I've seen a lot of information about this problem when $G$ is a $p$-group. But that need not be the case here.
Let $G$ be a group such that $G/Z(G)$ is abelian. Let $H$ be a non-trivial normal subgroup of $G$. Show $H\cap Z(G)$ is a non-trivial subgroup of $G$.
Clearly $H\cap Z(G)$ is a subgroup of $G$, but I am not sure how to show non-triviality. How does $G/Z(G)$ being abelian help us? Any hints?
We must assume $G$ is nontrivial, of course, though the problem does not mention it (I guess it is implied by the existence of a nontrivial normal subgroup...).
If $G/Z(G)$ is abelian, then $G$ is nilpotent (of class at most $2$). If $G$ is finite, then it is the product of its $p$-parts, and the problem is reduced to the problem for $p$-groups, which is solved here for nilpotent groups of any class.
But here’s an argument that does not require finiteness of $G$ or knowing about the nilpotency: note that because $G/Z(G)$ is abelian, then $[G,G]\subseteq Z(G)$.
Let $h\neq e$ be an element of $H$. If $h$ is central in $G$, there is nothing left to do. If $h$ is not central in $G$, let $g\in G$ be an element such that $gh\neq hg$. Then $$e\neq [h,g] = h^{-1}g^{-1}hg = h^{-1}(g^{-1}hg)\in H.$$ But we also have $[h,g]\in [G,G]\subseteq Z(G)$, thus exhibiting a non-trivial element of $H\cap Z(G)$.