# How to find all subgroups of $(\mathbb{Q},+)$

Exercise 1.6.13 from Scott's book Group Theory.

(Hard) Find all subgroups of $(\mathbb{Q},+)$. Hint: It is slightly easier to find those subgroups $H$ such that $1\in H.$

I've found some of those subgroups: $\mathbb{Z}$, $\mathbb{Q}$, $\langle 1,\frac{1}{2}\rangle$, $\langle 1,\frac{1}{2},\frac{1}{3}\rangle$, $\cdots$. But can't I find all of them. How to find them?

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The subgroup generated by 1 and 1/2 is generated by 1/2 alone. There are two cases: there is a smallest positive element in this subgroup (which you must show generates the group), or there is none. In the second case, look at the set of denomiators inside the integers and try to find its structure. – franz lemmermeyer Feb 1 '12 at 11:22
Any finitely generated subgroup of $\mathbb Q$ must be cyclic and, and a subgroup of $\mathbb Q$ is isomorphic to atleast one subgroup of $\mathbb Q$ containing $1$. I think these will essentially tell you all the subgroups, but I may be wrong. – user21436 Feb 1 '12 at 11:26
$\LaTeX$ tip: use \langle: $\langle$; and \rangle $\rangle$; for angle brackets, not < and >. The latter are operators, the former are delimiters. The spacing provided by $\LaTeX$ is different. – Arturo Magidin Feb 1 '12 at 17:24

## 1 Answer

Let $A$ be an additive subgroup of $\mathbb Q$. Let $D$ be the set of denominators occurring in $A$ when you consider reduced fractions only. Then the following are easy to prove:

• $b\in D, u \mid b \implies u\in D$
• $u,v \in D, (u,v)=1 \implies uv \in D$

The first property means that you cannot increase the power of a prime in $D$. The second property means that you can combine powers of different primes. That should give you a description of $D$.

$A \cap \mathbb Z$ is an additive subgroup of $\mathbb Z$ and these are easy to characterize. That should give you a description of the set $N$ of numerators in $A$.

For a complete solution, see the paper

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. MR0044522

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The solution can be simplified from the form in the paper -- you don't have to treat the numerator and denominator separately. Choose $k_i \in \mathbb{Z} \cup \{ -\infty \}$. The corresponding subgroup is the set of all rationals whose prime factorization has the property that the exponent on $p_i$ is bigger than or equal to $k_i$. – Hurkyl Mar 1 '12 at 19:24