Virtual nilpotent groups and lower central series

Question

By definition, a nilpotent group is one whose lower central series terminates in the trivial subgroup after finitely many steps. Now we want to consider the relation between the virtual nilpotent groups and its lower central series. Since virtual nilpotent group is a group which has a nilpotent subgroup with finite index, so I want to ask whether the lower central series of a virtual nilpotent group can teminate in a finite subgroup after finitely many steps. If not, what about the case if we add the condition that the group is finitely generated?

Answer 1

Proposition: The cardinality of the intersection of the lower nilpotent series of a virtually nilpotent group can be any arbitrary infinite cardinality, and any finite cardinality not of the form $4k+2$ (or 0).

Proof: Let K be an infinite field of characteristic not 2 or 5, and let G be generated by the following matrices over K, where t ranges over a group generating set of K: $$x = \begin{bmatrix} 1&0&0&0&0\\0&0&1&0&0\\0&1&0&0&0\\-1&-1&-1&-1&0\\0&0&0&0&1\end{bmatrix}, \quad y = \begin{bmatrix} 0&1&0&0&0\\0&0&0&1&0\\0&0&1&0&0\\1&0&0&0&0\\0&0&0&0&1\end{bmatrix}, \quad z(t) = \begin{bmatrix} 1&0&0&0&t\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\end{bmatrix}$$

G is the semi-direct product of a simple group of order 60 generated by x and y, and an abelian group V generated by the conjugacy classes of $z(t)$. The intersection of the the lower central series of G is G itself, because G is perfect. G has the same cardinality as K. (This is an explicit example of the kind mentioned by Derek Holt.)

If k is an odd number, then the dihedral group of order $2k$ has the intersection of its lower central series cyclic of order k. If k ≥ 4 is a power of 2, then the affine general linear group AGL(1, k) has the intersection of its lower central series elementary abelian of order k. The direct product of two such groups has the intersection of its lower central series of size the product of the cardinalities, so any positive integer not of the form $4k+2$.

If G is a group and N is a normal subgroup of size $4k+2$, then N has a characteristic subgroup K of order $2k+1$ by Cayley, and so G centralizes $N/K$, or in other words $[G,N] ≤ K$ and $N$ is not the last group in the lower central series of G. Obviously no subgroup has cardinality 0, so the claim has been shown.