Groups of infinite rank with finite conjugacy classes of subnormal subgroups

Abstract

A group is called a V -group if it has finite conjugacy classes of subnormal subgroups. It is proved here that if G is a periodic soluble group in which every subnormal subgroup of infinite rank has finitely many conjugates, then G is a V -group, provided that its Hirsch-Plotkin radical has infinite rank. Corresponding results for periodic soluble groups in which every subnormal subgroup of infinite rank has finite index in its normal closure and for those in which every subnormal subgroup of infinite rank is finite over its core, are also obtained. Moreover, it is shown that the assumption on the Hirsch-Plotkin radical can be avoided in the case of periodic groups with nilpotent commutator subgroup.

On groups in which subnormal subgroups of infinite rank are commensurable with some normal subgroup

“…In the last decade, several authors have studied the influence on a soluble group of the behavior of its subgroups of infinite rank (see for instance [5,10] or the bibliography in [9]). Recall that a group G is said to have finite rank r if every finitely generated subgroup of G can be generated by at most r elements, and r is the least positive integer with such property and infinite rank is there is no such r. For example, in [8] it was proved that if G is a periodic soluble group of infinite rank in which every subnormal subgroup of infinite rank is normal, then G is a T -group indeed.…”

“…Recall that a group G is said to have finite rank r if every finitely generated subgroup of G can be generated by at most r elements, and r is the least positive integer with such property and infinite rank is there is no such r. For example, in [8] it was proved that if G is a periodic soluble group of infinite rank in which every subnormal subgroup of infinite rank is normal, then G is a T -group indeed. Then in [9], authors have considered groups of infinite rank with properties T + (T + , resp. ), that is groups in which the condition of being nn (resp.…”

“…Replacing G by G/L π ′ it can be supposed that π = π(L). Then there exists an infinite subset π 0 of π such that G/L π ′ 0 contains a nilpotent normal subgroup of infinite rank (see [9], Corollary 11); hence G/L π ′ 0 is a T [ * ]-group by Theorem A. Therefore X π 0 L π ′ 0 is a cn-subgroup of G and so even a cf -subgroup (see [6], Proposition 1); hence X π 0 is cf and this is a contradiction because X p is not normal in G for each p ∈ π 0 .…”