Timeline for How to prove that index of second smallest index subgroup in $A_n$ is $n \choose 2$?
Current License: CC BY-SA 3.0
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| Jul 4, 2017 at 14:57 | comment | added | Nick Gill | The problem with using O'Nan-Scott is that you need a bound on the size of an almost simple group that embeds primitively into $A_n$. As far as I know if you want to avoid using CFSG, then the best bounds work for ANY primitive group and knowing that the group is almost simple isn't much of an advantage (so O'Nan--Scott hasn't really helped). | |
| Jul 2, 2017 at 16:54 | history | edited | Igor Rivin | CC BY-SA 3.0 |
gave a precise reference.
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| Jul 2, 2017 at 16:48 | history | answered | Igor Rivin | CC BY-SA 3.0 |