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Orbit-stabiliser theorem

Edited by Patrick Stevens last updated 20th Jun 2016
Requires: Stabiliser is a subgroup, Group action, Bijective function

Let G be a finite group, acting on a set X. Let x∈X. Writing StabG(x) for the stabiliser of x, and OrbG(x) for the orbit of x, we have |G|=|StabG(x)|×|OrbG(x)| 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 StabG(x) and the orbit OrbG(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 |OrbG(x)||StabG(x)|=|{left cosets of StabG(x)}||StabG(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 θ:OrbG(x)→{left cosets of StabG(x)}, by g(x)↦gStabG(x)

This map is well-defined: note that any element of OrbG(x) is given by g(x) for some g∈G, so we need to show that if g(x)=h(x), then gStabG(x)=hStabG(x). This follows: h−1g(x)=x so h−1g∈StabG(x).

The map is injective: if gStabG(x)=hStabG(x) then we need g(x)=h(x). But this is true: h−1g∈StabG(x) and so h−1g(x)=x, from which g(x)=h(x).

The map is surjective: let gStabG(x) be a left coset. Then g(x)∈OrbG(x) by definition of the orbit, so g(x) gets taken to gStabG(x) as required.

Hence θ is a well-defined bijection.

Parents:
Group action
Children:
Orbit-Stabiliser theorem: External Resources
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