## If the left regular representation of a finite group G contains an odd permutation, then G has a subgroup of index 2

Let $G$ be a finite group and $\pi : G \rightarrow S_G$ the left regular representation. Prove that if $\pi[G]$ contains an odd permutation, then $G$ has a subgroup of index 2.

If $\pi[G] \leq S_G$ contains an odd permutation, then $\pi[G] \not\leq A_G$. Now $A_G \vartriangleleft S_G$ has index 2, which is prime. By this previous exercise, we have $[ \pi[G] : A_G \cap \pi[G] ] = 2$. Since the action of $G$ on itself by left multiplication is faithful, $G \leq S_G$ can be identified with $\pi[G]$. Thus $G$ has a subgroup of index 2; namely the preimage of $A_G \cap \pi[G]$.