It is well known that the Frattini subgroups of any finite groups are nilpotent. If
a finite group is not nilpotent, it is not the Frattini subgroup of a finite group. In this paper,
we mainly discuss what kind of finite nilpotent groups cannot be the Frattini subgroup of
some finite groups and give some results. Moreover, we generalize Hobby’s Theorem.

1. Introduction

As we know, the Frattini subgroup of a finite group plays an important role in investigating the structure of finite groups. Many authors did this work, for example, the remarkable result of Burnside: let G be a finite p-group and let be N a G-invariant subgroup contained in the Frattini subgroup Φ(G) of G. If Z(N) is cyclic, then N is also cyclic. In particular, if G is a finite 2-group, then the Frattini subgroup Φ(G) cannot be a nonabelian group of order 8. Recently, in [1], Božikov studied the next possible case, where G is a finite 2-group and Φ(G) is nonabelian of order 16. He showed that in that case Φ(G)≅M×C2, where M≅D8 or M≅Q8 and classified all such groups (Theorem A). On the other hand, we also know that the Frattini subgroup of a finite group is nilpotent. If a finite group is not nilpotent, it is not the Frattini subgroup of a finite group. In this paper, we go into what kind of finite nilpotent groups cannot be the Frattini subgroup of some finite groups and give some results in terms of seminormal subgroups of a finite group. Moreover, we generalize Hobby’s Theorem: a nonabelian p-group with cyclic center cannot be the Frattini subgroup of any p-group.

Throughout the all groups mentioned are assumed to be finite groups. The terminology and notations employed agree with standard usage, as in [2] or [3]. We denote H⊴G to indicate that H is a normal subgroup of group G. H⊲⊲G denotes that H is a characteristic subgroup of group G. Φ(G) denotes the Frattini subgroup of group G. π(G) denotes the set of all primes dividing the order of group G.

2. Basic Definitions and Preliminary Results

In this section, we give one definition and some results that are needed in this paper.

Definition 1 (see [<xref ref-type="bibr" rid="B2">4</xref>, Definition 1]).

A subgroup A of G is seminormal in G if there exists a subgroup B such that AB=G and such that for every proper subgroup B1 of B, the product AB1 is a proper subgroup of G.

Lemma 2 (see [<xref ref-type="bibr" rid="B2">4</xref>, Proposition 5]).

Let G be a finite group. If p-subgroup P of G is seminormal in G, then P permutes with every q-subgroup of G, where p, q are primes dividing the order of G, p≠q.

Lemma 3 (see [<xref ref-type="bibr" rid="B1">5</xref>, Lemma <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M49"><mml:mn>2.2</mml:mn><mml:mo>.</mml:mo><mml:mn>2</mml:mn></mml:math></inline-formula>]).

Let M be a p-group. If there exists a normal subgroup N such that |N|=p2, then |M:CM(N)|≤p.

Lemma 4 (see [<xref ref-type="bibr" rid="B1">5</xref>, Lemma <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M55"><mml:mn>2.2</mml:mn><mml:mo>.</mml:mo><mml:mn>3</mml:mn></mml:math></inline-formula>]).

Let P be a p-group and let N=〈a〉×〈b〉 be a normal subgroup of P of order p3 with |a|=p2. If N≤Φ(P) and 〈a〉≤Z(Φ(P)), then N≤Z(Φ(P)).

Lemma 5.

Let G be a finite group. If there exists a normal subgroup N of G such that |N|=p2, moreover q∈π(G), (q,p2-1)=1, then |G:CG(N)|≤p.

Proof.

Since |N|=p2, we have that N is a cyclic group or an elementary abelian p-group.

If N=〈a〉, then |Aut(N)|=p2-p=p(p-1). Assume that Gq is a Sylow q-subgroup of G, q≠p. For x∈Gq, let (ai)σx=(ai)x, then σx∈Aut(N), so |σx|∣p(p-1). Since |σx|∣|x|, by the assumption we have that σx=1, so Gq≤CG(N). Let Gp be a Sylow p-subgroup of G. Set
(1)Ψ:Gp⟶Aut(N)x⟶σx:ai⟶(ai)x,
so Gp/kerΨ is isomorphic to a subgroup of Aut(N), and it follows that |Gp:kerΨ|≤p, that is, |Gp:CGp(N)|≤p, so 〈Gq,CGp(N)∣q≠p〉≤CG(N), and thus |G:CG(N)|≤p.

If N is an elementary abelian p-group, N=〈a〉×〈b〉, then |Aut(N)|=(p2-1)(p2-p)=p(p+1)(p-1). By the assumption we have that Gq≤CG(N), q≠p, Gq is a Sylow q-subgroup of G, |Gp:CGp(N)|≤p, and it follows that |G:CG(N)|≤p.

Lemma 6.

Let G be a finite group. Suppose that N≤Φ(G), N⊴G, N=〈a〉×〈b〉, |a|=p2, |b|=p, a∈Z(Φ(G)), and q∈π(G), (q,p2-1)=1, then N≤Z(Φ(G)).

Proof.

By the assumption we have that Np=〈ap〉, but Np ⊲⊲ N, so Np⊴G. Let T=Np, then N/T is an elementary abelian p-group; that is, N¯=〈a¯〉×〈b¯〉. By the assumption we have that Gq¯<CG¯(N¯), Gq is a Sylow q-group of G, q≠p. The preimage of 〈b¯〉 in N is 〈b,ap〉, so it is an elementary abelian p-group, and it follows that 〈b,ap〉 is centralized by Gq, so Gq≤CG(b). For g∈Gp, then bg must be: b,bap,…,bap(p-1), so |Gp:CGp(b)|≤p, and |G:CG(b)|≤p, thus Φ(G)≤CG(b). It follows that b∈Z(Φ(G)). So N≤Z(Φ(G)).

3. Main ResultsTheorem 7.

Let G be a finite solvable group. Suppose that a subgroup H with p3 order of G is not an abelian group, and what is more, any maximal subgroup of Sylow p-subgroup of G is seminormal in G, then H is not the Frattini subgroup of G.

Proof.

Suppose that the theorem is false, and let H=Φ(G), for any Gp∈Sylp(G), we have that Φ(G)≤Gp. It is clear that Φ(G)≠Gp. Since H⊴Gp, we can find a maximal subgroup N of H such that |N|=p2, N⊴Gp. By Lemma 3, we have that |Gp:CGp(N)|≤p, so CGp(N) is a maximal subgroup of Gp or equal to Gp. We can conclude that Φ(G)≤Φ(Gp), otherwise there exists a maximal subgroup M of Gp such that Gp=Φ(G)M. Since G is solvable, there exists a Sylow tower in G. Suppose that they are Sp1,Sp2,…,Spn, where p1=p. By the assumption and Lemma 2, we have that MSpi is a group i=2,…,n. It follows that MSp2⋯Spn is a subgroup of G, so G=Φ(G)MSp2⋯Spn, a contradiction. By Φ(G)≤Φ(Gp), so Φ(G)≤CGp(N), and thus N≤Z(Φ(G))=Z(H), but |Z(H)|=p, a contradiction, so H is not the Frattini subgroup of G.

Corollary 8.

Let G be a finite solvable group. Suppose that a subgroup H with p3 order of G is not an abelian group, and what is more, any maximal subgroup of Sylow p-subgroup of G is seminormal in G, and then H does not satisfy H⊴G, H≤Φ(G).

Corollary 9.

Suppose that H with p3 order is not an abelian p-group, then H is not the Frattini subgroup of any p-group.

Theorem 10.

Let G be a finite solvable group. Suppose that a subgroup H with cyclic centre of G is not an abelian p-group, and what is more, any maximal subgroup of Sylow p-subgroup of G is seminormal in G, and then H is not the Frattini subgroup of G.

Proof.

Let H=Φ(G). By Theorem 7, we have that |H|>p3. Suppose that the theorem is false, and let H be a counterexample of minimal order. Since H⊴G and Z(H) ⊲⊲ H, Z(H)⊴G. Since Z(H) is cyclic, we can suppose that 〈c〉≤Z(H) and |c|=p, and then 〈c〉 ⊲⊲ H, so 〈c〉⊴G, Φ(G/〈c〉)=Φ(G)/〈c〉=H/〈c〉, and it follows that H/〈c〉 is not abelian group with cyclic centre, a contradiction, since Z(H/〈c〉) is not cyclic group. Otherwise, we have that H/〈c〉 must be abelian group, it follows that H is an abelian group, a contradiction, since H/〈c〉 is not cyclic group. Let Q¯ be a group which is generated by elements with p order, so Q¯ is an elementary abelian p-group, |Q¯|≥p2, Q¯ ⊲⊲ H¯, and H¯⊴G¯, so Q¯⊴G¯. For any Gp∈Sylp(G), we have that H≤Gp. Let Gp¯ be a Sylow p-subgroup of G¯, and we can find two subgroups M¯, N¯ of Q¯ such that M¯, N¯⊴Gp¯, M¯<N¯, |M¯|=p, and |N¯|=p2, and let M, N be preimage of M¯, N¯ in H, respectively and so |M|=p2, |N|=p3. By Lemma 3, we have that |Gp:CGp(M)|≤p. By the proof of Theorem 7 we know that H≤Φ(Gp), but Φ(Gp)≤CGp(M), so M≤Z(H), it follows that M is cyclic, since |N/M|=p, M≤Z(N), and we can get that N is an abelian group. Since N¯ is an elementary abelian p-group, we have that N=〈a〉×〈b〉, |a|=p2, |b|=p, 〈a〉≤Z(Φ(Gp)), and by Lemma 4 we have that N≤Z(Φ(Gp)). Since H≤Φ(Gp), we know that N≤Z(H), this is contrary to that N is not cyclic group, so the counterexample of minimal order does not exist, and H is not the Frattini subgroup of G.

Corollary 11.

Let G be a finite solvable group. Suppose that a subgroup H with cyclic centre of G is not an abelian p-group, and what is more, any maximal subgroup of Sylow p-subgroup of G is seminormal in G, and then H does not satisfy H⊴G, H≤Φ(G).

Corollary 12.

Suppose that H with cyclic centre is not an abelian p-group, and then H is not the Frattini subgroup of any p-group.

Theorem 13.

Let G be a finite group. Suppose that a subgroup H with p3 order of G is not an abelian group, q∈π(G), (q,p2-1)=1, and then H is not the Frattini subgroup of G.

Proof.

Suppose that the theorem is false, and let H=Φ(G). Then, H⊴G, Z(H) ⊲⊲ H, so Z(H)⊴G. Let 〈c〉=Z(H), so |c|=p, and H/〈c〉 is not cyclic group. For any Gp∈Sylp(G), we have that H=Φ(G)≤Gp, so H/〈c〉⊴Gp/〈c〉, and then H¯⋂Z(Gp¯)≠1¯. Let 〈a¯〉≤Z(Gp¯), and then H¯=〈a¯〉×〈b¯〉 is an elementary abelian p-group, so |Aut(H¯)|=p(p-1)2(p+1). By the assumption we have that Gq¯<CG¯(H¯), q≠p, Gq is a Sylow q-subgroup of G, so Q=〈a,c〉⊴G, but |Q|=p2. By Lemma 5, we have that |G:CG(Q)|≤p, so Φ(G)=H≤CG(Q), and thus Q≤Z(H), but |Q|=p2, a contradiction, so H is not the Frattini subgroup of G.

Theorem 14.

Let G be a finite group. Suppose that a subgroup H with cyclic centre of G is not an abelian p-group, |H|=pn, q∈π(G), (q,pk-1)=1, and 1≤k≤n-1, and then H is not the Frattini subgroup of G.

Proof.

Let H=Φ(G), and by Theorem 13, we know that |H|>p3. Let H be a counterexample of minimal order. Since H=Φ(G)⊴G, Z(H) ⊲⊲ H, we get that Z(H)⊴G. Let c∈Z(H), and then |c|=p. Since Z(H) is cyclic group, we have that 〈c〉 ⊲⊲ Z(H), 〈c〉⊴G, and thus Φ(G/〈c〉)=Φ(G)/〈c〉=H/〈c〉, so Z(H/〈c〉) cannot be cyclic group; otherwise, by the assumption we have that H/〈c〉 must be an abelian group. Since H/〈c〉 is cyclic, we have that H is an abelian group, a contradiction, since Z(H/〈c〉) is not cyclic group. Let Q¯ be a group which is generated by elements with p order, then Q¯ is an elementary abelian p-group, |Q¯|≥p2, and Q¯⊴G¯. Let Q¯=〈a1¯〉×⋯×〈as¯〉, s≤n-1, and then |Aut(Q¯)|=(ps-1)(ps-p)⋯(ps-ps-1)=ps(s-1)/2(ps-1)⋯(p-1). For any g∈Gq, Gq is a Sylow q-subgroup of G, q≠p. Set xσg=xg, x∈Q¯, so σg∈Aut(Q¯), thus |σg ∣∣ qi,|σg| ∣∣ Aut(Q¯)|. By the assumption we have that σg=1. For any Gp∈Sylp(G), we have that H=Φ(G)<Gp. Since Q¯⊴Gp¯, we have that 1<M¯<N¯≤Q¯, M¯, N¯⊴Gp¯, |M¯|=p, and |N¯|=p2. Let M, N be pre-images of M¯, N¯ in H, respectively, so M,N⊴Gp. We know that Gq¯<CG¯(N¯), CG¯(M¯), so M,N⊴G. By Lemma 5, we have that |G:CG(M)|≤p, so Φ(G)≤CG(M), and thus M≤Z(Φ(G))=Z(H). It follows that M is cyclic group. Since N/M is cyclic, so N is abelian group. Since Q¯ is an elementary abelian group, we know that N is not cyclic group. By Lemma 6, we have that N≤Z(H), a contradiction, so the counterexample of minimal order does not exist, and H is not the Frattini subgroup of G.

Corollary 15.

Let G be a finite group and H≤G. Suppose that H=Sp1×⋯Spj×⋯×Spn, where Spi is Sylow pi-subgroup of H, Spj with cyclic centre is not an abelian group, q∈π(G), (q,pjk-1)=1, 1≤k≤pjlj-1, and |Spj|=pjlj, and then H is not the Frattini subgroup of G.

Acknowledgments

The research of the authors is supported by the National Natural Science Foundation of China (10771132), SGRC (GZ310), and the Research grant of Tianjin Polytechnic University and Shanghai Leading Academic Discipline Project (J50101).

BožikovZ.Finite 2-groups with a nonabelian Frattini subgroup of order 16RobinsonD. J. S.HuppertB.WangP. C.Some sufficient conditions of a nilpotent groupDeaconescuM.