AlgebraVolume 2013 (2013), Article ID 135045, 3 pageshttp://dx.doi.org/10.1155/2013/135045

Research Article

A Note on Hobby’s Theorem of Finite Groups

Qingjun Kong

Department of Mathematics, Tianjin Polytechnic University, Tianjin 300387, China

Received 10 March 2013; Accepted 27 May 2013

Academic Editor: Ricardo L. Soto

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.

Abstract.

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 be a finite -group and let be a -invariant subgroup contained in the Frattini subgroup of . If is cyclic, then is also cyclic. In particular, if is a finite 2-group, then the Frattini subgroup cannot be a nonabelian group of order 8. Recently, in [1], Božikov studied the next possible case, where is a finite 2-group and is nonabelian of order 16. He showed that in that case , where or 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 -group with cyclic center cannot be the Frattini subgroup of any -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 to indicate that is a normal subgroup of group . denotes that is a characteristic subgroup of group . denotes the Frattini subgroup of group . denotes the set of all primes dividing the order of group .

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 [4, Definition 1]). A subgroup of is seminormal in if there exists a subgroup such that and such that for every proper subgroup of , the product is a proper subgroup of .

Lemma 2 (see [4, Proposition 5]). Let be a finite group. If -subgroup of is seminormal in , then permutes with every -subgroup of , where , are primes dividing the order of , .

Lemma 3 (see [5, Lemma ]). Let be a -group. If there exists a normal subgroup such that , then .

Lemma 4 (see [5, Lemma ]). Let be a -group and let be a normal subgroup of of order with . If and , then .

Lemma 5. Let be a finite group. If there exists a normal subgroup of such that , moreover , , then .

Proof. Since , we have that is a cyclic group or an elementary abelian -group.(i)If , then . Assume that is a Sylow -subgroup of , . For , let , then , so . Since , by the assumption we have that , so . Let be a Sylow -subgroup of . Set so is isomorphic to a subgroup of , and it follows that , that is, , so , and thus .(ii)If is an elementary abelian -group, , then . By the assumption we have that , , is a Sylow -subgroup of , , and it follows that .

Lemma 6. Let be a finite group. Suppose that , , , , , , and , , then .

Proof. By the assumption we have that , but , so . Let , then is an elementary abelian -group; that is, . By the assumption we have that , is a Sylow -group of , . The preimage of in is , so it is an elementary abelian -group, and it follows that is centralized by , so . For , then must be: , so , and , thus . It follows that . So .

3. Main Results

Theorem 7. Let G be a finite solvable group. Suppose that a subgroup with order of is not an abelian group, and what is more, any maximal subgroup of Sylow -subgroup of is seminormal in , then is not the Frattini subgroup of .

Proof. Suppose that the theorem is false, and let , for any , we have that . It is clear that . Since , we can find a maximal subgroup of such that , . By Lemma 3, we have that , so is a maximal subgroup of or equal to . We can conclude that , otherwise there exists a maximal subgroup of such that . Since is solvable, there exists a Sylow tower in . Suppose that they are , where . By the assumption and Lemma 2, we have that is a group . It follows that is a subgroup of , so , a contradiction. By , so , and thus , but , a contradiction, so is not the Frattini subgroup of .

Corollary 8. Let be a finite solvable group. Suppose that a subgroup with order of is not an abelian group, and what is more, any maximal subgroup of Sylow -subgroup of is seminormal in , and then does not satisfy , .

Corollary 9. Suppose that with order is not an abelian -group, then is not the Frattini subgroup of any -group.

Theorem 10. Let be a finite solvable group. Suppose that a subgroup with cyclic centre of is not an abelian -group, and what is more, any maximal subgroup of Sylow -subgroup of is seminormal in , and then is not the Frattini subgroup of .

Proof. Let . By Theorem 7, we have that . Suppose that the theorem is false, and let be a counterexample of minimal order. Since and , . Since is cyclic, we can suppose that and , and then , so , , and it follows that is not abelian group with cyclic centre, a contradiction, since is not cyclic group. Otherwise, we have that must be abelian group, it follows that is an abelian group, a contradiction, since is not cyclic group. Let be a group which is generated by elements with order, so is an elementary abelian -group, , , and , so . For any , we have that . Let be a Sylow -subgroup of , and we can find two subgroups , of such that , , , , and , and let , be preimage of , in , respectively and so , . By Lemma 3, we have that . By the proof of Theorem 7 we know that , but , so , it follows that is cyclic, since , , and we can get that is an abelian group. Since is an elementary abelian -group, we have that , , , , and by Lemma 4 we have that . Since , we know that , this is contrary to that is not cyclic group, so the counterexample of minimal order does not exist, and is not the Frattini subgroup of .

Corollary 11. Let be a finite solvable group. Suppose that a subgroup with cyclic centre of is not an abelian -group, and what is more, any maximal subgroup of Sylow -subgroup of is seminormal in , and then does not satisfy , .

Corollary 12. Suppose that with cyclic centre is not an abelian -group, and then is not the Frattini subgroup of any -group.

Theorem 13. Let be a finite group. Suppose that a subgroup with order of is not an abelian group, , , and then is not the Frattini subgroup of .

Proof. Suppose that the theorem is false, and let . Then, , , so . Let , so , and is not cyclic group. For any , we have that , so , and then . Let , and then is an elementary abelian -group, so . By the assumption we have that , , is a Sylow -subgroup of , so , but . By Lemma 5, we have that , so , and thus , but , a contradiction, so is not the Frattini subgroup of .

Theorem 14. Let be a finite group. Suppose that a subgroup with cyclic centre of is not an abelian -group, , , , and , and then is not the Frattini subgroup of .

Proof. Let , and by Theorem 13, we know that . Let be a counterexample of minimal order. Since , , we get that . Let , and then . Since is cyclic group, we have that , , and thus , so cannot be cyclic group; otherwise, by the assumption we have that must be an abelian group. Since is cyclic, we have that is an abelian group, a contradiction, since is not cyclic group. Let be a group which is generated by elements with order, then is an elementary abelian -group, , and . Let , , and then . For any , is a Sylow -subgroup of , . Set , , so , thus . By the assumption we have that . For any , we have that . Since , we have that , , , , and . Let , be pre-images of , in , respectively, so . We know that , , so . By Lemma 5, we have that , so , and thus . It follows that is cyclic group. Since is cyclic, so is abelian group. Since is an elementary abelian group, we know that is not cyclic group. By Lemma 6, we have that , a contradiction, so the counterexample of minimal order does not exist, and is not the Frattini subgroup of .

Corollary 15. Let be a finite group and . Suppose that , where is Sylow -subgroup of , with cyclic centre is not an abelian group, , , , and , and then is not the Frattini subgroup of .

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).

References

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