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. $x=y$

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

$$y^{m+n}\leqq zw\leqq y^{-m-n}$$

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	2013-03-22 14:55:15
Last modified on	2013-03-22 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