---
tag: F78G4C4M6Z
kind: proof
name: Proof that the size of an orbit times the size of the stabiliser is the order of the group
msc: [20B05]
proves: 0S19CWFZM2
uses: [0E2HSCKZ6C, 04DC0SRN2K, 0RGGF8ZS6D, 00F2JFA41J]
sources:
  - url: https://en.wikipedia.org/wiki/Group_action
    locator: Orbit–stabilizer theorem
---

## Proof

We combine the orbit-stabiliser theorem with the index formula.

1. Let $G$ be finite, acting on $X$, and let $x \in X$. The [[stabiliser|0RGGF8ZS6D]] $G_x$ is a subgroup of $G$. Indeed, $e \cdot x = x$ gives $e \in G_x$, and $a, b \in G_x$ gives $(a b^{-1}) \cdot x = a \cdot (b^{-1} \cdot (b \cdot x)) = x$.
2. By the [[orbit-stabiliser theorem|0E2HSCKZ6C]], $\abs{G \cdot x} = [G : G_x]$.
3. Since $G$ is finite and $G_x$ is a subgroup, [[$\abs{G} = [G : G_x] \cdot \abs{G_x}$|04DC0SRN2K]].
4. Substituting step 2 into step 3 gives $\abs{G \cdot x} \cdot \abs{G_x} = \abs{G}$.
5. Since $\abs{G_x}$ is a positive integer, step 4 shows that $\abs{G \cdot x}$ divides $\abs{G}$. So the size of each [[orbit|00F2JFA41J]] divides $\abs{G}$.

## Idea

The orbit is in bijection with the cosets of the stabiliser. The cosets all have the size of the stabiliser and fill up the group, so the two sizes multiply to $\abs{G}$.