proof of basic theorem about ordered groups

Property 1:

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

ab-1S  ab-1=1  ab-1S-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 sS. Taking inversesMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath, this means that ba-1=s, so b<a, or equivalently a>b. Hence, one of the following three conditions must hold:

a<b  a=b  b<a

Property 2:

The hypotheses can be rewritten as

ab-1S  bc-1S

Multiplying, and remembering that S is closed under multiplicationPlanetmathPlanetmath,


In other words, a<c.

Property 3:

Suppose that a<b, so ab-1=sS. Then


so ac<bc.

By the defining property of S, we have csc-1S. Also,


hence (ca)(cb)-1S, 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 hypothesisMathworldPlanetmathPlanetmath, ab-1=sS. By the defining property, b-1sbS. Since b-1sb=b-1a, we have b-1aS. In other words, b-1<a-1.

Property 6:

By definition, a<1 means that a1-1S. Since 1-1=1 and a1=a, this is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath to stating that aS.

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