Let G be a finite group, $H \le G$, and $N\lhd G$. Suppose $|H|$ and $|G :N|$ are relatively prime. Is it true that $H \le N$?
Since $N$ is a normal subgroup, I know that $NH \le G \implies |NH|$ divides $|G|$. Also $|HN|=\frac{|H||N|}{H\cap N}$, so using the formula $|G :N|=\frac{|G|}{|N|}$ and manipulating I get,
$\frac{|G|}{|HN|}=\frac{|G||H\cap N|}{|H||N|}$
but I do not know how to continue(assuming I am on the right track to begin with.). Any help is much appreciated.