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



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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**](https://doi.org/10.1112/plms/s3-20.3.365), Proc. Lond. Math. Soc., III. Ser. 20, 365-392 (1970). [ZBL0194.03402](https://zbmath.org/?q=an:0194.03402).



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