Subgroups of index two are normal

Remarks

For $g \notin H$, the left cosets are $H$ and $gH = G \setminus H$, and the right cosets are $H$ and $Hg = G \setminus H$. Thus $gH = Hg$ for each $g \in G$, and the coset criterion applies. For example, the alternating group $A_n$ has index $2$ in the symmetric group $S_n$, and thus it is normal.

Going deeper

For a student

A subgroup $H$ is normal when $gH = Hg$ for each $g \in G$. In general this is a real condition. For a subgroup of index $2$ it is automatic.

The reason is that two cosets leave no choice:

1. The left cosets are $H$ and one other coset. Since the cosets partition $G$, the other coset is $G \setminus H$.
2. In the same way, the right cosets are $H$ and $G \setminus H$.
3. If $g \in H$, then $gH = H = Hg$.
4. If $g \notin H$, then $gH = G \setminus H = Hg$.

A worked example with $G = S_3$ and $H = A_3 = \{e, (1\,2\,3), (1\,3\,2)\}$:

1. $|S_3| = 6$ and $|A_3| = 3$, so the index is $2$.
2. For $g = (1\,2)$, the left coset $gA_3$ is the set of the three transpositions.
3. The right coset $A_3g$ is the same set. Thus $gA_3 = A_3g$.

The index two condition is sharp. The subgroup $\{e, (1\,2)\}$ of $S_3$ has index $3$, and it is not normal: $(1\,2\,3)H = \{(1\,2\,3), (1\,3)\}$, but $H(1\,2\,3) = \{(1\,2\,3), (2\,3)\}$.

Standard examples are $A_n$ in $S_n$, the rotations in a dihedral group, and $2\mathbb{Z}$ in $\mathbb{Z}$. A related fact is that a subgroup whose index is the smallest prime divisor of $|G|$ is also normal. The proof of that fact needs a group action.