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Publicly Available Published by De Gruyter May 19, 2019

On the 𝑝-closedness and 𝑝-nilpotency of finite groups

Shuqin Dong EMAIL logo , Hongfei Pan and Long Miao
From the journal Journal of Group Theory

Abstract

Let acd(G) and acs(G) denote the average irreducible character degree and the average conjugacy class size, respectively, of a finite group G. The object of this paper is to prove that if acd(G)<2(p+1)/(p+3), then G=Op(G)×Op(G), and that if acs(G)<4p/(p+3), then G=Op(G)×Op(G) with Op(G) abelian, where p is a prime.

1 Introduction

For a finite group G, let Irr(G) denote the set of all complex irreducible characters of G. As appeared in [9], we write

acd(G)=χIrr(G)χ(1)|Irr(G)|,acs(G)=|G||Irr(G)|

for the average irreducible character degree and the average conjugacy class size, respectively, of G. There are several results on the close connection between the characteristics of finite groups and the two invariants acd(G) and acs(G). For example, G is solvable when acd(G)<16/5 (see [8, Theorem A]) or acs(G)<12 (see [1, Theorem 11]). Furthermore, for a given prime p5, PSL(2,p) is exactly the non-p-solvable group with the smallest acd(G) and the smallest acs(G) (see [9, Theorems A and B]).

Recall that a group G is said to be p-closed if the Sylow p-subgroup of G is normal in G, and G is said to be p-nilpotent if its p-complement is normal in G. In this paper, we aim to characterize the p-closed and p-nilpotent groups in terms of acd(G) and acs(G), and we have the following results.

Theorem A.

Let G be a group and p a prime. Then we have the following:

  1. If acd(G)<2(p+1)/(p+3), then G=Op(G)×Op(G).

  2. If acs(G)<4p/(p+3), then G=Op(G)×Op(G) with Op(G) abelian, or, equivalently, G has a central Sylow p-subgroup.

Example B.

If G is the dihedral group of order 2p, where p is an odd prime, then acd(G)=2(p+1)/(p+3) and acs(G)=4p/(p+3), so any one of the bounds in Theorem A is best possible.

Example C.

Let G be an extraspecial p-group of order p2n+1. Then

acd(G)=(p2m+pm+1-pm)/(p2m+p-1),

so acd(G)1 as m. This example tells us that if

acd(G)2(p+1)/(p+3),

then Op(G) need not be abelian.

We are now able to prove the following corollary.

Corollary D ([2]).

Let G be a group. If acs(G)<8/5, then G is abelian.

Proof.

Note that the function f(p)=4p/(p+3) is monotonically increasing. This implies that f(p)f(2)=8/5 for all p2. So acs(G)<f(p) for all prime divisors of the order of G. By Theorem A (2), G is abelian. ∎

2 Lemmas

We first collect some results for the proof of Theorem A.

Lemma 2.1 ([9, Proposition 5.4 (2)]).

Let G be a group and p a prime divisor of |G|. Assume that acd(G)2p/(p+1). Then G has a normal Sylow p-subgroup.

Lemma 2.2.

Let G be a group and acd(G)<2. If NG, then

acd(G/N)acd(G).

Proof.

We may assume that G1 and N is a minimal normal subgroup of G.

Assume that NG. Then, for any χIrr(G)Irr(G/N), we have χ(1)2. If acd(G/N)2, then, by [6, Lemma 2.1], we derive that acd(G)2, a contradiction. Hence acd(G/N)<2, and again by [6, Lemma 2.1], acd(G/N)acd(G).

Assume that NG. Then NG=1 and NZ(G). For any λIrr(N), λ is extendible to G. By [3, Corollary 6.17],

χIrr(λG)χ(1)=βIrr(G/N)λ(1)β(1)=λ(1)βIrr(G/N)β(1).

Note that Irr(G)=λIrr(N)Irr(λG) and Irr(λ1G)Irr(λ2G)= when λ1λ2, where λ1,λ2Irr(N). Hence

acd(G)=χIrr(G)χ(1)|Irr(G)|=λIrr(N)χIrr(λG)χ(1)|Irr(G)|=|N|βIrr(G/N)β(1)|N||Irr(G/N)|=acd(G/N).

Lemma 2.3 ([4, Theorem 3.2]).

Let A be a normal subgroup of G, where A is abelian and G splits over A, and let e be the number of orbits in the action of G on the set of nonprincipal linear characters of A. Then the size t of one of these orbits satisfies acd(G)g(t,e), where g(t,e)=t(e+1)/(t+e).

Lemma 2.4 ([9, Lemma 5.2]).

Let M be a minimal normal subgroup of G with CG(M)=M. Let λIrr(M) be nonprincipal. If K=IG(λ), then KG=M.

Lemma 2.5.

Let G be a group. If HG and NG, then acs(H)acs(G) and acs(G/N)acs(N)acs(G). In particular, acs(G/N)acs(G).

Proof.

See [1, Lemma 2]. ∎

Lemma 2.6.

Let G be a group of p-prime power order. If acs(G)<4p/(p+3), then G is abelian.

Proof.

Let G be a counterexample with minimal order pa for some integer number a, then G is a nonabelian group in which every proper subgroup is abelian. By [7], we know that G has a maximal abelian normal subgroup of order pa-1 and |G|=p. By [3, Theorem 6.15], the degree of nonlinear irreducible character is p, and the number is (pa-pa-1)/p2. Then

acs(G)=|G||Irr(G)|=papa-1+(pa-pa-1)/p2=p3p2+p-14pp+3,

which is a contradiction. ∎

3 The proofs

We first prove Theorem A (1), which we restate below for convenience.

Theorem 3.1.

Let G be a group and p a prime. If acd(G)<2(p+1)/(p+3), then G=Op(G)×Op(G).

Proof.

By Lemma 2.1, G is p-closed because 2(p+1)/(p+3)<2p/(p+1). In order to prove p-nilpotency of G, we assume that G is a counterexample of minimal order. Since acd(G)<2, by Lemma 2.2, the hypothesis is inherited by all quotient groups. This implies by induction that all proper quotient groups of G are p-nilpotent. Hence G possesses a unique minimal normal subgroup M, and the Frattini subgroup Φ(G) of G is trivial.

Let F(G) be the Fitting subgroup of G. Then F(G)=M is the normal Sylow p-subgroup of G since G is p-closed. Further, G splits over M, and CG(M)=M. Write |M|=pr. Let e be the number of G-orbits of nonprincipal linear characters of M. By Lemma 2.3, we have acd(G)g(t,e), where t is the size of some G-orbit on the nonprincipal linear characters of M. Let nonprincipal λIrr(M) be such that t=|G:IG(λ)|. If t=1, then G=IG(λ)=M by Lemma 2.4, a contradiction. If t=2, then IG(λ)G and thus IG(λ)=M. Notice that if a group of order 2 acts irreducibly on an abelian group A, then A is of prime order. Thus |M|=p, G2.p. Then acd(G)=2(p+1)/(p+3), a contradiction. Hence t3.

Case 1. Assume that e3. Then acd(G)g(3,3)=2, a contradiction.

Case 2. Assume that e=2. Since g(t,e)acd(G)<2, we have t3 and thus t=3. Let nonprincipal μ,νIrr(M) be such that μ,ν lie in distinct G-orbits, and assume that T=IG(μ) satisfies |G:T|=3. Clearly, G/TG=G/MS3 and 3||G:M|.

Suppose that G/MS3. Then G is a Frobenius group of order 3|M| with kernel M. This implies that |G:IG(ν)|=3 or 6. In the former case, we have |M|=|Irr(M)|=1+3+3=7, and then G/MAut(M) is abelian, a contradiction. In the later case, we have |M|=|Irr(M)|=1+3+6=10, a contradiction.

Suppose that G/M3. Then G is a Frobenius group with kernel M. Hence |M|=|Irr(M)|=1+3+3=7=p, so

acd(G)=3+3+33+1+1=95<85=2(p+1)p+3,

a contradiction.

Case 3. Assume that e=1. Then t=|M|-1=pr-1. Since

2(p+1)p+3>acd(G)g(pr-1,1)=2(pr-1)pr,

we derive that r=1 and p=2, which is impossible. ∎

Now we are ready to prove Theorem A (2).

Theorem 3.2.

Let G be a group and p a prime. If acs(G)<4p/(p+3), then G=Op(G)×Op(G) with Op(G) abelian.

Proof.

Suppose the theorem is false, and let G be a counterexample of minimal order. Let P be a Sylow p-subgroup of G. Then P is abelian by Lemmas 2.5 and 2.6. Also, by Lemma 2.5, every proper subgroup or quotient group of G has a central Sylow p-subgroup. Assume that G has two different minimal normal subgroups, say, N1,N2. Since PNi/Ni is central in G/Ni for i=1,2, we have [P,G]N1N2=1. Then PZ(G), a contradiction. Hence we may assume that G has a unique minimal normal subgroup, say, N. We also assume that G is neither a p-group nor a p-group.

Case 1. Suppose that N is a p-group. Let |N|=pr, where r is an integer. Since G/N has a central Sylow p-subgroup, we see that G/N has a normal p-complement T/N. By the unique minimal normality of N, we have Op(G)=1.

Assume that T<G. Then T has a central abelian Sylow p-subgroup, and in particular, Op(T) is normal in G. This implies that Op(T)=1, and thus G is a p-group, a contradiction.

Assume that T=G. Then N=P, and G/N is a nontrivial p-group. If G/N has different maximal subgroups M1/N and M2/N, then P is central in M1 and M2. This implies that P is central in M1,M2=G, a contradiction. If G/N has a proper normal maximal subgroup M/N, then P is central in M. This implies that Op(G)Op(M)>1, a contradiction. Hence G/N is cyclic and of a prime order q, where qp. Then G=N. The degree of nonlinear irreducible character is q, and the number is (|G|-q)/q2=(pr-1)/q. Hence

acs(G)=|G||Irr(G)|=|G||G/G|+(|G|-q)/q2=q2prq2+pr-14pp+3,

a contradiction.

Case 2. Suppose that N is a p-group. By the unique minimal normality of N, we have Op(G)=1. Let UP be of order p. Since G/N has a central Sylow p-subgroup, UN is normal in G. If UN<G, then U is central in UN. This implies that Op(G)U>1, a contradiction. Thus G=UN.

Let N1 be a proper U-invariant subgroup of N. Then U is central in UN1. Since U acts nontrivially on N, by the Hall–Higman theorem [5, Theorem 8.5.1], we conclude that N is an elementary q-group with order qs, where q is a prime different from p and s is an integer. Then acs(G)=p2qs/(p2+qs-1)4p/(p+3), a contradiction.

Case 3. Suppose that N is neither a p-group nor a p-group. Assume that N<G. Then N has central Sylow p-subgroup, which is impossible. Assume that N=G. Then N is a nonabelian simple group of order divisible by p, and NG(P) has a central Sylow p-subgroup since NG(P) is a proper subgroup of G. Hence we have NG(P)=CG(P). By [5, Theorem 7.2.1], G has a normal p-complement, a contradiction. ∎

Clearly, Theorem A follows by Theorem 3.1 and Theorem 3.2.


Communicated by Robert Guralnick


Award Identifier / Grant number: 11801208

Award Identifier / Grant number: 11871062

Award Identifier / Grant number: BK20181451

Funding statement: Project supported by NSF of China (Nos. 11801208, 11871062), the Natural Science Foundation of Jiangsu Province (No. BK20181451) and the Postgraduate Innovation Project of Jiangsu Province (No. KYCX18_2360).

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Received: 2018-07-29
Revised: 2019-04-11
Published Online: 2019-05-19
Published in Print: 2019-09-01

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