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Version 4 |
| 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. |
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. |
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| \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 $y^n < x$. |
\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 $y^n < x$. |
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| {\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 $x > 1$. |
{\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 $x > 1$. |
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| Let $y$ be any element of $G$. There are five possibilities: |
Let $y$ be any element of $G$. There are five possibilities: |
| \begin{enumerate} |
\begin{enumerate} |
| \item $y = 1$ |
\item $y = 1$ |
| \item $x = y$ |
\item $x = y$ |
| \item $x < y < 1$ |
\item $x < y < 1$ |
| \item $y < x < 1$ |
\item $y < x < 1$ |
| \item $1 < y$ |
\item $1 < y$ |
| \end{enumerate} |
\end{enumerate} |
| We shall show that in each of these cases, $y \in F$. |
We shall show that in each of these cases, $y \in F$. |
| \begin{enumerate} |
\begin{enumerate} |
| \item Trivial --- 1 is an element of every group. |
\item Trivial --- 1 is an element of every group. |
| \item Trivial --- $x$ is assumed to belong to $F$ |
\item Trivial --- $x$ is assumed to belong to $F$ |
| \item Since $F$ is an isolated subgroup, $y \in G$ in case 3. |
\item Since $F$ is an isolated subgroup, $y \in G$ in case 3. |
| \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 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 $y \in F$. |
\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} |
\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. |
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. |
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| 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 |
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 |
| $$F_n = \{ z \in G \mid y^n \leqq z \leqq y^{-n} \}$$ |
$$F_n = \{ z \in G \mid y^n \leqq z \leqq y^{-n} \}$$ |
| and define $F = \bigcup_{n=1}^\infty F_n$. |
and define $F = \bigcup_{n=1}^\infty F_n$. |
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| We shall show that $F$ is a subgroup of $G$. First, note that, by a corrolary 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 |
We shall show that $F$ is a subgroup of $G$. First, note that, by a corrolary 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 |
| $$y^n \leqq z \leqq y^{-n}$$ |
$$y^n \leqq z \leqq y^{-n}$$ |
| and |
and |
| $$y^m \leqq w \leqq y^{-m}.$$ |
$$y^m \leqq w \leqq y^{-m}.$$ |
| Using conclusion 4 of the main theorem repeatedly, we conclude that |
Using conclusion 4 of the main theorem repeatedly, we conclude that |
| $$y^{m+n} \leqq z w \leqq y^{-m-n}$$ |
$$y^{m+n} \leqq z w \leqq y^{-m-n}$$ |
| 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 $G$. |
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 $G$. |
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| 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 $g \in F$. |
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 $g \in F$. |
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| 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 $F$. |
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.} |
\rightline{Q.E.D.} |
