The Inf l uence of Primitive Subgroups on the Structure of Finite Groups

LIU YU-FENG

(School of Mathematics and Informational Science,Shandong Institute of Business and Technology,Yantai,Shandong,264005)

Communicated by Du Xian-kun

The Inf l uence of Primitive Subgroups on the Structure of Finite Groups

LIU YU-FENG

(School of Mathematics and Informational Science,Shandong Institute of Business and Technology,Yantai,Shandong,264005)

Communicated by Du Xian-kun

A subgroup H of a group G is said to be primitive if it is a proper subgroup of the intersection of all subgroups of G containing H as its proper subgroup.The purpose of this note is to go further into the inf l uence of primitive subgroups on the structure of f i nite groups.Some new results are obtained.

primitive subgroup,supersoluble group,structure of group

1 Introduction

All groups considered in this paper are f i nite and G denotes a f i nite group.The generalized concept of maximal subgroups of a group G,namely,the primitive subgroup,was introduced by Johnson[1]in 1971.He called a subgroup H of a group G primitive if it is a proper subgroup of the intersection of all subgroups of G containing H as its proper subgroup.We denote by H＜primG that H is a primitive subgroup of G.It is interesting to note that every group G has a primitive subgroup and that every proper subgroup of G is the intersection of some primitive subgroups of G.Since the intersection of all primitive subgroups is the identity subgroup,we can easily see that the class of all primitive subgroups is obviously wider than the class of all maximal subgroups.Guo,Shum and Skiba[2]gave the structure of the f i nite group in which every primitive subgroup has a prime power index.They proved that every primitive subgroup of a f i nite group G has a prime-power index if and only if G=[D]M is a supersoluble group,where D and M are nilpotent Hall subgroups of G,D is the smallest term of the lower central series of G and G=DNG(D∩X)for every primitive subgroup X of G.

The purpose of this note is to go further into the inf l uence of primitive subgroups on the structure of f i nite groups.Some new results are obtained.

2 Preliminaries

Recall that the quaternion group is a 2-group with a unique element of order 2,and the generalized quaternion group Q2nof order 2nis the group with the following presentation of the form:

Note that a group G is said to be primary if the order of G is a prime power.

Lemma 2.1([3],Theorem III.8.2)If a p-group has a unique subgroup of order p,then G is a cyclic group or a generalized quaternion group.

Lemma 2.2[1]If every primitive subgroup of G has a prime power index,then G is supersoluble.

3 Main Results

Proposition 3.1If H＜primG,then NG(H)/H is either a cyclic p-group for some prime p or a generalized quaternion group.

Proof.Suppose that the factor group NG(H)/H had two dif f erent subgroups A/H and B/H of prime order.Then H were a proper subgroup of A and of B.H were a primitive subgroup of G,and

This implies that

The contradiction shows that NG(H)/H has a unique subgroup of prime order.By Sylow theorem,NG(H)/H is a primary group.Thus,by Lemma 2.1,we obtain that NG(H)/H is either a cyclic p-group for some prime p or a generalized quaternion group.

Proposition 3.2If the identity subgroup is primitive in G,then G is either a cyclic p-group for some prime p or a generalized quaternion group.

Proof.Obviously,G is a p-group for some prime p.If G had two subgroups P1and P2of order p,then 1＜P1,1＜P2and 1=P1∩P2,which contradicts the fact that 1 is a primitive subgroup.Hence G has only a subgroup of order p.By Lemma 2.1,we see that

G is either a cyclic p-group or a generalized quaternion group.This completes the proof.

As usual,a subgroup H of G is said to be non-trivial if H is neither an identity subgroup nor G.

Theorem 3.1Suppose that all non-trivial subgroups of a group G are primitive.

(1)If G is a p-group for some prime p,then G is a cyclic group or an elementary Abelian group of order p2;

(2)If G is a non-primary nilpotent group,then

where p,q are dif f erent primes;

(3)If G is a non-nilpotent group,then

where P is the minimal normal subgroup of G of order p or p2,Q is a subgroup of order q. Proof.Obviously,the conditions of the theorem hold for all subgroups of G.

(1)Since a p-group of order≤p2is either cyclic or an elementary Abelian group,we may assume that|G|≥p3.If G has only one subgroup of index p,then G has only one maximal subgroup.Hence G is cyclic(see Theorem III.3.16 in[3]).If G had two subgroups M1and M2with index p,then,obviously,their intersection were a non-trivial non-primitive subgroup of G.This is a contradiction.Hence(1)holds.

(2)Assume that G were a non-primary nilpotent group and

where P and Q are a Sylow p-subgroup and a Sylow q-subgroup of G,respectively.If X/=1, then

which contradicts the primitivity of P.Hence

If|P|is not a prime number,then there exists a subgroup P1such that

and

which contradicts the primitivity of P1.This shows that the subgroup P has prime order. Analogously,we can prove that|Q|is also a prime number.Thus

(3)Suppose that G is a non-nilpotent group.Then there exists a self-normalized maximal subgroup M in G.If x∈GM,then M∩Mxis not primitive,and so

This means that G=[N]M is a Frobenius group with a Frobenius complement M and the Frobenius kernel N(see(8.5.5)in[4]).By the maximality of M,we have that N is a minimal normal subgroup of G.By Thompsom theorem(see(10.5.6)in[4]),N is nilpotent. Hence N is an elementary Abelian p-group for some prime p.Let|N|=pn.By(1),we see that n≤2.If X is a subgroup of M of prime order,thenX is a primitive subgroup of G,and X=M.Thus,the order of M is a prime.This completes the proof.

Recall that a subgroup H of G is said to be permutable with a subgroup K if

Let p1,···,pkbe the distinct prime divisors of the order of G.A set of Sylow subgroups {P1,···,Pk}is said to be a Sylow system of G,if PiPj=PjPifor all i,j.

Proposition 3.3If a primitive subgroup V of G is permutable with every subgroup in some Sylow system of G,then the index|G:V|is of prime power.In particular,if every primitive subgroup of G is permutable with all subgroups in some Sylow system of G,then G is supersoluble.

Proof.Suppose that p,q were dif f erent prime divisors of the index|G:V|.By the hypothesis,there would exist a Sylow p-subgroup P and a Sylow q-subgroup Q of G such that

and

Obviously,V is a proper subgroup of V P and of V Q.Let U=V P∩V Q.Then

and

which simultaneously divides the order of P and the order of Q.This implies that

which contradicts the primitivity of V.Therefore|G:V|is a prime power.

If every primitive subgroup of G is permutable with all subgroup in some Sylow system of G,then every primitive subgroup has primary index in G.Hence,by Lemma 2.2,G is supersoluble.This completes the proof.

Proposition 3.4Let P be a p-group.Then every maximal subgroup of P is complemented in P if and only if Ω1(P)=P,where Ω1(P)is the subgroup generalized by all elements of P of order p.

Proof.Assume that Ω1(P)=P.If M is the maximal subgroup of P,then

Therefore there exists an element a of order p such that a/∈M.Obviously,the subgroup〈a〉is a complement of M in P.Conversely,assume that every maximal subgroup of P were complemented in P.If Ω1(P)/=P,then,obviously,every maximal subgroup M containing Ω1(P)were not complemented in P.This is a contradiction.Thus,the proposition is proved.

In the following,we denote by N the class of all nilpotent groups,and Nkdenotes the class of all soluble groups with nilpotent length k.It is well known that N and Nkare all saturated formations(see Theorem 3.1.20 in[5]).

Recall that the product of two non-empty formations F and H are def i ned as follows:

Lemma 3.1Let F be a formation of groups and G be a soluble group.If all primitive subgroups of G belong to F,then G∈NF.

Proof.If MG/=1 for every maximal subgroup M of G,then,obviously,G/MGsatisf i es the hypothesis.Hence G/MG∈NF by induction.It follows that G/Φ(G)∈NF.However NF is a saturated formation(see Theorem 3.1.21 in[5]),we obtain that G∈NF.We may, therefore,assume that there exists a maximal subgroup M such that MG=1.Then,since G is soluble,1/=F(G)is the unique minimal normal subgroup of G(see Theorem 15.2 in[6] or Lemma 2.3.3 in[5]).This implies that G=[F(G)]M,where M is a maximal subgroup of G.Since every maximal subgroup of G is a primitive subgroup of G,by the hypothesis, M∈F.This implies that G∈NF.

Theorem 3.2Let G be a soluble group.

(1)If the nilpotent length of every primitive subgroup of G is no more than a natural number n,then the nilpotent length of G is not more than 1+n;

(2)If the derived length of every primitive subgroup of G is no more than a natural number n,then the derived length of G/Φ(G)is not more than 1+n.

Proof.(1)It is clear that the nilpotent length of a soluble group X is equal to k if and only if X∈NkNk-1.By Lemma 3.1,we obtain that

This means that the nilpotent length of G is not more than 1+n.

(2)It is clear also that the derived length of a soluble group X is equal to k if and only if X∈AkAk-1,where A is the class of all Abelian groups.By Lemma 3.1,we see that G∈NAn.However,since G is soluble,F(X)/Φ(X)is an Abelian group.Hence the derived length of G/Φ(G)is not more than 1+n.

[1]Johnson D L.A note on supersoluble groups.Canad.J.Math.,1971,23:562–564.

[2]Guo W,Shum K P,Skiba A.On primitive subgroups of f i nite groups.Indian J.Pure Appl. Math.,2006,37(6):369–376.

[3]Huppert B.Endliche Gruppen I.Berlin-Heidelberg-New York:Springer-Verlag,1967.

[4]Robinson D J S.A Course in the Theory of Groups.New York-Heidelberg-Berlin:Springer-Verlag,1982.

[5]Guo W.The Theory of Class of Groups.Beijing-New York-Dordrecht-Boston:Science Press-Kluwer Academic Publishers,2000.

[6]Doerk K,Hawkes T.Finite Soluble Groups.Berlin-New York:Walter de Gruyter,1992.

tion:20D10,20D25

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1674-5647(2013)02-0179-05

Received date:Jan.15,2011.

The NSF(11071147)of China.

E-mail address:yf l iu@sdibt.edu.cn(Liu Y F).