[summary: A group action of a group $~$G$~$ acting on a set $~$X$~$ describes how $~$G$~$ sends elements of $~$X$~$ to other elements of $~$X$~$. Given a specific element $~$x \in X$~$, the stabiliser is all those elements of the group which send $~$x$~$ back to itself, and the orbit of $~$x$~$ is all the elements to which $~$x$~$ can get sent.

This theorem tells you that $~$G$~$ is divided into equal-sized pieces using $~$x$~$. Each piece "looks like" the stabilizer of $~$x$~$ (and is the same size), and the orbit of $~$x$~$ tells you how to "move the piece around" over $~$G$~$ in order to cover it.

Put another way, each element $~$y$~$ in the orbit of $~$x$~$ is transformed "in the same way" by $~$G$~$ relative to $~$y$~$.

This theorem is closely related to Lagrange's Theorem. ]

[summary(Technical): 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.]

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.