Prove that the center of a group is a normal subgroup 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 $G$.
 A: I have two different solutions. These are probably too fancy for this problem, but they might be interesting.
For the first solution, define the map $f: G \rightarrow \operatorname{Aut}(G)$ by $g \mapsto \phi_g$, where $\phi_g(x) = gxg^{-1}$ for all $x \in G$. The map $f$ is a homomorphism and $\operatorname{Ker}(f) = Z(G)$. Thus $Z(G)$ is a normal subgroup since the kernel of a homomorphism is always a normal subgroup. The image $\operatorname{Im}(\phi)$ is called the inner automorphism group of $G$ and is denoted $\operatorname{Inn}(G)$. 
For the second solution, recall that for a subgroup $H \leq G$ the normal core of $H$ in $G$ is defined as 
$$\operatorname{core}_G(H) = \bigcap_{g \in G} gHg^{-1}$$
The subgroup $\operatorname{core}_G(H)$ is always a normal subgroup. This can be seen directly or by noticing that it is the kernel of the coset action induced by $H$. For any conjugacy class $C$ of $G$, let $t_c \in C$. Let $\mathscr{C}$ be the family of all conjugacy classes of $G$. Then
\begin{align*}
Z(G) &= \bigcap_{g \in G} C_G(g) \\
&= \bigcap_{C \in \mathscr{C}} \bigcap_{t \in C} C_G(t_c) \\
&= \bigcap_{C \in \mathscr{C}} \bigcap_{g \in G} C_G(gt_cg^{-1}) \\
&= \bigcap_{C \in \mathscr{C}} \bigcap_{g \in G} gC_G(t_c)g^{-1} \\
&= \bigcap_{C \in \mathscr{C}} \operatorname{core}_G(C_G(t_c)) \\
\end{align*}
Since $Z(G)$ is the intersection of normal subgroups, it is a normal subgroup.
A: 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.
A: 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 $h^{-1} \in H$.
