Proposition 3

Let G be a totally ordered Commutative group and H a subgroup of G. If $h(G)$ and $h(H)$ denote the heights of G and H ($§ 4$, no. 4), then the inequality

holds.

In fact, let $G_0 \subset G_1 \subset \ldots \subset G_n$ be a strictly increasing sequence of isolated subgroups of G. It is necessary to establish the inequality

It is obvious for $n = 0$. Suppose $n \geq 1$ and let us argue by induction on $n$. Applying the induction hypothesis to the group $G_{n-1}$ and its subgroup $H \cap G_{n-1}$, we obtain

Then we distinguish two cases:

(a) $H \cap G_{n-1} = H$, in other words $H \subset G_{n-1}$. Inequality (7) may be written

Now $G/G_{n-1}$ is a totally ordered group not reduced to 0; it is therefore not a torsion group and $r(G/G_{n-1}) \geq 1$. Whence by (4), $r(G/H) \geq r(G_{n-1}/H) + 1$. Substituting in (8), we certainly obtain the desired inequality (6).

(b) $H \cap G_{n-1} \neq H$. As $H \cap G_{n-1}$ is an isolated subgroup of $H$, we conclude that $h(H) > h(H \cap G_{n-1}) + 1$. On the other hand, obviously $r(G/H) \geq r(G_{n-1}/(H \cap G_{n-1}))$. Substituting in (7), we again obtain (6).