# Finding a nilpotent, infinite, f.g., virtually abelian, irreducible integer matrix group

I was wondering if anyone here knows of an example of a group $$M \leq \mathrm{GL}_n(\mathbb{Z})$$ which is

• nilpotent,
• infinite,
• finitely generated,
• virtually abelian,
• irreducible (over $$\mathbb{Z}$$ or $$\mathbb{C}$$, shouldn't matter).

It is not hard to find examples that satisfy all but the last property (e.g. suitable affine groups). But so far we could not write down one with all five.

Motivation: A student of mine is developing an algorithm which computes for a given polycyclic group whether it is residually nilpotent. At this point, his algorithm can deal with all examples we have and any that colleagues provided. Yet it definitely has limits, but we have trouble finding an actual concrete example where the algorithm in its current form is not applicable. A group $$M$$ as described above could be used to construct a kind of "minimal counterexample" where his current algorithm does not work (and then he could either improve the algorithm, or at least discuss the obstacles in a more concrete setting).

• I think I can show that this does not exist if one requires irreducibility over $\mathbf{C}$. I can post details if necessary (this involves basics on algebraic groups).
– YCor
Apr 16 at 12:01
• I would be very much interested in those details :-) Apr 16 at 12:13

## 3 Answers

YCor's answer proving that there are no $$\mathbb{C}$$-irreducible examples is of course correct, but if you are coming at it using standard nilpotent group theory, you can streamline the proof somewhat. The fact that $$G\leq \mathrm{GL}_n(\mathbb{Z})$$ implies that the set of scalar matrices in $$G$$ is finite, and if $$G$$ is irreducible over $$\mathbb{C}$$ then the centre of $$G$$ acts as scalar matrices by Schur's lemma. Thus $$Z(G)$$ is finite.

But now it is a general fact that finitely generated nilpotent groups with finite centre are finite. To see this, remember that if $$G$$ is finitely generated so is each term of the upper/lower central series. Also, the exponent of $$Z_i(G)/Z_{i-1}(G)$$ divides the exponent of $$Z(G)$$.

• Oh, that's very nice. I actually knew the one about nilpotent with finite centre implying finite. That's definitely easier for my student to work with. But the concrete example which is irreducible over $\mathbb{R}$ is also very helpful. Apr 17 at 12:02

Consider the subgroup of $$\mathrm{GL}_4(\mathbf{Z})$$ generated by the matrices $$u=\begin{pmatrix}I_2 & 0\\0 & -I_2\end{pmatrix}$$ and $$v=\begin{pmatrix}0 & A\\I_2 & 0\end{pmatrix}$$, where $$A$$ is a matrix of trace $$\ge 3$$ (hence infinite order) in $$\mathrm{SL}_2(\mathbf{Z})$$. Its derived subgroup is central of order 2 (thus it is 2-step nilpotent), and it is infinite since $$v$$ has infinite order. If has an abelian subgroup of index 2.

It is not hard to show that it is irreducible over $$\mathbf{R}$$, using that $$A$$ is irreducible in dimension $$2$$ and that there are quite few $$u$$-invariant subspaces.

(But this is not irreducible over $$\mathbf{C}$$.)

There is no such group that is $$\mathbf{C}$$-irreducible.

Indeed, let $$G$$ be its Zariski closure. Let $$U$$ be its unipotent radical: by irreducibility, $$U$$ is trivial [this very point applies even assuming irreducibility over a subfield of $$\mathbf{C}$$].

Since $$G$$ is nilpotent, we deduce that $$G^0$$ is a torus, denote it as $$D$$. Every algebraic group over $$G$$ has a finite subgroup $$F$$ such that $$G^0F=G$$. Fixing such $$F$$, we have $$G=DF$$.

Identify the algebraic groups with their complex points.

Let $$T$$ be the torsion subgroup of $$D$$. Then $$F$$ acts on $$D/T$$, which is an infinite-dimensional vector space over $$\mathbf{Q}$$. Then the set of fixed points has a unique complement $$W$$, and $$[F,W]=W$$ (this holds for a action of an arbitrary finite group on an arbitrary vector space over $$\mathbf{Q}$$). Since $$F\ltimes (D/T)$$ is nilpotent, we deduce that $$W=0$$. So $$F$$ acts trivially on $$D/T$$. Hence $$[F,D]\subset T$$. Since $$D$$ is connected and $$T$$ is totally disconnected (it is countable), we deduce that $$[F,D]$$ is trivial. That is, $$D$$ is central.

Since $$D$$ is central and acts diagonalizably, it acts as scalars.

But $$G$$ has only the determinants $$\pm 1$$. So $$|D|$$ has cardinal $$\le 2n$$, and hence $$G=DF$$ is finite.

• Apologies, I don't quite follow at the end: why exactly is $|D|\leq 2n$ ? Apr 16 at 12:54
• @MaxHorn $D$ consists of determinant $\pm 1$ scalar matrices.
– YCor
Apr 16 at 12:57
• Ah of course (I missed the bit that $D$ acts as scalars; it must because it is central and the action is irreducible). Very nice. We'll have to digest this a bit, thank you very much Apr 16 at 13:36