= Conjugacy class count bound from the center
{title2=$k(G)\leq(|G|+|Z(G)|)/2$}
For a <finite group> $G$, write $k(G)$ for its number of <conjugacy classes>. Exactly the elements of the <center of a group> $Z(G)$ have <singleton> <conjugacy classes>; every other <conjugacy class> has at least two elements. Since the <conjugacy classes> form a <partition of a set> $G$, $k(G)-|Z(G)|\leq(|G|-|Z(G)|)/2$. If $G$ is a <nonabelian group>, the <cyclic quotient by the center> criterion and the fact that every <group> of <group order> at most three is a <cyclic group> give $[G:Z(G)]\geq4$, hence $k(G)\leq5|G|/8$. The <quaternion group> attains this bound, with five <conjugacy classes> among eight elements.
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