Proving that if $| G : N |<\infty$ then $|H: H\cap N| <\infty$, where $N$ is normal to $G$ and $H$ is subgroup of $G$

Question

Prove that if $|G : N |<\infty$ then $|H: H\cap N| <\infty$, where $N$ is normal to $G$ and $H$ is subgroup of $G$

So far I have proven that $H\cap N$ is a subgroup of $H, HN=\{ hn, h\in H,n\in N\} \vartriangleleft H$ in order to use that $|HN|=|H||N|/|H\cap N$|$ and I really don't know how to move on from here.

Some thoughts of mine are of using Lagrange's theorem in its general form:

If $G$ group with $K<H<G$ then $| G : K| = | G : H || H : K|$.

Answer 1

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Since $[G : N]$ is finite, we have $[G: N]=[G:HN]\cdot[HN : N]$. In particular $[G:HN]$, and $[HN : N]$ are finite.

Now, $N$ is normal in $G$, thus we have $N$ is normal in $HN$, $H\cap{N}$ is normal in $H$, and $\frac{HN}{N}\cong\frac{H}{H\cap{N}}$ from second isomorphism theorem. Then $[HN:N]=[H:H\cap N]$, and finally $$[G: N]=[G:HN]\cdot[HN : N]=[G:HN]\cdot[H:H\cap N]<\infty$$

In particular, $[H:H\cap N]$ is finite.