orbit-stabilizer theorem

Suppose that $G$ is a group acting (http://planetmath.org/GroupAction) on a set $X$ . For each $x\in X$ , let $G x$ be the orbit of $x$ , let $G_{x}$ be the stabilizer of $x$ , and let ${\cal L}_{x}$ be the set of left cosets of $G_{x}$ . Then for each $x\in X$ the function $f\colon Gx\to{\cal L}_{x}$ defined by $gx\mapsto gG_{x}$ is a bijection. In particular,

$|Gx|=[G:G_{x}]$

and

$|Gx|\cdot|G_{x}|=|G|$

for all $x\in X$ .

Proof:
If $y\in Gx$ is such that $y=g_{1}x=g_{2}x$ for some $g_{1},g_{2}\in G$ , then we have $g_{2}^{-1}g_{1}x=g_{2}^{-1}g_{2}x=1x=x$ , and so $g_{2}^{-1}g_{1}\in G_{x}$ , and therefore $g_{1}G_{x}=g_{2}G_{x}$ . This shows that $f$ is well-defined.

It is clear that $f$ is surjective. If $gG_{x}=g^{\prime}G_{x}$ , then $g=g^{\prime}h$ for some $h\in G_{x}$ , and so $gx=(g^{\prime}h)x=g^{\prime}(hx)=g^{\prime}x$ . Thus $f$ is also injective.

Title	orbit-stabilizer theorem
Canonical name	OrbitstabilizerTheorem
Date of creation	2013-03-22 12:23:10
Last modified on	2013-03-22 12:23:10
Owner	yark (2760)
Last modified by	yark (2760)
Numerical id	22
Author	yark (2760)
Entry type	Theorem
Classification	msc 20M30