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 ?

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 $G=HK$.
