Polycyclic groups

This is a guest post by Lawk Mineh

When approaching algebraic objects and structures, it is frequently useful to decompose the objects at hand into basic building blocks, each of which can be understood individually and then pieced together to understand the whole structure.

This approach has been famously prominent in the theory of finite groups, where decades of work has culminated in a complete classification of finite simple groups, with many powerful consequences.

The world of infinite groups, by contrast, is unfathomably more wild. General classification theorems are certainly out of the question, but we can often perform a sort of converse task and determine what we can say about groups that are constructed from some basic building blocks.

Perhaps the most basic of infinite groups is the infinite cyclic group, and polycyclic groups are those built out of cyclic groups. That is, \(G\) is polycyclic if it has a subnormal series

\[   \{1\} = G_n \lhd G_{n-1} \lhd \dots \lhd G_0 = G\]

with each factor group \(G_i/G_{i+1}\) cyclic.

The main aim of this post is to give a brief, theory-focused introduction to this class of groups. I hope that I can convey some of the cause of my interest in the topic: it happens that this very simple definition gives rise to quite a beautiful and rich theory, with deep connections to algebraic number theory, Lie theory, and computability.

The principal invariant of a polycyclic group \(G\) is its Hirsch length, \(h(G)\), which is the number of infinite cyclic factor groups appearing in subnormal series as above. Recall that a group is solvable if it has a subnormal series whose factor groups are abelian (i.e. it is a poly-abelian group). Of course, for finite groups, being polycyclic coincides with being solvable.

The situation is quite different for infinite groups, to the extent that one might think of polycyclic as the `correct' generalisation of solvability to the class of infinite groups. Let us make some basic observations.

Lemma 1:  Let \(G\) be a group. The following are equivalent:

1.  \(G\) is polycyclic;
2. \(G\) has a subnormal series whose factor groups are finitely generated abelian; 
3. \(G\) is solvable and each of its subgroups are finitely generated.

Proof:    That (i) implies (ii) is immediate.

 To see that (ii) implies (iii), observe that every subgroup of \(G\) also has a subnormal series whose factor groups are finitely generated (and abelian). An extension of a finitely generated group by another is again finitely generated, so by a finite induction any subgroup of \(G\) is finitely generated.

Lastly, (iii) implies (i) as each term appearing in the subnormal series for \(G\) will be finitely generated, so the factor groups will be finitely generated and abelian, by the definition of solvability. $\blacksquare$

The following we leave as a straightforward exercise: one can build polycyclic series in each case in more or less the obvious way.

Lemma 2: The class of polycyclic groups is closed under taking subgroups, quotients, and extensions.

There are many solvable groups containing subgroups that are not finitely generated. For a finitely generated example, consider the wreath product \(\mathbb{Z}_2 \wr \mathbb{Z}\) (often called the lamplighter group). This group solvable and is generated by the generators of the base and the exponent, which are both cyclic. However, it contains a copy of the infinite direct sum \(\bigoplus_{i=1}^\infty \mathbb{Z}_2\), which is not finitely generated.

As we will see, polycyclic groups enjoy many useful properties that solvable groups more generally do not.

Lemma 3:  Let \(G\) be a polycyclic group. Then \(G\) has a finite index subgroup which is poly-(infinite cyclic). In particular, \(G\) is virtually torsionfree.

Proof:    We proceed by an induction on the length of the subnormal series \(\{1\} = G_n \lhd G_{n-1} \lhd \dots \lhd G_0 = G\) for \(G\). By induction there is a poly-(infinite cyclic) subgroup \(H \leqslant_f G_1\) of finite index in \(G_1\). Since \(G_1\) is finitely generated, we may assume \(H \lhd G\) is a normal subgroup, replacing it by its characteristic core if necessary. If \(G/G_1\) is finite, \(G_1 \leqslant_f G\) is of finite index in \(G\), and so \(H \leqslant_f G\) is the desired subgroup in this case.

    Otherwise \(G/G_1\) is infinite cyclic. In this case, observe that \(G/G_1 \cong (G/H)/(G_1/H)\) and \(G_1/H\) is finite. Thus \(G/H\) is virtually infinite cyclic: take \(K/H \leqslant_f G/H\) to be a finite index infinite cyclic subgroup. Then \(K \leqslant_f G\) has finite index in \(G\) and is poly-(infinite cyclic) as required. $\blacksquare$

Many statements about polycyclic groups may be proved by induction on the Hirsch length: one reduces the complexity of the given group by cutting away infinite factors. To carry out the induction it is useful to have a well-behaved candidate either to take a quotient by, or pass to as a subgroup.

The following lemma provides us with a particularly useful example of such.

Lemma 4: Let \(G\) be a polycyclic group. Then \(G\) contains an nontrivial characteristic free abelian subgroup.

Proof:    By the previous lemma, \(G\) has a finite index torsionfree subgroup \(H \leqslant_f G\). As \(G\) is finitely generated, we may take \(H\) to be a characteristic subgroup.

    Since \(H\) is solvable, there is \(n\) such that \(H^{(n)} = \{1\}\), where \(H^{(k)}\) denotes the \(k^{th}\) term of the derived series of \(H\). It follows that \(A = H^{(n-1)}\) is a nontrivial abelian subgroup.

    Moreover, the terms of the derived series are characteristic subgroups of \(H\), and being a characteristic subgroup is a transitive property. Finally, \(A\) is free abelian as \(H\) is torsionfree. Thus \(A \leqslant G\) is the required subgroup. $\blacksquare$

As a sample result, we shall show that polycyclic groups are residually finite -- that each nontrivial group element is nontrivial in some finite quotient of the group.

Theorem 5: Polycyclic groups are residually finite.

Proof: We proceed by induction on the Hirsch length \(h(G) = n\) of the polycyclic group \(G\). Finite groups are certainly residually finite, so the statement holds for the base case of \(n = 0\).

 Suppose now that the statement holds for all polycyclic groups with Hirsch length less than \(n\). Let \(g \in G\) be a nontrivial element. We will show that there is a quotient \(\varphi \colon G \to Q\) with \(Q\) residually finite in which the image of \(g\) is nontrivial.

    To see that this suffices, see that this implies there is a finite quotient \(\psi \colon Q \to P\) with \(\psi(\varphi(g))\) nontrivial. Thus \(\psi \circ \varphi \colon G \to P\) is a finite quotient with \(\psi \circ \varphi (g)\) nontrivial, as required.

    Let \(A\) be a nontrivial characteristic free abelian subgroup of \(G\) as in the previous lemma. Write \(A_m = \{a^m \, | \, a \in A\} \leqslant A\) for each \(m \in \mathbb{N}\). Observe that each \(A_m\) is infinite and characteristic in \(A\), and so also characteristic in \(G\).

    Moreover, we can pick \(m\) sufficiently large so that \(g \notin A_m\) and set \(Q = G/A_m\) (if \(g \notin A\) then \(m = 1\) suffices). Now since \(A_m\) is infinite, \(h(Q) = h(G) - h(A_m) < h(G)\), and so \(Q\) is residually finite by induction. By construction the image of \(g\) is nontrivial in \(Q\), so we are done. $\blacksquare$

In fact, the `residual' properties of polycyclic groups -- that is to say, what can be seen of them via their finite quotients -- are very strong. It is a result of Grunewald, Pickel, and Segal that polycyclic groups can be (up to some finite error) distinguished from one another by their finite quotients. Of course, the proof is out of the scope of this post.

Linearity and algebraic structure

We will dedicate the remainder to beginning a deeper analysis of the algebraic structure of polycyclic groups, and in particular its connections with linearity. We will need a couple of preliminaries.

The first is a group-theoretic analogue to Lie's theorem on solvable Lie algebras, first proven by Mal'cev. The full proof is rather involved and technical, so we will give a very rough sketch, taking some assertions as facts and skipping most detail.

Theorem 6: There is a function \(f \colon \mathbb{N} \to \mathbb{N}\) such that the following holds.

    Let \(k\) be an algebraically closed field and let \(G\) be a finitely generated solvable subgroup of \(\operatorname{GL}_n(k)\). Then there is \(H \leqslant_f G\) with index \([G : H] < f(n)\) that is triangularisable.

Note that the function \(f\) appearing in the statement depends neither on the particular linear group, nor on the field over which it is linear, and rather only on the dimension.

Proof sketch:  Call \(G\) reducible if its action on \(k^n\) leaves a proper subspace invariant, and irreducible otherwise.

    The idea will be that, up to picking a suitable basis, matrices in \(G\) will be block upper triangular, with the diagonal blocks corresponding to irreducible representations of \(G\) in \(\operatorname{GL}_{m_i}(k)\) with \(\sum m_i = n\). To triangularise \(G\), then, we deal with such blocks independently, using an inductive argument.

    With this in mind, we may as well assume \(G\) is irreducible. We claim that in this case, \(G\) is has a scalar (i.e. consisting of matrices of the form \(xI_n\)) of uniformly bounded finite index.

    It is a fact that a soluble subgroup \(H\) of \(\operatorname{SL}_n(k)\) all of whose subgroups of index at most \(n\) are irreducible is itself finite. The key fact here is that abelian subgroups of \(\operatorname{SL}_n(k)\) are diagonalisable with entries being roots of unity.  One can find an appropriate abelian subgroup of \(H\) whose elements are matrices with entries \((n^2)^{th}\)-roots of unity, of which there are finitely many, then one can estimate the size of \(H\) using this. It is then possible to bound the index of a scalar subgroup of \(G\) by considering its overgroup \(G\cdot kI_n \cap \operatorname{SL}_n(k)\). $\blacksquare$

The other ingredient is Dirichlet's units theorem from algebraic number theory. The most commonly stated version of this result computes explicitly the rank of the group of units; here we just need to know that it is finitely generated.

Lemma 7:   Suppose \(k\) is an algebraic number field, and \(\mathcal{O_k}\) its ring of integers. Then \((\mathcal{O_k},+)\) and \((\mathcal{O_k}^\times,\cdot \,)\) are finitely generated as groups.

The first of our main results identifies certain solvable linear groups as being polycyclic, and is due to Mal'cev.

Theorem 8: If \(G\) is a finitely generated solvable subgroup of \(\operatorname{GL}_n(\mathbb{Z})\), then \(G\) is polycyclic.

Proof:    By Theorem 6, \(G\) has a finite index normal subgroup \(H\) that is triangularisable over \(\mathbb{C}\). Note \(G/H\) is solvable and finite, so polycyclic, and being polycyclic is closed under extensions. Hence \(G\) is polycyclic if and only if \(H\) is. Thus we may assume without loss of generality, after possibly conjugating, that \(G\) is itself triangular over \(\mathbb{C}\).

    Let \(S\) be a finite generating set of \(G\), and \(T \subseteq \mathbb{C}\) the set of entries of matrices in \(S\). Take \(k\) to be the smallest subfield of \(\mathbb{C}\) containing \(T\). As \(T\) is finite, \(k\) is a finite extension of \(\mathbb{Q}\): it is an algebraic number field.  Let \(\mathcal{O_k}\) be the ring of integers of \(k\). As  \(T\) is a finite sit of algebraic numbers, there is \(m \in \mathbb{Z}\) such that \(mT \subseteq \mathcal{O_k}\) is a set of algebraic integers.

    Let \(M \in \operatorname{GL}_n(\mathbb{Q})\) be the matrix whose \(i^{th}\) diagonal entry is \(m^{1-i}\) and whose other entries are zero. Then \(xGx^{-1} \leqslant \operatorname{GL}_n(k)\) and its entries over the diagonal are in \(\mathcal{O_k}\). Moreover, the roots of the characteristic polynomial \(p\) of any \(g \in G\) are algebraic integers, as \(p\) is invariant under conjugation and \(G\) is a subgroup of \(\operatorname{GL}_n(\mathbb{Z})\) by assumption.

    The roots of \(p\) are exactly the entries of \(g\) by the Cayley-Hamilton theorem, so the diagonal entries of \(g\) are also in \(\mathcal{O_k}\). Hence \(xGx^{-1}\) is a triangular subgroup of \(\operatorname{GL}_n(\mathcal{O_k})\).

    Write \(\operatorname{B}_n(\mathcal{O_k})\) for the (full) upper triangular subgroup, \(\operatorname{D}_n(\mathcal{O_k})\) for the diagonal subgroup, and \(\operatorname{UT}_n(\mathcal{O_k})\) for the upper unitriangular (i.e. with all diagonal entries equal to the identity) subgroup of \(\operatorname{GL}_n(\mathcal{O_k})\).

    There is a short exact sequence

    \[1 \to \operatorname{UT}_n(\mathcal{O_k}) \to \operatorname{B}_n(\mathcal{O_k}) \to \operatorname{D}_n(\mathcal{O_k}) \to 1\]

    where the map on the left is inclusion, and the map on the right kills all non-diagonal entries. By Lemma 7, \(\mathcal{O_k}^\times\) is a finitely generated (abelian) group. Of course, \(D \cong (\mathcal{O_k}^\times)^n\) is finitely generated abelian, and so polycyclic.

    We will see in a future post that \(\operatorname{UT}_n(\mathcal{O_k})\) is finitely generated nilpotent, and hence polycyclic. Since being polycyclic is closed under both taking extensions and  passing to subgroups, \(\operatorname{B}_n(\mathcal{O_k})\) -- and hence \(G\) -- is polycyclic. $\blacksquare$

Note that from the above proof, it can be extracted that solvable linear groups (over any field) are virtually nilpotent-by-abelian. The \(\mathbb{Z}\)-linearity comes into play only to ensure the finite generation of the upper unitriangular and diagonal subgroups.

This theorem in fact has a converse due to Auslander and Swan: polycyclic groups are exactly the class of solvable \(\mathbb{Z}\)-linear groups. The proof is a little out of the scope of this post, but we will see a simpler proof that finitely generated nilpotent groups are \(\mathbb{Z}\)-linear in a future post. This characterisation, however, can be seen to fail even when passing to fairly tame rings extending \(\mathbb{Z}\) -- there are finitely generated solvable groups linear over \(\mathbb{Z}[1/p]\), where \(p\) is a prime, that are not polycyclic.

The proof of the above gives somewhat of a strong an indication of the general structure of polycyclic groups. Indeed, every polycyclic group has a structure in some way resembling the upper triangular matrix group \(\operatorname{B}_n(\mathbb{Z})\) appearing above: it is virtually the extension of some finitely generated upper unitriangular matrix group by a finitely generated abelian group (acting like the diagonal).

We will show a purely group-theoretic sort of analogue to this fact, again due to Mal'cev. In fact (and perhaps surprisingly), the following is necessary when one (eventually) comes to prove linearity of polycyclic groups.

Theorem 9: Let \(G\) be a polycyclic group. Then \(G\) is has a finite index subgroup which is nilpotent-by-abelian.

Proof:  We again induct on the Hirsch length \(h(G)\) of \(G\). If \(h(G) = 0\), then \(G\) is finite and there is nothing to prove, so suppose \(n > 0\).  By Lemma 4, \(G\) contains a free abelian normal subgroup \(A\). Take such \(A \cong \mathbb{Z}^n\) of minimal rank in \(G\).

 We will first show that \(R = G/C_G(A)\) has a finite index abelian subgroup, and then use the induction hypothesis on \(G/A\) to construct a nilpotent-by-abelian subgroup of finite index.

The linear representation \(G \to \operatorname{Aut}(A)\) by conjugation has kernel \(C_G(A)\), and so induces an inclusion \(R \leqslant \operatorname{Aut}(A) \cong \operatorname{GL}_n(\mathbb{Z})\). Thus \(R\) is solvable and linear, and so has a finite index triangularisable normal subgroup \(S \leqslant_f R\) by Theorem 6.

It follows that the commutator subgroup \([S,S]\) of \(S\) is conjugate to an unitriangularisable subgroup of \(\operatorname{GL}_n(\mathbb{C})\) and so, after choosing an appropriate basis, preserves a standard flag in \(\mathbb{C}^n\) and hence also of \(V = \mathbb{Q}^n\). This means that the fixed subspace \(W \leq V\) under the action of \([S,S]\) is nontrivial. Moreover, \(W\) is a \(\mathbb{Q} R\)-submodule of \(V\) as \([S,S]\) is a normal subgroup of \(R\).

As \(A\) is of minimal rank in \(G\), every nontrivial \(\mathbb{Z} R\)-submodule of \(A\) is of finite index in \(A\). Thus \(V\) is a simple \(\mathbb{Q} R\)-module, so we must have \(W = V\). But no nontrivial subgroup of \(R\) fixes the whole of \(V\), so \([S,S]\) must be trivial.Therefore \(S \leqslant_f R\) is a finite index abelian subgroup of \(R\) as required.

Let \(H \lhd_f G\) be the finite index normal subgroup containing \(G_G(A)\) such that \(H/C_G(A)\) is abelian, so that \([H,H] \subseteq C_G(A)\). In other words, we have that \([A,[H,H]]\) is trivial.

 Also, \(h(G/A) < h(G)\) so by induction \(G/A\) has a finite index subgroup that is nilpotent-by-abelian, say \(K/A\). This means that the lower central series of \([K,K]/A\) terminates or, equivalently, that there is some \(m\) such that \(\gamma_m([K,K]) \subseteq A\). Now take \(L = H \cap K \lhd_f G\) and combine the above to obtain

    \[\gamma_{m+1}([L,L]) \subseteq [\gamma_m([K,K]),[H,H]] \subseteq [A,[H,H]] = 1\]

so that the lower central series of \([L,L]\) terminates. Thus \(L\) is a nilpotent-by-abelian finite index subgroup of \(G\) as required. $\blacksquare$