Orbit-Stabilizer Theorem (with proof)

Orbit-Stabilizer Theorem

Let $G$ be a group which acts on a finite set $X$ . Then $\displaystyle |\text{Orb}(x)|=[G:\text{Stab}(x)]=\frac{|G|}{|\text{Stab}(x)|}.$

Proof

Define $\phi:G/\text{Stab}(x)\to\text{Orb}(x)$ by $\displaystyle \phi(g\text{Stab}(x))=g\cdot x.$

Well-defined:

Note that $\text{Stab}(x)$ is a subgroup of $G$ . If $g\text{Stab}(x)=h\text{Stab}(x)$ , then $g^{-1}h\in\text{Stab}(x)$ . Thus $g^{-1}hx=x$ , which implies $hx=gx$ , thus $\phi$ is well-defined.

Surjective:

$\phi$ is clearly surjective.

Injective:

If $\phi(g\text{Stab}(x))=\phi(h\text{Stab}(x))$ , then $gx=hx$ . Thus $g^{-1}hx=x$ , so $g^{-1}h\in\text{Stab}(x)$ . Thus $g\text{Stab}(x)=h\text{Stab}(x)$ .

By Lagrange’s Theorem, $\displaystyle \frac{|G|}{|\text{Stab}(x)|}=|G/\text{Stab}(x)|=|\text{Orb}(x)|.$

Field Medallist Prof. Gowers has also written a nice post on the Orbit -Stabilizer Theorem and various proofs.

Author: mathtuition88

Math and Education Blog View all posts by mathtuition88

One thought on “Orbit-Stabilizer Theorem (with proof)”

Pingback: My 2018 Mathematics A To Z: Group Action – nebusresearch

Leave a Reply Cancel reply

Enter your comment here...

Fill in your details below or click an icon to log in:

Email (required) (Address never made public)

Name (required)

Website

You are commenting using your WordPress.com account. ( Log Out / Change )

You are commenting using your Twitter account. ( Log Out / Change )

You are commenting using your Facebook account. ( Log Out / Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

%d bloggers like this: