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A convex subgroup of an ordered group that is also closed under join and meet. A normal convex lattice subgroup is the kernel of the quotient lattice-ordered group homomorphism.

The free object on generators $x_1,\dots ,x_n$ has the universal property that an arbitrary assignment of these generators to an Abelian lattice-ordered group extends uniquely to a lattice-ordered group homomorphism. It can be constructed from terms using addition, negation, zero, join and meet, modulo identities valid in every Abelian lattice-ordered group. Its underlying group need not be finitely generated even when there are only two lattice generators.

For each $g\ne 1$, choose a value in a lattice-ordered group $P_g$, using Zorn's lemma. Such a value is a prime convex lattice subgroup. Otherwise the last construction in Question 1(i) would give two larger convex lattice subgroups whose intersection is $P_g$; maximality among subgroups omitting $g$ would put $g$ in both, hence in their intersection. Thus the right coset set ${\Omega }_g=P_g\backslash G$ is a totally ordered set.

Right multiplication defines an action ${\rho }_g(a):P_{g}h\mapsto P_{g}ha$. It is well-defined, preserves the total order, and has inverse ${\rho }_g(a^{-1})$. The coset formulas from Question 1(i) show pointwise preservation of both join and meet, becauseand likewise for the meet.

Take the disjoint union of these chains, arranged in any fixed order sum of the indices $g\ne 1$. Let every $a\in G$ act on each component by ${\rho }_g(a)$. This gives a lattice-ordered group homomorphism into the order automorphisms of one chain. It is injective: a nonidentity $a$ moves the identity coset in ${\Omega }_a$, since $P_{a}a\ne P_a$. Its image is closed under pointwise join and meet, giving exactly the required lattice-ordered permutation group. This is the Holland representation theorem.

No. The free Abelian lattice-ordered group on two generators is countable, since every element is represented by a finite term over a countable alphabet of operation symbols and the two generators. The additive group of all real-valued functions on the given half-plane is uncountable: it already contains a distinct constant function for every real number. A lattice-ordered group homomorphism that is an isomorphism would be a bijection, contradicting these cardinalities. The concrete free-group model consists of specially constrained integer-linear term functions, not all real-valued functions on a domain.

No. Assign the two free generators to $(1,0)$ and $(0,1)$ in the lexicographically ordered group ${\mathbb{Z}}^2$. The universal property gives a lattice-ordered group homomorphism onto this group: its image contains both ordinary group generators. It is not an Archimedean ordered group, sinceThus a lattice-homomorphic image can contain a nonzero positive infinitesimal even when the original free object has an Archimedean pointwise function representation.