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Apr 13, 2017 at 12:57 history edited CommunityBot
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May 7, 2013 at 2:43 comment added Qiaochu Yuan @Yemon: yes, and I don't remember. I think parts of the argument come up in the McKay correspondence.
May 7, 2013 at 0:19 comment added Yemon Choi Also, Qiaochu, do you know of a reference for the argument you give, or was it just something that emerged when you were learning representation theory? For a short note I'm typing up, I would like to both cite the result and indicate (since the note will probably be read by functional/harmonic analysts) that the proof doesn't require Feit-Thompson.
May 7, 2013 at 0:17 comment added Yemon Choi I am probably the only person who got confused, but just in case anyone else was also wondering: am I right in saying that the argument avoids Feit-Thompson by saying "assume G has an irrep of degree 2, then G must have even order, then G must have an involution," etc?
Aug 7, 2012 at 19:31 comment added Qiaochu Yuan @Mariano: well, in characteristic $0$ over a splitting field it only depends on the first step...
Aug 7, 2012 at 17:05 comment added Mariano Suárez-Álvarez In the second step of tour proof you mention Cauchy's theorem but that depends on the Feit-Thompson theorem, which is a few hundred orders of magnitude more complex... :-)
Aug 7, 2012 at 1:15 comment added Qiaochu Yuan @Geoff: thanks for the comment. The proof I'm familiar with fails but I didn't know if a different technique could work in positive characteristic.
Aug 7, 2012 at 0:40 comment added Geoff Robinson The first step does not usually work in prime characteristic, but does work for solvable groups by the fairly difficult Fong-Swan theorem.
Aug 7, 2012 at 0:03 history edited Qiaochu Yuan CC BY-SA 3.0
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Aug 6, 2012 at 23:58 history edited Qiaochu Yuan CC BY-SA 3.0
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Aug 6, 2012 at 23:50 history edited Qiaochu Yuan CC BY-SA 3.0
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Aug 6, 2012 at 23:35 history answered Qiaochu Yuan CC BY-SA 3.0