# Orbit-stabiliser theorem

https://arbital.com/p/orbit_stabiliser_theorem

by Patrick Stevens Jun 19 2016 updated Jul 1 2016

The Orbit-Stabiliser theorem tells us a lot about how a group acts on a given element.

[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.