Let $G$ be a group with center $Z(G)$ and let $\sim$ denote the conjugancy equivalence relation on $G$ (that is, $a \sim b$ iff $\exists g \in G$ s.t. $a = gbg^{-1}$). Finally let $[a]$ denote the conjugacy class which contains $a \in G$.
Now if $a \in Z(G)$ it is clear that $[a] = \{a\}$, for if $b = gag^{-1}$, then $b = a$ so that $[a] = \{a\}$ as desired.
Yet it is not clear to me why the converse is true (that is if $[a] = \{a\}$ that we necessarily have $a \in Z(G)$). If we assume that $[a] = \{a\}$, then clearly there exists a $g \in G$ s.t. $a = gag^{-1}$ (since $\sim$ is an equivalence relation which means $a \sim a$ $\forall a \in G$). But we cannot know that $a \in Z(G)$ until all such elements $g \in G$ satisfy this property, not merely a single one of them. How do I show this?