# proof of basic theorem about ordered groups

## Property 1:

Consider $ab^{-1}\in G$. Since $G$ can be written as a pairwise disjoint union, exactly one of the following conditions must hold:

 $ab^{-1}\in S\qquad ab^{-1}=1\qquad ab^{-1}\in S^{-1}$

By definition of the ordering relation, $a if the first condition holds. If the second condition holds, then $a=b$. If the third condition holds, then we must have $ab^{-1}=s^{-1}$ for some $s\in S$. Taking inverses, this means that $ba^{-1}=s$, so $b, or equivalently $a>b$. Hence, one of the following three conditions must hold:

 $a

## Property 2:

The hypotheses can be rewritten as

 $ab^{-1}\in S\qquad bc^{-1}\in S$

Multiplying, and remembering that $S$ is closed under multiplication,

 $ac^{-1}=(ab^{-1})(bc^{-1})\in S.$

In other words, $a.

## Property 3:

Suppose that $a, so $ab^{-1}=s\in S$. Then

 $s=ab^{-1}=a1b^{-1}=acc^{-1}b^{-1}=(ac)(bc)^{-1}$

so $ac.

By the defining property of $S$, we have $csc^{-1}\in S$. Also,

 $csc^{-1}=cab^{-1}c^{-1}=(ca)(cb)^{-1},$

hence $(ca)(cb)^{-1}\in S$, so $ca

## Property 4:

By property 3, $a implies $ac and likewise $c implies $bc. Then, by property 2, we conclude $ac.

## Property 5:

By the hypothesis, $ab^{-1}=s\in S$. By the defining property, $b^{-1}sb\in S$. Since $b^{-1}sb=b^{-1}a$, we have $b^{-1}a\in S$. In other words, $b^{-1}.

## Property 6:

By definition, $a<1$ means that $a1^{-1}\in S$. Since $1^{-1}=1$ and $a1=a$, this is equivalent to stating that $a\in S$.

Title proof of basic theorem about ordered groups ProofOfBasicTheoremAboutOrderedGroups 2013-03-22 14:54:46 2013-03-22 14:54:46 rspuzio (6075) rspuzio (6075) 14 rspuzio (6075) Proof msc 20F60 msc 06A05