Orbit

From AoPSWiki

Revision as of 20:44, 7 September 2008 by Jam (Talk | contribs)

(diff) ← Older revision | Current revision (diff) | Newer revision → (diff)

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.