Question: 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.

Answer: 
Remember, Langrange's theorem? There was an equivalence relation $a \sim b$ if $a^{-1}b \in H$
So $G\backslash H$ are the equivalence classes induced by $a \sim b$ if $a^{-1}b \in H$ In other words, the cosets represent the different equivalence classes induced by this equivalence relation. What do we know from equivalence classes? They $partition$ $G$, so if $aH$ and $bH$ are two different cosets then $aH \cap bH = \emptyset$ Therfore, if H has index 2 that means that $G\backslash H$ has two cosets. One coset you know is H and the other coset must have $a \in G$ such that $a \notin H$ which is just $H'$ These must also be the right cosets, hence $gH =Hg $ $\forall g \in G$
As you already know the number of left cosets is equal to the number of right ones, just take into account that

$$aH=H\iff a\in H$$

$$Ha=H\iff a\in H$$