Exercise 35, § 1

A major set A in an ordered group G is said to be finitely generated if there exists a finite set F such that $A = (F)$. We say that G is semi-archimedean if every finitely generated major set is invertible in the monoid $\mathscr{M}(G)$. Every lattice ordered (resp. archimedean) group is semi-archimedean (Ex. 31, a)). a) Show that every filtered semi-archimedean group is lattice ordered (if $nx \geq 0$, consider the major set $(F)$, where $F = \{0, x, ..., (n-1)x\}$, and use Ex. 30, g) and the fact that it is invertible). b) \* Let K be the lexicographic product $\mathbf{R} \times \mathbf{R}$, and G the (usual) product group $K \times \mathbf{R}$. Let 0 be an irrational number such that $0 < 0 < 1$, and let H be the subgroup generated by $((1, 0), 0), ((0, 0), 8)$ and $((0, x), 0)$ where $x$ runs through $\mathbf{R}$. Show that H, a subgroup of a product of two totally ordered groups, is not semi-archimedean (consider the major set generated by the two elements ((1, 0), 0) and ((0, 0), 0) of H, and show that it is not invertible). \* c) Let G be a semi-archimedean filtered group; show that the submonoid of $\mathfrak{M}(G)$ generated by the finitely generated major sets and their inverses is a lattice ordered group (cf. VI, p. 31, Ex. 13, b)). d) \* Let $0 > 0$ be an irrational number, and let G be the group $Z \times Z$ in which the set P of positive elements is taken to be the set of (x, y) such that $x \geq 0$ and $y \geq 0x$. Show that G is an archimedean group, but that the inverse in $\mathfrak{M}(G)$ of a finitely generated major set, not of the form (a), is not finitely generated. \*