Proposition 4

Let G be a totally ordered commutative group. Suppose that G is finitely generated and that $h(G) = r(G)$. Then G is isomorphic to $\mathbf{Z}^{r(G)}$ ordered lexicographically.

Let us write $r = r(G) = h(G)$. If $r = 0$, then $G = (0)$. If $r = 1$, the structure of finitely generated commutative groups shows that there is an isomorphism $j$ of G onto Z (Algebra, Chapter VII, § 4, no. 6, Theorem 3). Now Z has only two total orderings compatible with its group structure, namely the usual ordering and its opposite. Hence $j$ or $-j$ is an isomorphism of the ordered group G onto Z with the usual ordering.

Suppose now that $r \geq 2$ and let us argue by induction on $r$. Let H be an isolated subgroup of G of height $r - 1$. Then $r(H) + r(G/H) = r$ (formula (4)), $r(H) \geq h(H) = r - 1$ and $r(G/H) \geq h(G/H) = 1$ (Corollary to Proposition 3), whence $r(H) = r - 1$ and $r(G/H) = 1$. The induction hypothesis shows that H is isomorphic to $\mathbf{Z}^{r-1}$ ordered lexicographically and the case $r = 1$ shows that $G/H$ is isomorphic to Z. As Z is a free Z-module, H is a direct factor of G (Algebra, Chapter 11, § 1, no. 11, Proposition 21). The following lemma then shows that G is isomorphic (not canonically) to the lexicographical product $H \times (G/H)$, which completes the proof.