Classification of subgroups of $(\mathbb Q,+)$ that are not finitely generated. [closed]

Question

Proper subgroups of $(\mathbb Q,+)$ that are not finitely generated exist, e.g., Example of subgroup of $\mathbb Q$ which is not finitely generated.

Is there a classification of subgroups of $(\mathbb Q,+)$ that are not finitely generated?

Answer 1

Let $P$ be the set of all prime numbers and let $f: P\to\mathbb{N}\cup\infty$ be an arbitrary function. Let $$ N_f=\{n\in\mathbb{N}\mid n=\prod_{p\in P} p^{k_p}, k_p\leq f(p)\}. $$ Let $\mathbb{Q}_f=\{a/b\mid a\in\mathbb{Z},b\in N_f\}$.

Theorem. For any function $f$ the set $\mathbb{Q}_f$ is a subgroup.

If $H$ is a subgroup of $\mathbb{Q}$, then there exists a function $f$ and an non-negative integer $m$ that $H=m\mathbb{Q}_f$.

The subgroup $m\mathbb{Q}_f$ is finitely generated (and hence cyclic) if and only if $f(p)<\infty$ for every prime $p$ and $f(p)>0$ only for a finite number of prime.

Examples.

1. If $f(2)=\infty$, $f(p)=0$ for all $p\in P$, $p\neq2$, then we obtain a subgroup $\mathbb{Q}_2=\mathbb{Z}[1/2]$ (see Example of subgroup of $\mathbb{Q}$ which is not finitely generated).

2. If $f(p)=\infty$ for each $p\in P$, then $\mathbb{Q}_f=\mathbb{Q}$.

3. If $f(p)=1$ for each $p\in P$, then $\mathbb{Q}_f=\operatorname{gr}(1/p\mid p\in P)$.

4. If $f(2)=a>0$ and $f(p)=0$ for each $p\neq2$, then $\mathbb{Q}_f=\operatorname{gr}(1/2^a)$ is a cyclic group.

You can read about it in How to find all subgroups of $\mathbb{Q}$ and in the article Ross A. Beaumont and H. S. Zuckerman, A characterization of the subgroups of the additive rationals, Pacific J. Math. Volume 1, Number 2 (1951), 169-177.

Adding. In this connection it seems useful to read the chapter 'Torsion-Free Groups' of Laszlo Fuchs' book Abelian Groups. Any edition (and there are many) of this book has such a chapter.