Timeline for Bounding size of group by number of generators, order of elements, and nilpotency class (Restricted Burnside's)
Current License: CC BY-SA 4.0
10 events
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May 25, 2023 at 22:58 | history | became hot network question | |||
May 25, 2023 at 16:51 | vote | accept | Zach Hunter | ||
May 25, 2023 at 16:43 | comment | added | LSpice | @SeanEberhard, re, $\lvert G\rvert$ also makes sense for a Lie group $G$ over $\mathbb F_p$, in the sense of the group $\mathbb G(\mathbb F_p)$ of rational points of an affine algebraic $\mathbb F_p$-group $\mathbb G$. But I don't even know enough about the question to see this claim in either form on p. 4 = p. 264, so possibly my reading is absurd in context. (OP's comment, which I hadn't noticed, related to your answer suggests that it probably is.) | |
May 25, 2023 at 16:37 | comment | added | Sean Eberhard | @LSpice Well he refers to the order $|G|$, so Lie algebra makes sense. It would also make sense to assume $G$ is a finite $p$-group. It hardly makes a difference. | |
May 25, 2023 at 16:35 | comment | added | LSpice | Do you really mean that $G$ is a Lie algebra, not a Lie group? (Also, physical page 4 is logical page 264.) | |
May 25, 2023 at 16:34 | history | edited | LSpice | CC BY-SA 4.0 |
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May 25, 2023 at 16:31 | answer | added | Sean Eberhard | timeline score: 3 | |
May 25, 2023 at 15:10 | comment | added | Zach Hunter | ah, okay, so morally $G$ bijects to things generated by words of length $\le C$ in the original generators. okay, that matched my vague hunch, glad to know! | |
May 25, 2023 at 14:46 | comment | added | Sean Eberhard | The $k$th term of the lower central series modulo the $(k+1)$th term is generated by the left-normed commutators in $g_1, \dots, g_m$ of weight $k$, of which there are at most $m^{k-1}(m-1)$ for $k > 1$. Now just sum a geometric series and do some estimation, I think. It's obvious it should be $O(m^C)$ anyway. | |
May 25, 2023 at 14:18 | history | asked | Zach Hunter | CC BY-SA 4.0 |