What is known about the classification of ntransitive group actions for n large without using the classification of finite simple groups? With the classification of finite simple groups a complete list of all 2transitive group actions is known, in particular there are no 6transitive groups other than the symmetric groups and the alternating groups. I want something like "there are no interesting ntransitive group actions for n sufficiently large" but without the classification theorem (however, I'd be happy if the n in that statement was obscenely large). Even any partial (but unconditional) results would interest me (like any ntransitive group for n sufficiently large needs to have properties X, Y, and Z).

$\begingroup$ You're assuming they're finite? $\endgroup$ – Autumn Kent Nov 18 '09 at 19:15

$\begingroup$ Yes, finite. Of course if n is an actual number then G might as well be finite. I suppose you could ask questions like this for n a cardinal, but I'm not so interested in that. $\endgroup$ – Noah Snyder Nov 18 '09 at 19:25
Marshall Hall's The Theory of finite groups only cites an asymptotic bound: a permutation group of degree n that isn't S_{n} or A_{n} can be at most ttransitive for t less than 3 log n. I suppose that was the state of the art at the time (late 1960s). There is an earlier paper by G.A. Miller on JSTOR that you can find by searching for "multiply transitive group".
There is a classical theorem of Jordan that classified sharply quadruply transitive permutation groups, i.e., those for which only the identity stabilizes a given set of 4 elements (from Wikipedia).


$\begingroup$ I'm speculating wildly here, but there may be intermediate structural results in the classification that give you bounds without proceeding through the caseelimination. $\endgroup$ – S. Carnahan♦ Nov 18 '09 at 19:31
There is a classical result of Wielandt that if you assume the Schreier conjecture (that the outer automorphism group of an finite nonabelian simple groups is solvable), then a group of degree n other that A_n or S_n is at most 7transitive. Unfortunately, the only known proof of the Schreier conjecture uses the classification of finite simple groups.
Well, the theorem that $M_{11}$ and $M_{12}$ are the only sharply $k$transitive groups for $k> 3$ is about 100 years older than the classification theorem...
Also, if $k>3$ (without the sharply hypothesis), $G$ must be simple, unless it is in $S_4$. Of course, that's why the classification is useful.

$\begingroup$ Yeah, I almost remarked that I already knew this in the question and for some reason decided against it. $\endgroup$ – Noah Snyder Nov 18 '09 at 19:23
