Does center contains commutator subgroup imply group is nilpotent? 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?

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