Let $G$ be a finite group, $H\le G$ and $N\unlhd G$ such that $\gcd(|H|,[G:N])=1$. Prove that $H\le N$.

Question

Task I'm trying to solve is:

Let $G$ be a finite group, $H\le G$ and $N\unlhd G$ such that $\gcd(|H|,[G:N])=1$. Prove that $H\le N$.

I have the following proof but I'm really not confident with it:

Since $\gcd(|H|, [G : N]) = 1$, this implies that $H\cap G/N = \{e\}$. We also know from the 2nd isomorphism theorem that $HN<G$. However, this implies that $H<N$ since otherwise $H\cap G/N = \{e\}$ would not hold.

Am I making sense?

Answer 1

By the Second Isomorphism Theorem, we have that $HN$ is a subgroup of $G$ with order $$|HN| = \frac{|H|\, |N|}{|H \cap N|}.$$ By Lagrange's Theorem, we have that $$|G| = [G : HN] \, |HN| = \frac{[G : HN] \, |H| \, |N|}{|H \cap N|} = \frac{[G : HN] \, |H| \, |G|}{[G : N] \, |H \cap N|}.$$ Cancelling the factor of $|G|$ from both sides and clearing denominators gives us that $[G : N] \, |H \cap N| = [G : HN] \, |H|.$ By hypothesis that $\gcd(|H|, [G : N]) = 1,$ we must have that $|H|$ divides $|H \cap N|$ so that $|H| \leq |H \cap N|.$ But this implies that $|H| = |H \cap N|.$ We conclude therefore that $|HN| = |N|$ so that $HN = N,$ from which it follows that $H \leq N.$ QED.

Answer 2

$(|H|,|G:N|)=1$, so there exist integers $m$ and $n$ such that $$m|G:N|+n|H|=1.$$

Next we only need to verify that $H\subset N$. For every $h\in H$, \begin{align*} hN & =h^{m|G:N|+n|H|}N \\ & =h^{m|G:N|}N \\ & =(hN)^{m|G:N|} \\ & =N, \end{align*} which means that $h\in N$, therefore $H\subset N$.