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.

Children:

Parents:

• Group action

“Groups, as men, will be known by their actions.”