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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.

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1 Answer 1

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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 $G$.

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