How is it possible that $\textrm{HK}=\textrm{G}$? For the following problem:

If $\textrm{H}$ and $\textrm{K}$ are distinct subgroups of $\textrm{G}$ of index $2$, then $\textrm{H}\cap\textrm{K}$ is a normal subgroup of $\textrm{G}$ of index $4$.

It is obvious that both $\textrm{H}$ and $\textrm{K}$ are normal subgroups since both have index $2$. Hence their intersection $\textrm{H}\cap\textrm{K}$ is also normal.
The main issue comes with proving that it has an index $4$, using 2nd Isomorphism Theorem. Clearly $\displaystyle\frac{\textrm{H}}{\textrm{H}\cap\textrm{K}}$ is isomorphic to $\displaystyle\frac{\textrm{HK}}{\textrm{K}}$ and using this we can show that $\textrm{H}\cap\textrm{K}$ that index $4$. But on this forum, I have seen another proof of this problem where a person wrote that $\textrm{H}\textrm{K} = \textrm{G}$.
How the heck are they equal? I'm having extreme difficulty in  proving this.
Has this got something to do with both $\textrm{H}$ and $\textrm{K}$ being distinct ?
 A: It has to do with $H$ and $K$ being distinct and having index $2$:  these hypotheses imply that none of them is contained in the other. So take an element $h\in H\smallsetminus K$. Then $\;G=hK\cup K$ since $K$ has index $2$. Also, since both $hK,K\subset HK$, we have $G=HK$.
A: $H,K$ have index $2$. This means that:


*

*$H,K$ are normal.

*$H,K$ are maximal.
Indeed, let $M$ be a proper subgroup of $G$ containing $H$. Then $M/H$ is a proper subgroup of $G/H$. Since
$G/H$ is the group with two elements, it follows that $M/H$ is trivial, i.e. $M=H$. The same holds for $K$.

*By maximality of $H$ and $K$, you have that either $HK=G$ or $HK=H$ or $HK=K$. But the latter two can hold only for $H=HK=K$ (and this contradicts the hypothesis).
A: We first remark, in general, if $G$ is a group (not necessarily finite) and $H$ is a subgroup of index $2$, $H$ must be normal. This is because the left and right cosets of $H$ necessarily coincide: there is one and only one partition of $G$  into $2$ parts such that one of the parts is $H$. 
It follows that for this problem, $H$, $K$, and $H \cap K$ are all normal. In particular, this means $HK$ is a subgroup. It suffices to show that $H \cap K$ has index $2$ in $H$. This will imply that $H \cap K$ has index $4$ in $G$. 
Observe that $HK = G$, since $H, K$ are distinct maximal subgroups. Moreover, by one of the isomorphism theorems, we have that $H/(H \cap K) \cong (HK)/K \cong G/K$.  Hence $H \cap K$ has index $2$ in $H$, as desired. 
