I am trying to prove the following result:
Let $G$ be a primitive permutation group on $\Omega$ of degree $n$ that contains a cycle $g$ fixing $k \geq 3$ points. Then, $A_n \leq G$ where $A_n$ is the alternating subgroup of degree $n$.
This is a generalisation of Jordan's theorem in which the cycle is of prime length. The generalisation has been achieved in a paper by Gareth Jones (2013) in a more general theorem that also deals with the cases of $k \in \{0,1,2 \}$. The proof, however relies on the classification of finite simple groups, which I would like to avoid using if possible. I have managed to show that the theorem is implied by induction if I can prove the base case of $k = 3$, and also that this base case is true if such a group must contain any element of degree strictly less than 5 (where the degree of a group element is the number of points moved by the element). This is where I am stuck. Of course, since this theorem is true, clearly such an element must exist but I am not sure how to show this. I might be hitting a dead end here if the classification of finite simple groups is an essential tool in proving even this simplified version, but I still want to give it a shot.
Any suggestions or insight will help. Also, if anybody has a good reason why this would be a dead end without application of the classification of finite simple groups, this would also be good to know! Specifically I am trying to show that either the case $k=3$ is true or that if $k = 3$, then $G$ contains an element that moves at most four points. I know that such a group is 4-fold transitive and I also know how to show that it is true when the length of the cycle $g$ is less than 5. I have tried a lot of the methods used in the proofs of theorems in Wielandt's book Finite Permutation Groups, specifically those in chapter 13, but nothing seems to work.
