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.

  • $\begingroup$ I expect Example 2 can be adapted (by forcing the whole group to be metabelian) to answer your question. Anyway I posted an example which looks simpler to me. $\endgroup$
    – YCor
    Commented Nov 8, 2020 at 12:13
  • 2
    $\begingroup$ I am confused. There are infinite $p$-groups of nilpotency class $2$ with centre of order $p$, i.e. infinite extraspecial groups. $\endgroup$
    – Derek Holt
    Commented Nov 8, 2020 at 12:32
  • $\begingroup$ oh yes...I messed up the question. So I was initially thinking of the torsion-free case... Ycor mentioned in his answer another way to make the the question less trivial (requiring that the centre of any quotient by a finite group to be finite). Either way, Ycor's answer covers them both. $\endgroup$
    – ARG
    Commented Nov 8, 2020 at 13:15

1 Answer 1



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.

  • $\begingroup$ Since you mention the FC-center: in the 1st case it's all of $V$ (when $R$ is finite); in the additional example $\mathbf{Q}\ltimes \mathbf{Q}[t^{>0}]$, the FC-center is trivial. $\endgroup$
    – YCor
    Commented Nov 8, 2020 at 12:15
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    $\begingroup$ PS the obvious nilpotent examples provided by Derek Holt suggest that the requirement "finite center" (for counterexamples) should at least be strengthened to "the quotient by every finite normal subgroup has finite center". This already discards infinite nilpotent groups. $\endgroup$
    – YCor
    Commented Nov 8, 2020 at 12:54

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