Show that the center $\mathcal{Z}(G) =\{g \in G ∶ ga=ag \text{ for all a} \in G\}$ is abelian

Question

I see help for showing $G/\mathcal{Z}(G)$ is abelian, but not $\mathcal{Z}(G)$. By definition isn't $\mathcal{Z}(G)$ abelian? How can I use the fact that $\mathcal{Z}(G)$ is a subgroup of $G$, because that is all I can think of to use.

Answer 1

The centre of $G$ is the set of all elements of $G$ that commute with all other elements of $G$; in particular, these elements commute with all elements of the centre of $G$, since $\mathcal{Z}(G)\subseteq G$. Hence $\mathcal{Z}(G)$ is abelian.

Answer 2

$g \in Z$, $a \in Z \implies g \in Z$, $a \in G \implies ga=ag$.