- #1

wheezyg

- 5

- 0

Problem

Let G be a finite group operating on a set S ( |S| >=2 ). Suppose there exists only one orbit. Prove there exists an x \in G which has no fixed point (ie xs \neq s for all s \in S)

Relevant theorem

Let G be a group operating on a set S and s \in S . Then the order of the orbit Gs is equal to the index of G_s (stabalizer group or isotropy group in Lang) in G

My attempt

Suppose every point in G has a fixed point. Then for every x \in G, xs=s \in S. From this (G:G_s) =1. **Since there is only one orbit, (G:G_s)=|S| which implies |S|=1, a contradiction.

I have a feeling that my argument falls apart at **. Any guidance to the flaws inmky logic and/or understanding would be useful. I am working from Lang's Algebra (graduate level).