## If a finite group has order 2k where k is odd, then it has a subgroup of index 2

Prove that if $|G| = 2k$ where $k$ is odd, then $G$ has a subgroup of index 2.

$G$ contains an element $x$ of order 2 by Cauchy’s Theorem. Let $\pi : G \rightarrow S_G$ be the left regular representation of $G$. By this previous exercise, $\pi(x)$ is a product of $k$ disjoint 2-cycles. Since $k$ is odd, $\pi(x)$ is an odd permutation. By this previous exercise, $\pi[G]$ has a subgroup of index 2.