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Cyclically presented groups with length four positive relators

  • William A. Bogley EMAIL logo and Forrest W. Parker
Published/Copyright: June 15, 2018

Abstract

We classify cyclically presented groups of the form G=Gn(x0xjxkxl) for finiteness and, modulo two unresolved cases, we classify asphericity for the underlying presentations. We relate finiteness and asphericity to the dynamics of the shift action by the cyclic group of order n on the nonidentity elements of G and show that the fixed point subgroup of the shift is always finite.

1 Cyclically presented groups and shift dynamics

One purpose of this article is to provide classifications of finiteness and asphericity (modulo two unresolved cases) for cyclic group presentations where the relators are positive words of length four. More generally, we relate finiteness and asphericity to the dynamics of the shift action that arises from the cyclic symmetry in these presentations, thus extending results from [3].

Let n be a positive integer and let F be the free group of rank n with basis x0,,xn-1. The shift automorphism is given by θF(xi)=xi+1, where subscripts are taken modulo n. A word wF determines the cyclic (group) presentation

𝒫n(w)=x0,,xn-1|w,θF(w),,θFn-1(w)

which in turn defines the cyclically presented groupG=Gn(w). The shift determines an automorphism θGAut(G). Viewed in Aut(G), the shift θG has order dividing n and may or may not be an inner automorphism of G.

The following two classification problems have received attention in the literature. See [6, 10, 22, 23], for example.

Finiteness.

For which n and w is the cyclically presented group Gn(w) finite?

Asphericity.

For which n and w is the cyclic presentation 𝒫n(w) (combinatorially) aspherical?

See Section 2 for a discussion of combinatorial asphericity. A group presentation is called aspherical if its two-dimensional cellular model K is aspherical, or equivalently the second homotopy module π2K is trivial. Combinatorial asphericity is a more general concept that allows for the presence of proper power and repeated relators, in which case π2K0. See [7] for a comprehensive treatment of asphericity concepts for group presentations. Asphericity and finiteness are related in that any finite subgroup of a group defined by a combinatorially aspherical presentation is finite cyclic [12]. This generalizes the well-known fact that if a two-complex K satisfies π2K=0, then the fundamental group π1K is torsion-free.

In [6, 10], the finiteness and asphericity classification problems were completely solved for cyclic presentations 𝒫n(w) where the defining relator is a positive word of length three. In this article we deal with the length four case. Thus we consider presentations of the form

(1.1)𝒫n(x0xjxkxl)=x0,,xn-1|xixi+jxi+kxi+l,0i<n,

where the integer parameters j,k,l are to be considered modulo the positive integer n. (Note that the presentation 𝒫n(w) is unchanged if we replace w by any of its shifts.) Modeling an approach developed in [10], our classifications are framed in terms of conditions on the parameters (n,j,k,l) starting with A, B, C as follows; all congruences are considered modulo n:

  1. 2k0 or 2j2l,

  2. k2j or k2l or j+l2k or j+l0,

  3. lj+k or lj-k,

Table 1 provides an overview of our main results regarding asphericity and finiteness in terms of truth values for conditions A, B, C.

Table 1

Asphericity and finiteness for 𝒫n(x0xjxkxl).

ABCAspherical [I]Group [II]W [5.25.4]
TTTyesfinite n=14 [6.1 (b)]z4
TFTC(4)-T(4)finite TTT(z2α)2
TFF[6.2, 6.3]i=14xαi
FFFi=14xαi
FTTno [6.3]finite 4 [7.2]z4α
FFTsee [II (b)], [7.2 (iv)]z2αz2β
TTFno [8.1]finite γ=1 [8.1]u3αuα±1
finite solvable [8.1]
FTF(I): no [9.7]finite (I5), (I6), or (I6′′) [9.8]u3αuβ
(U): unresolved
else: yes [9.2]

We now discuss the specialized notation in Table 1, where bracketed references [-] are internal to this article. The cyclic group of order n acts on a cyclically presented group G=Gn(w) by automorphisms and so there is the shift extensionEn(w)=Gn(w)θGn. This group admits a two-generator two-relator presentation En(w)a,x|an,W, where W is obtained from w via the substitutions xi=aixa-i, so the shift θGAut(G) arises via conjugation by a in En(w). When w=x0xjxkxl, the relator W takes the form

(1.2)W=xajxak-jxal-kxa-l.

Conditions A, B, C enable simplifications of relation (1.2) as shown in the right-hand column of Table 1. For example, if k2jmodn, as in B, then the substitution u=xaj transforms W to u3al-3jua-l-j. Simplifications of this sort are developed in Section 5. Section 2 discusses the concept of combinatorial asphericity and its relation to the small cancellation property C(4)-T(4) [15, Chapter V.2]. Section 3 discusses isomorphisms among groups of the form Gn(x0xjxkxl) that arise from an action of the finite group Γn=D4×(nn) on the set of length four positive words in the generators x0,,xn-1. Here, D4 is the dihedral group of order 8 and n is the multiplicative group of units acting naturally on the additive group n. Thus the group Γn has order 8nϕ(n) where ϕ is the Euler totient function. The secondary divisor

γ=gcd(n,k-2j,j-2k+l,k-2l,j+l)

is introduced in Section 4.

The various combinations of conditions represented in the left-hand column of Table 1 are analyzed individually in Sections 69. As in [3, 6, 10], we depend on previous work concerning asphericity of relative presentations, in this case [2]. There, the nonaspherical cases fall into well-defined infinite families, along with a handful of isolated cases. In addition, there are two cases treated in [2] where the asphericity status remains unresolved even when we incorporate recent progress due to Aldwaik and Edjvet [1] and Williams [24]. See the proof of Theorem 9.2 for these updates.

The bottom row of Table 1 is the most complex, where we encounter eight isolated conditions and the two unresolved conditions, which are collectively referred to as (I) and (U), respectively. As we now describe, these conditions are defined in terms of exemplars that are listed in Tables 2 and 3.

Table 2

Exemplars for the eight isolated conditions (I5)–(I24).

ConditionExemplarShift extension
(I5)𝒫5(x0x3x1x1)a,x|a5,xa3xa-2xxa-1
u=xa3a,u|a5,u3a2ua
(I6)𝒫6(x0x4x2x3)a,x|a6,xa4xa-2xaxa-3
u=xa4a,u|a6,u3a-3ua-1
(I6′′)𝒫6(x0x0x1x2)a,x|a6,xxaxaxa-2
u=xaa,u|a6,u3a-3ua-1
(I10)𝒫10(x0x3x6x1)a,x|a10,xa3xa3xa-5xa-1
u=xa3a,u|a10,u3a-8ua-4
(I12)𝒫12(x0x1x2x9)a,x|a12,xaxaxa7xa-9
u=xaa,u|a12,u3a6ua2
(I16)𝒫16(x0x3x6x1)a,x|a16,xa3xa3xa-5xa-1
u=xa3a,u|a16,u3a-8ua-4
(I20)𝒫20(x0x3x6x1)a,x|a20,xa3xa3xa-5xa-1
u=xa3a,u|a20,u3a-8ua-4
(I24)𝒫24(x0x1x2x15)a,x|a24,xaxaxa13xa-15
u=xaa,u|a24,u3a12ua8
Table 3

Exemplars for the two unresolved conditions (U24) and (U24′′).

ConditionExemplarShift extension
(U24)𝒫24(x0x3x6x1)a,x|a24,xa3xa3xa-5xa-1
u=xa3a,u|a24,u3a-8ua-4
(U24′′)𝒫24(x0x1x2x19)a,x|a24,xaxaxa17xa-19
u=xaa,u|a24,u3a16ua4

Types (I) and (U).

As discussed in Section 3, the orbit of a positive word xixjxkxl under the action of the finite group Γn consists of all shifts of cyclic permutations of words of the form xiuxjuxkuxlu or xulxukxujxui, where un is a multiplicative unit modulo n. In Theorem 3.2 we show that each such orbit gives rise to a single group up to isomorphism, asphericity status, and shift dynamics. Applying this to the exemplar listed under (I5) in Table 2, we say that a cyclic presentation 𝒫n(w) is of type (I5) if and only if n=5 and w is a length four positive word in the orbit of x0x3x1x1 under the action of the group Γ5 of order 160, which means that w is a shift of a cyclic permutation of a word of the form x0x3uxuxu or xuxux3ux0, where u5. In the same way, the remaining seven exemplars listed in Table 2 determine corresponding isolated types (I6), (I6′′), (I10), (I12), (I16), (I20), and (I24). Any cyclic presentation that adheres to any one of these eight isolated types is said to be of type (I). In the same way, the exemplars listed in Table 3 determine the presentations of type (U24) and (U24′′) in terms of the action of the group Γ24 of order 1536. Any presentation of type (U24) or (U24′′) is of type (U).

Types (I) and (U).

It is easy to check that every presentation 𝒫n(x0xjxkxl) of type (I) or (U) satisfies gcd(n,j,k,l)=1. For arbitrary (n,j,k,l), we say that 𝒫n(x0xjxkxl) is of type (I) (resp. (U)) if upon setting c=gcd(n,j,k,l), the presentation 𝒫n/c(x0xj/cxk/cxl/c) is of type (I) (resp. (U)). The presentations of type (I) are all nonaspherical (see Lemma 9.1 and Theorem 9.7), but it is unknown whether the presentations of type (U) are aspherical. Resolution of this ambiguity reduces to consideration of the exemplars 𝒫24(x0x3x6x1) and 𝒫24(x0x1x2x19); see Lemma 9.1 and Section 10. The presentations of type (I) and (U) all satisfy FTF relative to conditions A, B, and C. See Theorem 9.2.

Our main results regarding asphericity and finiteness are Theorems III. Note that the finiteness classification is complete, with no unresolved cases remaining.

Theorem I (Asphericity).

Assume that P=Pn(x0xjxkxl) is not of type (U). Then P is combinatorially aspherical if and only if at least one of the following holds:

  1. A and C are both true,

  2. B and C are both false, or

  3. B is true, A and C are false, and 𝒫 is not of type (I).

The presentation Pn(x0xjxkxl) is aspherical if and only if one of (b) or (c) holds. If Pn(x0xjxkxl) is combinatorially aspherical and either k0 or jlmodn, then Gn(x0xjxkxl) is a torsion-free infinite group of geometric dimension at most two.

Theorem II (Finiteness).

Given (n,j,k,l), the cyclically presented group G=Gn(x0xjxkxl) is finite if and only if at least one of the following holds:

  1. gcd(n,2k)=1 and either

    1. lj+kmodn and gcd(n,j)=1, or

    2. lj-kmodn and gcd(n,l)=1.

  2. A and B are true, C is false, and the secondary divisor γ=1.

  3. The presentation 𝒫n(x0xjxkxl) is of type (I5), (I6), or (I6′′).

In part (a), which includes the case n=1, we have GZ4. In part (b), G is solvable. In type (I5), G is metacyclic and nonnilpotent of order 220. The presentations of type (I6) and (I6′′) define nonisomorphic nonsolvable groups G of order 4,088,448=27337132, each containing the simple group PSL(3,3).

These classifications are supported by simultaneous consideration of shift dynamics, which concern the action of the cyclic group n via powers of the shift θGn(w) on (the nonidentity elements of) Gn(w). For arbitrary n and w, shift dynamics encompass at least these two general problems.

Fixed points.

For which n and w does the shift θGn(w) have a nonidentity fixed point?

Freeness.

For which n and w does the shift determine a free n-action on the nonidentity elements of Gn(w)?

In detail, freeness means that if θGn(w)d(g)=g, then either g=1 in Gn(w) or else d is divisible by n. In particular, if Gn(w) is nontrivial and the shift action is free, then θGn(w) has order n in the outer automorphism group Out(Gn(w)). (Compare [13].) This is because every inner automorphism of a nontrivial group has a nonidentity fixed point.

Asphericity and freeness are related by a general result from [3]. A cyclic presentation 𝒫n(w) is orientable if w is not a cyclic permutation of the inverse of any of its shifts. In the general case, a cyclic presentation 𝒫n(w) fails to be orientable if and only if n=2m is even and w=uθFm(u)-1 for some word u (see [3, Lemma 3.6]) (in which case uFix(θGm)). Note that any cyclic presentation defined by a positive word is orientable. In [3, Theorem A] it was shown that if 𝒫n(w) is orientable and combinatorially aspherical, then n acts freely via the shift on the nonidentity elements of the cyclically presented group Gn(w). We note that there are just a few known examples of orientable cyclic presentations that are not combinatorially aspherical and yet have free shift action. Examples include the Fibonacci groups F(2,n)=Gn(x0x1x2-1) for n=5,7, which are finite cyclic having orders 11 and 29, respectively (see [22]).

In [3, Theorems B,C] it was shown that if w=x0xkxl is a positive word of length three, then the relationships between finiteness, asphericity, and shift dynamics are sharp:

  1. 𝒫n(x0xkxl) is combinatorially aspherical if and only if n acts freely via the shift on the nonidentity elements of Gn(x0xkxl).

  2. G=Gn(x0xkxl) is finite if and only if the shift θG has a nonidentity fixed point.

One of the principal aims of this article is to extend these connections to the case of length four relators. The relation between asphericity and freeness is not as tidy due to the unresolved cases in the asphericity classification.

Theorem III.

If P=Pn(x0xjxkxl) is not of type (U), then P is combinatorially aspherical if and only if the cyclic group Zn acts freely on the nonidentity elements of Gn(x0xjxkxl).

It is not unreasonable to hypothesize that the identification between asphericity and free action will persist once the asphericity status is resolved for type (U).

When combined with Theorem II, the next result explicitly determines those cyclic presentations 𝒫n(x0xjxkxl) for which the shift has a nonidentity fixed point.

Theorem IV.

The shift on G=Gn(x0xjxkxl) has a nonidentity fixed point if and only if G is finite or else C is true and gcd(n,2k)=1.

Instances where G=Gn(x0xjxkxl) is infinite and Fix(θG)1 occur when condition C is true, gcd(n,2k)=1, and either lj+k but gcd(n,j)1 or else lj-k but gcd(n,l)1 (see Theorem 7.2). By analyzing the infinite cases in detail, we are able to show that nonidentity fixed points for the shift always derive from finite cyclically presented subgroups.

Theorem V.

The shift on G=Gn(x0xjxkxl) has a nonidentity fixed point if and only if G possesses a finite cyclically presented subgroup of the form H=Gn(v) where θG|H=θH and Fix(θG)=Fix(θH)1.

Combining with the results of [3], we conclude the following.

Corollary 1.1.

If G=Gn(w), where n is a positive integer and w is a positive word of length at most four, then the fixed point subgroup Fix(θG) is finite. Moreover, if G is finite, then Fix(θG) is nontrivial.

We have already outlined the contents of the preliminary Sections 25, so we conclude this introduction with a comment on the structure of the remainder of the article. For each combination of truth values for conditions A, B, C, we treat finiteness, asphericity, and shift dynamics simultaneously. In this way we take advantage of the intrinsic connections between these concepts and avoid needless repetition of common arguments. Section 6 treats four of the eight possible combinations of A, B, C, all of which lead to small cancellation conditions. Section 7 deals with the case where A is false and C is true (FTT or FFT). Section 8 deals with the case where A and B are true and C is false (TTF). Section 9 deals with the most interesting case, where B is true and both A and C are false (FTF). A concluding section wraps up the details of Theorems IV.

The results of this article are drawn from the PhD dissertation [18] of the second named author. We are grateful to Gerald Williams for helpful comments and for granting permission to discuss his computations regarding the group L with presentation t,u|t6,u3t3ut, which plays key roles in Lemma 9.6 and in Theorem II.

2 Combinatorial asphericity

Given a presentation 𝒫=𝐱|𝐫 for a group G, the free group F=F(𝐱) with basis 𝐱 acts by automorphisms on the free group 𝔽=F(F×𝐫) with basis F×𝐫 by v(u,r)=(vu,r). The homomorphism :𝔽F given by (u,r)=uru-1 is F-equivariant where F acts on itself by (left) conjugation. The image of is the normal closure of 𝐫 in F so the cokernel of is isomorphic to G. Elements of the kernel 𝕀=ker are called identities (or identity sequences) for 𝒫. Then we have the exact sequence

(2.1)1𝕀𝔽FG1

for the presentation 𝒫. Among the identities are the so-called Peiffer identities having the form (u,r)ϵ(v,s)δ(u,r)-ϵ(urϵu-1v,s)-δ, where u,vF, r,s𝐫, and δ,ϵ=±1. If denotes the normal closure of the Peiffer identities in 𝔽, then the F-action on 𝔽 descends to a G-action on the abelian quotient group 𝕀/ and there is the exact fundamental sequence

(2.2)0𝕀/𝔽/FG1

that provides a combinatorial description for the long exact homotopy sequence of the skeleton pair (K,K1), where K is the two-complex modeled on the presentation 𝒫. In particular, π2K𝕀/ as G-modules. See [7, Section 1], [20, Section 2], or [21] for details.

As in [7, Proposition 1.4], a presentation 𝒫=𝐱|𝐫 for a group G is combinatorially aspherical if the homotopy module π2K𝕀/ is generated as a G-module by the classes of identities that have length two in the free group 𝔽 with basis F(𝐱)×𝐫. The two-dimensional cellular model K of the presentation 𝒫 is aspherical, in the sense that π2K=0, if and only if the presentation 𝒫 is combinatorially aspherical and no relator of 𝒫 is a proper power, nor is conjugate to any another relator or its inverse. See [7], [3, Section 3], or the survey [19], which contains a thorough discussion of identity sequences and their relation to pictures under the blanket assumption that no relator is conjugate to any another relator or its inverse. From the point of view of spherical pictures (or dually, spherical diagrams), an arbitrarily given presentation is combinatorially aspherical if and only if its second homotopy module is generated by the classes of spherical pictures that contain exactly two relator discs. See [19, Section 2.3].

Presentations satisfying any of the C(p)-T(q) small cancellation conditions are combinatorially aspherical if 1/p+1/q1/2 (see [7, 14]). In fact, they satisfy even stronger conditions such as diagrammatic asphericity [8]. Here we shall be interested in the C(4)-T(4) condition. Recall that a word is a piece for a presentation if it occurs as a common initial subword of two distinct cyclic permutations of the relators or their inverses.

Lemma 2.1.

The presentation Pn(x0xjxkxl) satisfies the C(4)-T(4) small cancellation condition if and only if no length two cyclic subword of w=x0xjxkxl is a piece.

Proof.

Since the relators of 𝒫n(x0xjxkxl) have length four, the C(4) condition is satisfied if and only if each piece has length one. On the other hand, if all pieces have length one, the T(4) condition is also satisfied because the relators are all positive words, see [11]. The cyclic symmetry in the presentation implies that it suffices to verify that no length two cyclic subword of w=x0xjxkxl is a piece. ∎

3 Isomorphisms

Classifications involving cyclic presentations are complicated by the fact that there are many isomorphisms that must be accounted for, not all of which are obvious or easily catalogued. See [6, Section 2] and [10, Lemma 2.1], for example.

For a fixed positive integer n, let Φn denote the set of all positive words of length four in the generators x0,,xn-1 for the free group F of rank n. Thus every element wΦn has the form w=xixjxkxl, where integers i,j,k,l are defined modulo n. We consider the following bijective transformations on Φn:

σ(xixjxkxl)=xlxkxjxi,τ(xixjxkxl)=xlxixjxk,
θF(xixjxkxl)=xi+1xj+1xk+1xl+1,u(xixjxkxl)=xuixujxukxul.

Here the element u is taken from the multiplicative group n of units modulo n. All subscripts are interpreted modulo n.

Lemma 3.1.

Within the group Sym(Φn) of permutations of Φn, the transformations σ,τ,θF and uZn generate a subgroup that is a homomorphic image of Γn=D4×(ZnZn), where D4 is the dihedral group of order 8 generated by σ and τ and the semidirect product involves the natural action of ZnAut(Zn) acting on the additive group Zn by multiplication.

Proof.

Working in Sym(Φn), one checks that σ2=τ4=(στ)2=θFn=1 and uθF=θFuu for all un. Moreover, the transformations σ,τ commute pairwise with θF,u. ∎

The orbit of the word w=xixjxkxlΦn under the action of Γn consists of all words that arise as cyclic permutations of shifts of words of the form u(w)=xuixujxukxul or σ(u(w))=xulxukxujxui, where un.

Theorem 3.2.

For each cΓn=D4×(ZnZn) and w=xixjxkxlΦn, there is a group isomorphism cw:Gn(w)Gn(c(w)) such that

(3.1)cwθGn(w)=θGn(c(w))ψ(c)cw,

where ψ:ΓnZn is the natural projection. Moreover, Pn(w) is combinatorially aspherical if and only if Pn(c(w)) is combinatorially aspherical.

Proof.

Let F=F(𝐱) be the free group with basis 𝐱={xp:0p<n}. Given cΓn and w=xixjxkxlΦn, we denote the relator sets for the presentations 𝒫n(w) and 𝒫n(c(w)) by

𝐫w={θFq(w):0q<n}={xi+qxj+qxk+qxl+q:0q<n}

and

c(𝐫w)={θFq(c(w)):0q<n},

respectively. We first construct homomorphisms cw:FF, c^w:𝔽w𝔽c(w), where 𝔽w=F(F×𝐫w) and similarly 𝔽c(w)=F(F×c(𝐫w)). For c=σ,τ,θF or un and w=xixjxkxlΦn, the assignments

σw(xp)=xp-1,σ^w(v,θFq(w))=(σw(v),θFqσ(w))-1,
τw=1F,τ^w(v,θFq(w))=(vθFq(xl)-1,θFqτ(w)),
(θF)w=θF,(θF)^w(v,θFq(w))=(θF(v),θFq+1(w)),
uw(xp)=xup,u^w(v,θFq(w))=(uw(v),θFuq(u(w)))

define length-preserving homomorphisms cw:FF, c^w:𝔽w𝔽c(w). These homomorphisms are constructed to satisfy

c^w=cw

for c=σ,τ,θF,uΓn=D4×(nn). Viewed in Aut(F), the isomorphisms σw,τw,θF,uw are compatible with the relations of Γn=D4×(nn). Specifically, one notes that

σw2=τw4=(σwτw)2=θFn=1,twuw=(tu)w(t,un)

and that the homomorphisms σw,τw commute pairwise with θF and uw (un). Since Γn acts on Φn, this shows that Γn acts on F by automorphisms and in fact this action is independent of w. (We retain the notation cw in order to distinguish between cΓn and cwAut(F).) For each cΓn we now have an automorphism cwAut(F) that is length-preserving and which is compatible with the boundary maps from the exact sequences (2.1) for the presentations 𝒫n(w) and 𝒫n(c(w)):

It follows that for any cΓn, the automorphism cwAut(F) induces an isomorphism Gn(w)Gn(c(w)), which we also denote by cw. As for shift equivariance, we have

σwθGn(w)=θGn(σ(w))σw,
τwθGn(w)=θGn(τ(w))τw,
uwθGn(w)=θGn(u(w))uuw.

Moreover, Gn(θF(w))=Gn(w) and (θF)w=θGn(w). This verifies equivariance as in (3.1).

It remains to verify that combinatorial asphericity is preserved under the action of Γn on the set of presentations of the form 𝒫n(w), wΦn. Note that the homomorphism τ^w depends on w. It turns out that the homomorphisms σ^w, τ^w, (θF)^w, u^w are compatible with some of the relations of Γn but not all. The fact that σ^w2=1𝔽w is straightforward:

σ^w2(v,θFq(w))=σ^w(σw(v),θFqσ(w))-1
=((σw2(v),θFq(σ2(w))-1)-1
=(v,θFq(w)).

However, τ^w4:𝔽w𝔽w is not the identity. For example,

τ^w4(1,xixjxkxl)=τ^w3(xl-1,xlxixjxk)
=τ^w2(xl-1xk-1,xkxlxixj)
=τ^w(xl-1xk-1xj-1,xjxkxlxi)
=(xl-1xk-1xj-1xi-1,xixjxkxl).

Nevertheless, the fact that

xl-1xk-1xj-1xi-1=1

in the group Gn(w)=Gn(xixjxkxl) implies that

(1,xixjxkxl)(xl-1xk-1xj-1xi-1,xixjxkxl)

modulo the group w of Peiffer identities for the presentation 𝒫n(w). See, for instance, [21, Lemma 2.4]. In fact, one can verify that the homomorphisms σ^w, τ^w, (θF)^w,u^w are compatible with all of the relations of Γn if we work modulo Peiffer identities. For each cΓn and wΦn, we obtain a commutative diagram that compares the fundamental sequences (2.2) for the presentations 𝒫n(w) and 𝒫n(c(w)); the vertical arrows all represent isomorphisms:

It remains to note that the isomorphism 𝕀w/w𝕀c(w)/c(w) is length-preserving and so it follows that 𝕀w/w is generated by classes of length two identity sequences if and only if the same is true for 𝕀c(w)/c(w). ∎

Corollary 3.3.

If w=xixjxkxlΦn and cΓn=D4×(ZnZn), then

Gn(w)Gn(c(w))

via an isomorphism that preserves shift dynamics. Thus:

  1. The shift on Gn(w) has a nonidentity fixed point if and only if the shift on Gn(c(w)) has a nonidentity fixed point.

  2. The shift action by n on the nonidentity elements of Gn(w) is free if and only if the shift action by n is free on the nonidentity elements of Gn(c(w)).

Proof.

The group isomorphisms of Theorem 3.2 are shift equivariant and so define isomorphisms of left n-sets, where the shift action of n is twisted as necessary via an automorphism un=Aut(n). ∎

In closing this section, we note that a similar discussion can be carried out for positive words of any given length, where positive words of length L in the free group of rank n are acted upon by the group DL×(nn). The actions give rise to group isomorphisms that preserve asphericity status and shift dynamics as in Corollary 3.3. In particular, the isomorphisms of [10, Lemma 2.1] for the case L=3 all arise in this way and so they too preserve shift dynamics.

4 Divisor criteria

To analyze cyclic presentations of the form 𝒫n(x0xjxkxl), it is well known [9] that one can generally restrict to the case where gcd(n,j,k,l)=1. This includes considerations of (combinatorial) asphericity [7, Theorem 4.2], finiteness [6, Lemma 2.4], and shift dynamics [3, Lemma 5.1]. The reason for this is that if c=gcd(n,j,k,l), then the cyclic presentation 𝒫n(x0xjxkxl) is a disjoint union of cyclic subpresentations and G=Gn(x0xjxkxl) decomposes as a free product

Gi=1cGn/c(x0xj/cxk/cxl/c),

where the shift transitively permutes the free factors. In particular, if c1, then G is infinite and the shift has no nonidentity fixed points. A more general but equally checkable criterion involves the secondary divisor

γ=gcd(n,k-2j,l-2k+j,k-2l,j+l),

which arises as follows.

As described in Section 1, the shift action by the cyclic group n on G=Gn(x0xjxkxl) determines the shift extensionE=GθGn, which admits a two-generator two-relator presentation of the form

E=GθGna,x|an,xajxak-jxal-kxa-l,

where the relator W=xajxak-jxal-kxa-l is obtained from the cyclic relation x0xjxkxl via the substitutions xi=aixa-i. The terms k-2j, l-2k+j, -2l+k, j+l in the definition of the secondary divisor γ are the differences of consecutive exponents on the coefficient a in the cyclic relator W=xajxak-jxal-kxa-l for the shift extension. The dependence relation

(k-2j)+(l-2k+j)+(k-2l)+(j+l)=0

implies that any one of the four bracketed terms can be deleted when calculating γ. Thus γ=gcd(n,k-2j,k-2l,j+l), for example. It is not difficult to show that if the primary divisor c=gcd(n,j,k,l)=1, then the secondary divisor γ=gcd(n,k-2j,k-2l,j+l) is equal to 1,2, or 4. We omit the details.

Since the shift θG arises from conjugation by a in the shift extension

E=GθGna,x|an,W,

the fixed point set for any power θGd of the shift is expressible in terms of a centralizer in E:

Fix(θGd)=GCentE(ad).

In the proof of Theorem 4.2 below, we show that if γ1, then the shift extension E admits an amalgamated free product decomposition with an as a vertex group. We then apply the following general result, which is derived from [16, Theorem 4.5].

Lemma 4.1 (Centralizer Lemma).

Suppose that Π=ACB is an amalgamated free product. If aA and gCentΠ(a), then either gA or there exists αA such that aαCα-1.

Theorem 4.2 (Secondary Divisor Criterion).

Given integers n,j,k,l, let G=Gn(x0xjxkxl) and let γ=gcd(n,k-2j,l-2k+j,k-2l,j+l) be the secondary divisor.

  1. If Fix(θGd)1, then γd.

  2. If the secondary divisor γ>1, then G is infinite and Fix(θG)=1.

Proof.

The substitution u=xaj reveals an amalgamated free product decomposition for the shift extension:

E=GθGn=a,x|an,xajxak-jxal-kxa-l
a,u|an,u2ak-2jual-k-jua-l-j
a,b,u|b=aγ,an,u2ak-2jua-(j+l)-(k-2l)ua-(j+l)
nn/γD,

where γ=gcd(n,k-2j,k-2l,j+l) and

D=b,u|bn/γ,u2b(k-2j)/γub(l-k-j)/γub-(j+l)/γ.

We have D/b4, so the amalgamating subgroup b=aγn/γ is a proper subgroup of D. Now suppose that Fix(θGd)1. Then GCentE(ad)1 so using the fact that Ga=1, the Centralizer Lemma implies that adaγ. Since γn, this implies that γd. If we now suppose γ1, then Fix(θG)=1, as above, and moreover n/γ=aγ is a proper subgroup of n=a, so E is infinite, whence the index n subgroup G is infinite. ∎

Corollary 4.3 (Primary Divisor Criterion).

If gcd(n,j,k,l)1, then the cyclically presented group G=Gn(x0xjxkxl) is infinite and Fix(θG)=1. Moreover, if Fix(θGd)1, then gcd(n,j,k,l)d.

Proof.

One simply notes that any common divisor of n,j,k,l is also a divisor of γ. ∎

Example 4.4.

For the following tuples (n,j,k,l),

(10,3,6,1),(12,1,2,9),(16,3,6,1),(20,3,6,1),(24,1,2,15),(24,3,6,1),(24,1,2,19),

the primary divisor gcd(n,j,k,l)=1 gives no information, but the secondary divisor γ=gcd(n,k-2j,k-2l,j+l) is greater than one, so the conclusions of Theorem 4.2 apply. Compare Tables 2 and 3.

5 Conditions A, B, C

Consider a cyclic presentation 𝒫n(x0xjxkxl) for the group G=Gn(x0xjxkxl) as in (1.1). We now use conditions A, B, C to simplify presentations for the shift extension E=GθGn. The following facts are easily verified.

Lemma 5.1.

Given (n,j,k,l), conditions A and C are true if and only if the word w=x0xjxkxl is a cyclic permutation of θFi(w)=xixi+jxi+kxi+l for some 0<i<n, in which case both conditions in A are true and both conditions in C are true. The word w=x0xjxkxl is a proper power if and only if k0 and ljmodn.

We first consider the situation where condition C is true.

Lemma 5.2.

Given (n,j,k,l), let G=Gn(x0xjxkxl) and assume that C is true. There exists an integer p such that the shift extension E=GθGZn admits a presentation of the form

Ea,z|an,z2ak-2pz2a-k-2pa,u,z|an,uakua-k,ua2p=z2,

where G=kerν for a retraction ν:EZn=a satisfying ν(a)=a, ν(u)=1, and ν(z)=ap. If lj+kmodn in C, then we take p=j. If lj-kmodn in C, we take p=-l.

Proof.

If lj+kmodn. then we have

Ea,x|an,xajxak-jxal-kxa-l
a,x|an,xajxak-jxajxa-k-j
a,x,z|an,z=xaj,z2ak-2jz2a-k-2j
a,z,u|an,u=z2a-2j,uakua-k

and so we take p=j in this case. If lj-k, then a similar approach where z=xa-l and u=z2a2l yields the desired result with p=-l. Since ν(a)=a and ν(x)=1, we have ν(u)=1 and ν(z)=ap in both cases. ∎

Using the action of the group Γn described in Section 3, the four cases in condition B can be reduced to a single case as follows.

Lemma 5.3.

If the parameters (n,j,k,l) satisfy condition B, then there exists cΓn such that c(x0xjxkxl)=x0xjxkxl, where

  1. k2jmodn,

  2. gcd(n,j,k,l)=gcd(n,j,k,l),

  3. gcd(n,k-2j,l-2k+j,k-2l,j+l)=gcd(n,k-2j,l-2k+j,k-2l,j+l),

  4. the parameters (n,j,k,l) satisfy A if and only if (n,j,k,l) satisfy A,

  5. the parameters (n,j,k,l) satisfy C if and only if (n,j,k,l) satisfy C.

Proof.

All congruences are interpreted modulo n. If the parameters (n,j,k,l) satisfy B, then k2j or k2l or j+l0 or j+l2k. If k2j, then there is nothing to prove and we take c=1Γn.

If k2l, we take c=τσ, so that x0xjxkxl=τσ(x0xjxkxl)=x0xlxkxj so by setting (n,j,k,l)=(n,l,k,j) we have k=k2l=2j. In addition gcd(n,j,k,l)=gcd(n,l,k,j) and

γ=gcd(n,k-2j,l-2k+j,k-2l,j+l)
=gcd(n,k-2l,j-2k+l,k-2j,l+j)
=γ.

One quickly checks that (n,j,k,l)=(n,l,k,j) satisfies A if and only if (n,j,k,l) satisfies A, and similarly for C.

If j+l0, then we apply elements of Γn as follows:

x0xjxkxl=x0xjxkx-jθFjxjx2jxj+kx0𝜏x0xjx2jxj+k.

Thus by setting (n,j,k,l)=(n,j,2j,j+k) we obtain k=2j=2j. In addition, using j+l0 we have

gcd(n,j,k,l)=gcd(n,j,2j,j+k)=gcd(n,j,k)=gcd(n,j,k,l)

and

γ=gcd(n,k-2j,l-2k+j,k-2l,j+l)
=gcd(n,0,j+k-4j+j,2j-2(j+k),2j+k)
=gcd(n,k-2j,2k,2j+k)

so that

γ=gcd(n,k-2j,l-2k+j,k-2l,j+l)
=gcd(n,k-2j,2k,k+2j,0)
=γ.

Again using j+l0, we note that

2k04j02j-2l02j2l,
2j2l2j2j+2k2k0,
lj+kj+k3jk2jlj-k,
lj-kj+k-jlj+k.

Thus (n,j,k,l) satisfies A (resp. C) if and only if (n,j,k,l) satisfies A (resp. C).

And finally, if j+l2k, then we apply elements of Γn as follows:

x0xjxkxl𝜎xlxkxjx0
τ-1xkxjx0xl
θF-kx0xj-kx-kxl-k.

Thus by setting (n,j,k,l)=(n,j-k,-k,l-k), and by way of reducing to the previous case, we have j+l=j-k+l-kj+l-2k0. In addition, we have gcd(n,j,k,l)=gcd(n,j-k,-k,l-k)=gcd(n,j,k,l) and

γ=gcd(n,k-2j,l-2k+j,k-2l,j+l)
=gcd(n,-k-2(j-k),l-k+2k+j-k,-k-2(l-k),j-k+l-k)
=gcd(n,,k-2j,j+l,k-2l,0)=γ.

Again we note that

2k02k0,
2j2l2j-2k2l-2k2j2l,
lj+kl-kj-k-klj-k,
lj-kl-kj-k+klj+k.

Thus (n,j,k,l) satisfies A (resp. C) if and only if (n,j,k,l) satisfies A (resp. C). Thus the case j+l2k reduces to the case j+l0 and the proof is complete. ∎

Lemma 5.4.

Given a tuple (n,j,k,l), let G=Gn(x0xjxkxl) and assume that k2jmodn as in B. Let γ=gcd(n,k-2j,l-2k+j,k-2l,j+l) be the secondary divisor. With the substitution u=xaj, the standard presentation for the shift extension E=GθGZn takes the form a,u|an,u3αuβ, where α=al-3j and β=a-l-j in aZn. Considering the parameters (n,j,k,l) we have:

  1. Condition A is true if and only if α=β±1.

  2. If A is true, then the element αan has order n/γ.

  3. Condition C is true if and only if α=1 or β=1.

Proof.

Until otherwise stated, all congruences are to be considered modulo n. Using the substitution u=xaj, we transform the standard presentation for the shift extension E as follows:

Ea,x|an,xajxak-jxal-kxa-l
k2ja,x|an,xajxajxal-2jxa-l
u=xaja,u|an,u3al-3jua-l-j.

Thus α=al-3j and β=a-l-j and we verify (a) as follows:

α=β-1l-3jj+l4j0k2j2k0,
α=βl-3j-l-j2l2j.

For statement (b), we claim that if either 2k0 or 2j2l, then the secondary divisor γ has the following simplified description:

γ=gcd(n,k-2j,l-2k+j,j+l)=gcd(n,j+l).

This follows by noting that if k2j and 2k0modn, then working modulo any common divisor of n and j+l, we have l-2k+j0. And similarly, if k2j and 2j2lmodn, then working modulo any common divisor of n and j+l, we have l-2k+jl-3jl-2l-j-l-j0. From this it follows that the order of α=β±1 is n/gcd(n,j+l)=n/γ.

For statement (c),

α=1l3jlj+2jj+k,

as in C. Also j-kj-2j-j, so

β=1j+l0l-jj-k,

as in C. This completes the proof. ∎

6 C(4)-T(4) cases

In this section we consider cases that satisfy the C(4)-T(4) small cancellation conditions. We begin with the observation that with just one exception, finiteness and combinatorial asphericity are mutually exclusive.

Lemma 6.1.

Assume that P=Pn(x0xjxkxl) is combinatorially aspherical and let G=Gn(x0xjxkxl).

  1. If G has nontrivial torsion, then k0 and jlmodn, so A and C are true.

  2. If G is finite, then n=1, so A, B, and C are true, 𝒫=x0|x04, and G4.

Proof.

Since 𝒫 is combinatorially aspherical and G has torsion, [12, Theorem 3] implies that the relator w=x0xjxkxl or one of its shifts must be a proper power in the free group F with basis x0,x1,,xn-1. This implies that k0 and jlmodn so that GGn((x0xj)2). This proves (a). If G is finite, then since G is nontrivial (having a 4 homomorphic image), we know that k0modn and C is true so by Lemma 5.2, the shift extension has a presentation

E=GθGna,u,z|an,u2,ua2p=z2

and so E contains the free product a,u|an,u2n2 as a subgroup. If G is finite, then E must be finite and so n=1, as in (b). ∎

Lemma 6.2.

If B and C are false, then Pn(x0xjxkxl) satisfies the C(4)-T(4) small cancellation conditions and so is combinatorially aspherical.

Proof.

If conditions B and C are false, then j,k-j,l-k,-l are pairwise distinct modulo n. It follows easily from Lemma 2.1 that 𝒫n(x0xjxkxl) satisfies the C(4)-T(4) small cancellation conditions. ∎

Theorem 6.3.

Given (n,j,k,l), if C is true, then the following are equivalent:

  1. A is true.

  2. 𝒫n(x0xjxkxl) satisfies the C(4)-T(4) small cancellation conditions.

  3. 𝒫n(x0xjxkxl) is combinatorially aspherical.

  4. n acts freely via the shift on the nonidentity elements of Gn(x0xjxkxl).

Proof.

Assuming that condition C is true, the word w=x0xjxkxl takes the form w=x0xjxkxj±k. We first show that (i) implies (ii), so suppose that A is true. Then 2k0modn and so w=x0xjxkxj+k. Up to cyclic shifts, the only length two cyclic subwords of w are x0xj and xjxk, so by Lemma 2.1 it suffices to show that neither of these words is a piece. So suppose that the initial subword x0xj of w=x0xjxkxj+k is equal to an initial subword of a cyclic permutation v of a shift θFi(w)=xixi+jxi+kxi+j+k of w. We are to show that w=v. There are four cases to check. If v=θFi(w)=xixi+jxi+kxi+j+k, then i=0 so v=x0xjxkxj+k=w. If v=xi+jxi+kxi+j+kxi, then xi+jxi+k=x0xj, so i+j0 and i+kj (congruences modulo n). Then i+j+kk and the fact that 2k0modn implies that ii+2kj+k, so w=v. Similar arguments apply to the remaining two cyclic permutations of θFi(w) to show that x0xj is not a piece. One argues similarly to show that xjxk is not a piece by comparing the cyclic permutation xjxkxj+kx0 of w with the cyclic permutations v of the shift θFi(w)=xixi+jxi+kxi+j+k. It follows that no length two cyclic subword of w is a piece and so 𝒫n(x0xjxkxl) satisfies C(4)-T(4) by Lemma 2.1.

As previously noted, C(4)-T(4) presentations are combinatorially aspherical, so (ii) implies (iii), and [3, Theorem A] provides that (iii) implies (iv), so it remains to show that if condition A is false and condition C is true, then some power of the shift θG on the group G=Gn(x0xjxkxl) has a nonidentity fixed point. In the presentation for the shift extension E=GθGn from Lemma 5.2 we have two elements a,uE satisfying uakua-k=1, where u1 in E because u1 in E/aEz,u|u2,u=z24. Further, G is the kernel of a retraction ν:En for which ν(u)=1. Thus it follows that 1uGCentE(a2k)=Fix(θG2k). Since A is false, we also have 2k0modn and so the group an does not act freely via the shift. ∎

7 FTT or FFT

In this section we consider the case where C is true and A is false. Note that 𝒫n(x0xjxkxl) is not combinatorially aspherical and the shift action is not free by Theorem 6.3.

Theorem 7.1.

Given (n,j,k,l), let G=Gn(x0xjxkxl). If C is true and A is false, then the following are equivalent:

  1. The shift θG has a nonidentity fixed point.

  2. The group D=a,u|an,uakua-k is finite.

  3. n is odd and gcd(n,k)=1 (i.e. gcd(n,2k)=1).

Proof.

By Lemma 5.2 the shift extension E=GθGn has a presentation

Ea,u,z|an,uakua-k,z2=ua2pD,z|z2=ua2p.

We have the retraction ν:En=a with kernel G=kerν, where ν(a)=a, ν(u)=1, and ν(z)=ap. The restriction ν|D:Dn has cyclically presented kernel H=kerνDGn(u0uk). (See [3, Theorem 2.3].) Letting κ=gcd(n,k), it is routine to verify that

Hi=1κGn(u0u1){i=1κif n is even,i=1κ2if n is odd.

Since D is finite if and only if H if finite, this establishes that (ii)  (iii).

Next we show that (i) (iii). The split extension E=GθGn decomposes as E=Dua2p=z2z=DYz, where Y=ua2p=z2=Dz. If we assume that 1gFix(θG)=GCentE(a), as in (i), then since aD, the Centralizer Lemma 4.1 implies that either gD or there exists αD such that aαYα-1. Either way, we claim that n is odd and gcd(n,k)=1. If gD, then 1gFix(θH) and so gcd(n,k)=1 by an obvious length two analogue of Corollary 4.3. Further, n must be odd because otherwise HGn(u0u1) and θH maps a generator to its inverse and so is fixed point free. So suppose that aαYα-1, where αD. In this case a lies in the normal closure YD of Y in D and so the fact that

D/YDa,u|an,uakua-k,u=a-2p
a|an,a-4p
a|agcd(n,4p)

implies that gcd(n,4p)=1, whence n is odd. Working in D, the element a is a power of αua2pα-1, where αD, so that αua2pα-1CentD(a). Letting b=ak, we have

D=a,u|an,uakua-k
a|anak=bb,u|bn/gcd(n,k),ubub-1
=nBC,

where

C=b,u|bn/gcd(n,k),ubub-1,B=b=akn/gcd(n,k).

From the Centralizer Lemma 4.1 it follows that either αua2pα-1a or else aβBβ-1=B=ak for some βa. Now the first of these conditions cannot be true because D/aDu|u2 so uaD and hence ua2paD. Thus it follows that aak and so gcd(n,k)=1.

It remains to show that if n is odd and gcd(n,k)=1, then θG has a nonidentity fixed point. For this we have 1ukerν=G and uakua-k=1 and hence a2kua-2k=u. Since gcd(n,2k)=1, it follows that

uGCentE(a2k)=GCentE(a)=Fix(θG).

In fact, D2×n2n. ∎

Theorem 7.2.

Given (n,j,k,l), let G=Gn(x0xjxkxl). If C is true and A is false, then the following are equivalent:

  1. E=GθGn4n.

  2. G4.

  3. G is finite.

  4. gcd(n,2k)=1 and either

    1. lj+kmodn and gcd(n,j)=1, or

    2. lj-kmodn and gcd(n,l)=1.

Proof.

The implications (i) (ii) (iii) being obvious, assume that G is finite, so that by Lemma 5.2 the shift extension

E=GθGn
a,u,z|an,uakua-k,z2=ua2p
Dua2p=z2z

is finite, where Da,u|an,uakua-k. Thus D is finite and so gcd(n,2k)=1 by Theorem 7.1, as in (iv). Further,

Ea,z|an,z2ak-2pz2a-k-2p
a,v,z|an,v2a-2k,v=z2ak-2p
a,v|an,v2a-2kva2p-k=z2z.

Since gcd(n,2k)=1, it follows that v is a generator of a,v|an,v2a-2k=v2n and so E2nva2p-k=z2z. Now G is finite if and only if the element va2p-kv2n has order 2n, in which case E4n. It remains to verify that G is finite if and only if one of the itemized conditions in (iv) is true.

Let k be a multiplicative inverse for 2k modulo n so that 2kk1modn. Then v2k=a2kk=a and hence the order of the element

va2p-k=v1+2k2p-2kkv2n

is 2n/g, where

g=gcd(2n,1+4pk-2kk)
=gcd(n,1+4pk-2kk)
=gcd(n,1+4pk-1)=gcd(n,p).

Thus the group G is finite if and only if gcd(n,p)=1. Since C is true, we have lj±kmodn. If lj+k, then by Lemma 5.2 we have p=j, whereas if lj-k, then p=-l. The result follows. ∎

Example 7.3.

If C is true and A is false, then GZ4 can occur regardless of whether B is true or not, as the following examples demonstrate.

  1. If G=Gn(x0x1x2x3) and n>4, then A is false, B and C are true, and G is finite if and only if G4 if and only if n is odd.

  2. If G=Gn(x0x1x3x4) and n>6, then A and B are false, C is true, and G is finite if and only if G4 if and only if gcd(n,6)=1.

8 TTF

In this section we consider the case where A and B are true and C is false.

Theorem 8.1.

Given (n,j,k,l), assume that A and B are true and C is false.

  1. The shift action on G=Gn(x0xjxkxl) is not free, so 𝒫n(x0xjxkxl) is not combinatorially aspherical (by [3, Theorem A]).

  2. The following are equivalent:

    1. G is finite.

    2. The secondary divisor γ=gcd(n,k-2j,j-2k+l,k-2l,j+l)=1.

    3. The shift θG has a nonidentity fixed point.

  3. If G is finite, then G is solvable.

Proof.

Let w=x0xjxkxl. By Lemma 5.3, there exists an element cΓn such that c(w)=x0xjxkxl, where k2jmodn and where the primary and secondary divisors, as well as the status of conditions A and C, are identical for w and for c(w). Further, the group structure, asphericity status, and shift dynamics are identical for the presentations 𝒫n(w) and 𝒫n(c(w)) by Theorem 3.2. Thus we may assume that the parameters (n,j,k,l) are such that k2jmodn, A is true, and C is false.

By Lemma 5.4 the shift extension E=GθGn has a presentation

E=a,u|an,u3αuβ,

where 1α=β±1 has order n/γ in an, and where γ is the secondary divisor. We verify claims (a)–(c) in the cases α=β-1 and α=β separately. Recall that G is the kernel of a retraction ν:Ean.

Suppose first that 1α=β-1 so that we have an amalgamated free product decomposition

Ea,u|an,u3αuα-1nαM,

where M is the group with presentation

Mα,u|αn/γ,u3αuα-1.

By [5, Lemma 2.2], the group M is the split metacyclic group Muα, where u has order μ=3n/γ-(-1)n/γ and uα=1 in E. The order μ is divisible by four and since C is false, Lemma 5.4(c) provides that n/γ>1, so μ>4, which means that the element v=uμ/4M is a nontrivial element of E that lies outside of α=aM. Now

αvα-1=αuμ/4α-1=u-3μ/4=u(1-4)μ/4=vu-μ=v,

and so vCentE(α). Since va, it follows that g=vν(v)-1GCentE(α) is nontrivial. Since α1 in an, we have shown that a nonidentity power of a centralizes a nonidentity element of G in the shift extension E. Thus the shift action on G is not free, as in (a). Since α is a proper subgroup of M, we have that

Gis finiteEis finiteE=Ma=αγ=1,

in which case G is a subgroup of the metacyclic group M and so G is metacyclic, hence solvable, as in (c). If γ=1, then

1gGCentE(α)=GCentE(a)=Fix(θG),

whereas if γ1, then Fix(θG)=1 by Theorem 4.2. This completes the proof of the theorem in the case where β=α-1.

Assume that 1α=β has order n/γ in the amalgamated free product

Ea,u|an,u3αuαnαΔ,

where Δ=α,u|αn/γ,u3αuα. Working in Δ we have

u2αuαu=1,
αu2uαu=1,

which implies that u2CentE(α) and so g=u2ν(u2)-1GCentE(α). Since u2aE, we know that 1gG. Again the fact that C is false means that 1αan, so we conclude that a nonidentity power of a centralizes a nonidentity element of G. Thus the shift action on G is not free, as in (a).

The central quotient

Δ/u2=α,u|αn/γ,u2,u3αuαα,u|αn/γ,u2,(uα)2

is the dihedral group Dn/γ of order 2n/γ and we have the five term homology sequence

H2ΔH2Dn/γu2H1ΔH1Dn/γ0.

The group Δ has a 2×2 relation matrix with determinant 4n/γ0 so H2Δ=0 and |H1Δ|=4n/γ. For the dihedral group we have |H1Dn/γ|-|H2Dn/γ|=2 and so the five term sequence implies that

0=|H2Δ|-|H2Dn/γ|+|u2|-|H1Δ|+|H1Dn/γ|=|u2|+2-4nγ.

Thus Δ is a central extension of the dihedral group of order 2n/γ with cyclic kernel u2 of order -2+4n/γ. In particular, Δ is solvable. As before, α is a proper subgroup of Δ and so

Gis finiteEis finiteE=Δa=αγ=1,

in which case G is a subgroup of the solvable group Δ and so G is solvable, as in (c). If γ=1, then 1gGCentE(α)=GCentE(a)=Fix(θG), whereas if γ1, then Fix(θG)=1 by Theorem 4.2. This completes the proof of the theorem in the case where β=α. ∎

9 FTF

Thus far we have considered seven of the eight possible combinations of conditions A, B, C. The remaining possibility, when B is true and both A and C are false, is the most complex and the most interesting. Here we will apply results from [2, 3] to conclude that in most cases, the cyclic presentation 𝒫n(x0xjxkxl) is combinatorially aspherical (see Theorem 9.2), in which case we know all that we need to know. Fortunately, the remaining nonaspherical cases are few in number and so it is possible to analyze these individually, in some cases via computational means.

Exemplars for isolated (I) and unresolved (U) conditions appear in Tables 2 and 3. In order to deal with cases where the primary divisor c=gcd(n,j,k,l) is not equal to one, we introduced the generalized conditions (I) and (U). The asphericity status for 𝒫n(x0xjxkxl) in types (I) and (U) is entirely determined by that of (I) and (U) because of the following well-known result.

Lemma 9.1 (Free product).

The cyclic presentation Pnd(x0xjdxkdxld) is combinatorially aspherical if and only if the cyclic presentation Pn(x0xjxkxl) is combinatorially aspherical.

Proof.

As in [9], 𝒫nd(x0xjdxkdxld) decomposes as a disjoint union of d subpresentations that can each be identified with 𝒫n(x0xjxkxl), so this follows from [7, Theorem 4.2]. ∎

The main result of this section is the following.

Theorem 9.2.

Assume that gcd(n,j,k,l)=1. Assume that B is true and that both A and C are false. If Pn(x0xjxkxl) is not combinatorially aspherical, then Pn(x0xjxkxl) is of type (I) or of type (U).

Proof.

By Lemma 5.3 and Theorem 3.2, it suffices to prove the theorem in the case where k2jmodn and the parameters (n,j,k,l) satisfy gcd(n,j,k,l)=1 and both A and C are false. Let G=Gn(x0xjxkxl). By Lemma 5.4, the shift extension E=GθGn admits a relative presentation of the form n,u|u3αuβ with na, where

α=al-3jandβ=a-l-j

are such that α,β1 and αβ±1.

Let K=K(n,1) be a connected aspherical CW complex with fundamental group π1Kna and let

M=KSu1c2

be the cellular model of the relative presentation n,u|u3αuβ. Thus the two-cell c2 is attached via a based loop ϕ˙:S1KSu1 representing

[ϕ˙]u3αuβaunπ1(KSu1).

By [3, Theorem 4.1], the cyclic presentation 𝒫n(x0xjxkxl) is combinatorially aspherical if and only if π2M=0.

Since K was chosen to be aspherical, [2, Theorem 4] provides a (partial) classification, in terms of the elements α,βn, of those cases where π2M=0, in which case the cellular model M is (topologically) aspherical. For five exceptional cases labeled E1E5, the asphericity status of M was left unresolved. Since then, Aldwaik and Edjvet [1] have shown that the cellular model is aspherical in the previously unresolved cases E4 and E5. Using computational methods, Williams [24] has shown that π2M0 in case E2. (See Lemma 9.6 below for additional details.) Combining these updated results with [2, Theorem 4], it follows that π2M=0, and so 𝒫n(x0xjxkxl) is combinatorially aspherical, unless one of the following conditions is satisfied. (Here o(-) refers to the order of a group element.)

  1. α=β2 and 4o(β)6,

  2. β=α2 and 4o(α)6,

  3. α=β-2 and o(β)=6,

  4. β=α-2 and o(α)=6,

  5. α=β3 and o(β)=6,

  6. β=α3 and o(α)=6,

  7. {o(α),o(β)}={2,3} and α,β2×3,

  8. 1o(α)+1o(β)+1o(αβ-1)>1.

Using the fact that gcd(n,j,k,l)=1, we show that if any of these conditions hold, then (I) or (U) is satisfied.

First note that it suffices to consider just one from each of the paired conditions (a)–(a), (b)–(b), and (c)–(c). This is because switching the roles of α and β arises from the Γn-transformations

x0xjxkxl𝜎xlxkxjx0
τ-1xkxjx0xl
θF-kx0xj-kx-kxl-k
×-1x0xk-jxkxk-l=x0xjxkxl,

where

α=al-3j=ak-l-3(k-j)=a3j-2k-l=k2ja-j-l=β,
β=a-l-j=al-k-(k-j)=al-2k+j=k2jal-3j=α,

and

2j=2(k-j)=2k-2j=k=k.

Next, condition (e) does not apply because α,βn is finite cyclic and o(α), o(β), o(αβ-1)1; for if 1o(α)+1o(β)+1o(αβ-1)>1, then α,β is a cyclic quotient of either a finite dihedral group, the alternating group A4, the symmetric group S4, or the alternating group A5, which means that α,β has order at most three. But this is contrary to the fact that α,β,αβ±11. Thus we consider cases (a), (b), (c), and (d) individually. All congruences are modulo n.

Case (a): α=β2 and 4o(β)6. Here we have l-3j-2l-2j and 4o(β)=n/gcd(n,j+l)6. That is, we have

j3landn=o(β)gcd(n,j+l).

Since k2j, we also have k6l. Now the fact that

1=gcd(n,j,k,l)=gcd(n,3l,6l,l)

implies that gcd(n,l)=1 and so w=x0x3lx6lxl, where ln is a unit. In particular, w lies in the orbit of the word x0x3x6x1 under the action of the group Γn=Dn×(nn).

Now j+l3l+l so n=o(β)gcd(n,4l)=o(β)gcd(n,4). If o(β)=4, then n=4gcd(n,4) and it follows that n=16, so w is in the Γ16-orbit of the exemplar for the isolated type (I16). See Table 2.

If o(β)=5, then n=5gcd(n,4) and it follows that n=5,10 or 20. If n=5, then x0x3x6x1=x0x3x1x1 so we are in the isolated type (I5). If n=10, then this is the isolated type (I10). If n=20, then this is the isolated type (I20).

If o(β)=6, then n=6gcd(n,4) and it follows that n=24. This is the unresolved type (U24).

Case (b): α=β-2 and o(β)=6. In this case we have l-3j2j+2l and n=6gcd(n,j+l). Thus l-5j and k2j so

1=gcd(n,j,k,l)=gcd(n,j,2j,-5j)

and so we have gcd(n,j)=1. It follows that jn and w=x0xjx2jx-5j. Next, n=6gcd(n,j+l)=6gcd(n,-4j)=6gcd(n,4) and it follows that n=24. So w=x0xjx2jx19j is in the Γ24-orbit of x0x1x2x19, which means that 𝒫n(w) is of the unresolved type (U24′′). See Table 3.

Case (c): α=β3 and o(β)=6. Here l-3j3(-l-j) so that 4l0, and n=6gcd(n,j+l). The fact that gcd(n,j,k,l)=1 and k2j provides that

1=gcd(n,j,2j,l)=gcd(n,j,l)
=gcd(n,j+l,l)=gcd(l,gcd(n,j+l))=gcd(l,n/6).

Since 4l0, it follows that n22l and so n62l. The fact that gcd(l,n/6)=1 thus implies that n/6=1 or 2, so n=6 or 12.

The element α=al-3j has order two in an so n=2gcd(n,l-3j) and so the fact that 6n means that 3l-3j and so 3l.

Now suppose that n=6, so that l0 or 3. If l0, then

1=gcd(n,j,l)=gcd(n,j)

so jn and w=x0xjx2jx0 is a cyclic permutation of x0x0xjx2j as in (I6′′). If l3, then n=6gcd(n,j+l)=6gcd(n,j+3)=6 so gcd(n,j+3)=1, which implies that j2 or 4. Thus w=x0x2x4x3=x0x-4x-2x3 or x0x4x2x3 as in (I6).

Suppose that n=12, so l0,3,6, or 9. We have 12=6gcd(n,j+l), so j+l is even and the fact that gcd(n,j,k,l)=gcd(n,j,l)=1 implies that j and l are both odd, so that l3 or 9. Neither 3 nor 4 is a divisor of j+l, so we have j+l2 or 10. The only arrangements meeting these requirements are (j,l)(7,3),(11,3),(1,9), or (5,9). In each of these four cases we have l9j so w=x0xjx2jx9j, where j12 as in (I12).

Case (d): {o(α),o(β)}={2,3} and α,βZ2×Z3. It suffices to consider the case where o(α)=o(al-3j)=2 and o(β)=o(a-l-j)=3. Here, 2l6j and 3l-3j, so l3l-2l-3j-6j-9j. Coupled with the fact that k2j, we have 1=gcd(n,j,k,l)=gcd(n,j,2j,-9j), so gcd(n,j)=1 and hence jn with w=x0xjx2jx-9j. Now from o(α)=o(al-3j)=2 we have

n=2gcd(n,l-3j)=2gcd(n,-12j)=2gcd(n,12),

which implies that 8 is a divisor of n and so with o(β)=o(a-l-j)=3, it follows that

n=3gcd(n,-l-j)=3gcd(n,8j)=3gcd(n,8)=24.

Thus w=x0xjx2jx15j as in (I24). This completes the proof. ∎

The following four computational results deal with the isolated types.

Lemma 9.3.

The group J4=t,y|t4,y3t2yt is finite metacyclic of order 272. The centralizer CentJ4(t) is cyclic of order 16.

Proof.

Using the GAP [25] commands Size,DerivedSubgroup,AbelianInvariants, and Centralizer, we find that

|J4|=272,J4/J416,J417,CentJ4(t)16.

We remark that J4 is the group J4(4,1) of [4], where it is shown that J4 is a semidirect product 1716. ∎

Lemma 9.4.

The group K=t,u|t5,u3t2ut is finite metacyclic of order 1100. The centralizer CentK(t) is an elementary abelian group of order 25.

Proof.

Computing as in Lemma 9.3, we find that

|K|=1100,K/K20,K55,CentK(t)52.

Lemma 9.5.

The group J6=t,u|t6,u3t3ut2 is finite metacyclic of order 4632. The centralizer CentJ6(t) is cyclic of order 24.

Proof.

Here |J6|=4632, J6/J624, J6193, and CentJ6(t)24. We remark that J6 is the group J6(4,1) of [4], where it is shown that J6 is a semidirect product 19324. ∎

Lemma 9.6.

The group L=t,u|t6,u3t3ut is a nonsolvable group of order 24,530,688=28347132 with second derived subgroup L′′ isomorphic to the simple group PSL(3,3) of order 5616. The centralizer CentL(t) contains an element that does not lie in the normal closure tL of t in L.

Proof.

Although the computations are somewhat lengthy, the order and nonsolvability claims for L can be verified directly in GAP using Size,DerivedSubgroup, and GroupId commands, and this was originally done by Williams [24]. The centralizer claim is more difficult and is resistant to brute force application of the Centralizer command. The group L is the common shift extension for the groups G6(x0x4x2x3) of (I6) and G6(x02x1x2) of (I6′′). One easily checks that

LE=a,x|a6,x2axaxa-2G6(x02x1x2)θ6

via a=t-1 and x=ua-1=ut. A direct coset enumeration using Size returns |E|=24,530,688. Now consider the word

(9.1)v=a3x3axa3xa5xa5xa2xa4xaxa3xa2xa5xa2x.

A second coset enumeration reveals that the quotient group

Q=a,x|a6,x2axaxa-2,ava-1v-1

has the identical order |Q|=24,530,688 and so vCentE(a). The fact that x occurs in the word v with exponent sum 14 implies that the element vE is not in the normal closure aE of a in E because E/aEx|x44. The element vE translates back to L to provide the desired centralizing element. ∎

Remark.

The word v at equation (9.1) was discovered using two separate processes in GAP that rely on a stored coset table for the subgroup an in Ea,x|a6,x2axaxa-2. This table has columns numbered 1 to 4,088,448, one for each (right) coset aw, with column 1 corresponding to the coset a1. The column corresponding to the coset aw has numbered entries corresponding to the cosets awa,awa-1,awx,awx-1 that arise from the action of the generators a,x of E and their inverses. The first process identifies those columns j for which the entry in the first (and second) row is also j, as this corresponds to a coset for which av=ava, so that vCentE(a). The second process uses a spanning tree algorithm to extract a word v from the column index j. The existing GAP command

TracedCosetFpGroup(tablename,word,columnindex)

produces a column index from a word and a column index. (See 47.6-2 in the GAP Manual.) The spanning tree algorithm does the reverse, outputting a word v that solves the equation

j=TracedCosetFpGroup(tablename,v,1),

where j is the column index corresponding to a centralizing word found in the first process. The tree search is such that the generators a and x both occur with only positive exponents in the output word v and so that the exponent sum of x in v is as small as possible. For code and documentation, see [17]. The validity of the algorithm is independently verified by the computations in the proof of Lemma 9.6.

Theorem 9.7.

If Pn(x0xjxkxl) is of type (I), then the shift action by Zn on the nonidentity elements of G=Gn(x0xjxkxl) is not free, so Pn(x0xjxkxl) is not combinatorially aspherical. In types (I5), (I6), and (I6′′), G is finite and the shift θG has a nonidentity fixed point.

Proof.

Let c=gcd(n,j,k,l) be the primary divisor. If (I) is true, the parameters (n/c,j/c,k/c,l/c) satisfy condition (I) and G is the free product Gi=1cH, where H=Gn/c(x0xj/cxk/cxl/c). Further, as described in [3, Lemma 5.1] for the case of length three relators, the shift on H arises as a restriction of a power of the shift on G: θH=θGc|H. It follows that if n/c does not act freely via the shift on H, then n does not act freely via the shift on G. This means that it suffices to prove the theorem under the assumption that condition (I) is true for the parameters (n,j,k,l).

Taking cases in turn, if n=5,10,20, or 16, then for type (In) we consider G=Gn(x0x3x6x1), which has shift extension

Ea,x|an,xa3xa3xa-5xa-1
u=xa3a,u|an,u3a-8ua-4

and where G is the kernel of a retraction ν:En onto the cyclic group a of order n.

If n=5, then Ea,u|a5,u3a2uaK as in Lemma 9.4, so the centralizer CentE(a) has order 25. Since aCentE(a), the retraction ν restricts to a surjection of CentE(a) onto 5 and so Fix(θG)=GCentE(a) is cyclic of order 5.

If n=10 or 20, then

Ea,u|an,u3a-8ua-4na-4=tt,u|t5,u3t2ut,

where t,u|t5,u3t2ut=K as in Lemma 9.4, so the centralizer CentK(t) has order 25. Since t=a-4CentK(t), the retraction ν maps CentK(t) onto the subgroup a4 of order 5 in n and so Fix(θG4)=GCentE(a4) contains a cyclic group of order 5.

If n=16, then

Ea,u|a16,u3a-8ua-416a-4=tt,u|t4,u3t2ut,

where t,u|t4,u3t2ut=J4 as in Lemma 9.3, so the centralizer CentJ4(t) has order 16. Since t=a-4CentJ4(t), the retraction ν maps CentJ4(t) onto the subgroup a4 of order 4 in 16 and so Fix(θG4)=GCentE(a4) contains a cyclic group of order 4.

For the types (I6) and (I6′′), consider the group L=a,u|a6,u3a3ua as in Lemma 9.6, for which we can choose an element wCentL(a) that is not a power of a. For f=2,5, there are retractions νf:La6 satisfying νf(a)=a and νf(u)=af. Using the rewriting scheme described in [3, Theorem 2.3] together with isomorphisms from Theorem 3.2, we have

kerν2G6(u0u2u4u3)×-1G6(u0u4u2u3),
kerν5G6(u0u5u4u0)𝜏G6(u02u5u4)×-1G6(u02u1u2),

so these are the groups from (I6) and (I6′′), respectively. For f=2,5, the element gf=wνf(w)-1 is a nontrivial element of kerνfCentL(a) and so the shifts on the groups kerν2G6(u0u4u2u3) and kerν5G6(u02u1u2) each have nonidentity fixed points.

For type (I12), the group G=G12(x0x1x2x9) is the kernel of a retraction ν:E12 defined on the shift extension

Ea,x|a12,xaxaxa7xa-9
u=xaa,u|a12,u3a6ua-10
12a2=tt,u|t6,u3t3ut,

where L=t,u|t6,u3t3ut is the group from Lemma 9.6. An element w in CentL(t) that is not a power of t=a2 determines a nonidentity element

g=wν(w)-1GCentE(a2)=Fix(θG2).

For type (I24), the group G=G24(x0x1x2x15) is the kernel of a retraction ν:E24 defined on the shift extension

Ea,x|a24,xaxaxa13xa-15
u=xaa,u|a24,u3a12ua8
12a4=tt,u|t6,u3t3ut2,

where J6=t,u|t6,u3t3ut2 is the group from Lemma 9.5. Since t=a4 lies in CentJ6(t), the retraction ν maps CentJ6(t) onto the subgroup a4 of order 6 in 24 and so Fix(θG4)=GCentE(a4) contains a cyclic group of order 4. ∎

Theorem 9.8.

Given (n,j,k,l), let G=Gn(x0xjxkxl). Assume that B is true and A and C are both false.

  1. If 𝒫n(x0xjxkxl) is not of type (U), then the following are equivalent:

    1. 𝒫n(x0xjxkxl) is combinatorially aspherical.

    2. n acts freely via the shift on the nonidentity elements of G.

    3. 𝒫n(x0xjxkxl) is not of type (I).

  2. The following are equivalent:

    1. G is finite.

    2. 𝒫n(x0xjxkxl) is of type (I) or (U) and the secondary divisor γ=1.

    3. 𝒫n(x0xjxkxl) is of type (I5) or (I6) or (I6′′).

    4. The shift θG has a nonidentity fixed point.

Proof.

For part (a), assume that 𝒫n(x0xjxkxl) is not of type (U). The implications (i) (ii) (iii) follow from [3, Theorem A] and Theorem 9.7, respectively. If 𝒫n(x0xjxkxl) is not of type (I) or (U), then 𝒫n/c(x0xj/cxk/cxl/c) is combinatorially aspherical by Theorem 9.2, and so 𝒫n(x0xjxkxl) is combinatorially aspherical by Lemma 9.1. This shows that (iii) (i) and so completes the proof of part (a).

For part (b), assume first that G is finite. The primary and secondary divisors c=gcd(n,j,k,l) and γ are both equal to one by Corollary 4.3 and Theorem 4.2, respectively. Since A is false, it follows that n1 and so 𝒫n(x0xjxkxl) is not combinatorially aspherical by Lemma 6.1. That (I) or (U) must be true then follows from Theorem 9.2. This shows that (i) (ii). The secondary divisor calculations in Example 4.4 show that (ii) (iii). That (iii) (iv) and (iii) (i) follow directly from Theorem 9.7. And finally, if the shift θG has a nonidentity fixed point, then the primary and secondary divisors are both equal to one by Theorem 4.2 and 𝒫n(x0xjxkxl) is not combinatorially aspherical by [3, Theorem A], whence (I) or (U) is true by Theorem 9.2. Thus (iv) (ii) and the proof is complete. ∎

10 Concluding remarks

In this concluding section we distill the proofs of Theorems IV from the results in Sections 69. This includes a discussion of the finite groups Gn(x0xjxkxl) that occur in Theorem II. We conclude with a discussion of the unresolved cases from the asphericity classification Theorem I.

Avoiding the unresolved situation (U), the proof of Theorem I relies on Lemmas 6.1, 6.2 and Theorems 6.3, 8.1 (a), 9.2, and 9.7 to characterize combinatorial asphericity. A combinatorially aspherical presentation is aspherical if and only if there are no proper power or repeated relators up to cyclic permutation and inversion; the characterization of asphericity (still avoiding (U)) therefore follows from Lemma 5.1. If k0 or jlmodn, the presentation 𝒫=𝒫n(x0xjxkxl) has no proper power relators, in which case combinatorial asphericity implies that after removing repeated relators from the two-dimensional model of 𝒫, the group Gn(x0xjxkxl) is the fundamental group of an aspherical two-complex, and so is torsion-free with geometric dimension at most two.

The finiteness classification Theorem II follows from Lemma 6.1 (b) and Theorems 7.2, 8.1 (b), and 9.8 (b). These results also detail the structural claims for the finite groups that occur, as for example in Theorem 8.1 (c), with the exception of those occurring in types (I5), (I6), and (I6′′). The group G5(x0x3x12) of (I5) is a normal subgroup of index 5 in its shift extension, which is the group K of Lemma 9.4. Thus G5(x0x3x12) is metacyclic of order 220 and in fact is a semidirect product 544. As a nonabelian group with cyclic abelianization, this group is not nilpotent. As in the proof of Lemma 9.6, the groups

G1=G6(x0x4x2x3)andG2=G6(x02x1x2)

of (I6) and (I6′′) share the common shift extension L=a,u|a6,u3a3ua of order 24,530,688, which has second derived subgroup L′′ isomorphic to the simple group PSL(3,3) of order 5616. Both G1 and G2 therefore are nonsolvable of order 4,088,448 and contain L′′PSL(3,3). Calculations in GAP show that the abelianizations G1ab8 and G2ab78 are not isomorphic, so G1G2.

For the proof of Theorem III, since cyclic presentations with positive relators are orientable, [3, Theorem A] provides that combinatorial asphericity always implies free action. For the converse, and avoiding the unresolved type (U), in each nonaspherical case we have shown that the shift action is not free. See Lemma 6.2 and Theorems 6.3, 8.1 (a), and 9.7.

For the proof of Theorem IV, for each finite group Gn(x0xjxkxl), either the shift action is trivial (and so has a nonidentity fixed point), as when n=1 and in Lemma 6.1 (b) and Theorem 6.3, or else we have shown that the shift has a nonidentity fixed point, as in Theorems 8.1 (b) and 9.7. If condition C is true and gcd(n,2k)=1, then A is false and Fix(θG)1 by Theorem 7.1. Conversely, if Fix(θG)1, then either n=1 so that G4 is finite (Lemma 6.1 (b)), C is true and gcd(n,2k)=1 (Theorem 7.1), or G is finite (Theorems 8.1 (b), 9.8 (b)).

In order to prove Theorem V, suppose that Fix(θG)1, where G is infinite (for otherwise we may take G=H). By Theorem IV, C is true and gcd(n,2k)=1. By Lemma 5.2, there exists an integer p such that the shift extension

E=GθGn

admits a presentation of the form

Ea,u,z|an,uakua-k,ua2p=z2,

where G=kerν for a retraction ν:En=a satisfying ν(a)=a, ν(u)=1, and ν(z)=ap. Finally, as G is infinite, Theorem 7.2 implies that gcd(n,p)1. Now we have EDua2p=z2z, where D=a,u|an,uakua-kn×2. As in [3, Theorem 2.3], it follows that G=kerν contains the finite group H=kerν|DGn(u0uk)2 and so Fix(θG) contains Fix(θH)=21. It remains to show that Fix(θG)=Fix(θH). So assume that gFix(θG)=CentE(a)G. By the Centralizer Lemma 4.1, either gDG=H or else there exists dD such that adua2pd-1. In the former case we have gFix(θH), as desired, and the latter case cannot occur due to the fact that gcd(n,p)1, so ν(ua2p) generates a proper subgroup of ν(E)=ν(a)=n. Thus Fix(θG)=Fix(θH) and the proof is complete.

We conclude with a discussion of the unresolved type (U), which boils down to consideration of the two exemplar presentations of type (U) from Table 3:

𝒫24(x0x3x6x1)and𝒫24(x0x1x2x19).

In both cases, the secondary divisor is equal to four so by Theorem 4.2, the groups defined by these presentations are infinite and their shifts are fixed point free. However, it is unknown whether these presentations are combinatorially aspherical or whether either shift action is free. It would be of interest to determine, for example, whether these groups are torsion-free. At this point we can note that the groups

G1=G24(x0x3x6x1)andG2=G24(x0x1x2x19)

are not isomorphic. Routine calculations in GAP provide that they abelianize to

G1/G13527andG2/G23373.

It is possible that G1 and G2 are commensurable within a suitable common overgroup, although their shift extensions

E1a,x|a24,xa3xa3xa-5xa-1,
E2a,x|a24,xaxaxa17xa-19

are not isomorphic, having second derived quotients

E1/E1′′95874,andE2/E2′′934734.


Communicated by Alexander Olshanskii


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Received: 2017-10-03
Revised: 2018-04-17
Published Online: 2018-06-15
Published in Print: 2018-09-01

© 2018 Walter de Gruyter GmbH, Berlin/Boston

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