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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 !

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  • $\begingroup$ Please suggest that how can I get the result. $\endgroup$ – Priya Jul 8 '14 at 6:32
  • $\begingroup$ @Amudhan please guide me in proving this result.. $\endgroup$ – Priya Jul 12 '14 at 11:50
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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 $\mathbb{R}$.

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