suppose that $H \leq G$ and $N \unlhd G$ and $(|H|,|G:N|) = 1$ then prove that $H \leq N$

Question

suppose that $H \leq G$ and $N \unlhd G$ and $(|H|,|G:N|) = 1$ then prove that $H \leq N$ I don't know how to start ! I know that I should prove that $H \subseteq N$!Can N be infinite subgroup?Can any one help me to prove this?

Answer 1

Let $n=|H|$, $m=|G:N|$. Then for $h\in H$,

$$(hN)^n=h^nN=1N=N$$

so the order of $hN$ divides $n$, and it also divides $m$ by Lagrange's Theorem because $|G/N|=m$. But the only positive integer dividing both of these is $1$, so $|hN|=1$, i.e. $hN=N$ and $h\in N$. Therefore $H\le N$.

Answer 2

Hint: Consider the image of $H$ under the canonical projection $G\to G/N$.