In the previous section, we have shown that all of the inclusions from the diagrams in Figs. 1, 2 and 3 hold. In this section, we show that every non-empty area in the diagram does indeed contain a group.
Correctness of Fig. <xref rid="Fig1" ref-type="fig">1</xref>We will exhibit the following examples: A group with commuting conjugates that does not have commuting cyclic conjugates (Proposition 3.6).
A group with commuting cyclic conjugates that does not have commuting Z/n-conjugates, nor commuting Z-conjugates (Proposition 3.3).
A group with commuting Z/n-conjugates that does not have commuting Z-conjugates (Proposition 3.3).
A group with commuting Z-conjugates that does not have commuting Z/n-conjugates for any n (Proposition 3.5).
A group which has commuting Z-conjugates and commuting Z/n-conjugates (Example 3.1).
The last item is the most straightforward:
Example 3.1(c. Z c. and c. Z/n c.) Consider the group Homeoc(R2). It has commuting Z-conjugates [11, Corollary 5.19], as well as commuting Z/2-conjugates [11, Corollary 5.11 and its proof]. It is straightforward to adapt the proof to show that it has commuting Z/n-conjugates for every n≥2.
We proceed with a general construction that allows to treat the second and third items of the previous list, and moreover shows that the parameter n in the definition of commuting Z/n-conjugates does indeed make a difference.
We fix a sequence n=(n1,n2,…), where each ni∈Z≥2. Set Γ0 to be an arbitrary finitely generated non-abelian group, and define by induction Γi+1:=Γi≀Z/ni+1. Recall that this denotes the standard restricted wreath product ⊕ni+1Γi⋊Z/ni+1. The standard embeddings Γi→Γi+1 make this into a directed system of groups, and we denote by Γn their directed union.
Lemma 3.2For every prime p, the group Γn has commuting Z/p-conjugates if and only if p divides ni for infinitely many indices i. In all cases, Γn has commuting cyclic conjugates.
ProofFirst, suppose that p divides ni for infinitely many indices i. Let H≤Γ be a finitely generated subgroup. Then H≤Γj for some j. By hypothesis there exists i>j j$$\end{document}]]> such that p divides ni, say ni=kp for some 1≤k<ni. We consider Γi=Γi-1≀Z/ni, and let t denote the generator of Z/ni. We claim that tk satisfies the conditions in the definition of commuting Z/p-conjugates for H. First, for all 1≤q<p we have [H,tkqH]≤[Γi-1,tkqΓi-1]=1, since 1≤kq<kp=ni. Secondly, tkp=tni=1, and so clearly [H,tkp]=1. This shows that Γn has commuting Z/p-conjugates.
Conversely, suppose that Γn has commuting Z/p-conjugates. We will show that, for all i, there exists j>i i$$\end{document}]]> such that p divides nj. Since Γi is finitely generated, there exists t∈Γn that satisfies the conditions in the definition of commuting Z/p-conjugates for Γi. Let j be such that t∈Γj; we pick t so that j is minimal. Since Γi is not abelian, t cannot belong to Γi, otherwise [Γi,tΓi]=[Γi,Γi]≠1. This shows that j>i i$$\end{document}]]>. We claim that p divides nj.
We consider the group Γj=Γj-1≀Z/nj, and notice that Γi≤Γj-1. The element t can be written uniquely as t=skb, where b∈Γj-1nj, s is the generator of Z/nj and 0≤k<nj. If k=0, then the action of t=b by conjugacy on Γj-1 is the same as the action of the 0-th component of b, that we call t′. Since Γi≤Γj-1, the element t′ satisfies the conditions in the definition of commuting Z/p-conjugates for Γi, and t′∈Γj-1, which contradicts the choice of t with minimal j. It follows that k≠0. Now tp can be written uniquely as skpb′, where again b′∈Γj-1nj. If skp≠1, then tpΓj-1=skpΓj-1, which is a different copy of Γj-1, and thus has trivial intersection with it. It follows that also tpΓi has trivial intersection with Γi; therefore [Γi,tp]≠1. So skp=1, which implies that p is not a unit modulo nj, and since p is prime, it must divide nj.
We have shown that, if Γn has commuting Z/p-conjugates, then for all i there exists j>i i$$\end{document}]]> such that p divides nj, i.e. p divides infinitely many nj, which concludes the proof.
The fact that Γn has commuting cyclic conjugates follows directly from the construction: Every finitely generated subgroup H is contained in some Γi, and we can choose n=ni+1 and t to be the generator of Z/ni+1≤Γi+1. □
This allows to provide the following examples:
Proposition 3.3(c. c. c. vs. c. Z/n c. and c. Z c.) There exist groups with the following properties:For every non-empty set of primes P there exists a group that has commuting Z/p-conjugates if and only if p∈P;
There exists a group that has commuting cyclic conjugates, but not commuting Z/n-conjugates for any n≥2;
Both cases can be chosen not to have commuting Z-conjugates.
ProofWe enumerate P as p1<p2<⋯, and set ni:=∏j<ipj. By Lemma 3.2, the resulting group Γn has commuting Z/p-conjugates if and only if p divides ni for infinitely many i, which happens exactly when p=pi for some i.
If instead we take ni to be an increasing sequence of primes, then by Lemma 3.2 the resulting group Γn does not have commuting Z/p-conjugates for any prime p. Note that if a group has commuting Z/n-conjugates, then it has commuting Z/m-conjugates for every divisor m of n, and so the more general statement follows.
For the last item, we claim that if we start the construction with a finitely generated non-abelian torsion group Γ0, then the resulting groups Γn will not have commuting Z-conjugates. By induction, each Γi is torsion, and therefore so is Γn. Suppose by contradiction that there exists t∈Γn such that [Γ0,tpΓ0]=1 for all p≥1. Since Γn is torsion, taking p to be the order of t we obtain [Γ0,Γ0]=1, which contradicts the choice of Γ0 as non-abelian. □
Remark 3.4(Non-abelian torsion groups do not have c. Z c.) The previous argument shows that in fact non-abelian torsion groups cannot have commuting Z-conjugates.
In particular, the previous proposition shows examples of groups with commuting Z/2-conjugates that do not have commuting Z-conjugates; we will see more naturally occurring examples in the next section. Next, we show:
Proposition 3.5(c. Z c. not c. Z/n c.) The group Homeoc(R) has commuting Z-conjugates but does not have commuting Z/n-conjugates for any n≥2.
ProofThe group Homeoc(R) is dissipated and hence has commuting Z-conjugates by Example 2.15 (see also Example 3.8 below). We show now that it does not have commuting Z/n-conjugates for any n≥2.
Let H<Homeoc(R) be the copy of Thompson’s group F acting on (0, 1) and fixing the rest of R pointwise. Note that H is finitely generated and acts minimally on (0, 1). If H commutes with tn, it follows in particular that tn(0,1)=(0,1). Since t is compactly supported it preserves the orientation, and thus tn(0)=0,tn(1)=1. Similarly we deduce that t(0)=0 and t(1)=1. Thus tH is also supported on (0,1)=t(0,1) and acts minimally on it. Let now h∈H be an element whose only fixed point in (0, 1) is 1/2. Because tH commutes with h, it must preserve its fixed point set and thus fix 1/2. This contradicts the minimality of the action of tH. It follows that H and tH do not commute. □
Note that this result is in striking contrast with the case of Homeoc(R2) discussed before in Example 3.1.
Finally, we provide an example of a group with commuting conjugates that does not have commuting cyclic conjugates.
Proposition 3.6(c. c. not c. c. c.) Let Γ1:=Homeoc(R), and let Γ2 be a group with commuting conjugates that does not have commuting Z-conjugates. Then Γ1×Γ2 has commuting conjugates but does not have commuting cyclic conjugates.
For example, Γ2 may be chosen as one of the groups in Proposition 3.3.
ProofThe group Γ:=Γ1×Γ2 has commuting conjugates because each factor does.
Now the proof of Proposition 3.5 shows that there exists a finitely generated subgroup H1<Γ1, such that if [H1,t1n]=1 for some n≥2 and some t1∈Γ1, then [H1,t1H1]≠1. The assumption on Γ2 implies that there exists a finitely generated subgroup H2<Γ2 such that for all t2∈Γ2 there exists p≥1 such that [H2,t2pH2]≠1.
Suppose by contradiction that Γ has commuting cyclic conjugates, and let H:=H1×H2. Since Γ has commuting cyclic conjugates, there exists t=(t1,t2)∈Γ and n∈Z≥2∪{∞} such that [Hi,tipHi]=1 for all 1≤p<n, and [Hi,tin]=1, for i=1,2. If n<∞, the statement with i=1 contradicts the property of H1; if n=∞, the statement with i=2 contradicts the property of H2. □
More examples with commuting <inline-formula id="IEq419"><alternatives><mml:math id="IEq419_Math"><mml:mrow><mml:mi mathvariant="double-struck">Z</mml:mi><mml:mo stretchy="false">/</mml:mo><mml:mn>2</mml:mn></mml:mrow></mml:math><tex-math id="IEq419_TeX"><![CDATA[\documentclass[12pt]{minimal}
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\begin{document}$$\mathbb {Z}/2$$\end{document}]]></tex-math><inline-graphic specific-use="web" mime-subtype="GIF" xlink:href="229_2024_1604_Article_IEq419.gif"/></alternatives></inline-formula>-conjugates but not commuting <inline-formula id="IEq420"><alternatives><mml:math id="IEq420_Math"><mml:mi mathvariant="double-struck">Z</mml:mi></mml:math><tex-math id="IEq420_TeX"><![CDATA[\documentclass[12pt]{minimal}
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The stable mapping class group Γ∞1 (conservation of genus).
For linear groups note that our argument needs an appropriate notion of dimension or rank. For example, the general linear group of the cone of a ring is dissipated, and in particular it has commuting Z-conjugates [4, pp. 84–85].
Proposition 3.7(c. Z/2 c. not c. Z c.) The group GL(R) has commuting Z/2-conjugates but not commuting Z-conjugates.
ProofThe fact that GL(R) has commuting Z/2-conjugates has already been proved [11, Corollary 5.45 and its proof]. We are left to show here that GL(R) does not have commuting Z-conjugates.
Let {e1,e2,…} denote the standard basis elements of ∪n∈NRn, and let H be the copy of GL2(Z) acting on R2:=⟨e1,e2⟩ and fixing all other ei. Note that H≤GL(R) is finitely generated. We compute via block multiplication the conjugation of an element of GL(R) by an element of H:X00IABCDX-100I=XAX-1XBCX-1D,![]()
where X∈GL2(Z). Testing with bothX=-1001and100-1![]()
we see that an element in the centralizer of H must have B=C=0 and A diagonal. Moreover, testing withX=1011![]()
we see that A must be a scalar multiple of the identity. It follows that for every g in the centralizer of H, the action of g on R2=⟨e1,e2⟩ is by scalar multiplication.
Let t∈GL(R) be such that [H,tH]=1. Then the action of tH on R2 is by scalar multiplication. For the same reason, the action of H on tR2 is also by scalar multiplication. Since the action of H on R2 has no one-dimensional invariant subspace, this shows that R2∩tR2={0}.
Now suppose that there exists t∈GL(R) such that [H,tpH]=1 for all p≥1. Then the previous paragraph implies that R2∩tpR2={0} for all p≥1. It follows that tp1R2∩tp2R2={0} for all p1≠p2. Therefore the span of {tpe1,tpe2∣p≥1} is infinite-dimensional. But this is not possible since t∈GLn(R) for some n. Thus GL(R) does not have commuting Z-conjugates. □
Correctness of Fig. <xref rid="Fig2" ref-type="fig">2</xref>We will exhibit the following examples: A group with commuting Z-conjugates that is not dissipated and does not have conjugates in commuting Z-conjugate (Proposition 3.15).
A group with conjugates in commuting Z-conjugate that is not dissipated (Example 3.9).
A dissipated group that does not have conjugates in commuting Z-conjugate (Proposition 3.13).
A group with conjugates in commuting Z-conjugate that is also dissipated (Example 3.8).
We start with two well-known examples:
Example 3.8(c. c. Z c. and dissipated) The group Homeoc(R) is dissipated [5, 25] and has conjugates in commuting Z-conjugate [28, 30].
Example 3.9(c. c. Z c. not dissipated) The derived subgroup F′ of Thompson’s group F [12] has conjugates in commuting Z-conjugate [30]. However it is not dissipated, in fact it is not binate. This is a combination of the computations of its homology [19] (that is non trivial) and the fact that binate groups are acyclic [5, 33]. In fact, it can be argued that F′ is not binate analogously to the proof of Proposition 3.15 below, without appealing to homological results.
We now move on to the two remaining examples, which will stem from the same construction. We construct two groups Γ<G inside Homeoc(R). We start with I1:=(0,1) and Γ1=G1 the copy of Thompson’s group F (which is finitely generated) acting minimally on I1 and fixing its complement.
Suppose by induction that we have defined a group Gi supported on a bounded interval Ii and containing a finitely generated subgroup Γi. We choose an element ti+1∈Homeoc(R) with connected support, such that ti+1(Ii)∩Ii=∅. For simplicity, we denote Iip:=ti+1p(Ii) for all p∈Z, and note that since ti+1 is orientation-preserving, the intervals Iip are pairwise disjoint. We denote by Giω the group of homeomorphisms g supported on the disjoint union of intervals {Iip}p∈Z such that the restriction of g on Iip coincides with the action of an element of ti+1pGi. Note that we are allowing elements of Giω to act simultaneously on infinitely many intervals, and therefore Giω is naturally isomorphic to ∏ZGi. We set Gi+1:=⟨Giω,ti+1⟩ which is naturally isomorphic to the unrestricted wreath product GiWrZ. We also set Γi+1:=⟨Γi,ti+1⟩, which is naturally isomorphic to Γi≀Z, seen as a subgroup of Gi≀Z≤GiWrZ=Gi+1. Finally, we set Ii+1:=supp(ti+1)=supp(Γi+1)=supp(Gi+1).
Let Γ<G<Homeoc(R) be the directed unions of the Γi and the Gi respectively. It follows from the construction that:
Lemma 3.10The group G is dissipated, and the subgroup Γ<G has commuting Z-conjugates.
The following lemma will be useful:
Lemma 3.11For each i∈N, the subgroup Γi<Gi is normal.
ProofWe proceed by induction on i∈N, the case i=1 being tautological. Suppose by induction that Γi is normal in Gi. ConsiderΓi+1=⨁ZΓi⋊⟨ti+1⟩≤Gi+1=Giω⋊⟨ti+1⟩.![]()
Note that it suffices to show that ⊕ZΓi is normal in Giω and invariant under the action of ti+1. The latter is clear. Now a direct sum is normal in its corresponding direct product, and by induction hypothesis Γi is normal in Gi. Thus⨁ZΓi⊴∏ZΓi⊴∏ZGi=Giω.![]()
□
The key observation is the following:
Lemma 3.12For every g∈G and every i∈N, we have that eitherg(Ii)∩Ii=∅, or
g(Ii)=Ii, gGi=Gi and gΓi=Γi.
ProofWe prove the claim by induction on d=(j-i)∈Z where i is as in the statement, and j is such that g∈Gj. If j≤i, then g is supported on Ii and belongs to Gi, so g(Ii)=Ii and gGi=Gi are trivial, and gΓi=Γi follows from Lemma 3.11.
Now suppose that g∈Gi+1, i.e. d=1. By construction Gi+1 is isomorphic to GiWrZ, with ti+1 representing the stable letter. The copy of Gi sitting at the lamp labelled by k is supported on Iik and fixes pointwise its complement, in particular it fixes pointwise Iiℓ for all ℓ≠k. We write g=ti+1kb for some b∈Giω and k∈Z. If k≠0, then g(Ii)=ti+1kb(Ii)=ti+1k(Ii)=Iik which is disjoint from Ii. Otherwise, g=b preserves Ii, and the action of g by conjugacy on Gi is the same as the action of the 0-th component of g, which belongs to Gi. Therefore gGi=Gi and gΓi=Γi, as in the first paragraph.
Now suppose that d>1 1$$\end{document}]]>, i.e. j-1>i i$$\end{document}]]>. Since Ii⊂Ij-1, if g(Ij-1)∩Ij-1=∅, then we are done. Otherwise, the above argument shows that g(Ij-1)=Ij-1, and there exists an element h∈Gj-1 such that g|Ij-1=h|Ij-1, and therefore the actions of g and h by conjugacy on Gj-1 coincide. Then we can apply the induction hypothesis to h and conclude. □
We are now ready to provide one more example for Fig. 2:
Proposition 3.13(Dissipated not c. c. Z c.) The group G is dissipated, but does not have conjugates in commuting Z-conjugate.
ProofLet Γi,Gi,ti,Γ and G be as in the construction above. We have already seen in Lemma 3.10 that G is dissipated.
Now suppose by contradiction that G has conjugates in commuting Z-conjugate. So there exist Λ≤G and t∈G with the following properties: [Λ,tpΛ]=1 for all p≥1;
Every finite subset of G is conjugate into Λ.
Let n∈N be such that t∈Gn. The second property implies that there exists s∈G such that sΓn≤Λ, since Γn is finitely generated. By Lemma 3.12, we split into two cases. First, suppose that s(In)∩In=∅. Then t commutes with sΓn because they are supported on disjoint intervals, and so [sΓn,t(sΓn)]=[sΓn,sΓn]≠1, because Γn is not abelian. In the other case, s(In)=In and sΓn=Γn and so again [sΓn,t(sΓn)]=[Γn,tΓn]=[Γn,Γn]≠1, where the last equality follows from Lemma 3.11. In both cases, [sΓn,t(sΓn)]≠1 and thus [Λ,tΛ]≠1, which contradicts the defining property of t. □
Finally, we will show that Γ fits the last remaining non-empty region in Fig. 2: It has commuting Z-conjugates, but it is neither dissipated nor it has conjugates in commuting Z-conjugate.
Lemma 3.14Let g∈Γ. Then the support of g is a disjoint union of finitely many intervals.
ProofWe prove this by induction on the index i∈N such that g∈Γi. If i=1, then g is an element of Thompson’s group F acting on the interval, and so the statement is well-known [12]. Now suppose that this is true up to Γi, and let g∈Γi+1. Then we can write g=ti+1kb for some b∈⊕ZΓi and some k∈Z. If k≠0, then supp(g)=supp(ti+1)=Ii+1, which is an interval. Otherwise, g=b is the product of finitely many elements in conjugates of Γi supported on disjoint sets, and so the induction hypothesis applies. □
Proposition 3.15(c. Z c. not dissipated and not c. c. Z c.) The group Γ has commuting Z-conjugates, but is not dissipated (in fact, it is not even binate) and does not have conjugates in commuting Z-conjugate.
ProofWe have already seen in Lemma 3.10 that Γ has commuting Z-conjugates. It follows from (a shorter version of) the argument of Proposition 3.13 that Γ does not have conjugates in commuting Z-conjugate.
Suppose by contradiction that Γ is binate. Consider Γ1<Γ, which is finitely generated and supported on I1, and let g∈Γ1 be an element of full support. By assumption, there exists a homomorphism f:Γ1→Γ such that [Γ1,f(Γ1)]=1 and t∈Γ such that tf(g)=g·f(g). By Lemma 3.14, there exist finitely many disjoint intervals I2,…,In whose union equals the support of f(g). Now, Γ1 acts minimally on I1, so f(Γ1) fixes I1 pointwise; in particular I1 is disjoint from Ik for 2≤k≤n. It follows that g·f(g) is supported on the disjoint union I1∪⋯∪In. This gives t(I2∪⋯∪In)=I1∪⋯∪In, a contradiction. □
Correctness of Fig. <xref rid="Fig3" ref-type="fig">3</xref>We need to exhibit the following examples: A binate group that does not have commuting conjugates (Proposition 3.19).
A group with commuting conjugates that is not binate (Example 3.16).
A binate group with commuting conjugates that is not dissipated (Corollary 3.18).
We leave the first item for the end, and start with the other two examples, which are well-known.
Example 3.16(c. c. not binate) We have already seen in Example 3.9 that F′ has conjugates in commuting Z-conjugate but is not binate. In particular F′ is an example of a group with commuting conjugates that is not binate.
For the next example, we go a little further:
Proposition 3.17Hall’s universal locally finite group has the following combination of properties: It is mitotic;
It has commuting Z/n-conjugates for every n;
It does not have commuting Z-conjugates.
We immediately deduce:
Corollary 3.18(Binate and c. c. not dissipated) Hall’s universal locally finite group is binate, has commuting conjugates, but is not dissipated.
Proof of Proposition 3.17Recall that Hall’s universal locally finite group U is a countable locally finite group such that every finite group embeds into U, and such an embedding is unique up to conjugacy. It is easy to deduce from this universal property that U is mitotic [5].
Let H≤U be a finitely generated subgroup, which is therefore finite. By the first property, there exists an embedding ι:H≀Z/n→U. By the second property, there exists s∈U such that h=sι(h) for all h∈H. Let t0 denote the generator of Z/n. Then t:=sι(t0) satisfies the required properties. First [H,tpH]=[sι(H),sι(t0p)(sι(H))]=sι[H,t0pH]=1 for all 1≤p<n; and tn=sι(t0n)=1, so clearly [H,tn]=1. This shows that U has commuting Z/n-conjugates for every n.
On the other hand, U does not have commuting Z-conjugates because of being a non-abelian torsion group (Remark 3.4). □
Finally, we construct a binate group that does not have commuting conjugates. In particular, this will be an example of a binate group that does not fit into either of the main categories of binate groups, namely mitotic and dissipated groups, since both of these have commuting conjugates (for different reasons).
We use a standard construction to embed groups into binate groups, called binate towers [5]. For a group Γ, we letb(Γ):=⟨Γ×Γ,d∣d(1,g)=(g,g)for allg∈Γ⟩.![]()
We can describe b(Γ) as an HNN-extension with base Γ×Γ and associated subgroups Γ+:={1}×Γ and ΔΓ:={(g,g)∈Γ×Γ∣g∈Γ}. We denote by Γ-:=Γ×{1} and note that Γ- embeds into b(Γ). This allows to iterate this construction by setting b0(Γ):=Γ and bi(Γ):=b(bi-1(Γ)) for all i≥1, and to construct the group β(Γ) as the directed union of {bi(Γ)}i∈N. The group β(Γ) is called the universal binate tower overΓ, and it is the basic example of a binate group. We will show that any such construction gives the desired example:
Proposition 3.19(Binate towers do not have c. c.) For every finitely generated group Γ, the binate tower β(Γ) does not have commuting conjugates.
The key claim is the following structural result about the functor b.
Lemma 3.20Let Γ be a group, identified with its image Γ-≤Γ×Γ≤b(Γ). Then every element 1≠h∈Γ- fixes a unique vertex in the Bass–Serre tree of b(Γ), which is the vertex fixed by Γ×Γ.
ProofLet T be the Bass–Serre tree for the HNN-extension expression of b(Γ). It has a unique orbit of vertices and exactly two orbits of oriented edges. For a suitable choice of base vertex v, incoming edge e and outgoing edge f, the stabilizer of v is Γ×Γ, the stabilizer of e is ΔΓ, and the stabilizer of f is Γ+. Moreover, all incoming edges are permuted by Γ×Γ, therefore their stabilizers are conjugates of ΔΓ in Γ×Γ, and similarly for outgoing edges and Γ+. It follows that a non-trivial element h∈Γ- does not belong to the stabilizer of any edge adjacent to v, and therefore fixes no other vertex of T. □
Proof of Proposition 3.19Let Γ be a finitely generated group. Up to replacing Γ by b(Γ), and noticing that β(b(Γ))≅β(Γ), we may assume that Γ is non-abelian. Suppose by contradiction that β(Γ) has commuting conjugates. Then there exists t∈β(Γ) such that [Γ,tΓ]=1, and by construction there exists i≥0 such that t∈bi(Γ). We choose t so that i is minimal. Then i≥1, for otherwise t∈Γ and [Γ,tΓ]=[Γ,Γ]≠1.
We see bi(Γ) as b(bi-1(Γ)) and notice that Γ is identified with a subgroup of bi-1(Γ)-≤bi-1(Γ)×bi-1(Γ)≤bi(Γ). By Lemma 3.20, if we denote by v the vertex in the Bass–Serre tree that is fixed by bi-1(Γ)×bi-1(Γ), then v is the unique vertex fixed by Γ. Therefore tΓ fixes the vertex tv and no other vertex. On the other hand tΓ centralizes Γ, and therefore must preserve the fixed point set of Γ. It follows that tv=v, and so t belongs to the stabilizer of v in bi(Γ), that is bi-1(Γ)×bi-1(Γ), which allows to write t=t-t+ as a product of its projections onto the two factors. Now the conjugation by t+ on Γ≤bi-1(Γ)- has no effect, and therefore t- satisfies [Γ,t-Γ]=1. But t-∈bi-1(Γ)-, which is the standard copy of bi-1(Γ) in β(Γ). This contradicts the minimality of i and concludes the proof. □
Mitotic groupsMitotic groups form an important class of binate groups; However they are fundamentally different from the examples of boundedly supported groups with displacement properties that we have been mostly interested in. For example, one can show that if Γ is a boundedly supported group acting on a space X which exhibits commuting conjugates as in Example 2.10, and Γ is moreover mitotic, then the support of every finitely generated subgroup of Γ can be written as the disjoint union of an arbitrarily large number of open sets. However, we are unable to show that this situation cannot occur. In particular, the following problem is an obstacle for fitting mitotic groups into Fig. 3:
Problem 3.21(Dissipated and mitotic group) Does there exist a dissipated group that is also mitotic?
As for the interactions with other properties, it is clear that mitotic groups have commuting conjugates, and Hall’s universal locally finite group witnesses that mitotic groups need not have commuting Z-conjugates (Proposition 3.17). However, the following remains open:
Problem 3.22(Mitotic implies c. c. c.) Let Γ be a mitotic group. Does Γ have commuting cyclic conjugates?
We propose a natural candidate for a counterexample to Problem 3.22. Following Baumslag–Dyer–Heller [3, Section 5], for a group Γ, we denote by m(Γ) its standard mitosis, namelym(Γ):=⟨Γ×Γ,s,d∣s(g,1)=(1,g),d(1,g)=(g,g)⟩.![]()
We can describe m(Γ) as the fundamental group of a graph of groups whose underlying graph is a wedge of two circles, with unique vertex group Γ×Γ and with both edge groups Γ. Now Γ embeds into m(Γ) by first embedding into the first factor of Γ×Γ. We can thus iterate this construction by setting mi(Γ)=m(mi-1(Γ)), and we denote by μ(Γ) the directed union of {mi(Γ)}i∈N, which is called the standard mitotic embedding of Γ. Then μ(Γ) is mitotic [3, Lemma 5.4], and thus binate. This construction can be seen as the mitotic analogue of the binate tower construction that we used for Proposition 3.19.
Problem 3.23(Mitotic tower of free group has c. c. c.) Let Fk denote a non-abelian free group. Does μ(Fk) have commuting cyclic conjugates?
We suspect that it does not. At least, it is easy to show that μ(Fk) does not have commuting Z-conjugates:
ProofIf μ(Fk) has commuting Z-conjugates, then there exists t∈μ(Fk) such that [Fk,tpFk]=1 for all p≥1. Since Fk has trivial center, it follows that ⟨Fk,t⟩ contains a copy of ⊕NFk, which has infinite cohomological dimension. Hence, if i∈N is such that t∈mi(Fk), it follows that mi(Fk) has infinite cohomological dimension as well.
However it is easy to see that cd(m(Γ))=2cd(Γ) for every group Γ [7, Proposition 6.12]. Therefore cd(mi(Fk))=2i, which contradicts the previous paragraph. □
Despite the similarity of the problem, adapting the argument from Proposition 3.19 to show that mitotic towers do not have commuting cyclic conjugates presents a challenge. Namely, a crucial step in the proof that β(Γ) does not have commuting conjugates is the fact that the image bi-1(Γ)→bi(Γ) intersects trivially all edge stabilizers in the corresponding Bass–Serre tree. This is however not the case with mi-1(Γ)→mi(Γ), where the image is identified with an edge stabilizer, and in fact can be shown to fix infinitely many edges.