The weak minimal condition on subgroups which fail to be close to normal subgroups

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

“…Note that if G = A ⋊ B is the holomorph group of the additive group A of the rational numbers by the multiplicative group B of positive rationals (acting by usual multiplication), then, as noticed in [8], the only subnormal non-normal subgroups of G are those contained in A (which has rank 1) so that G is T [+]. However all proper non-trivial subgroups of A are not cn, since if they were cn then they were cf (see [6], Proposition 1) contradicting the fact that A is minimal normal in G.…”

“…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 .…”

“…Among all counterexamples choose G in such a way that X has the smallest possible derived length. Then the derived subgroup Y = X ′ of X is a cn-subgroup by the minimal choice on the derived length of X; on the other hand, Y has finite rank and so Y /Y G is finite (see [6], Proposition 1). Then X/Y G is a finite-by-abelian group of finite rank, and hence its centre Z/Y G = Z(X/Y G ) has finite index in X/Y G by Lemma 4.…”

“…Then X/Y G is a finite-by-abelian group of finite rank, and hence its centre Z/Y G = Z(X/Y G ) has finite index in X/Y G by Lemma 4. Thus Z is a subnormal subgroup of G which has finite index in X, so that the index |Z : Z G | is infinite and hence Z cannot be a cn-subgroup of G (see [6], Proposition 1). Since Y G has finite rank, the Hirsch-Plotkin radical of G/Y G has infinite rank and so G/Y G is also a counterexample; thus replacing G with G/Y G and X with Z/Y G it can be supposed that X is abelian.…”

“…Thus Theorem A yields that G/L p ′ is a T [ * ]-group. Therefore X p L p ′ is a cn-subgroup of G, and hence it is even cf because it has finite rank (see [6], Proposition 1). The p-component of the core (X p L p ′ ) G of X p L p ′ in G is G-invariant, it coincides with the subgroup X p ∩ (X p L p ′ ) G and so has finite index in X p , therefore X p is cf .…”