Solution
= Solution
The map $g\mapsto g^{-1}$ partitions $S$ into disjoint two-element sets $\{g,g^{-1}\}$, so $|S|$ is even. Since $|G|$ is even, $|G\setminus S|$ is even as well. This complement contains the identity and consists of elements with $g^2=e$. It therefore contains at least one further element $h\ne e$, and that element has order two. Thus \b[every finite group of even order contains an involution], without using <Cauchy theorem for groups>.