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From this question, we know that subgroups of the rationals is either generated by one element or infinitely generated. The former case is easy to classify: $\left\langle \frac p q \right\rangle$ with $p,q \in \Bbb Z$ and $q \ne 0$.

(Note that if $p=0$, then the subgroup generated is trivial.)

Now, my conjecture is that all infinitely generated proper subgroups of $\Bbb Q$ is of the form $\Bbb Z\left[\frac 1q\right]$ with $q \in \Bbb Z \setminus \{0\}$, i.e. polynomials in $\frac 1q$ with integer coefficients, i.e. $\frac n{q^m}$ with $n \in \Bbb Z$ and $m \in \Bbb N$.

Is my conjecture correct?

Kenny Lau
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  • What about $\mathbb{Z}[\frac{1}{q}, \frac{1}{p}]$ ? – Maxime Ramzi Sep 03 '17 at 21:49
  • @Max I suspect that's isomorphic to some $\Bbb Z \left[\frac1r\right]$, but I might be wrong. – Kenny Lau Sep 03 '17 at 21:50
  • Possible duplicate of https://math.stackexchange.com/questions/104571/how-to-find-all-subgroups-of-mathbbq. – lhf Sep 04 '17 at 00:37
  • Oh so you're looking for a classification up to isomorphism ? That's a very different job then. But even though, I'm not convinced my example would be isomorphic to one of these. In $\mathbb{Z}[\frac{1}{p}, \frac{1}{q}]$, the equations $p^n x = y, q^n z=t$ have a solution for all $n, y, t$, which would not be the case in a $\mathbb{Z}[\frac{1}{r}]$. – Maxime Ramzi Sep 04 '17 at 05:32
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    @Max no 'isomorphic' is actually a red-herring there. Just take $r = pq$. The groups are equal. – quid Sep 04 '17 at 18:25
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    @quid : right, I didn't notice that, my bad. I "thought" too quickly, thanks for the correction – Maxime Ramzi Sep 04 '17 at 19:41

1 Answers1

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This is not correct. Consider for example the subgroup generated by the set $\{p^{-1} \colon p \text{ prime number} \}$.

quid
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