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4
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'ordered group'
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| Title of object: |
ordered group |
| Canonical Name: |
OrderedGroup |
| Type: |
Definition |
| Created on: |
2004-12-27 12:58:11 |
| Modified on: |
2004-12-27 14:03:17 |
| Classification: |
msc:06A05, msc:20F60 |
| Defines: |
ordered group equipped with zero |
Revision comment (for changes between this and next version):
| Changes for correction #5483 ('Please number the list'). |
Preamble:
% 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}
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%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
%\usepackage{amsthm}
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%\usepackage{xypic}
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Content:
\textbf{Definition 1.} \,We say that the subsemigroup $S$ of the group $G$ (with the operation denoted multiplicatively) defines an {\em order for the group} $G$, if
\begin{enumerate}
\item $a^{-1}Sa \subseteq S \quad \forall a\in G,$
\item $G = S\cup \{1\} \cup S^{-1}$ \,\,where \,$S^{-1} = \{s^{-1}: \,s\in S\}$\, and the members of the union are pairwise disjoint.
\end{enumerate}
The order ``$<$'' of the group $G$ is explicitely given by setting in $G$:
$$a < b \,\, \Leftrightarrow \,\,ab^{-1}\in S$$
Then we speak of the {\em ordered group} $(G,\,<)$ or simply $G$.
\textbf{Theorem.} \,The order ``$<$'' defined by the subsemigroup $S$ of the group $G$ has the following properties.
\begin{itemize}
\item Exactly one of the conditions \,\,$a < b,\,\,a = b,\,\,b < a$\,\, holds.
\item $a < b \,\land\, b < c \,\,\Rightarrow\,\,a < c$
\item $a < b \,\,\Rightarrow\,\, ac < bc \,\land\, ca < cb$
\item $a < b \,\land\, c < d \,\,\Rightarrow\,\, ac < bd$
\item $a < b \,\,\Leftrightarrow\,\, b^{-1} < a^{-1}$
\item $a < 1 \,\,\Leftrightarrow\,\, a\in S$
\end{itemize}
\textbf{Definition 2.} \,The set $G$ is an {\em ordered group equipped with zero} 0, if the set $G^*$ of its elements distinct from its element 0 form an ordered group \,$(G^*,\,<)$\, and if
\begin{enumerate}
\item $0a = a0 = 0 \quad\forall a\in G,$
\item $0 < a \quad\forall a\in G^*.$
\end{enumerate} |
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