# Product of Supremum and Infimum in Lattice-Ordered Group

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

$\blacksquare$