Is a finitely-generated torsion-free wreath product of an abelian group and a residually nilpotent group itself residually nilpotent?

Question

Suppose that $A$ is finitely-generated torsion-free abelian and $H$ is torsion-free, finitely-generated and residually nilpotent.

Is the restricted wreath product $A \wr H$ necessarily residually nilpotent?

Residual nilpotency of a group $G$ means $\bigcap_n G_n = 1$ where $G_n$ is the lower central series of $G$.

Because $A$ is abelian, for any quotient $\pi : H \to K$ we have a corresponding natural quotient $\hat\pi : A \wr H \to A \wr K$. If $g \in A \wr H$ is nontrivial, then we can clearly find a nilpotent quotient $\pi : H \to K$ such that $\hat\pi(g)$ is nontrivial.

If we could find a nilpotent torsion-free $K$, then Theorem B2 in [Hartley] says that $A \wr K$ is residually nilpotent, and then we would we able to map $g$ to a nontrivial element of a nilpotent group.

So it would suffice to show that a torsion-free residually nilpotent group is residually torsion-free nilpotent. I know that the quotients $H/H_n$ are not always torsion-free even if $H$ is, but the examples I know where this happens are nilpotent, so not very useful.

On the other hand, one could attempt to use Theorem B1 of [Hartley], namely it also suffices that $K$ is a finite $p$-group, or is infinite but not torsion-free and for some $p$ is residually (bounded exponent $p$-group). I don't see why a torsion-free residually nilpotent group would have to admit such quotients, but I don't see a counterexample either (I rarely work with nilpotent groups).

The paper [Hartley] does not seem to give the exact conditions for residual nilpotency of wreath products, and I did not find other papers discussing this problem.

Reference:

[Hartley] Hartley, B., The residual nilpotence of wreath products, Proc. Lond. Math. Soc., III. Ser. 20, 365-392 (1970). ZBL0194.03402.

Answer 1

If $A$ is f.g. torsion-free abelian and $H$ is residually-$p$, then $A\wr H$ is residually-$p$ (easy) and hence residually nilpotent. Thus covers many cases (including the case when $H$ is residually torsion-free nilpotent, but many more).

But the answer is no in general.

Fact. There are torsion-free residually nilpotent groups $H_2,H_3$ and for $p=2,3$, $1\neq c_p\in H_p$, such that for each $p\in\{2,3\}$ and each prime $q\neq p$, the image of $u_p$ in every quotient of $H_p$ that is a finite $q$-group is trivial.

If so then $G=\mathbf{Z}\wr H$ is not residually nilpotent for $H=H_2\times H_3$. Namely, one easily sees that for every prime $q$, denoting by $\delta$ the delta at $1_H$ in the wreath product, the element $$\delta.{}^{u_3^2}\delta.{}^{u_3^2u_2}\delta^{-1}.{}^{u_2u_3}\delta^{-1}$$ is killed in every finite quotient of $G$ that is a $q$-group. So it is killed in every nilpotent quotient of $G$.

The fact is obtained, for instance, by defining $H_p$ as the solvable Baumslag-Solitar group $\mathrm{BS}(1,p+1)=\langle t,x:txt^{-1}x^{-1}=x^p\rangle$ and $u_p=x$.

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Added: for $p$ prime, it is a straightforward observation that $\mathrm{BS}(1,p+1)=\mathbf{Z}\ltimes_{1+p}\mathbf{Z}[1/(1+p)]$ is residually a finite $p$-group. Indeed we can consider its quotients $$\mathrm{BS}(1,p+1)\to\mathbf{Z}\ltimes_{1+p}\mathbf{Z}[1/(1+p)]/p^n\mathbf{Z}[1/(1+p)]$$ $$=\mathbf{Z}\ltimes_{1+p}\mathbf{Z}/p^n\mathbf{Z}\to (\mathbf{Z}/p^n\mathbf{Z})\ltimes_{1+p}\mathbf{Z}/p^n\mathbf{Z}:$$ every nontrivial element of $\mathrm{BS}(1,p+1)$ survives in one of these quotients. For a reference, see Ashot Minasyan's comment. This is also a particular case of Theorem 2.4 in Raptis E. Raptis and D. Varsos: Residual properties of HNN-extensions with base group an abelian group. J. Pure Appl. Algebra, vol. 59, n ̊ 3, 1989, pp. 285–290 ([DOI link](https://doi.org/10.1016/0022-4049(89)90098-4))