Order Preserving Isomorphism

Following your notation, for $\beta\in G$, being $(C,C')$ the corresponding Dedekind-cut, then $C=\{r\in\mathbb{Q}:r<\phi(\beta)\}$ and $C'=\{r\in\mathbb{Q}:r\geq\phi(\beta)\}$.

For $\beta$ and $\gamma$, for $x<\phi(\beta)+\phi(\gamma)$, then let $y=\frac{x+\phi(\beta)-\phi(\gamma)}{2}$ and $z=\frac{x-\phi(\beta)+\phi(\gamma)}{2}$, then $x=y+z$ and $y<\phi(\beta)$ and $z<\phi(\gamma)$, so there are $y<\frac{m}{m'}<\phi(\beta)$ and $z<\frac{n}{n'}<\phi(\gamma)$ with $m'>0$ and $n'>0$, so $m\alpha<m'\beta$ and $n\alpha<n'\gamma$, so $(mn'+m'n)\alpha<m'n'(\beta+\gamma)$, so $x=y+z<\frac{m}{m'}+\frac{n}{n'}=\frac{mn'+m'n}{m'n'}<\phi(\beta+\gamma)$; thus $\phi(\beta)+\phi(\gamma)\leq\phi(\beta+\gamma)$.

For $\beta$ and $\gamma$, for $x>\phi(\beta)+\phi(\gamma)$, then let $y=\frac{x+\phi(\beta)-\phi(\gamma)}{2}$ and $z=\frac{x-\phi(\beta)+\phi(\gamma)}{2}$, then $x=y+z$ and $y>\phi(\beta)$ and $z>\phi(\gamma)$, so there are $y>\frac{m}{m'}>\phi(\beta)$ and $z>\frac{n}{n'}>\phi(\gamma)$ with $m'>0$ and $n'>0$, so $m\alpha\geq m'\beta$ and $n\alpha\geq n'\gamma$, so $(mn'+m'n)\alpha\geq m'n'(\beta+\gamma)$, so $x=y+z>\frac{m}{m'}+\frac{n}{n'}=\frac{mn'+m'n}{m'n'}\geq \phi(\beta+\gamma)$; thus $\phi(\beta)+\phi(\gamma)\geq\phi(\beta+\gamma)$.

For $\beta$ and $\gamma$, if $\beta\leq\gamma$, then for $x<\phi(\beta)$, there is $x<\frac{m}{m'}<\phi(\beta)$ with $m'>0$, so $m\alpha<m'\beta\leq m'\gamma$, so $x<\frac{m}{m'}<\phi(\gamma)$; thus $\phi(\beta)\leq\phi(\gamma)$.

For $\beta$, if $\phi(\beta)=0$, then for any $n>0$ we have $\frac{-1}{n}<\phi(\beta)\leq\frac{1}{n}$, so $-\alpha<n\beta$ and $\alpha\geq n\beta$; thus, by the Archimedean property, we have $\beta=0$.

Therefore, $\phi$ is an order-preserving isomorphism of $G$ into a subgroup of $\mathbb{R}$.