Return to Question

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 $zx \le zy$.

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.)