= Solution
{c}
A <conjugacy class> $C(x)$ is a <singleton> exactly when $gxg^{-1}=x$ for every $g\in G$, equivalently when $gx=xg$ for every $g$. Thus the <singleton> <conjugacy classes> correspond exactly to elements of the <center of a group> $A$, and there are $|A|$ of them.
Every other <conjugacy class> has at least two elements. Since the <conjugacy classes> form a <partition of a set> $G$, if $r$ is the number of these remaining classes then $2r\leq|G|-|A|$. Consequently
$$
\boxed{m(G)=|A|+r\leq|A|+\frac{|G|-|A|}{2}
=\frac{|G|+|A|}{2}.}
$$
This <conjugacy class count bound from the center> needs no classification of the <finite group>.
Back to article page
