Proposition 3

Under the hypotheses of Definition 4, if the set of strictly positive elements of G has no least element, then $\varepsilon(G, H) = 1$ for all H. If there exists a least element > O of G and $G'$ denotes the subgroup it generates, then

In the first case, let x be an element > 0 in G. The set of $y \in G$ such that $0 < y < x$ is infinite and hence there exist two elements of this set which are distinct and congruent modulo H; their difference is an element z of H such that $0 < z < x$. Hence every major subset which contains all the strictly positive elements of H contains x and hence all the elements > 0 of G.

In the second case, let $x$ be the least element >0 of G and let $n$ be the least integer >0 such that $nx \in H$. Clearly $n = (G : (G \cap H))$. On the other hand, writing $M(y)$ for the set of $z \in G$ such that $y \leq z \ (y \in G)$, it is immediately seen that the major sets of Definition 4 are just $M(x), M(2x), \ldots, M(nx)$.