Timeline for Finite groups generated by 3 involutions interchanging disjoint residue classes of the integers
Current License: CC BY-SA 3.0
11 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Apr 20, 2016 at 13:15 | comment | added | Stefan Kohl♦ | @DimaPasechnik: Let $a := \tau_{1(7),6(7)}$, $b := \tau_{0(5),3(5)}$ and $c := \tau_{0(4),5(6)}$ be the generators of the group $G$ from the question. Then we have $|ab| = 4$, $|ac| = 12$ and $|bc| = 60$. | |
| Apr 20, 2016 at 13:08 | history | edited | Stefan Kohl♦ | CC BY-SA 3.0 |
Fixed notation (was just pasted from GAP before).
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| Apr 20, 2016 at 13:04 | comment | added | Stefan Kohl♦ | @NickGill: For each $m \leq 9$, up to conjugacy and choice of generators there is only one group of maximal finite order (that means in particular that the groups for $m = 7, 8, 9$ are the same), and these groups are not solvable. | |
| Apr 20, 2016 at 11:26 | comment | added | Dima Pasechnik | It's a quotient of a Coxeter group with 3 generators, in each case, right? Could you specify the order of the product of each pair of generators? | |
| Apr 20, 2016 at 11:20 | comment | added | Nick Gill | Do you know if, in cases 7,8 and 9, there is only one group turning up with that maximal order? Any information about, say, solvability of the group(s) in question? | |
| Apr 20, 2016 at 10:58 | comment | added | Stefan Kohl♦ | I have revised, updated and undeleted this question from 2012. In particular I have added data computed since then. | |
| Apr 20, 2016 at 10:57 | history | edited | Stefan Kohl♦ | CC BY-SA 3.0 |
Revised and updated this old question from 2012.
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| Apr 20, 2016 at 10:54 | history | undeleted | Stefan Kohl♦ | ||
| Sep 1, 2013 at 10:21 | history | deleted | Stefan Kohl♦ | via Vote | |
| Nov 19, 2012 at 0:15 | history | asked | Stefan Kohl♦ | CC BY-SA 3.0 |