distributivity in po-groups

Let G be a po-group and A be a set of elements of G. Denote the supremumMathworldPlanetmathPlanetmath of elements of A, if it exists, by A. Similarly, denote the infimumMathworldPlanetmath of elements of A, if it exists, by A. Furthermore, let A-1={a-1aA}, and for any gG, let gA={gaaA} and Ag={agaA}.

  1. 1.

    If A exists, so do gA and Ag.

  2. 2.

    If 1. is true, then gA=gA=Ag.

  3. 3.

    A exists iff A-1 exists; when this is the case, A-1=(A)-1.

  4. 4.

    If A exists, so do gA, and Ag.

  5. 5.

    If 4. is true, then gA=gA=Ag.

  6. 6.

    If 1. is true and A={a,b}, then ab exists and is equal to a(ab)-1b.


Suppose A exists.

  • (1. and 2.) Clearly, for each aA, aA, so that gagA, and therefore elements of gA are bounded from above by gA. To show that gA is the least upper bound of elements of gA, suppose b is the upper bound of elements of gA, that is, gab for all aA, this means that ag-1b for all aA. Since A is the least upper bound of the a’s, Ag-1b, so that gAb. This shows that gA is the supremum of elements of gA; in other words, gA=gA. Similarly, Ag exists and gA=Ag as well.

  • (3.) Write c=A. Then ac for each aA. This means c-1a-1. If ba-1 for all aA, then ab-1 for all aA, so that cb-1, or bc-1. This shows that c-1 is the greatest lower bound of elements of A-1, or (A)-1=A-1. The converseMathworldPlanetmath is proved likewise.

  • (4. and 5.) This is just the dual of 1. and 2., so the proof is omitted.

  • (6.) If A={a,b}, then aA-1b=A, and the existence of A is the same as the existence of (aA-1b), which is the same as the existence of a(A-1)b by 4 and 5 above. Since A exists, so does A-1, and hence a(A-1)b, by 3 above. Also by 3, we have the equality a(A-1)b=a(A)-1b. Putting everything together, we have the result: ab=a(ab)-1b.

This completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof. ∎

Remark. From the above result, we see that group multiplication distributes over arbitrary joins and meets, if these joins and meets exist.

One can use this result to prove the following: every Dedekind complete po-group is an Archimedean po-group.


Suppose anb for all integers n. Let A={ann}. Then A is bounded from above by b so has least upper bound A. Then aA=aA=A, since aA=A. As a result, multiplying both sides by (A)-1, we get a=e. ∎

Remark. The above is a generalizationPlanetmathPlanetmath of a famous property of the real numbers: has the Archimedean property.

Title distributivity in po-groups
Canonical name DistributivityInPogroups
Date of creation 2013-03-22 17:05:12
Last modified on 2013-03-22 17:05:12
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 6
Author CWoo (3771)
Entry type Definition
Classification msc 06F05
Classification msc 06F20
Classification msc 06F15
Classification msc 20F60