Transitive group action on finite set has fixed point-free permutation

Statement
Suppose $$G$$ is a group with a transitive group action on a finite set $$S$$ and $$S$$ has more than one element. Then, there exists an element of $$G$$ such that the action on $$S$$ induced by that element is a fixed point-free permutation, i.e., the permutation moves every element of $$S$$.

Facts used

 * 1) uses::Fundamental theorem of group actions
 * 2) uses::Union of all conjugates is proper (for a subgroup of finite index)

Proof
Given: A group $$G$$ with a transitive action on a finite set $$S$$ of size more than one.

To prove: There is an element of $$G$$ that does not fix any element of $$S$$.

Proof: Denote, for $$s \in S$$, the subgroup of $$G$$ that fixes $$s$$ as $$G_s$$ (this is called the stabilizer or isotropy subgroup of $$s$$.