Coset (source code)
= Coset
{title2=$gH,\ Hg$}
{wiki}
For a <subgroup> $H\leq G$ of a <group> $G$, a left <coset> is $gH=\{gh:h\in H\}$; a right <coset> is $Hg=\{hg:h\in H\}$. The name refers to the side on which the representative $g$ multiplies $H$. Think of a <coset> as a translated copy of a <subgroup>: it remembers the relative differences allowed by $H$, while moving the whole collection to another location in the <group>.
Two left <cosets> of the same <subgroup> are either equal or disjoint. Thus distinct left <cosets> form a <partition of a set> covering $G$; the same statement holds separately for right <cosets>. The useful membership test is
$$
aH=bH\quad\Longleftrightarrow\quad b^{-1}a\in H.
$$
This also defines an <equivalence relation>: changing a representative by right multiplication with a member of $H$ leaves its left <coset> unchanged. These facts hold even for infinite <groups>. See https://ocw.mit.edu/courses/18-703-modern-algebra-spring-2013/df580757da0149c376f1fda29bf82a91_MIT18_703S13_pra_l_3.pdf[MIT's coset notes, definitions 3.6 and 3.9].
For an explicit additive example, take the <cyclic group> $G=\mathbb Z/6\mathbb Z$, with <modular arithmetic>, and the <subgroup> $H=\{0,3\}$. Adding each representative to both members gives exactly
$$
0+H=\{0,3\},\qquad 1+H=\{1,4\},\qquad 2+H=\{2,5\}.
$$
Representatives $3,4,5$ repeat these three <cosets>, respectively. For example, $4+H=\{4,1\}=1+H$, since $4-1=3\in H$. Every residue appears once in the displayed <partition of a set>. Each <coset> has two members, so the <group order> of $H$ is $2$ and its <index of a subgroup> is $3$. Left and right <cosets> coincide here because $G$ is an <abelian group>.
\Image[/cyclic-coset-partition.png]
{title=Three cosets of the order-two subgroup of the additive cyclic group of order six}
A <coset> need not be a <subgroup>. In this example, $\{1,4\}$ lacks the <identity element> $0$, and $1+1=2$ leaves it. In fact, the only left or right <coset> of $H$ that is a <subgroup> is $H$ itself: a <subgroup> must contain the <identity element>, and the <coset> containing that element is $H$. Equal sizes come instead from the <bijection> $h\mapsto gh$, whose inverse is $x\mapsto g^{-1}x$; for right <cosets>, use $h\mapsto hg$ and $x\mapsto xg^{-1}$.
The <nonabelian group> $S_3$, the <symmetric group> on three letters, separates the two sides. Use <function composition> with the rightmost <permutation> applied first, and let $H=\{e,(12)\}$. Then $(13)(12)=(123)$ but $(12)(13)=(132)$, giving the differing pair
$$
(13)H=\{(13),(123)\},\qquad H(13)=\{(13),(132)\}.
$$
The complete <partition of a set> into left <cosets> consists of $H$, $\{(13),(123)\}$ and $\{(23),(132)\}$; the complete <partition of a set> into right <cosets> consists of $H$, $\{(13),(132)\}$ and $\{(23),(123)\}$. Each lists all six <permutations> exactly once. Both have three parts of size two, although their parts differ.
\Image[/permutation-left-right-cosets.png]
{title=Left and right coset partitions in S3 with rightmost-first permutation composition}
Left <cosets> support a <group action> by $a\cdot(gH)=(ag)H$ for every <subgroup>, as in the <coset action>. Multiplication of representatives defines a <quotient group> operation $(aH)(bH)=(ab)H$ precisely when $H$ is a <normal subgroup>. Otherwise the answer can depend on the representatives: in this $S_3$ example, $e$ and $(12)$ represent the same <coset> $H$, yet $e(13)H=(13)H$ and $(12)(13)H=(23)H$ differ. In the additive example, $H$ is a <normal subgroup>, making $(1+H)+(2+H)=3+H=H$ valid. See https://ocw.mit.edu/courses/18-703-modern-algebra-spring-2013/ecdbd211f736b3ac92379dea05b34491_MIT18_703S13_pra_l_9.pdf[MIT's quotient-group notes, theorem 9.1].
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