ALGEBRA Algebra 2314-4114 2314-4106 Hindawi Publishing Corporation 135045 10.1155/2013/135045 135045 Research Article A Note on Hobby’s Theorem of Finite Groups Kong Qingjun Soto Ricardo L. Department of Mathematics Tianjin Polytechnic University Tianjin 300387 China tjpu.edu.cn 2013 12 6 2013 2013 10 03 2013 27 05 2013 2013 Copyright © 2013 Qingjun Kong. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 , 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 MD8 or MQ8 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  or . We denote HG 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, pq.

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 NΦ(P) and aZ(Φ(P)), then NZ(Φ(P)).

Lemma 5.

Let G be a finite group. If there exists a normal subgroup N of G such that |N|=p2, moreover qπ(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, qp. For xGq, let (ai)σx=(ai)x, then σxAut(N), so |σx|p(p-1). Since |σx||x|, by the assumption we have that σx=1, so GqCG(N). Let Gp be a Sylow p-subgroup of G. Set (1)Ψ:GpAut(N)xσx: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)qpCG(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 GqCG(N), qp, 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 NΦ(G), NG, N=a×b, |a|=p2, |b|=p, aZ(Φ(G)), and qπ(G), (q,p2-1)=1, then NZ(Φ(G)).

Proof.

By the assumption we have that Np=ap, but Np  ⊲⊲  N, so NpG. 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, qp. 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 GqCG(b). For gGp, 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 bZ(Φ(G)). So NZ(Φ(G)).

3. Main Results Theorem 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 GpSylp(G), we have that Φ(G)Gp. It is clear that Φ(G)Gp. Since HGp, we can find a maximal subgroup N of H such that |N|=p2, NGp. 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 MSp2Spn is a subgroup of G, so G=Φ(G)MSp2Spn, a contradiction. By Φ(G)Φ(Gp), so Φ(G)CGp(N), and thus NZ(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 HG, HΦ(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 HG and Z(H)  ⊲⊲  H, Z(H)G. Since Z(H) is cyclic, we can suppose that cZ(H) and |c|=p, and then c  ⊲⊲  H, so cG, 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 GpSylp(G), we have that HGp. 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 HΦ(Gp), but Φ(Gp)CGp(M), so MZ(H), it follows that M is cyclic, since |N/M|=p, MZ(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, aZ(Φ(Gp)), and by Lemma 4 we have that NZ(Φ(Gp)). Since HΦ(Gp), we know that NZ(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 HG, HΦ(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, qπ(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, HG, Z(H)  ⊲⊲  H, so Z(H)G. Let c=Z(H), so |c|=p, and H/c is not cyclic group. For any GpSylp(G), we have that H=Φ(G)Gp, so H/cGp/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¯), qp, Gq is a Sylow q-subgroup of G, so Q=a,cG, but |Q|=p2. By Lemma 5, we have that |G:CG(Q)|p, so Φ(G)=HCG(Q), and thus QZ(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, qπ(G), (q,pk-1)=1, and 1kn-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 cZ(H), and then |c|=p. Since Z(H) is cyclic group, we have that c  ⊲⊲  Z(H), cG, 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¯, sn-1, and then |Aut(Q¯)|=(ps-1)(ps-p)(ps-ps-1)=ps(s-1)/2(ps-1)(p-1). For any gGq, Gq is a Sylow q-subgroup of G, qp. Set xσg=xg, xQ¯, so σgAut(Q¯), thus |σg  ∣∣  qi,|σg|  ∣∣  Aut(Q¯)|. By the assumption we have that σg=1. For any GpSylp(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,NGp. We know that Gq¯<CG¯(N¯), CG¯(M¯), so M,NG. By Lemma 5, we have that |G:CG(M)|p, so Φ(G)CG(M), and thus MZ(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 NZ(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 HG. Suppose that H=Sp1×Spj××Spn, where Spi is Sylow pi-subgroup of H, Spj with cyclic centre is not an abelian group, qπ(G), (q,pjk-1)=1, 1kpjlj-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).

Božikov Z. Finite 2-groups with a nonabelian Frattini subgroup of order 16 Archiv der Mathematik 2006 86 1 11 15 10.1007/s00013-005-1474-z MR2201290 ZBL1091.20011 Robinson D. J. S. A Course in the Theory of Groups 1993 80 New York, NY, USA Springer xviii+481 MR1261639 Huppert B. Endliche Gruppen. I 1967 Berlin, Germany Springer xii+793 MR0224703 Wang P. C. Some sufficient conditions of a nilpotent group Journal of Algebra 1992 148 2 289 295 10.1016/0021-8693(92)90194-Q MR1163736 Deaconescu M. Frattini-Like Subgroups of Finite Groups 1986 2, part 4 Mathematical Reports