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Let $G$ be a finitely generated group and $\varphi:G\to \operatorname{Aut}(\mathbb C)$ a homomorphism, where $\operatorname{Aut}(\mathbb C)$ is the group of complex affine transfromations $a z+b$.

Can we find a torsion free-subgroup $H$ of $G$ with finite index? And can we find a normal subgroup $H$ which is torsion-free with finite index?

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    $\begingroup$ You probably want $\varphi$ to be injective. $\endgroup$
    – abx
    Feb 26, 2022 at 4:48
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    $\begingroup$ What do you mean by $Aut(\mathbb{C})$, automorphisms preserving which structure? $\endgroup$
    – Antoine Labelle
    Feb 26, 2022 at 5:15
  • $\begingroup$ @Antoine Labelle: From the title it seems safe to assume that it is the group of affine transformations $z\mapsto az+b$. $\endgroup$
    – abx
    Feb 26, 2022 at 5:27
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    $\begingroup$ Having a finite-index torsion-free subgroup and a finite-index torsion-free normal subgroup are equivalent conditions. $\endgroup$
    – YCor
    Feb 26, 2022 at 6:39
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    $\begingroup$ If $\varphi(G)={1}$, your group is an arbitrary finitely generated group. $\endgroup$
    – abx
    Feb 26, 2022 at 7:36

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Yes. More generally, for any field $K$ we have an embedding of $\operatorname{Aff}(K^n)$ in $\operatorname{GL}_{n+1}(K)$, and so if $K$ has characteristic zero we can apply Selberg's lemma to conclude that a finitely generated group of affine transformations of $K^n$ is virtually torsion-free. The normal core of any finite index torsion-free subgroup will be a normal finite index torsion-free subgroup.

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    $\begingroup$ Of course in such a case the projection of $G$ in $K^*$ if a f.g. abelian group, so is virtually torsion-free, and hence so is $G$ (since the kernel $K$ is torsion-free), so the fact is straightforward without use of the (non-trivial) Selberg lemma. $\endgroup$
    – YCor
    Feb 26, 2022 at 22:49
  • $\begingroup$ Yes, the case $n=1$ is easier, but in general you need Selberg. $\endgroup$
    – Giles Gardam
    Feb 27, 2022 at 6:57
  • $\begingroup$ Yes of course for the affine group in every dimension it's equivalent to Selberg. $\endgroup$
    – YCor
    Feb 27, 2022 at 8:07

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