Solvable subgroups in groups with self-normalizing subgroup

Abstract

We study the structure of some solvable finite subgroups in groups with self-normalizing subgroup.The investigation of groups with given properties of a system of subgroups is one of the main directions in general group theory.In the present paper, we study the structure of solvable finite subgroups containing a fixed element of prime odd order in groups with self-normalizing subgroup.Groups with self-normalizing subgroups are often encountered in the works of Shunkov and his disciples (see [1,2]). Infinite Frobenius groups are a special case of these groups. Therefore, we consider the problem of obtaining information on the structure of solvable subgroups of the form T a l( ), which are the main tool for the investigation of groups with self-normalizing subgroups. Theorem 1. Suppose that G is a group, H is its subgroup that has a finite periodic part, N H G ( ) = H is an element of prime order p ≠ 2 from H , and the normalizer of any nontrivial ( ) a -invariant finite subgroup from H is contained in H.Then any finite solvable subgroup K of the form T a l( ) from G that contains a and does not belong to H has the form K nilpotent radical of the group K and M is a Sylow 2-subgroup from K.To prove the theorem, we first prove several lemmas. Let G be a group, let H and K be its subgroups, and let a be an element from H for which the conditions of the theorem are satisfied. Assume that K = K L K / ( ) and L K ( ) is the nilpotent radical of the group K . A group has a finite periodic part if the set of all its elements of finite order forms a finite subgroup. Lemma 1. The intersection L K ( ) ∩ H is trivial.Proof. The nilpotent radical L K ( ) of the group K is not contained in the subgroup H because otherwise, by virtue of the the conditions of the theorem and the fact that L K ( ) is an ( ) a -invariant subgroup from H, we get K < H, which contradicts the fact that the subgroup K does not belong to H.Let D = L K ( ) ∩ H ≠ 1. It is obvious that D is an ( ) a -invariant subgroup and, according to the result proved above, D ≠ L K ( ). Since L K ( ) is a nilpotent group, by virtue of the normalizer condition in nilpotent groups each proper subgroup of it differs from its normalizer. Since, according to the conditions of the theorem, N D L K ( ) ( ) < H, we conclude that H intersects L K ( ) along a subgroup greater than the group D. Thus, we arrive at a contradiction. Therefore, L K ( ) ∩ H = 1. The lemma is proved.