Let $G$ be a group. We define
$H=\{h\in G\mid \forall g\in G: hg=gh\},$
the center of $G$.
Prove that $H$ is a (normal) subgroup of $G$.
Let $G$ be a group. We define
$H=\{h\in G\mid \forall g\in G: hg=gh\},$
the center of $G$.
Prove that $H$ is a (normal) subgroup of $G$.
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.
i think the kernel argument (Mikko's no 1) is the most elegant approach. but sxd shows with thoroughness that the centre is a group, and it is perhaps worth a passing mention that once we know it is a group its normality is implied by the fact that it consists of complete conjugacy classes, albeit they are all singletons.