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when toggle format what by license comment
May 17, 2021 at 14:20 history edited Martin Sleziak CC BY-SA 4.0
http -> https (the question was bumped anyway)
May 7, 2014 at 18:52 comment added Jesper M. Moller @GeoffRobinson: Yes, the displayed formula shows that Brown's theorem and the $n=|G|_p$ case of Frobenius' theorem are equivalent. Thanks to your work with Marty Isaacs we know how to get from this special case to the general case of Frobenius' Theorem. Thus the theorems of Frobenius and Brown are equivalent. Let me mention the general relation $\sum \widetilde{\chi}(\mathcal{S}^*_{N_G(H)/H})=-1$ related to the displayed formula.
May 7, 2014 at 16:06 comment added Geoff Robinson Actually, I find the displayed formula more interesting that Brown's Theorem. And it would appear that Brown's theorem implies the case of $n = |G|_{p}$ of Frobenius's theorem as well. In fact, in a paper of Marty Isaacs and myself, we showed that that case essentially is sufficient to do every $n.$
S May 7, 2014 at 14:07 review Late answers
May 7, 2014 at 14:08
S May 7, 2014 at 14:07 review First posts
May 7, 2014 at 14:10
May 7, 2014 at 13:48 history answered Jesper M. Moller CC BY-SA 3.0