Past exam of the algebra course of the University of Oxford / 2009 / B / 4 / a

For the Abelian case of Cauchy theorem for groups, use induction on $|G|$. Choose $a\ne e$. If $p$ divides $m=|a|$, then $a^{m/p}$ has order $p$. Otherwise $⟨a⟩$ is a nontrivial normal subgroup of order prime to $p$, and $G/⟨a⟩$ has smaller order still divisible by $p$. Induction gives a coset of order $p$; any lift $b$ has order divisible by $p$, because the order of its image divides its own order. A suitable power of $b$ then has order $p$. This uses no classification of finite Abelian groups.

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 every finite group of even order contains an involution, without using Cauchy theorem for groups.