Let $G$ be a group and let $N$ be a normal subgroup of $G$ of finite index. Show that if $H$ is a finite subgroup of $G$ whose order is coprime with $[G:N]$, then $H$ is a subgroup of $N$.
I don't know what to do in this exercise so I would appreciate hints and suggestions to have an idea of how could I solve it.
$HN/N \cong H/(H\cap N)$. Hence the finite number $|HN/N|$ divides index$[G:N]=|G/N|$. But being equal to $|H/(H \cap N)|$ is also divides $|H|$. Since gcd($|G/N|,|H|)=1$, we must have $HN=N$, that is $H \subseteq N$.