Recipe

Build the quotient group of an index-2 subgroup

Use when: you're given a subgroup $H$ with $[G:H]=2$ , exactly two cosets, and need both a normality proof and a description of $G/H$

Goal: a proof that $H$ is normal, plus the complete operation table of the two-element group $G/H$

Taught in Quotient groups of permutation groups

Quick recipe

Confirm the index is exactly 2
Conclude H is normal, with no conjugation check
Compute the one product that isn't forced by the group axioms

1
Compute $[G:H]=|G|/|H|$ , or count directly. Confirm it equals $2$ before doing anything else. This shortcut only works at index $2$ , never any larger index.

Example

$|S_3|=6$ and $|A_3|=3$ , so $[S_3:A_3]=6/3=2$ .

Why: the whole reasoning below only holds because a partition into exactly two blocks leaves no room for $gH$ and $Hg$ to disagree.
2
An index-2 subgroup is automatically normal. Its only two left cosets, $H$ and $G\setminus H$ , are also its only two right cosets. That forces $gH=Hg$ for every $g$ .

Example

$A_3\trianglelefteq S_3$ follows immediately from Step 1's division, with no conjugation computation needed.

Why: coset agreement is equivalent to being normal, and index 2 forces that agreement automatically, with no exceptions.
3
$G/H$ has exactly two elements, $H$ itself and $O=G\setminus H$ . $H$ acts as the identity automatically. So the only genuinely new computation is $O\cdot O$ .

Example

In $S_3/A_3$ , representative $(1\,2)\in O$ gives $(1\,2)(1\,2)=e\in A_3$ , so $O\cdot O=A_3$ directly from one representative.
- $G$ is some $S_n$ and $H=A_n$ — skip the representative computation entirely, using the shortcut below.
- $G$ is some other group with an index-2 subgroup — pick any representative from $O$ , multiply it by itself in $G$ , and see which coset catches the result.
⚡ Shortcut

For $G=S_n$ and $H=A_n$ specifically, two odd permutations always compose to an even one, by chapter 4's parity rule. So $O\cdot O=A_n$ is guaranteed without picking any representative at all. The same shortcut works unchanged from $S_3/A_3$ up to $S_n/A_n$ for any $n$ .

Why: a two-element set with $H$ as the identity has only one product left undetermined by the group axioms. Finding it completes the whole table.

Common mistakes

Trying to apply this shortcut to a subgroup whose index isn't exactly $2$ . The reasoning breaks down completely; it doesn't just get harder.
Building a full multiplication table by hand anyway. This misses that three of the four entries are forced for free once $H$ is confirmed to be the identity.
Forgetting to check the index equals $2$ first, and assuming any roughly-half-sized split works the same way.
Confusing $H\trianglelefteq G$ , a subgroup property, with $G/H$ having the same two-row multiplication table as $\mathbb{Z}_2$ . That second fact is about the resulting quotient, not about $H$ itself.