If abelian group $G$ has an archimedean order then there is an order preserving isomorphism $\phi$ of $G$ onto a subgroup of $\mathbb{R}$. Here we can say that $G$ is archimedean totally ordered abelian group.
I try this by :-
Suppose $G$ has an archimedean order. Fix $x_0\in G$, $x_0>0$, and if $x\in G, x>0$, let $E(x)$ be the set of all rational numbers $m/n$ ($m,n\in \mathbb{N}$) such that $nx>mx_0$.
If $\phi(x)$ is the least upper bound of $E(x)$, and if $\phi(-x) = -\phi(x)$, then for $\epsilon>0$, $\exists r\in E(x)$ such that $\phi(x)<r+\epsilon$.
But I am unable to prove that $\phi$ is an isomorphism of $G$ onto a subgroup of $\mathbb{R}$ and $\phi$ preserves the order of $G$.
I tried it also by using dedekind cut. The proof is given as :- Fix an positive element $\alpha$ of $G$. If $\beta$ is any element of $G$, we divide the set of all rational numbers $m/n$ ($m\in \mathbb{Z}$ and $n \in \mathbb{N}$) into two classes $C$ and $C'$, as follows :
$m/n\in C$ if $m\alpha < n\beta$ , and
$m/n\in C'$ if $m\alpha \geq n\beta$.
Since $G$ is archimedean, so neither $C$ nor $C'$ is empty.
By definition of Dedekind cut, the pair of classes $C$ and $C'$ defines a Dedekind cut in the set of rational numbers. If $b$ is the real number defined by this Dedekind cut, we set $\phi(\beta)=b$.
Now I have to prove that $\phi$ is an order-preserving isomorphism of $G$ into a subgroup of the set of real numbers $\mathbb{R}$ but I am unable to prove this.
Any suggestions will be appreciated !