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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?

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up vote 2 down vote accepted

Use the fact that $[G:K] = [G:H][H:K]$ and that every group of order $15$ is cyclic.

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