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Jun 1, 2015 at 3:19 comment added Qiaochu Yuan Related: mathoverflow.net/questions/530/…
May 31, 2015 at 19:54 vote accept BHZ
May 31, 2015 at 13:07 answer added Geoff Robinson timeline score: 8
May 31, 2015 at 10:32 history edited Derek Holt
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May 31, 2015 at 9:00 comment added BHZ Thanks for the comments and hints. I asked the above question for the complex irreducible characters.
May 31, 2015 at 8:53 comment added Duchamp Gérard H. E. @GeoffRobinson Ok for the terminology. A Jordan block of size greater than one never acts irreducibly (in particular, as you noted, in characteristic $p$). Can the OP make precise whether it is real, complex or any characteristic ?
May 31, 2015 at 8:43 comment added Geoff Robinson @DuchampGérardH.E.: I interpreted this as a question about complex irreducible characters. Even in characteristic $p$, a Jordan block of size greater than $1$ does not act irreducibly (it does act indecomposably).
May 31, 2015 at 8:33 comment added Duchamp Gérard H. E. For $p$, I think it's $p$ itself as, if you denote $l(p)$ this lower bound, one must have $p\leq l(p)$ (because of the regular representation) and, contrariwise, $l(p)\leq p$ because of the existence of the group in $Gl(p)$ generated by a maximal Jordan block with $1$ as eigenvalue. For $p^k, k\geq 2$, I do not know.
May 31, 2015 at 8:30 comment added Geoff Robinson In the case $p$, the lowest possible order, given the orthogonality relations would be $p^{2}+p$. This order can be attained, but only when $p=2$ or $p$ is a Mersenne prime, in which case there is a Frobenius group of order $p(p+1)$ with kernel of order $p+1$ and complement of order $p$.
May 31, 2015 at 8:22 history edited BHZ CC BY-SA 3.0
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May 31, 2015 at 8:09 history asked BHZ CC BY-SA 3.0