Coset ($gH,Hg$)

For a subgroup $H\le G$ 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 isThis 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 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 exactlyRepresentatives $3,4,5$ 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.

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 $x\mapsto x{g}^{-1}$.

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 pairThe 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.

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 MIT's quotient-group notes, theorem 9.1.

A nonempty subset $B$ of any group $G$ is a left coset of a subgroup if and only if any two left translates $gB,hB$ are equal or disjoint. No finiteness assumption is needed.

For the forward direction, write $B=bH$ with $H$ a subgroup. Then $gB=gbH$, and distinct left cosets of $H$ are disjoint. For the converse, first assume $e\in B$. For every $b\in B$, $b\in B\cap bB$, so $bB=B$. This proves closure under multiplication. Since $e\in B=bB$, there is $c\in B$ with $bc=e$, showing $b^{-1}\in B$. Thus $B$ passes the subgroup test.

For a general nonempty $B$, choose $b\in B$ and set $H=b^{-1}B$. It contains $e$, and all its left translates have the assumed property because $gH=(g{b}^{-1})B$. The previous argument makes $H$ a subgroup, and $B=bH$ is a left coset. This gives a recognition criterion for cosets from their translate intersections alone.

For a subgroup $H$ of a group $G$ and $g\in G$, its left coset is $gH=\{gh:h\in H\}$. Left multiplication $h\mapsto gh$ is a bijection from $H$ onto $gH$. Left and right cosets agree for every $g$ exactly when $H$ is a normal subgroup.

For a subgroup $H$ of a group $G$ and $g\in G$, its right coset is $Hg=\{hg:h\in H\}$. The map $h\mapsto hg$ is a bijection from $H$ onto $Hg$, with inverse $x\mapsto x{g}^{-1}$; in particular all right cosets have the same cardinality as $H$. Multiplication by subgroup elements is on the left in this expression because $g$ is the factor on the right.

A right coset transversal for a subgroup $A\le G$ chooses one representative $r$ for each right coset $Ar$. Every $g\in G$ then has a unique expression $g=ar$ with $a\in A$ and $r$ in the transversal. Choosing $1$ as the representative of $A$ makes membership in $A$ equivalent to having representative $1$.

For subgroups $H,P\le G$, the double coset of $x\in G$ isThe double cosets partition $G$. The action of $H$ on the left cosets of $P$ givesby the orbit-stabilizer theorem.

For the order-eight dihedral group generated by $r$ of order four and $s$ of order two with $srs=r^{-1}$, let $H=K=⟨s⟩$. There are exactly three double cosets: $H$, $H{r}^{2}H$, and $HrH$, of sizes two, two, and four. Their number does not divide eight. This differs from the number of ordinary cosets, which divides group order by Lagrange's theorem.

A double coset is generally not a subgroup, so Lagrange's theorem does not force its size to divide the ambient group order. In $S_3$, take $H=K=⟨(12)⟩$. Then $HeH$ has size two, while $H(13)H$ has size four, which does not divide six. The general size formula $|HgK|=|H||K|/|H\cap gK{g}^{-1}|$ explains why different double cosets can have different sizes.

For a group $G$ and subgroup $\Gamma$ such that every double coset has finitely many orbits under left multiplication by $\Gamma$, this algebra consists of complex $\Gamma$-bi-invariant functions on $G$ supported on finitely many double cosets. Its convolution isThe double coset indicator functions form a basis. On invariant modular functions it acts on the right by $f\ast h={\sum }_{\Gamma x}h(x)f{|}_{k}x$. For $G=G{L}_2(\mathbb{Q}{)}^{+}$ and $\Gamma =S{L}_2(\mathbb{Z})$, rational conjugation of finite-index modular subgroups verifies the finiteness condition. The determinant-$n$ Hecke operator uses the indicator of all integral determinant-$n$ matrices, not only a single double coset when $n$ is composite.

If $P$ is a Sylow subgroup of a finite group $G$ and $H\le G$, some intersection $H\cap xP{x}^{-1}$ is a Sylow subgroup of $H$. Otherwise every double-coset sizewould be divisible by a larger power of $p$ than $|G|$, as would their sum $|G|$.