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It is given in Regular left-orders on groups that the solvable Baumslag-Solitar group $\mathrm{BS}(1,n)=\langle a, b\mid aba^{-1}=b^n\rangle $ is isomorphic to $\mathbb{Z}[1/n]\rtimes \mathbb{Z}$ for $n\in \mathbb{Z}.$ Can we find an isomorphism $\phi:\mathbb{Z}[1/n]\rtimes \mathbb{Z}\rightarrow \mathrm{BS}(1,n) $ such that if $u<v$ implies $\phi(u)<\phi(v),$ where $u,v\in\mathbb{Z}[1/n]\rtimes \mathbb{Z}$.

Let $u=(r,m)$ and $v=(r',m')$ so $u<v$ means $(r,m)<(r',m')$ which happens if either $m'>m$ or if $m=m'$ then $r'>r.$ In BS(1,n) every element can be written as can be written in the form $a^n(a^{-m}b^ka^m)$with $m\geq 0$, $n, k \in \mathbb{Z}$. So $a^{n_1}(a^{-{m_1}}b^{k_1}a^{m_1})>a^{n_2}(a^{-{m_2}}b^{k_2}a^{m_2})$ if either $n_1>n_2$ or if $n_1=n_2$ then $m_1>m_2.$

It is given in Regular left-orders on groups that the solvable Baumslag-Solitar group $\mathrm{BS}(1,n)=\langle a, b\mid aba^{-1}=b^n\rangle $ is isomorphic to $\mathbb{Z}[1/n]\rtimes \mathbb{Z}$ for $n\in \mathbb{Z}.$ Can we find an isomorphism $\phi:\mathbb{Z}[1/n]\rtimes \mathbb{Z}\rightarrow \mathrm{BS}(1,n) $ such that if $u<v$ implies $\phi(u)<\phi(v),$ where $u,v\in\mathbb{Z}[1/n]\rtimes \mathbb{Z}$.

Let $u=(r,m)$ and $v=(r',m')$ so $u<v$ means $(r,m)<(r',m')$ which happens if either $m'>m$ or if $m=m'$ then $r'>r.$ 

In BS(1,n) every element can be written as can be written in the form $a^n(a^{-m}b^ka^m)$with $m\geq 0$, $n, k \in \mathbb{Z}$. So I guess $a^{n_1}(a^{-{m_1}}b^{k_1}a^{m_1})>a^{n_2}(a^{-{m_2}}b^{k_2}a^{m_2})$ if either $n_1>n_2$ or if $n_1=n_2$ then $m_1>m_2.$ I am not sure about the ordering in $BS(1,n)$. When can we determine that one element is lesser than another in $BS(1,n)$?)

It is given in Regular left-orders on groups that the solvable Baumslag-Solitar group $\mathrm{BS}(1,n)=\langle a, b\mid aba^{-1}=b^n\rangle $ is isomorphic to $\mathbb{Z}[1/n]\rtimes \mathbb{Z}$ for $n\in \mathbb{Z}.$ Can we find an isomorphism $\phi:\mathbb{Z}[1/n]\rtimes \mathbb{Z}\rightarrow \mathrm{BS}(1,n) $ such that if $a<b$ implies $\phi(a)<\phi(b),$ where $\mathbb{Z}[1/n]\rtimes \mathbb{Z}$.

Let $a=(r,m)$ and $b=(r',m')$ so $a<b$ means $(r,m)<(r',m')$ which happens if either $m'>m$ or if $m=m'$ then $r'>r.$ In BS(1,n) every element can be written as can be written in the form $a^n(a^{-m}b^ka^m)$with $m\geq 0$, $n, k \in \mathbb{Z}$. So $a^{n_1}(a^{-{m_1}}b^{k_1}a^{m_1})>a^{n_2}(a^{-{m_2}}b^{k_2}a^{m_2})$ if either $n_1>n_2$ or if $n_1=n_2$ then $m_1>m_2.$

It is given in Regular left-orders on groups that the solvable Baumslag-Solitar group $\mathrm{BS}(1,n)=\langle a, b\mid aba^{-1}=b^n\rangle $ is isomorphic to $\mathbb{Z}[1/n]\rtimes \mathbb{Z}$ for $n\in \mathbb{Z}.$ Can we find an isomorphism $\phi:\mathbb{Z}[1/n]\rtimes \mathbb{Z}\rightarrow \mathrm{BS}(1,n) $ such that if $u<v$ implies $\phi(u)<\phi(v),$ where $u,v\in\mathbb{Z}[1/n]\rtimes \mathbb{Z}$.

Let $u=(r,m)$ and $v=(r',m')$ so $u<v$ means $(r,m)<(r',m')$ which happens if either $m'>m$ or if $m=m'$ then $r'>r.$ In BS(1,n) every element can be written as can be written in the form $a^n(a^{-m}b^ka^m)$with $m\geq 0$, $n, k \in \mathbb{Z}$. So $a^{n_1}(a^{-{m_1}}b^{k_1}a^{m_1})>a^{n_2}(a^{-{m_2}}b^{k_2}a^{m_2})$ if either $n_1>n_2$ or if $n_1=n_2$ then $m_1>m_2.$

It is given in Regular left-orders on groups that the solvable Baumslag-Solitar group $\mathrm{BS}(1,n)=\langle a, b\mid aba^{-1}=b^n\rangle $ is isomorphic to $\mathbb{Z}[1/n]\rtimes \mathbb{Z}$ for $n\in \mathbb{Z}.$ Can we find an isomorphism $\phi:\mathbb{Z}[1/n]\rtimes \mathbb{Z}\rightarrow \mathrm{BS}(1,n) $ such that if $a<b$ implies $\phi(a)<\phi(b),$ where $\mathbb{Z}[1/n]\rtimes \mathbb{Z}$.

It is given in Regular left-orders on groups that the solvable Baumslag-Solitar group $\mathrm{BS}(1,n)=\langle a, b\mid aba^{-1}=b^n\rangle $ is isomorphic to $\mathbb{Z}[1/n]\rtimes \mathbb{Z}$ for $n\in \mathbb{Z}.$ Can we find an isomorphism $\phi:\mathbb{Z}[1/n]\rtimes \mathbb{Z}\rightarrow \mathrm{BS}(1,n) $ such that if $a<b$ implies $\phi(a)<\phi(b),$ where $\mathbb{Z}[1/n]\rtimes \mathbb{Z}$.

Let $a=(r,m)$ and $b=(r',m')$ so $a<b$ means $(r,m)<(r',m')$ which happens if either $m'>m$ or if $m=m'$ then $r'>r.$ In BS(1,n) every element can be written as can be written in the form $a^n(a^{-m}b^ka^m)$with $m\geq 0$, $n, k \in \mathbb{Z}$. So $a^{n_1}(a^{-{m_1}}b^{k_1}a^{m_1})>a^{n_2}(a^{-{m_2}}b^{k_2}a^{m_2})$ if either $n_1>n_2$ or if $n_1=n_2$ then $m_1>m_2.$