The 𝑅∞-property for nilpotent quotients of Baumslag–Solitar groups

Karel Dekimpe; Daciberg Lima Gonçalves

doi:10.1515/jgth-2018-0182

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The 𝑅_∞-property for nilpotent quotients of Baumslag–Solitar groups

Karel Dekimpe and Daciberg Lima Gonçalves

Published/Copyright: December 19, 2019

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From the journal Journal of Group Theory Volume 23 Issue 3

Abstract

A group G has the R∞-property if the number R⁢(φ) of twisted conjugacy classes is infinite for any automorphism φ of G. For such a group G, the R∞-nilpotency degree is the least integer c such that G/γc+1⁢(G) still has the R∞-property. In this paper, we determine the R∞-nilpotency degree of all Baumslag–Solitar groups.

1 Introduction

Any endomorphism φ of a group G determines an equivalence relation on G by setting x∼y⇔there exists⁢z∈G:x=z⁢y⁢φ⁢(z)-1. The equivalence classes of this relation are called Reidemeister classes or twisted conjugacy classes, and their number is denoted by R⁢(φ). We are most interested in this number when φ is an automorphism.

For information on the development, historical aspects and the relation of this concept with other topics in mathematics such as fixed-point theory, we refer the reader to the introduction of [3] and its references. An important concept in this context is that of groups having the R∞-property.

Definition 1.

A group G is said to have the R∞-property if, for every automorphism φ:G→G, the number R⁢(φ) is infinite.

A central problem is to decide which groups have the R∞-property. The study of this problem has been a quite active research topic in recent years. Several families of groups have been studied by many authors. A non-exhaustive list of references is [1, 2, 3, 4, 5, 6, 7, 8, 9, 13, 17, 15].

Of particular interest for this paper is the fact that, in [5], it was proved that the Baumslag–Solitar groups BS⁢(m,n) have the R∞-property except for m=n=1 (or m=n=-1, which is the same group). Recently, in [3], motivated by the results of [1], new examples of groups which have the R∞-property were obtained by looking at quotients of a group which has the R∞-property by the terms of the lower central series as well the derived central series. So it is natural to ask the same question for the groups BS⁢(m,n).

Related to this approach, we introduced in [3] the following notion.

Definition 2.

Let G be a group. The R∞-nilpotency degree of a group G is the least integer c such that G/γc+1⁢(G) has the R∞-property. If no such integer exists, then we say that G has R∞-nilpotency degree infinite.

In this work, we determine the R∞-nilpotency degree for all the Baumslag–Solitar groups BS⁢(m,n). The main results of this work are the following two theorems.

Theorem \ref{main1}.

Let m,n be integers with 0<m≤|n| and gcd⁡(m,n)=1. Let p denote the largest integer such that 2p∣2m+2. Then the R∞-nilpotency degree r of BS⁢(m,n) is given by the following conditions.

If n<0 and n≠-1, then r=2.
If n=-1 (so m=1), then r=∞.
If n=m (so n=m=1), then r=∞.
If n-m=1, then r=∞.
If n-m=2, then r=p+2.
If n-m≥3, then r=2.

Theorem \ref{main2}.

Let 0<m≤|n| with m≠n, and take d=gcd⁡(m,n). Let p denote the largest integer such that 2p∣2md+2. Then the R∞-nilpotency degree r of BS⁢(m,n) is given by the following conditions.

If n<0 and n≠-m, then r=2.
If n=-m, then r=∞.
If n=m, then r=∞.
If n-m=d, then r=∞.
If n-m=2⁢d, then 2≤r≤p+2.
If n-m≥3⁢d, then r=2.

At this point, we would also like to mention one interesting family of groups which naturally extends the class of Baumslag–Solitar groups, namely the family of GBS groups, the generalized Baumslag–Solitar groups. In [14], the following strong result about the Reidemeister number of a homomorphism of such groups is proved.

Proposition ([14, Proposition 2.7]).

Let α:G→G be an endomorphism of a non-elementary GBS group. If one of the following conditions holds, then R⁢(α) is infinite.

α is surjective.
α is injective, and G is not unimodular.
G=BS⁢(m,n) with |m|≠|n|, and the image of α is not cyclic.

Recently, other families of groups, which also naturally extend the class of Baumslag–Solitar groups, were considered. One generalization goes as follows. The class of GBS groups coincides with the class of fundamental groups of graphs all of whose vertex and edge groups are infinite cyclic. So one can generalize this to the class of fundamental groups of graphs where the vertex and edge groups are virtually infinite cyclic. In [11], it was shown by Taback and Wong that any group which is quasi-isometric to a group in this family has the R∞-property.

Another family was considered by Taback and Whyte in [10], generalizing the solvable Baumslag–Solitar groups BS⁢(1,n) to another class of groups that are also solvable. These are split extensions fitting into a short exact sequence 1→ℤ⁢[1n]→Γ→ℤk→1. For this second family, Taback and Wong showed in [12] that any group quasi-isometric to one of these group has the R∞-property.

As a generalization of the results of this paper, it would be natural to study the R∞-nilpotency degree for the families of groups above.

This work is divided into three sections besides the introduction. In Section 2, we provide some preliminary results about the description of the terms of the lower central series and the corresponding quotients of BS⁢(m,n) when the integers (m,n) are coprime. In Section 3, we construct certain specific nilpotent groups in a format which is convenient for our study. Then we identify these groups with the ones that we want to study, namely the quotients BS⁢(m,n)/γc+1⁢(BS⁢(m,n)). In Section 4, we then show the main result for the BS⁢(m,n) groups, where m and n are coprime. Finally, in Section 5, we provide a proof for the remaining cases.

2 Baumslag–Solitar groups

Let BS⁢(m,n)=〈a,b∣a-1⁢bm⁢a=bn〉 for m,n integers. It suffices to consider 1≤m≤|n|. We will use the notation [x,y]=x-1⁢y-1⁢x⁢y.

Lemma 1.

Consider a Baumslag–Solitar group BS⁢(m,n). For all positive integers k, we have b(m-n)k∈γk+1⁢(BS⁢(m,n)).

Proof.

Since a-1⁢b-m⁢a=b-n, we have bm-n=[a,bm]∈γ2⁢(BS⁢(m,n)), which proves the lemma for k=1.

Now, we assume that k≥1 and that b(m-n)k∈γk+1⁢(BS⁢(m,n)). Then we find

b-m⁢(m-n)k∈γk+1⁢(BS⁢(m,n))⟹a-1⁢b-m⁢(m-n)k⁢a⁢bm⁢(m-n)k∈γk+2⁢(BS⁢(m,n))⟹(a-1⁢bm⁢a)-(m-n)k⁢bm⁢(m-n)k∈γk+2⁢(BS⁢(m,n))⟹b-n⁢(m-n)k⁢bm⁢(m-n)k=b(m-n)k+1∈γk+2⁢(BS⁢(m,n)),

which proves the lemma, by induction. ∎

As we will be dealing with nilpotent quotients of the Baumslag–Solitar groups, we introduce the notation

BSc⁢(m,n)=BS⁢(m,n)γc+1⁢(BS⁢(m,n)).

For a nilpotent group N, we use τ⁢N to indicate its torsion subgroup.

Lemma 2.

Let m≠n. For all positive integers c, the nilpotent group BSc⁢(m,n) has Hirsch length 1, and if we denote by b¯ the natural projection of b in BSc⁢(m,n), we have τ⁢BSc⁢(m,n)=〈b¯,γ2⁢(BSc⁢(m,n))〉.

Proof.

We first consider the case c=1. Note that

BS1⁢(m,n)=〈a¯,b¯∣[a¯,b¯]=1,b¯m-n=1〉≅ℤ⊕ℤ|m-n|.

So τ⁢BS1⁢(m,n)=〈b¯〉.

Now, let c>1. From the case c=1, it follows that

τ⁢BSc⁢(m,n)⊆〈b¯,γ2⁢(BSc⁢(m,n))〉,

and hence it suffices to show that γ2⁢(BSc⁢(m,n)) is a torsion group. To obtain this result, we prove by induction on i≥2 that

γi⁢(BSc⁢(m,n))/γi+1⁢(BSc⁢(m,n))=γi⁢(BS⁢(m,n))/γi+1⁢(BS⁢(m,n))

is finite.

The group γ2⁢(BS⁢(m,n))/γ3⁢(BS⁢(m,n)) is generated by [a,b]⁢γ3⁢(BS⁢(m,n)). By the previous lemma, we know that bm-n∈γ2⁢(BS⁢(m,n)), using this we find

[a,b]m-n⁢γ3⁢(BS⁢(m,n))=[a,bm-n]⁢γ3⁢(BS⁢(m,n))=1⁢γ3⁢(BS⁢(m,n)),

and so [a,b]⁢γ3⁢(BS⁢(m,n)) is of finite order (≤|m-n|) in

γ2⁢(BS⁢(m,n))/γ3⁢(BS⁢(m,n)).

Now, assume that γi⁢(BS⁢(m,n))/γi+1⁢(BS⁢(m,n)) is finite. The group

γi+1⁢(BS⁢(m,n))/γi+2⁢(BS⁢(m,n))

is generated by all elements of the form [x,y]⁢γi+2⁢(BS⁢(m,n)) for x∈BS⁢(m,n) and y∈γi⁢(BS⁢(m,n)). By our assumption, there is some k>0 such that we have yk∈γi+1⁢(BS⁢(m,n)). As before, it then follows that

[x,y]k⁢γi+2⁢(BS⁢(m,n))=[x,yk]⁢γi+2⁢(BS⁢(m,n))=1⁢γi+2⁢(BS⁢(m,n)),

from which we deduce that γi+1⁢(BS⁢(m,n))/γi+2⁢(BS⁢(m,n)) is finite.

The fact that BSc⁢(m,n) has Hirsch length 1 follows from the fact that

BSc⁢(m,n)/γ2⁢(BSc⁢(m,n))≅BS1⁢(m,n)

has Hirsch length 1 and γ2⁢(BSc⁢(m,n)) has Hirsch length 0. ∎

In this paper, the situation where gcd⁡(m,n)=1 will play a rather crucial role. For these groups, the structure of BSc⁢(m,n) is easier to understand than in the general case. For example, we have the following lemma.

Lemma 3.

Suppose that gcd⁡(m,n)=1 and m≠n. For any c>1 and k>1, we have γk⁢(BSc⁢(m,n))=〈b¯(m-n)k-1〉. Again, b¯ denotes the projection of b in BSc⁢(m,n).

Proof.

For sake of simplicity, we will write Γi instead of γi⁢(BSc⁢(m,n)) in the rest of this proof. We will prove by induction on k≥2 that b¯(m-n)k-1⁢Γk+1 generates Γk/Γk+1.

For k=2, we have that [a¯,b¯]⁢Γ3 generates Γ2/Γ3, and from Lemma 1, we know that [a¯,b¯]m-n∈Γ3; hence the order of [a¯,b¯]⁢Γ3 in Γ2/Γ3 is a divisor of m-n. As gcd⁡(m,n)=1, also gcd⁡(m,m-n)=1, and therefore also [a¯,b¯]m⁢Γ3 is a generator of Γ2/Γ3. Now, [a¯,b¯]m⁢Γ3=[a¯,b¯m]⁢Γ3=b¯m-n⁢Γ3, from which we find that b¯m-n⁢Γ3 generates Γ2/Γ3.

Now, we assume that k>2 and that Γk-1/Γk is generated by b¯(m-n)k-2⁢Γk. The next quotient Γk/Γk+1 is then generated by [a¯,b¯(m-n)k-2]⁢Γk+1. Again, by Lemma 1, we have

[a¯,b¯(m-n)k-2]m-n⁢Γk+1=[a¯,b¯(m-n)k-1]⁢Γk+1=1⁢Γk+1,

and so the order of the generator [a¯,b¯(m-n)k-2]⁢Γk+1 divides m-n. As before, it follows that also [a¯,b¯(m-n)k-2]m⁢Γk+1 generates Γk/Γk+1. In BS⁢(m,n), we have [a,bk⁢m]=bk⁢(m-n), which we now use to obtain

[a¯,b¯(m-n)k-2]m⁢Γk+1=[a¯,b¯m⁢(m-n)k-2]⁢Γk+1=b¯(m-n)k-1⁢Γk+1,

which finishes the proof. ∎

Corollary 4.

Suppose that gcd⁡(m,n)=1 and m≠n. Then, for all c≥1, we have τ⁢BSc⁢(m,n)=〈b¯〉.

3 Some nilpotent quotients of Baumslag–Solitar groups

For the rest of this section, we assume that m≠n. For any positive integer c, we will construct a nilpotent group Gc⁢(m,n) of class ≤c which can be seen as a quotient of BS⁢(m,n). To construct this group, we fix m, n and c and consider the morphism φ:ℤc→ℤc, which is represented by the matrix

(3.1)(n-m00⋯00-mn-m0⋯000-mn-m⋯00⋮⋮⋮⋱⋮⋮000⋯n-m0000⋯-mn-m).

Here we use the convention that elements of ℤc are written as columns, so also in the matrix above, the image of the i-th standard generator of ℤc is given by the i-th column of that matrix. We now consider the abelian group

Ac⁢(m,n)=ℤcIm⁡φ.

So Ac⁢(m,n) is a finite group of order |n-m|c.

We consider also the morphism ψ:ℤc→ℤc, which is represented by

M=(100⋯00110⋯00011⋯00⋮⋮⋮⋱⋮⋮000⋯10000⋯11).

We have φ⁢ψ=ψ⁢φ, and therefore ψ induces an automorphism of Ac⁢(m,n), which we will also denote by the symbol ψ.

Now, we are ready to define the group Gc⁢(m,n), which is given as a semi-direct product Gc⁢(m,n)=Ac⁢(m,n)⋊〈t〉, where 〈t〉 is the infinite cyclic group and where the semi-direct product structure is given by the requirement that, for all a∈Ac⁢(m,n), we have t-1⁢a⁢t=ψ⁢(a).

For any z∈ℤc, let z¯=z+Im⁡φ denote its natural projection in Ac⁢(m,n). We use e1,e2,…,ec to denote the standard generators of ℤc, so ei is the column vector having a 1 on the i-th spot and 0’s on all other positions. Obviously, we have that e1¯,e2¯,…,ec¯ generate Ac⁢(m,n). For sake of simplicity, sometimes, we will write G instead of Gc⁢(m,n).

Remark.

It is easy to see that from the fact that

e2¯=t-1⁢e1¯⁢t⁢e1¯-1,e3¯=t-1⁢e2¯⁢t⁢e2¯-1,…

follows

γ2⁢(G)⊆〈e2¯,e3¯,…,ec¯〉

γ3⁢(G)⊆〈e3¯,e4¯,…,ec¯〉

⋮

γc⁢(G)⊆〈ec¯〉

γc+1⁢(G)=1.

Hence Gc⁢(m,n) is nilpotent of class ≤c.

Lemma 1.

There is a surjective morphism of groups f:BS⁢(n,m)→Gc⁢(m,n) which is determined by f⁢(a)=t and f⁢(b)=e1¯.

Proof.

In order for f to be a morphism, we need to check that f preserves the defining relation of BS⁢(n,m), that is, the relation t-1⁢e1¯m⁢t=e1¯n should hold. This follows from the computation

t-1⁢e1¯m⁢t=ψ⁢(e1¯m)=e1¯m⁢e2¯m=e1¯n⁢(e1¯m-n⁢e2¯m)=e1¯n⁢φ⁢(-e1)¯=e1¯n.

To prove that f is a surjective map, it is enough to show that e1¯ and t generate Gc⁢(m,n). This follows from the fact that

e2¯=t-1⁢e1¯⁢t⁢e1¯-1,e3¯=t-1⁢e2¯⁢t⁢e2¯-1,…∎

As Gc⁢(m,n) is nilpotent of class ≤c, f induces a surjective morphism

BSc⁢(m,n)=BS⁢(m,n)γc+1⁢(BS⁢(m,n))→Gc⁢(m,n).

For Gc⁢(m,n), we have τ⁢G=A, and so [τ⁢G,τ⁢G]=1.

Proposition 2.

The morphism f:BS⁢(m,n)→Gc⁢(m,n) induces an isomorphism

μ:BSc⁢(m,n)[τ⁢BSc⁢(m,n),τ⁢BSc⁢(m,n)]→Gc⁢(m,n).

Proof.

As already explained, f induces a morphism ν:BSc⁢(m,n)→Gc⁢(m,n). Of course, ν⁢(τ⁢BSc⁢(m,n))⊆τ⁢G, and so

ν⁢[τ⁢BSc⁢(m,n),τ⁢BSc⁢(m,n)]⊆[τ⁢G,τ⁢G]=1.

Therefore, there is an induced morphism

μ:BSc⁢(m,n)[τ⁢BSc⁢(m,n),τ⁢BSc⁢(m,n)]→Gc⁢(m,n).

As f is surjective, we know that μ is surjective too. In Lemma 2, we showed that BSc⁢(m,n) has Hirsch length 1. Then also the quotient

BSc⁢(m,n)/[τ⁢BSc⁢(m,n),τ⁢BSc⁢(m,n)]

has Hirsch length 1 since we take the quotient by a finite subgroup. As also, by construction, Gc⁢(m,n) has Hirsch length 1 and μ is surjective, we must have that the kernel of μ has Hirsch length 0, i.e., the kernel of μ has to be finite. For sake of simplicity, we introduce the notation

H=BSc⁢(m,n)[τ⁢BSc⁢(m,n),τ⁢BSc⁢(m,n)].

We already know, by Lemma 2, that τ⁢H is generated by b¯ and γ2⁢(H). (Here b¯ denotes the image of b in H.) As μ is surjective and has finite kernel (so Ker⁡(μ)⊆τ⁢H), we know that μ⁢(τ⁢H)=τ⁢Gc⁢(m,n). Therefore, in order to prove that μ is injective, it is enough to show that #⁢τ⁢H≤#⁢τ⁢Gc⁢(m,n)=|m-n|c.

To be able to find a bound on #⁢τ⁢H, we look at the quotients γi⁢(H)/γi+1⁢(H).

γ2⁢(H)/γ3⁢(H) is generated by [a¯,b¯]⁢γ3⁢(H).
Then γ3⁢(H)/γ4⁢(H) is generated by [a¯,[a¯,b¯]]⁢γ4⁢(H) and [b¯,[a¯,b¯]]⁢γ4⁢(H). In H, however, we have [b¯,[a¯,b¯]]=1 (since we divide out [τ⁢BSc⁢(m,n),τ⁢BSc⁢(m,n)]). So γ3⁢(H)/γ4⁢(H) is generated by [a¯,[a¯,b¯]]⁢γ4⁢(H).
Continuing by induction, we find that γi⁢(H)/γi+1⁢(H) is a cyclic group generated by
[a¯,[a¯,[a¯,…,[a¯,b¯]]]]⁢γi+1⁢(H) (with⁢i-1⁢times⁢a¯).

We already know that #⁢τ⁢H/γ2⁢(H)=|m-n| (so b¯m-n⁢γ2⁢(H)=1⁢γ2⁢(H); see the proof of Lemma 2). Let c1=b¯, and for i>1, we let ci=[a¯,[a¯,[a¯,…,[a¯,b¯]]]] (with i-1 times a¯). Then γi⁢(H)/γi+1⁢(H) is generated by ci⁢γi+1⁢(H) for i>1, and τ⁢(H)/γ2⁢(H) is generated by c1⁢γ2⁢(H).

We now show by induction on i that cim-n⁢γi+1⁢(H)=1⁢γi+1⁢(H) and hence #⁢γi⁢(H)/γi+1⁢(H)≤|m-n|. We already obtained the case i=1. Now, assume the result holds for ci-1 (with i>1). Then ci=[a¯,ci-1], and we have

cim-n⁢γi+1⁢(H)=[a¯,ci-1]m-n⁢γi+1⁢(H)=[a¯,ci-1m-n]⁢γi+1⁢(H)=1⁢γi+1⁢(H).

As a conclusion, we find that

#⁢τ⁢(H)=#⁢τ⁢Hγ2⁢(H)×#⁢γ2⁢(H)γ3⁢(H)×⋯×#⁢γc⁢(H)γc+1⁢(H)≤|m-n|c=#⁢τ⁢Gc⁢(m,n).

We can conclude that μ is injective (and hence an isomorphism). ∎

Corollary 3.

If gcd⁡(m,n)=1, the morphism f:BS⁢(m,n)→Gc⁢(m,n) induces an isomorphism BSc⁢(m,n)≅Gc⁢(m,n).

Proof.

It follows from Corollary 4 that, in this case,

[τ⁢BSc⁢(m,n),τ⁢BSc⁢(m,n)]=1.∎

4 The case where gcd⁡(m,n)=1

In the next lemma, we will make use of Smith normal form, details about which can be found e.g. in [16]. We remind the reader that Smith normal form is a useful tool in dealing with quotients of free modules (over PIDs).

Lemma 1.

Let a,b∈Z with gcd⁡(a,b)=1. Then the Smith normal form of the n×n-matrix

An=(a000⋯00ba00⋯000ba0⋯0000ba⋯00⋮⋮⋮⋮⋱⋮⋮0000⋯a00000⋯ba) 𝑖𝑠 (100⋯00010⋯00001⋯00⋮⋮⋮⋱⋮⋮000⋯10000⋯0an).

Proof.

We will prove a slightly more general version of this lemma and consider for any positive integer k the matrix An⁢(k), which is the same matrix as An, except that the first entry of An⁢(k) (so on the first row and the first column) is ak instead of a. So An=An⁢(1). We will now show, by induction on n, that the Smith normal form of An⁢(k) is

(10⋯0001⋯00⋮⋮⋱⋮⋮00⋯1000⋯0an-1+k).

When n=1, there is nothing to show, so we assume that n>1. As gcd⁡(a,b)=1, there exist integers α,β∈ℤ such that α⁢ak+β⁢b=1. Now, consider

P=(αβ-bak)∈GL⁡(2,ℤ) and Q=(1-a⁢β01)∈GL⁡(2,ℤ).

It is now easy to compute that (with In-2 the (n-2)×(n-2) identity matrix)

(P00In-2)⁢An⁢(k)⁢(Q00In-2)=(100An-1⁢(k+1)).

By induction, we know the Smith normal form of An-1⁢(k+1) and hence also of

(100An-1⁢(k+1)),

which is then exactly

(10⋯0001⋯00⋮⋮⋱⋮⋮00⋯1000⋯0an-1+k)

as claimed. ∎

Corollary 2.

Let m,n be two integers with gcd⁡(m,n)=1 and m≠n. Then

Ac⁢(m,n)≅ℤ|m-n|c.

Proof.

Recall that Ac⁢(m,n)=ℤcIm⁡φ, where φ:ℤc→ℤc is represented by the matrix (3.1). The lemma above shows that the Smith normal form of this matrix is

(10⋯0001⋯00⋮⋮⋱⋮⋮00⋯1000⋯0|n-m|c),

from which the result follows. ∎

For the rest of this section, we will assume that m≠n and that gcd⁡(m,n)=1 (so BSc⁢(m,n)≅Gc⁢(m,n); see Corollary 3). Moreover, we will use s=e1¯ to denote the canonical projection of the first standard generator of ℤc in the group Ac⁢(m,n). As Ac⁢(m,n) is a cyclic group and Ac⁢(m,n) is generated as a 〈t〉-module by s, it follows that s is also a generator of Ac⁢(m,n) as a cyclic group. So

〈s〉=Ac⁢(m,n)≅ℤ|n-m|c and Gc⁢(m,n)=〈s〉⋊〈t〉.

Proposition 3.

Let m≠n and gcd⁡(m,n)=1. Then Gc⁢(m,n)=〈s〉⋊〈t〉, with t-1⁢s⁢t=sν, where ν∈Z is an integer satisfying

gcd⁡(ν,n-m)=1,
ν⁢m≡nmod|n-m|c.

Proof.

We already explained that Gc⁢(m,n)=〈s〉⋊〈t〉. As s is a generator of the cyclic group of order |n-m|c, we must have that also t-1⁢s⁢t is a generator of 〈s〉, which implies that t-1⁢s⁢t=sν for some integer ν with gcd⁡(ν,n-m)=1.

As we already saw in the proof of Lemma 1 (recall s=e1¯), we also have t-1⁢sm⁢t=sn. As t-1⁢sm⁢t=sν⁢m, it follows that sν⁢m=sn; hence

ν⁢m≡nmod|n-m|c.∎

Let φ:Gc⁢(m,n)→Gc⁢(m,n) be an automorphism. Since 〈s〉=τ⁢Gc⁢(m,n), φ induces an automorphism

φ¯:Gc⁢(m,n)/〈s〉=〈t〉≅ℤ→Gc⁢(m,n)/〈s〉=〈t〉≅ℤ.

So φ¯⁢(t)=t±1. The following lemma is easy to check.

Lemma 4.

With the notation above, we have R⁢(φ)<∞⇔φ¯⁢(t)=t-1.

It follows that Gc⁢(m,n) does not have the R∞-property if and only if there exists an automorphism φ of Gc⁢(m,n) such that φ¯⁢(t)=t-1. We are now ready to prove the main theorem of this section which gives us the R∞-nilpotency degree of any Baumslag–Solitar group which is determined by coprime parameters m and n.

Theorem 5.

Let m,n be integers with 0<m≤|n| and gcd⁡(m,n)=1. Let p denote the largest integer such that 2p∣2m+2. Then the R∞-nilpotency degree r of BS⁢(m,n) is given by the following conditions.

If n<-1, then r=2.
If n=-1 (so m=1), then r=∞.
If n=m (so n=m=1), then r=∞.
If n-m=1, then r=∞.
If n-m=2, then r=p+2.
If n-m≥3, then r=2.

Proof.

Let m and n be as in the statement of the theorem.

Let m=n. Then the fact that gcd⁡(m,n)=1 and m>0 implies that m=n=1. We have that BS⁢(1,1)=ℤ2=BSc⁢(m,n) (for all c) does not have the R∞-property, from which it follows that, in this case, the R∞-nilpotency index is ∞.

So, from now onwards, we assume that m≠n. We have to examine for which c the group Gc⁢(m,n) has the R∞-property. So we have to investigate when Gc⁢(m,n) admits an automorphism φ with φ¯⁢(t)=t-1 (Lemma 4). Such a morphism φ satisfies

(4.1)φ⁢(s)=sμ and φ⁢(t)=sβ⁢t-1 for some⁢μ,β∈ℤ.

In fact, given μ,β∈ℤ, the expressions of (4.1) above determine an endomorphism of Gc⁢(m,n) if and only if the relation t-1⁢s⁢t=sν (where ν is as in Proposition 3) is preserved, i.e., it must hold that

φ⁢(t)-1⁢φ⁢(s)⁢φ⁢(t)=φ⁢(s)ν

⇕

t⁢s-β⁢sμ⁢sβ⁢t-1=sμ⁢ν

⇕

sμ=t-1⁢sμ⁢ν⁢t=sμ⁢ν2.

Moreover, such a φ is an automorphism if sμ is a generator of 〈s〉, i.e. when gcd⁡(μ,|n-m|)=1. In this case, the last condition is equivalent to

ν2≡1mod|n-m|c.

Moreover, as we also have gcd⁡(m,|n-m|)=1, this is also equivalent to the requirement that

ν2⁢m2≡m2mod|n-m|c.

Finally, using Proposition 3, which says that ν⁢m≡nmod|n-m|c, we find that

Gc⁢(m,n)⁢does not have the⁢R∞⁢-property

⇕

n2≡m2mod|n-m|c

⇕

n+m≡0mod|n-m|c-1

So, from now on, we have to examine when the condition

(4.2)n+m≡0mod|n-m|c-1

is satisfied.

When c=1, the equation is always satisfied (reflecting the fact that finitely generated abelian groups do not have the R∞-property). So, from now onwards, we consider the case c>1.

Let n=-m. In this case, n=-1 and m=1 since gcd⁡(m,n)=1; then equation (4.2) is always satisfied. This shows that BS⁢(-1,1) (which is the fundamental group of the Klein bottle) has an infinite R∞-nilpotency degree (although BS⁢(-1,1) does have the R∞-property [9, Theorem 2.2]).

Let n<-1. Then |n-m|c-1=(|n|+m)c-1>|n+m|≠0. This implies that equation (4.2) is never satisfied. This means that, in this case, the R∞-nilpotency degree of BS⁢(n,m) is 2.

Now, we consider the case of positive n, where we already treated the case when n=m. So we have n=m+k for k>0. Moreover, as gcd⁡(n,m)=1, we also have gcd⁡(k,m)=1. If equation (4.2) is satisfied, then |n-m|=k divides n+m=2⁢m+k, so k∣2m, and as gcd⁡(k,m)=1, we must have k∣2, so k=1 or k=2.

Let n=m+k for k≥3. From the considerations of the paragraph above, we have that the R∞-nilpotency degree of BS⁢(m+k,k) is 2.

Let n=m+1. In this case, equation (4.2) is again satisfied for all c, and hence the R∞-nilpotency degree of BS⁢(m+1,m) is ∞.

Let n=m+2. Then equation (4.2) is of the form 2⁢m+2≡0mod2c-1. This equation is satisfied exactly when c≤p+1. It follows that the R∞-degree of BS⁢(m+2,m) (where m is odd) is p+2. This finishes the proof. ∎

5 The case where gcd⁡(m,n)≠1

Lemma 1.

Let m,n be non-zero integers with m≠n. If d=gcd⁡(m,n), then

Ac⁢(m,n)≅ℤdc⊕ℤ|n-md|c.

Proof.

Note that the matrix (3.1) is

d⁢(n-md00⋯00-mdn-md0⋯000-mdn-md⋯00⋮⋮⋮⋱⋮⋮000⋯n-mm0000⋯-mdn-md).

with gcd⁡(md,n-md)=1. It follows that the Smith normal form of (3.1) is

(d0⋯000d⋯00⋮⋮⋱⋮⋮00⋯d000⋯0d⁢|n-md|c),

from which the result follows. ∎

It follows that d⁢Ac⁢(m,n)≅ℤ|n-md|c is a cyclic subgroup of Ac⁢(m,n), and since this subgroup is invariant under the action of 〈t〉, the semi-direct product (d⁢Ac⁢(m,n))⋊〈t〉 is a subgroup of Gc⁢(m,n).

Lemma 2.

Let 0<m≤|n| with m≠n, and take d=gcd⁡(m,n). Then we have that (d⁢Ac⁢(m,n))⋊〈t〉 is a subgroup of Gc⁢(m,n) and

(d⁢Ac⁢(m,n))⋊〈t〉≅Gc⁢(md,nd).

Proof.

The fact that (d⁢Ac⁢(m,n))⋊〈t〉 is a subgroup of Gc⁢(m,n) was already discussed before the statement of the lemma. Let φ′:ℤc→ℤc be the morphism represented by the matrix

(n-md00⋯00-mdn-md0⋯000-mdn-md⋯00⋮⋮⋮⋱⋮⋮000⋯n-mm0000⋯-mdn-md);

then φ=d⁢φ′. We have

d⁢Ac⁢(m,n)=d⁢ℤcIm⁡φ=(d⁢ℤ)cd⁢Im⁡φ′≅ℤcIm⁡φ′=Ac⁢(md,nd).

It is now easy to see that, under the identification d⁢Ac⁢(m,n)≅Ac⁢(md,nd), the action of t is still the same as what we had before, and so

(d⁢Ac⁢(m,n))⋊〈t〉≅Gc⁢(md,nd).∎

Lemma 3.

Let 0<m≤|n| with m≠n, and take d=gcd⁡(m,n). If Gc⁢(md,nd) has property R∞, then also Gc⁢(m,n) has property R∞.

Proof.

First let us remark that, for any α∈Ac⁢(m,n), there is an automorphism ψα of Gc⁢(m,n)=Ac⁢(m,n)⋊〈t〉 with

ψα⁢(a)=a for all⁢a∈Ac⁢(m,n) and ψα⁢(t)=t⁢α.

Now, suppose that ψ∈Aut⁡(Gc⁢(m,n)) is an automorphism with R⁢(ψ)<∞. This means that ψ⁢(t)=α⁢t-1 for some α∈Ac⁢(m,n). After composing ψ with ψα, we may assume that ψ⁢(t)=t-1. Since we also have ψ⁢(d⁢Ac⁢(m,n))=d⁢Ac⁢(m,n) (since Ac⁢(m,n) is a characteristic subgroup of Gc⁢(m,n)), we have that ψ restricts to an automorphism of (d⁢Ac⁢(m,n))⋊〈t〉≅Gc⁢(md,nd) with finite Reidemeister number. This shows that if Gc⁢(m,n) does not have property R∞, then also Gc⁢(md,nd) does not have this property. ∎

The main result of this section is the following result.

Theorem 4.

Let 0<m≤|n|, and take d=gcd⁡(m,n). Let p denote the largest integer such that 2p∣2md+2. Then the R∞-nilpotency degree r of BS⁢(m,n) is given by the following conditions.

If n<0 and n≠-m, then r=2.
If n=-m, then r=∞.
If n=m, then r=∞.
If n-m=d, then r=∞.
If n-m=2⁢d, then 2≤r≤p+2.
If n-m≥3⁢d, then r=2.

Remark.

As d=gcd⁡(m,n), the difference n-m is a multiple of d, so the theorem above does treat all possible cases.

Proof.

We will first deal with a few special cases and then treat the general case.

Let n=m. In this case, there is an automorphism ψ of BS⁢(m,m) mapping a to a-1 and b to b-1. This automorphism induces minus the identity map on BS1⁢(m,m)=γ1⁢(BS⁢(m,m))γ2⁢(BS⁢(m,m))≅ℤ2. It now follows that the induced map ψ¯ on any quotient BSc⁢(m,m) has -1 as an eigenvalue, and hence R⁢(ψ¯)<∞ (see [3]). It follows that the R∞-nilpotency index of BS⁢(m,m) is ∞.

Let n=-m. Now, consider the automorphism ψ of BS⁢(m,-m) mapping a to a-1 and b to b. Then ψ induces a map on BS1⁢(m,-m)≅ℤ⊕ℤ2⁢m, which is minus the identity on the ℤ-factor and the identity on the ℤ2⁢m factor. The same argument as in the previous case now allows us to conclude that the R∞-nilpotency index of BS⁢(m,-m) is ∞.

Let n=m+d. So m=k⁢d and n=(k+1)⁢d for some positive integers k and d. We claim that bd∈γc⁢(BS⁢(k⁢d,(k+1)⁢d)) for all c≥1. This claim is certainly correct for c=1. Now, fix c≥1, and assume that

bd∈γc⁢(BS⁢(k⁢d,(k+1)⁢d)).

Then also bk⁢d∈γc⁢(BS⁢(k⁢d,(k+1)⁢d)), and hence

[a,bk⁢d]∈γc+1⁢(BS⁢(k⁢d,(k+1)⁢d)).

But as a-1⁢bk⁢d⁢a=b(k+1)⁢d, we have

[a,bk⁢d]=a-1⁢b-k⁢d⁢a⁢bk⁢d=b-d∈γc+1⁢(BS⁢(k⁢d,(k+1)⁢d)).

By induction, this finishes the proof of the claim.

It follows that the group BS⁢(k⁢d,(k+1)⁢d) and the group

C⁢(d)=〈a,b∣a-1⁢bk⁢d⁢a=b(k+1)⁢d,bd〉=〈a,b∣bd〉

have isomorphic nilpotent quotients, i.e.,

BS⁢(k⁢d,(k+1)⁢d)γc+1⁢(BS⁢(k⁢d,(k+1)⁢d))≅C⁢(d)γc+1⁢(C⁢(d)).

Now, it is easy to see that C⁢(d) has an automorphism ψ mapping a to a-1 and b to b such that ψ induces an automorphism ψ¯ on C⁢(d)γc+1⁢(C⁢(d)) with finite Reidemeister number. It follows that the R∞-nilpotency degree of BS⁢(k⁢d,(k+1)⁢d) is infinite.

All the other cases. As a finitely generated abelian group never has the R∞-property, we have r≥2. Now, assume that ψ is an automorphism of BSc⁢(m,n) with R⁢(ψ)<∞. Then ψ induces an automorphism ψ¯ of Gc⁢(m,n) (since we divide out a characteristic subgroup to go from BSc⁢(m,n) to Gc⁢(m,n) by Proposition 2) with R⁢(ψ)<∞. It follows that the R∞-nilpotency degree is bounded above by the smallest c for which Gc⁢(m,n) has property R∞. In turn, this number is bounded above by the smallest c such that Gc⁢(md,nd) has the R∞-property. This is exactly what we determined in the proof of Theorem 5, which finishes the proof. ∎

Communicated by Dessislava H. Kochloukova

Funding source: Fundação de Amparo à Pesquisa do Estado de São Paulo

Award Identifier / Grant number: 2016/24707-4

Funding statement: K. Dekimpe is supported by long term structural funding – Methusalem grant of the Flemish Government. D. L. Gonçalves is partially supported by Projeto Temático Topologia Algébrica, Geométrica e Diferencial FAPESP no. 2016/24707-4.

Acknowledgements

We would like to thank Peter Wong for drawing our attention to the recent results on the R∞-property for several families of groups generalizing the class of Baumslag–Solitar groups.

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Received: 2018-09-09

Revised: 2019-11-18

Published Online: 2019-12-19

Published in Print: 2020-05-01

Articles in the same Issue

https://doi.org/10.1515/jgth-2018-0182