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Homedistributivity in pogroups
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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 $a\in A$, $a\leq\bigvee A$, so that $ga\leq g\bigvee A$, 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, $ga\leq b$ for all $a\in A$, this means that $a\leq g^{{1}}b$ for all $a\in A$. Since $\bigvee A$ is the least upper bound of the $a$’s, $\bigvee A\leq g^{{1}}b$, so that $g\bigvee A\leq b$. 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 $a\leq c$ for each $a\in A$. This means $c^{{1}}\leq a^{{1}}$. If $b\leq a^{{1}}$ for all $a\in A$, then $a\leq b^{{1}}$ for all $a\in A$, so that $c\leq b^{{1}}$, or $b\leq c^{{1}}$. 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 $aA^{{1}}b=A$, and the existence of $\bigwedge A$ is the same as the existence of $\bigwedge(aA^{{1}}b)$, 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.
Remark. The above is a generalization of a famous property of the real numbers: $\mathbb{R}$ has the Archimedean property.
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