$\fbox{1}$ Prove that any finitely generated subgroup of $(\mathbb{Q},+)$ is cyclic.

$\fbox{2}$ Prove that $\mathbb{Q}$ is not isomorphic to $\mathbb{Q}\times \mathbb{Q}$.

Any hints would be appreciated.

  • 5
    $\begingroup$ A hint that almost always applies: organize and present your own thoughts on the question. If you don't figure out the answer yourself in the process, the information you provide will not only allow others to write more helpful answers, but it will give them confidence you're looking to learn how to solve problems rather than to copy answers. $\endgroup$
    – user14972
    Commented Jul 20, 2012 at 0:40
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    $\begingroup$ I don't think the downvote is appropriate here. The votes should represent the quality of the question, not of the person who asked it. If the downvote came for a different reason, please explain. There are many questions on the site whose quality is more or less as this one without any downvotes. $\endgroup$
    – user12014
    Commented Jul 20, 2012 at 0:49
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    $\begingroup$ @PZZ: One of the hover legends on the downvote button is "this question does not show any research effort", and that certainly does describe this question. It's rather uncharitable to assume that the downvoter voted out of a dislike of the asker rather than because of the actual problems that do exist with the question. $\endgroup$ Commented Jul 20, 2012 at 0:54
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    $\begingroup$ @HenningMakholm Fair enough. However, I still stand by the fact that there are many questions showing just as little research effort that are not downvoted. $\endgroup$
    – user12014
    Commented Jul 20, 2012 at 0:56
  • $\begingroup$ @PZZ I try to hold off on downvoting if the user is very new, and to leave a message explaining the general sentiment if I have time. In felipeuni's case I think that Zev has done this a few times. I do hope that he or she does not take any downvotes too hard — it's just a signal that one could do better. $\endgroup$ Commented Jul 20, 2012 at 1:14

4 Answers 4


Hint 1. Given any two elements $a,b\in\mathbb{Q}$, can you find an element $r\in\mathbb{Q}$ such that $a=mr$ and $b=nr$ for some integers $m$ and $n$? If so, then $\langle a,b\rangle\subseteq \langle r\rangle$; what do we know about subgroups of cyclic groups.

Hint 2. Is there a finitely generated subgroup of $\mathbb{Q}\times\mathbb{Q}$ that is not cyclic?

  • $\begingroup$ I see that your answer proves what needs to be proved. But what is the generator of the subgroup generated by $\{a/b, c/d\}$? Is it accurate to say that it is $gcd(a,c)/lcm(b,d)$? I'm assuming $(a, b) = (c, d) = 1$ are relative primes, and I think the claimed expression is contained in the generated group by Bezout's identity, and it clearly generates the generating set. Is this correct? $\endgroup$
    – gen
    Commented Oct 7, 2021 at 0:20
  • $\begingroup$ @gen, $$\begin{aligned}q_1':&=\frac{\gcd(q_1,q_2)}{q_1},&q_2':=\frac{\gcd(q_1,q_2)}{q_2}\\\frac{p_1}{q_1}&=\frac{p_1q_1'}{\gcd(q_1,q_2)},&\frac{p_2}{q_2}\color{white}: =\frac{p_2q_2'}{\gcd(q_1,q_2)}\\m:&=p_1q_1'&n:=p_2q_2' \\r:&=\frac1{\gcd(q_1,q_2)}\end{aligned}$$ $\endgroup$
    – PinkyWay
    Commented Oct 20, 2023 at 20:10

Let $\frac{m_1}{n_1},...,\frac{m_k}{n_k}$ be the generators of $G\leq\Bbb Q$ and $g\in G$. Then we have $g=a_1\frac{m_1}{n_1}+...+a_k\frac{m_k}{n_k}=\frac{a_1m_1n_2...n_{k}+...+a_km_kn_1...n_{k-1}}{n_1...n_k}$ for some integers $a_1,...,a_k$. Thus $G$ is a subgroup of the cyclic group $<\frac{1}{n_1...n_k}>$. Hence $G$ must be cyclic.


Hint 1. Suppose $\frac{m_1}{n_1},\frac{m_2}{n_2},\ldots,\frac{m_k}{n_k}$ are the generators of $G\leq\Bbb Q$. We can use these generators to construct an element $\frac mn$ which already generates $G$.

Hint 2. Suppose $f:\Bbb Q\to\Bbb Q\times\Bbb Q$ is an isomorphism. How are $f(1)$ and $f(\frac mn)$ related?


If you are stuck, then it often helps to look at special cases. Here it turns out that if you can prove that subgroups generated by two elements are cyclic, then this will quickly imply the full result. To be even more concrete, let's consider the subgroup generated by $\frac12$ and $\frac13$.

How is this going to work? The subgroup consists of elements of the form, for $n, m \in \mathbb Z$, \[ \frac n2 + \frac m3 = \frac{3n + 2m}6. \] The numerator here should set off some bells, and I think you can now prove that the subgroup is generated by $\frac16$. Can you make this work for two general rational numbers?

One way to do the second part is to look at the finitely generated subgroup $\mathbb Z \times \mathbb Z$ of $\mathbb Q \times \mathbb Q$.


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