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I was looking for a proper subgroup of $\mathbb Q$ which is not finitely generated under the addition operation.

We know every finitely generated subgroup of $\mathbb Q$ is cyclic. For a proper subgroup I am just thinking about the subgroup $H$ generated by $\{\frac{1}{p} : p \text{ prime }\}$ may work. It seems $1/4$ is not in $H.$ Is this a correct example? Thanks

user26857
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CAA
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    @CAA: Your example is correct. It consists of fractions whose reduced form has a square free denominator. – orangeskid Jul 02 '15 at 05:05

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Try $\mathbb Z[\frac12]$, that is, the set of binary fractions.

More generally, $\mathbb Z[\frac1p]$, that is, the set of fractions whose denominators are powers of the prime $p$.

lhf
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  • Thank you lhf . Is $\mathbb Z[\frac{1}{2}] = {a + b/2 : a, b \in \mathbb Z}$? – CAA Jul 02 '15 at 03:42
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    No, it's the set of all fractions whose denominators are a power of $2$. Your ${a + b/2 : a, b \in \mathbb Z}$ is finitely generated (by $1$ and $1/2$, and so by $1/2$ alone). – lhf Jul 02 '15 at 03:43
  • This makes sense. Thank you so much @lhf. – CAA Jul 02 '15 at 03:44
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    That would be then $\Bbb Z [ \frac 1 2]={\frac 1 {2^k}: k \in \Bbb Z}$? – YoTengoUnLCD Jul 02 '15 at 04:19
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    @YoTengoUnLCD Actually, it would be ${\frac{j}{2^k}:j,k\in\mathbb{Z}}$. These are commonly known as the dyadic rationals and you can find more information on them searching under that name. – Steven Stadnicki Jul 03 '15 at 21:21