Product of Supremum and Infimum in Lattice-Ordered Group

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Theorem

Let $\struct {G, \odot}$ be a group.

Let $\preccurlyeq$ be a lattice ordering on $G$.

Let $x, y \in G$ such that $x$ commutes with $y$.


Then:

$\sup \set {x, y} \odot \inf \set {x, y} = x \odot y$


Proof

First we show that $x$ commutes with $\sup \set {x, y}$.

Indeed:

\(\ds x \odot \sup \set {x, y}\) \(=\) \(\ds \sup \set {x \odot x, x \odot y}\) Suprema in Ordered Group
\(\ds \) \(=\) \(\ds \sup \set {x \odot x, y \odot x}\) as $x$ and $y$ commute
\(\ds \) \(=\) \(\ds \sup \set {x, y} \odot x\) Suprema in Ordered Group

Similarly:

$y$ commutes with $\sup \set {x, y}$

Then using Infima in Ordered Group in the same way as above:

$x$ commutes with $\inf \set {x, y}$
$y$ commutes with $\inf \set {x, y}$


Then we have:

\(\ds x \odot \paren {\sup \set {x, y} }^{-1} \odot y\) \(=\) \(\ds x \odot \inf \set {x^{-1}, y^{-1} } \odot y\)
\(\ds \) \(=\) \(\ds \inf \set {x \odot x^{-1} \odot y, x \odot y^{-1} \odot y}\) Inverse of Supremum in Ordered Group is Infimum of Inverses
\(\ds \) \(=\) \(\ds \inf \set {y, x}\)
\(\ds \) \(=\) \(\ds \inf \set {x, y}\)
\(\ds \leadsto \ \ \) \(\ds x \odot y \odot \paren {\sup \set {x, y} }^{-1}\) \(=\) \(\ds \inf \set {x, y}\) from the commutativity proved a priori
\(\ds \leadsto \ \ \) \(\ds x \odot y\) \(=\) \(\ds \sup \set {x, y} \odot \inf \set {x, y}\)

$\blacksquare$


Sources

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