The generalized hypercenter and the structure of finite groups

Introduction

All groups considered in this paper will be finite. For a p-group P, we denote

. Two subgroups H and K of a group G are said to permute if HK =KH. It is easily seen that H and K permute if and only if the set HK is a subgroup of G. A subgroup of G is quasinormal in G if it permutes with every subgroup of G. We say, following Kegel [7] , that a subgroup of G is S-quasinormal in G if it permutes with every Sylow subgroup of G. Agrawal [1] defined the generalized center, genz (G), of G to be the subgroup generated by all elements g of G such that g is S-quasinormal in G. The generalized hypercenter, genz  (G), is the largest term of the series

) for all i  0. More recently, Asaad and Ezzat [3] have given a new characterization of genz  (G) by introducing the following definition: A normal subgroup H of a group G is generalized supersolvably embedded, GSE, in G if there exists a

It is easily verified that if H and K are normal GSE subgroups of G, then HK is GSE in G. From this, every group G has a unique maximal generalized supersolvably embedded subgroup of G and it is denoted by GSE (G). The generalized hypercenter, genz  (G), is exactly the maximal generalized supersolvably embedded subgroup GSE (G) (see [3]). Furthermore, they proved the following: (1) Let p be the smallest prime dividing the order of G and let P be a Sylow p-

The present paper represents an attempt to extend and improve the above mentioned results.

Preliminaries

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

Proof: Suppose the result is false and let G be a counter example of minimal order. By [6; I, Satz 18.1], G/P  K and so G is solvable. Then it is easy to show that the hypotheses are inherited by all subgroups of G, so we may assume that G is a minimal non-supersolvable group. Then by Lemma 2.5(iii), P= ) (P  and so G=

. Thus G is supersolvable, a Contradiction.

Main Result

Theorem 3.1 Let p be the smallest prime dividing the order of G and let P be a Sylow p-subgroup of G. If P is quaternionfree and

Proof Suppose the result is false and let G be a counterexample of minimal order. Then G is not p-nilpotent and so G contains a minimal non-p-nilpotent subgroup, K, say. By [6; IV, Satz 5.4], K is a minimal non nil potent subgroup of G. By [6; III, Satz 5.2], |K|= m n q p for a prime q p  , K has a normal Sylow p-subgroup p K of exponent p at p>2 or at most 4 at p=2 and a non-normal cyclic Sylow q-subgroup q K . Without loss of generality, we assume that

is supersolvable. Since p is the smallest prime divisor of |G|, it follows that

, and hence p=2. So, by [6; III, Satz 5.2],

. Now, by applying Lemma 2.2, we conclude that K is 2-nilpotent; a final contradiction. As a corollary of the proof of Theorem 3.1, we have: Corollary 3.2 Let p be the smallest prime dividing the order of G and let P be a Sylow p-subgroup of G. If

genz  (G), then G is p-nilpotent. Now we can prove:

genz  (G) for all Sylow subgroups P of G, then G is supersolvable or G has a section isomorphic to the quaternion group of order 8.

Proof If G has a section isomorphic to the quaternion group of order 8, then the result holds. Thus we can assume that G has no section isomorphic to the quaternion group of order 8. Theorem 3.1 implies that G is r-nilpotent, where r is the smallest prime dividing the order of G. Then G = RK, where R is a Sylow r-subgroup of G and K is a normal Hall

for all Sylow subgroups P of K. By hypothesis and Lemma 2.3,

for all Sylow subgroups P of K. Then K is supersolvable by induction on the order of G. Hence K possesses an ordered Sylow tower and so Q is a normal Sylow q-subgroup of K, where q is the largest prime dividing the order of K. Since Q char K and K  G, we have that Q  G. By

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