If H is a subgroup of index 2 in $G$. Is it true that the set of all elements of $G$ that are not in $H$ is a coset of $H$?

Question

We discussed this in class and it was seemingly just mentioned in passing. Is there a proof available to this?

Answer 1

For any subgroup $H$, $G$ is a disjoint union of all the cosets $g_iH$ (where $g_i$ are representatives). Since $|G:H|=2$, there are only two cosets: $H$ and $gH$, i.e. 'the other one'. Therefore $G\setminus H=gH$, the other coset.

Answer 2

Hint: can you find a bijection, even in the infinite case? [take an element not in $H$ ...]

Answer 3

If $H$ is a subgroup of $G$ of index $2$ then one of its cosets is $H$. Since all cosets have equal size there can only be another coset which is?