Prove that the torsion subgroup of a finitely generated nilpotent group is finite. More generally, in any group with "almost" no torsion all periodic subgroups are finite.

Here "almost" means that there is a subgroup of finite index which has no torsion elements.

I tried to "apply" the main theorem that a f.g. NP group is polycyclic. That is probably the first step, but I don't see how to proceed from there.

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    $\begingroup$ A hint: the torsion subgroup is itself finitely generated nilpotent, hence polycyclic. Consider its subnormal series with cyclic factors. Can the infinite cyclic group appear as a factor here? $\endgroup$ – Miha Habič Aug 18 '13 at 15:23
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    $\begingroup$ For your more general query, if $G$ has a torsion-free subgroup of index $n$, then any torsion subgroup has order at most $n$, since otherwise it would intersect the torsion-free subgroup nontrivially. $\endgroup$ – Derek Holt Aug 18 '13 at 16:21
  • $\begingroup$ The main difficulty is to know that f.g. nilpotent implies polycyclic. This being granted, $T$ denoting the torsion subgroup, both $T\cap G'$ (by induction, since $G'$ is also finitely generated) and the image of $T$ in $G/G'$ are finite, so $T$ is finite. $\endgroup$ – YCor Jul 12 '18 at 21:08

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