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?
Embedding in a finitely generated group, up to finite index
0
$\begingroup$
group-theory
-
0$G/H$ is a finite group, hence it trivially embeds in a finitely generated group (in itself). This does not give any information. – 2017-01-26
-
0Oops, sorry, $H$ is supposed to be finite, not finite index! (Question edited.) – 2017-01-26
-
1If 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. – 2017-01-26
1 Answers
1
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?
-
0Wow, can't believe I never heard this before! Thanks. – 2017-01-26
-
0@Mark: This is what the HNN extension was invented for. – 2017-01-26