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, 
$$H/H\cap K\simeq HK/H=G/H.$$
A: For a slightly more constructive approach:
Suppose that the cosets of $H$ in $G$ are:
$$
H\quad\text{and}\quad g_1H
$$
and the cosets of $K$ in $G$ are:
$$
K\quad\text{and}\quad g_2K.
$$
We can assume, wlog that $g_1\in K$ and $g_2\in H$ (otherwise $K=H$).
I claim that the cosets of $H\cap K$ in $G$ are
$$
H\cap K, g_1(H\cap K), g_2(H\cap K),\text{ and }g_1g_2(H\cap K).
$$
We can see that $g_1(H\cap K)=(g_1H)\cap K$ as follows: 


*

*Suppose that $a\in g_1(H\cap K)$, then there exists $b\in H\cap K$ so that $g_1b=a$.  Since $g_1\in K$, $g_1b=a\in K$, and we can see, directly, that $g_1b\in g_1H$.  

*On the other hand, suppose that $c\in (g_1H)\cap K$, then $c\in K$ and $c=g_1d$ for some $d\in H$.  Since $g_1\in K$, $g_1^{-1}c=d$ is also in $K$.  Hence $d\in H\cap K$ and $g_1d=c$.
Similarly, $g_1g_2(H\cap K)=(g_1H)\cap (g_2K)$ as follows:


*

*Let $a\in g_1g_2(H\cap K)$, then there is some $b\in H\cap K$ such that $a=g_1g_2b$.  Since $g_2\in H$, $g_2b\in H$, so there is some $c\in H$ such that $a=g_1c$.  Hence, $a\in g_1H$.  On the other hand, since $H$ is normal, since $g_2\in H$, there is some $g_1'$ such that $g_1g_2=g_2g_1'$.  Moreover, $g_2^{-1}g_1g_2=g_1'$.  Since $K$ is normal, we know that $g_1'$ is in $K$.  Therefore, $a=g_2g_1'b$.  Since both $g_1'$ and $b$ are in $K$, there is some $d\in K$ such that $a=g_2d$, so $a\in g_2K$.

*On the other hand, suppose that $a\in(g_1H)\cap(g_2K)$.  Then, there are $b\in H$ and $c\in K$ so that $a=g_1b$ and $a=g_2c$.  Since $g_2\in H$, there is some $d$ such that $b=g_2b$.  Hence $a=g_1g_2d$.  On the other hand, since $K$ is normal, we know that $g_2^{-1}g_1g_2\in K$.  Moreover, there is some $e\in K$ so that $c=g_2^{-1}g_1g_2e$.  Therefore, $a=g_2g_2^{-1}g_1g_2e=g_1g_2e$.  By cancellation, $d=e$, so $d\in H\cap K$.  Hence $a\in g_1g_2(H\cap K)$.
Continuing in this way, we can write all of these cosets in this way.  Using the fact that the cosets for $H$ and $K$ partition $G$, we get an explicit partition of $G$ into cosets of $H\cap K$.
