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 $G$.

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.
