7
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '14 at 14:31
  • $\begingroup$ I believe this. Thanks again Ycor, this really helped me out $\endgroup$ – Caleb Eckhardt Aug 3 '14 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 '16 at 7:22

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.