Prove that any subgroup of the center $Z(G)$ of $G$ is a normal subgroup of $G$.

Let $H$ be such a subgroup and let $h \in H$. Then, for all $g \in G$, $g h g^{-1} = h g g^{-1} = h \in H$. Therefore, $H$ is normal.

Let $H < Z(G)$. Recall that $Z(G) = \{g \in G : gh = hg \hspace{0.1cm} \forall \hspace{0.1cm}h \in G\}$.

(You can multiply by $g^{-1}$ in the definition to have $ghg^{-1}=h$ if you prefer to see conjugancy here)

Now take $g \in G$, for any $h \in H$, in particular $H \in Z(G)$ hence $ghg^{-1}= h$, in other words you said that $gHg^{-1} = H$ for every $g \in G$ and $h \in H$, hence $H$ is normal in $G$.

Everything in the center commutes with every other element of the group. So if $H\le Z (G)$, we easily get for any $g\in G $ that $gHg^{-1}=gg^{-1}H=H $.