If $H\leq Z (G)$ and $G/H$ is nilpotent, then $G$ is nilpotent. Let $G$ be a group. If $H\leq Z(G)$ and $G/H$ is nilpotent, then $G$ is nilpotent. To prove this proposition, first, I have tried to show that $G/Z (G)$ has non-trivial center. I can see that all cosets of $H$ in $Z (G)$ belongs to $Z (G/H)$ but this does not satisfy $G/Z (G)$ has non-trivial center. Can any one give me a hint for the proposition. Thanks.

So you want to show that $G/Z(G)$ has a non-trivial center. First, assume that $G$ is not abelian, or you would be done anyway. Next, use that $G/Z(G)$ is a quotient of $G/H$ since $H$ is contained in $Z(G)$. This tells you that $G/Z(G)$ is abelian. Now you should be able to complete the proof.
