Timeline for Groups whose derived length is logarithmic in the order?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Apr 30 at 13:24 | answer | added | David E Speyer | timeline score: 2 | |
| Apr 30 at 11:10 | comment | added | Emil Jeřábek | I disagree with labelling this misuse of $O$ as being "computer science" usage. The definition of $O$ in computer science is the same as elsewhere in mathematics. I'm sure many people get it wrong, especially in informal contexts, but this is not limited to computer science. As for what was, actually, the question intended by the OP, I think this is hard to guess without further input. Your interpretation may or may not be correct, I won't speculate on that. In any case, Dave Benson's answer clarifies the situation. | |
| Apr 30 at 10:43 | comment | added | YCor | @EmilJeřábek cf the comment below Dave Benson's answer. I don't know if you disagree with this use of $O$ (I disagree too) or with my interpretation of the question. | |
| Apr 30 at 10:16 | history | edited | Emil Jeřábek | CC BY-SA 4.0 |
remove ignorant bashing of computer science
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| Apr 30 at 9:50 | comment | added | Yiftach Barnea | @YCor my guess would be that the meaning is the standard meaning as in subgroup growth type, namely, the $c_1v_n<u_n$ is not for all $n$, but for infinitely many $n$'s (the other inequality is for all $n$). I should point out that this would not be an equivalent relation. Informally, it is the "best upper bound". (I added the not to this comment). | |
| Apr 30 at 9:40 | comment | added | Emil Jeřábek | I don’t understand the edit: “$c_1v_n<u_n<c_2v_n$ for positive constants $c_1,c_2$ and $n\gg 1$” is the computer science meaning of $u_n=\Theta(v_n)$, not of $u_n=O(v_n)$. In fact, the $\Theta$ notation was invented by Knuth, a computer scientist, for this very purpose. | |
| Apr 30 at 9:25 | history | edited | YCor | CC BY-SA 4.0 |
added context
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| Apr 30 at 8:33 | history | became hot network question | |||
| Apr 29 at 20:34 | history | edited | LSpice | CC BY-SA 4.0 |
Link syntax
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| Apr 29 at 19:54 | comment | added | Dave Benson | @KentaSuzuki You might be thinking of nilpotence class. The derived length is logarithmic in $n$, so it's much smaller than $\sqrt{\log|G|}$. I guess it's still big-O of it, but it's possible that's not what you intended, nor the original poster. | |
| Apr 29 at 19:10 | comment | added | Derek Holt | Yes, (nontrivial) abelian groups. Their derived length is $1$, which is $O(\log |G|)$. | |
| Apr 29 at 19:09 | answer | added | Dave Benson | timeline score: 10 | |
| Apr 29 at 18:19 | comment | added | Kenta Suzuki | $G$ being the group of strictly upper triangular $n\times n$ matrices over $\mathbb F_p$ gives $O(\sqrt{\log|G|})$. | |
| Apr 29 at 17:25 | history | asked | User01 | CC BY-SA 4.0 |