distributivity in pogroups
Let $G$ be a pogroup and $A$ be a set of elements of $G$. Denote the supremum^{} of elements of $A$, if it exists, by $\bigvee A$. Similarly, denote the infimum^{} of elements of $A$, if it exists, by $\bigwedge A$. Furthermore, let ${A}^{1}=\{{a}^{1}\mid a\in A\}$, and for any $g\in G$, let $gA=\{ga\mid a\in A\}$ and $Ag=\{ag\mid a\in A\}$.

1.
If $\bigvee A$ exists, so do $\bigvee gA$ and $\bigvee Ag$.

2.
If 1. is true, then $g\bigvee A=\bigvee gA=\bigvee Ag$.

3.
$\bigvee A$ exists iff $\bigwedge {A}^{1}$ exists; when this is the case, $\bigwedge {A}^{1}={(\bigvee A)}^{1}$.

4.
If $\bigwedge A$ exists, so do $\bigwedge gA$, and $\bigwedge Ag$.

5.
If 4. is true, then $g\bigwedge A=\bigwedge gA=\bigwedge Ag$.

6.
If 1. is true and $A=\{a,b\}$, then $a\wedge b$ exists and is equal to $a{(a\vee b)}^{1}b$.
Proof.
Suppose $\bigvee A$ exists.

•
(1. and 2.) Clearly, for each $a\in A$, $a\le \bigvee A$, so that $ga\le g\bigvee A$, and therefore elements of $gA$ are bounded from above by $g\bigvee A$. To show that $g\bigvee A$ is the least upper bound of elements of $gA$, suppose $b$ is the upper bound of elements of $gA$, that is, $ga\le b$ for all $a\in A$, this means that $a\le {g}^{1}b$ for all $a\in A$. Since $\bigvee A$ is the least upper bound of the $a$’s, $\bigvee A\le {g}^{1}b$, so that $g\bigvee A\le b$. This shows that $g\bigvee A$ is the supremum of elements of $gA$; in other words, $g\bigvee A=\bigvee gA$. Similarly, $\bigvee Ag$ exists and $g\bigvee A=\bigvee Ag$ as well.

•
(3.) Write $c=\bigvee A$. Then $a\le c$ for each $a\in A$. This means ${c}^{1}\le {a}^{1}$. If $b\le {a}^{1}$ for all $a\in A$, then $a\le {b}^{1}$ for all $a\in A$, so that $c\le {b}^{1}$, or $b\le {c}^{1}$. This shows that ${c}^{1}$ is the greatest lower bound of elements of ${A}^{1}$, or ${(\bigvee A)}^{1}=\bigwedge {A}^{1}$. The converse^{} 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 $a{A}^{1}b=A$, and the existence of $\bigwedge A$ is the same as the existence of $\bigwedge (a{A}^{1}b)$, which is the same as the existence of $a(\bigwedge {A}^{1})b$ by 4 and 5 above. Since $\bigvee A$ exists, so does $\bigwedge {A}^{1}$, and hence $a(\bigwedge {A}^{1})b$, by 3 above. Also by 3, we have the equality $a(\bigwedge {A}^{1})b=a{(\bigvee A)}^{1}b$. Putting everything together, we have the result: $a\wedge b=a{(a\vee b)}^{1}b$.
This completes^{} 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 pogroup is an Archimedean pogroup.
Proof.
Suppose ${a}^{n}\le b$ for all integers $n$. Let $A=\{{a}^{n}\mid n\in \mathbb{Z}\}$. Then $A$ is bounded from above by $b$ so has least upper bound $\bigvee A$. Then $a\bigvee A=\bigvee aA=\bigvee A$, since $aA=A$. As a result, multiplying both sides by ${(\bigvee A)}^{1}$, we get $a=e$. ∎
Remark. The above is a generalization^{} of a famous property of the real numbers: $\mathbb{R}$ has the Archimedean property.
Title  distributivity in pogroups 

Canonical name  DistributivityInPogroups 
Date of creation  20130322 17:05:12 
Last modified on  20130322 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 