4

All the proofs I’ve seen of this are very combinatorial and at least for me not very memorable. Is there a cleaner or more conceptual proof? For instance, I wondered if representation theory might yield something easier. $S_n$ is such a ubiquitous group, that I imagine that looking at an appropriate action and employing techniques from the category in which the action lives, we might find a cleaner proof.

Shaun
  • 44,997
Vik78
  • 3,877
  • 2
    If you are familiar with the proof of the simplicity of $PSL_n(\Bbb{F}_q)$ there is a parallel. This is an exercise in e.g. Alperin & Bell. My estimate is that the overall structure of that proof of the simplicity of $A_n$ is more complicated than the ones you have seen. In theory you could also use the full character table of $S_n$ or $A_n$ to check that no non-trivial normal subgroups are there, but the same remark applies. – Jyrki Lahtonen Aug 13 '22 at 07:06
  • 3
    But, I'm not an expert on this. I do think that if a simpler proof existed, one of the books we both have read, would have it. – Jyrki Lahtonen Aug 13 '22 at 07:08
  • 3
    See for example this post for several elegant proofs, and the linked proofs from Conrad's text. – Dietrich Burde Aug 13 '22 at 14:32
  • 1
    Concerning the title: once we know that $A_n$ is simple, the second claim is trivial, see for example here. – Dietrich Burde Aug 13 '22 at 18:20

0 Answers0