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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 :)

  • 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. –  Oct 15 '16 at 19:08
  • @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? –  Oct 15 '16 at 19:50

3 Answers3

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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

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

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

  • 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? –  Oct 15 '16 at 20:11
  • @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. –  Oct 15 '16 at 20:21
  • 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. –  Oct 15 '16 at 20:33
  • 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. –  Oct 15 '16 at 20:40
  • 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? –  Oct 15 '16 at 20:53
  • How do you get one element out of two in the integers? $\langle 6/1, 15/1 \rangle = \langle 3/1 \rangle$, and yes it will be a linear combination of the two elements. You describe a cyclic group, $\langle 1/m \rangle$, which contains your group, so which element of $\langle 1/m \rangle$ actually generates your given group? For abelian groups the group is all linear combinations of its generators. –  Oct 15 '16 at 20:58
  • Okay. I realize you probably about want to kill me by now. But I am desperate to understand this problem and get a correct solution. So, as I said, $\left \langle \frac{n_{1}}{m_{1}}, \frac{n_{2}}{m_{2}}\right \rangle \leqslant \left \langle \frac{1}{m}\right \rangle$. The elements of $\left \langle \frac{1}{m} \right \rangle$ are the integer multiples of $\frac{1}{m}$. Elements of $\left \langle \frac{n_{1}}{m_{1}}, \frac{n_{2}}{m_{2}} \right \rangle$ are linear combinations of those two generators, so $a \frac{n_{1}}{m_{1}} + b \frac{n_{2}}{m_{2}}$. So, the element of... –  Oct 15 '16 at 21:30
  • $\displaystyle \left \langle \frac{1}{m} \right \rangle$ that generates my group is $\displaystyle \frac{an_{1}m_{2}+bn_{2}m_{1}}{m_{1}m_{2}}$ where $m_{1}m_{2} = m$? –  Oct 15 '16 at 21:32
  • @JessyCat That isn't one element though, that is all linear combinations... Seriously do a couple examples and see what you got to do to get a generator. You are on the right track. Anyways you got more than enough to figure it out. –  Oct 15 '16 at 21:53
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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.

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Dietrich Burde
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  • 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. –  Oct 15 '16 at 20:27
  • Why did it confuse you? Did you know it already? Cyclic is already a simpler way. – Dietrich Burde Oct 15 '16 at 20:28
  • because things like isomorphisms and enumerating the primes are beyond the scope of what we have studied in class at this point. –  Oct 15 '16 at 20:29
  • 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. – Dietrich Burde Oct 15 '16 at 20:30
  • 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. –  Oct 15 '16 at 20:30
  • I know that Proposition 1 is fine. It's the rest of it that I need help with. And I did not find that other question, its solutions, or that paper helpful in that respect because (see above). –  Oct 15 '16 at 20:31
  • Why do you need anything more? $2$-generated subgroups are in fact $1$-generated. More would be probably beyond the scope of the class. – Dietrich Burde Oct 15 '16 at 20:33
  • I just want to know what the generating elements look like. I.e., how to write them correctly. –  Oct 15 '16 at 20:34
  • I see. Just try with an example: $\langle 1/2,1/3\rangle$. Actually, this is what Paul Plummer wants to say in his answer, I guess. We want to find $n,m$ such that $\langle 1/2,1/3\rangle=\langle n/m\rangle$. – Dietrich Burde Oct 15 '16 at 20:36
  • oh so like a linear combination or something? –  Oct 15 '16 at 20:37