The center of a group is an abelian subgroup

Let $(G,\circ)$ be a group and let $Z(G):=\{x \in G : ax=xa \ \forall \ a \in G\}$ be the center of $G$.

How can I show that $Z(G)$ is an abelian subgroup of $G$?

What I did so far:

$Z(G)$ is a subgroup if $$a,b \ \in Z(G) \implies a\circ b^{-1} \in Z(G)$$ But I don't know/understand how I can show this. And still there is missing the commutativity. Maybe someone can help me out with this!

Here is one part:

An element of the center commutes with all elements of $G$. In particular, an element of the center commutes with all elements of the center. Hence, the center is abelian.

Let $a,b \in Z(G)$, to prove $ab^{-1}\in Z(G)$, we have to prove that $$ (ab^{-1})c = c(ab^{-1}), \qquad \text{all }c \in G$$ So let $c \in G$, we have \begin{align*} ab^{-1}c &= a(c^{-1}b)^{-1}\\ &= a(bc^{-1})^{-1} & \text{as $b \in Z(G)$}\\ &= acb^{-1}\\ &= cab^{-1} & \text{as $a \in Z(G)$} \end{align*} For the "abelian" part, see @lhf's answer.