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.

I have the solution in my old notes:
$G/H$ is nilpotent, for example of class $m$, so $L_{m+1}(G/H)=H/H$. We can probe this fact that $L_{m+1}(G/H)=L_{m+1}(G)H/H$ so,$$L_{m+1}(G)H/H=H/H$$ or $L_{m+1}(G)\subset H$. Since $H\subset Z(G)$ then $L_{m+1}(G)=[L_m(G),G]\leq[H,G]=\{e\}$. This is what we need
