Let be a transitive permutation group on the finite set with . Show there is some such that for all . (Such an element is called fixed point free.)
First we prove a lemma.
Lemma: . Proof: if , note that . Thus . Let . Then , so that ; hence , so that .
Suppose by way of contradiction that for all there exists an element such that . Then .
Now because this action is transitive, if we fix , then as ranges over , is arbitrary in . So in fact, . Now because the action is transitive and , we know that is a proper subgroup. Finally, is finite. By a previous theorem, we have a contradiction. Thus there exists an element that is fixed point free.