$\mathbb{Z}\rtimes\mathbb{Z}$ is left-orderable but not right-orderable.

I was working in the following problem:

Prove that $\mathbb{Z}\rtimes\mathbb{Z}=\langle x,y\mid x^{-1}yx=y^{-1}\rangle$ is left-orderable but not right-orderable.

Where a group $G$ is right-orderable if there exists a total order $<$ such that $\forall~x,y,z\in G~x<y\Rightarrow xz<yz$, and it is left-orderable if there exists a total order $<$ such that $\forall~x,y,z\in G~x<y\Rightarrow zx<zy$.

Proving that $\mathbb{Z}\rtimes\mathbb{Z}$ is not right-orderable is easy:

• $1<y\Rightarrow y^{-1}<1\Rightarrow x^{-1}yx<1\Rightarrow yx<x\Rightarrow y<1$, which is a contradiction.

• $1>y\Rightarrow y^{-1}>1\Rightarrow x^{-1}yx>1\Rightarrow yx>x\Rightarrow y>1$, which is a contradiction.

However I can't prove that $\mathbb{Z}\rtimes\mathbb{Z}$ is left-orderable. I know that there are some results that solve this as an immediate consequence, but I want to find the explicit order (if possible) or to know what is the easiest result (the one which needs less previous results) that solves this.

Thanks for your help.

Notice that $yx=xy^{-1}$ implies that every element of $\mathbb{Z}\rtimes\mathbb{Z}$ is of the form $x^\alpha y^\beta$ with $\alpha,\beta\in\mathbb{Z}$.

Hence we can define the order $x^{\alpha_1} y^{\beta_1}>x^{\alpha_2} y^{\beta_2}$ if $\alpha_1>\alpha_2$ or $\alpha_1=\alpha_2$ and $\beta_1>\beta_2$, let us see that this order makes $\mathbb{Z}\rtimes\mathbb{Z}$ a left-ordered group. Let $a,b,c\in \mathbb{Z}\rtimes\mathbb{Z}$, then $a=x^{a_1}y^{a_2}$, $b=x^{b_1}y^{b_2}$ and $c=x^{c_1}y^{c_2}$ for some $a_1,a_2,b_1,b_2,c_1,c_2\in\mathbb{Z}$.

We have to prove that $a<b$ implies $ca<cb$. Notice that $a<b\Leftrightarrow x^{a_1}y^{a_2}<x^{b_1}y^{b_2}\Leftrightarrow a_1<b_1$ or $a_1=b_2$ and $a_2<b_2$. On the other hand $ca=x^{c_1}y^{c_2}x^{a_1}y^{a_2}=x^{c_1}x^{a_1}y^{(-1)^{a_1}c_2}y^{a_2}=x^{a_1+c_1}y^{a_2+(-1)^{a_1}c_2}$ and $cb=x^{b_1+c_1}y^{b_2+(-1)^{b_1}c_2}$. Then, if $a_1<b_1$, we have $a_1+c_1<b_1+c_1$ which implies $ca<cb$ and if $a_1=b_2$ and $a_2<b_2$ then $a_1+c_1=b_1+c_1$ and $a_2+(-1)^{a_1}c_2<b_2+(-1)^{b_1}c_2$, which implies $ca<cb$.