Prove that the center of a group is a normal subgroup

As you said in a comment you already showed that it is normal. So I will only show that it is a subgroup.

Clearly it contains $e$, since $eg = ge$.

Now, we will show that it is closed. Let $a,b \in H$, we know that $\forall g: ag = ga$ and $gb = gb$. Thus, $gab = agb = abg$ and thus $ab \in H$.

Now we only have to show that every $h \in H$ has an inverse and we are done. Let $h \in H$, we know that $\forall g \in G: gh = hg$, thus $$\begin{align*}h^{-1}(gh)h^{-1} &= h^{-1}(hg)h^{-1}\\ h^{-1}g(hh^{-1}) &= (h^{-1}h)gh^{-1}\\ h^{-1}(ge) &= (eg)h^{-1}\\ h^{-1}g &= gh^{-1} \end{align*}$$

Which implies that $h^{-1} \in H$.

sxd has shown that it is a subgroup.

That it is normal follows from here:

Let $x\in Z(G)$ (center of $G$).

Then for any $g\in G$, $gxg^{-1}=gg^{-1}x=x\in Z(G)$.

This proves $Z(G)$ is a normal subgroup.