Prove that if $H$ is a subgroup of index $2$ in a finite group $G$, then $gH = Hg \; \forall \; g \in G$.
I know that $H$ itself is one coset of the subgroup and the other is the compliment of the subgroup, but I don't really understand why the second coset is the compliment. I know that the union of the cosets must be $G$, but how do we know that we can't say, for example, $2H \cup H \equiv G$? Why do we know for sure that $H'\cup H$ is $G$?
I also know that the number of left and right cosets are identical.