distributivity in po-groups

Let $G$ be a po-group 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 po-group is an Archimedean po-group.

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: $ℝ$ has the Archimedean property.