Let $n = p^k$ be a power of a prime. Let $G$ be a transitive permutation group on $n$ letters. Show that there exists a fixed-point free element of order $p$.
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 $p$:
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 $x$ is $p^k$. By the orbit-stabilizer theorem, the order of the stabilizer of some element $x$ is equal to the order of the group divided by $p^k$. Since the order of a stabilizer is an integer, $p^k$ divides the order of the group. By Cauchy's theorem, there is an element of order $p$ in the group.
I cannot, however, show that there is an element of order $p$ with no fixed points. Does anyone have any hints?
Best Answer
As you ask only for hints, here are two:
Step 1: Show that some/any $p$-Sylow subgroup $P$ of $G$ acts transitively on the $n$ letters.
Step 2: Then consider the action of a central element of $P$ of order $p$.