Subgroup of Index 2 is Normal

Theorem
A subgroup of index $2$ is always normal.

Proof
Suppose $H \le G$ such that $\left[{G : H}\right] = 2$.

Thus $H$ has two left cosets (and two right cosets) in $G$.

If $g \in H$, then $g H = H = H g$.

If $g \notin H$, then $g H = G \setminus H$ as there are only two cosets and the cosets partition $G$.

For the same reason, $g \notin H \implies H g = G \setminus H$.

That is, $g H = H g$.

The result follows from the definition of normal subgroup.

Also see

 * Subgroup of Index Least Prime Divisor is Normal