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May 3, 2010 at 17:40 comment added Ben Green Jim: I agree completely. I'd still like an actual reference to the argument I sketched above though, if there is one earlier than Tao's blog.
May 3, 2010 at 17:10 comment added Jim Humphreys @Ben: The "simplest" argument that gives some bound may be one of those already mentioned, unless somebody finds a really new approach. Jordan's result is impressively general in scope, but the tools relevant to proving it seem quite limited. A more transparent proof would obviously be welcome. But after Frobenius and Schur, the published treatments (Curtis-Reiner, Isaacs,...) are basically similar apart from the way they are integrated with other theorems on finite groups and linear groups. Leaving aside the question of best bounds, it's hard to find a fresh approach.
May 2, 2010 at 21:32 comment added Ben Green I should also point out that by $j(n)$ I think you mean the index of the biggest normal abelian subgroup of $A$; so your $j(n)$ is at most $F(n)!$, but they need not be the same.
May 2, 2010 at 21:28 comment added Ben Green Igor, I believe this issue is comprehensively despatched in this paper of M. J. Collins: On Jordan's theorem for complex linear groups. J. Group Theory 10 (2007), no. 4, 411--423. He evaluates $j(n)$ for all $n$ and shows that $(n+1)!$ is the truth for $n \geq 71$ (and not for $n = 70$). But my interest is more in finding the simplest argument that gives some bound.
May 2, 2010 at 21:18 history answered Igor Pak CC BY-SA 2.5