Prove that G contains at least |X| − 1 fixed-point-free elements, i.e., elements g such that Fix(g) is empty.

Question

Let $G$ be a finite group acting transitively on a finite set $X$ where $|X| > 1$. Prove that $G$ contains at least $|X| − 1$ fixed-point-free elements, i.e., elements $g$ such that $Fix(g)$ is empty.

Answer 1

The average number of fixed points is 1. The identity fixes $|X|$ points. The rest is arithmetic.... How many 0's do you need to have any chance of bringing the average down to 1?

How do you know the average number of fixed points is one? Each element of $X$ is in an orbit of length $|X|$ and so fixed by $|G|/|X|$ elements of $G$. There are $|X|$ elements of $X$, so there are $|X|\cdot\frac{|G|}{|X|}=|G|$ pairs $(g,x)$ with $x^g=x$. Thus in total, the $|G|$ elements of $G$ have $|G|$ fixed points.