Even character degrees and normal Sylow 2-subgroups

Theorem A.

Let G be a finite group with S2⁢(G)<32. Then G has a normal Sylow 2-subgroup.

Theorem B.

Let G be a finite group with S2,+⁢(G)<32. Then G has a normal Sylow 2-subgroup.

Corollary C.

Let G be a finite group with S2,R⁢(G)<32. Then G has a normal Sylow 2-subgroup.

Theorem D.

Let G be a finite group. Suppose that S2⁢(G)<175. Then G is solvable.

Lemma 2.1.

Let N be a normal subgroup of a group G such that N⊆G′.

1. If S2⁢(G)≤2, then S2⁢(G/N)≤S2⁢(G).

2. If S2,+⁢(G)≤2, then S2,+⁢(G/N)≤S2,+⁢(G).

Proof.

(i) Write A:=S2(G). Then

and thus

From the hypothesis, we know that n1⁢(G)=n1⁢(G/N). Also, A≤2, and for k≥2, we have nk⁢(G)≥nk⁢(G/N). It follows that

implying that

Hence S2⁢(G/N)≤A=S2⁢(G).

(ii) Similar to (i). ∎

Lemma 2.2.

Let N be a normal subgroup of a group G such that G/N has odd order. If S2,+⁢(G)≤2, then S2,+⁢(N)≤S2,+⁢(G).

Proof.

Clearly, every strongly real linear character of G restricts to a strongly real linear character of N. Also, by [7, Lemma 2.1], every strongly real linear character of N lies under a unique strongly real linear character of G. Hence we have n1,+⁢(N)=n1,+⁢(G). We now have

Write a:=S2,+(G). Then, from

we get that

It follows that S2,+⁢(N)≤a=S2,+⁢(G), as desired. ∎

Theorem 3.1.

Let G be a finite group. Suppose that S2⁢(G)<175. Then G is solvable.

Proof.

Let G be a counterexample of minimal order. Let M be a normal subgroup of G such that M is minimal nonsolvable, and let N be a minimal normal subgroup of G contained in M. Then N≤M=M′≤G′. If further [M,Rad⁡(M)]≠1, then we suppose N⊆[M,Rad⁡(M)], where Rad⁡(M) is the solvable radical of M.

Suppose that N is abelian. Then G/N is nonsolvable. By the choice of G, we have S2⁢(G)<175≤S2⁢(G/N). Hence we have

Observe that nk⁢(G/N)≤nk⁢(G) for k≥1 and n1⁢(G/N)=n1⁢(G), so we have n2⁢(G/N)<n2⁢(G). Therefore, we can find a χ∈Irr⁡(G)-Irr⁡(G/N) such that χ⁢(1)=2. Let L=ker⁡(χ). Then N⊈L, and thus M⁢L/L is a nontrivial nonsolvable subgroup of G/L.

We claim that χ is a primitive character of G¯:=G/L. Suppose χ=ψG¯ for some character ψ of a proper subgroup K¯ of G¯. Then 2=χ(1)=|G¯:K¯|ψ(1), and thus G¯/K¯≅ℤ2 and ψ⁢(1)=1. Now all the irreducible constituents of χ|K¯ are linear. Note that χ is a faithful character of G¯, and therefore [K¯,K¯]=1, implying that K¯ is abelian. We now have that G¯ is solvable, which is impossible. This proves the claim.

Now G/L is a nonsolvable primitive linear group of degree 2. Let

By Lemma [4, Theorem 14.23], G/C≅𝖠5. By the proof of [5, Theorem 2.2], we can show that G=M⁢C is a central product. Here we give this argument again for the reader’s convenience. Note that G/C is simple and M⁢C⊴G. If G>M⁢C, then M⁢C=C, and thus M⁢L/L≤C/L=𝐙⁢(G/L), a contradiction. Hence G=M⁢C and M⊈C. Now M∩C<M and M∩C⊴G. By the choice of M, we have M∩C⊆Rad⁡(M). Since M/(M∩C)≅G/C is simple, we have M∩C=Rad⁡(M), and thus Rad⁡(M)⊆C. Since C/L=𝐙⁢(G/L), we have [M,Rad⁡(M)]⊆L. Hence N⊈[M,Rad⁡(M)]. By the choice of N, we have [M,Rad⁡(M)]=1, and thus Rad⁡(M)≤𝐙⁢(M). Clearly, 𝐙⁢(M)≤Rad⁡(M), and N is abelian, so N≤Rad⁡(M)=M∩C=𝐙⁢(M). Observe that

and hence [M,C,M]=[C,M,M]=1. Using the three subgroups lemma, we then have [M,C]=[M,M,C]=1. It follows that G=M⁢C is a central product. Here G/C≅M/(M∩C)≅A5, N=Z⁢(M)=M∩C≅ℤ2 and M≅SL⁢(2,5).

Let λ∈Irr⁡(N)-1N; we have a bijection

Notice that χ∈Irr⁡(G|λ), so there are α∈Irr⁡(M|λ) (=Irr⁡(M)-Irr⁡(M/N)) and β∈Irr⁡(C|λ) such that χ⁢(1)=α⁢(1)⁢β⁢(1). Then α⁢(1)=2, β⁢(1)=1, and thus λ extends to C. By Gallagher’s theorem [4, Corollary 6.17], we get a bijection Irr⁡(C/N)→Irr⁡(C|λ). Then n1(G)=n1(C/N)=n1(C|λ). By ($*$), we also have

where

Similarly,

Now

Hence S2⁢(G)≥175, a contradiction.

Suppose that N is not abelian and N≇A5. By [2, Theorem 2.2], there exists ϕ∈Irr⁡(N) of even degree such that ϕ⁢(1)≥8 and ϕ is extendible to IG⁢(ϕ). By [2, Proposition 2.3], we have

where d=ϕ(1)|G:IG(ϕ)|≥8|G:IG(ϕ)|. Now

leading to a contradiction again.

Finally, assume that N≅A5. By [1], the irreducible character of N of degree 4 is extendible to G since G≤Aut⁡(N). Then, by [2, Proposition 2.3] again, we have n1⁢(G)≤n4⁢(G) and n2⁢(G)≤n8⁢(G). Now

and thus S2⁢(G)≥175. This final contradiction completes the proof. ∎

Lemma 4.1.

Let G=M⋉N, where N≤G′ is an abelian group. Assume that no nontrivial irreducible character of N is invariant under M. If S2⁢(G)<32, then there is no orbit of even size in the action of M on the set of irreducible characters of N.

Proof.

Let {α0=1N,α1,…,αt} be a set of representatives of the action M on Irr⁡(N). Let Ii=IG⁢(αi), 1≤i≤t. Since no nontrivial irreducible character of N is invariant under M, we have Ii<G for i≥1. Suppose that there is some orbit of even size in the action of M on the set of irreducible characters of N. Then there is some j such that 2∣|G:Ij|. For 0≤i≤t, we set

Let βi be an extension of αi to Ii. By Gallagher’s theorem and the Clifford correspondence, we have bijections

Note that (λβi)G(1)=|G:Ii|λ(1) is even if and only if |G:Ii| is even or |G:Ii| is odd and λ⁢(1) is even. Hence we have

On the other hand,

Therefore, we derive that

Since 2∣|G:Ij|, we deduce that

Observe that

so n1(G/N)≤|G:Ij|nj,1. Also, T⁢(Ij/N)≥n1⁢(Ij/N). Hence we get that

It then follows that

which is a contradiction. ∎

Theorem 4.2.

Let G be a finite group. Suppose that S2⁢(G)<32. Then G has a normal Sylow 2-subgroup.

Proof.

Let G be a counterexample of minimal order, and let N be a minimal normal subgroup of G with N≤G′. Since S2⁢(G)<32<175, by Theorem 3.1, G is solvable, and then N is an elementary abelian group. By Lemma 2.1, we have S2⁢(G/N)≤S2⁢(G). Due to the choice of G, we know that G/N has a normal Sylow 2-subgroup T/N. If N is a 2-group, then T is a normal Sylow p-subgroup of G, a contradiction. Hence we may assume that N is an elementary abelian 2′-group. Using the Schur–Zassenhaus theorem [3, Chapter 1, Theorem 18.1], we deduce that T=P⋉N, where P is a Sylow 2-subgroup of T. By the Frattini argument, we then have G=T⁢NG⁢(P)=N⁢NG⁢(P).

If N≤NG⁢(P), then G=NG⁢(P), and thus P is a normal Sylow 2-subgroup of G, a contradiction. Hence we have N≰NG⁢(P). Now

Then, by the minimality of N, NG⁢(P)∩N=1, and thus G=NG⁢(P)⋉N. If N≤Z⁢(G), then T=P×N, and thus P⊴G, a contradiction. Hence N≰Z⁢(G), and then [N,G]=N. This means that no nontrivial irreducible character of N is invariant under NG⁢(P). By Lemma 4.1, there is no orbit of even size in the action of P on N, that is, P acts trivially on N, and hence T=P×N. It follows that P⊴G, completing the proof. ∎

Theorem 5.1.

Let G be a finite group. Suppose that S2,+⁢(G)<32. Then G has a normal Sylow 2-subgroup.

Proof.

Let G be a counterexample of minimal order, and let N be a minimal normal subgroup of G with N≤G′. According to the Cauchy–Schwarz inequality, we have

so, by [2, Theorem 5.1], we know that G is solvable. By Lemma 2.1 and the choice of G, G/N has a normal Sylow 2-subgroup T/N, T=P⋉N, where P∈Syl2⁢(G) and N is an elementary abelian 2′-group. If T is a proper subgroup of G, then, by Lemma 2.2, we have T=P×N, and thus P⊴G, a contradiction.

Hence G=T=P⋉N. Since N has odd order, 1N is the only real irreducible character of N. Then N⊆ker⁡(χ) for any strongly real linear character χ of G. So

By [2, Lemma 5.4], we have n1,+⁢(G)≤|N|-1. Let Ω1,…,Ωs be the orbits of the action of P on Irr⁡(N)-1N. Write |Ωi|=di,i=1,…,s. Then we have n1,+⁢(G)≤d1+⋯+ds. Notice that P acts irreducibly on Irr⁡(N). Let a∈Z⁢(P) be of order 2, and let M={ϕ∈Irr⁡(N)∣ϕa=ϕ-1}. Then M is a P-invariant subgroup of Irr⁡(N). Also, since o⁢(a)=2, M is nontrivial. Then M=Irr⁡(N), and thus, for ϕ∈Irr⁡(N)-1N, ϕ and ϕ¯ are P-conjugate. By [10, Lemma 2.5], there exists a strongly real character χ∈Irr⁡(G) lying above ϕ.

Now let ϕi∈Ωi for 1≤i≤s. There is a strongly real character χi∈Irr⁡(G) with degree χi(1)=|G:IG(ψi)|ψi(1)=|G:IG(ψi)|, where ψi is an extension of ϕi to IG⁢(ϕi). Then χi(1)≥|G:IG(ϕi)|=di, and thus

Now

For any χ∈Irr2,+⁡(G) with χ⁢(1)≥2, we have χ⁢(1)2/χ⁢(1)≥2≥32, so, by [9, Lemma 2.1], we get that S2,+⁢(G)≥32, and this contradiction completes the proof. ∎