Timeline for size of smallest generating set of a group
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Mar 29, 2022 at 6:32 | comment | added | David Roberts♦ | The link in Ian's comment is broken, here's a replacement: arxiv.org/abs/math/0407438 | |
| Jun 3, 2014 at 19:06 | comment | added | argentpepper | The rank of a finite group given as its multiplication table is efficiently computable. I answered a similar question, and I'm working on a brief explanation of why the rank is efficiently computable here. | |
| S Jul 26, 2013 at 20:07 | history | suggested | Dominik | CC BY-SA 3.0 |
Added $ for correct matrix depiction
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| Jul 26, 2013 at 20:04 | review | Suggested edits | |||
| S Jul 26, 2013 at 20:07 | |||||
| Jul 16, 2012 at 11:49 | answer | added | Venkataramana | timeline score: 5 | |
| Sep 28, 2011 at 12:59 | history | edited | Sam Nead | CC BY-SA 3.0 |
Tried to tidy the latex a bit
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| Sep 28, 2011 at 11:04 | history | edited | Igor Rivin | CC BY-SA 3.0 |
added greater wisdom.
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| Sep 27, 2011 at 9:20 | answer | added | kassabov | timeline score: 5 | |
| Sep 27, 2011 at 4:58 | comment | added | Ian Agol | Kapovich and Weidmann proved that the rank is computable for Kleinian groups. front.math.ucdavis.edu/0407.5438 Their method should apply more generally to hyperbolic groups which are locally quasi-convex. | |
| Sep 27, 2011 at 2:45 | answer | added | Igor Pak | timeline score: 13 | |
| Sep 26, 2011 at 23:54 | answer | added | user6976 | timeline score: 7 | |
| Sep 26, 2011 at 16:02 | comment | added | user9072 | @Andy and Igor: Thank you for the explanation! | |
| Sep 26, 2011 at 15:49 | comment | added | Igor Rivin | @quid: in addition, what Richard really meant, i think, is that in some cases there are heuristics for UNDECIDABLE problems which work reasonably well in practice (Todd-Coxeter is the poster child for this) | |
| Sep 26, 2011 at 15:45 | comment | added | Andy Putman | @quid : Algorithms in combinatorial group theory often have absurdly long run times. For instance, to solve the word problem for residually finite groups, you have one computer enumerating all finite groups and all possible homomorphisms to those finite groups and one computer systematically enumerating relations in the group. Eventually one of these computers will win (the first one showing that a word in the generators is nontrivial, the second one showing that the word is a relation), but the mind boggles trying to estimate the runtime... | |
| Sep 26, 2011 at 15:10 | comment | added | user9072 | Now, I am totally confused. | |
| Sep 26, 2011 at 14:21 | comment | added | Andy Putman | Sacrificing a goat would be more effective than running some algorithms in combinatorial group theory :) | |
| Sep 26, 2011 at 13:36 | history | edited | Igor Rivin | CC BY-SA 3.0 |
added a possible clarification
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| Sep 26, 2011 at 13:35 | comment | added | Igor Rivin | You mean, sacrificing a goat and dancing around a campfire is better than an algorithm? If not, what do you mean? I am somewhat confused... | |
| Sep 26, 2011 at 13:24 | comment | added | Autumn Kent | Do you actually want to know, or do you want an algorithm? | |
| Sep 26, 2011 at 13:09 | history | asked | Igor Rivin | CC BY-SA 3.0 |