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Let $G=\operatorname{BS}(m,n)$ denote the Baumslag–Solitar group defined by the presentation $\langle a,b: b^m a=a b^n\rangle$.

We assume that $G$ is non-abelian, i.e., $m,n\in\mathbb{Z}\smallsetminus\{0\}$ and $(m,n)\neq \pm (1,1)$.

Question: Find $m,n$ such that $G$ is an ordered group (in a nontrivial way), i.e. $G$ is a group on which a (nontrivial) partial order relation $\le $ is given such that for any elements $x,y,z \in G$, from $x \le y$ it follows that $xz \le yz$ and $zx \le zy$.

(The trivial partial order is the equality relation.)

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    $\begingroup$ Crossposted at math.SE: math.stackexchange.com/questions/365660/ordered-groups-examples $\endgroup$ Apr 18, 2013 at 20:02
  • $\begingroup$ Do you mean a non-trivial partial order? $\endgroup$ Apr 18, 2013 at 20:34
  • $\begingroup$ Taking $m, n=1$ has a natural non-trivial ordering. I think you should sharpen your question. $\endgroup$ Apr 18, 2013 at 21:26
  • $\begingroup$ No exercises here, please. $\endgroup$ Apr 19, 2013 at 2:26
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    $\begingroup$ 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. $\endgroup$
    – Misha
    Apr 24, 2013 at 20:57

1 Answer 1

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The Baumslag-Solitar groups $B(1,n)$ for $n\ge 2$ can be ordered, even with a total order. Indeed, all orderings have been classified in this case, see C. Rivas: On spaces of Conradian group orderings. Here is one way to obtain an ordering for $B(1,n)$. Start with an exact sequence $$ 0 \rightarrow \mathbb{Z}\rightarrow BS(1,n) \rightarrow \mathbb{Z}[1/n]\rightarrow 0, $$ and define bi-orderings from this: $(k,r/n^j)\succ id$ iff either $k>0$ or $k=0$ and $r/n^j>0$. Another way to see the result is to note that $B(1,n)$ embedds into the affine group, and hence obtains such an ordering (see Alain Valette's remark).

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  • $\begingroup$ Side remark: for $n\le -1$, $BS(1,n)$ is left-orderable but not bi-orderable. $\endgroup$
    – YCor
    Apr 24, 2013 at 15:51

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