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: $$a < b \qquad a = b \qquad b < a$$

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, $a < c$ .

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 $ca < cb$

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, $a b^{-1} = s \in S$ . 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, $b^{-1} < a^{-1}$ .

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 $a \in S$ .