# Torsion-free subgroup of affine group

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?

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

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.
• 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.
• Yes, the case $n=1$ is easier, but in general you need Selberg. Feb 27 at 6:57