we easily check that the Glauberman correspondence ∗ commutes with the action of
Gal(Q̄/Q) and with the action of the group of automorphisms of the semidirect
product LQ that fix Q. In particular, we have that Q(χ) = Q(χ∗ ).
The next deep result is key in character theory. Its proof, in the case where Z = 1,
is due to E. C. Dade. (Other proofs are due to L. Puig.) The following useful
strengthening is due to A. Turull, to whom we thank for useful conversations on this
subject.
Theorem 4.1. Suppose that G is a finite group, LQ ⊳ G, where L ⊳ G, (|L|, |Q|) = 1,
and Q is a p-group for some prime p. Suppose that LQ ⊆ N ⊳ G, and Z ⊳ G, is
contained in Q and in Z(N). Let λ ∈ Irr(Z). Let H = NG (Q) and C = CL (Q).
Then for every τ ∈ IrrQ (L) there is a bijection
π(N, τ ) : Irr(N|τ ) → Irr(N ∩ H|τ ∗ ) ,
where τ ∗ ∈ Irr(C) is the Q-Glauberman correspondent of τ , such that:
(a) For γ ∈ Irr(N|τ ), h ∈ H we have that
π(N, τ h )(γ h ) = (π(N, τ )(γ))h .
(b) ρ ∈ Irr(N|τ ) lies over λ if and only if π(N, τ )(ρ) lies over λ.
Proof. It follows from the proofs of Theorem 7.12 of [T1] and Theorem 6.5 of [T2].
Specifically, we make ψ = θ in Theorem 7.12 of [T1], and G, H, θ in Theorem 7.12 of
[T1], correspond to G, L, τ ; while G′ , H ′ , θ′ correspond to H, C and τ ∗ , respectively.
Now, Theorem 7.12 (1) and (2) predicts a bijection
[
[
′
:
Irr(N|τ x ) →
Irr(N ∩ H|(τ ∗ )x ) ,
x∈H

x∈H

which commutes with the action of H (part (7) of Theorem 7.12). By parts (4), (1)
and (2) of the same theorem, writing R = L and S = N, we have that γ ∈ Irr(N|τ )

CERTAIN IRREDUCIBLE CHARACTERS OVER A NORMAL SUBGROUP

if and only if γ ′ ∈ Irr(N ∩ H|τ ∗ ). We call π(N, τ ) the restriction of the map
Irr(N|τ ). Part (b) follows from Theorem 10.1 of [T3].

7
′

to

The following is easy.
Lemma 4.2. Suppose that LQ ⊳ G, where L ⊳ G, (|L|, |Q|) = 1, and Q is a p-group
for some prime p. Suppose that τ ∈ IrrQ (L), and let τ ∗ ∈ Irr(C) be the Glauberman
correspondent, where C = CL (Q). Suppose that Z ⊳ G is contained in C. Let
λ ∈ Irr(Z) be L-invariant. Let H = NG (Q). Suppose that
λL = f (τ h1 + . . . + τ hs ) ,
for some hi ∈ H, and some integer f . Then
λC = f ∗ ((τ ∗ )h1 + . . . + (τ ∗ )hs ) ,
for some integer f ∗ .
Proof. We know by Theorem (13.29) of [Is] that if ν ∈ IrrQ (L), then ν ∗ lies above λ
if and only if ν lies above λ. Let ρ ∈ Irr(C|λ). Then ρ = ν ∗ for some ν ∈ Irr(L|λ).
Thus ν = τ h for some h ∈ H, by hypothesis. Then
ρ = ν ∗ = (τ h )∗ = (τ ∗ )h ,
because H commutes with Glauberman correspondence. Since λ is C-invariant, then
we easily conclude the proof of the lemma.

5. Theorem A
In this section, we prove Theorem A.
Theorem 5.1. Assume Hypothesis 3.1. Then IN (λ)/Z is solvable.
Proof. We argue by induction on |N : Z|. Let S/Z be the largest solvable normal
subgroup of N/Z. Let T = IG (λ) be the stabilizer of λ in G.
Step 1. We may assume that λ is G-invariant.
We claim that (IG (λ), IN (λ), λ) satisfies Hypothesis 3.1. Indeed, let τi ∈ Irr(IN (λ)|λ)
for i = 1, 2.
By Lemma 3.2(c) (with K = Z), there is g ∈ IG (λ) such that (τ1 )g = τ2 . Hence, by
working in IG (λ), we see that it is no loss to assume that λ is invariant in G. Hence,
we wish to prove that N/Z is solvable, that is, that S = N.
Step 2. If Z ≤ K < N, with K ⊳ G, then K/Z is solvable. Also N/S is isomorphic
to a direct product of a non-abelian simple group X.

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GABRIEL NAVARRO AND NOELIA RIZO

By Lemma 3.2 (a) and induction, we have that if Z ≤ K < N, with K ⊳ G, then
K/Z is solvable. Then N/S is a chief factor of G/Z, and it is isomorphic to a direct
product of a non-abelian simple group X.
Step 3. We may assume that Z is cyclic and that λ is faithful.
This follows by using character triple isomorphisms.
Step 4. If Z < K ⊆ N is a normal subgroup of G, and τ ∈ Irr(K|λ), then IN (τ )/K
is solvable. Also S > Z.
The first part is a direct consequence of Lemma 3.2(c) and induction. If S = Z,
then by Step 2, we have that N/Z is a minimal normal non-abelian subgroup of G/Z.
Then N/Z is a direct product of non-abelian simple groups isomorphic to X, and
Z = Z(N). Also, N ′ Z = N. By Theorem 3.4, there is a prime p dividing |X| such
that p does not divide |M(X)|. By Corollary 7.2 of [HI], we have that p does not
divide |N ′ ∩ Z|. Since p divides |N ′ |, this contradicts Theorem 3.3 with U = N ′ .
Step 5. F = F(N) = S.
Otherwise, let R/F be a solvable chief factor of G inside N. Thus R/F is a q-group
for some prime q. Let L be the Sylow q-complement of F . Let Zq′ = L ∩ Z. Let Q be
a Sylow q-subgroup of R, so that R = LQ. Let Zq = Q ∩ Z, so that Z = Zq′ × Zq . We
have that G = LH, where H = NG (Q), by the Frattini argument. Let C = CL (Q).
Write λ = λq′ × λq , where λq′ = λZq′ , and λq = λZq . By coprime action and
counting, we see that Q fixes some τq′ ∈ Irr(L|λq′ ). Let τ = τq′ × λq ∈ Irr(LZ). By
hypothesis and Lemma 3.2(a), we can write
λLZ = f (τ h1 + . . . + τ hs ) ,
where hi ∈ H, and λhi = λ, because λ is G-invariant. Hence
λLq′ = f (τqh′1 + . . . + τqh′s ) .
By Lemma 4.2, we have that
∗
∗ h1
+ . . . + (τq∗′ )hs ) .
λC
q ′ = f ((τq ′ )

By Theorem 4.1, we know that there is a bijection
π(N, τq′ ) : Irr(N|τq′ ) → Irr(NN (Q)|τq∗′ ))
that commutes with H-action.
We claim that (NG (Q), NN (Q), λ) satisfies Hypothesis 3.1. If this is the case, since
NN (Q) < N, we will have that |NN (Q) : Z| < |N : Z|, and by induction, we will
conclude that NN (Q)/Z is solvable. This implies that N/Z is solvable, and the proof
of the theorem would be complete. Suppose now that ψi ∈ Irr(NN (Q)|λ) for i = 1, 2.
We are going to show that there exists x ∈ H such that ψ1x = ψ2 . Since ψi lies over
λq′ , then we have that ψ1 lies over some (τq∗′ )hj , and ψ2 lies over some (τq∗′ )hk for some
−1
hj , hk ∈ H. Conjugating by h−1
j and by hk , we may assume that ψ1 and ψ2 lie over
τq∗′ .

CERTAIN IRREDUCIBLE CHARACTERS OVER A NORMAL SUBGROUP

9

Now, we know that there exists µi ∈ Irr(N|τq′ ) such that π(N, τq′ )(µi ) = ψi . In fact,
since ψi lies over λq , we have that µi ∈ Irr(N|λq ) by Theorem 4.1(b) (with the role of λ
in that theorem being played now here by λq ), and therefore µi ∈ Irr(N|τ ) ⊆ Irr(N|λ).
By hypothesis, there is h ∈ H such that µh1 = µ2 . Now, τqh′ and τq′ are below µ2 , so
1
there is h1 ∈ N ∩ H such that τqhh
= τq′ . Replacing h by hh1 , we may assume that
′
h
(τq′ ) = τq′ . Now
ψ1h = π(N, τq′ )(µ1 )h = π(N, τqh′ )(µh1 ) = π(N, τq′ )(µ2 ) = ψ2 ,
as desired. By induction, N ∩ H is solvable, so N is solvable. This proves Step 5.
Step 6. If p divides |F : Z|, then N has a solvable subgroup of p-power index.
Therefore, so do the simple groups factors in the direct product of N/S.
Suppose that Q/Z is a non-trivial normal p-subgroup of G/Z, where Q is contained
in N. Then the irreducible constituents of λQ all have the same degree by Lemma
3.2(a), for instance. So we can write
λQ = f (τ1 + · · · + τk ) ,
where τi ∈ Irr(Q|λ) are all the different constituents. Write τ = τ1 . Notice that f =
τ (1). Thus we deduce that k is a power of p. Now, since G acts on Ω = {τ1 , . . . , τk }
transitively by conjugation by Lemma 3.2(a), we have that |G : IG (τ )| = k is a power
of p. Hence, |N : IN (τ )| is a power of p. If Q > Z, then we know by induction that
IN (τ )/Q is solvable. In this case, we deduce that that N has a solvable subgroup
with p-power index. The same happens for factors of N.
Step 7. Final contradiction.
We know by Step 2 that N/S is isomorphic to a direct product of a non-abelian
simple group X. By Theorem (2.1), there exists a prime q dividing |X|, such that
q does not divide the order of the Schur multiplier of X, and such that no solvable
subgroup of X has q-power index. By Step 6, we have that q does not divide |F : Z|.
Let W be the normal q-complement of F . Hence F = W Z. Also F/W = Z(N/W ).
By Corollary 7.2 of [HI], we have that q does not divide |(N/W )′ ∩ F/W |. But F/W
is a q-group, so (N/W )′ ∩F/W = W/W . In particular, N ′ ∩F ⊆ W . Thus q does not
divide |N ′ ∩ F |. Thus q does not divide |N ′ ∩ Z|. Since N/F is perfect, we have that
N ′ F = N, so that q divides |N ′ |. But this contradicts Theorem 3.3 with U = N ′ .
Next is Theorem A of the introduction.
Corollary 5.2. Suppose that Z ⊳ G, and let λ ∈ Irr(Z). Assume that if χ, ψ ∈
Irr(G|λ), then there exists a ∈ Aut(G) stabilizing Z such that χa = ψ. If T is the
stabilizer of λ in G, then T /Z is solvable.

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GABRIEL NAVARRO AND NOELIA RIZO

Proof. Let A = Aut(G)Z be the group of automorphisms of G that stabilize Z. Let
Γ = GA be the semidirect product. We have that Z ⊳ Γ. By hypothesis, (Γ, G, λ)
satisfies Hypothesis 3.1. By Theorem 5.1, we have that T /Z is solvable.
6. Theorem B
We begin by giving another proof of a result of U. Riese ([5]) that we shall need
later.
Lemma 6.1. Let H ⊆ G and α ∈ Irr(H). Suppose that αG = χ ∈ Irr(G) and that
every irreducible constituent of χH has degree equal to α(1). Then χ vanishes on
G − H.
Proof. By hypothesis, χH is the sum of χ(1)/α(1) = |G : H| irreducible characters,
and thus [χH , χH ] ≥ |G : H|. Then |H|[χH , χH ] ≥ |G|[χ, χ], and so χ vanishes on
G − H, as claimed.
Next is the proof of Riese’s theorem (by M. Isaacs).
Theorem 6.2. Let A ⊆ G, where A is abelian, and assume that λG is irreducible,
where λ ∈ Irr(A). Then A ⊳ ⊳ G.
Proof. We prove the theorem by induction on |G|. Write χ = λG ∈ Irr(G), and let
V be the subgroup of G generated by the elements g ∈ G with χ(g) 6= 0. Since A is
abelian, each irreducible constituent of χA has degree 1 = λ(1), and thus by Lemma
6.1, we have V ⊆ A. Also, writing Z = Z(G), we have Z ⊆ V .
If A ⊆ H < G, then since λH is irreducible, the inductive hypothesis yields A⊳ ⊳ H.
Assuming that A is not subnormal in G, then Wielandt’s zipper lemma (Theorem 2.9
of [Is4]) guarantees that there is a unique maximal subgroup M of G with A ⊆ M.
Also, if the normal closure AG < G then A ⊳ ⊳ AG ⊳ G, and we are done. We can
thus supppose that AG = G, and so Ag 6⊆ M for some element g ∈ G. By the
uniqueness of M, therefore, we have < A, Ag >= G. But V ⊳ G and V ⊆ A, and
thus V ⊆ A ∩ Ag ⊆ Z, and we have V = Z = A ∩ Ag . Thus χ vanishes off Z, and
so χ is fully ramified with respect to Z. In particular |G : A|2 = χ(1)2 = |G : Z|,
and we have |G : A| = |A : Z|. Thus |G : A| = |Ag : Ag ∩ A|, and it follows that
AAg = G. This implies that A = Ag , and thus A = G. This is a contradiction since
A was assumed to be not subnormal.
Corollary 6.3. Let θ ∈ Irr(N), where N ⊳ G and θ is G-invariant. Let N ⊆ A ⊆ G,
where A/N is abelian, and suppose that θ has an extension φ ∈ Irr(A) such that φG
is irreducible. Then A ⊳ ⊳ G.

CERTAIN IRREDUCIBLE CHARACTERS OVER A NORMAL SUBGROUP

11

Proof. By character triple isomorphisms (see Chapter 11 of [Is]), we can assume that
θ is linear and faithful. Then φ is linear and A′ ⊆ N ∩ ker(φ) = ker(θ) = 1. Then A
is abelian, and since φG is irreducible, Theorem 6.2 yields the result.
Now, we prove an extension of Theorem B. We should mention that the Classification of Finite Simple Groups is implicitely used in the proof. Specifically, in the
Howlett-Isaacs ([HI]) theorem on central type groups.
Theorem 6.4. Let N ⊳ G. Suppose that θ ∈ Irr(N) is G-invariant and that o(θ)θ(1)
is a π-number. Assume that G/N is π-separable and that Oπ (G/N) = 1. Then all
members of Irr(G|θ) have equal degrees if and only if G/N is an abelian π ′ -group.
Proof. If G/N is an abelian π ′ -group, then θ extends to G by Corollary 8.16 of [Is],
and we are done by Gallagher’s Corollary 6.17 of [Is]. To prove the converse, we
argue by induction on |G/N| and assume that |G/N| > 1. We argue first that the
common degree d of the characters in Irr(G|θ) is a π-number. To see this, let q ∈ π ′
and let Q/N ∈ Sylq (G/N). Then θ extends to Q, and the induction to G of such
an extension has degree θ(1)|G : Q|, which is a q ′ -number. Since this degree is a
multiple of d, it follows that d is a q ′ -number, and since q ∈ π ′ was arbitrary, we see
that d is a π-number.
Let U/N = Oπ′ (G/N) and note that U > N. All degrees of characters in Irr(U|θ)
divide d, and so are π-numbers. But since U/N is a π ′ -number, it follows that all
degrees of characters in Irr(U|θ) equal θ(1), and so all of these characters extend θ.
It follows that U/N is abelian by Gallagher Corollary 6.17 of [Is]. If U = G, we
are done, and so we suppose that U < G and we let V /U = Oπ (G/U). Note that
V > U. By Corollary 8.16 of [Is], there exists a unique extension θ̂ ∈ Irr(U) of θ with
determinantal π-order. By uniqueness, θ̂ is G-invariant. Now, let φ ∈ Irr(V |θ̂). Since
V /U is a π-group, φU is a multiple of θ̂ and o(θ̂) is a π-number, we easily have that
o(φ) is a π-number. Write T = Gφ for the stabilizer of φ in G. Then all members
of Irr(T |φ) induce irreducibly to G, yielding characters of degree d, and thus these
characters all have degree d/|G : T |. We claim that T satisfies the hypotheses of the
theorem with respect to the character φ and the normal subgroup V . To see this, we
need to check that Oπ (T /V ) is trivial.
Let W/V = Oπ′ (G/V ). We argue that W stabilizes φ. This is because the G/V orbit of φ has size dividing d, and so is a π-number, and W/V is a normal π ′ subgroup of G/V . Thus W ⊆ T and Oπ (T /V ) centralizes the normal π ′ -subgroup
W/V = Oπ′ (G/V ). But Oπ (G/V ) is trivial, and Lemma 1.2.3 applies to show that
Oπ (T /V ) = 1, as wanted.
By the inductive hypothesis, we conclude that T /V is a π ′ -group. Also, by the
Clifford correspondence, |G : T | divides d, which we know is a π-number. Thus T /V
is a full Hall π ′ -subgroup of G/V . Also, φ extends to T , and so φ(1) = d/|G : T | =

12

GABRIEL NAVARRO AND NOELIA RIZO

d/|G/V |π is constant for φ ∈ Irr(V |θ). It follows that the hypotheses are satisfied in
the group V with respect to θ. If V < G, the inductive hypothesis yields that V /N
is a π ′ -group, and this is a contradiction.
It follows that V = G and G/U is a π-group. Also, G/U acts faithfully on U/N
because Oπ (G/N) is trivial. Now let λ ∈ Irr(U/N), so that λ is linear. Let S = Gλ ,
and note that λ extends to S since S/U is a π-group. Write a = |G : S|.
Note that S is the stabilizer of λθ̂ in G, and thus all characters in Irr(S|λθ̂) have
degree d/a. If r is the number of such characters, this yields r(d/a)2 = |S : U|θ(1)2 .
Also, since λ extends to S, by Theorem 6.16 of [Is] there is a degree-preserving
bijection between Irr(S|λθ̂) and Irr(S|θ̂), and hence the latter set contains exactly r
characters, and each has degree d/a. Each of these must therefore induce irreducibly
to G, and it follows that each member of Irr(G|θ̂) is induced from a member of
Irr(S|θ̂).
Note that the number of different members of Irr(S|θ̂) that can have the same
induction to G is at most |G : S| = a.
Now let t = |Irr(G|θ̂)| so that td2 = |G : U|θ(1). If we divide this equation by
our previous one, we get ta2 /r = |G : S| = a, and so t = r/a. It follows that each
of the t members of Irr(G|θ̂) is induced from exactly a characters in Irr(S|θ̂). In
other words, if χ ∈ Irr(G|θ̂), then χS has exactly a distinct irreducible constituents,
each with degree d/a, and so by Lemma 6.1, it follows that χ vanishes on G − S. In
other words, the only elements of G on which χ can have a nonzero value lie in the
stabilizer of λ for every linear character λ of U/N. But G/U acts faithfully on this
set of linear chararacters, and thus χ vanishes on G − U. In other words, θ̂ is fully
ramified in G. It follows that d = θ(1)|G : U|1/2 .
Also, aθ(1) divides d, and so a must divide |G : U|1/2 . Write s = |S : U|, so that
as = |G : U|. Then a2 divides as, and thus a divides s. In particular, we have a ≤ s,
so |G : S| ≤ |S : U|. Thus
|G : U| = |G : S||S : U| ≤ |S : U|2 .
Now, by the Howlett-Isaacs theorem we have that G/U is solvable. This group
acts faithfully on the group of linear characters of U/N, and so by the main result in
[D], there exist character stabilizers T and R such that T ∩ R = U. By the result of
the previous paragraph, each of T /U and R/U has order at least |G : U|1/2 . Now
|G : U| = |G : T ||T : U| ≥ |R : U||T : U| ≥ |G : U| .
Then T R = G, and that each of |T : U| and |R : U| has order |G : U|1/2 . Therefore all
characters in Irr(T |θ̂) are extensions of θ̂ and induce irreducibly to G. In particular,
T /U is abelian, and similarly R/U is abelian.
By Corollary 6.3, it follows that R⊳ ⊳ G, and since R/U is abelian, R/U ⊆ F(G/U).
Similarly, T /U ⊆ F(G/U) and thus G/U is nilpotent. But then, since G/U acts

CERTAIN IRREDUCIBLE CHARACTERS OVER A NORMAL SUBGROUP

13

faithfully on the group of linear characters of U/N, it follows that if G/U is nontrivial,
then some character λ ∈ Irr(U/N) has a stabilizer S in G such that
|S : U| < |G : U|1/2
by Theorem B of [Is3]. But then |G : U| = |G : S||S : U| ≤ |S : U|2 < |G : U|. This
contradiction completes the proof.
Acknowledgment. We thank the referee for a very careful reading of this manuscript.
References
[D]
[H]
[HI]
[Is2]
[Is3]
[Is]
[Is4]
[LY]
[N]
[NST]
[R]
[T1]
[T2]
[T3]

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Departament d’Àlgebra, Universitat de València, 46100 Burjassot, València, Spain
E-mail address: gabriel.navarro@uv.es
Departament d’Àlgebra, Universitat de València, 46100 Burjassot, València, Spain
E-mail address: noelia.rizo@uv.es

