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'orbit-stabilizer theorem'
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| Title of object: |
orbit-stabilizer theorem |
| Canonical Name: |
OrbitStabilizerTheorem |
| Type: |
Theorem |
| Created on: |
2002-02-19 06:41:46 |
| Modified on: |
2008-04-15 08:52:27 |
| Classification: |
msc:20M30 |
Revision comment (for changes between this and next version):
Preamble:
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts} |
Content:
Suppose that $G$ is a group \PMlinkname{acting}{GroupAction} on a set $X$.
For each $x\in X$, let $Gx$ 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 $x\in X$.
{\bf Proof}:\\
If $y\in Gx$ is such that $y=g_1x=g_2x$ for some $g_1,g_2\in G$,
then we have $g_2^{-1}g_1x=g_2^{-1}g_2x=1x=x$, and so $g_2^{-1}g_1\in G_x$,
and therefore $g_1G_x=g_2G_x$.
This shows that $f$ is well-defined.
It is clear that $f$ is surjective.
If $gG_x = g'G_x$, then $g = g'h$ for some $h \in G_x$,
and so $gx = (g'h)x= g'(hx) = g'x$.
Thus $f$ is also injective. |
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