This is related to this question. It is wellknown that $S_n$ and $A_n$ are the only six transitive permutation groups, and it is likewise wellknown 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).

3$\begingroup$ In his book The Theory of Groups, Marshall Hall proves (Theorem 5.7.2): "Let the integer $n=kp+r$, where $p$ is a prime, $p\gt k$, $r\gt k$. Except for $k=1$, $r=2$, a group of degree $n$ cannot be as much as $(r+1)$fold transitive unless it is $S_n$ or $A_n$." He also proves (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." Does not use the classification, of course. $\endgroup$ – Arturo Magidin Mar 24 '14 at 18:52

3$\begingroup$ See the posts: Highly transitive groups (without assuming the classification of finite simple groups) and Is there an easy proof for the classification of 6transitive finite groups? $\endgroup$ – Sebastien Palcoux Mar 24 '14 at 19:26

4$\begingroup$ This is most definitely not known  it would be a huge deal in finite group theory if it were. One way to solve this problem would be to prove the Schreier Conjecture, since Wielandt has shown that, given the Schreier Conjecture, any 7transitive group contains $A_n$. You might be interested in this  dropbox.com/s/7abiro0h13jndob/Neumann%20on%20Wielandt.pdf  where Peter Neumann takes one paragraph to outline a weak version of Wielandt's argument which gives the result for 8transitive rather than 7transitive... $\endgroup$ – Nick Gill Mar 24 '14 at 20:39

1$\begingroup$ ... Of course the Schreier Conjecture is, as far as anyone can tell, impossibly difficult. (It asserts that the outer automorphism group of a finite simple group is solvable.)... $\endgroup$ – Nick Gill Mar 24 '14 at 20:39

1$\begingroup$ ... There are also some classical results of Bocherdt, Manning and others, that directly bear on this. They give the conclusion you seek but require extra conditions, namely the presence of an element 'of small support'. (The ancestor of these kinds of results is Jordan's classification of primitive groups containing a 3cycle.)... I can write down references if you want. $\endgroup$ – Nick Gill Mar 24 '14 at 20:42
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:
Burnside (in his 1897 book) proved that a 2transitive group is affine or almost simple. Note that for an affine group $AGL(n,q)$ to be 3transitive, one requires that $GL(n,q)$ acts 2transitively on the set of nonzero vectors in $n$dimensional vector space over $\mathbb{F}_q$. This (pretty much) never happens, so to classify the 3transitive 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 3transitive groups is due to Cameron and Kantor (without CFSG), but I'm not sure which of their papers is the relevant one.
Wielandt showed that the Schreier conjecture implies that any 7transitive group contains $A_n$. A weak version of Wielandt's argument which gives the result for 8transitive rather than 7transitive can be found here. (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 6transitivity:
 O'Nan, Michael E. Normal structure of the onepoint stabilizer of a doublytransitive permutation group. I, II. Trans. Amer. Math. Soc. 214 (1975), 1–42; ibid. 214 (1975), 43–74.
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 3cycle, then $G$ contains $A_n$.
 (Jordan, 1873) Suppose that $G$ is a primitive subgroup of degree $n$ and contains a $p$cycle with $p\leq n3$. Then $G$ contains $A_n$.
 (Bochert, 1892) Suppose that $G$ is a 2transitive 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 $$ (3, n/31), (4, (n1)/2), (5, n/2), (6, 3n/5), (25, 25n/31). $$ The relevant papers are called The degree and class of multiply transitive groups I, II and III.
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 $A_n$.
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.
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.

2$\begingroup$ Here is one more reference. Michael O'Nan proved,using the Schreier Conjecture, that any finite $6$transitive group is symmetric or alternating. O'Nan, Michael E. Normal structure of the onepoint stabilizer of a doublytransitive permutation group. I, II. Trans. Amer. Math. Soc. 214 (1975), 1–42; ibid. 214 (1975), 43–74. $\endgroup$ – Derek Holt Mar 25 '14 at 19:44

$\begingroup$ 2 contains an offbyone error: Wielandt proved it for 8 (and Neumann's oneparagraph proof is about 9). This was reduced to 7 by Nagao (mathscinet.ams.org/mathscinetgetitem?mr=178053). How does one read the 6 result out of the O'Nan paper? $\endgroup$ – Sean Eberhard Jul 22 '20 at 9:47