proof of basic theorem about ordered groups

Property 1:

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

$$a{b}^{-1}\in S\mathit{\hspace{1em}\hspace{1em}}a{b}^{-1}=1\mathit{\hspace{1em}\hspace{1em}}a{b}^{-1}\in S^{-1}$$

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

Property 2:

The hypotheses can be rewritten as

$$a{b}^{-1}\in S\mathit{\hspace{1em}\hspace{1em}}b{c}^{-1}\in S$$

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

$$a{c}^{-1}=(a{b}^{-1})(b{c}^{-1})\in S.$$

In other words, .

Property 3:

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

$$s=a{b}^{-1}=a1b^{-1}=ac{c}^{-1}{b}^{-1}=(ac){(bc)}^{-1}$$

so .

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

$$cs{c}^{-1}=ca{b}^{-1}{c}^{-1}=(ca){(cb)}^{-1},$$

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

Property 4:

By property 3, implies and likewise implies . Then, by property 2, we conclude .

Property 5:

By the hypothesis, $a{b}^{-1}=s\in S$. 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, .

Property 6:

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