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.