G must have an element with no fixed point when there is only one orbit

[wheezyg]

I am studying for a modterm on Monday and asking for help on the homework questions I got WRONG on my problem sets (so I can hopefully improve my understanding and see my mistake). This is my reworked version of the incorrect HW problem and I would like to know if I am on the right track...

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).

 

[boboYO]

I don't think the stabilizer group will be too useful here.
G_s = g \in G such that gs=s.
This statement of yours is wrong:
"suppose every point in G has a fixed point. Then for every g \in G, gs=s \in S."

you are getting your quantifiers mixed up. it should be

"suppose every point in G has a fixed point. Then for every g \in G, there exists s\in S such that gs=s \in S."