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Suppose that $G$ is a group with a finite subgroup $H$ such that $G/H$ embeds into a finitely generated group. Does $G$ itself embed into a finitely generated group?

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  • $\begingroup$ $G/H$ is a finite group, hence it trivially embeds in a finitely generated group (in itself). This does not give any information. $\endgroup$ – Crostul Jan 26 '17 at 20:10
  • $\begingroup$ Oops, sorry, $H$ is supposed to be finite, not finite index! (Question edited.) $\endgroup$ – Mark Jan 26 '17 at 20:21
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    $\begingroup$ If you want a direct proof, you could embed $G$ in the the wreath product $H \wr G/H$, which embeds in $H \wr N$ when $G/H < N$, and $H \wr N$ is finitely generated if $N$ is. $\endgroup$ – Derek Holt Jan 26 '17 at 21:18
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A group embeds in a finitely generated group if and only if it is countable. You can find a proof in Lyndon and Schupp. Can you finish the proof now?

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  • $\begingroup$ Wow, can't believe I never heard this before! Thanks. $\endgroup$ – Mark Jan 26 '17 at 21:58
  • $\begingroup$ @Mark: This is what the HNN extension was invented for. $\endgroup$ – Moishe Kohan Jan 26 '17 at 22:49

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