Let be a finite group, acting on a set $X$ . Let $x \in X$ . Writing ${S t a b}_{G} (x)$ for the stabiliser of $x$ , and ${O r b}_{G} (x)$ for the orbit of $x$ , we have $| G | = | {S t a b}_{G} (x) | \times | {O r b}_{G} (x) |$ where $| \cdot |$ refers to the size of a set.

This statement generalises to infinite groups, where the same proof goes through to show that there is a bijection between the left cosets of the group ${S t a b}_{G} (x)$ and the orbit ${O r b}_{G} (x)$ .

Proof

Recall that the stabiliser is a subgroup of the parent group.

Firstly, it is enough to show that there is a bijection between the left cosets of the stabiliser, and the orbit. Indeed, then $| {O r b}_{G} (x) | | {S t a b}_{G} (x) | = | {left cosets of {S t a b}_{G} (x)} | | {S t a b}_{G} (x) |$ but the right-hand side is simply $| G |$ because an element of $G$ is specified exactly by specifying an element of the stabiliser and a coset. (This follows because the cosets partition the group.)

Finding the bijection

Define $θ : {O r b}_{G} (x) \to {left cosets of {S t a b}_{G} (x)}$ , by $g (x) \mapsto g {S t a b}_{G} (x)$

This map is well-defined: note that any element of ${O r b}_{G} (x)$ is given by $g (x)$ for some $g \in G$ , so we need to show that if $g (x) = h (x)$ , then $g {S t a b}_{G} (x) = h {S t a b}_{G} (x)$ . This follows: $h^{- 1} g (x) = x$ so $h^{- 1} g \in {S t a b}_{G} (x)$ .

The map is injective: if $g {S t a b}_{G} (x) = h {S t a b}_{G} (x)$ then we need $g (x) = h (x)$ . But this is true: $h^{- 1} g \in {S t a b}_{G} (x)$ and so $h^{- 1} g (x) = x$ , from which $g (x) = h (x)$ .

The map is surjective: let $g {S t a b}_{G} (x)$ be a left coset. Then $g (x) \in {O r b}_{G} (x)$ by definition of the orbit, so $g (x)$ gets taken to $g {S t a b}_{G} (x)$ as required.

Hence $θ$ is a well-defined bijection.