This is related to [this question.](https://mathoverflow.net/questions/136012/multiply-transitive-groups) It is well-known that $S\_n$ and $A\_n$ are the only six transitive permutation groups, and it is likewise well-known that the proof of this requires the classification of finite simple groups. The question is: is there some number $k \gg 6$ such that one can prove that every $k$-transitive group is either $S\_n$ or $A\_n$ *without* using the classification? (this is not an existential question of whether such a thing is possible, but rather "has it been done", though obviously if there is a deep reason why it is unlikely to ever be done, that's of interest).



Multiply transitive groups, continued
*Since @Igor asked for some references, I'll make this an answer and summarise some of the things mentioned above.*


No one knows how to prove this result without resort to CFSG. It would be a huge result if someone could manage it. There are a number of results that head in the same general direction, as follows:


1. Burnside (in his 1897 book) proved that a 2-transitive group is affine or almost simple. Note that for an affine group $AGL(n,q)$ to be 3-transitive, one requires that $GL(n,q)$ acts 2-transitively on the set of non-zero vectors in $n$-dimensional vector space over $\mathbb{F}\_q$. This (pretty much) never happens, so to classify the 3-transitive groups one need only consider almost simple groups (and one immediately sees why CFSG is so vital). References:


	* Burnside, *Theory of groups of finite order*, 1897, Cambridge University Press.
	* I believe the classification of affine 3-transitive groups is due to Cameron and Kantor (without CFSG), but I'm not sure which of their papers is the relevant one.
2. Wielandt showed that the Schreier conjecture implies that any 7-transitive group contains $A\_n$. A weak version of Wielandt's argument which gives the result for 8-transitive rather than 7-transitive can be found [here](https://www.dropbox.com/s/7abiro0h13jndob/Neumann%20on%20Wielandt.pdf).
(The Schreier conjecture asserts that the outer automorphism group of a finite simple group is solvable.) Derek's comment below references a stronger result due to Michael O'Nan where the same conclusion is deduced assuming only 6-transitivity:


	* O'Nan, Michael E. *Normal structure of the one-point stabilizer of a doubly-transitive permutation group. I, II.* Trans. Amer. Math. Soc. 214 (1975), 1–42; ibid. 214 (1975), 43–74.
3. There are a number of results that suppose that some group $G$ has certain transitivity properties **and** contains an element of a certain kind... and then conclude that $G$ contains $A\_n$. Some examples:


	* (Jordan I think, classical) If $G$ is primitive and contains a transposition, then $G=S\_n$. If $G$ is primitive and contains a 3-cycle, then $G$ contains $A\_n$.
	* (Jordan, 1873) Suppose that $G$ is a primitive subgroup of degree $n$ and contains a $p$-cycle with $p\leq n-3$. Then $G$ contains $A\_n$.
	* (Bochert, 1892) Suppose that $G$ is a 2-transitive subgroup of degree $n$ and containing an element of support $<\frac{n}{4}-1$. Then $G$ contains $A\_n$.
	* (Manning, 1917 - 1933) Manning proved variants of Bochert's result of the following variety: Suppose that $G$ is a $k$-transitive group of degree $n$ containing an element of support $<c$. Then $G$ contains $A\_n$. You can take $(k,c)$ to be any of 
	$$
	(3, n/3-1), (4, (n-1)/2), (5, n/2), (6, 3n/5), (25, 25n/31).
	$$
	The relevant papers are called *The degree and class of multiply transitive groups* I, II and III.
4. Asymptotic results. The best asymptotic result is, I think, due to Wielandt and can be found in Marshall Hall's *The Theory of finite groups*: If $G$ is a $t$-transitive group of degree $n$ with $t>3\log n$, then $G$ contains $A\_n$.
5. Results mentioned by Arturo above - refer to Marshall Hall's book again:


	* (Theorem 5.7.2): Let the integer $n=kp+r$, where $p$ is a prime, $p>k$, $r>k$. Except for $k=1, r=2$, a group of degree $n$ cannot be as much as $(r+1)$-fold transitive unless it contains $A\_n$.
	* (Theorem 5.8.1): A group $G$ quadruply transitive on a set of letters, finite or infinite, in which a subgroup $H$ fixing four letters is of finite odd order, must be one of the following groups: $S\_4$, $S\_5$, $A\_6$, $A\_7$, or the Mathieu group on 11 letters.
6. There are a bunch of results that give upper bounds for the order of multiply transitive groups not containing $A\_n$. Bochert proved one of the first, and his result is outlined on p.41 of Wielandt. Babai and Pyber both have results in this area and I'd recommend you look here:


	* Pyber, L. On the orders of doubly transitive permutation groups, elementary estimates Journal of Combinatorial Theory, Series A
	Volume 62, Issue 2, March 1993, Pages 361–366


A good general source that discusses some of the results above is Wielandt's *Finite permutation groups* which can be found [here](https://www.dropbox.com/s/rmmzoj5kcqey3e9/Wielandt%20-%20Finite%20permutation%20groups.pdf).


