finetely generated Abelien -by-nilpotent group Let G finetely generated Abelien -by-nilpotent group (i.e there existe a  abelien subgroup H  in G and G/H is nilpotent )With each of its two-generator is nilpotent-by-finite show that
 G is nilpotent-by-finite (i.e there existe a  nilpotent  subgroup M  in G and G/M is finite ).
 I could'nt  continue this argument  using"  two-generator sub-group of G is nilpotent-by-finite".
 A: Here are some thoughts. We are given $H \unlhd G$ with $H$ abelian and $G/H$ nilpotent. Choose $N \lhd G$ with $H \le N$ and $G/N$ infinite cyclic. (If there is no such $N$ then $G/H$ is finite and we are done.)
First show that $N$ is finitely generated. Since $G$ is finitely generated, there is a finite generating set $a,n_1,n_2,\ldots,n_k$ of $G$ with each $n_i \in N$. By applying the 2-generator condition, each subgroup $G_i = \langle a,n_i \rangle$ is nilpotent by finite, so $G_i \cap N$ is finitely generated, and since these subgroups generate $N$, so is $N$.
So by induction on the Hirsch length of $G/H$ (i.e. the number of infinite factors in a polycyclic series), $N$ is nilpotent by finite, and then, by passing to a finite index subgroup of $G$, we can assume that $N$ is nilpotent.
Now we use induction on the Hirsch length of $N$. Let $b \in Z(N)$ have infinite order (if there is no such $b$ then $N$ is finite and we are done). Now, let $K$ be a normal nilpotent subgroup of finite index in $\langle a,b \rangle$, and choose an element $c \in Z(K) \cap N$ of infinite order. Then (since $c \in Z(N)$), $C_G(c)$ has finite index in $G$, so we can assume $c \in Z(G)$ and now we can aply induction to $G/\langle c \rangle$, completing the proof.
