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I am currently working on a problem trying to prove that there is no subgroup $ H $ of $\mathbb{Q}$ such that $H \cong \mathbb{Z} \times \mathbb{Z}$.

I was able to show that $\mathbb{Q}$ cannot be isomorphic to $ \mathbb{Z} \times \mathbb{Z}$, since $ \mathbb{Z} \times \mathbb{Z} $ can be generated by two elements, where as $\mathbb{Q}$ cannot be finitely generated. Also the subgroup $\mathbb{Z}$ of $\mathbb{Q}$ can also not be isomorphic since it can be generated by a single element.

However, I am somewhat stuck showing, that there cannot exist any subgroup of $\mathbb{Q}$ isomorphic to $ \mathbb{Z} \times \mathbb{Z} $. Some help would be very much appreciated.

I am also wondering, if it is also true that $\mathbb{R}$ and $\mathbb{C}$ have no subgroup isomorphic to $ \mathbb{Z} \times \mathbb{Z} $. I would guess that they do not have one since $\mathbb{Q}$ has none.

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    Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. – Community Oct 21 '23 at 11:14
  • I don't know how $ \mathbb{Q} $ being divisible helps. But I have a different idea. Can I say that any subgroup that is finitely generated in $ \mathbb{Q} $ has to be cyclic. Since $ \mathbb{Z} \times \mathbb{Z} $ can be generated by at least two elements we would need a subgroup that can also be generated by at least two elements. But this subgroup of $ \mathbb{Q} $ would be cyclic. Since $ \mathbb{Z} \times \mathbb{Z} $ is not cyclic we are done with the proof? –  Oct 21 '23 at 11:20
  • @MathMaestro sure, but how would you show that every finitely generated subgroup of $\mathbb{Q}$ is cyclic? – freakish Oct 21 '23 at 11:28
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    here is a general version of the question that was actually being asked. – lulu Oct 21 '23 at 11:44
  • Yes. And to comment on $G=\mathbb{R}$ or $G=\mathbb{C}$. For each $n$ our $G$ has a subgroup isomorphic to $\mathbb{Z}^n$. That's because $G$ is a vector space over $\mathbb{Q}$ of infinite dimension (by the cardinality argument). And so a choice of $n$ linearly independent numbers will generate such subgroup. – freakish Oct 21 '23 at 11:47

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Hint: Suppose that $f:\Bbb Z\times \Bbb Z \to H$ is an isomorphism. Take $\frac{a}{b}=f(1,0)$ and $\frac{c}{d}=f(0,1)$. Then $\langle \frac{a}{b},\frac{c}{d}\rangle = \{n \frac{a}{b}+m\frac{c}{d}:n,m\in\Bbb Z\}$ would be isomorphic to $\Bbb Z \times \Bbb Z$, but $\langle \frac{a}{b},\frac{c}{d}\rangle$ is cyclic.

jjagmath
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  • Is it correct, that $ \frac{1}{bd} $ would generate the same group? –  Oct 21 '23 at 20:09
  • No, but it's obviously a subgroup of $\langle \frac{1}{bd}\rangle$, which is enough to prove it's cyclic. Take for example $\langle \frac{2}{3},\frac{2}{5}\rangle$. It's evident that $\frac{1}{3\cdot 5}$ is not in the group. – jjagmath Oct 22 '23 at 04:36
  • Thank you very much –  Oct 22 '23 at 05:45