Recipe

Compute a coset of a subgroup

Use when: you're given a subgroup H, a representative g, and a side (left or right), and asked for the coset gH or Hg

Goal: the complete coset as a set of |H| elements, with none skipped and none miscomputed

Taught in Left and right cosets

Quick recipe

List every element of H
Multiply g by each element of H, on the correct side
Collect the results into a set
Compare to the opposite-side coset, if the problem asks for it

1
Write out all of H's elements explicitly; the coset will have exactly this many entries, no more and no fewer.

Example

$H=\{e,(1\ 2)\}$ in $S_3$ : two elements to multiply through.

Why: a coset $gH$ is defined by multiplying g by every element of H, so missing an element of H directly means missing an element of the coset.
2
For a left coset $gH$ , compute $g\cdot h$ for every $h\in H$ , applying h first then g. For a right coset $Hg$ , compute $h\cdot g$ instead, applying g first.

Example

Left coset $(1\ 2\ 3)H$ : $(1\ 2\ 3)\cdot e=(1\ 2\ 3)$ , and $(1\ 2\ 3)\cdot(1\ 2)$ traces to $(1\ 3)$ , applying $(1\ 2)$ first, then $(1\ 2\ 3)$ .

Why: $gH$ and $Hg$ genuinely differ in a non-abelian group, so multiplying on the wrong side produces the wrong set entirely, not just a relabeling.
3
Gather every product from Step 2 into one set; a coset always has exactly $|H|$ elements, since $h\mapsto gh$ (or $h\mapsto hg$ ) is a bijection from H onto it.

Example

$(1\ 2\ 3)H=\{(1\ 2\ 3),(1\ 3)\}$ , two elements, matching $|H|=2$ .
- every element of H was actually multiplied through — the set has $|H|$ elements, and the coset is complete.
- only the identity of H was multiplied through, stopping early — the "coset" found has just 1 element, a sign a step was skipped.
Why: stopping before every element of H is processed produces a subset of the true coset instead of the whole thing.
4
Repeat the process on the other side ( $Hg$ if you computed $gH$ , or vice versa), and check whether the two sets are identical.

Example

$H(1\ 2\ 3)=\{(1\ 2\ 3),(2\ 3)\}$ , which differs from $(1\ 2\ 3)H=\{(1\ 2\ 3),(1\ 3)\}$ found above, since $S_3$ is non-abelian.

Why: the two sides always agree automatically in an abelian group, so genuinely comparing both is the only way to confirm which situation applies.

Common mistakes

Stopping after multiplying g by just the identity of H, producing a "coset" of size 1.
Multiplying on the wrong side, using $gh$ when the problem asked for $Hg$ , or vice versa.
Assuming left and right cosets always agree, which only holds in an abelian group.
Forgetting that g itself is one element of any coset containing it, never the entire coset alone.