Permutation groups: Burnsides Lemma and the Fixed point free theorem

Definition If $G$ acts on $\Omega$ and $g \in G$ then the fixed point set $\operatorname{fix}(g)$ is $\{\alpha \in \Omega | \alpha g = \alpha \}$. If $\operatorname{fix}(g) = \{\}$ then $g$ is FPF (fixed point free).

Lemma ("Burnsides lemma" except it was actually proved by Frobenius) If $G$ acts on $\Omega$ then the number of orbits is $$\frac{1}{|G|}\sum_{g \in G}|\operatorname{fix}(g)|.$$
proof: Consider the set $S = \{(\alpha,g) \in \Omega\times G | \alpha g = \alpha \}$. We shall count it in two different ways:
First given $\alpha$, consider the number of ways it can occur as the first component of the pair: that's just $|G_\alpha|$ the size of its stabilizer, from the fact $|G_{\alpha g}| = |G_{\alpha}^g| = |G_\alpha|$ proved in the previous post we get that the contribution to $S$ from the orbit is $\alpha G$ is $|G_\alpha| |\alpha G| = |G|$ therefore $|S| = |G| \cdot \text{number of orbits}$.
Secondly given $g$, consider the number of ways it can occur as the second component of the pair: that's just $\operatorname{fix}(g)$.
Putting these together gives the formula.

Corollary If $G$ acts transitively on $\Omega$ (and $|\Omega| > 1$) some element of $g$ is fixed point free. proof: There is only 1 orbit, so the average number (over $G$) of elements fixed is 1.. but the identity has $|G|$ fixed points - every element!