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

  • #1
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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|##.
 

Answers and Replies

  • #2
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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.
 
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