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An orbit is part of a set on which a group acts.

Let $G$ be a group, and let $S$ be a $G$-set. The orbit of an element $x\in S$ is the set $Gx$, i.e., the set of conjugates of $x$, or the set of elements $y$ in $S$ for which there exists $\alpha \in G$ for which $\alpha\cdot x = y$.

For $x\in S$, the mapping $\alpha \mapsto \alpha\cdot x$ is sometimes known as the orbital mapping defined by $x$; it is a homomorphism of the $G$-set $G$ (with action on itself, by left translation) into $S$; the image of $G$ is the orbit of $x$. We say that $G$ acts freely on $S$ if the orbital mapping defined by $x$ is injective, for all $x \in S$.

The set of orbits of $S$ is the quotient set of $S$ under the relation of conjugation. This set is denoted $G\backslash S$, or $S/G$. (Sometimes the first notation is used when $G$ acts on the left, and the second, when $G$ acts on the right.)

Let $G$ be a set acting on $S$ from the right, and let $H$ be a normal subgroup of $G$. Then $G$ acts on $S/H$ from the right, under the action $xH\cdot g = xHg = xgH$, for $x\in S$. ($H$ acts trivially on this set, so $(E/H)/G = (E/H)/(G/H)$.) Consider the canonical mapping $\phi : E/H \to E/G$. The inverse images of elements of $E/G$ under $\phi$ are the orbits of $E/H$ under action of $G$; thus on passing to the quotient, $\phi$ defines an isomorphism from $(E/H)/G$ to $E/G$.

Suppose $G$ and $H$ are groups, and $G$ acts on $S$ on the left, and $H$ on the right; suppose furthermore that the operations of $G$ and $H$ commute, i.e., for all $g\in G$, $h\in H$, $x\in S$, \[(g\cdot x)\cdot h = g\cdot (x\cdot h) .\] Let $H^0$ be the opposite group of $G$; then the actions of $G$ and $H$ on $S$ define a left action of $G \times H^0$ on $S$. The set $(G\times H^0)\backslash S$ is denoted $G \backslash S /H$. Since $G$ and $H^0$ are normal subgroups of $G \times H^0$, by the previous paragraph, the $G$-sets $(G \backslash S)/H$, $G \backslash ( S/H)$, $G \backslash S /H$ are isomorphic and identitfied with each other.

Let $G$ be a group, and $H$ a subgroup of $G$; let it act on $G$ from the right. Then the set $G/H$ is the set of left cosets mod $H$.

If $G$ is a group and $H,K$ are subgroups of $G$, then the set $H \backslash G /K$ is called the set of double cosets mod $H$ and $K$.

See also