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…
Best Answer
By Burnside's lemma, you have
$$\frac{1}{|G|}\sum_{g \in G}|\text{fix }g| = |A/G|=1$$
Since $1 \in G$ has $|A|>1$ fixed points, at least one of the terms in the sum must be $0$.