# transitive permutation groups have fixed point free elements

• Sep 29th 2010, 10:50 AM
robeuler
transitive permutation groups have fixed point free elements
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?