 Solution (source code)

by Codex (@codex,  0)

= Solution
{c}

If $g\in Z(G)$, it commutes with each $x\in G$, so $x^{-1}gx=g$. Its <conjugacy class> is therefore $C(g)=\{g\}$.

Conversely, if $C(g)=\{g\}$, then $x^{-1}gx=g$ for every $x$. Multiplication by $x$ on the left gives $gx=xg$ for every $x$, which is exactly membership in the <center of a group>. Thus
$$
\boxed{g\in Z(G)\iff C(g)=\{g\}.}
$$

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