Timeline for Is there a left-orderable profinite group?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Feb 16, 2020 at 2:02 | comment | added | YCor | I added the condition that the graph of $<$ is open, since the question is not very interesting without it (cf the remark on torsion-free abelian groups) and because it's the assumption made in the accepted answer. | |
| Feb 16, 2020 at 2:00 | history | edited | YCor | CC BY-SA 4.0 |
added missing assumption, added tag
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| Dec 16, 2014 at 23:53 | vote | accept | Pablo | ||
| Dec 16, 2014 at 19:38 | answer | added | Jan-Christoph Schlage-Puchta | timeline score: 7 | |
| Dec 15, 2014 at 17:09 | comment | added | Benjamin Steinberg | The partial order is a set of ordered pairs. If this is set is closed in the product topology the ordering will be trivial. | |
| Dec 15, 2014 at 16:24 | comment | added | Pablo | Benjamin, I do not understand your comment. The profinite group is not necessarily a product of finite groups. | |
| Dec 15, 2014 at 15:47 | review | Close votes | |||
| Dec 16, 2014 at 7:03 | |||||
| Dec 15, 2014 at 15:25 | comment | added | Benjamin Steinberg | If the order is closed in the topology on the direct product the answer is no by reduction to the finite case I believe. | |
| Dec 15, 2014 at 14:26 | comment | added | jmc | Ok, yes, a relation with the topology is very reasonable. (I kind of assumed it, but indeed, you did not state it.) | |
| Dec 15, 2014 at 14:20 | comment | added | Pablo | @jmc maybe continuous orders should be considered since by a result of Levi every abelian group with no torsion is orderable and this works for the additive profinite p-adic groups. | |
| Dec 15, 2014 at 13:08 | comment | added | jmc | My guess would be that only $G = \{1\}$ works (finite groups are profinite). I don't have any proofs though. | |
| Dec 15, 2014 at 13:03 | history | asked | Pablo | CC BY-SA 3.0 |