# 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

$$s=a{b}^{-1}=a1{b}^{-1}=ac{c}^{-1}{b}^{-1}=(ac){(bc)}^{-1}$$ |

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 $a\in S$.

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 |