Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 4 / 6

No: there are at most three, including zero and the whole group. A two-generator ordered Abelian group is torsion-free and has rank of an abelian group at most two. A convex subgroup of an ordered group is pure: if $ng\in C$ with $n>0$, then $g$ lies between zero and $ng$ or their negatives, so $g\in C$. Therefore a proper inclusion $C<D$ of convex subgroups increases rational rank: an element of $D\setminus C$ cannot have a nonzero integer multiple in $C$. Convex subgroups form a chain, so a rank-two group permits at most two strict inclusions. This is the bound for convex subgroups of a finite-rank ordered Abelian group. It is sharp: lexicographically ordered group ${\mathbb{Z}}^2$ has exactly $0$, $\{0\}\times \mathbb{Z}$, and ${\mathbb{Z}}^2$.

A concrete model of the two-generated free Abelian lattice-ordered group is the lattice subgroup of real-valued functions on ${\mathbb{R}}^2$ generated by the coordinate projections $X(x,y)=x$, $Y(x,y)=y$. Its elements are finite joins of finite meets of integer linear forms $ax+by$. Thus they are continuous positively homogeneous piecewise linear functions, with pointwise addition and lattice operations. This is the piecewise integer-linear representation of a free Abelian lattice-ordered group.

For the construction, first form the term algebra on two symbols using addition, negation, zero, join and meet, then identify terms that agree in every Abelian lattice-ordered group. Evaluation under any assignment of the two generators factors through this quotient, proving its universal property. To obtain the concrete function model, use the identity-detection theorem that the real line generates the variety of Abelian lattice-ordered groups. Its proof passes a failed identity to a prime ordered quotient, chooses its finitely many active linear pieces, and uses the rational-inequality alternative underlying Farkas' lemma to find a real witness. This makes evaluation on ${\mathbb{R}}^2$ injective as well as surjective onto the generated function group.

Its underlying group is not ${\mathbb{Z}}^n$ for any finite $n$. The integer-linear hinge functions have infinite independent rank: $h_k(x,y)=(x-ky)\vee 0$ are integer-linearly independent for distinct integers $k$. On $y=1$, only $h_k$ has a slope jump at $x=k$, so each coefficient in a finite relation must vanish. The underlying group therefore has infinite rank of an abelian group despite having two lattice generators.

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.