CharacterizationOfOrderedGroupsOfRankOne 2004-12-29 15:40:53 2005-02-01 01:59:59 Theorem isolated subgroup % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here For an ordered group, having \PMlinkname{rank}{IsolatedSubgroup} one is \PMlinkescapetext{equivalent} to an Archimedean property. In this entry, we use multiplicative notation for groups. \textbf{Lemma} An ordered group has rank one if and only if, for every two elements $x$ and $y$ such that $x < y < 1$, there exists an integer $n > 1$ such that $y^n < x$. {\it 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 $x' \in F$ such that $x' \neq 1$. By conclusion 1 of the basic theorem on ordered groups, either $x' < 1$, or $x' > 1$ (since we assumed that the case $x' = 1$ is excluded). If $x' < 1$, set $x = x'$. If not, by conclusion 5, if $x' > 1$, then we will have $x'^{-1} < 0$ and therefore will set $x = x'^{-1}$ when $x > 1$. Let $y$ be any element of $G$. There are five possibilities: \begin{enumerate} \item $y = 1$ \item $x = y$ \item $x < y < 1$ \item $y < x < 1$ \item $1 < y$ \end{enumerate} We shall show that in each of these cases, $y \in F$. \begin{enumerate} \item Trivial --- 1 is an element of every group. \item Trivial --- $x$ is assumed to belong to $F$ \item Since $F$ is an isolated subgroup, $y \in G$. \item By the Archimedean property,there exists an integer $n$ such that $x^n < y < 1$. Since $x^n \in F$ and $F$ is \PMlinkname{isolated}{IsolatedSubgroup}, it follows that $y \in F$. \item $1 < y$ By conclusion 5 of the basic theorem on ordered groups, $y^{-1} < 1$. By conclusion 1 of the same theorem, either $y^{-1} < x$ or $y^{-1} = 1$ or $x < y$. 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 $y \in F$. \end{enumerate} 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 $x < y^n < 1$ 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, $y^n < 1 < y$, 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 $$y^{m+n} \leqq z w \leqq y^{-m-n}$$ so $z w \in F_{m+n}$. Hence, if $z \in F$ and $w \in F$, then $zw \in F$. this \PMlinkescapetext{completes} 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 $x < y^n$ 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 $F$. \rightline{Q.E.D.}