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 ?

$H,K$ 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).
