Context: Self-study.
Seth Warner's Modern Algebra (1965), question $15.11$ gives:
If $(G, \circ \preccurlyeq)$ is a lattice-ordered group and if $x$ and $y$ are commuting elements of $G$, then $$\sup \{x, y\} \circ \inf \{x, y\} = x \circ y$$ [Use Exercise $15.10$.]
Exercise $15.10$ gives that:
if any of $\{x, y\}$ or $\{x \circ z, y \circ z\}$ or $\{z \circ x, z \circ y\}$ admit a supremum, then they all do, and $$\sup \{x \circ z, y \circ z\} = \sup \{x, y\} \circ z$$ $$\sup \{z \circ x, z \circ y\} = z \circ \sup \{x, y\}$$
along with: $$(\inf \{x, y\})^{-1} = \sup \{x^{-1}, y^{-1}\}$$ (and its obvious dual), if these exist.
I managed $15.10$ with no difficulty, but $15.11$ stumps me.
I've established by judicious substitutions and commutativity of $x$ and $y$ that:
$$\sup \{x, y\} \circ \inf \{x, y\} = \sup \{\inf \{x \circ x, x \circ y\}, \inf \{x \circ y, y \circ y\} \}$$
and:
$$\sup \{x, y\} \circ \inf \{x, y\} = \inf \{\sup \{x \circ x, x \circ y\}, \sup \{x \circ y, y \circ y\} \}$$
but I can't see where to go from there. As $\preccurlyeq$ is only a lattice ordering, and not necessarily a total ordering, we can't even exploit any relation compatibility properties because we can't assert that either $x \preccurlyeq y$ or $y \preccurlyeq x$.
I've even popped my head over the top of the trench to see what would happen if we explored substituting the identities involving the inverses, but the impenetrable tangle of notation doesn't seem to offer a path through it.
It would be a wonderful result to prove, because then we'd have a super-slick abstract-algebraical way to prove $\gcd \{x, y\} \times \operatorname{lcm}\{x, y\} = x y$.
I will write $x\wedge y$ for $\textrm{inf}\{x,y\}$, $x\vee y$ for $\textrm{sup}\{x,y\}$, and $xy$ for $x\circ y$.
Claim 1. If $x$ commutes with $y$, then $x$ also commutes with $x\wedge y$.
Reason. $x(x\wedge y) = (x^2\wedge xy) = (x^2\wedge yx) = (x\wedge y)x$. $\Box$
Claim 2. If $x$ commutes with $y$, then $xy = (x\wedge y)(x\vee y)$.
Reason. $x(x\wedge y)^{-1}y = x(x^{-1}\vee y^{-1})y = (xx^{-1}y\vee xy^{-1}y) = (y\vee x) = (x\vee y)$.
Multiplying both sides of $x(x\wedge y)^{-1}y = (x\vee y)$ with $(x\wedge y)$ on the left and using Claim 1 yields $xy=(x\wedge y)(x\vee y)$. $\Box$