# 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 $x, there exists an integer $n>1$ such that $y^{n}.

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 $x<1$. By assumption  , there must exist an element $x^{\prime}\in F$ such that $x^{\prime}\neq 1$. By conclusion  1 of the basic theorem on ordered groups, either $x^{\prime}<1$, or $x^{\prime}>1$ (since we assumed that the case $x^{\prime}=1$ is excluded). If $x^{\prime}<1$, set $x=x^{\prime}$. If not, by conclusion 5, if $x^{\prime}>1$, then we will have $x^{\prime-1}<0$ and therefore will set $x=x^{\prime-1}$ when $x>1$.

Let $y$ be any element of $G$. There are five possibilities:

1. 1.

$y=1$

2. 2.

$x=y$

3. 3.

$x

4. 4.

$y

5. 5.

$1

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

1. 1.

Trivial — 1 is an element of every group.

2. 2.

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

3. 3.

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

4. 4.

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

5. 5.

$1 By conclusion 5 of the basic theorem on ordered groups, $y^{-1}<1$. By conclusion 1 of the same theorem, either $y^{-1} or $y^{-1}=1$ or $x. 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 $x 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 $F=\bigcup_{n=1}^{\infty}F_{n}$.

We shall show that $F$ is a subgroup  of $G$. First, note that, by a corollary of the basic theorem on ordered groups, $y^{n}<1, 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 $x 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 CharacterizationOfOrderedGroupsOfRankOne 2013-03-22 14:55:15 2013-03-22 14:55:15 rspuzio (6075) rspuzio (6075) 11 rspuzio (6075) Theorem msc 06A05 msc 20F60