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Sep 26, 2022 at 12:27 history edited Ian Gershon Teixeira CC BY-SA 4.0
deleted 172 characters in body
Sep 26, 2022 at 12:20 history edited Ian Gershon Teixeira CC BY-SA 4.0
deleted 172 characters in body
Mar 8, 2022 at 12:32 history edited Ian Gershon Teixeira CC BY-SA 4.0
deleted 4 characters in body
Feb 2, 2022 at 17:46 history edited Ian Gershon Teixeira CC BY-SA 4.0
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Jan 31, 2022 at 18:14 vote accept Ian Gershon Teixeira
Jan 31, 2022 at 17:18 history edited LSpice CC BY-SA 4.0
Name of reference; 25{,}920
Jan 31, 2022 at 17:14 comment added LSpice $\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\SL{SL}\DeclareMathOperator\Z{Z}$I'm not sure if it's reasonable to call $\PSL_n(q)$ "of Lie type $A_{n - 1}(q)$". I'm not up on the definitions, but I'd expect this to refer only to $\mathbb F_q$-rational points of reductive groups over $\mathbb F_q$, whereas your $\PSL_n(q)$ is not of that type. (If you take not $\SL_n(q)/\Z(\SL_n(q))$ but $(\SL_n/\Z(\SL_n))(q)$, then you get $\operatorname{PGL}_n(q)$.) Similarly for $\operatorname{PSU}$. Also, TeX: $25,920$ needs curly brackets: $25{,}920$ 25{,}920. I edited accordingly.
Jan 31, 2022 at 16:47 answer added David A. Craven timeline score: 8
Jan 20, 2022 at 20:13 comment added Nate Not an answer, but I suspect there are too many small values of $m$ for this to be true. In particular the smallest non-trivial representation of $PSL_n(q)$ grows like $q^n$. So varying $n$ and $q$ it's roughly like counting all numbers vs counting prime powers.
Jan 20, 2022 at 17:57 history edited YCor CC BY-SA 4.0
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Jan 20, 2022 at 17:53 history asked Ian Gershon Teixeira CC BY-SA 4.0