Does center contains commutator subgroup imply group is nilpotent?

Question

Assume that $G^{'}\leq Z(G)$. Show that G is nilpotent.

I show that if $G^{'}\leq Z(G)$ then $G/Z(G)$ is abelian and in particular $G/G^{'}$ is the largest abelian quotient group of $G$. but what I can do after this?

Answer 1

As you know $G/Z(G)$ is abelian; in particular $G/Z(G)$ is nilpotent. Now you can use the result of this question.

Answer 2

Nilpotence is generally defined in terms of either the ascending or descending central series.

Descending: If $[G,G]$ is central in $G$ then what is the next term, $[[G,G],G]$?

Ascending: If $G/Z(G)$ is abelian then what is $Z(G/Z(G))$? Is it proper in $G/Z(G)$? So then what does it correspond to in $G$ (via lattice theorem)? So what's the next term in this series?

Answer 3

\begin{equation*} \gamma^3 G = [\gamma^2 G, G] = [G', G] \leq [Z(G), G] = 1 \end{equation*}