Let $\displaystyle G$ be a permutation group on $\displaystyle n$ letters. So a subgroup of $\displaystyle S_{n}$.

There is a theorem of Fein, Kantor and Schacher (1981) that if $\displaystyle G$ is transitive, then $\displaystyle G$ has a fix point free element of prime order.

The proof is very hard and relies on the classification of finite simple groups. I would like to prove a special case: let $\displaystyle n=p^{k}$, then if the group is transitive there exists an element of order $\displaystyle p$ with no fixed points.

I can show that there must be a fixed point:

The transitivity of the group implies that there is only 1 orbit. Burnside's lemma then implies that the average number of fixed points for an element in the group is 1. Since the identity fixes everything, there is a fixed point free element.

I can show that there exists an element of order $\displaystyle p$:

Again we use that the group is transitive and that there is therefore only one orbit. Since there is only one orbit, the order of the orbit of $\displaystyle x$ is $\displaystyle p^{k}$. By the orbit-stabilizer theorem, the order of the stabilizer of some element $\displaystyle x$ is equal to the order of the group divided by $\displaystyle p^{k}$. Since the order of a stabilizer is an integer, $\displaystyle p^{k}$ divides the order of the group. By Cauchy's theorem, there is an element of order $\displaystyle p$ in the group.

I cannot, however, show that there is an element of order $\displaystyle p$ with no fixed points. Does anyone have any hints?