|Orbit(s)| = |G| when action is fixed-point free

[Mr Davis 97]

Homework Statement

Prove that if $G$ is a finite group and the action of $G$ on $S$ is fixed-point free, then for any $s\in S$ we have $| \operatorname{Orbit}_G(s)|=|G|$.

Homework Equations

The Attempt at a Solution

We are trying to show that two finite sets have the same number of elements. By set theory, this means that we must exhibit a bijection between the two sets.

Let $s \in S$, and consider the following map: $\phi : G \to \operatorname{Orbit}_G(s)$ such that $\phi (g) = g \cdot s$.

This map is injective: Let $a,b \in G$ and suppose that $\phi (a) = \phi (b)$. Then $a \cdot s = b \cdot s$. But then $b^{-1} \cdot (a \cdot s) = b^{-1} \cdot (b \cdot s) = (b^{-1} b) \cdot s = e_g \cdot s = s$. So $(b^{-1} a)\cdot s = s$. However, the action is fixed-point free, meaning that $b^{-1} a = e_G \implies a = b$.

This map is clearly surjective: Since every element in the codomain is of the form $g \cdot s$ and since we can always choose $g$ such that we get $g \cdot s$ back, $\phi$ maps to every element in the codomain.

Since $\phi$ is a bijection between $G$ and $\operatorname{Orbit}_G(s)$, we conclude that $|\operatorname{Orbit}_G(s)|=|G|$.

 

[fresh_42]

Looks o.k. The general formula (standard and very useful) is (for finite groups)
$$
|G|=|G.x|\cdot |G_x| \text{ where } G.x=\operatorname{Orbit}_G(x) \text{ and } G_x=\operatorname{Stab}_G(x) = \operatorname{Fix}_G(x)=\{\,g\in G \,|\,g.x=x\,\}
$$
The fixed points form a subgroup which is called stabilizer of $x$. The proof is basically the same, only that we don't have injectivity and thus cosets $G/G_x$ instead.