Let \(G\) be a finite group, acting on a set \(X\). Let \(x \in X\). Writing \(\mathrm{Stab}_G(x)\) for the stabiliser of \(x\), and \(\mathrm{Orb}_G(x)\) for the orbit of \(x\), we have \($|G| = |\mathrm{Stab}_G(x)| \times |\mathrm{Orb}_G(x)|\)$ 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 \(\mathrm{Stab}_G(x)\) and the orbit \(\mathrm{Orb}_G(x)\).

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 \($|\mathrm{Orb}_G(x)| |\mathrm{Stab}_G(x)| = |\{ \text{left cosets of} \ \mathrm{Stab}_G(x) \}| |\mathrm{Stab}_G(x)|\)$ 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 \(\theta: \mathrm{Orb}_G(x) \to \{ \text{left cosets of} \ \mathrm{Stab}_G(x) \}\), by \($g(x) \mapsto g \mathrm{Stab}_G(x)\)$

This map is well-defined: note that any element of \(\mathrm{Orb}_G(x)\) is given by \(g(x)\) for some \(g \in G\), so we need to show that if \(g(x) = h(x)\), then \(g \mathrm{Stab}_G(x) = h \mathrm{Stab}_G(x)\). This follows: \(h^{-1}g(x) = x\) so \(h^{-1}g \in \mathrm{Stab}_G(x)\).

The map is injective: if \(g \mathrm{Stab}_G(x) = h \mathrm{Stab}_G(x)\) then we need \(g(x)=h(x)\). But this is true: \(h^{-1} g \in \mathrm{Stab}_G(x)\) and so \(h^{-1}g(x) = x\), from which \(g(x) = h(x)\).

The map is surjective: let \(g \mathrm{Stab}_G(x)\) be a left coset. Then \(g(x) \in \mathrm{Orb}_G(x)\) by definition of the orbit, so \(g(x)\) gets taken to \(g \mathrm{Stab}_G(x)\) as required.

Hence \(\theta\) is a well-defined bijection.