Every Transitive Permutation Group Has a Fixed Point Free Element If $G$ acts transitively by permutations on a finite set $A$ with more than one element (i.e. $G$ is a transitive permutation subgroup of the symmetric group $S_A$). Why does $G$ necessarily contain an element which has no fixed points (i.e. $g$ such that $g \cdot a \neq a$ for any $a \in A$)?



The hint I have is to think about, given $a \in A$, what fraction of elements of $G$ fixes $a$. I'm not sure how to go about this hint...
By Burnside's lemma, you have

$$\frac{1}{|G|}\sum_{g \in G}|\text{fix }g| = |A/G|=1$$

Since $1 \in G$ has $|A|&gt;1$ fixed points, at least one of the terms in the sum must be $0$.
It suffices to show that $\bigcup_{a \in A} \operatorname{Stab}_a \neq G$ where $G$ is the transitive permutation group on $A$, so there is some $\sigma \in G$ not in $\bigcup_{a \in A} \operatorname{Stab}_a$, meaning $\sigma(a) \neq a$ for all $a \in A$.
As $G$ is transitive, the Orbit-Stabilizer theorem says that $|\operatorname{Stab}_a| = \frac{|G|}{|A|}$ for all $a$, which gives $$\left| \bigcup_{a \in A} \operatorname{Stab}_a \right| \le 1 + |A|\left(\frac{|G|}{|A|} - 1\right) &lt; |G|,$$
which is what we want.