Hindawi Publishing Corporation Algebra Volume 2013, Article ID 135045, 3 pages http://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 Correspondence should be addressed to Qingjun Kong; [email protected] Received 10 March 2013; Accepted 27 May 2013 Academic Editor: Ricardo L. Soto Copyright © 2013 Qingjun Kong. is 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 eorem. 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 Φ() of . If () is cyclic, then is also cyclic. In particular, if is a finite 2-group, then the Frattini subgroup Φ() cannot be a nonabelian group of order 8. Recently, in [1], Boˇ zikov studied the next possible case, where is a finite 2-group and Φ() is nonabelian of order 16. He showed that in that case Φ() ≅ × 2 , where ≅ 8 or ≅ 8 and classified all such groups (eorem 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 eorem: a nonabelian -group with cyclic center cannot be the Frattini subgroup of any -group. roughout the all groups mentioned are assumed to be finite groups. e 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 . Φ() 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 1 of , the product 1 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 2.2.2]). Let be a -group. If there exists a normal subgroup such that || = 2 , then | : ()| ≤ . Lemma 4 (see [5, Lemma 2.2.3]). Let be a -group and let = ⟨⟩ × ⟨⟩ be a normal subgroup of of order 3 with || = 2 . If ≤ Φ() and ⟨⟩ ≤ (Φ()), then ≤ (Φ()). Lemma 5. Let be a finite group. If there exists a normal subgroup of such that || = 2 , moreover ∈ (), (, 2 − 1) = 1, then | : ()| ≤ . Proof. Since || = 2 , we have that is a cyclic group or an elementary abelian -group.
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Hindawi Publishing CorporationAlgebraVolume 2013 Article ID 135045 3 pageshttpdxdoiorg1011552013135045
Research ArticleA Note on Hobbyrsquos Theorem of Finite Groups
Qingjun Kong
Department of Mathematics Tianjin Polytechnic University Tianjin 300387 China
Correspondence should be addressed to Qingjun Kong kqj2929163com
Received 10 March 2013 Accepted 27 May 2013
Academic Editor Ricardo L Soto
Copyright copy 2013 Qingjun KongThis is an open access article distributed under theCreativeCommonsAttribution License whichpermits 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 Frattinisubgroup of a finite group In this paper we mainly discuss what kind of finite nilpotent groups cannot be the Frattini subgroup ofsome finite groups and give some results Moreover we generalize Hobbyrsquos Theorem
1 Introduction
As we know the Frattini subgroup of a finite group plays animportant role in investigating the structure of finite groupsMany authors did this work for example the remarkableresult of Burnside let 119866 be a finite 119901-group and let be119873 a 119866-invariant subgroup contained in the Frattini subgroup Φ(119866)
of 119866 If 119885(119873) is cyclic then 119873 is also cyclic In particular if119866 is a finite 2-group then the Frattini subgroupΦ(119866) cannotbe a nonabelian group of order 8 Recently in [1] Bozikovstudied the next possible case where119866 is a finite 2-group andΦ(119866) is nonabelian of order 16 He showed that in that caseΦ(119866) cong 119872 times 119862
2 where119872 cong 119863
8or119872 cong 119876
8and classified all
such groups (Theorem A) On the other hand we also knowthat the Frattini subgroup of a finite group is nilpotent If afinite group is not nilpotent it is not the Frattini subgroup ofa finite group In this paper we go into what kind of finitenilpotent groups cannot be the Frattini subgroup of somefinite groups and give some results in terms of seminormalsubgroups of a finite group Moreover we generalize HobbyrsquosTheorem a nonabelian 119901-group with cyclic center cannot bethe Frattini subgroup of any 119901-group
Throughout the all groups mentioned are assumed to befinite groupsThe terminology and notations employed agreewith standard usage as in [2] or [3] We denote 119867 ⊴ 119866 toindicate that 119867 is a normal subgroup of group 119866 119867 ⊲⊲ 119866
denotes that119867 is a characteristic subgroup of group 119866Φ(119866)
denotes the Frattini subgroup of group 119866 120587(119866) denotes theset of all primes dividing the order of group 119866
2 Basic Definitions and Preliminary Results
In this section we give one definition and some results thatare needed in this paper
Definition 1 (see [4 Definition 1]) A subgroup 119860 of 119866 isseminormal in 119866 if there exists a subgroup 119861 such that 119860119861 =
119866 and such that for every proper subgroup 1198611of 119861 the
product 1198601198611is a proper subgroup of 119866
Lemma 2 (see [4 Proposition 5]) Let 119866 be a finite group If119901-subgroup 119875 of 119866 is seminormal in 119866 then 119875 permutes withevery 119902-subgroup of119866 where 119901 119902 are primes dividing the orderof 119866 119901 = 119902
Lemma 3 (see [5 Lemma 222]) Let119872 be a 119901-group If thereexists a normal subgroup 119873 such that |119873| = 119901
2 then |119872
119862119872(119873)| le 119901
Lemma 4 (see [5 Lemma 223]) Let 119875 be a 119901-group and let119873 = ⟨119886⟩ times ⟨119887⟩ be a normal subgroup of 119875 of order 119901
3 with|119886| = 119901
2 If119873 le Φ(119875) and ⟨119886⟩ le 119885(Φ(119875)) then119873 le 119885(Φ(119875))
Lemma 5 Let 119866 be a finite group If there exists a normalsubgroup 119873 of 119866 such that |119873| = 119901
2 moreover 119902 isin 120587(119866)(119902 1199012minus 1) = 1 then |119866 119862
119866(119873)| le 119901
Proof Since |119873| = 1199012 we have that119873 is a cyclic group or an
elementary abelian 119901-group
2 Algebra
(i) If119873 = ⟨119886⟩ then |Aut(119873)| = 1199012minus119901 = 119901(119901minus1) Assume
that 119866119902is a Sylow 119902-subgroup of 119866 119902 = 119901 For 119909 isin 119866
119902
let (119886119894)120590119909 = (119886119894)119909 then 120590
119909isin Aut(119873) so |120590
119909| | 119901(119901minus1)
Since |120590119909| | |119909| by the assumption we have that 120590
119909=
1 so 119866119902le 119862119866(119873) Let 119866
119901be a Sylow 119901-subgroup of
119866 Set
Ψ 119866119901997888rarr Aut (119873)
119909 997888rarr 120590119909 119886119894997888rarr (119886
119894)119909
(1)
so 119866119901 kerΨ is isomorphic to a subgroup of Aut(119873)
and it follows that |119866119901
kerΨ| le 119901 that is |119866119901
119862119866119901
(119873)| le 119901 so ⟨119866119902 119862119866119901
(119873) | 119902 = 119901⟩ le 119862119866(119873) and
thus |119866 119862119866(119873)| le 119901
(ii) If119873 is an elementary abelian 119901-group119873 = ⟨119886⟩times ⟨119887⟩then |Aut(119873)| = (119901
2minus1)(119901
2minus119901) = 119901(119901+1)(119901minus1) By
the assumption we have that 119866119902le 119862119866(119873) 119902 = 119901 119866
119902
is a Sylow 119902-subgroup of 119866 |119866119901 119862119866119901
(119873)| le 119901 and itfollows that |119866 119862
119866(119873)| le 119901
Lemma 6 Let 119866 be a finite group Suppose that 119873 le Φ(119866)119873 ⊴ 119866 119873 = ⟨119886⟩ times ⟨119887⟩ |119886| = 119901
2 |119887| = 119901 119886 isin 119885(Φ(119866)) and119902 isin 120587(119866) (119902 1199012 minus 1) = 1 then119873 le 119885(Φ(119866))
Proof By the assumptionwe have that119873119901 = ⟨119886119901⟩ but119873119901 ⊲⊲
119873 so 119873119901
⊴ 119866 Let 119879 = 119873119901 then 119873119879 is an elementary
abelian 119901-group that is119873 = ⟨119886⟩times⟨119887⟩ By the assumption wehave that 119866
119902lt 119862119866(119873) 119866
119902is a Sylow 119902-group of 119866 119902 = 119901 The
preimage of ⟨119887⟩ in119873 is ⟨119887 119886119901⟩ so it is an elementary abelian119901-group and it follows that ⟨119887 119886119901⟩ is centralized by 119866
119902 so
119866119902le 119862119866(119887) For 119892 isin 119866
119901 then 119887
119892must be 119887 119887119886119901 119887119886119901(119901minus1)so |119866119901
119862119866119901
(119887)| le 119901 and |119866 119862119866(119887)| le 119901 thus Φ(119866) le
119862119866(119887) It follows that 119887 isin 119885(Φ(119866)) So119873 le 119885(Φ(119866))
3 Main Results
Theorem 7 Let G be a finite solvable group Suppose that asubgroup 119867 with 119901
3 order of 119866 is not an abelian group andwhat is more any maximal subgroup of Sylow 119901-subgroup of119866is seminormal in 119866 then119867 is not the Frattini subgroup of 119866
Proof Suppose that the theorem is false and let 119867 = Φ(119866)for any 119866
119901isin Syl119901(119866) we have thatΦ(119866) le 119866
119901 It is clear that
Φ(119866) = 119866119901 Since 119867 ⊴ 119866
119901 we can find a maximal subgroup
119873 of 119867 such that |119873| = 1199012 119873 ⊴ 119866
119901 By Lemma 3 we have
that |119866119901
119862119866119901
(119873)| le 119901 so 119862119866119901
(119873) is a maximal subgroupof 119866119901or equal to 119866
119901 We can conclude that Φ(119866) le Φ(119866
119901)
otherwise there exists a maximal subgroup119872 of119866119901such that
119866119901= Φ(119866)119872 Since 119866 is solvable there exists a Sylow tower
in 119866 Suppose that they are 1198781199011
1198781199012
119878119901119899
where 1199011= 119901 By
the assumption and Lemma 2 we have that 119872119878119901119894
is a group119894 = 2 119899 It follows that119872119878
1199012
sdot sdot sdot 119878119901119899
is a subgroup of 119866 so119866 = Φ(119866)119872119878
1199012
sdot sdot sdot 119878119901119899
a contradiction By Φ(119866) le Φ(119866119901)
so Φ(119866) le 119862119866119901
(119873) and thus 119873 le 119885(Φ(119866)) = 119885(119867) but|119885(119867)| = 119901 a contradiction so119867 is not the Frattini subgroupof 119866
Corollary 8 Let 119866 be a finite solvable group Suppose that asubgroup 119867 with 119901
3 order of 119866 is not an abelian group andwhat is more any maximal subgroup of Sylow 119901-subgroup of119866 is seminormal in 119866 and then 119867 does not satisfy 119867 ⊴ 119866119867 le Φ(119866)
Corollary 9 Suppose that 119867 with 1199013 order is not an abelian
119901-group then119867 is not the Frattini subgroup of any 119901-group
Theorem 10 Let 119866 be a finite solvable group Suppose that asubgroup 119867 with cyclic centre of 119866 is not an abelian 119901-groupand what is more any maximal subgroup of Sylow 119901-subgroupof119866 is seminormal in119866 and then119867 is not the Frattini subgroupof 119866
Proof Let 119867 = Φ(119866) By Theorem 7 we have that |119867| gt
1199013 Suppose that the theorem is false and let 119867 be a
counterexample ofminimal order Since119867 ⊴ 119866 and119885(119867) ⊲⊲
119867 119885(119867) ⊴ 119866 Since 119885(119867) is cyclic we can suppose that⟨119888⟩ le 119885(119867) and |119888| = 119901 and then ⟨119888⟩ ⊲⊲ 119867 so ⟨119888⟩ ⊴ 119866Φ(119866⟨119888⟩) = Φ(119866)⟨119888⟩ = 119867⟨119888⟩ and it follows that 119867⟨119888⟩
is not abelian group with cyclic centre a contradiction since119885(119867⟨119888⟩) is not cyclic group Otherwise we have that119867⟨119888⟩
must be abelian group it follows that 119867 is an abelian groupa contradiction since 119867⟨119888⟩ is not cyclic group Let 119876 be agroup which is generated by elements with 119901 order so 119876 isan elementary abelian 119901-group |119876| ge 119901
2 119876 ⊲⊲ 119867 and119867 ⊴ 119866 so 119876 ⊴ 119866 For any 119866
119901isin Syl
119901(119866) we have that
119867 le 119866119901 Let 119866
119901be a Sylow 119901-subgroup of 119866 and we can
find two subgroups119872119873 of119876 such that119872119873 ⊴ 119866119901119872 lt 119873
|119872| = 119901 and |119873| = 1199012 and let119872119873 be preimage of119872119873 in
119867 respectively and so |119872| = 1199012 |119873| = 119901
3 By Lemma 3 wehave that |119866
119901 119862119866119901
(119872)| le 119901 By the proof of Theorem 7 weknow that119867 le Φ(119866
119901) but Φ(119866
119901) le 119862119866119901
(119872) so119872 le 119885(119867)it follows that 119872 is cyclic since |119873119872| = 119901 119872 le 119885(119873)and we can get that 119873 is an abelian group Since 119873 is anelementary abelian 119901-group we have that 119873 = ⟨119886⟩ times ⟨119887⟩|119886| = 119901
2 |119887| = 119901 ⟨119886⟩ le 119885(Φ(119866119901)) and by Lemma 4 we
have that 119873 le 119885(Φ(119866119901)) Since 119867 le Φ(119866
119901) we know that
119873 le 119885(119867) this is contrary to that 119873 is not cyclic group sothe counterexample of minimal order does not exist and 119867
is not the Frattini subgroup of 119866
Corollary 11 Let 119866 be a finite solvable group Suppose that asubgroup 119867 with cyclic centre of 119866 is not an abelian 119901-groupand what is more any maximal subgroup of Sylow 119901-subgroupof 119866 is seminormal in 119866 and then 119867 does not satisfy 119867 ⊴ 119866119867 le Φ(119866)
Corollary 12 Suppose that 119867 with cyclic centre is not anabelian 119901-group and then119867 is not the Frattini subgroup of any119901-group
Theorem 13 Let 119866 be a finite group Suppose that a subgroup119867with 119901
3 order of119866 is not an abelian group 119902 isin 120587(119866) (119902 1199012minus1) = 1 and then119867 is not the Frattini subgroup of 119866
Algebra 3
Proof Suppose that the theorem is false and let 119867 = Φ(119866)Then119867 ⊴ 119866 119885(119867) ⊲⊲ 119867 so 119885(119867) ⊴ 119866 Let ⟨119888⟩ = 119885(119867) so|119888| = 119901 and119867⟨119888⟩ is not cyclic group For any 119866
119901isin Syl119901(119866)
we have that 119867 = Φ(119866) le 119866119901 so 119867⟨119888⟩ ⊴ 119866
119901⟨119888⟩ and then
119867⋂119885(119866119901) = 1 Let ⟨119886⟩ le 119885(119866
119901) and then119867 = ⟨119886⟩times⟨119887⟩ is an
elementary abelian 119901-group so |Aut(119867)| = 119901(119901 minus 1)2(119901 + 1)
By the assumption we have that 119866119902lt 119862119866(119867) 119902 = 119901 119866
119902is a
Sylow 119902-subgroup of 119866 so 119876 = ⟨119886 119888⟩ ⊴ 119866 but |119876| = 1199012 By
Lemma 5 we have that |119866 119862119866(119876)| le 119901 so Φ(119866) = 119867 le
119862119866(119876) and thus 119876 le 119885(119867) but |119876| = 119901
2 a contradiction so119867 is not the Frattini subgroup of 119866
Theorem 14 Let 119866 be a finite group Suppose that a subgroup119867 with cyclic centre of 119866 is not an abelian 119901-group |119867| = 119901
119899119902 isin 120587(119866) (119902 119901119896 minus1) = 1 and 1 le 119896 le 119899minus1 and then119867 is notthe Frattini subgroup of 119866
Proof Let 119867 = Φ(119866) and by Theorem 13 we know that|119867| gt 119901
3 Let119867 be a counterexample of minimal order Since119867 = Φ(119866) ⊴ 119866 119885(119867) ⊲⊲ 119867 we get that 119885(119867) ⊴ 119866 Let119888 isin 119885(119867) and then |119888| = 119901 Since 119885(119867) is cyclic groupwe have that ⟨119888⟩ ⊲⊲ 119885(119867) ⟨119888⟩ ⊴ 119866 and thus Φ(119866⟨119888⟩) =
Φ(119866)⟨119888⟩ = 119867⟨119888⟩ so 119885(119867⟨119888⟩) cannot be cyclic groupotherwise by the assumption we have that 119867⟨119888⟩ must bean abelian group Since 119867⟨119888⟩ is cyclic we have that 119867 is anabelian group a contradiction since 119885(119867⟨119888⟩) is not cyclicgroup Let119876 be a groupwhich is generated by elementswith119901
order then119876 is an elementary abelian 119901-group |119876| ge 1199012 and
119876 ⊴ 119866 Let119876 = ⟨1198861⟩timessdot sdot sdottimes⟨119886
119904⟩ 119904 le 119899minus1 and then |Aut(119876)| =
(119901119904minus1)(119901
119904minus119901) sdot sdot sdot (119901
119904minus119901119904minus1
) = 119901119904(119904minus1)2
(119901119904minus1) sdot sdot sdot (119901minus1) For
any119892 isin 119866119902119866119902is a Sylow 119902-subgroup of119866 119902 = 119901 Set119909120590119892 = 119909
119892119909 isin 119876 so 120590
119892isin Aut(119876) thus |120590
119892|| 119902119894 |120590119892| || Aut(119876)| By
the assumption we have that 120590119892= 1 For any 119866
119901isin Syl119901(119866)
we have that 119867 = Φ(119866) lt 119866119901 Since 119876 ⊴ 119866
119901 we have that
1 lt 119872 lt 119873 le 119876119872119873 ⊴ 119866119901 |119872| = 119901 and |119873| = 119901
2 Let119872119873 be pre-images of 119872 119873 in 119867 respectively so 119872119873 ⊴ 119866
119901
We know that 119866119902
lt 119862119866(119873) 119862
119866(119872) so 119872119873 ⊴ 119866 By
Lemma 5 we have that |119866 119862119866(119872)| le 119901 so Φ(119866) le 119862
119866(119872)
and thus 119872 le 119885(Φ(119866)) = 119885(119867) It follows that 119872 is cyclicgroup Since 119873119872 is cyclic so 119873 is abelian group Since 119876
is an elementary abelian group we know that 119873 is not cyclicgroup By Lemma 6 we have that119873 le 119885(119867) a contradictionso the counterexample of minimal order does not exist and119867 is not the Frattini subgroup of 119866
Corollary 15 Let119866 be a finite group and119867 le 119866 Suppose that119867 = 119878
1199011
times sdot sdot sdot 119878119901119895
times sdot sdot sdot times 119878119901119899
where 119878119901119894
is Sylow 119901119894-subgroup
of 119867 119878119901119895
with cyclic centre is not an abelian group 119902 isin 120587(119866)
(119902 119901119896
119895minus 1) = 1 1 le 119896 le 119901
119897119895minus1
119895 and |119878
119901119895
| = 119901119897119895
119895 and then 119867 is
not the Frattini subgroup of 119866
Acknowledgments
The research of the authors is supported by the National Nat-ural Science Foundation of China (10771132) SGRC (GZ310)
and the Research grant of Tianjin Polytechnic University andShanghai Leading Academic Discipline Project (J50101)
References
[1] Z Bozikov ldquoFinite 2-groups with a nonabelian Frattini sub-group of order 16rdquo Archiv der Mathematik vol 86 no 1 pp11ndash15 2006
[2] D J S Robinson A Course in the Theory of Groups vol 80Springer New York NY USA 1993
[3] B Huppert Endliche Gruppen I Springer Berlin Germany1967
[4] P C Wang ldquoSome sufficient conditions of a nilpotent grouprdquoJournal of Algebra vol 148 no 2 pp 289ndash295 1992
[5] M Deaconescu Frattini-Like Subgroups of Finite Groups vol 2part 4 ofMathematical Reports 1986
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2 Algebra
(i) If119873 = ⟨119886⟩ then |Aut(119873)| = 1199012minus119901 = 119901(119901minus1) Assume
that 119866119902is a Sylow 119902-subgroup of 119866 119902 = 119901 For 119909 isin 119866
119902
let (119886119894)120590119909 = (119886119894)119909 then 120590
119909isin Aut(119873) so |120590
119909| | 119901(119901minus1)
Since |120590119909| | |119909| by the assumption we have that 120590
119909=
1 so 119866119902le 119862119866(119873) Let 119866
119901be a Sylow 119901-subgroup of
119866 Set
Ψ 119866119901997888rarr Aut (119873)
119909 997888rarr 120590119909 119886119894997888rarr (119886
119894)119909
(1)
so 119866119901 kerΨ is isomorphic to a subgroup of Aut(119873)
and it follows that |119866119901
kerΨ| le 119901 that is |119866119901
119862119866119901
(119873)| le 119901 so ⟨119866119902 119862119866119901
(119873) | 119902 = 119901⟩ le 119862119866(119873) and
thus |119866 119862119866(119873)| le 119901
(ii) If119873 is an elementary abelian 119901-group119873 = ⟨119886⟩times ⟨119887⟩then |Aut(119873)| = (119901
2minus1)(119901
2minus119901) = 119901(119901+1)(119901minus1) By
the assumption we have that 119866119902le 119862119866(119873) 119902 = 119901 119866
119902
is a Sylow 119902-subgroup of 119866 |119866119901 119862119866119901
(119873)| le 119901 and itfollows that |119866 119862
119866(119873)| le 119901
Lemma 6 Let 119866 be a finite group Suppose that 119873 le Φ(119866)119873 ⊴ 119866 119873 = ⟨119886⟩ times ⟨119887⟩ |119886| = 119901
2 |119887| = 119901 119886 isin 119885(Φ(119866)) and119902 isin 120587(119866) (119902 1199012 minus 1) = 1 then119873 le 119885(Φ(119866))
Proof By the assumptionwe have that119873119901 = ⟨119886119901⟩ but119873119901 ⊲⊲
119873 so 119873119901
⊴ 119866 Let 119879 = 119873119901 then 119873119879 is an elementary
abelian 119901-group that is119873 = ⟨119886⟩times⟨119887⟩ By the assumption wehave that 119866
119902lt 119862119866(119873) 119866
119902is a Sylow 119902-group of 119866 119902 = 119901 The
preimage of ⟨119887⟩ in119873 is ⟨119887 119886119901⟩ so it is an elementary abelian119901-group and it follows that ⟨119887 119886119901⟩ is centralized by 119866
119902 so
119866119902le 119862119866(119887) For 119892 isin 119866
119901 then 119887
119892must be 119887 119887119886119901 119887119886119901(119901minus1)so |119866119901
119862119866119901
(119887)| le 119901 and |119866 119862119866(119887)| le 119901 thus Φ(119866) le
119862119866(119887) It follows that 119887 isin 119885(Φ(119866)) So119873 le 119885(Φ(119866))
3 Main Results
Theorem 7 Let G be a finite solvable group Suppose that asubgroup 119867 with 119901
3 order of 119866 is not an abelian group andwhat is more any maximal subgroup of Sylow 119901-subgroup of119866is seminormal in 119866 then119867 is not the Frattini subgroup of 119866
Proof Suppose that the theorem is false and let 119867 = Φ(119866)for any 119866
119901isin Syl119901(119866) we have thatΦ(119866) le 119866
119901 It is clear that
Φ(119866) = 119866119901 Since 119867 ⊴ 119866
119901 we can find a maximal subgroup
119873 of 119867 such that |119873| = 1199012 119873 ⊴ 119866
119901 By Lemma 3 we have
that |119866119901
119862119866119901
(119873)| le 119901 so 119862119866119901
(119873) is a maximal subgroupof 119866119901or equal to 119866
119901 We can conclude that Φ(119866) le Φ(119866
119901)
otherwise there exists a maximal subgroup119872 of119866119901such that
119866119901= Φ(119866)119872 Since 119866 is solvable there exists a Sylow tower
in 119866 Suppose that they are 1198781199011
1198781199012
119878119901119899
where 1199011= 119901 By
the assumption and Lemma 2 we have that 119872119878119901119894
is a group119894 = 2 119899 It follows that119872119878
1199012
sdot sdot sdot 119878119901119899
is a subgroup of 119866 so119866 = Φ(119866)119872119878
1199012
sdot sdot sdot 119878119901119899
a contradiction By Φ(119866) le Φ(119866119901)
so Φ(119866) le 119862119866119901
(119873) and thus 119873 le 119885(Φ(119866)) = 119885(119867) but|119885(119867)| = 119901 a contradiction so119867 is not the Frattini subgroupof 119866
Corollary 8 Let 119866 be a finite solvable group Suppose that asubgroup 119867 with 119901
3 order of 119866 is not an abelian group andwhat is more any maximal subgroup of Sylow 119901-subgroup of119866 is seminormal in 119866 and then 119867 does not satisfy 119867 ⊴ 119866119867 le Φ(119866)
Corollary 9 Suppose that 119867 with 1199013 order is not an abelian
119901-group then119867 is not the Frattini subgroup of any 119901-group
Theorem 10 Let 119866 be a finite solvable group Suppose that asubgroup 119867 with cyclic centre of 119866 is not an abelian 119901-groupand what is more any maximal subgroup of Sylow 119901-subgroupof119866 is seminormal in119866 and then119867 is not the Frattini subgroupof 119866
Proof Let 119867 = Φ(119866) By Theorem 7 we have that |119867| gt
1199013 Suppose that the theorem is false and let 119867 be a
counterexample ofminimal order Since119867 ⊴ 119866 and119885(119867) ⊲⊲
119867 119885(119867) ⊴ 119866 Since 119885(119867) is cyclic we can suppose that⟨119888⟩ le 119885(119867) and |119888| = 119901 and then ⟨119888⟩ ⊲⊲ 119867 so ⟨119888⟩ ⊴ 119866Φ(119866⟨119888⟩) = Φ(119866)⟨119888⟩ = 119867⟨119888⟩ and it follows that 119867⟨119888⟩
is not abelian group with cyclic centre a contradiction since119885(119867⟨119888⟩) is not cyclic group Otherwise we have that119867⟨119888⟩
must be abelian group it follows that 119867 is an abelian groupa contradiction since 119867⟨119888⟩ is not cyclic group Let 119876 be agroup which is generated by elements with 119901 order so 119876 isan elementary abelian 119901-group |119876| ge 119901
2 119876 ⊲⊲ 119867 and119867 ⊴ 119866 so 119876 ⊴ 119866 For any 119866
119901isin Syl
119901(119866) we have that
119867 le 119866119901 Let 119866
119901be a Sylow 119901-subgroup of 119866 and we can
find two subgroups119872119873 of119876 such that119872119873 ⊴ 119866119901119872 lt 119873
|119872| = 119901 and |119873| = 1199012 and let119872119873 be preimage of119872119873 in
119867 respectively and so |119872| = 1199012 |119873| = 119901
3 By Lemma 3 wehave that |119866
119901 119862119866119901
(119872)| le 119901 By the proof of Theorem 7 weknow that119867 le Φ(119866
119901) but Φ(119866
119901) le 119862119866119901
(119872) so119872 le 119885(119867)it follows that 119872 is cyclic since |119873119872| = 119901 119872 le 119885(119873)and we can get that 119873 is an abelian group Since 119873 is anelementary abelian 119901-group we have that 119873 = ⟨119886⟩ times ⟨119887⟩|119886| = 119901
2 |119887| = 119901 ⟨119886⟩ le 119885(Φ(119866119901)) and by Lemma 4 we
have that 119873 le 119885(Φ(119866119901)) Since 119867 le Φ(119866
119901) we know that
119873 le 119885(119867) this is contrary to that 119873 is not cyclic group sothe counterexample of minimal order does not exist and 119867
is not the Frattini subgroup of 119866
Corollary 11 Let 119866 be a finite solvable group Suppose that asubgroup 119867 with cyclic centre of 119866 is not an abelian 119901-groupand what is more any maximal subgroup of Sylow 119901-subgroupof 119866 is seminormal in 119866 and then 119867 does not satisfy 119867 ⊴ 119866119867 le Φ(119866)
Corollary 12 Suppose that 119867 with cyclic centre is not anabelian 119901-group and then119867 is not the Frattini subgroup of any119901-group
Theorem 13 Let 119866 be a finite group Suppose that a subgroup119867with 119901
3 order of119866 is not an abelian group 119902 isin 120587(119866) (119902 1199012minus1) = 1 and then119867 is not the Frattini subgroup of 119866
Algebra 3
Proof Suppose that the theorem is false and let 119867 = Φ(119866)Then119867 ⊴ 119866 119885(119867) ⊲⊲ 119867 so 119885(119867) ⊴ 119866 Let ⟨119888⟩ = 119885(119867) so|119888| = 119901 and119867⟨119888⟩ is not cyclic group For any 119866
119901isin Syl119901(119866)
we have that 119867 = Φ(119866) le 119866119901 so 119867⟨119888⟩ ⊴ 119866
119901⟨119888⟩ and then
119867⋂119885(119866119901) = 1 Let ⟨119886⟩ le 119885(119866
119901) and then119867 = ⟨119886⟩times⟨119887⟩ is an
elementary abelian 119901-group so |Aut(119867)| = 119901(119901 minus 1)2(119901 + 1)
By the assumption we have that 119866119902lt 119862119866(119867) 119902 = 119901 119866
119902is a
Sylow 119902-subgroup of 119866 so 119876 = ⟨119886 119888⟩ ⊴ 119866 but |119876| = 1199012 By
Lemma 5 we have that |119866 119862119866(119876)| le 119901 so Φ(119866) = 119867 le
119862119866(119876) and thus 119876 le 119885(119867) but |119876| = 119901
2 a contradiction so119867 is not the Frattini subgroup of 119866
Theorem 14 Let 119866 be a finite group Suppose that a subgroup119867 with cyclic centre of 119866 is not an abelian 119901-group |119867| = 119901
119899119902 isin 120587(119866) (119902 119901119896 minus1) = 1 and 1 le 119896 le 119899minus1 and then119867 is notthe Frattini subgroup of 119866
Proof Let 119867 = Φ(119866) and by Theorem 13 we know that|119867| gt 119901
3 Let119867 be a counterexample of minimal order Since119867 = Φ(119866) ⊴ 119866 119885(119867) ⊲⊲ 119867 we get that 119885(119867) ⊴ 119866 Let119888 isin 119885(119867) and then |119888| = 119901 Since 119885(119867) is cyclic groupwe have that ⟨119888⟩ ⊲⊲ 119885(119867) ⟨119888⟩ ⊴ 119866 and thus Φ(119866⟨119888⟩) =
Φ(119866)⟨119888⟩ = 119867⟨119888⟩ so 119885(119867⟨119888⟩) cannot be cyclic groupotherwise by the assumption we have that 119867⟨119888⟩ must bean abelian group Since 119867⟨119888⟩ is cyclic we have that 119867 is anabelian group a contradiction since 119885(119867⟨119888⟩) is not cyclicgroup Let119876 be a groupwhich is generated by elementswith119901
order then119876 is an elementary abelian 119901-group |119876| ge 1199012 and
119876 ⊴ 119866 Let119876 = ⟨1198861⟩timessdot sdot sdottimes⟨119886
119904⟩ 119904 le 119899minus1 and then |Aut(119876)| =
(119901119904minus1)(119901
119904minus119901) sdot sdot sdot (119901
119904minus119901119904minus1
) = 119901119904(119904minus1)2
(119901119904minus1) sdot sdot sdot (119901minus1) For
any119892 isin 119866119902119866119902is a Sylow 119902-subgroup of119866 119902 = 119901 Set119909120590119892 = 119909
119892119909 isin 119876 so 120590
119892isin Aut(119876) thus |120590
119892|| 119902119894 |120590119892| || Aut(119876)| By
the assumption we have that 120590119892= 1 For any 119866
119901isin Syl119901(119866)
we have that 119867 = Φ(119866) lt 119866119901 Since 119876 ⊴ 119866
119901 we have that
1 lt 119872 lt 119873 le 119876119872119873 ⊴ 119866119901 |119872| = 119901 and |119873| = 119901
2 Let119872119873 be pre-images of 119872 119873 in 119867 respectively so 119872119873 ⊴ 119866
119901
We know that 119866119902
lt 119862119866(119873) 119862
119866(119872) so 119872119873 ⊴ 119866 By
Lemma 5 we have that |119866 119862119866(119872)| le 119901 so Φ(119866) le 119862
119866(119872)
and thus 119872 le 119885(Φ(119866)) = 119885(119867) It follows that 119872 is cyclicgroup Since 119873119872 is cyclic so 119873 is abelian group Since 119876
is an elementary abelian group we know that 119873 is not cyclicgroup By Lemma 6 we have that119873 le 119885(119867) a contradictionso the counterexample of minimal order does not exist and119867 is not the Frattini subgroup of 119866
Corollary 15 Let119866 be a finite group and119867 le 119866 Suppose that119867 = 119878
1199011
times sdot sdot sdot 119878119901119895
times sdot sdot sdot times 119878119901119899
where 119878119901119894
is Sylow 119901119894-subgroup
of 119867 119878119901119895
with cyclic centre is not an abelian group 119902 isin 120587(119866)
(119902 119901119896
119895minus 1) = 1 1 le 119896 le 119901
119897119895minus1
119895 and |119878
119901119895
| = 119901119897119895
119895 and then 119867 is
not the Frattini subgroup of 119866
Acknowledgments
The research of the authors is supported by the National Nat-ural Science Foundation of China (10771132) SGRC (GZ310)
and the Research grant of Tianjin Polytechnic University andShanghai Leading Academic Discipline Project (J50101)
References
[1] Z Bozikov ldquoFinite 2-groups with a nonabelian Frattini sub-group of order 16rdquo Archiv der Mathematik vol 86 no 1 pp11ndash15 2006
[2] D J S Robinson A Course in the Theory of Groups vol 80Springer New York NY USA 1993
[3] B Huppert Endliche Gruppen I Springer Berlin Germany1967
[4] P C Wang ldquoSome sufficient conditions of a nilpotent grouprdquoJournal of Algebra vol 148 no 2 pp 289ndash295 1992
[5] M Deaconescu Frattini-Like Subgroups of Finite Groups vol 2part 4 ofMathematical Reports 1986
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Algebra 3
Proof Suppose that the theorem is false and let 119867 = Φ(119866)Then119867 ⊴ 119866 119885(119867) ⊲⊲ 119867 so 119885(119867) ⊴ 119866 Let ⟨119888⟩ = 119885(119867) so|119888| = 119901 and119867⟨119888⟩ is not cyclic group For any 119866
119901isin Syl119901(119866)
we have that 119867 = Φ(119866) le 119866119901 so 119867⟨119888⟩ ⊴ 119866
119901⟨119888⟩ and then
119867⋂119885(119866119901) = 1 Let ⟨119886⟩ le 119885(119866
119901) and then119867 = ⟨119886⟩times⟨119887⟩ is an
elementary abelian 119901-group so |Aut(119867)| = 119901(119901 minus 1)2(119901 + 1)
By the assumption we have that 119866119902lt 119862119866(119867) 119902 = 119901 119866
119902is a
Sylow 119902-subgroup of 119866 so 119876 = ⟨119886 119888⟩ ⊴ 119866 but |119876| = 1199012 By
Lemma 5 we have that |119866 119862119866(119876)| le 119901 so Φ(119866) = 119867 le
119862119866(119876) and thus 119876 le 119885(119867) but |119876| = 119901
2 a contradiction so119867 is not the Frattini subgroup of 119866
Theorem 14 Let 119866 be a finite group Suppose that a subgroup119867 with cyclic centre of 119866 is not an abelian 119901-group |119867| = 119901
119899119902 isin 120587(119866) (119902 119901119896 minus1) = 1 and 1 le 119896 le 119899minus1 and then119867 is notthe Frattini subgroup of 119866
Proof Let 119867 = Φ(119866) and by Theorem 13 we know that|119867| gt 119901
3 Let119867 be a counterexample of minimal order Since119867 = Φ(119866) ⊴ 119866 119885(119867) ⊲⊲ 119867 we get that 119885(119867) ⊴ 119866 Let119888 isin 119885(119867) and then |119888| = 119901 Since 119885(119867) is cyclic groupwe have that ⟨119888⟩ ⊲⊲ 119885(119867) ⟨119888⟩ ⊴ 119866 and thus Φ(119866⟨119888⟩) =
Φ(119866)⟨119888⟩ = 119867⟨119888⟩ so 119885(119867⟨119888⟩) cannot be cyclic groupotherwise by the assumption we have that 119867⟨119888⟩ must bean abelian group Since 119867⟨119888⟩ is cyclic we have that 119867 is anabelian group a contradiction since 119885(119867⟨119888⟩) is not cyclicgroup Let119876 be a groupwhich is generated by elementswith119901
order then119876 is an elementary abelian 119901-group |119876| ge 1199012 and
119876 ⊴ 119866 Let119876 = ⟨1198861⟩timessdot sdot sdottimes⟨119886
119904⟩ 119904 le 119899minus1 and then |Aut(119876)| =
(119901119904minus1)(119901
119904minus119901) sdot sdot sdot (119901
119904minus119901119904minus1
) = 119901119904(119904minus1)2
(119901119904minus1) sdot sdot sdot (119901minus1) For
any119892 isin 119866119902119866119902is a Sylow 119902-subgroup of119866 119902 = 119901 Set119909120590119892 = 119909
119892119909 isin 119876 so 120590
119892isin Aut(119876) thus |120590
119892|| 119902119894 |120590119892| || Aut(119876)| By
the assumption we have that 120590119892= 1 For any 119866
119901isin Syl119901(119866)
we have that 119867 = Φ(119866) lt 119866119901 Since 119876 ⊴ 119866
119901 we have that
1 lt 119872 lt 119873 le 119876119872119873 ⊴ 119866119901 |119872| = 119901 and |119873| = 119901
2 Let119872119873 be pre-images of 119872 119873 in 119867 respectively so 119872119873 ⊴ 119866
119901
We know that 119866119902
lt 119862119866(119873) 119862
119866(119872) so 119872119873 ⊴ 119866 By
Lemma 5 we have that |119866 119862119866(119872)| le 119901 so Φ(119866) le 119862
119866(119872)
and thus 119872 le 119885(Φ(119866)) = 119885(119867) It follows that 119872 is cyclicgroup Since 119873119872 is cyclic so 119873 is abelian group Since 119876
is an elementary abelian group we know that 119873 is not cyclicgroup By Lemma 6 we have that119873 le 119885(119867) a contradictionso the counterexample of minimal order does not exist and119867 is not the Frattini subgroup of 119866
Corollary 15 Let119866 be a finite group and119867 le 119866 Suppose that119867 = 119878
1199011
times sdot sdot sdot 119878119901119895
times sdot sdot sdot times 119878119901119899
where 119878119901119894
is Sylow 119901119894-subgroup
of 119867 119878119901119895
with cyclic centre is not an abelian group 119902 isin 120587(119866)
(119902 119901119896
119895minus 1) = 1 1 le 119896 le 119901
119897119895minus1
119895 and |119878
119901119895
| = 119901119897119895
119895 and then 119867 is
not the Frattini subgroup of 119866
Acknowledgments
The research of the authors is supported by the National Nat-ural Science Foundation of China (10771132) SGRC (GZ310)
and the Research grant of Tianjin Polytechnic University andShanghai Leading Academic Discipline Project (J50101)
References
[1] Z Bozikov ldquoFinite 2-groups with a nonabelian Frattini sub-group of order 16rdquo Archiv der Mathematik vol 86 no 1 pp11ndash15 2006
[2] D J S Robinson A Course in the Theory of Groups vol 80Springer New York NY USA 1993
[3] B Huppert Endliche Gruppen I Springer Berlin Germany1967
[4] P C Wang ldquoSome sufficient conditions of a nilpotent grouprdquoJournal of Algebra vol 148 no 2 pp 289ndash295 1992
[5] M Deaconescu Frattini-Like Subgroups of Finite Groups vol 2part 4 ofMathematical Reports 1986
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
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