There is a theorem that states: "Every polycyclic group has a normal poly-infinite cyclic subgroup of finite index. "
I just read the proof of it and honestly found some difficulties in it. The main part of the proof is as follows:
Let {$G_{i}$}, 0 ≤ $i$ ≤ $k$ be a cyclic series in a polycyclic group G. If $i$ < 1, then G is cyclic and the result is obvious. Let $i$ > I and put N = $G_{n-1}$· By induction on $i$ there is a normal subgroup $M$ of $N$ such that $M$ is poly-infinite cyclic and $N$/$M$ is finite. Now $N$/$M_{G}$ is finite because it is a finitely generated torsion group.Thus nothing is lost if we assume that $M$ is normal in $G$. If $G$/$N$ is finite, so is $G$/$M$ and we are finished. Assume therefore that $G$/$N$ is infinite cyclic, generated by $xN$ say and ....
My question is refered to the proof when it starts with $N$/$M_{G}$. Why $N$/$M_{G}$ would be finitely generated? Am I right that normalithy of $N$ in $G$ makes $N$/$M_{G}$ to be torsion group?
$M_{G}$ is intersection of all $g^{-1}M$$g$ where $g$ is in $G$.