Nilpotent and torsion-free not implies torsion-free abelianization

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This article gives the statement, and possibly proof, of the fact that for a group, the subgroup obtained by applying a given subgroup-defining function (i.e., derived subgroup) does not always satisfy a particular subgroup property (i.e., quotient-torsion-freeness-closed subgroup)
View subgroup property satisfactions for subgroup-defining functions | View subgroup property dissatisfactions for subgroup-defining functions

Statement

It is possible to have a torsion-free nilpotent group G such that the abelianization of G has p-torsion for every prime p.

In particular, this shows that the derived subgroup in a nilpotent group is not necessarily a quotient-torsion-freeness-closed subgroup.

Related facts

Dual fact

Somewhat surprisingly, the dual fact to this is not true. The dual fact, if true, would state that the center of a divisible nilpotent group need not be divisible (and in particular, that the center need not be divisibility-closed in a nilpotent group). This is false. In fact, upper central series members are completely divisibility-closed in nilpotent group.

Converse

Proof

Further information: central product of UT(3,Z) and Q

Let G be the central product of unitriangular matrix group:UT(3,Z) with the group of rational numbers, where the center of the former is identified with a copy of \mathbb{Z} in the latter. Then,

  • G is torsion-free.
  • G' is isomorphic to \mathbb{Z}, and G/G' is isomorphic to \mathbb{Z} \times \mathbb{Z} \times \mathbb{Q}/\mathbb{Z}. This has p-torsion for all primes p.