Proposition 3

Let $G$ be an ordered group and $P$ the set of its positive elements.

(a) The kernel of an increasing homomorphism of $G$ to an ordered group is an isolated subgroup of $G$.

(b) Conversely, let $H$ be an isolated subgroup of $G$ and $g$ the canonical homomorphism of $G$ onto $G/H$. Then $g(P)$ is the set of positive elements of an ordered group structure on $G/H$. Moreover, if $G$ is totally ordered, so is $G/H$.

(a) Let $f$ be an increasing homomorphism from $G$ to an ordered group; let $H$ denote the kernel off. If $0 \leq y \leq x$ and $x \in H$, then $0 \leq f(y) \leq f(x) = 0$, whence $f(y) = 0$, that is $y \in H$. Hence $H$ is isolated.

(b) Let $H$ be an isolated subgroup of $G$ and $g : G \to G/H$. Let $P' = g(P)$. Clearly $P' + P' \subset P'$. Also

for, if $x$ and $y$ are elements of $P$ such that $g(x) = -g(y)$, then $x + y \in H$, whence $x \in H$ and $y \in H$ since $H$ is isolated; hence $g(x) = g(y) = 0$. Thus $P'$ is the set of positive elements of an ordered group structure on $G/H$ (Algebra, Chapter VI, § 1, no. 3, Proposition 3). Finally, if $G$ is totally ordered, then $P \cap (-P) = G$, whence $P' \cup (-P') = G/H$ and therefore $G/H$ is totally ordered (loc. cit.).