I was trying to understand the costruction of the Heineken and Mohamed groups.
(II) Every proper subgroup of $G$ is subnormal and nilpotent.
Lemma 1. If $G$ is a non-nilpotent soluble group satisfying condition (II) and $G'\neq G$, then
(a) If $G=<U, V>$, then $G=V$ or $G=U$
(b) $G/G'=C_{p^{\infty}}$
(c) $G$ is a countable $p$-group.
(d) $(G')^p\neq G'$
Lemma 2. Assume that $G$ is a non-abelian group possessing an abelian normal subgroup $N$ of finite exponent $p^k$ such that $G/N=C_{p^{\infty}}$. Then
(a) $G$ is not hypercentral.
(b) $G$ satisfies condition (II) iff there is no subgroup $U\neq G$ s.t. $UN=G$.
The problem I have is on the following
Corollary 2. If the hypercentral group $G$ satisfies condition (II), then $G$ is nilpotent.
Proof. This follows from statements (b), (d) of Lemma 1 together with statement (a) of Lemma 2.
We assume for a contradiction that $G$ is not nilpotent. By Lemma 1 we have to find a subgroup $N$ with which we can apply Lemma 2 and deducing the contradiction. My question is: how we can find $N$ with the property required by Lemma 2? (When we use the condition (d)?)
Notation
$p$ is a prime.
Heineken & Mohamed, "Trivial Centre and normalizer Condition", Journal of Algebra 10. 368-376 (1968)