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

We assume that G is non-abelian.

Question: Find $m,n$ such that $G$ is an ordered group, i.e. $G$ is a group on which a 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 $zx \le zy$.

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Crossposted at math.SE: math.stackexchange.com/questions/365660/ordered-groups-examples –  Qiaochu Yuan Apr 18 '13 at 20:02
    
Do you mean a non-trivial partial order? –  Wlodzimierz Holsztynski Apr 18 '13 at 20:34
    
Taking $m, n=1$ has a natural non-trivial ordering. I think you should sharpen your question. –  Noah S Apr 18 '13 at 21:26
    
No exercises here, please. –  Fernando Muro Apr 19 '13 at 2:26
    
Of course, I mean a non-trivial partial order. –  bsog Apr 19 '13 at 5:19
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1 Answer 1

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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Side remark: for $n\le -1$, $BS(1,n)$ is left-orderable but not bi-orderable. –  YCor Apr 24 '13 at 15:51
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