Centre of solvable locally nilpotent groups

Question

This question is motivated by two examples of locally nilpotent groups which I came across (see below).

Question: Given an infinite solvable and locally nilpotent group $G$, does $G$ have an infinite centre?

The question concerns groups which are not finitely generated (otherwise the answer is well-known [yes]). To see that the hypothesis "solvable" is required (and the answer is negative even if one relaxes to hypersolvable), here are the examples I stumbled upon:

Example 1: Look at the group of infinite matrices (with rows and columns indexed by $\mathbb{Z}$) with 1 on the diagonal and only finitely many non-zero integer entries above the diagonal. This group is locally nilpotent (because a finite number of elements will remain in a finite range of indices). This group is hypersolvable but not solvable and has trivial centre (or FC-centre)

Example 2: Consider the group generated by the elements $n_i$ for $i \in \mathbb{Z}$. Add relations so that $n_i$ and $n_j$ generate a free $p$-nilpotent group of rank $c_{|i-j|}$ (where $c_k$ is a sequence of [strictly increasing] positive integers). Again this group is hypersolvable but has trivial centre.

I came across both examples in the literature; making an extension by $\mathbb{Z}$ of these groups give finitely generated [elementarily amenable] groups with amusing properties.

Answer 1

No.

For a scalar (= unital associative commutative) ring $R$, consider $V=V(R)=R[X]$, the polynomial ring, and $q$ the operator $X^n\mapsto X^{n-1}$, $X\mapsto 0$. Let $V_n=V_n(R)$ be the $R$-submodule of degree $\le n$ polynomials. Then $q$ stabilizes $V_n$ and is nilpotent on $V_n$. Hence $1+q$ is invertible on $V_n$, and hence on $V$, and $G(R)=\mathbf{Z}\ltimes_{1+q} V$ is locally nilpotent and metabelian.

The center is reduced to $V_0(R)\simeq R$. Indeed the center is obviously contained in $V$ (since $1+q$ has infinite order) and, on $V$, the fixed points of $1+q$, that is, the kernel of $q$, is reduced to $V_0$. Hence choosing $R$ finite this center is finite (and $G(R)$ is infinite as soon as $R\neq 0$, so $R=\mathbf{Z}/2\mathbf{Z}$ is fine).

One can produce similar examples with trivial center and torsion-free. Here take $V=\mathbf{Q}[X^{\mathbf{Q}_{>0}}]:=\bigcup_nX^{1/n!}\mathbf{Q}[X^{1/n!}]$. Define $q$ similarly (divide by $X$ and eliminate terms with non-positive exponents). So $q$ is locally nilpotent and $1+q$ is invertible. In addition, one can define $(1+q)^a$ for arbitrary $a\in\mathbf{Q}$, using the usual power series (using that $q^nx=0$ for all large enough $n\ge n_0(x)$). This defines an action of $\mathbf{Q}$ on $V$, whose fixed points are reduced to $\{0\}$ (this is why I removed constants). Hence the semidirect product $\mathbf{Q}\ltimes_{((1+q)^a)_a} V$ is metabelian, locally nilpotent, infinite, has trivial center, and is torsion-free.