characterization of ordered groups of rank one
For an ordered group, having rank (http://planetmath.org/IsolatedSubgroup) one is to an Archimedean property. In this entry, we use multiplicative notation for groups.
Lemma An ordered group has rank one if and only if, for every two elements $x$ and $y$ such that $$, there exists an integer $n>1$ such that $$.
Proof Suppose that the Archimedean property is satisfied and that $F$ is an isolated subgroup of $G$. We shall show that if $F$ contains any element other than the identity^{}, then $F=G$. First note that there must exist an $x\in F$ such that $$. By assumption^{}, there must exist an element ${x}^{\prime}\in F$ such that ${x}^{\prime}\ne 1$. By conclusion^{} 1 of the basic theorem on ordered groups, either $$, or ${x}^{\prime}>1$ (since we assumed that the case ${x}^{\prime}=1$ is excluded). If $$, set $x={x}^{\prime}$. If not, by conclusion 5, if ${x}^{\prime}>1$, then we will have $$ and therefore will set $x={x}^{\prime 1}$ when $x>1$.
Let $y$ be any element of $G$. There are five possibilities:

1.
$y=1$

2.
$x=y$

3.
$$

4.
$$

5.
$$
We shall show that in each of these cases, $y\in F$.

1.
Trivial — 1 is an element of every group.

2.
Trivial — $x$ is assumed to belong to $F$

3.
Since $F$ is an isolated subgroup, $y\in G$.

4.
By the Archimedean property,there exists an integer $n$ such that $$. Since ${x}^{n}\in F$ and $F$ is isolated (http://planetmath.org/IsolatedSubgroup), it follows that $y\in F$.

5.
$$ By conclusion 5 of the basic theorem on ordered groups, $$. By conclusion 1 of the same theorem, either $$ or ${y}^{1}=1$ or $$. In each of these three cases, it follows that ${y}^{1}\in F$ from what we have already shown. Since $F$ is a group, ${y}^{1}\in F$ implies $y\in F$.
This shows that the only isolated subgroups of $G$ are the two trivial subgroups (i.e. the group $\{1\}$ and $G$ itself), and hence $G$ has rank one.
Next, suppose that $G$ does not enjoy the Archimedean property. Then there must exist $x\in G$ and $y\in G$ such that $$ for all integers $n>0$. Define the sets ${F}_{n}$ as
$${F}_{n}=\{z\in G\mid {y}^{n}\leqq z\leqq {y}^{n}\}$$ 
and define $F={\bigcup}_{n=1}^{\mathrm{\infty}}{F}_{n}$.
We shall show that $F$ is a subgroup^{} of $G$. First, note that, by a corollary of the basic theorem on ordered groups, $$, so $1\in {F}_{n}$ for all $n$, hence $1\in F$. Second, suppose that $z\in {F}_{n}$. Then ${y}^{n}\leqq z\leqq {y}^{n}$. By conclusion 5 of the basic theorem, ${y}^{n}\leqq z$ implies ${z}^{1}\leqq {y}^{n}$ and $z\leqq {y}^{n}$ implies ${y}^{n}\leqq {z}^{1}$. Thus, ${y}^{n}\leqq {z}^{1}\leqq {y}^{n}$, so ${z}^{1}\in {F}_{n}$. Hence, if $z\in F$, then ${z}^{1}\in F$. Third, suppose that $z\in F$ and $w\in F$. Then there must exist integers $m$ and $n$ such that $z\in {F}_{n}$ and $w\in {F}_{m}$, so
$${y}^{n}\leqq z\leqq {y}^{n}$$ 
and
$${y}^{m}\leqq w\leqq {y}^{m}.$$ 
Using conclusion 4 of the main theorem repeatedly, we conclude that
$${y}^{m+n}\leqq zw\leqq {y}^{mn}$$ 
so $zw\in {F}_{m+n}$. Hence, if $z\in F$ and $w\in F$, then $zw\in F$. this the proof that $F$ is a subgroup of $G$.
Not only is $F$ a subgroup of $G$, it is an isolated subgroup. Suppose that $f\in F$ and $g\in G$ and $f\leqq g\leqq 1$. Since $f\in F$, there must exist an $n$ such that $f\in {F}_{n}$, hence ${y}^{n}\leqq f$. By conclusion 2 of the basic theorem on ordered groups, ${y}^{n}\leqq f$ and $f\leqq g$ imply ${y}^{n}\leqq g$. Combining this with the facts that $g\leqq 1$ and $1\leqq {y}^{n}$, we conclude that ${y}^{n}\leqq g\leqq {y}^{n}$, so $g\in {F}_{n}$. Hence $g\in F$.
Note that $F$ is not trivial since $y\notin F$. The reason for this is that $x\notin {F}_{n}$ for any $n$ because we assumed that $$ for all $n$. Hence, the order of the group $G$ must be at least 2 because $F$ and $\{1\}$ are two examples of isolated subgroups of $F$.
Q.E.D.
Title  characterization of ordered groups of rank one 

Canonical name  CharacterizationOfOrderedGroupsOfRankOne 
Date of creation  20130322 14:55:15 
Last modified on  20130322 14:55:15 
Owner  rspuzio (6075) 
Last modified by  rspuzio (6075) 
Numerical id  11 
Author  rspuzio (6075) 
Entry type  Theorem 
Classification  msc 06A05 
Classification  msc 20F60 