# Ordered groups - examples

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.)

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

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).
• Side remark: for $n\le -1$, $BS(1,n)$ is left-orderable but not bi-orderable.