If $G / Z(G)$ nilpotent then G is nilpotent.

Question

Let $G$ be a group. If $G / Z(G)$ nilpotent, then prove that $G$ is nilpotent.

The way I have worked out this question is fairly tedious, and I would be interested if there is a more straightforward approach here.

I solved this problem by showing that if $G / Z(G)$ is of nilpotence class $k$, then $G / Z(G) = Z_{k}(G / Z(G)) = Z_{k+1}(G) / Z(G)$, the latter equality shown using a tedious induction proof. Then the conclusion readily follows.

Is there a different approach to this problem, or is there a short way to prove that $Z_{k}(G / Z(G)) = Z_{k+1}(G) / Z(G)$?

Answer 1

Hint: given a central series $$ Z(G)/Z(G)\lhd H_1/Z(G)\lhd H_2/Z(G)\lhd \dots \lhd H_{k-1}/Z(G)\lhd H_k/Z(G)=G/Z(G) $$ of $G/Z(G)$, then $$ \{1\}\lhd Z(G)\lhd H_1\lhd H_2\lhd \dots \lhd H_{k-1}\lhd H_k=G, $$ is a central series for $G$.

Answer 2

Along the lines of egreg's answer, you might consider proving (and using) the more general statement:

A group $G$ is nilpotent if and only if, for every normal subgroup $H$, both $H$ and $G/H$ are nilpotent.

The argument is the one he suggested.