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Jun 24, 2021 at 13:58 vote accept Master Gang
Jun 21, 2021 at 13:12 comment added Master Gang @Jemery Rickard Very thanks for your nice counterexample. To be honest, I am considering the general case here. As there are some references for the case when char(k) does not divide |G|. I am not really interested in that situation.
Jun 21, 2021 at 10:53 history edited Jeremy Rickard CC BY-SA 4.0
changed notation for vertices
Jun 21, 2021 at 10:43 comment added Jeremy Rickard Possibly the answer is different if you add the requirement that $\operatorname{char}(k)$ does not divide $|G|$, although I've not really thought about it. Lots of statements of this kind require that condition.
Jun 21, 2021 at 10:41 history answered Jeremy Rickard CC BY-SA 4.0