Given a a subgroup $H$ of $G$ with index $3$, we have to show there is a subgroup $K$ of $G$ with index $2$, assuming that $H$ is not a normal subgroup of $G$.
My line of thinking was the following:
So $[G:H]=3$ and $H$ is not normal. This means there is a smaller prime which can divide $G$, thus, $2$ divides $G$. We then use Cauchy's theorem to say that there is an element $g$ in $G$ with order $2$.
Will it work to then use the permutation representation of left multiplication on $G$ and the existence of an odd permutation?
Any hint in the right direction will be much appreciated. Thanks.