Certain irreducible characters over a normal subgroup

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 an automorphism a∈Aut⁢(G) stabilizing Z, such that χa=ψ. If T is the stabilizer of λ in G, then T/Z is solvable.

Theorem B.

Suppose that 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.

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 d⁢λ⁢(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 [6, Chapter 11]), 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). Since by hypothesis, all of the degrees χj⁢(1) are equal, we can write

where d=χj⁢(1) for all j. Also, we have sd2=|G:Z|. Write

and observe that because λ⁢(1)=1, we have

where di=δi⁢(1) and ∑di2=|P:Z|. We can write

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

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

Taking p-parts in sd2=|G:Z|, we obtain that sp≥t. ∎

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), B=Irr(P|λ). Suppose that A is a finite group acting on A and B in such a way that

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 d⁢λ⁢(1). Hence

By Lemma 2.1, we have that dp⁢λ⁢(1) is the minimum of the degrees in ℬ and that t≤sp. Write

so that

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

Hence

and therefore p does not divide

Therefore there is some 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

Thus t=sp, |B:Bδj|=t and everything follows. ∎

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.

Lemma 3.2.

Suppose that (G,N,λ) satisfies Hypotheses 3.1. Let Z⊆K⊆N, where K◁G. Then the following hold.

1. Let τi∈Irr(K|λ) for i=1,2. Then there exists g∈G such that τ1g=τ2.

2. Suppose that L◁G is contained in K. Let ϵ∈Irr⁢(L), and suppose that γi∈Irr(IK(ϵ)|ϵ) are such that (γi)K lie over λ for i=1,2. Then there is some g∈IG⁢(ϵ)such that γ1g=γ2.

3. Let τ∈Irr(K|λ). Let γi∈Irr(IN(τ)|τ) for i=1,2. Then there exists some g∈IG⁢(τ) such that γ1g=γ2.

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 some n∈N such that τ1x⁢n=τ2. Set g=x⁢n.

(b) By part (a), there is some g∈G such that ((γ1)K)g=(γ2)K. Now, ϵg and ϵ are under (γ2)K, so by replacing g by gk for some k∈K, we may assume also that ϵg=ϵ. Then g∈IG⁢(ϵ). 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 some g∈G such that ((γ1)N)g=(γ2)N. Now, τg and τ are N-conjugate by Clifford’s theorem, so by replacing g by gn, for some n∈N, we may assume that τg=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=U⁢Z◁G. If q does not divide |K:Z|=|U:U∩Z|,we are done. Let 1≠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=K⁢𝐍Gλ⁢(Q). Notice then that A=𝐍Gλ⁢(Q) acts transitively on Irr(K|λ). Also A acts on Irr(Q|λ) and [(χa)Q,δa]=[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 ν^∈Irr⁢(Q∩U) of q′-order. By [4, Corollary (4.2)], we know that restriction defines a natural bijection

Let ρ∈Irr(Q|λ) be such that ρQ∩U=ν^. 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 (ν^)a=ν^. 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’s Corollary (6.17) of [6], we know that

We conclude that Q=Z, and this is the final contradiction. ∎

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 [3, Theorem (2.1)]. ∎

Theorem 4.1.

Suppose that G is a finite group, L⁢Q◁G, where L◁G, (|L|,|Q|) is equal to 1, and Q is a p-group for some prime p. Suppose that L⁢Q⊆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

where τ*∈Irr⁢(C) is the Q-Glauberman correspondent of τ, such that:

1. For γ∈Irr(N|τ), h∈H we have that

   π⁢(N,τh)⁢(γh)=(π⁢(N,τ)⁢(γ))h.

2. 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 [12, Theorem 7.12] and [11, Theorem 6.5]. Specifically, we make ψ=θ in [12, Theorem 7.12], and G, H, θ in [12, Theorem 7.12], correspond to G, L, τ; while G′, H′, θ′ correspond to H, C and τ*, respectively. Now, parts (1) and (2) of [12, Theorem 7.12] predict a bijection

which commutes with the action of H (part (7) of [12, 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) if and only if γ′∈Irr(N∩H|τ*). We call π⁢(N,τ) the restriction of the map ′ to Irr(N|τ). Part (b) follows from [13, Theorem 10.1]. ∎

Lemma 4.2.

Suppose that L⁢Q◁G, where L◁G, (|L|,|Q|) is equal to 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

for some hi∈H and some integer f. Then

for some integer f*.

Proof.

We know by [6, Theorem (13.29)] 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=τh for some h∈H, by hypothesis. Then

because H commutes with Glauberman correspondence. Since λ is C-invariant, we easily conclude the proof of the lemma. ∎

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.

We claim that the triple (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 an element 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.

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.

This follows by using character triple isomorphisms.

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 [3, Corollary 7.2], we have that p does not divide |N′∩Z|. Since p divides |N′|, this contradicts Theorem 3.3 with U=N′.

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, and let Zq′=L∩Z. Let Q be a Sylow q-subgroup of R, so that R=L⁢Q, and let Zq=Q∩Z, so that Z=Zq′×Zq. We have that G=L⁢H, 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⁢(L⁢Z). By hypothesis and Lemma 3.2 (a), we can write

where hi∈H, and λhi=λ, because λ is G-invariant. Hence

By Lemma 4.2, we have that

By Theorem 4.1, we know that there is a bijection

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 |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′, we have that ψ1 lies over some (τq′*)hj, and ψ2 lies over some (τq′*)hk for some hj,hk∈H. Conjugating by hj-1 and by hk-1, we may assume that ψ1 and ψ2 lie over τq′*.

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 some h∈H such that μ1h=μ2. Now, τq′h and τq′ are below μ2, so there is some element h1∈N∩H such that τq′h⁢h1=τq′. Replacing h by h⁢h1, we may assume that (τq′)h=τq′. Now

as desired. By induction, N∩H is solvable, so N is solvable. This proves Step 5.

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 N has a solvable subgroup with p-power index. The same happens for factors of N.

We know by Step 2 that N/S is isomorphic to a direct product of a non-abelian simple group X. By [3, 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 [3, Corollary 7.2], 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′. ∎

Corollary 5.2.

Suppose that Z◁G, and let λ∈Irr⁢(Z). Assume that if we have 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.

Proof.

Let A=Aut⁢(G)Z be the group of automorphisms of G that stabilize Z. Let Γ=G⁢A be the semidirect product. We have Z◁Γ. By hypothesis, (Γ,G,λ) satisfies Hypothesis 3.1. By Theorem 5.1, we have that T/Z is solvable. ∎

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. ∎

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)≠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 [7, Theorem 2.9] 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 suppose that AG=G, and so Ag⊈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| equals |Ag:Ag∩A|, and it follows that A⁢Ag=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 N◁G and let θ∈Irr⁢(N) be 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.

Proof.

By character triple isomorphisms (see [6, Chapter 11]), 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. ∎

Theorem 6.4.

Let N◁G. Suppose that θ∈Irr⁢(N) is G-invariant and 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 [6, Corollary 8.16], and we are done by Gallagher’s Corollary 6.17 of [6]. 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’s Corollary 6.17 of [6]. 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 [6, Corollary 8.16], there exists a unique extension θ^∈Irr⁢(U) of θ with determinantal π-order. By uniqueness, θ^ is G-invariant. Now, let ϕ∈Irr(V|θ^). Since V/U is a π-group, ϕU is a multiple of θ^ and o⁢(θ^) 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 Hall–Higman’s 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|=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 λ⁢θ^ in G, and thus all characters in Irr(S|λθ^) have degree d/a. If r is the number of such characters, this yields

Also, since λ extends to S, by [6, Theorem 6.16] there is a degree-preserving bijection between Irr(S|λθ^) and Irr(S|θ^), 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|θ^) is induced from a member of Irr(S|θ^).

Note that the number of different members of Irr(S|θ^) that can have the same induction to G is at most |G:S|=a.

Now let t=|Irr(G|θ^)| 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|θ^) is induced from exactly a characters in Irr(S|θ^). In other words, if χ∈Irr(G|θ^), 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 characters, and thus χ vanishes on G-U. In other words, θ^ is fully ramified in G. It follows that d=θ(1)|G:U|1/2.

Also, a⁢θ⁢(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