# Nilpotent quotient of semidirect product of a nilpotent group and a free abelian group

Let $$N$$ be a finitely generated infinite nilpotent group and let us denote by $$G$$ the semidirect product $$N \rtimes \mathbb{Z}^n$$ for some $$n\in \mathbb{N}$$.

I would like to know if there is an epimorphism $$\varphi \colon G \mapsto N_f$$, where $$N_f$$ is a torsion-free non-abelian nilpotent group. More generally, for any non-abelian polycyclic group $$H$$ is there an epimorphism onto a torsion-free non-abelian nilpotent group?

My attempt is the following:

Suppose that $$\mathbb{Z}^n$$ is freely generated by the elements $$x_1\dots,x_n$$. We take $$\langle \langle x_1\dots,x_n \rangle \rangle$$ the normal closure of the subgroup $$\langle x_1\dots,x_n \rangle= \mathbb{Z}^n$$ in $$G$$. Then, I would say that $$G / \langle \langle x_1\dots,x_n \rangle \rangle$$ is nilpotent? Let us denote $$G / \langle \langle x_1\dots,x_n \rangle \rangle$$ by $$G_0$$ and the quotient homomorphism $$G \mapsto G_0$$ by $$\varphi_0$$.

If $$G_0$$ is an infinite nilpotent group, $$G_0= N_t \times N_f$$, where $$N_t$$ is a finite group and $$N_f$$ is a torsion-free nilpotent group. Then, we take $$G_1$$ to be $$N_f$$ and $$\varphi_1 \colon G_0 \mapsto N_f$$.

Then, $$\varphi_1 \circ \varphi_0$$ is an epimorphism $$G \mapsto N_f$$ where $$N_f$$ is a torsion-free nilpotent group.

However, if $$G_0$$ is a finite group, I cannot define the $$\varphi_1$$ epimorphism, I would just get that $$\varphi_0 \colon G \mapsto G_0$$ is an epimorphism with $$G_0$$ a finite nilpotent group.

There are counterxamples of the form $${\mathbb Z}^2 \rtimes {\mathbb Z}$$.
For example, we could take the group $$G$$ defined by the presentation $$\langle x,y,t \mid xy=yx, t^{-1}xt = y, t^{-1}yt = xy \rangle,$$ with $$N = \langle x,y \rangle$$.
Since $$[G,N] = N$$, $$G/N \cong {\mathbb Z}$$ is the largest nilpotent quotient of $$G$$.