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, $a<b$ 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 $b<a$ , or equivalently $a>b$ . Hence, one of the following three conditions must hold:

a<b\qquad a=b\qquad b<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, $a<c$ .

Property 3:

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

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

so $ac<bc$ .

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 $ca<cb$

Property 4:

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

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, $b^{-1}<a^{-1}$ .

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
Canonical name	ProofOfBasicTheoremAboutOrderedGroups
Date of creation	2013-03-22 14:54:46
Last modified on	2013-03-22 14:54:46
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	14
Author	rspuzio (6075)
Entry type	Proof
Classification	msc 20F60
Classification	msc 06A05