Show that if a group $G$ has odd order, any subgroup $H$ of index 3 in $G$ is normal in $G$.
I think this is equivalent to the following: Let $H$ and $K$ be subgroups of a group $G$, with $K \leq H$. We can then show that $[G:K]=[G:H] \cdot [H:K]$. However, I'm not sure what the best approach is here. What's a good way to prove this theorem?
This question is different from Normal subgroup of prime index because we are not assuming that $3$ divides the order of $G$.