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?

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    $\begingroup$ What happens in the infinite dihedral group? $\endgroup$ – user641 Feb 10 '12 at 8:20
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    $\begingroup$ Well yes it can do, but it doesn't always! It is even possible for a virtually nilpotent group to be perfect. $\endgroup$ – Derek Holt Feb 10 '12 at 9:00

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.

  • $\begingroup$ Since K is infinite,the number of elements in the conjugacy class of z may be infinite, in this case G is not finitely generated. For general nilpotent group I think this is a nice example, but I am not sure that G is finitely generated. $\endgroup$ – Siming Tu Feb 10 '12 at 13:48
  • $\begingroup$ @Siming Tu: You are correct. A subgroup of a finitely generated group is always countable. Taking K to be the integers (not a very good field, but it works nonetheless) gives a finitely generated example with countably infinite intersection of the lower central series. The (corrected) result is then sharp. $\endgroup$ – Jack Schmidt Feb 10 '12 at 14:00
  • $\begingroup$ Just to be clear, the group generated by {x,y,z(1)} is a finitely generated, virtually nilpotent, perfect group and so provides a (negative) answer to the question we think you wanted to ask: no, the lower central series of a finitely generated, virtually nilpotent group need not terminate after finitely many steps in a finite subgroup. Indeed, it can terminate after only a single step in an infinite subgroup. Steve's comment shows it need not terminate after finitely many steps. $\endgroup$ – Jack Schmidt Feb 10 '12 at 14:16

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