# 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

https://mathtuition88.com/

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