Timeline for How can you order a free group?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jun 8, 2022 at 17:33 | answer | added | marco de manccini | timeline score: 3 | |
| Jun 8, 2022 at 8:33 | history | edited | Ville Salo | CC BY-SA 4.0 |
trivial clarification in formula: $a \neq c$ (since this is on the front page now)
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Feb 9, 2020 at 10:06 | history | edited | YCor |
edited tags; edited tags
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| Jan 22, 2020 at 5:14 | comment | added | Ville Salo | I mean simple in the usual sense. With the first question I meant, does everything act on a scattered left order? If yes, then your observation is not very interesting. With the second question, I figured if you have a simple torsion-free non-orderable group, then that will be a counterexample. But then I realized $\mathbb{Z}[1/2]$ is an easier example. I agree you can also f.g.ify it. | |
| Jan 21, 2020 at 23:07 | comment | added | YCor | Don't understand your last 2 questions. You're using "simple" in the meaning "non-complicated"? Anyway every torsion-free abelian group is orderable. Nevertheless I agree that for $p\ge 2$ $\mathbf{Z}[1/p]$ (and hence $BS(1,p)$) cannot act faithfully on any scattered totally ordered space. | |
| Jan 21, 2020 at 20:45 | comment | added | Ville Salo | Erm wait, dyadic rationals. | |
| Jan 21, 2020 at 20:42 | comment | added | Ville Salo | Does everything? Is there a simple torsion-free non-orderable group? | |
| Jan 21, 2020 at 19:20 | comment | added | YCor | Weaker than your first question, but $F_2$ acts faithfully on a scattered left order. Indeed, for each quotient by a term of the lower central series one get such a (non-faithful) action, and then one can concatenate. | |
| Jan 21, 2020 at 14:10 | history | asked | Ville Salo | CC BY-SA 4.0 |