characterization of ordered groups of rank one

For an ordered group, having rank ( one is to an Archimedean property. In this entry, we use multiplicative notation for groups.

Lemma An ordered group has rank one if and only if, for every two elements x and y such that x<y<1, there exists an integer n>1 such that yn<x.

Proof Suppose that the Archimedean property is satisfied and that F is an isolated subgroup of G. We shall show that if F contains any element other than the identityPlanetmathPlanetmathPlanetmathPlanetmath, then F=G. First note that there must exist an xF such that x<1. By assumptionPlanetmathPlanetmath, there must exist an element xF such that x1. By conclusionMathworldPlanetmath 1 of the basic theorem on ordered groups, either x<1, or x>1 (since we assumed that the case x=1 is excluded). If x<1, set x=x. If not, by conclusion 5, if x>1, then we will have x-1<0 and therefore will set x=x-1 when x>1.

Let y be any element of G. There are five possibilities:

  1. 1.


  2. 2.


  3. 3.


  4. 4.


  5. 5.


We shall show that in each of these cases, yF.

  1. 1.

    Trivial — 1 is an element of every group.

  2. 2.

    Trivial — x is assumed to belong to F

  3. 3.

    Since F is an isolated subgroup, yG.

  4. 4.

    By the Archimedean property,there exists an integer n such that xn<y<1. Since xnF and F is isolated (, it follows that yF.

  5. 5.

    1<y By conclusion 5 of the basic theorem on ordered groups, y-1<1. By conclusion 1 of the same theorem, either y-1<x or y-1=1 or x<y. In each of these three cases, it follows that y-1F from what we have already shown. Since F is a group, y-1F implies yF.

This shows that the only isolated subgroups of G are the two trivial subgroups (i.e. the group {1} and G itself), and hence G has rank one.

Next, suppose that G does not enjoy the Archimedean property. Then there must exist xG and yG such that x<yn<1 for all integers n>0. Define the sets Fn as


and define F=n=1Fn.

We shall show that F is a subgroupMathworldPlanetmath of G. First, note that, by a corollary of the basic theorem on ordered groups, yn<1<y, so 1Fn for all n, hence 1F. Second, suppose that zFn. Then ynzy-n. By conclusion 5 of the basic theorem, ynz implies z-1y-n and zy-n implies ynz-1. Thus, ynz-1y-n, so z-1Fn. Hence, if zF, then z-1F. Third, suppose that zF and wF. Then there must exist integers m and n such that zFn and wFm, so




Using conclusion 4 of the main theorem repeatedly, we conclude that


so zwFm+n. Hence, if zF and wF, then zwF. this the proof that F is a subgroup of G.

Not only is F a subgroup of G, it is an isolated subgroup. Suppose that fF and gG and fg1. Since fF, there must exist an n such that fFn, hence ynf. By conclusion 2 of the basic theorem on ordered groups, ynf and fg imply yng. Combining this with the facts that g1 and 1y-n, we conclude that yngy-n, so gFn. Hence gF.

Note that F is not trivial since yF. The reason for this is that xFn for any n because we assumed that x<yn for all n. Hence, the order of the group G must be at least 2 because F and {1} are two examples of isolated subgroups of F.


Title characterization of ordered groups of rank one
Canonical name CharacterizationOfOrderedGroupsOfRankOne
Date of creation 2013-03-22 14:55:15
Last modified on 2013-03-22 14:55:15
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 11
Author rspuzio (6075)
Entry type Theorem
Classification msc 06A05
Classification msc 20F60