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Timeline for Ordered groups - examples

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Feb 9, 2020 at 12:20 review Close votes
Feb 17, 2020 at 3:01
Feb 9, 2020 at 11:58 comment added YCor (...) In particular, a group admits a nontrivial bi-invariant partial order iff it has a nonempty conjugation-invariant subsemigroup not containing $1$. This is obviously satisfied if it has a $\mathbf{Z}$ quotient, as Misha already said. On the other hand, this does not exist in the infinite dihedral group $D_\infty$. (Side remark: it also follows that a group admits a nontrivial left-invariant partial order iff it is not torsion.)
Feb 9, 2020 at 11:58 comment added YCor In a group $G$, there is a canonical bijection between the set of left-invariant partial preorders $\le$ and the set of submonoids of $G$. It is given by $\le\mapsto S_\le$ and $S\mapsto \le_S$, where $S_\le=\{g:g\ge 1_G\}$ and $g\le_S h$ iff $g^{-1}h\in S$. Then (a) $\le$ is nontrivial iff $S_\le\neq\{1_G\}$; (b) $\le$ is a order (i.e., antissymetric) iff $S\cap S^{-1}=\{1_G\}$; (c) $\le$ is bi-invariant iff $S$ is conjugation-invariant; (d) $\le$ is total iff $S\cup S^{-1}=G$. (...)
Feb 9, 2020 at 11:45 history edited YCor CC BY-SA 4.0
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May 7, 2013 at 14:21 history edited Lee Mosher
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Apr 24, 2013 at 20:57 comment added Misha Every BS-group $G$ admits an epimorphism $f$ to ${\mathbb Z}$. Pull-back of the order on ${\mathbb Z}$ will give a partial order on $G$ after you declare that distinct elements $x, y\in G$ satisfying $f(x)=f(y)$ are not comparable.
Apr 24, 2013 at 15:34 comment added Alain Valette Oops, thank you Yves, indeed my remark on $Homeo^+(\mathbb{R})$ is false. One day it will be possible to edit comments on MO.
Apr 24, 2013 at 15:25 comment added YCor @Alain: a countable group admits a left-invariant ordering iff it embeds in $Homeo^+(\mathbf{R})$. This proves that the (positive!) affine group of $\mathbf{R}$ is left-orderable. But it is indeed bi-orderable by a general fact on group extensions, here using the fact that the action of the multiplicative group of positive reals on the additive group of reals preserves an ordering.
Apr 24, 2013 at 14:39 comment added YCor It is safe to write bi-orderable group instead of ordered group. Because orderable groups denote, according to authors, either left-orderable or bi-orderable groups and is thus blatantly ambiguous. Also left/bi-ordered group is more suitable to mean a group endowed with a (left or bi...)-invariant total order. For instance, there is a Chabauty space of bi-ordered groups on $k$ generators, which surjects onto the Chabauty space of bi-orderable groups on $k$ generators, which is a closed subset of the space of marked groups on $k$ generators.
Apr 24, 2013 at 14:12 history edited Nick Gill
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Apr 24, 2013 at 13:01 answer added Dietrich Burde timeline score: 2
Apr 21, 2013 at 16:54 comment added Alain Valette Follow-up to Qiaochu's comment: two answers posted on math.stackexchange.com/questions/365660/ordered-groups-examples I found the one by Tournesol interesting. Recall that a group of orientation-preserving homeo's of $\mathbb{R}$ admits a total, bi-invariant ordering (with $f\leq g$ iff $f(x)\leq g(x)$ for every $x$). Since $B(m,1)$ embeds in the affine group of $\mathbb{R}$, it admits such an ordering.
Apr 19, 2013 at 5:34 comment added bsog To Noah: Note that I assume that G is non-abelian. If m=n=1, then G is abelian and the question is trivial.
Apr 19, 2013 at 5:19 comment added bsog Of course, I mean a non-trivial partial order.
Apr 19, 2013 at 2:26 comment added Fernando Muro No exercises here, please.
Apr 18, 2013 at 21:26 comment added Noah Schweber Taking $m, n=1$ has a natural non-trivial ordering. I think you should sharpen your question.
Apr 18, 2013 at 20:34 comment added Włodzimierz Holsztyński Do you mean a non-trivial partial order?
Apr 18, 2013 at 20:02 comment added Qiaochu Yuan Crossposted at math.SE: math.stackexchange.com/questions/365660/ordered-groups-examples
Apr 18, 2013 at 18:39 history asked bsog CC BY-SA 3.0