proof of embedding theorem for ordered abelian groups of rank one

To prove the theoremMathworldPlanetmath, we shall use the characterization of ordered groups of rank one. Using this result, the theorem to be be proved can be reformulated as follows: (Note that we are now using additive notation for groups, so the formulation of the Archimedean property appears slightly differently than it did in the entry characterizing the ordered groups of rank one.)

Theorem.  Let G be an AbelianMathworldPlanetmath ordered group with order ( at least 2.  Then G enjoys the Archimedean property


iff there is an order-preserving isomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath from G onto some subgroupMathworldPlanetmathPlanetmath of the additive groupMathworldPlanetmath of real numbers.

Proof   One of the two implicationsMathworldPlanetmath is rather easy. Since the real numbers enjoy the Archimedean property and it is obvious that any subset of a set enjoying the Archimedean property also enjoys this property, it follows that every subgroup of the additive group of the real numbers enjoys the Archimedean property. It only remains to show that every group enjoying this property is isomorphic to a subgroup of additive group of the real numbers.

To accomplish this, we shall use the technique of Dedekind cuts to map our group into the real numbers. Since the order of our group is at least 2, there exists an element xG such that x0. By the basic theorem on ordered groups, it follows that either x>0 or x<0 and that, if x>0 then -x<0. Thus there exists an element yG such that -y<0 — if x<0, set y=x, else set y=-x.

We shall now construct a map μ:G. Define the sets A(z) and B(z) as follows:


By Corollary 2 to the basic theorem on ordered groups, it follows that, for any strictly positive integer k, we have (m,n)A(z) if and only if (km,kn)A(z). Likewise, (m,n)B(z) if and only if (km,kn)B(z). This means that A(z) and B(z) may be regarded as sets of rational numbers, and we can rewrite the defining equations as


We shall now show that (A(z),B(z)) is a Dedekind cut. This requires us to verify the three defining properties of a Dedekind cut. First, we need to check that neither A(z) nor B(z) are empty. We shall only present the proof for A(z) since the proof for B(z) can easily be obtained from it by reversing the inequality signs suitably. By the basic theorem, either -y<z or z=-y or z<-y. In the first case, 1/1A(z). In the second case, 2/1A(z). In the third case, the Archimedean property implies that there exists n such that -1<nz<0, so 1/nA(z). Second, we need to check that every element of A(z) is less than every element of B(z). Suppose that m/nA(z) and m/nB(z), so that my<nz and nz<my. Then, by Corollary 2 of the basic theorem, mny<nnz and nnz<mny. By conclusionMathworldPlanetmath 2 of the basic theorem, this implies that mny<mny. By Corollary 2, this means that mn<mn. Dividing by nn, we conclude that m/n<m/n. Third, we need to show that at most one rational number does not belong to either A(z) or to B(z). Suppose that m/n does not belong to either A(z) or B(z). Then it cannot be the case that either




By the first conclusion of the basic theorem, it follows that my=nz. Now suppose that m/n also does not belong to either A(z) or to B(z). By the same line of reasoning, we must have my=nz. This would imply that mny=nnz. Similarly, mny=nnz. Combining these two facts using conclusion 2 of the basic theorem, we would have mn=mn, which implies that m/n=m/n.

Since (A(z),B(z)) is a Dedekind cut, it defines a real number. We shall define μ(z) to be this number.

We will now show that μ is a homomrphism. To accomplish this, it suffices to show that μ(a-b)=μ(a)-μ(b) for every a,bG. Suppose that m1/n1A(a) and m2/n2B(b) (as before, we assume n1,n2>0. Then, by definition,




By Corollary 2 of the basic theorem, we obtain




By conclusion 4 of the basic theorem, we have


Therefore, m1n2-m2n1n1n2A(a-b). Thus, we have shown that, if m1/n1A(a) and m2/n2B(b), then m2/n2-m2/n2A(a-b). Likewise, if m1/n1B(a) and m2/n2A(b), then m2/n2-m2/n2B(a-b). By the definition of subtraction for Dedekind cuts, this implies that μ(a-b)=μ(a)-μ(b).

Next, we shall show that μ preserves the order. Since we already know that μ is a homomorphismMathworldPlanetmathPlanetmathPlanetmath, it suffices to show that μ(z)0 when z0. By definition of μ(z) as a Dedekind cut, this is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath to demanding that, whenever z0, then m/n0 for all m/nA(z). This follows readily from the fact that, if z<0 and m<0 and n>0, then nz<0<my, so nz<my and hence m/n could not possibly belong to A(z).

To the proof, we must show that μ is in fact an isomorphism. This may be accomplished by showing that the kernel of μ is trivial. Suppose that μ(z)=0. Then, by the definition of Dedekind cut, A(z) consists of the negative rational numbers and B(z) consists of the positive rational numbers. By the definitions of A and B this, in turn, means that my<nz whenever m>0 and n>0 and my>nz whenever m<0 and n>0. By the basic theorem, either z<0 or z=0 or z>0. If z<0 then, by the Archimedean property, there must exist an n such that nz<y. However, this would contradict the assertion that my<nz whenever m>0 and n>0, so it is not possible to have z<0. Likewise, if z>0, then there must exist an n>0 such that -y<nz, contradicting the assertion that my>nz whenever m<0 and n>0. The only remaining possibility is to have z=0. Therefore, the homomorphism μ is, in fact, an isomorphism, so G is isomorphic to a subset of the additive group of the real numbers.


Title proof of embedding theorem for ordered abelian groups of rank one
Canonical name ProofOfEmbeddingTheoremForOrderedAbelianGroupsOfRankOne
Date of creation 2013-03-22 14:55:33
Last modified on 2013-03-22 14:55:33
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 22
Author rspuzio (6075)
Entry type Proof
Classification msc 20F60
Classification msc 06A05