It seems to be well-known that the six-transitive finite groups are the symmetric and alternating groups, and the only other four-transitive finite groups are the Mathieu groups (the statement can be found in Cameron's 1999 "Permutation Groups", p 110), but I can't seem to figure out where the result was first stated (Cameron is not much help, though he does give a vague sketch of how one might go about proving this result).
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asked Jul 7, 2013 at 16:16
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1$\begingroup$ As I recall, Marshal Hall's "Theory of Groups" attributes the result to Jordan, and includes a proof. But I'm away from home and I cannot confirm. $\endgroup$– Arturo MagidinCommented Jul 7, 2013 at 18:35
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2$\begingroup$ Since the result depends on the classification of finite simple groups, Jordan seems a little unlikely, though I would not be surpise if he asked the question... $\endgroup$– Igor RivinCommented Jul 7, 2013 at 18:37
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4$\begingroup$ I found a reference on line. Jordan proved it for sharply transitive groups, and Hall extended the result to groups in which the stabilizers of sufficiently many elements are finite of odd order. Perhaps extending it to simply "transitive" (as opposed to sharply transitive) requires the classification? $\endgroup$– Arturo MagidinCommented Jul 7, 2013 at 18:45
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3 Answers
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In Pacific Journal of Math 4 (1954), pp 219-226, Marshall Hall, Jr. writes in the paper "On a theorem of Jordan" that:
In 1872, Jordan showed that a finite quadruply transitive group in which only the identity fixes four letters must be one of the following: the symmetric group on four or five letters, the alternating group on six letters, or the Mathieu group on eleven letters.
(Hall's paper is available from Project Euclid ).
The citation is: C. Jordan, Recherches sur les substitutions, J. Math. Pures Appl. (2) 17 (1872), 351-363.
Hall generalizes the result to any group (finite or infinite) in which the stabilizer of four letters is finite of odd order. As I recall, he presents the result extending it from 4-transitive to $k$-transitive (with $k\gt 3$) in his book The Theory of Groups, but I'm away from the office this week and unable to verify this.
answered Jul 7, 2013 at 18:42
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Peter Cameron has an article "Finite permutation groups and finite simple groups. He states that Wielandt showed that any 6-transitve group is alternating or symmetric, given the Schreier's conjecture is true. Schreier's conjecture is that the outer automorphism group of a finite simple group is solvable, and the truth of this is a consequence of the classification.
I found Cameron's article by googling on "Wielandt Schreier". Wielandt's result was published as H. Wielandt, "Uber den Transitivitatsgrad von Permutationsgruppen", Math. Z. 74 (1960), 297-298.
answered Jul 7, 2013 at 20:44
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On page 218 of
John D. Dixon, Brian Mortimer: Permutation Groups, Springer GTM 163, 1996
it is stated:
It is a consequence of the classification of finite simple groups that a finite permutation group which does not contain the alternating group is at most 5-transitive. Except for the alternating and symmetric groups, the only finite groups which are 4- or 5-transitive are the Mathieu groups $M_{11}$, $M_{12}$, $M_{23}$ and $M_{24}$. The proof of this strong statement involves a case-by-case analysis of the finite simple groups.
Though Dixon / Mortimer also do not say where the result has been first stated.
answered Jul 7, 2013 at 16:56