I have already proven that $\mathbb{Q}$ is not finitely generated by showing that every finitely generated subgroup of $\mathbb{Q}$ is cyclic and then using proof by contradiction to show that it is not finitely generated. I was then asked to describe all subgroups of $\mathbb{Q}$ that are generated by 2 elements. Isn't that contradictory or am I perhaps misunderstanding the question?

Thanks in advance for any insight.

  • 2
    $\begingroup$ As you already proved, every subgroup generated by $2$ elements can be in fact generated by $1$ element. $\endgroup$ – Nishant Nov 21 '15 at 23:18
  • $\begingroup$ Are we talking about generated as an abelian group? $\endgroup$ – Gregory Grant Nov 21 '15 at 23:20
  • $\begingroup$ @Nishant, so then would that suffice for describing all the said subgroups? $\endgroup$ – user0990 Nov 21 '15 at 23:25
  • $\begingroup$ @GregoryGrant, we are not assuming abelian-ness here $\endgroup$ – user0990 Nov 21 '15 at 23:27
  • $\begingroup$ What group action do you have? $\endgroup$ – Rico Nov 21 '15 at 23:32

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