Let $\Gamma$ be a finitely-generated, torsion-free, nilpotent group, of nilpotency class $n\ge 2$. Is there an $N \lhd \Gamma$, such that (i) $N$ is two-step nilpotent, (ii) $\Gamma / N$ is torsion-free, and (iii) $C_\Gamma(N) = Z(\Gamma)$?
(By "two-step nilpotent" I mean nilpotency class $2$.)
Edit: Derek Holt has shown that there might not even be any normal subgroups with nilpotency class smaller than $n - 1$. Hence, a better question is the following:
Question: Let $\Gamma$ be finitely-generated, torsion-free, and nilpotent, with nilpotency class $n > 2$. Is there an $N\lhd \Gamma$ such that (i) the nilpotency class of $N$ is smaller than $n$, (ii) $\Gamma / N$ is torsion-free, and (iii) $C_\Gamma(N) = Z(\Gamma)$?
I don't believe that this is true. In fact I don't even believe that there is always a class $2$ normal nilpotent subgroup. Let $N = {\mathbb Z}^n$ be free abelian with generators $x_1,\ldots,x_n$, let $y$ act on $N$ by $x_i^y = x_ix_{i+1}$ ($i<n$), $x_n^y=x_n$, and let $G = N \rtimes \langle y\rangle$ be the semidirect product. Then $G$ is nilpotent of class $n$.
Any normal subgroup of $G$ that is not contained in $N$ contains an element $y^kx$ for some $k>0$, $x \in N$, and then it must contain $[y^k,x_i]$ for all $i$, and so it contains the subgroup $\langle x_2^k,x_3^k,\ldots,x_n^k \rangle$ of $N$ and has class at least $n-1$. So if we choose $n>2$, then there are no normal subgroups of class $2$.