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

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)<4⁢p/(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 2⁢p, where p is an odd prime, then acd⁡(G)=2⁢(p+1)/(p+3) and acs⁡(G)=4⁢p/(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 p2⁢n+1. Then

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

then Op⁢(G) need not be abelian.

Corollary D ([2]).

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

Proof.

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

Lemma 2.1 ([9, Proposition 5.4 (2)]).

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

Lemma 2.2.

Let G be a group and acd⁡(G)<2. If N⊴G, then

Proof.

We may assume that G′≠1 and N is a minimal normal subgroup of G.

Assume that N≤G′. Then, for any χ∈Irr⁡(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 N≰G′. Then N∩G′=1 and N≤Z⁢(G). For any λ∈Irr⁡(N), λ is extendible to G. By [3, Corollary 6.17],

Note that Irr⁡(G)=⋃λ∈Irr⁡(N)Irr⁡(λG) and Irr⁡(λ1G)∩Irr⁡(λ2G)=∅ when λ1≠λ2, where λ1,λ2∈Irr⁡(N). Hence

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 λ∈Irr⁡(M) be nonprincipal. If K=IG⁢(λ), then KG=M.

Lemma 2.5.

Let G be a group. If H≤G and N⊴G, 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)<4⁢p/(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

which is a contradiction. ∎

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)<2⁢p/(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 λ∈Irr⁡(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, G≅2.ℤp. Then acd⁡(G)=2⁢(p+1)/(p+3), a contradiction. Hence t≥3.

Case 1. Assume that e≥3. Then acd⁡(G)≥g⁢(3,3)=2, a contradiction.

Case 2. Assume that e=2. Since g⁢(t,e)≤acd⁡(G)<2, we have t≤3 and thus t=3. Let nonprincipal μ,ν∈Irr⁡(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/M≲S3 and 3||G:M|.

Suppose that G/M≅S3. 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/M≲Aut⁡(M) is abelian, a contradiction. In the later case, we have |M|=|Irr⁡(M)|=1+3+6=10, a contradiction.

Suppose that G/M≅ℤ3. Then G is a Frobenius group with kernel M. Hence |M|=|Irr⁡(M)|=1+3+3=7=p, so

a contradiction.

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

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

Theorem 3.2.

Let G be a group and p a prime. If acs⁡(G)<4⁢p/(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 P⁢Ni/Ni is central in G/Ni for i=1,2, we have [P,G]≤N1∩N2=1. Then P≤Z⁢(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 q≠p. Then G′=N. The degree of nonlinear irreducible character is q, and the number is (|G|-q)/q2=(pr-1)/q. Hence

a contradiction.

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

Let N1 be a proper U-invariant subgroup of N. Then U is central in U⁢N1. 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)=p2⁢qs/(p2+qs-1)≥4⁢p/(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. ∎