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proof of basic theorem about ordered groups

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\subsubsection*{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:
 $$a b^{-1} \in S \qquad a b^{-1} = 1 \qquad a b^{-1} \in S^{-1}$$
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 $a b^{-1} = s^{-1}$ for some $s \in S$.  Taking inverses, this means that $b a^{-1} = s$, so $b < a$, or equivalently $a > b$.  Hence, one of the following three conditions must hold:
 $$a < b \qquad a = b \qquad b < a$$

\subsubsection*{Property 2:}

The hypotheses can be rewritten as
 $$a b^{-1} \in S \qquad b c^{-1} \in S$$
Multiplying, and remembering that $S$ is closed under multiplication,
 $$a c^{-1} = ( a b^{-1} )( b c^{-1} ) \in S.$$
In other words, $a < c$.

\subsubsection*{Property 3:}

Suppose that $a < b$, so $a b ^{-1} = s \in S$.   Then
 $$s = a b^{-1} = a 1 b^{-1} = a c c^{-1} b^{-1} = (ac) (bc)^{-1}$$
so $ac < bc$.

By the defining property of $S$, we have $c s c^{-1} \in S$.  Also,
 $$c s c^{-1} = c a b^{-1} c^{-1} = (ca) (cb)^{-1},$$
hence $(ca) (cb)^{-1} \in S$, so $ca < cb$

\subsubsection*{Property 4:}

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

\subsubsection*{Property 5:}

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

\subsubsection*{Property 6:}

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