Nilpotent group with rationally powered abelianization need not be rationally powered

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-powering-faithful subgroup)
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Statement

It is possible to have a nilpotent group $G$ such that the abelianization of $G$ is a rationally powered group (which in this case means it is a vector space over $\mathbb {Q}$) but such that $G$ is not rationally powered over any prime. In fact, we can select $G$ to be such that it has $p$-torsion for every prime $p$.

This also shows that the derived subgroup of a nilpotent group need not be a quotient-powering-faithful subgroup.

Related facts

Opposite facts

• The group is forced to be divisible for all primes that the abelianization is divisible by. See Nilpotent group is divisible by a prime iff its abelianization is and iff all lower central series quotients are.

Proof

Further information: quotient of UT(3,Q) by a central Z

Let $G$ be the quotient group of $UT(3,\mathbb {Q} )$ (the unitriangular matrix group of degree three over the field of rational numbers) by a subgroup $\mathbb {Z}$ inside its center (which is a copy of $\mathbb {Q}$). Explicitly, we can think of $G$ as matrices of the form:

$\{{\begin{pmatrix}1&a_{12}&{\overline {a_{13}}}\\0&1&a_{23}\\0&0&1\\\end{pmatrix}}\mid a_{12},a_{23}\in \mathbb {Q} ,{\overline {a_{13}}}\in \mathbb {Q} /\mathbb {Z} \}$

with the matrix multiplication defined as:

${\begin{pmatrix}1&a_{12}&{\overline {a_{13}}}\\0&1&a_{23}\\0&0&1\\\end{pmatrix}}{\begin{pmatrix}1&b_{12}&{\overline {b_{13}}}\\0&1&b_{23}\\0&0&1\\\end{pmatrix}}={\begin{pmatrix}1&a_{12}+b_{12}&{\overline {a_{12}b_{23}}}+{\overline {a_{13}}}+{\overline {b_{13}}}\\0&1&a_{23}+b_{23}\\0&0&1\\\end{pmatrix}}$

where ${\overline {a_{12}b_{23}}}$ is understood to be the image of $a_{12}b_{23}$ under the quotient map $\mathbb {Q} \to \mathbb {Q} /\mathbb {Z}$.

The center of $G$ coincides with its derived subgroup, and is:

$\left\{{\begin{pmatrix}1&0&{\overline {a_{13}}}\\0&1&0\\0&0&1\\\end{pmatrix}}\mid {\overline {a_{13}}}\in \mathbb {Q} /\mathbb {Z} \right\}$

The inner automorphism group and the abelianization are therefore both isomorphic to $\mathbb {Q} \times \mathbb {Q}$, which is rationally powered. However, the group as a whole has $p$-torsion for all primes $p$.