Group is equal to the product of two distinct subgroups of index $2$. Problem
Suppose $H$ and $K$ are two distinct subgroups of index 2 . Prove $G = HK$
Attempt
Since both $H$ and $K$ have indices 2 , both are normal subgroups. Because 
$H$ and $K$ are distinct ,assume $x\in K-H$
. Since $H$ has index 2, $G = H \cup xH $
How to proceed after this ? 
$Edit$
As H and K are normal subgroup , HK is also  normal subgroup and it contains both H and K . Therefore [G:HK]= 1 ,since it cannot be more than 1. 
 A: Let it be that $h\in H$ and $h\notin K$ is true for some $k\in G$.
Then $G=K\cup hK$ because the index of $K$ is $2$, leading directly to $G\subseteq HK$.
If no $h$ exists with $h\in H$ and $h\notin K$ then - because the subgroups are distinct - some $k\in K$ will exist with $k\notin H$.
Then $G=H\cup Hk$ because the index of $H$ is $2$, leading directly to $G\subseteq HK$.
A: If $K$ is normal in $G$ then $HK\leq G$ for all subgroups $H\leq G$. ( in fact, more is true: if $N_G(K)$, then $HK\leq G$, where $N_G(K)=\{g\in G: gKg^{-1}=K\}$).
Then either $G= H\sqcup xH$ or $G= K\sqcup yK$ gives $G\leq HK$, where either $x\in K-H$ or $y\in H-K$. 
Indeed, if $g\in G$ then either $g\in H\leq HK$ or $g=xh\in KH=HK$. In the other case, either $g\in K\leq HK$, or $g=yK\in HK$.
A: As $H$ and $K$ are normal subgroups, then $KH$ id a normal subgroup that contains both $H$ and $K$. If the order of $G$ is finite, by Lagrange's theorem (analogue for $K$),
$$\text{ord}(G)=\text{ord}(HK)[G:HK]=\text{ord}(HK)\frac{[G:H]}{[HK:H]}.$$
Now, we have that $H\subsetneq HK$ (if they are equal, then hypothesis of $H$ and $K$ are different is not satisfied), so $[HK:H]>1$. As $[G:HK]\in \mathbb{N}$ (because is the order of a group) and $[G:H]=2$, then $[HK:H]=2$ (the options were $1$ or $2$, but we desestimate $1$), so 
$$\text{ord}(G)=\text{ord}(HK).$$
Finally, as $HK$ is a subgroup of $G$ with the same order, they're equal. 
