Lemma 3

Let $G'$ be a totally ordered group, $G$ a subgroup of $G$ and $\mathscr{G}'$ (resp. $\mathscr{G}$) the set of isolated subgroups of $G'$ (resp. $G$). The mapping $H' \mapsto H' \cap G$ maps $\mathscr{G}'$ onto $\mathscr{G}$. This mapping is bijective if $G'/G$ is a torsion group.

Clearly $H' \in \mathscr{G}'$ implies $H' \cap G \in \mathscr{G}$. Now let $H \in \mathscr{G}$; let $H'$ denote the set of $x' \in G'$ such that there exists $h \in H$ satisfying $-h \leq x' \leq h$; it is immediately verified that $H'$ is an isolated subgroup of $G'$; then $H' \cap G = H$ since $H$ is isolated; hence the mapping $H' \mapsto H' \cap G$ is surjective. Suppose finally that $G'/G$ is a torsion group; let $H'_1$ and $H'_2$ be two isolated subgroups of $G'$ such that $H'_1 \cap G = H'_2 \cap G$; then, for example, $H'_1 \supset H'_2$ (cf. § 4, no. 4); then $H'_1/H'_2$ is a totally ordered group and is isomorphic to a quotient group of $H'_1/(H'_1 \cap G)$ which itself is identified with a subgroup of $G'/G$; hence $H'_1/H'_2$ is a torsion group and therefore reduces to 0.