Orbit-stabiliser theorem

Written by Patrick Stevens last updated

Summaries

Let be a finite group, acting on a set . Let . Writing for the stabiliser of , and for the orbit of , we have where 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 and the orbit .

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 but the right-hand side is simply because an element of 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 , by

This map is well-defined: note that any element of is given by for some , so we need to show that if , then . This follows: so .

The map is injective: if then we need . But this is true: and so , from which .

The map is surjective: let be a left coset. Then by definition of the orbit, so gets taken to as required.

Hence is a well-defined bijection.