Let $G$ be a torsion-free, finitely-generated, nilpotent group of nilpotency class at least 3. Does there exist a normal subgroup 
$N\leq G$ such that $G/N\cong \mathbb{Z}$ and $Z(G)=Z(N)$? (By $Z(H)$ I mean the center of the group $H.$) 


The basic examples I've played with have this property, but I'm no group theorist (this question arose in an operator algebraic setting) so that's the only evidence I have one way or the other.



Subgroups of Nilpotent groups with prescribed center
Here are two examples. I describe it as Lie algebras (over any field $K$).


(1) The 7-dimensional, 3-step nilpotent Lie algebra with basis $(X\_1,\dots,X\_7)$ and nonzero brackets


$$ [X\_1,X\_2]=X\_4,[X\_1,X\_3]=X\_5,[X\_2,X\_3]=X\_6,[X\_1,X\_4]=[X\_1,X\_5]=[X\_2,X\_4]=[X\_3,X\_6]=X\_7$$


(2) The 6-dimensional, 4-step nilpotent Lie algebra with basis $(X\_1,\dots,X\_6)$ and nonzero brackets


$$[X\_1,X\_2]=X\_3, [X\_1,X\_3]=X\_4, [X\_2,X\_3]=X\_5, [X\_1,X\_5]=[X\_2,X\_4]=X\_6.$$


Here's a common proof. In (1), let $V$ be the subspace generated by $X\_1,X\_2,X\_3$, $W$ the subspace generated by $X\_4,X\_5,X\_6$, and $Z$ the subspace generated by $X\_7$. In (2), let $V$ be the subspace generated by $X\_1,X\_2$, $W$ the subspace generated by $X\_4,X\_5$, and $Z$ the subspace generated by $X\_6$.


Then in each $\mathfrak{g}$ of these two: $Z$ is the 1-dimensional center, and the bracket $V\times W\to Z$ defines a non-degenerate pairing. Thus for each hyperplane $X$ in $V$, its centralizer in $W$ is equal to its orthogonal $X^\bot$ with respect to this pairing.


Each codimension 1 ideal of $\mathfrak{g}$ has the form $I\_X=X\oplus \mathfrak{g}$ where $X$ is some hyperplane, and the center of $I\_X$ is equal to the 2-dimensional ideal $X^\bot\oplus Z$.


Both Lie algebra being defined over $\mathbf{Q}$, we can consider the corresponding lattice, which therefore answers your question.


