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Give a finitely generated group that is not virtually torsion-free.

What I know: The lamplighter group ( The Wreath product of $\mathbb{Z}_2$ and $\mathbb{Z}$) is a finitely generated group that is not virtually torsion-free.

Does this mean that it has a finite index subgroup that is not torsion-free? I could not get any idea.

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  • $\begingroup$ See the wikipedia article on virtual properties of groups. $\endgroup$ – Rob Arthan Aug 1 '18 at 1:38
  • $\begingroup$ Free product of any two ( non trivial) finite group $\endgroup$ – Anubhav Mukherjee Aug 1 '18 at 1:40
  • $\begingroup$ @AnubhavMukherjee That's virtually torsion-free. $\endgroup$ – Derek Holt Aug 1 '18 at 7:57
  • $\begingroup$ Every finite group is not virtually torsion-free and finitely generated, right? Or do we treat a trivial group as torsion-free? $\endgroup$ – freakish Aug 1 '18 at 10:12
  • $\begingroup$ @freakish A group is torison-free if it had no nontrivial torsion elements, so yes the trivial group is torsion-free. $\endgroup$ – Derek Holt Aug 1 '18 at 13:29
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By the definition $G$ is virtually torsion-free if it has a subgroup of finite index (possibly itself) which is torsion-free.

So $G$ is not virtually torsion-free if there is no subgroup of finite index which is torsion-free.

I don't know about the Wreath product, but consider this example:

Let $p$ be a prime number and $G$ be a Tarski monster group for $p$, i.e. $G$ is infinite and if $H$ is a proper, nontrivial subgroup of $G$ then $|H|=p$. Such groups exist if $p > 10^{75}$ and are always finitely generated (in fact if $g,h\in G$ are nontrivial and such that $\langle g\rangle\neq\langle h\rangle$ then $\langle g,h\rangle=G$) . And it is not virtually torsion-free since it is infinite (meaning the trivial subgroup is not of finite index) and every element is of finite order (meaning it doesn't have a nontrivial torsion-free subgroup to begin with).

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    $\begingroup$ Proving the existence of Tarski monsters is horrendously difficult. Wreath products are relatively elementary. A torsion-free subgroup of $C_2 <wr {\mathbb Z}$ must intersect the base group trivially. Since the base group is infinite, this group is not virtually torsion-free. $\endgroup$ – Derek Holt Aug 1 '18 at 13:33

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