Let $H$ be a subgroup of a group $G$ of index $3$. Prove that either $H$ is normal, or
that $H$ has a subgroup $N$ of index $2$ in $H$ such that $N$ is normal in $G$.
All I could show is that if $H$ is not normal then $2||G|$ and so $2| |H|$. I am not sure how to proceed
Best Answer
Let $G$ be a group and let $H$ be a subgroup of index $3$. Let $\phi:G\to S_3$ be the permutation action on the cosets of $H$. The kernel $N$ of this action is a normal subgroup of $G$ contained in $H$ (the core of $H$ in $G$), and has index dividing $3!=6$. If $N=H$ then $H\lhd G$ and we are done. If $|G:N|=6$ (the only other option as $\mathrm{im}(\phi)$ is a subgroup of $S_3$ of order a multiple of $3$) then $G/N\cong S_3$. In particular, $G/N$, and hence $G$, have a subgroup $N_0$ of index $2$. The subgroup $N$ is then $H\cap N_0$.