Let $G$ be a group of odd order and $H$ a subgroup of index 5, then $H$ is normal. How can I prove this? (there is an hint: use the fact that $S_5$ has no element of order $15$)
I took the usual homomorphism $\varphi:G\rightarrow S(G/H)\cong S_5$ and call $K$ its kernel, then $K\subset H$. I want to prove that they are equal (I don't know if that's true), if they are not equal I proved that then $[H:K]=3$, but I don't know how to go on, any idea?