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Mar 15, 2020 at 15:51 comment added Geoff Robinson @LSpice : See this question |: mathoverflow.net/questions/240317/…
Mar 15, 2020 at 15:37 comment added LSpice @GeoffRobinson, do you have a reference for that result?
Mar 15, 2020 at 15:22 history edited Denis Serre CC BY-SA 4.0
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Mar 15, 2020 at 15:05 comment added Geoff Robinson Specifically, if $n_{1},n_{2}, \ldots, n_{t}$ are all the distinct values taken by ${\rm trace}(w(A,B))$ as $w(A,B)$ runs over all non-identity elements of $G$, then , assuming $G$ is finite, its order divides $(n-n_{1})(n-n_{2}) \ldots (n-n_{t})$.
Mar 15, 2020 at 14:45 comment added Geoff Robinson Not only does $w(A,B)$ have integer trace, but also non-negative integer trace. This is actually enough to show that if $G$ is finite, then its order does indeed divide $n!$, by a result of Blichfeldt, which has appeared on MO in relation to several questions.
Mar 15, 2020 at 14:09 history answered Denis Serre CC BY-SA 4.0