Timeline for Conditions for a group to be lattice-ordered
Current License: CC BY-SA 3.0
14 events
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| Feb 13, 2020 at 17:45 | history | edited | YCor |
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Sep 15, 2013 at 14:00 | answer | added | Tobias Schlemmer | timeline score: 3 | |
| Sep 15, 2013 at 13:21 | comment | added | Boris Novikov | Taking into account the comment of Tobias Schlemmer I corrected the answer. | |
| Sep 14, 2013 at 19:52 | vote | accept | Xodarap | ||
| Sep 15, 2013 at 16:12 | |||||
| Sep 14, 2013 at 17:15 | comment | added | Tobias Schlemmer | Is every right-ordered group with a lattice order already an ordered group? If not there are counterexamples as the implications $x≤1 ⇒ xz≤z$ and similarly for $x≥1$ is a conclusion of being right-ordered. Otherwise it would be sufficient to prove that it is a right-ordered group. | |
| Sep 14, 2013 at 16:42 | comment | added | Boris Novikov | I added the answer. | |
| Sep 14, 2013 at 11:56 | history | edited | Xodarap | CC BY-SA 3.0 |
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| Sep 14, 2013 at 11:56 | comment | added | Xodarap | @Boris: You are right! The group of functions $\mathbb R\to\mathbb R$ under addition with pointwise ordering is a counter example to this subset being closed. But the overall group still is lattice ordered. I will update the question. | |
| Sep 13, 2013 at 19:47 | comment | added | Boris Novikov | Please explain me: 1) why $\{g\in G: g\not\parallel 1\}$ is a subgroup? 2) why $\{g\in G: g\not\parallel 1\}$ is lattice-ordered? It don't seem following from your condition. | |
| Sep 13, 2013 at 13:13 | history | edited | Xodarap | CC BY-SA 3.0 |
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| Sep 13, 2013 at 12:30 | answer | added | Boris Novikov | timeline score: 2 | |
| Sep 13, 2013 at 11:40 | review | First posts | |||
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| Sep 13, 2013 at 11:21 | history | asked | Xodarap | CC BY-SA 3.0 |