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It is well known that $\mathbb{Q}$ (under the operation of addition) is not finitely generated. However, a subgroup of a finitely generated group need not be finitely generated.

So I'm curious: Is there a finitely generated group $G$ with a subgroup $S$ such that $S$ is isomorphic to $\mathbb{Q}$?

MartianInvader
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    https://mathoverflow.net/questions/23352/an-explicit-example-of-a-finitely-presented-group-containing-a-subgroup-isomorph – Eric Towers Oct 12 '24 at 18:56
  • @EricTowers Seems to be an answer, not a comment. – Martin Brandenburg Oct 12 '24 at 19:01
  • I'm not sure what the policy is for questions here that are answered on mathoverflow, however if there is a simpler example that is only finitely generated (not finitely presented) I would accept that as well. – MartianInvader Oct 12 '24 at 19:22

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