Nonsolvable groups with few primitive character degrees

Theorem A1.

Let G be a nonsolvable group and assume that all nonlinear primitive characters of G are of the same degree, say d. Then there exists a chief factor N/Sol⁢(G) of G satisfying the following conditions:

1. G/N is solvable,

2. N/Sol⁢(G) is a direct product of t copies of L2⁢(q), where q=pf≥4,

3. d=4t if q=4,5, d=6t if q=7, and d=qt if q≥8.

Theorem A2 ([17]).

Assume that all nonmonomial characters of G are of the same degree. Then G is solvable.

Theorem B1.

Let G be a nonsolvable group and assume that all primitive characters of G are of square-free degree. Then G/Sol⁢(G) is isomorphic to one of the following groups: A7, S7, L2⁢(p) or PGL2⁢(p), where p≥7 is a prime, p-1 is square-free.

Theorem B2.

Let G be a nonsolvable group and assume that all nonmonomial characters of G are of square-free degree. Then G/Sol⁢(G) is isomorphic to A7, L2⁢(p) or PGL2⁢(p), where p≥7 is a prime, p-1 is square-free.

Theorem C1.

Let G be a nonsolvable group and assume that all primitive characters of G are of prime power degree. Let K1,…,Ks be all the nonabelian composition factors of G. Then there is a prime p such that for all i, Ki≅L2⁢(pfi), where pfi=4 or ≥8.

Theorem C2.

Let G be a nonsolvable group and assume that all nonmonomial characters of G are of prime power degree. Then G/Sol⁢(G) is one of the following types:

1. L2⁢(9)≤G/Sol⁢(G)≤Aut⁢(L2⁢(9)),

2. L2⁢(p), where p=1+2k is a prime,

3. S5,

4. L2⁢(2f), where 2f-1 is a prime.

Theorem D1.

Assume that all primitive characters of G are of odd degree. Then every nonabelian composition factor of G is of type L2⁢(q), where q is odd and q=pf≥9.

Theorem D2.

Assume that all nonmonomial characters of G are of odd degree. Then G is solvable.

Hypothesis 1.1.

Let Ω={S1,…,St}, where S1,…,St are isomorphic to a nonabelian simple group S. Assume that N is a minimal normal subgroup of G and that

Lemma 2.1 ([2, Lemma 2]).

Let N=S1×⋯×St be a direct product of isomorphic nonabelian simple groups, and let M be a maximal subgroup of N with S1≰M. Then one of the following assertions holds:

1. M=D×S2×⋯×St, where D is maximal in S1.

2. If one of the subgroups S2,…,St, say S2, is not contained in M, then

   M=D×S3×⋯×St,

   where D∩S1=D∩S2=1 and S1≅S2≅D<S1×S2.

Lemma 2.2.

Assume that Si are nonabelian simple groups and N=S1×⋯×Sm is normal in G. Then N can be written as a direct product of some minimal G-invariant subgroups, and each of which is a direct product of some members of {Si:1≤i≤m}.

Proof.

Let L be a minimal G-invariant subgroup of N. Since every composition factor of L is nonabelian by the Jordan–Hölder theorem, we have Z⁢(L)=1. Without loss of generality, we may assume that S1∩L=⋯=Sk∩L=1, k≤m, and that Sj∩L>1 for all j>k. Since Sj and L are normal in N, the simplicity of Sj implies that Sj≤L for all j>k. Observe that S1×⋯×Sk≤CN⁢(L) and CN⁢(L)∩L=Z⁢(L)=1. We have N=L×CN⁢(L), where L=Sk+1×⋯×Sm and CN⁢(L)=S1×⋯×Sk. As CN⁢(L) is also normal in G, by induction we get the required result. ∎

Lemma 2.3.

Let Ω={S1,…,St}, where Si are isomorphic to a nonabelian simple group S. Let

and H≤N. Assume that for every i=1,…,t, Si is equal to the ith component Hi of H under the decomposition (D2). Then H and N can be written as

where every Bj is a direct product of rj members of Ω with ∑1≤j≤krj=t, S≅Aj≤Bj, and lΩ⁢(aj)=rj for every nonidentity aj∈Aj.

Proof.

Working by induction on t, we may assume that H≤E for some maximal subgroup E of N. Since Hi=Si for all i, by Lemma 2.1 we may assume that

where Ej=Sj for all j≤t-2 and one has that S≅Et-1<St-1×St with Et-1∩St-1=Et-1∩St=1. In particular, E is a direct product of t-1 copies of S. Set Δ={E1,…,Et-1}. For every h∈H, write

Then ej=hj for all j≤t-2, and et-1=ht-1⁢ht. Let Hj′ be the jth component of H under the decomposition (D3). Clearly we have Hj′=Ej=Sj if j≤t-2. Observe that Ht-1′≤Et-1 and that

We have Ht-1′=Et-1. By induction, we may write

where Cj is a direct product of rj members of Δ, S≅Aj≤Cj, and lΔ⁢(aj)=rj for all 1≠aj∈Aj. Without loss of generality, we may assume that

where U is a direct product of (rk-1) members of Δ-{Et-1}. Now we obtain N=B1×⋯×Bk, where Bk=U×St-1×St, and Bj=Cj for all j≤k-1. Clearly Bk is a direct product of (rk+1) members of Ω, and

for all 1≠ak∈Ak. The result follows.∎

Lemma 2.4.

Let G and N satisfy Hypothesis 1.1 and assume that H is a maximal subgroup of G such that G=H⁢N. Set K=H∩N. Then one of the following holds:

1. K=K1×⋯×Kt, where Ki<Si for all i, and H acts transitively on the set {K1,…,Kt}.

2. K=A1×⋯×Ak is minimal normal in H, where A1≅⋯≅Ak≅S, k∣t, k<t, and every nonidentity element of Ai has length t/k under the decomposition ((D)).

Proof.

Clearly K⁢⊴⁢H and K<N. Let Ki be the ith component of K under the decomposition (D). Clearly G=H⁢N and G acts transitively on Ω, thus H acts transitively on Ω. We claim that K1 and Ki are H-conjugate. Since S1a=Si for some a∈H, we have K1a⊆S1a=Si. Note that if k1a∈Si and k1 is the first component of an element k∈K, then k1a is the ith component of ka∈K. This implies that K1a⊆Ki. Similarly, we have Ki⊆K1b for some b∈H. Hence |K1|=|Ki| and the claim follows. Now we see that H acts transitively on {K1,…,Kt}. In particular, K1×⋯×Kt is H-invariant.

Assume that Ki<Si for all i. Let Δ=K1×⋯×Kt. It is easy to see that Δ<N, K≤Δ, and H⁢Δ≤G because Δ is H-invariant. Since

we get by the maximality of H that Δ≤H. Therefore Δ=K, and (1) follows.

Assume that Ki=Si for some i. Then Ki=Si for all i because H acts transitively on {K1,…,Kt}. By Lemma 2.3, we may write K=A1×⋯×Ak and N=B1×⋯×Bk, where every Bj is a direct product of rj members of Ω, S≅Aj≤Bj, and lΩ⁢(aj)=rj for all nonidentity aj∈Aj. Let Ωi be the set of direct simple factors of Bi. Then |Ωi|=ri, where ∑1≤i≤kri=t. Note that if 1≠ai∈Ai, then

where lΩi⁢(ai) is the length of ai under the decomposition of Bi. Assume that r:=r1=min⁡{ri:i=1,…,k}. Let 1≠a1∈A1. Since H acts transitively on Ω and K=A1×⋯×Ak, for each i≤k, there exist yi∈H and 1≠a1⁢i∈Ai such that a1yi=∏1≤j≤ka1⁢j, where a1⁢j∈Aj. Clearly

It follows that

and that

Since A1yi and Ai are simple and normal in K, we have A1yi=Ai. This implies that H acts transitively on {A1,…,Ak}, and so K is minimal normal in H. Since t=∑1≤i≤kri=r⁢k and K<N, we get that k<t, k∣t, and (2) follows.∎

Lemma 3.1 ([4, Lemma 5]).

Let G and N satisfy Hypothesis 1.1, and assume that σ∈Irr⁢(S1) extends to NG⁢(S1). Then σG^ extends to G.

Proof.

Note that in the original version of [4, Lemma 5], σ is assumed to be extendible to Aut⁢(S1). Repeating its proof, we get the result. ∎

Definition 3.2.

An irreducible character χ of G is called to be super-primitive, provided that for every H<G and for every ψ∈Irr⁢(H), ψG does not equal a multiple of χ.

Proposition 3.3.

Let G and N satisfy Hypothesis 1.1, and let θ∈Irr⁢(N) be such that the first component σ1 of θ is primitive. Assume that χ∈Irr(G|θ) is induced from a character of a proper subgroup H of G. Then H⁢N<G whenever the following conditions hold:

1. If τ1 is the first component of some G-conjugate of θ with τ1⁢(1)=σ1⁢(1), then τ1=σ1.

2. Either σ1 is super-primitive or [χN,θ]=1.

Proof.

Write χ=ξG, where ξ∈Irr⁢(H) and H<G. Assume that the conditions hold but the result is false. Then G=H⁢N and K:=H∩N<N. To find a contradiction, we may assume that H is maximal in G. Since χ∈Irr(G|θ) and

by the reciprocity, all irreducible components of (ξK)N are conjugate to θ in G, and in particular are of the same degree θ⁢(1). By Proposition 2.4, we need only to investigate the following cases.

Suppose that the first case in Proposition 2.4 holds. Then K=K1×⋯×Kt, where Ki<Si for all i. Since χ∈Irr(G|θ), there exists an irreducible constituent λ of ξK such that [θ,λN]>0. Write λ=∏1≤i≤tλi, where λi∈Irr⁢(Ki). Note that (see, for example, [2, Lemma 1])

Since all irreducible constituents of λN have the same degree, we get by (3.2) that all members of Irr(S1|λ1) have the same degree, say d. Clearly σ1∈Irr(S1|λ1) as θ∈Irr(N|λ), it follows that d=σ1⁢(1). Observe that by (3.1) and (3.2) every member of Irr(S1|λ1) is the first component of some G-conjugate of θ. Now condition (1) implies that (λ1)S1=m⁢σ1 for some m∈ℤ+. So σ1 is not super-primitive, and it follows by condition (2) that [χN,θ]=1. Now

we get that m=1 and that σ1=(λ1)S1 is imprimitive, a contradiction.

Suppose that the second case in Proposition 2.4 holds. We may assume that K=A1×A* and N=B1×B*, where S≅A1<B1, B1=S1×⋯×Sl with l≥2, and 1≤A*≤B*. Let U=S2×⋯×Sl. Since every nonidentity element of A1 has length l under the decomposition B1=S1×⋯×Sl, we have

Let λ be an irreducible constituent of ξK. Write λ=λ1×λ*, where λ1∈Irr⁢(A1), λ*∈Irr⁢(A*). Since all irreducible constituents of λN are of the same degree θ⁢(1) and λN=(λ1)B1×(λ*)B*, all irreducible constituents of (λ1)B1 have the same degree, say d. Observe that by (3.3) λ1∈Irr⁢(A1) extends to an irreducible character of B1 with kernel U, it follows that d=λ1⁢(1). Since A1≅S and U is a direct product of l-1 copies of S, there exists τ∈Irr⁢(U) such that τ⁢(1)=d. As B1=S1×U, we may write

where a⁢(ψ⁢τ),a⁢(ψ⁢ξ)∈ℕ.

Suppose that a⁢(ψ⁢τ)>0. Then (ψ⁢τ)⁢(1)=d, so ψ=1S1. Since λ1 is irreducible and λ1⁢(1)=((1S1⁢τ)A1)⁢(1), we have [λ1,(1S1⁢τ)A1]∈{0,1}. By (3.4) and the reciprocity, we get that

Since ((λ1)B1)U=((λ1)A1∩U)U=d⁢∑ξ∈Irr⁢(U)ξ⁢(1)⁢ξ, we have

By (3.6), (3.4) and (3.5), we get that

Hence d=1, so λ1=1A1. As A1 is a proper subgroup of B1, (1A1)B1 has an irreducible constituent of degree exceeding d=1, a contradiction. ∎

Corollary 3.4.

Assume that G and N satisfy Hypothesis 1.1 and ψ∈Pri⁢(G/N). Let σ∈Priext⁢(G)⁢(S1), and let χ be an extension of θ:=σG^∈Irr⁢(N) to G (see Lemma 3.1). Then any one of the following conditions guarantees the primitivity of χ⁢ψ:

1. ψ is linear.

2. σ is super-primitive.

Proof.

By [12, Corollary 6.17], χ⁢ψ is irreducible. Assume that either ψ is linear or σ is super-primitive, but that χ⁢ψ is imprimitive. Then χ⁢ψ=ηG, where η∈Irr⁢(H), H is maximal in G. Observe that χψ∈Irr(G|θ), σ is the first component of θx=θ for all x∈G, and that either σ is super-primitive or, ψ is linear and so [(χ⁢ψ)N,θ]=1. Hence Proposition 3.3 is applied, and we get that

Write χH=λ. Clearly λ is an extension of θ to H. Note that if ν1,ν2∈Irr⁢(H/N), then λ⁢ν1=λ⁢ν2 if and only if ν1=ν2 (see [12, Corollary 6.17]). Let

where bi∈ℤ+, μ1,…,μm∈Irr⁢(H/N) are distinct. Then

where η=λ⁢μ1 and λ⁢μi are distinct irreducible characters of H. Put

where cj∈ℤ+, ψ1=ψ,ψ2,…,ψs∈Irr⁢(G/N) are distinct. Clearly

Since θ extends to χ∈Irr⁢(G), applying [12, Corollary 6.17] again we get that χ⁢ψj are irreducible and distinct. Observe that [χ⁢ψi,ηG]=[(χ⁢ψi)H,η]≥[λ⁢μ1,η]=1 and ηG∈Irr⁢(G). It follows that s=1 and that μ1G=c1⁢ψ1=ψ is imprimitive, a contradiction. ∎

Corollary 3.5.

Let G and N satisfy Hypothesis 1.1. Let χ∈Irr(G|θ) and let θ=∏1≤i≤tσi∈Irr⁢(N), where σi∈Irr⁢(Si), 1S1≠σ1∈Priext⁢(G)⁢(S1) is super-primitive, and σj⁢(1)≠σ1⁢(1) holds for every j≥2. Assume that χ=λG, where λ∈Irr⁢(H) and H≤G. Then S1≤Hg for some g∈G, σ1(1)∣λ(1), and in particular χ is nonmonomial.

Proof.

Suppose that H⁢N=G. Let x∈G and let τ1 be the first component of θx. If x∈NG⁢(S1), then σ1x=σ1, so τ1=σ1; and if x∉NG⁢(S1), then τ1=σix for some i≥2, so τ1⁢(1)≠σ1⁢(1). Thus conditions (1) and (2) in Proposition 3.3 hold, so H=G, and the result follows.

Suppose that H⁢N<G. Let ψ=λH⁢N. Since ψG=χ and N⁢⊴⁢G, some G-conjugate of θ is a constituent of ψN. Replacing H, λ, HN and ψ by their suitable G-conjugates if necessary, we may assume that [ψN,θ]>0. By Lemma 2.2, we may assume that N=N1×N2, where N1=S1×⋯×Sk is minimal normal in HN and N2=Sk+1×⋯×St is normal in HN. Let θ1=∏1≤i≤kσi. Then we have ψ∈Irr(HN|θ1). Replacing G,N,χ,θ,H,λ by H⁢N,N1,ψ,θ1,H,λ, respectively, we get the required result by induction. ∎

Lemma 4.1.

Let S be one of the sporadic simple groups or F42⁢(2)′. Then two members θ1,θ2 of Priext⁢(S) and their degrees are listed in Table 1, where χi are labeled as [6].

Proof.

We make some remarks on the choice of θ1 and θ2. Since Aut⁢(S)/S⪅ℤ2, every Aut⁢(S)-invariant θ∈Irr⁢(S) extends to Aut⁢(S).

Assume that S does not belong to {Fi22,Th,Fi23,J4,Fi24′,B,M}. By [6], we may take Aut⁢(S)-invariant θ1,θ2∈Irr⁢(S) such that |S:M| divides neither θ1⁢(1) nor θ2⁢(1) for every maximal subgroup M of S. Then θ1 and θ2 are primitive, so θ1,θ2∈Priext⁢(S).

Assume that S∈{Fi22,Th,Fi23,J4,Fi24′,B,M}. For these cases, there are only some maximal subgroups of S listed in [6]. The characters θ1,θ2 in the table are of “small” degree. Actually, θ1⁢(1)<θ2⁢(1), θ2⁢(1) does not exceed the third minimal degree (excluding 1) of S unless S=Fi22. Note that if θ1 is of the first minimal degree, then θ1 is primitive (see [17, Lemma 2.8]). We can prove the primitivity of θ1 and θ2 directly. For example, we show that θ2 is primitive when S=B. The first three minimal character degrees of B are