If $N\lhd G$, $|N|$ finite, $H < G$, $[G: H]$ finite, and $[G: H]$ and $|N|$ are relatively prime, then $N < H$.

Question

The following is Exercise I.5.20. from Hungerford's Algebra (GTM 73).

If $N\lhd G$, $|N|$ finite, $H < G$, $[G: H]$ finite, and $[G: H]$ and $|N|$ are relatively prime, then $N < H$.

I only found the problem here, where $G$ is assumed to be finite. But when $G$ is infinite I don't know what to do. The second isomorphism theorem does not apply here. Any hints will be appreciated!

Answer 1

We claim that $[N: H\cap N] =1$, but it follows from $[N:H\cap N]$ is a divisor of $|N|$ and $[G:H]$.

The assertion $[N: H\cap N]\mid [G:H]$ may need a proof, and it is a consequence of $[NH : H] = [N: H\cap N]$.