The center of a group is an abelian subgroup

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.