Need solution verification: Describe all $2$-generated subgroups of $\mathbb{Q}$

Question

I need to describe all $2$-generated subgroups of the group of rational numbers under addition, $\mathbb{Q}$ (i.e. all subgroups of $\mathbb{Q}$ that are generated by $2$ elements). I have come up with a proof, and would like somebody to please take a look at it, tell me if it's correct, and if not, what I need to do in order to make it so.

To start my proof, I began with the following proposition:

Proposition 1: All finitely generated subgroups of $\mathbb{Q}$ are cyclic.

Proof: Suppose that $H$ is a finitely generated subgroup of $\mathbb{Q}$, say $ \displaystyle H = \left\langle \frac{n_{1}}{m_{1}}, \cdots , \frac{n_{r}}{m_{r}} \right\rangle $ where $n_{i}, m_{i} \in \mathbb{Z}$ $\forall i \in \mathbb{N}$. Define $m = m_{1}m_{2}\cdots m_{r}$.

Then, $\forall i$ such that $1 \leq i \leq r$, we have that $\displaystyle \frac{n_{i}}{m_{i}}=n_{i} \cdot (m_{1}m_{2} \cdots m_{i-1}m_{i+1}\cdots m_{r}) \cdot \frac{1}{m} \in \left \langle \frac{1}{m} \right \rangle$, as $\displaystyle \left \langle \frac{1}{m} \right \rangle$ contains all multiples of $\displaystyle n \cdot \frac{1}{m}$, $n \in \mathbb{Z}$, and clearly $n_{i} \cdot (m_{1}m_{2}\cdots m_{i-1}m_{i+1}\cdots m_{r}) \in \mathbb{Z}$.

Since $\displaystyle \left \langle \frac{n_{1}}{m_{1}}, \cdots \frac{n_{r}}{m_{r}}\right \rangle$ is the smallest subgroup of $\mathbb{Q}$ containing all of the $\displaystyle \frac{n_{i}}{m_{i}}$, by definition, we have that $\displaystyle H = \left \langle \frac{n_{1}}{m_{1}} \cdots \frac{n_{r}}{m_{r}}\right \rangle \leqslant \left \langle \frac{1}{m} \right \rangle$, and since every subgroup of a cyclic group is cyclic, $H$ must be cyclic.

So, we have that all finitely-generated subgroups of $\mathbb{Q}$ must be cyclic.

From this proposition, I have the following conclusion as to the nature of all $2$-generated subgroups of $\mathbb{Q}$:

The $2$-generated subgroups of $\mathbb{Q}$ are merely subgroups of the form $\displaystyle \left \langle \frac{n}{m}\right \rangle = \left \langle \frac{n}{m},\frac{n}{m} \right \rangle$, where the two generating elements are not distinct, and $\gcd(n,m)=1$.

Could somebody please tell me if this is correct? Especially the last part - somehow it doesn't seem like enough. And if it's not correct, what do I do to make it correct?

Thank you :)

Answer 1

Let's say you want an algorithm to simplify

$$\left\langle \frac{m_1}{n_1},\cdots,\frac{m_r}{n_r}\right\rangle= \left\langle\frac{m}{n}\right\rangle.$$

It suffices to do it for $r=2$, because for larger $r$ we can work two-at-a-time.

Compute

$$\begin{array}{ll} \displaystyle\left\langle\frac{m_1}{n_1},\frac{m_2}{n_2}\right\rangle & \displaystyle =\frac{1}{n_1n_2}\langle m_1n_2,m_2n_1\rangle \\ & \displaystyle =\frac{1}{n_1n_2}\langle \gcd(m_1n_2,m_2n_1)\rangle \\ & = \displaystyle \left\langle \frac{\gcd(m_1n_2,m_2n_1)}{n_1n_2} \right\rangle. \end{array}$$

Answer 2

Your proof seems fine, in that you show two generated subgroups are cyclic, but it may be the case that the question is looking for a bit more information. That is you have shown $\langle n_1/m_1, n_2/m_2 \rangle = \langle n/m \rangle$ for some $n/m$, but maybe you can actually describe what $n$ and $m$ are. For example in the case of the integers you can say $\langle a,b \rangle= \langle \gcd(a,b) \rangle$.

Answer 3

Proposition $1$ is fine; it has been proved here at MSE already. More generally, all subgroups of $\mathbb{Q}$ have been described here, see " How to find all subgroups of $(\mathbb{Q},+)$". For particular cases, like $2$ generators, this description gives some more information.