0
$\begingroup$

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?

$\endgroup$
7
  • 3
    $\begingroup$ You probably want $\varphi$ to be injective. $\endgroup$
    – abx
    Feb 26 at 4:48
  • 1
    $\begingroup$ What do you mean by $Aut(\mathbb{C})$, automorphisms preserving which structure? $\endgroup$ Feb 26 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 at 5:27
  • 1
    $\begingroup$ Having a finite-index torsion-free subgroup and a finite-index torsion-free normal subgroup are equivalent conditions. $\endgroup$
    – YCor
    Feb 26 at 6:39
  • 2
    $\begingroup$ If $\varphi(G)={1}$, your group is an arbitrary finitely generated group. $\endgroup$
    – abx
    Feb 26 at 7:36

1 Answer 1

6
$\begingroup$

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.

$\endgroup$
3
  • 2
    $\begingroup$ 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 at 22:49
  • $\begingroup$ Yes, the case $n=1$ is easier, but in general you need Selberg. $\endgroup$ Feb 27 at 6:57
  • $\begingroup$ Yes of course for the affine group in every dimension it's equivalent to Selberg. $\endgroup$
    – YCor
    Feb 27 at 8:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.