If $G/Z(G)$ is Abelian and $\{e\}\ne H\triangleleft G$, then $H\cap Z(G)\ne\{e\}$ I am somewhat unfamiliar with commutator subgroups, so I am not sure about the last couple of lines of this proof.

Let $G$ be a group such that $G/Z(G)$ is Abelian, and let $\{e\}\ne H\triangleleft G.$  Show that $H\cap Z(G)\ne\{e\}.$

Consider the commutator subgroup $G'<G.$ For any $N\triangleleft G$ with $G/N$ Abelian, we have that $G'\subset N.$ Thus since $Z(G)$ is a normal subgroup of $G$ and the quotient group is Abelian by assumption, $G'\subset Z(G).$ Now let $g\in G$ and $h\in H$. Then 
$$ghg^{-1}h^{-1} = (ghg^{-1})h^{-1}\in H\cap G'.$$ Now if $ghg^{-1}h^{-1}\ne e$ for some $g,h,$ then we're done. But if $ghg^{-1}h^{-1} = e$ for all $g\in G,\; h\in H$, then this implies $H\subset Z(G)$. 
 A: 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 $H \cap Z(G) \neq 1$.
A: The commutator subgroup $G'$ of $G$ is defined as $$G' = \{ghg^{-1}h^{-1}: g, h \in G\}.$$ The commutator of two elements measures how non-commutative that they are. For instance, if $g$ and $h$ commute then their commutator $ghg^{-1}h^{-1}$ is trivial.
So, to discuss the proof. (We will verify each claim.)
If $N$ is normal and $G/N$ is abelian then $G' < N$. This is true since if $a, b \in G$ then $aNbN = bNaN$ so $a^{-1}bNaN = NbN = NNb$ so letting the elements in the first two $N$'s be the identity gives that there is $x \in N$ such that $a^{-1}ba = xb$, so $N \ni x = a^{-1}bab^{-1}$. However, $a$ and $b$ were arbitrary so by replacing $a$ with $a^{-1}$ we see that $N$ contains all the commutators.
$(ghg^{-1})h^{-1} \in H\cap G' \subset H \cap Z(G)$ since $H$ is normal and the entire product is a commutator.
If $ghg^{-1}h^{-1} \neq e$ for some choice of $h \in H$ and $g\in G$ then we have that $H \cap Z(G) \neq \{e\}$.
If $ghg^{-1}h^{-1} = e$ for all $h \in H$ and $g\in G$, then this means that $H$ commutes with every element of $G$ so $H \subset Z(G)$, so $H \cap Z(G) \neq \{e\}$ since $H$ is non-trivial.
