If $G/Z(G)$ is Abelian and $\{e\}\ne H\triangleleft G$, then $H\cap Z(G)\ne\{e\}$

We start with a

Lemma Let $H \unlhd G$ with $H \cap G’=1$. Then $H \subseteq Z(G)$.
Proof Let $h \in H$ and $g \in G$ be arbitrary. Then the commutator $[g,h]=g^{-1}h^{-1}gh=(g^{-1}h^{-1}g)h \in H$, since $H$ is normal. But also $[g,h] \in G’$, hence since $H$ and $G’$ intersect trivially we get $[g,h]=1$ and $H$ must be central.

Now assume that $G/Z(G)$ is abelian. This is equivalent to $G'\subseteq Z(G)$. If $1 \neq H \unlhd G$, then we have two cases: either $H \cap G' = 1$ and then by the Lemma we have $H \subseteq Z(G)$, so certainly $H \cap Z(G)=H \neq 1$. If $H \cap G' \neq 1$, then since $G’\subseteq Z(G)$, we get $1 \neq H \cap G’\subseteq H \cap Z(G)$ and we are also done.

Note that $G/Z(G)$ being abelian implies that $G$ is nilpotent. In general, if $G$ is nilpotent and $1 \neq H \unlhd G$, then $H \cap Z(G) \neq 1$.