Let $G$ be a group, $H\le G$, $N\unlhd G$ with $\gcd(|N|, |G:N|)=1$. Show that $H\le N$ iff $|H|$ divides $|N|$.
So, obviously we can see that if $H\le N$ we have that $|H|$ divides $|N|$. Proving the other way, that $|H|$ divides $|N|$ implies $H\le N$, seems tricker.
I see that I've got that $|G:N|$ is coprime to $|N|,$ so if $|H|$ divides $|N|,$ it can't be in my quotient group of $G/N$ ... but I don't see why I even care about that. There is most likely some connection between $H$ not being contained in the quotient $G/N$ and therefore needing to be contained within $N$ that I can't find (although thats a longshot guess).
Thanks.