Q:

If $H, K$ both have index 2 in $G$, prove that $[G: H \cap K] = 4$.

I am helping someone out in abstract algebra and I was stumped by the following prompt:
Suppose $H$ and $K$ are distinct subgroups of $G$ such that both $H$ and $K$ have index 2. Prove that $H \cap K$ is normal in $G$ and $\left[ G: H \cap K \right] = 4.$
We understand that subgroups of index two are normal and the intersection of normal subgroups is again normal in $G$, but the rest of the prompt is not as obvious. My first thought is to use the multiplicative properties of the index, i.e.,
$\left[G: H \cap K \right] = [G:H][H:H \cap K] = 2 [H: H \cap K],$
and one could just as easily use $K$ as their subgroup of choice here. In the few examples I have tried it has worked out that $[H: H \cap K] =2$ but I do not yet see a particular reason why, yet this would give the result. 
Could anyone please fill me in why $[H : H\cap K] =2$ is true under these conditions?

A:

It results  from the second isomorphism theorem: as $H$ (and $K$) is normal in $G$, HK is a subgroup of $G$ which contains (strictly) $H$, hence $HK=G$. By the isomorphism theorem, 
$$H/H\cap K\simeq HK/H=G/H.$$

