2
$\begingroup$

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 :)

$\endgroup$
  • $\begingroup$ Do you mean $H$ is a subgroup of $\langle 1/m \rangle$ rather than equal (it seems that way given what you say after). Also I think there may be an implication in the problem that you want to describe the generator in terms of the two given elements. $\endgroup$ – Paul Plummer Oct 15 '16 at 19:08
  • $\begingroup$ @PaulPlummer, yes to $H$ a subgroup. I'll fix that. The implication in the problem where I want to describe the generator in terms of the two given elements is something I have been confused about for over a week now. I have asked a few people and have gotten no answers. Could you please elaborate on this further, perhaps in the form of an answer? $\endgroup$ – ALannister Oct 15 '16 at 19:50
2
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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 $\langle a,b \rangle= \langle \gcd(a,b) \rangle$.

$\endgroup$
  • $\begingroup$ er...okay... I'm a little confused. I'm not talking about the integers, I'm talking about the rationals. Can you translate what you just said at the end in terms of rationals? $\endgroup$ – ALannister Oct 15 '16 at 20:11
  • $\begingroup$ @JessyCat You asked to check your proof, not to do the problem. I just gave an example to highlight what I think the question is asking (describing the generator in terms of the two given generators), Maybe the question is not asking that, and it is worth getting clarification. $\endgroup$ – Paul Plummer Oct 15 '16 at 20:21
  • $\begingroup$ the proof is the entire thing I posted, not just the proof of Proposition 1. I'm not asking anyone to do the problem. To do the problem, I first needed to do proposition 1. But now I am confused as to what the generators should look like. As in I am not sure how to correctly write them. I asked my prof. He said I should be able to extract that from my proof of Proposition 1 but I don't see how. $\endgroup$ – ALannister Oct 15 '16 at 20:33
  • $\begingroup$ I read the whole thing and said it was fine... You showed that two generator subgroups can be written as $\langle n/m \rangle$. If you want me to describe the generator, I am not going to do that, and would not have bothered answering the question if that is what you actually asked. If you are haveing a hard time I recommend trying some explicit examples, like finding the generator for $\langle 3/4, 6/7 \rangle$ (really anything), maybe you will get an idea of how to generalize that. $\endgroup$ – Paul Plummer Oct 15 '16 at 20:40
  • $\begingroup$ I'm not entirely sure how generators consisting of two elements generate sets. I think that's the issue. I know how cyclic ones work. For example, if $\langle 5/6 \rangle$ was a generator for a subset of $\mathbb{Q}$, then you'd take additive powers of 5/6 to find the elements of that subset. How do you do it for generating sets with two elements? Can you at least show me a worked out example of that? $\endgroup$ – ALannister Oct 15 '16 at 20:53
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ which then redirects me to a paper that confused me more. It would be nice to see some of that stuff explained in a simpler way. $\endgroup$ – ALannister Oct 15 '16 at 20:27
  • $\begingroup$ Why did it confuse you? Did you know it already? Cyclic is already a simpler way. $\endgroup$ – Dietrich Burde Oct 15 '16 at 20:28
  • $\begingroup$ because things like isomorphisms and enumerating the primes are beyond the scope of what we have studied in class at this point. $\endgroup$ – ALannister Oct 15 '16 at 20:29
  • $\begingroup$ If it is for a class, then we just have to know what else is expected for a homework solution. Then I would say, we are done with proposition $1$. More is beyond the scope of the class, I think. These groups are in fact cyclic, there is no simpler way to say this. $\endgroup$ – Dietrich Burde Oct 15 '16 at 20:30
  • $\begingroup$ what I would like to see, and hence, why I included the proof verification tag, is for someone to maybe help me deconstruct what I have done so far and add in the relevant details for showing exactly what the 2-generators of $\mathbb{Q}$ look like without adding in a bunch of things that are more advanced than the simple group theory that we have done in class thus far. $\endgroup$ – ALannister Oct 15 '16 at 20:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.