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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.

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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.

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  • $\begingroup$ You're right, sorry for the mess. I finally removed the general construction, which was correct, and left 2 examples, for the sake of clarity. $\endgroup$
    – YCor
    Aug 2, 2014 at 14:31
  • $\begingroup$ I believe this. Thanks again Ycor, this really helped me out $\endgroup$
    – Caleb Eckhardt
    Aug 3, 2014 at 19:58
  • $\begingroup$ For clarity, I forgot to say from the beginning that I'm giving examples of nilpotent Lie algebras in which the center of every codimension 1 ideal strictly contains the center of the whole Lie algebra (in order to show that $N$ in the question does not always exist). $\endgroup$
    – YCor
    Dec 4, 2016 at 7:22

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