For the background, see the post: Classification of triply transitive finite groups

Thanks to the classification of finite simple groups (CFSG), we know that if $G$ is a finite $6$-transitive groups, then $G=S_n$ or $A_n$.

Question: Is there an easy proof of this result (i.e. without using CFSG) ?

I'm also interesting by such a proof for $k$-transitive groups with $k$ sufficiently large.
If such proof doesn't exist (yet), are there people working on ?

I hope such an easy proof (will) exist.

  • $\begingroup$ mathoverflow.net/questions/5993/… $\endgroup$ – Myself Mar 5 '14 at 12:39
  • $\begingroup$ @Myself: Thank you ! The question you link is similar, but it is more than $4$ years old, and it does not answer the question. In particular it does not ask about people working on an easy proof. Does the community of finite groups theorists thinks that such a proof exists ? $\endgroup$ – Sebastien Palcoux Mar 5 '14 at 12:49
  • $\begingroup$ I'm not an expert but I have the impression that most specialists are pessimistic about such proofs. I have always assumed this is because it would overcome certain difficulties in the classification too easily and with all the hard work done, someone should have found it a long time ago. $\endgroup$ – Myself Mar 5 '14 at 12:56
  • 3
    $\begingroup$ The answer to the question is no, and I am not aware of anyone working on it, mainly because nobody has any idea how to do it, even for large values of $k$. $\endgroup$ – Derek Holt Mar 5 '14 at 16:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.