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Publicly Available Published by De Gruyter November 30, 2023

Virtual planar braid groups and permutations

  • Tushar Kanta Naik , Neha Nanda and Mahender Singh EMAIL logo
From the journal Journal of Group Theory
https://doi.org/10.1515/jgth-2023-0010
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Abstract

Twin groups and virtual twin groups are planar analogues of braid groups and virtual braid groups, respectively. These groups play the role of braid groups in the Alexander–Markov correspondence for the theory of stable isotopy classes of immersed circles on orientable surfaces. Motivated by the general idea of Artin and recent work of Bellingeri and Paris [P. Bellingeri and L. Paris, Virtual braids and permutations, Ann. Inst. Fourier (Grenoble) 70 (2020), 3, 1341–1362], we obtain a complete description of homomorphisms between virtual twin groups and symmetric groups, which as an application gives us the precise structure of the automorphism group of the virtual twin group VT n on n ≥ 2 strands. This is achieved by showing the existence of an irreducible right-angled Coxeter group KT n inside VT n . As a by-product, it also follows that the twin group T n embeds inside the virtual twin group VT n , which is an analogue of a similar result for braid groups.

1 Introduction

Doodles on a 2-sphere first appeared in the work [15] of Fenn and Taylor as finite collections of simple closed curves on a 2-sphere without triple or higher intersections. Allowing self-intersections of curves, Khovanov [23] extended the idea to finite collections of closed curves without triple or higher intersections on a closed oriented surface. Khovanov also introduced an analogue of the link group for doodles and constructed several infinite families of doodles whose fundamental groups have infinite centre. Recently, Bartholomew–Fenn–Kamada–Kamada [3, 4] extended the study of doodles to immersed circles without triple or higher intersection points on closed oriented surfaces, which can be thought of as a planar analogue of virtual knot theory with the sphere case corresponding to classical knot theory. It is a natural problem to look for invariants for these topological objects. In [5], colouring of diagrams using a special type of algebra has been used to construct an invariant for virtual doodles. Further, an Alexander type invariant for oriented doodles which vanishes on unlinked doodles with more than one component has been constructed in a recent paper [11].

In tandem with classical knot theory, the study of doodles on surfaces is structured around a suitable group theory framework. The role of groups for doodles on a 2-sphere is played by a class of right-angled Coxeter groups called twin groups (also called planar braid groups), which first appeared in the work of Shabat and Voevodsky [35]. Twin groups have been brought to attention by Khovanov [23] who gave a topological interpretation of these groups. For each n ≥ 2 , the twin group T n is the set of homotopy classes of configurations of 𝑛 arcs in the infinite strip R × [ 0 , 1 ] connecting fixed 𝑛 marked points on each of the parallel boundary lines such that each arc is monotonic and no three arcs have a point in common. The group structure on T n is given by the natural stacking operation. Taking the one point compactification of the plane, one can define the closure of a twin on a 2-sphere analogous to the closure of a geometric braid in the 3-space. While Khovanov proved that every oriented doodle on a 2-sphere is the closure of a twin, an analogue of Markov’s theorem for doodles on a 2-sphere is known due to Gotin [18]. Recent work by Nanda and Singh [33] established Alexander and Markov theorems for the virtual case. It is proved that a new class of groups called virtual twin groups, introduced in [2] and denoted by VT n , plays the role of groups in the theory of virtual doodles. These correspondences can be summarised as

⋃ n ≥ 2 T n / Markov equivalence ↔ homotopy classes of doodles on 2-sphere , ⋃ n ≥ 2 VT n / Markov equivalence ↔ stable equivalence classes of doodles on surfaces .

Analogues of pure braid groups and pure virtual braid groups can be defined for twin groups and virtual twin groups as well. The pure twin group PT n is defined as the kernel of the natural surjection from T n onto the symmetric group S n . The structure of PT n is completely known for small number of strands. Bardakov, Singh and Vesnin [2] proved that PT n is free for n = 3 , 4 and not free for n ≥ 6 . González, León-Medina and Roque [17] showed that PT 5 is a free group of rank 31. A precise description of PT 6 has been obtained by Mostovoy and Roque-Márquez [29] who proved that PT 6 ≅ F 71 * 20 ( Z ⊕ Z ) . Recently, a minimal presentation of PT n for all 𝑛 has been announced by Mostovoy [28]. Automorphisms, (twisted) conjugacy classes and centralisers of involutions in twin groups have been explored in recent work of the authors [30, 31]. In a recent preprint [13], Farley has shown that PT n is always a diagram group, in the sense of Guba and Sapir. It is worth noting that (pure) twin groups are also used by physicists in the study of three-body interactions and topological exchange statistics in one dimension [19, 20]. The pure virtual twin group PVT n is defined analogously as the kernel of the natural surjection from VT n onto S n . A precise presentation of PVT n has been obtained in recent work [32] of the authors, where it has been shown to be an irreducible right-angled Artin group. Further, a complete description of the automorphism group of PVT n has been given.

The present paper contributes to our understanding of virtual twin groups and is motivated by recent work of Bellingeri and Paris [8] on virtual braid groups. We show that there exists an irreducible right-angled Coxeter group KT n inside the virtual twin group VT n and that KT n contains T n . As a consequence, it follows that the twin group T n embeds inside the virtual twin group VT n , which is an analogue of a similar but non-obvious result on embedding of braid groups inside virtual braid groups [14, 16, 22, 24]. The group KT n is further used to obtain a complete description of homomorphisms between virtual twin groups and symmetric groups. It is worth pointing out that the study of homomorphisms from braid groups to symmetric groups goes back to Artin [1] and was later used by Dyer and Grossman [12] to determine the automorphism groups of braid groups. The paper [8] and our paper follow this general idea, although the techniques involved are quite different.

We begin by recalling the definition and topological interpretation of virtual twin groups in Section 2. In Section 3, we give a presentation of KT n showing that it is an irreducible right-angled Coxeter group. More precisely, we prove the following result (Theorem 3.3).

Theorem

For each n ≥ 2 , the group KT n is generated by

S = { α i , j ∣ 1 ≤ i ≠ j ≤ n } ,

where α i , i + 1 = s i and α i + 1 , i = ρ i ⁢ s i ⁢ ρ i . Further, the defining relations are the following:

  1. α i , j 2 = 1 for all 1 ≤ i ≠ j ≤ n ,

  2. α i , j ⁢ α k , l = α k , l ⁢ α i , j for distinct integers i , j , k , l .

In Section 4, we give a complete description of homomorphisms from VT n to S m (Theorem 4.2). Section 5 contains many technical results and occupies the main chunk of this paper. The main result of this section gives a complete description of homomorphisms from S n to VT m (Theorem 5.22). Finally, in Section 6, building upon the preceding sections, we give a complete description of homomorphisms from VT n to VT m . To be more specific, we establish the following result (Theorem 6.2).

Theorem

Let 𝑛 and 𝑚 be integers such that n ≥ m , n ≥ 5 and m ≥ 2 , and let ϕ : VT n → VT m be a homomorphism. Then, up to conjugation of homomorphisms, one of the following assertions holds:

  1. the image of 𝜙 is abelian,

  2. n = m and ϕ ∈ { λ ⁢ π , λ ⁢ θ , ϕ m , ζ ⁢ ϕ m , where ⁢ m ∈ Z } ,

  3. n = m = 6 and ϕ ∈ { λ ⁢ ν ⁢ θ , λ ⁢ ν ⁢ π } .

As a consequence, we obtain the structure of the automorphism group of VT n (Theorem 6.5) and prove that Aut ⁡ ( VT n ) ≅ VT n ⋊ Z 2 for n ≥ 5 . As an application, we deduce that VT n is not co-Hopfian for n ≥ 2 (Corollary 6.4). We conclude the paper by tabulating the status of some structural properties of braid groups, virtual braid groups, twin groups, virtual twin groups and their pure subgroups.

2 Preliminaries

Consider the group VT n with generators { s 1 , s 2 , … , s n − 1 , ρ 1 , ρ 2 , … , ρ n − 1 } and defining relations

s i 2 = 1 for ⁢ i = 1 , 2 , … , n − 1 ,
s i ⁢ s j = s j ⁢ s i for ⁢ | i − j | ≥ 2 ,
(2.1) ρ i 2 = 1 for ⁢ i = 1 , 2 , … , n − 1 ,
(2.2) ρ i ⁢ ρ j = ρ j ⁢ ρ i for ⁢ | i − j | ≥ 2 ,
(2.3) ρ i ⁢ ρ i + 1 ⁢ ρ i = ρ i + 1 ⁢ ρ i ⁢ ρ i + 1 for ⁢ i = 1 , 2 , … , n − 2 ,
ρ i ⁢ s j = s j ⁢ ρ i for ⁢ | i − j | ≥ 2 ,
ρ i ⁢ ρ i + 1 ⁢ s i = s i + 1 ⁢ ρ i ⁢ ρ i + 1 for ⁢ i = 1 , 2 , … , n − 2 .

Elements of the group VT n can be topologically interpreted as follows [33]. Consider a subset 𝐷 of R × [ 0 , 1 ] consisting of 𝑛 intervals called strands with ∂ ( D ) = Q n × { 0 , 1 } , where Q n is a fixed set of 𝑛 points in ℝ. The set 𝐷 is called a virtual twin diagram on 𝑛 strands if it satisfies the following conditions.

  1. Every strand is monotonic; more precisely, each strand maps homeomorphically onto the unit interval [ 0 , 1 ] by the natural projection R × [ 0 , 1 ] → [ 0 , 1 ] .

  2. The set V ⁢ ( D ) of all crossings of the diagram 𝐷 consists of transverse double points of 𝐷, where each crossing has the pre-assigned information of being a real or a virtual crossing as depicted in Figure 1. A virtual crossing is depicted by a crossing encircled with a small circle.

Figure 1 Real and virtual crossings
Figure 1

Real and virtual crossings

We say that the two virtual twin diagrams on 𝑛 strands are equivalent if one can be obtained from the other by a finite sequence of isotopies of the plane and the moves as in Figure 2. Such an equivalence class is called a virtual twin. It turns out that VT n is isomorphic to the group of virtual twins on 𝑛 strands with the operation of concatenation [33, Proposition 3.3]. The generators s i and ρ i of VT n can be represented by configurations shown in Figure 3.

Figure 2 Reidemeister moves for virtual twin diagrams
Figure 2

Reidemeister moves for virtual twin diagrams

Figure 3 Generator s i s_{i} and ρ i \rho_{i}
Figure 3

Generator s i and ρ i

Let τ i denote the transposition ( i , i + 1 ) . The symmetric group S n on 𝑛 symbols is generated by τ 1 , τ 2 , … , τ n − 1 . There is a natural surjective homomorphism π : VT n → S n given by π ⁢ ( s i ) = π ⁢ ( ρ i ) = τ i for all 𝑖. The kernel PVT n of this surjection is called the pure virtual twin group on 𝑛 strands.

There is another surjective group homomorphism θ : VT n → S n given by

θ ⁢ ( s i ) = 1 and θ ⁢ ( ρ i ) = τ i

for all 𝑖. We denote the kernel of this surjection by KT n . This group plays a crucial role in the rest of this paper. The map λ : S n → VT n given by λ ⁢ ( τ i ) = ρ i is a splitting of the short exact sequence

1 → KT n → VT n → S n → 1 ,

and hence VT n = KT n ⋊ S n .

The twin group T n has generators { s 1 , s 2 , … , s n − 1 } and defining relations

s i 2 = 1 for ⁢ i = 1 , 2 , … , n − 1 , s i ⁢ s j = s j ⁢ s i for ⁢ | i − j | ≥ 2 .

It is not clear immediately whether T n is a subgroup of VT n . We show later in Corollary 3.5 that this is indeed the case.

Throughout, x ̂ denotes the inner automorphism of a group 𝐺 induced by an element x ∈ G . To be precise, x ̂ ⁢ ( y ) = x ⁢ y ⁢ x − 1 for all y ∈ G . As usual, the commutator x ⁢ y ⁢ x − 1 ⁢ y − 1 is denoted by [ x , y ] . The centraliser of a subgroup 𝐻 of 𝐺 is denoted by C G ⁡ ( H ) .

3 Presentation of KT n

In this section, we give a presentation of KT n . We use the standard presentation of VT n from Section 2 and the Reidemeister–Schreier method [27, Theorem 2.6]. We take the set

M n = { m 1 , i 1 m 2 , i 2 … m n − 1 , i n − 1 ∣ m k , i k = ρ k ⁢ ρ k − 1 ⁢ … ⁢ ρ i k + 1 for each 1 ≤ k ≤ n − 1 and 0 ≤ i k < k }

as the Schreier system of coset representatives of KT n in VT n . We set m k ⁢ k = 1 for 1 ≤ k ≤ n − 1 . For an element w ∈ VT n , let w ̄ denote the unique coset representative of the coset of 𝑤 in the Schreier set M n . By Reidemeister–Schreier method, the group KT n is generated by the set

{ γ ⁢ ( μ , a ) = ( μ ⁢ a ) ⁢ ( μ ⁢ a ̄ ) − 1 ∣ μ ∈ M n ⁡ and ⁢ a ∈ { s 1 , … , s n − 1 , ρ 1 , … , ρ n − 1 } }

with defining relations

{ τ ⁢ ( μ ⁢ r ⁢ μ − 1 ) ∣ μ ∈ M n ⁡ and ⁢ r ⁢ is a defining relation in ⁢ VT n } ,

where 𝜏 is the rewriting process. More precisely, we have

τ ⁢ ( g ) = γ ⁢ ( 1 , g 1 ) ⁢ γ ⁢ ( g 1 ̄ , g 2 ) ⁢ … ⁢ γ ⁢ ( g 1 ⁢ g 2 ⁢ … ⁢ g k − 1 ̄ , g k )

for an element g = g 1 ⁢ g 2 ⁢ … ⁢ g k ∈ VT n .

For each 1 ≤ i ≤ n − 1 , we set

α i , i + 1 = s i and α i + 1 , i = ρ i ⁢ s i ⁢ ρ i .

For each 1 ≤ i < j ≤ n and j ≠ i + 1 , we set

α i , j = ( ρ j − 1 ⁢ ρ j − 2 ⁢ … ⁢ ρ i + 1 ) ⁢ α i , i + 1 ⁢ ( ρ i + 1 ⁢ … ⁢ ρ j − 2 ⁢ ρ j − 1 ) = ( ρ j − 1 ⁢ ρ j − 2 ⁢ … ⁢ ρ i + 1 ) ⁢ s i ⁢ ( ρ i + 1 ⁢ … ⁢ ρ j − 2 ⁢ ρ j − 1 ) ,
α j , i = ( ρ j − 1 ⁢ ρ j − 2 ⁢ … ⁢ ρ i + 1 ) ⁢ α i + 1 , i ⁢ ( ρ i + 1 ⁢ … ⁢ ρ j − 2 ⁢ ρ j − 1 ) = ( ρ j − 1 ⁢ ρ j − 2 ⁢ … ⁢ ρ i + 1 ) ⁢ ρ i ⁢ s i ⁢ ρ i ⁢ ( ρ i + 1 ⁢ … ⁢ ρ j − 2 ⁢ ρ j − 1 ) .
For each n ≥ 2 , let us define

S = { α i , j ∣ 1 ≤ i ≠ j ≤ n } .

For each pair ( i , j ) with 1 ≤ i < j ≤ n , the generators α i , j and α j , i can be topologically represented as in Figure 4.

Figure 4 Generators α i , j \alpha_{i,j} and α j , i \alpha_{j,i} of KT n \mathrm{KT}_{n}
Figure 4

Generators α i , j and α j , i of KT n

Theorem 3.1

For each n ≥ 2 , the group KT n is generated by 𝒮.

Proof

The case n = 2 is immediate, and hence we assume n ≥ 3 . Note that KT n is generated by elements γ ⁢ ( μ , a ) = ( μ ⁢ a ) ⁢ ( μ ⁢ a ̄ ) − 1 , where μ ∈ M n and

a ∈ { s 1 , … , s n − 1 , ρ 1 , … , ρ n − 1 } .

Let w = w 1 ⁢ w 2 ⁢ … ⁢ w k , where w i ∈ { s 1 , … , s n − 1 , ρ 1 , … , ρ n − 1 } . Then we have w ̄ = w 1 * ⁢ w 2 * ⁢ … ⁢ w k * , where

w i * = { w i if ⁢ w i ∈ { ρ 1 , … , ρ n − 1 } , 1 if ⁢ w i ∈ { s 1 , … , s n − 1 } .

Thus, for each μ ∈ M n and 1 ≤ i ≤ n − 1 , we have

γ ⁢ ( μ , ρ i ) = ( μ ⁢ ρ i ) ⁢ ( μ ⁢ ρ i ̄ ) − 1 = ( μ ⁢ ρ i ) ⁢ ( μ ⁢ ρ i ) − 1 = 1 , γ ⁢ ( μ , s i ) = ( μ ⁢ s i ) ⁢ ( μ ⁢ s i ̄ ) − 1 = μ ⁢ s i ⁢ μ − 1 = μ ⁢ α i , i + 1 ⁢ μ − 1 .

We claim that each γ ⁢ ( μ , s i ) lies in 𝒮 and that the conjugation action of

⟨ ρ 1 , … , ρ n − 1 ⟩ ≅ S n

on 𝒮 is transitive.

First consider α i , i + 1 and α i + 1 , i for a fixed 1 ≤ i ≤ n − 1 .

  • [leftmargin=*]

  • 1 ≤ k ≤ i − 2 or i + 2 ≤ k ≤ n − 1 :

    ρ k ⁢ α i , i + 1 ⁢ ρ k = α i , i + 1 , ρ k ⁢ α i + 1 , i ⁢ ρ k = α i + 1 , i .

  • k = i − 1 :

    ρ k ⁢ α i , i + 1 ⁢ ρ k = ρ i − 1 ⁢ s i ⁢ ρ i − 1 = ρ i ⁢ s i − 1 ⁢ ρ i = α i − 1 , i + 1 , ρ k ⁢ α i + 1 , i ⁢ ρ k = ρ i ⁢ ρ i − 1 ⁢ ( ρ i ⁢ ρ i − 1 ⁢ s i ⁢ ρ i − 1 ⁢ ρ i ) ⁢ ρ i − 1 ⁢ ρ i = α i + 1 , i − 1 .

  • k = i :

    ρ k ⁢ α i , i + 1 ⁢ ρ k = ρ i ⁢ s i ⁢ ρ i = α i + 1 , i , ρ k ⁢ α i + 1 , i ⁢ ρ k = ρ i ⁢ ρ i ⁢ s i ⁢ ρ i ⁢ ρ i = s i = α i , i + 1 .

  • k = i + 1 :

    ρ k ⁢ α i , i + 1 ⁢ ρ k = ρ i + 1 ⁢ s i ⁢ ρ i + 1 = α i , i + 2 , ρ k ⁢ α i + 1 , i ⁢ ρ k = ρ i + 1 ⁢ ρ i ⁢ s i ⁢ ρ i ⁢ ρ i + 1 = α i + 2 , i .

Next, we consider α i , j and α j , i for some fixed 1 ≤ i < j ≤ n with j ≠ i + 1 .

  • 1 ≤ k ≤ i − 2 or j + 1 ≤ k ≤ n − 1 :

    ρ k ⁢ α i , j ⁢ ρ k = α i , j , ρ k ⁢ α j , i ⁢ ρ k = α j , i .

  • k = i − 1 :

    ρ k ⁢ α i , j ⁢ ρ k = ρ i − 1 ⁢ ( ρ j − 1 ⁢ ρ j − 2 ⁢ … ⁢ ρ i + 1 ) ⁢ s i ⁢ ( ρ i + 1 ⁢ … ⁢ ρ j − 2 ⁢ ρ j − 1 ) ⁢ ρ i − 1 = ( ρ j − 1 ⁢ ρ j − 2 ⁢ … ⁢ ρ i + 1 ) ⁢ ρ i ⁢ s i − 1 ⁢ ρ i ⁢ ( ρ i + 1 ⁢ … ⁢ ρ j − 2 ⁢ ρ j − 1 ) = α i − 1 , j ,
    ρ k ⁢ α j , i ⁢ ρ k = ρ i − 1 ⁢ ( ρ j − 1 ⁢ ρ j − 2 ⁢ … ⁢ ρ i + 1 ) ⁢ ρ i ⁢ s i ⁢ ρ i ⁢ ( ρ i + 1 ⁢ … ⁢ ρ j − 2 ⁢ ρ j − 1 ) ⁢ ρ i − 1 = ( ρ j − 1 ⁢ ρ j − 2 ⁢ … ⁢ ρ i + 1 ) ⁢ ρ i ⁢ ρ i − 1 ⁢ ( ρ i ⁢ ρ i − 1 ⁢ s i ⁢ ρ i − 1 ⁢ ρ i ) ⁢ ρ i − 1 ⁢ ρ i ( ρ i + 1 ⁢ … ⁢ ρ j − 2 ⁢ ρ j − 1 ) = ( ρ j − 1 ⁢ ρ j − 2 ⁢ … ⁢ ρ i + 1 ) ⁢ ρ i ⁢ ρ i − 1 ⁢ s i − 1 ⁢ ρ i − 1 ⁢ ρ i ⁢ ( ρ i + 1 ⁢ … ⁢ ρ j − 2 ⁢ ρ j − 1 ) = α j , i − 1 .

  • k = i :

    ρ k ⁢ α i , j ⁢ ρ k = ρ i ⁢ ( ρ j − 1 ⁢ ρ j − 2 ⁢ … ⁢ ρ i + 1 ) ⁢ s i ⁢ ( ρ i + 1 ⁢ … ⁢ ρ j − 2 ⁢ ρ j − 1 ) ⁢ ρ i = ( ρ j − 1 ⁢ ρ j − 2 ⁢ … ⁢ ρ i + 2 ) ⁢ ρ i ⁢ ρ i + 1 ⁢ s i ⁢ ρ i + 1 ⁢ ρ i ⁢ ( ρ i + 2 ⁢ … ⁢ ρ j − 2 ⁢ ρ j − 1 ) = ( ρ j − 1 ⁢ ρ j − 2 ⁢ … ⁢ ρ i + 2 ) ⁢ s i + 1 ⁢ ( ρ i + 2 ⁢ … ⁢ ρ j − 2 ⁢ ρ j − 1 ) = α i + 1 , j ,
    ρ k ⁢ α j , i ⁢ ρ k = ρ i ⁢ ( ρ j − 1 ⁢ ρ j − 2 ⁢ … ⁢ ρ i + 1 ) ⁢ ρ i ⁢ s i ⁢ ρ i ⁢ ( ρ i + 1 ⁢ … ⁢ ρ j − 2 ⁢ ρ j − 1 ) ⁢ ρ i = ( ρ j − 1 ⁢ ρ j − 2 ⁢ … ⁢ ρ i + 2 ) ⁢ ρ i ⁢ ρ i + 1 ⁢ ρ i ⁢ s i ⁢ ρ i ⁢ ρ i + 1 ⁢ ρ i ⁢ ( ρ i + 2 ⁢ … ⁢ ρ j − 2 ⁢ ρ j − 1 ) = ( ρ j − 1 ⁢ ρ j − 2 ⁢ … ⁢ ρ i + 2 ) ⁢ ρ i + 1 ⁢ s i + 1 ⁢ ρ i + 1 ⁢ ( ρ i + 2 ⁢ … ⁢ ρ j − 2 ⁢ ρ j − 1 ) = α j , i + 1 .

  • k = j − 1 :

    ρ k ⁢ α i , j ⁢ ρ k = ρ j − 1 ⁢ ρ j − 1 ⁢ … ⁢ ρ k + 1 ⁢ ρ k ⁢ … ⁢ ρ i + 1 ⁢ s i ⁢ ρ i + 1 ⁢ … ⁢ ρ k ⁢ ρ k + 1 ⁢ … ⁢ ρ j − 1 ⁢ ρ j − 1 = ρ j − 2 ⁢ … ⁢ ρ k + 1 ⁢ ρ k ⁢ … ⁢ ρ i + 1 ⁢ s i ⁢ ρ i + 1 ⁢ … ⁢ ρ k ⁢ ρ k + 1 ⁢ … ⁢ ρ j − 2 = α i , j − 1
    ρ k ⁢ α j , i ⁢ ρ k = ρ j − 1 ⁢ ρ j − 1 ⁢ … ⁢ ρ k + 1 ⁢ ρ k ⁢ … ⁢ ρ i + 1 ⁢ ρ i ⁢ s i ⁢ ρ i ⁢ ρ i + 1 ⁢ … ⁢ ρ k ⁢ ρ k + 1 ⁢ … ⁢ ρ j − 1 ⁢ ρ j − 1 = ρ j − 2 ⁢ … ⁢ ρ k + 1 ⁢ ρ k ⁢ … ⁢ ρ i + 1 ⁢ ρ i ⁢ s i ⁢ ρ i ⁢ ρ i + 1 ⁢ … ⁢ ρ k ⁢ ρ k + 1 ⁢ … ⁢ ρ j − 2 = α j − 1 , i .

  • k = j :

    ρ k ⁢ α i , j ⁢ ρ k = α i , j + 1 , ρ k ⁢ α j , i ⁢ ρ k = α j + 1 , i .

  • i + 1 ≤ k ≤ j − 2 :

    ρ k ⁢ α i , j ⁢ ρ k = ρ k ⁢ ρ j − 1 ⁢ … ⁢ ρ k + 1 ⁢ ρ k ⁢ … ⁢ ρ i + 1 ⁢ s i ⁢ ρ i + 1 ⁢ … ⁢ ρ k ⁢ ρ k + 1 ⁢ … ⁢ ρ j − 1 ⁢ ρ k = ( ρ j − 1 ⁢ … ⁢ ρ k + 2 ) ⁢ ρ k ⁢ ρ k + 1 ⁢ ρ k ⁢ ( ρ k − 1 ⁢ … ⁢ ρ i + 1 ) ⁢ s i ( ρ i + 1 ⁢ … ⁢ ρ k − 1 ) ⁢ ρ k ⁢ ρ k + 1 ⁢ ρ k ⁢ ( ρ k + 2 ⁢ … ⁢ ρ j − 1 ) = ( ρ j − 1 ⁢ … ⁢ ρ k + 2 ) ⁢ ρ k + 1 ⁢ ρ k ⁢ ρ k + 1 ⁢ ( ρ k − 1 ⁢ … ⁢ ρ i + 1 ) ⁢ s i ( ρ i + 1 ⁢ … ⁢ ρ k − 1 ) ⁢ ρ k + 1 ⁢ ρ k ⁢ ρ k + 1 ⁢ ( ρ k + 2 ⁢ … ⁢ ρ j − 1 ) = ( ρ j − 1 ⁢ … ⁢ ρ k + 2 ⁢ ρ k + 1 ⁢ ρ k ⁢ ρ k − 1 ⁢ … ⁢ ρ i + 1 ) ⁢ ρ k + 1 ⁢ s i ⁢ ρ k + 1 ( ρ i + 1 ⁢ … ⁢ ρ k − 1 ⁢ ρ k ⁢ ρ k + 1 ⁢ ρ k + 2 ⁢ … ⁢ ρ j − 1 ) = ( ρ j − 1 ⁢ … ⁢ ρ k + 2 ⁢ ρ k + 1 ⁢ ρ k ⁢ ρ k − 1 ⁢ … ⁢ ρ i + 1 ) ⁢ s i ( ρ i + 1 ⁢ … ⁢ ρ k − 1 ⁢ ρ k ⁢ ρ k + 1 ⁢ ρ k + 2 ⁢ … ⁢ ρ j − 1 ) = α i , j ,
    ρ k ⁢ α j , i ⁢ ρ k = ρ k ⁢ ρ j − 1 ⁢ … ⁢ ρ k + 1 ⁢ ρ k ⁢ … ⁢ ρ i + 1 ⁢ ρ i ⁢ s i ⁢ ρ i ⁢ ρ i + 1 ⁢ … ⁢ ρ k ⁢ ρ k + 1 ⁢ … ⁢ ρ j − 1 ⁢ ρ k = ( ρ j − 1 ⁢ … ⁢ ρ k + 2 ) ⁢ ρ k ⁢ ρ k + 1 ⁢ ρ k ⁢ ( ρ k − 1 ⁢ … ⁢ ρ i + 1 ) ⁢ ρ i ⁢ s i ⁢ ρ i ( ρ i + 1 ⁢ … ⁢ ρ k − 1 ) ⁢ ρ k ⁢ ρ k + 1 ⁢ ρ k ⁢ ( ρ k + 2 ⁢ … ⁢ ρ j − 1 ) = ( ρ j − 1 ⁢ … ⁢ ρ k + 2 ⁢ ρ k + 1 ⁢ ρ k ⁢ ρ k − 1 ⁢ … ⁢ ρ i + 1 ) ⁢ ρ k + 1 ⁢ ρ i ⁢ s i ⁢ ρ i ⁢ ρ k + 1 ( ρ i + 1 ⁢ … ⁢ ρ k − 1 ⁢ ρ k ⁢ ρ k + 1 ⁢ ρ k + 2 ⁢ … ⁢ ρ j − 1 ) = ( ρ j − 1 ⁢ … ⁢ ρ k + 2 ⁢ ρ k + 1 ⁢ ρ k ⁢ ρ k − 1 ⁢ … ⁢ ρ i + 1 ) ⁢ ρ i ⁢ s i ⁢ ρ i ( ρ i + 1 ⁢ … ⁢ ρ k − 1 ⁢ ρ k ⁢ ρ k + 1 ⁢ ρ k + 2 ⁢ … ⁢ ρ j − 1 ) = α j , i .

Hence, each generator γ ⁢ ( μ , s i ) lies in 𝒮. Conversely, if 1 ≤ i < j ≤ n , then we see that conjugation by ( ρ i − 1 ⁢ ρ i − 2 ⁢ … ⁢ ρ 2 ⁢ ρ 1 ) ⁢ ( ρ j − 1 ⁢ ρ j − 2 ⁢ … ⁢ ρ 3 ⁢ ρ 2 ) maps α 1 , 2 and α 2 , 1 to α i , j and α j , i , respectively, whereas conjugation by ρ 1 maps α 1 , 2 to α 2 , 1 . That is, the conjugation action of S n on the set 𝒮 is transitive. Hence, we have proved that 𝒮 generates KT n . ∎

Remark 3.2

We can summarise the (left) action of S n on the set 𝒮 as

τ k ⋅ α i , j := ρ k ⁢ α i , j ⁢ ρ k = α ρ k ⁢ ( i ) , ρ k ⁢ ( j )

for every 1 ≤ i ≠ j ≤ n and 1 ≤ k ≤ n − 1 .

Theorem 3.3

For each n ≥ 2 , the group KT n is generated by 𝒮 with the following defining relations:

  1. α i , j 2 = 1 for all 1 ≤ i ≠ j ≤ n , and

  2. α i , j ⁢ α k , l = α k , l ⁢ α i , j for distinct integers i , j , k , l .

Proof

Theorem 3.1 already shows that 𝒮 generates KT n . The defining relations are given by τ ⁢ ( μ ⁢ r ⁢ μ − 1 ) , where 𝜏 is the rewriting process, μ ∈ M n and 𝑟 is a defining relation in VT n .

Let us take μ = ρ i 1 ⁢ ρ i 2 ⁢ … ⁢ ρ i k ∈ M n and g = g 1 ⁢ g 2 ⁢ … ⁢ g t a relation of VT n . Note that, since γ ⁢ ( μ , ρ i ) = 1 for all 𝑖, we have

τ ⁢ ( μ ⁢ g ⁢ μ − 1 ) = γ ⁢ ( μ ̄ , g 1 ) ⁢ γ ⁢ ( μ ⁢ g 1 ̄ , g 2 ) ⁢ … ⁢ γ ⁢ ( μ ⁢ g 1 ⁢ g 2 ⁢ … ⁢ g t − 1 ̄ , g t ) .

Further, no non-trivial relations for KT n can be obtained from relations (2.1)–(2.3) of VT n . Next, we consider the remaining relations one by one.

  • First we consider the relations s i 2 = 1 for 1 ≤ i ≤ n − 1 . In this case, we have

    τ ⁢ ( μ ⁢ s i 2 ⁢ μ − 1 ) = γ ⁢ ( μ ̄ , s i ) ⁢ γ ⁢ ( μ ⁢ s i ̄ , s i ) = γ ⁢ ( μ , s i ) ⁢ γ ⁢ ( μ , s i ) = ( μ ⁢ α i , i + 1 ⁢ μ − 1 ) 2 .

    Since the conjugation action of S n on 𝒮 is transitive, it follows that all the generators are involutions.

  • Next we consider the relations ( s i ⁢ ρ j ) 2 = 1 for | i − j | > 1 . We have

    τ ⁢ ( μ ⁢ s i ⁢ ρ j ⁢ s i ⁢ ρ j ⁢ μ − 1 ) = γ ⁢ ( μ ̄ , s i ) ⁢ γ ⁢ ( μ ⁢ s i ⁢ ρ j ̄ , s i ) = γ ⁢ ( μ , s i ) ⁢ γ ⁢ ( μ ⁢ ρ j , s i ) = ( μ ⁢ s i ⁢ μ − 1 ) ⁢ ( μ ⁢ ρ j ⁢ s i ⁢ ρ j ⁢ μ − 1 ) = ( μ ⁢ s i ⁢ μ − 1 ) 2 ,

    which again shows that the generators are of order two.

  • Now, we consider the relations ρ i ⁢ s i + 1 ⁢ ρ i ⁢ ρ i + 1 ⁢ s i ⁢ ρ i + 1 = 1 , where 1 ≤ i ≤ n − 2 . Computing

    τ ⁢ ( μ ⁢ ρ i ⁢ s i + 1 ⁢ ρ i ⁢ ρ i + 1 ⁢ s i ⁢ ρ i + 1 ⁢ μ − 1 ) = γ ⁢ ( μ ⁢ ρ i ̄ , s i + 1 ) ⁢ γ ⁢ ( μ ⁢ ρ i ⁢ s i + 1 ⁢ ρ i ⁢ ρ i + 1 ̄ , s i ) = γ ⁢ ( μ ⁢ ρ i , s i + 1 ) ⁢ γ ⁢ ( μ ⁢ ρ i + 1 , s i ) = ( μ ⁢ ρ i ⁢ s i + 1 ⁢ ρ i ⁢ μ − 1 ) ⁢ ( μ ⁢ ρ i + 1 ⁢ s i ⁢ ρ i + 1 ⁢ μ − 1 ) = ( μ ⁢ ρ i ⁢ α i + 1 , i + 2 ⁢ ρ i ⁢ μ − 1 ) ⁢ ( μ ⁢ ρ i + 1 ⁢ α i , i + 1 ⁢ ρ i + 1 ⁢ μ − 1 ) = ( μ ⁢ α i , i + 2 ⁢ μ − 1 ) ⁢ ( μ ⁢ α i , i + 2 ⁢ μ − 1 ) ,

    we see that the generators are of order two.

  • Finally, we consider the relations ( s i ⁢ s j ) 2 = 1 for | i − j | > 1 . If μ = 1 , then we have

    τ ⁢ ( s i ⁢ s j ⁢ s i ⁢ s j ) = γ ⁢ ( 1 , s i ) ⁢ γ ⁢ ( s i ̄ , s j ) ⁢ γ ⁢ ( s i ⁢ s j ̄ , s i ) ⁢ γ ⁢ ( s i ⁢ s j ⁢ s i ̄ , s j ) = γ ⁢ ( 1 , s i ) ⁢ γ ⁢ ( 1 , s j ) ⁢ γ ⁢ ( 1 , s i ) ⁢ γ ⁢ ( 1 , s j ) = ( α i , i + 1 ⁢ α j , j + 1 ) 2 .

    For μ ≠ 1 , we have

    (3.1) τ ⁢ ( μ ⁢ s i ⁢ s j ⁢ s i ⁢ s j ⁢ μ − 1 ) = γ ⁢ ( μ ̄ , s i ) ⁢ γ ⁢ ( μ ⁢ s i ̄ , s j ) ⁢ γ ⁢ ( μ ⁢ s i ⁢ s j ̄ , s i ) ⁢ γ ⁢ ( μ ⁢ s i ⁢ s j ⁢ s i ̄ , s j ) = γ ⁢ ( μ , s i ) ⁢ γ ⁢ ( μ , s j ) ⁢ γ ⁢ ( μ , s i ) ⁢ γ ⁢ ( μ , s j ) = ( μ ⁢ s i ⁢ μ − 1 ) ⁢ ( μ ⁢ s j ⁢ μ − 1 ) ⁢ ( μ ⁢ s i ⁢ μ − 1 ) ⁢ ( μ ⁢ s j ⁢ μ − 1 ) = ( ( μ ⁢ α i , i + 1 ⁢ μ − 1 ) ⁢ ( μ ⁢ α j , j + 1 ⁢ μ − 1 ) ) 2 .

Note that, for n = 2 , 3 , these types of relations do not occur. For n ≥ 4 , we set

D = { ( α i , j , α k , l ) ∣ i , j , k , l ⁢ are distinct integers between ⁢ 1 ⁢ and ⁢ n } .

Remark 3.2 and the theory of symmetric groups gives an induced transitive action of S n on 𝒟 given by

ρ ⋅ ( α i , j , α k , l ) = ( α ρ ⁢ ( i ) , ρ ⁢ ( j ) , α ρ ⁢ ( k ) , ρ ⁢ ( l ) )

for all ρ ∈ S n . Thus, the defining relations of KT n obtained from (3.1) are precisely of the form α i , j ⁢ α k , l = α k , l ⁢ α i , j , where i , j , k , l are distinct integers between 1 and 𝑛. This completes the proof of the theorem. ∎

Corollary 3.4

For each n ≥ 2 , the group KT n is an irreducible right-angled Coxeter group of rank n ⁢ ( n − 1 ) with trivial centre.

Proof

That KT n is a right-angled Coxeter group of rank n ⁢ ( n − 1 ) follows from Theorem 3.3. Irreducibility follows from the corresponding Coxeter graph of KT n . By Bourbaki [9, p. 137], the centre of an infinite irreducible Coxeter group is trivial. ∎

Recall that, if ( W , S ) is a Coxeter system and 𝑋 a subset of 𝑆, then the subgroup of 𝑊 generated by 𝑋 is called a standard parabolic subgroup of 𝑊 and is denoted by W ⁢ [ X ] . It is well known that the Artin braid group B n embeds inside the virtual braid group VB n (see [14, 16, 22, 24]). As another consequence of Theorem 3.3, we obtain a similar result for twin and virtual twin groups.

Corollary 3.5

T n is a subgroup of VT n for each n ≥ 2 .

Proof

The standard parabolic subgroup of KT n generated by

{ α i , i + 1 = s i ∣ 1 ≤ i ≤ n − 1 }

is precisely the twin group T n . Hence, T n sits inside KT n , and consequently inside VT n . ∎

Recall that the pure twin group PT n is the kernel of the natural surjection from T n onto S n given by s i ↦ τ i . It follows from Corollary 3.5 that PT n is a subgroup of PVT n , where it has been proved recently that PVT n is a right-angled Artin group [32, Corollary 3.4]. As noted in the introduction, PT n is free for n = 3 , 4 , 5 , and PT 6 is isomorphic to the free product of F 71 and 20 copies of Z ⊕ Z , which are all right-angled Artin groups. Though a presentation of PT n has been given in [28], it is not clear whether PT n is a right-angled Artin group for n ≥ 7 , but we believe that it is the case.

Conjecture 3.6

PT n is a right-angled Artin group for each n ≥ 3 .

4 Homomorphisms from VT n to S m

A group homomorphism ψ : G → H is said to be abelian if ψ ⁢ ( G ) is an abelian subgroup of 𝐻. Two homomorphisms ψ 1 , ψ 2 : G → H are said to be conjugate if there exists x ∈ H such that ψ 2 = x ̂ ⁢ ψ 1 , where x ̂ is the inner automorphism induced by 𝑥, as defined at the end of Section 2. It is to be noted that Out ⁡ ( S n ) is trivial for all n ≠ 6 and Out ⁡ ( S 6 ) ≅ Z 2 . The latter group is generated by the class of a non-inner automorphism ν : S 6 → S 6 of order two.

The following result is well known from the works of Artin [1] and Lin [26, 25] and is crucial for the proof of Theorem 4.2.

Proposition 4.1

Let n , m be integers such that n ≥ m , n ≥ 5 and m ≥ 2 . Let ϕ : S n → S m be a homomorphism. Then, up to conjugation of homomorphisms, one of the following assertions holds:

  1. 𝜙 is abelian,

  2. n = m and ϕ = id ,

  3. n = m = 6 and ϕ = ν .

Let θ : VT n → S n and λ : S n → VT n be as defined in Section 2. We prove the following result.

Theorem 4.2

Let 𝑛 and 𝑚 be integers such that n ≥ m , n ≥ 5 and m ≥ 2 . Let ψ : VT n → S m be a homomorphism. Then, up to conjugation of homomorphisms, one of the following assertions holds:

  1. 𝜓 is abelian,

  2. n = m and ψ = π or 𝜃,

  3. n = m = 6 and ψ = ν ⁢ π or ν ⁢ θ .

Proof

Consider the composition S n → λ VB n → ψ S m . By Proposition 4.1, one of the following holds for ψ ⁢ λ :

  1. ψ ⁢ λ is abelian,

  2. n = m and ψ ⁢ λ = id ,

  3. n = m = 6 and ψ ⁢ λ = ν .

Case (1): Let ψ ⁢ λ be abelian. We claim that there exists w ∈ S m such that ψ ⁢ λ ⁢ ( τ i ) = w for all 1 ≤ i ≤ n − 1 . Suppose on the contrary that there exist 𝑖 and w 1 ≠ w 2 in S m such that ψ ⁢ λ ⁢ ( τ i ) = w 1 and ψ ⁢ λ ⁢ ( τ i + 1 ) = w 2 . The braid relation τ i ⁢ τ i + 1 ⁢ τ i = τ i + 1 ⁢ τ i ⁢ τ i + 1 gives w 1 ⁢ w 2 ⁢ w 1 = w 2 ⁢ w 1 ⁢ w 2 . Since ψ ⁢ λ is abelian, we must have w 1 = w 2 , a contradiction. This proves the claim. Next, we find ψ ⁢ ( s i ) . The relation ρ i ⁢ s i + 1 ⁢ ρ i = ρ i + 1 ⁢ s i ⁢ ρ i + 1 gives ψ ⁢ ( s i ) = ψ ⁢ ( s i + 1 ) = z (say) for all 𝑖. Finally, the relation s 1 ⁢ ρ 3 = ρ 3 ⁢ s 1 gives z ⁢ w = w ⁢ z , and hence 𝜓 is abelian.

Case (2): Suppose that n = m and ψ ⁢ λ = id . In this case, we have ψ ⁢ ( ρ i ) = τ i for all 1 ≤ i ≤ n − 1 . Next, we need to find ψ ⁢ ( s i ) . Recall the relation s 1 ⁢ ρ i = ρ i ⁢ s 1 for 3 ≤ i ≤ n − 1 . It follows that ψ ⁢ ( s 1 ) ∈ ⟨ τ 1 ⟩ , the centraliser of the subgroup ⟨ τ 3 , τ 4 , … , τ n − 1 ⟩ in S n . Thus, either ψ ⁢ ( s 1 ) = 1 or ψ ⁢ ( s 1 ) = τ 1 . If ψ ⁢ ( s 1 ) = 1 , then the relation ρ 1 ⁢ s 2 ⁢ ρ 1 = ρ 2 ⁢ s 1 ⁢ ρ 2 gives ψ ⁢ ( s 2 ) = 1 , and consequently ψ ⁢ ( s i ) = 1 for all 𝑖. Thus, we obtain ψ = θ . And if ψ ⁢ ( s 1 ) = τ 1 , then ρ 1 ⁢ s 2 ⁢ ρ 1 = ρ 2 ⁢ s 1 ⁢ ρ 2 gives τ 1 ⁢ ψ ⁢ ( s 2 ) ⁢ τ 1 = τ 2 ⁢ τ 1 ⁢ τ 2 = τ 1 ⁢ τ 2 ⁢ τ 1 . Thus, we get ψ ⁢ ( s 2 ) = τ 2 , and consequently ψ ⁢ ( s i ) = τ i for all 𝑖. Thus, in this case, ψ = π .

Case (3): Suppose that n = m = 6 and ψ ⁢ λ = ν . Then we have ν − 1 ⁢ ψ ⁢ λ = id . By case (2), we have ν − 1 ⁢ ψ = π or ν − 1 ⁢ ψ = θ , and hence ψ = ν ⁢ π or ψ = ν ⁢ θ . ∎

5 Homomorphisms from S n to VT m

This section occupies most of the remaining part of the paper. For notational convenience, for the rest of the paper, we set K n := KT n for each n ≥ 2 . Recall that K n is a right-angled Coxeter group with a Coxeter generating set

S = { α i , j ∣ 1 ≤ i ≠ j ≤ n }

and defining relations

  1. α i , j 2 = 1 for all 1 ≤ i ≠ j ≤ n , and

  2. α i , j ⁢ α k , l = α k , l ⁢ α i , j for distinct integers 1 ≤ i , j , k , l ≤ n .

We have VT n = K n ⋊ S n , where the conjugation action of S n on K n is given as

ρ ⁢ α i , j ⁢ ρ − 1 = α ρ ⁢ ( i ) , ρ ⁢ ( j )

for all 1 ≤ i ≠ j ≤ n and ρ ∈ S n .

We begin by recalling some general results. The following three results are well known [21].

Lemma 5.1

Let ( W , S ) be a Coxeter system, and 𝑋 and 𝑌 two subsets of 𝑆. Then

W ⁢ [ X ] ∩ W ⁢ [ Y ] = W ⁢ [ X ∩ Y ] .

Lemma 5.2

Let ( W , S ) be a Coxeter system. Let 𝑋 and 𝑌 be two subsets of 𝑆 such that S = X ∪ Y and the exponents m s , t = ∞ for each s ∈ X ∖ Y and t ∈ Y ∖ X . Then

W = W ⁢ [ X ] * W ⁢ [ X ∩ Y ] W ⁢ [ Y ] .

A cyclic permutation of a word w = x i 1 ⁢ x i 2 ⁢ … ⁢ x i k (not necessarily reduced) is a word w ′ (not necessarily distinct from 𝑤) of the form

x i t ⁢ x i t + 1 ⁢ x i t + 2 ⁢ … ⁢ x i k ⁢ x i 1 ⁢ x i 2 ⁢ ⋯ ⁢ x i t − 1

for some 1 ≤ t ≤ k . A word is called cyclically reduced if each of its cyclic permutations is reduced. It is immediate that a cyclically reduced word is reduced, but the converse is not true.

Lemma 5.3

Let 𝑊 be a right-angled Coxeter group and g ∈ W a cyclically reduced word. Then 𝑔 is of order two if and only if [ s , t ] = 1 for every pair of generators 𝑠 and 𝑡 occurring in 𝑔.

The following result on the normal form for amalgamated free products is due to Serre [34, Section 1.1, Theorem 1].

Lemma 5.4

Let G 1 , G 2 , … , G r , H be a collection of groups such that 𝐻 is a subgroup of G j for each 1 ≤ j ≤ r . Consider the amalgamated free product G = G 1 ∗ H G 2 ∗ H ⋯ ∗ H G r . For each 1 ≤ j ≤ r , choose a set T j of representatives of left cosets of 𝐻 in G j such that T j contains the identity element 1. Then each element g ∈ G can be written in a unique way in the form g = t 1 ⁢ t 2 ⁢ … ⁢ t l ⁢ h such that

  1. h ∈ H and, for each i ∈ { 1 , 2 , … , l } , there exists j = j ⁢ ( i ) ∈ { 1 , 2 , … , r } such that t i ∈ T j ∖ { 1 } ,

  2. j ⁢ ( i ) ≠ j ⁢ ( i + 1 ) for all i ∈ { 1 , 2 , … , l − 1 } .

In particular, we have g ∈ H if and only if l = 0 and g = h .

Given a group 𝐺 and an automorphism 𝜙 of 𝐺, let G ϕ = { g ∈ G ∣ ϕ ⁢ ( g ) = g } denote the group of fixed-points of 𝜙. The following lemma is due to Bellingeri and Paris [8, Lemma 3.6].

Lemma 5.5

Let 𝐻 be a common subgroup of groups G 1 , G 2 and G = G 1 ∗ H G 2 their amalgamated free product. Let ϕ : G → G be an automorphism of order two such that ϕ ⁢ ( G 1 ) = G 2 and ϕ ⁢ ( G 2 ) = G 1 . Then G ϕ is a subgroup of 𝐻.

We also need the following result [8, Lemma 3.9].

Lemma 5.6

Let 𝐻 be a common subgroup of groups G 1 , G 2 and G = G 1 ∗ H G 2 their amalgamated free product. Let ϕ : G → G be an automorphism of order two such that ϕ ⁢ ( G 1 ) = G 2 and ϕ ⁢ ( G 2 ) = G 1 . Let x ∈ G such that ϕ ⁢ ( x ) = x − 1 . Then there exist y ∈ G and z ∈ H such that ϕ ⁢ ( z ) = z − 1 and x = y ⁢ z ⁢ ϕ ⁢ ( y ) − 1 .

The next three subsections consisting of quite technical results occupy the rest of this section.

5.1 Technical results I

For the rest of this section, we set

X k = { α i , j ∈ S ∣ i , j ∉ { k , k + 1 } }

for each 1 ≤ k ≤ n − 1 . Note that the conjugation action of ρ k is an order two automorphism of K n and its action on K n ⁢ [ X k ] is trivial for each 1 ≤ k ≤ n − 1 .

Lemma 5.7

Suppose w ∈ ⟨ ρ 1 , … , ρ n − 1 ⟩ such that w ⁢ ρ k ⁢ w − 1 = ρ ℓ . Then we have w ⁢ X k ⁢ w − 1 = X ℓ , and consequently w ⁢ K n ⁢ [ X k ] ⁢ w − 1 = K n ⁢ [ X ℓ ] .

Proof

For all 1 ≤ t ≤ n − 1 , we set X ̄ t = S ∖ X t , the complement of X t in 𝒮. Then we have

S = w ⁢ S ⁢ w − 1 = w ⁢ ( X k ∪ X ̄ k ) ⁢ w − 1 = ( w ⁢ X k ⁢ w − 1 ) ∪ ( w ⁢ X ̄ k ⁢ w − 1 ) = ( w ⁢ X k ⁢ w − 1 ) ∪ X ̄ ℓ ,

which gives w ⁢ X k ⁢ w − 1 = S ∖ X ̄ ℓ = X ℓ . ∎

Proposition 5.8

Let 1 ≤ k ≤ n − 1 be a fixed integer and 𝑋 a subset of 𝒮 invariant under the conjugation action of ρ k . Then

K n ⁢ [ X ] ρ k ̂ = K n ⁢ [ X ∩ X k ] .

Proof

We first prove the proposition for k = 1 .

The fact that K n ⁢ [ X ∩ X 1 ] ⊆ K n ⁢ [ X ] ρ 1 ̂ is obvious. We now prove the reverse inclusion. Let

V = { α i , j ∈ X ∣ ( i , j ) ∉ { ( 1 , 2 ) , ( 2 , 1 ) } } .

First, we prove that K n ⁢ [ X ] ρ 1 ̂ ⊆ K n ⁢ [ V ] . If X = V , then there is nothing to prove. Otherwise, since 𝑋 is invariant under the conjugation action of ρ 1 , we must have X = V ∪ { α 1 , 2 , α 2 , 1 } . We set V ′ = V ∪ { α 1 , 2 } and V ′′ = V ∪ { α 2 , 1 } . Then, by Lemma 5.2, we get K n ⁢ [ X ] = K n ⁢ [ V ′ ] * K n ⁢ [ V ] K n ⁢ [ V ′′ ] . Also,

ρ 1 ⁢ ( K n ⁢ [ V ′ ] ) ⁢ ρ 1 = K n ⁢ [ V ′′ ] and ρ 1 ⁢ ( K n ⁢ [ V ′′ ] ) ⁢ ρ 1 = K n ⁢ [ V ′ ] .

Thus, by Lemma 5.5, we get that K n ⁢ [ X ] ρ 1 ̂ ⊆ K n ⁢ [ V ] .

More generally, for 2 ≤ k ≤ n , we set

V k = { α i , j ∈ X ∣ ( i , j ) ∉ { 1 , 2 } × { 1 , 2 , … , k } } .

We prove by induction on 𝑘 that K n ⁢ [ X ] ρ 1 ̂ ⊆ K n ⁢ [ V k ] . The case k = 2 holds since V 2 = V . Suppose that the induction hypothesis holds for k − 1 , that is,

K n ⁢ [ X ] ρ 1 ̂ ⊆ K n ⁢ [ V k − 1 ] .

Now, if V k = V k − 1 , there is nothing to prove. So we suppose that V k ≠ V k − 1 . Since the set V k − 1 is invariant under the conjugation action of ρ 1 , we have

V k − 1 = V k ∪ { α 1 , k , α 2 , k } .

Set V k ′ = V k ∪ { α 1 , k } and V k ′′ = V k ∪ { α 2 , k } . By Lemma 5.2, we have

K n ⁢ [ V k − 1 ] = K n ⁢ [ V k ′ ] * K n ⁢ [ V k ] K n ⁢ [ V k ′′ ] .

Also,

ρ 1 ⁢ ( K n ⁢ [ V k ′ ] ) ⁢ ρ 1 = K n ⁢ [ V k ′′ ] and ρ 1 ⁢ ( K n ⁢ [ V k ′′ ] ) ⁢ ρ 1 = K n ⁢ [ V k ′ ] .

Hence, by Lemma 5.5, we get K n ⁢ [ X ] ρ 1 ̂ ⊆ K n ⁢ [ V k ] .

Next, for 2 ≤ k ≤ n , we consider the set

W k = { α i , j ∈ X ∣ ( i , j ) ∉ { 1 , 2 } × { 1 , 2 , … , n } ∪ { 1 , 2 , … , k } × { 1 , 2 } } .

We show by induction on 𝑘 that K n ⁢ [ X ] ρ 1 ̂ ⊆ K n ⁢ [ W k ] . The case k = 2 holds as W 2 = V n . We now suppose that the induction hypothesis holds for k − 1 , that is, K n ⁢ [ X ] ρ 1 ̂ ⊆ K n ⁢ [ W k − 1 ] . If W k = W k − 1 , there is nothing to prove. So we suppose that W k ≠ W k − 1 . Since the set W k − 1 is invariant under the conjugation action of ρ 1 , we have W k − 1 = W k ∪ { α 1 , k , α 2 , k } . Set W k ′ = W k ∪ { α 1 , k } and W k ′′ = W k ∪ { α 2 , k } . Again, by Lemma 5.2, we have

K n ⁢ [ W k − 1 ] = K n ⁢ [ W k ′ ] * K n ⁢ [ W k ] K n ⁢ [ W k ′′ ] .

Also,

ρ 1 ⁢ ( K n ⁢ [ W k ′ ] ) ⁢ ρ 1 = K n ⁢ [ W k ′′ ] and ρ 1 ⁢ ( K n ⁢ [ W k ′′ ] ) ⁢ ρ 1 = K n ⁢ [ W k ′ ] .

Thus, by Lemma 5.5, we get K n ⁢ [ X ] ρ 1 ̂ ⊆ K n ⁢ [ W k ] .

Finally, we note that W k = X ∩ X 1 , and hence K n ⁢ [ X ] ρ 1 ̂ ⊆ K n ⁢ [ X ∩ X 1 ] . This proves the proposition for k = 1 .

Now, we consider k ≥ 2 . Choose an element w ∈ ⟨ ρ 1 , … , ρ n − 1 ⟩ such that w ⁢ ρ 1 ⁢ w − 1 = ρ k . Given that the set 𝑋 is invariant under the action of ρ k . Then the set Y = w − 1 ⁢ X ⁢ w is invariant under the action of ρ 1 . By the earlier case, we have K n ⁢ [ Y ] ρ 1 ̂ = K n ⁢ [ Y ∩ X 1 ] .

It is easy to check that

w ⁢ ( K n ⁢ [ Y ] ρ 1 ̂ ) ⁢ w − 1 = K n ⁢ [ w ⁢ Y ⁢ w − 1 ] ρ k ̂ = K n ⁢ [ X ] ρ k ̂ , w ⁢ ( K n ⁢ [ Y ∩ X 1 ] ) ⁢ w − 1 = K n ⁢ [ w ⁢ Y ⁢ w − 1 ∩ X k ] = K n ⁢ [ X ∩ X k ] .

Thus, we get K n ⁢ [ X ] ρ k ̂ = K n ⁢ [ X ∩ X k ] , as desired. ∎

It follows from Proposition 5.8 that

K n ρ k ̂ = K n ⁢ [ S ] ρ k ̂ = K n ⁢ [ X k ]

for each 1 ≤ k ≤ n − 1 .

Corollary 5.9

For each n ≥ 3 , C VT n ⁡ ( S n ) = C K n ⁡ ( S n ) = 1 .

Proof

Recall that VT n = K n ⋊ S n . Let x ⁢ y ∈ C VT n ⁡ ( S n ) , where x ∈ K n , y ∈ S n . Then x ⁢ y ⁢ ρ = ρ ⁢ x ⁢ y , that is, x ⁢ y = ( ρ ⁢ x ⁢ ρ − 1 ) ⁢ ( ρ ⁢ y ⁢ ρ − 1 ) for each ρ ∈ S n . This implies that y = ρ ⁢ y ⁢ ρ − 1 for all ρ ∈ S n , that is, y ∈ Z ⁡ ( S n ) = 1 as n ≥ 3 . Thus, we have C VT n ⁡ ( S n ) = C K n ⁡ ( S n ) . But note that

C K n ⁡ ( S n ) ≤ ⋂ k = 1 n − 1 C K n ⁡ ( ρ k ) = ⋂ k = 1 n − 1 K n ρ k ̂ = ⋂ k = 1 n − 1 K n ⁢ [ X k ] = 1 ,

which is desired. ∎

Following is an analogue of [8, Lemma 3.10].

Lemma 5.10

Let 𝑋 be a subset of 𝒮 invariant under the conjugation action of ρ 1 . Let α ∈ K n ⁢ [ X ] such that ρ 1 ⁢ α ⁢ ρ 1 = α − 1 . Then there exist α ′ ∈ K n ⁢ [ X ] and β ∈ K n ⁢ [ X ∩ X 1 ] such that

α = α ′ ⁢ β ⁢ ρ 1 ⁢ α ′ ⁣ − 1 ⁢ ρ 1 and β 2 = 1 .

Proof

We complete the proof in the following three steps.

Step (1): For 2 ≤ k ≤ n , we set

V k = { α i , j ∈ X ∣ ( i , j ) ∉ { 1 , 2 } × { 1 , 2 , … , k } } .

We prove by induction on 𝑘 that there exist α ′ ∈ K n ⁢ [ X ] and β ′ ∈ K n ⁢ [ V k ] such that ρ 1 ⁢ β ′ ⁢ ρ 1 = β ′ ⁣ − 1 and α = α ′ ⁢ β ′ ⁢ ρ 1 ⁢ α ′ ⁣ − 1 ⁢ ρ 1 .

Note that V 2 = { α i , j ∈ X ∣ ( i , j ) ∉ { ( 1 , 2 ) , ( 2 , 1 ) } . If V 2 = X , take α ′ = 1 , β ′ = α , and we are done. So we assume that V 2 ≠ X . Since 𝑋 is invariant under the conjugation action of ρ 1 , we have X = V 2 ∪ { α 1 , 2 , α 2 , 1 } . Set V 2 ′ = V 2 ∪ { α 1 , 2 } and V 2 ′′ = V 2 ∪ { α 2 , 1 } . By Lemma 5.2, we have

K n ⁢ [ X ] = K n ⁢ [ V 2 ′ ] * K n ⁢ [ V 2 ] K n ⁢ [ V 2 ′′ ] .

Also, note that ρ 1 ⁢ K n ⁢ [ V 2 ′ ] ⁢ ρ 1 = K n ⁢ [ V 2 ′′ ] and ρ 1 ⁢ K n ⁢ [ V 2 ′′ ] ⁢ ρ 1 = K n ⁢ [ V 2 ′ ] . Recall that the conjugation action of ρ 1 on K n ⁢ [ X ] is an order two automorphism of K n ⁢ [ X ] . Thus, we are done for the case k = 2 by Lemma 5.6.

Suppose that k ≥ 3 and that the induction hypothesis holds, i.e., there exist α 1 ′ ∈ K n ⁢ [ X ] and β 1 ′ ∈ K n ⁢ [ V k − 1 ] such that ρ 1 ⁢ β 1 ′ ⁢ ρ 1 = β 1 ′ ⁣ − 1 and α = α 1 ′ ⁢ β 1 ′ ⁢ ρ 1 ⁢ α 1 ′ ⁣ − 1 ⁢ ρ 1 . Now, if V k = V k − 1 , we are done. So we assume that V k ≠ V k − 1 . Since both V k and V k − 1 are invariant under the conjugation action of ρ 1 , we have

V k − 1 = V k ∪ { α 1 , k , α 2 , k } .

Set V k ′ = V k ∪ { α 1 , k } and V k ′′ = V k ∪ { α 2 , k } . By Lemma 5.2, we have

K n ⁢ [ V k − 1 ] = K n ⁢ [ V k ′ ] * K n ⁢ [ V k ] K n ⁢ [ V k ′′ ] .

Again, the conjugation action of ρ 1 on K n ⁢ [ V k − 1 ] is an order two automorphism of K n ⁢ [ V k − 1 ] . We also have ρ 1 ⁢ K n ⁢ [ V k ′ ] ⁢ ρ 1 = K n ⁢ [ V k ′′ ] , ρ 1 ⁢ K n ⁢ [ V k ′′ ] ⁢ ρ 1 = K n ⁢ [ V k ′ ] and ρ 1 ⁢ β 1 ′ ⁢ ρ 1 = β 1 ′ ⁣ − 1 , where β 1 ′ ∈ K n ⁢ [ V k − 1 ] . By Lemma 5.6, there exist α 2 ′ ∈ K n ⁢ [ V k − 1 ] and β ′ ∈ K n ⁢ [ V k ] such that

ρ 1 ⁢ β ′ ⁢ ρ 1 = β ′ ⁣ − 1 and β 1 ′ = α 2 ′ ⁢ β ′ ⁢ ρ 1 ⁢ α 2 ′ ⁣ − 1 ⁢ ρ 1 .

Now, set α ′ = α 1 ′ ⁢ α 2 ′ . From the induction hypothesis, we have α = α 1 ′ ⁢ β 1 ′ ⁢ ρ 1 ⁢ α 1 ′ ⁣ − 1 ⁢ ρ 1 . Putting β 1 ′ = α 2 ′ ⁢ β ′ ⁢ ρ 1 ⁢ α 2 ′ ⁣ − 1 ⁢ ρ 1 , we have

α = α 1 ′ ⁢ α 2 ′ ⁢ β ′ ⁢ ρ 1 ⁢ α 2 ′ ⁣ − 1 ⁢ ρ 1 ⁢ ρ 1 ⁢ α 1 ′ ⁣ − 1 ⁢ ρ 1 = α 1 ′ ⁢ α 2 ′ ⁢ β ′ ⁢ ρ 1 ⁢ α 2 ′ ⁣ − 1 ⁢ α 1 ′ ⁣ − 1 ⁢ ρ 1 = α ′ ⁢ β ′ ⁢ ρ 1 ⁢ α ′ ⁣ − 1 ⁢ ρ 1 .

We already have ρ 1 ⁢ β ′ ⁢ ρ 1 = β ′ ⁣ − 1 . This completes the proof of step (1).

Step (2): For 2 ≤ k ≤ n , we set

W k = { α i , j ∈ X ∣ ( i , j ) ∉ { 1 , 2 } × { 1 , 2 , … , n } ∪ { 1 , 2 , … , k } × { 1 , 2 } } .

We now prove by induction on 𝑘 that there exist α ′ ∈ K n ⁢ [ X ] and β ′ ∈ K n ⁢ [ W k ] such that ρ 1 ⁢ β ′ ⁢ ρ 1 = β ′ ⁣ − 1 and α = α ′ ⁢ β ′ ⁢ ρ 1 ⁢ α ′ ⁣ − 1 ⁢ ρ 1 .

Since W 2 = V n , the base case k = 2 is done by step (1). Suppose that k ≥ 3 and the induction hypothesis holds, i.e., there exist α 1 ′ ∈ K n ⁢ [ X ] and β 1 ′ ∈ K n ⁢ [ W k − 1 ] such that ρ 1 ⁢ β 1 ′ ⁢ ρ 1 = β 1 ′ ⁣ − 1 and α = α 1 ′ ⁢ β 1 ′ ⁢ ρ 1 ⁢ α 1 ′ ⁣ − 1 ⁢ ρ 1 . Now, if W k = W k − 1 , we are done. So we assume that W k ≠ W k − 1 . Since both W k and W k − 1 are invariant under the conjugation action of ρ 1 , we have

W k − 1 = W k ∪ { α k , 1 , α k , 2 } .

Set W k ′ = W k ∪ { α k , 1 } and W k ′′ = W k ∪ { α k , 2 } . By Lemma 5.2, we have

K n ⁢ [ W k − 1 ] = K n ⁢ [ W k ′ ] * K n ⁢ [ W k ] K n ⁢ [ W k ′′ ] .

Again, the conjugation action of ρ 1 on K n ⁢ [ W k − 1 ] is an order two automorphism of K n ⁢ [ W k − 1 ] . We also have ρ 1 ⁢ K n ⁢ [ W k ′ ] ⁢ ρ 1 = K n ⁢ [ W k ′′ ] , ρ 1 ⁢ K n ⁢ [ W k ′′ ] ⁢ ρ 1 = K n ⁢ [ W k ′ ] and ρ 1 ⁢ β 1 ′ ⁢ ρ 1 = β 1 ′ ⁣ − 1 , where β 1 ′ ∈ K n ⁢ [ W k − 1 ] . By Lemma 5.6, there exist α 2 ′ ∈ K n ⁢ [ W k − 1 ] and β ′ ∈ K n ⁢ [ W k ] such that ρ 1 ⁢ β ′ ⁢ ρ 1 = β ′ ⁣ − 1 and β 1 ′ = α 2 ′ ⁢ β ′ ⁢ ρ 1 ⁢ α 2 ′ ⁣ − 1 ⁢ ρ 1 . Now, set α ′ = α 1 ′ ⁢ α 2 ′ . From the induction hypothesis, we have α = α 1 ′ ⁢ β 1 ′ ⁢ ρ 1 ⁢ α 1 ′ ⁣ − 1 ⁢ ρ 1 . Putting β 1 ′ = α 2 ′ ⁢ β ′ ⁢ ρ 1 ⁢ α 2 ′ ⁣ − 1 ⁢ ρ 1 , we have

α = α 1 ′ ⁢ α 2 ′ ⁢ β ′ ⁢ ρ 1 ⁢ α 2 ′ ⁣ − 1 ⁢ ρ 1 ⁢ ρ 1 ⁢ α 1 ′ ⁣ − 1 ⁢ ρ 1 = α 1 ′ ⁢ α 2 ′ ⁢ β ′ ⁢ ρ 1 ⁢ α 2 ′ ⁣ − 1 ⁢ α 1 ′ ⁣ − 1 ⁢ ρ 1 = α ′ ⁢ β ′ ⁢ ρ 1 ⁢ α ′ ⁣ − 1 ⁢ ρ 1 .

We already have ρ 1 ⁢ β ′ ⁢ ρ 1 = β ′ ⁣ − 1 . This completes the proof of step (2).

Step (3): Note that W n = X ∩ X 1 . Thus, we have

α ′ ∈ K n ⁢ [ X ] and β ∈ K n ⁢ [ X ∩ X 1 ]

such that

α = α ′ ⁢ β ⁢ ρ 1 ⁢ α ′ ⁣ − 1 ⁢ ρ 1 and ρ 1 ⁢ β ⁢ ρ 1 = β − 1 .

It only remains to be shown that β 2 = 1 . But, by Proposition 5.8, ρ 1 ⁢ β ⁢ ρ 1 = β . Thus, β = ρ 1 ⁢ β ⁢ ρ 1 = β − 1 , and this completes the proof. ∎

Next, we generalise the preceding lemma.

Lemma 5.11

Let 1 ≤ k ≤ n − 1 be a fixed integer and 𝑋 a subset of 𝒮 invariant under the conjugation action of ρ k . Let α ∈ K n ⁢ [ X ] such that ρ k ⁢ α ⁢ ρ k = α − 1 . Then there exist α ′ ∈ K n ⁢ [ X ] and β ∈ K n ⁢ [ X ∩ X k ] such that

α = α ′ ⁢ β ⁢ ρ k ⁢ α ′ ⁣ − 1 ⁢ ρ k and β 2 = 1 .

Proof

The case k = 1 follows from Lemma 5.10. So we suppose that k ≥ 2 . Choose an element

w ∈ ⟨ ρ 1 , … , ρ n − 1 ⟩ such that w − 1 ⁢ ρ k ⁢ w = ρ 1 ,

i.e., w ⁢ ρ 1 ⁢ w − 1 = ρ k . Note that

ρ 1 ⁢ ( w − 1 ⁢ X ⁢ w ) ⁢ ρ 1 = w − 1 ⁢ ( w ⁢ ρ 1 ⁢ w − 1 ) ⁢ X ⁢ ( w ⁢ ρ 1 ⁢ w − 1 ) ⁢ w = w − 1 ⁢ ρ k ⁢ X ⁢ ρ k ⁢ w = w − 1 ⁢ X ⁢ w ,

i.e., w − 1 ⁢ X ⁢ w ⊆ S is invariant under the conjugation action ρ 1 . Similarly, we have

ρ 1 ⁢ ( w − 1 ⁢ α ⁢ w ) ⁢ ρ 1 = w − 1 ⁢ ( w ⁢ ρ 1 ⁢ w − 1 ) ⁢ α ⁢ ( w ⁢ ρ 1 ⁢ w − 1 ) ⁢ w = w − 1 ⁢ ρ k ⁢ α ⁢ ρ k ⁢ w = w − 1 ⁢ α − 1 ⁢ w .

Now, the proof follows from a direct application of Lemma 5.10. ∎

5.2 Technical results II

Let θ : VT n → S n and λ : S n → VT n be as defined in Section 2.

Proposition 5.12

Let 1 ≤ k ≤ n − 1 be a fixed integer and let ϕ : S n → VT n be a homomorphism such that θ ⁢ ϕ is the identity on S n . Suppose that ϕ ⁢ ( τ k ) = β ⁢ ρ k for some β ∈ K n ⁢ [ X k ] with β 2 = 1 . Then β = 1 .

Proof

We first prove the assertion for k = 1 . If n ≤ 3 , then K n ⁢ [ X 1 ] = 1 and the assertion is vacuously true. Thus, we assume that n ≥ 4 .

Suppose that β ≠ 1 . We proceed to obtain a contradiction. By Lemma 5.3, we have β = x ⁢ g ⁢ x − 1 for some x , g ∈ K n ⁢ [ X 1 ] with g ≠ 1 cyclically reduced of the form

g = α i 1 , j 1 ⁢ α i 2 , j 2 ⁢ … ⁢ α i k , j k ,

where 3 ≤ i 1 , j 1 , i 2 , j 2 , … , i k , j k ≤ n are all distinct integers. Without loss of generality, we assume that i 1 = 3 . Since θ ⁢ ϕ is the identity on S n , we can write ϕ ⁢ ( τ 3 ) = x 3 ⁢ ρ 3 for some x 3 ∈ K n . As τ 1 and τ 3 commute with each other, we have β ⁢ ρ 1 ⁢ x 3 ⁢ ρ 3 = x 3 ⁢ ρ 3 ⁢ β ⁢ ρ 1 , which can be rewritten as

(5.1) β ⁢ ( ρ 1 ⁢ x 3 ⁢ ρ 1 ) ⁢ ( ρ 3 ⁢ β ⁢ ρ 3 ) ⁢ x 3 − 1 = 1 .

Note that

β , ρ 3 ⁢ β ⁢ ρ 3 , ρ 1 ⁢ x 3 ⁢ ρ 1 , x 3 − 1 ∈ K n .

We define an epimorphism η : K n → { 1 , − 1 } by setting

η ⁢ ( α i , j ) = { − 1 if ⁢ ( i , j ) = ( 3 , j 1 ) , 1 otherwise.

It follows from the construction of 𝑔 that η ⁢ ( g ) = − 1 , and hence

η ⁢ ( β ) = η ⁢ ( x ) ⁢ η ⁢ ( g ) ⁢ η ⁢ ( x ) − 1 = η ⁢ ( g ) = − 1 .

The conditions on the indices j 1 , i 2 , j 2 , i 3 , j 3 , … , i k , j k imply that α 3 , j 1 does not appear in the expression ρ 3 ⁢ g ⁢ ρ 3 , and hence η ⁢ ( ρ 3 ⁢ g ⁢ ρ 3 ) = 1 . Thus, we see that

η ⁢ ( ρ 3 ⁢ β ⁢ ρ 3 ) = η ⁢ ( ρ 3 ⁢ x ⁢ g ⁢ x − 1 ⁢ ρ 3 ) = η ⁢ ( ρ 3 ⁢ x ⁢ ρ 3 ) ⁢ η ⁢ ( ρ 3 ⁢ g ⁢ ρ 3 ) ⁢ η ⁢ ( ρ 3 ⁢ x − 1 ⁢ ρ 3 ) = η ⁢ ( ρ 3 ⁢ g ⁢ ρ 3 ) = 1 .

Since ρ 1 ⁢ α 3 , j 1 ⁢ ρ 1 = α 3 , j 1 , it follows that η ⁢ ( ρ 1 ⁢ x 3 ⁢ ρ 1 ) = η ⁢ ( x 3 ) = η ⁢ ( x 3 − 1 ) . Now, applying 𝜂 on (5.1) gives

1 = η ⁢ ( β ⁢ ( ρ 1 ⁢ x 3 ⁢ ρ 1 ) ⁢ ( ρ 3 ⁢ β ⁢ ρ 3 ) ⁢ x 3 − 1 ) = η ⁢ ( β ) ⁢ η ⁢ ( ρ 1 ⁢ x 3 ⁢ ρ 1 ) ⁢ η ⁢ ( ρ 3 ⁢ β ⁢ ρ 3 ) ⁢ η ⁢ ( x 3 − 1 ) = − 1 ,

a contradiction. Hence, 𝛽 must be trivial and the proposition is proved for k = 1 .

Now, we assume that k ≥ 2 . Choose an element w ∈ ⟨ ρ 1 , … , ρ n − 1 ⟩ such that w ⁢ ρ 1 ⁢ w − 1 = ρ k . Setting g = θ ⁢ ( w ) , we get g ⁢ τ 1 ⁢ g − 1 = τ k . Consider the composition w ̂ − 1 ⁢ ϕ ⁢ g ̂ : S n → VT n . It is easy to check that θ ⁢ w ̂ − 1 ⁢ ϕ ⁢ g ̂ is the identity on S n . Further, note that

w ̂ − 1 ⁢ ϕ ⁢ g ̂ ⁢ ( τ 1 ) = w ̂ − 1 ⁢ ϕ ⁢ ( g ⁢ τ 1 ⁢ g − 1 ) = w ̂ − 1 ⁢ ϕ ⁢ ( τ k ) = w ̂ − 1 ⁢ ( β ⁢ ρ k ) = ( w − 1 ⁢ β ⁢ w ) ⁢ ( w − 1 ⁢ ρ k ⁢ w ) = ( w − 1 ⁢ β ⁢ w ) ⁢ ρ 1 .

By Lemma 5.7, ( w − 1 ⁢ β ⁢ w ) ∈ K n ⁢ [ X 1 ] is an involution. Thus, by the case k = 1 , we get w − 1 ⁢ β ⁢ w = 1 , and hence β = 1 . ∎

We note that if ϕ : S n → VT n is a homomorphism such that θ ⁢ ϕ is the identity on S n , then θ ⁢ w ̂ ⁢ ϕ is also the identity on S n for all w ∈ K n , where w ̂ is the inner automorphism of VT n induced by 𝑤. This, together with Proposition 5.12, yields the following.

Corollary 5.13

Let 1 ≤ k ≤ n − 1 be a fixed integer and let ϕ : S n → VT n be a homomorphism such that θ ⁢ ϕ is the identity on S n . Suppose ϕ ⁢ ( τ k ) = w − 1 ⁢ β ⁢ ρ k ⁢ w for some w ∈ K n and β ∈ K n ⁢ [ X k ] with β 2 = 1 . Then β = 1 .

Corollary 5.14

Let ϕ : S n → VT n be a homomorphism such that θ ⁢ ϕ is the identity on S n . Then, for each 1 ≤ k ≤ n − 1 , there exists

x k ∈ K n such that ϕ ⁢ ( τ k ) = x k ⁢ ρ k ⁢ x k − 1 .

Proof

Since θ ⁢ ϕ is the identity on S n , for each 1 ≤ k ≤ n − 1 , there exists α k ∈ K n such that ϕ ⁢ ( τ k ) = α k ⁢ ρ k . This gives α k ⁢ ρ k ⁢ α k ⁢ ρ k = 1 , i.e., ρ k ⁢ α k ⁢ ρ k = α k − 1 . Now, we are done by Lemma 5.11 and Corollary 5.13. ∎

Corollary 5.15

Let 1 ≤ k ≤ n − 1 be a fixed integer. Suppose that ρ k = w − 1 ⁢ β ⁢ ρ k ⁢ w for some w ∈ K n and β ∈ K n ⁢ [ X k ] such that β 2 = 1 . Then β = 1 .

Proof

This follows by taking ϕ = λ in Corollary 5.13. ∎

5.3 Technical results III

We say that an element α ∈ K n satisfies condition (C) if

(C) α ⁢ ( ρ 2 ⁢ α − 1 ⁢ ρ 2 ) ⁢ ( ρ 2 ⁢ ρ 1 ⁢ α ⁢ ρ 1 ⁢ ρ 2 ) ( ρ 2 ⁢ ρ 1 ⁢ ρ 2 ⁢ α − 1 ⁢ ρ 2 ⁢ ρ 1 ⁢ ρ 2 ) ⁢ ( ρ 1 ⁢ ρ 2 ⁢ α ⁢ ρ 2 ⁢ ρ 1 ) ⁢ ( ρ 1 ⁢ α − 1 ⁢ ρ 1 ) = 1 ,

or equivalently

[ α , ρ 2 ] ⁢ ρ 2 ⁢ ρ 1 ̂ ⁢ ( [ α , ρ 2 ] ) ⁢ ρ 1 ̂ ⁢ ( [ ρ 2 , α ] ) = 1 ,

or equivalently

α ⁢ ( ρ 2 ⁢ α − 1 ⁢ ρ 1 ⁢ α ⁢ ρ 1 ⁢ ρ 2 ) ⁢ ρ 2 ⁢ ρ 1 ̂ ⁢ ( ( ρ 2 ⁢ α − 1 ⁢ ρ 1 ⁢ α ⁢ ρ 1 ⁢ ρ 2 ) ) ⁢ ( ρ 1 ⁢ α − 1 ⁢ ρ 1 ) = 1 .

Let α = x ⁢ z ⁢ y for some x ∈ K n ⁢ [ X 1 ] , y ∈ K n ⁢ [ X 2 ] and z ∈ K n . It is easy to see that if 𝛼 satisfies condition (C), then 𝑧 also satisfies condition (C).

For the rest of this section, we assume that n ≥ 4 and fix a subset 𝑋 of 𝒮 which is invariant under the conjugation action of both ρ 1 and ρ 2 . We set U 3 = X and

U k = { α i , j ∈ X ∣ ( i , j ) ∉ { 1 , 2 , 3 } × { 4 , 5 , … , k } }

for 4 ≤ k ≤ n . Note that

U n ⊆ U n − 1 ⊆ ⋯ ⊆ U k ⊆ U k − 1 ⊆ ⋯ ⊆ U 3 = X

and each U k is invariant under the conjugation action of both ρ 1 and ρ 2 . The next three lemmas are analogues of [8, Lemma 3.11] for virtual twin groups.

Lemma 5.16

Let z ∈ K n ⁢ [ U k − 1 ] satisfy condition (C) for some 4 ≤ k ≤ n . Then there exist x ∈ K n ⁢ [ X 1 ∩ X ] , y ∈ K n ⁢ [ X 2 ∩ X ] and z 1 ∈ K n ⁢ [ U k ] such that z = x ⁢ z 1 ⁢ y and z 1 satisfies condition (C).

Proof

If z ∈ K n ⁢ [ U k ] , then we are done by taking x = 1 , y = 1 and z 1 = z . So we assume that z ∈ K n ⁢ [ U k − 1 ] ∖ K n ⁢ [ U k ] . This implies that U k ≠ U k − 1 . Since both U k − 1 and U k are invariant under the conjugation action of ρ 1 and ρ 2 , we have U k − 1 = U k ∪ { α 1 , k , α 2 , k , α 3 , k } . We set

G j := K n ⁢ [ U k ∪ { α j , k } ] , 1 ≤ j ≤ 3 , H := G 1 ∩ G 2 ∩ G 3 = K n ⁢ [ U k ] , G := G 1 * H G 2 * H G 3 = K n ⁢ [ U k − 1 ] .

Due to Lemma 5.4, we can write z = a 1 ⁢ a 2 ⁢ … ⁢ a l for some integer l ≥ 1 such that

  1. for each 1 ≤ i ≤ l , there exists j = j ⁢ ( i ) ∈ { 1 , 2 , 3 } such that a i ∈ G j ∖ H , and

  2. j ⁢ ( i ) ≠ j ⁢ ( i + 1 ) for all 1 ≤ i ≤ l − 1 .

We now argue by induction on the length 𝑙 of 𝑧. Suppose that l = 1 . Then either a 1 ∈ G 1 ∖ H or a 1 ∈ G 2 ∖ H or a 1 ∈ G 3 ∖ H . First suppose that z = a 1 ∈ G 1 ∖ H . We now set

b 1 = a 1 ∈ G 1 ∖ H , b 2 = ρ 2 ⁢ a 1 − 1 ⁢ ρ 2 ∈ G 1 ∖ H , b 3 = ρ 2 ⁢ ρ 1 ⁢ a 1 ⁢ ρ 1 ⁢ ρ 2 ∈ G 3 ∖ H , b 4 = ρ 2 ⁢ ρ 1 ⁢ ρ 2 ⁢ a 1 − 1 ⁢ ρ 2 ⁢ ρ 1 ⁢ ρ 2 ∈ G 3 ∖ H , b 5 = ρ 1 ⁢ ρ 2 ⁢ a 1 ⁢ ρ 2 ⁢ ρ 1 ∈ G 2 ∖ H , b 6 = ρ 1 ⁢ a 1 − 1 ⁢ ρ 1 ∈ G 2 ∖ H .

Since z = a 1 satisfies condition (C), we have b 1 ⁢ b 2 ⁢ b 3 ⁢ b 4 ⁢ b 5 ⁢ b 6 = 1 . In view of Lemma 5.4 and the fact that 𝐻 is invariant under the conjugation action of ρ 1 and ρ 2 , this is possible only if b 1 ⁢ b 2 , b 3 ⁢ b 4 and b 5 ⁢ b 6 all lie in 𝐻. Set

c = a 1 ⁢ ρ 2 ⁢ a 1 − 1 ⁢ ρ 2 ∈ H .

Note that

ρ 2 ⁢ c ⁢ ρ 2 = ρ 2 ⁢ a 1 ⁢ ρ 2 ⁢ a 1 − 1 ⁢ ρ 2 ⁢ ρ 2 = ρ 2 ⁢ a 1 ⁢ ρ 2 ⁢ a 1 − 1 = c − 1 .

Hence, by Lemma 5.11, there exist α ∈ H = K n ⁢ [ U k ] and β ∈ K n ⁢ [ U k ∩ X 2 ] such that c = α ⁢ β ⁢ ρ 2 ⁢ α − 1 ⁢ ρ 2 with β 2 = 1 . This gives us a 1 ⁢ ρ 2 ⁢ a 1 − 1 ⁢ ρ 2 = c = α ⁢ β ⁢ ρ 2 ⁢ α − 1 ⁢ ρ 2 , and consequently ρ 2 = a 1 − 1 ⁢ α ⁢ β ⁢ ρ 2 ⁢ α − 1 ⁢ a 1 , where a 1 − 1 ⁢ α ∈ K n . By Corollary 5.15, we have β = 1 . So we have c = α ⁢ ρ 2 ⁢ α − 1 ⁢ ρ 2 . Set

y = α − 1 ⁢ a 1 , x = 1 and z 1 = α ∈ H = K n ⁢ [ U k ] .

This gives us z = a 1 = x ⁢ z 1 ⁢ y and

ρ 2 ⁢ y ⁢ ρ 2 = ρ 2 ⁢ α − 1 ⁢ a 1 ⁢ ρ 2 = ( ρ 2 ⁢ α − 1 ⁢ ρ 2 ) ⁢ ( ρ 2 ⁢ a 1 ⁢ ρ 2 ) = ( α − 1 ⁢ c ) ⁢ ( c − 1 ⁢ a 1 ) = y ,

i.e., y ∈ K n ⁢ [ X 2 ∩ X ] . So we are done for the case a 1 ∈ G 1 ∖ H .

Now, consider a 1 ∈ G 2 ∖ H . We again set

b 1 = a 1 ∈ G 2 ∖ H , b 2 = ρ 2 ⁢ a 1 − 1 ⁢ ρ 2 ∈ G 3 ∖ H , b 3 = ρ 2 ⁢ ρ 1 ⁢ a 1 ⁢ ρ 1 ⁢ ρ 2 ∈ G 1 ∖ H , b 4 = ρ 2 ⁢ ρ 1 ⁢ ρ 2 ⁢ a 1 − 1 ⁢ ρ 2 ⁢ ρ 1 ⁢ ρ 2 ∈ G 2 ∖ H , b 5 = ρ 1 ⁢ ρ 2 ⁢ a 1 ⁢ ρ 2 ⁢ ρ 1 ∈ G 3 ∖ H , b 6 = ρ 1 ⁢ a 1 − 1 ⁢ ρ 1 ∈ G 1 ∖ H .

Since z = a 1 satisfies condition (C), we have b 1 ⁢ b 2 ⁢ b 3 ⁢ b 4 ⁢ b 5 ⁢ b 6 = 1 . But this leads to a contradiction, thanks to Lemma 5.4. Finally, suppose that a 1 ∈ G 3 ∖ H . We again set

b 1 = a 1 ∈ G 3 ∖ H , b 2 = ρ 2 ⁢ a 1 − 1 ⁢ ρ 2 ∈ G 2 ∖ H , b 3 = ρ 2 ⁢ ρ 1 ⁢ a 1 ⁢ ρ 1 ⁢ ρ 2 ∈ G 2 ∖ H , b 4 = ρ 2 ⁢ ρ 1 ⁢ ρ 2 ⁢ a 1 − 1 ⁢ ρ 2 ⁢ ρ 1 ⁢ ρ 2 ∈ G 1 ∖ H , b 5 = ρ 1 ⁢ ρ 2 ⁢ a 1 ⁢ ρ 2 ⁢ ρ 1 ∈ G 1 ∖ H , b 6 = ρ 1 ⁢ a 1 − 1 ⁢ ρ 1 ∈ G 3 ∖ H .

Since a 1 satisfies condition (C), again we have b 1 ⁢ b 2 ⁢ b 3 ⁢ b 4 ⁢ b 5 ⁢ b 6 = 1 . It follows that b 2 ⁢ b 3 and b 4 ⁢ b 5 lie in 𝐻, and consequently a 1 − 1 ⁢ ρ 1 ⁢ a 1 ⁢ ρ 1 = c (say) ∈ H . Note that ρ 1 ⁢ c ⁢ ρ 1 = c − 1 . Hence, by Lemma 5.11, there exist

d ∈ H = K n ⁢ [ U k ] and t ∈ K n ⁢ [ U k ∩ X 1 ]

such that c = d ⁢ t ⁢ ρ 1 ⁢ d − 1 ⁢ ρ 1 with t 2 = 1 . Thus, we have

a 1 − 1 ⁢ ρ 1 ⁢ a 1 ⁢ ρ 1 = c = d ⁢ t ⁢ ρ 1 ⁢ d − 1 ⁢ ρ 1 ,

and consequently ρ 1 = a 1 ⁢ d ⁢ t ⁢ ρ 1 ⁢ d − 1 ⁢ a 1 − 1 , where a 1 ⁢ d ∈ K n . By Corollary 5.15, we have t = 1 . Now, set x = a 1 ⁢ d , y = 1 and z 1 = d − 1 ∈ K n ⁢ [ U k ] . It follows that z = x ⁢ z 1 ⁢ y and

ρ 1 ⁢ x ⁢ ρ 1 = ρ 1 ⁢ a 1 ⁢ d ⁢ ρ 1 = ( ρ 1 ⁢ a 1 ⁢ ρ 1 ) ⁢ ( ρ 1 ⁢ d ⁢ ρ 1 ) = ( a 1 ⁢ c ) ⁢ ( c − 1 ⁢ d ) = a 1 ⁢ d = x ,

i.e., x ∈ K n ⁢ [ X 1 ∩ X ] . Hence, we are done for the case l = 1 .

Let us now suppose that l ≥ 2 and that the induction hypothesis holds. Since z = a 1 ⁢ a 2 ⁢ … ⁢ a l satisfy condition (C), we have

( a 1 ⁢ … ⁢ a l ) ⁢ ρ 2 ̂ ⁢ ( a l − 1 ⁢ … ⁢ a 1 − 1 ) ⁢ ρ 2 ⁢ ρ 1 ̂ ⁢ ( a 1 ⁢ … ⁢ a l ) ⁢ ρ 2 ⁢ ρ 1 ⁢ ρ 2 ̂ ( a l − 1 ⁢ … ⁢ a 1 − 1 ) ⁢ ρ 1 ⁢ ρ 2 ̂ ⁢ ( a 1 ⁢ … ⁢ a l ) ⁢ ρ 1 ̂ ⁢ ( a l − 1 ⁢ … ⁢ a 1 − 1 ) = 1 .

Then, by Lemma 5.4, we have either of the following:

  1. a l ⁢ ρ 2 ̂ ⁢ ( a l − 1 ) ∈ H ,

  2. ρ 2 ̂ ⁢ ( a 1 − 1 ) ⁢ ρ 2 ⁢ ρ 1 ̂ ⁢ ( a 1 ) ∈ H ,

  3. ρ 2 ⁢ ρ 1 ̂ ⁢ ( a l ) ⁢ ρ 2 ⁢ ρ 1 ⁢ ρ 2 ̂ ⁢ ( a l − 1 ) = ρ 2 ⁢ ρ 1 ̂ ⁢ ( a l ) ⁢ ρ 1 ⁢ ρ 2 ⁢ ρ 1 ̂ ⁢ ( a l − 1 ) ∈ H ,

  4. ρ 2 ⁢ ρ 1 ⁢ ρ 2 ̂ ⁢ ( a 1 − 1 ) ⁢ ρ 1 ⁢ ρ 2 ̂ ⁢ ( a 1 ) = ρ 1 ⁢ ρ 2 ⁢ ρ 1 ̂ ⁢ ( a 1 − 1 ) ⁢ ρ 1 ⁢ ρ 2 ̂ ⁢ ( a 1 ) ∈ H ,

  5. ρ 1 ⁢ ρ 2 ̂ ⁢ ( a l ) ⁢ ρ 1 ̂ ⁢ ( a l − 1 ) ∈ H .

Since 𝐻 is invariant under the conjugation action of ρ 1 and ρ 2 , we note that all of the above possibilities can be boiled down to the fact that either a l ⁢ ρ 2 ̂ ⁢ ( a l − 1 ) ∈ H or a 1 − 1 ⁢ ρ 1 ̂ ⁢ ( a 1 ) ∈ H .

We first suppose that c = a l ⁢ ρ 2 ⁢ a l − 1 ⁢ ρ 2 ∈ H . Then we have ρ 2 ⁢ c ⁢ ρ 2 = c − 1 . Repeating the above arguments, we get c = d ⁢ ρ 2 ⁢ d − 1 ⁢ ρ 2 for some d ∈ H = K n ⁢ [ U k ] .

Set x l = 1 , w l = a l − 1 ⁢ d , y l = d − 1 ⁢ a l and z l = a 1 ⁢ a 2 ⁢ … ⁢ a l − 2 ⁢ w l . Then we see that z = x l ⁢ z l ⁢ y l . It is easy to check that y l ∈ K n ⁢ [ X 2 ∩ X ] , and hence z l satisfies condition (C). Note that the length of z l is l − 1 , whereas the length of 𝑧 is 𝑙. Thus, by the induction hypothesis, there exist x ′ , y ′ ∈ K n and z ′ ∈ H = K n ⁢ [ U k ] such that z l = x ′ ⁢ z ′ ⁢ y ′ with ρ 1 ⁢ x ′ ⁢ ρ 1 = x ′ and ρ 2 ⁢ y ′ ⁢ ρ 2 = y ′ . Now, set x = x l ⁢ x ′ , y = y ′ ⁢ y l and z = z ′ ∈ H = K n ⁢ [ U k ] . It follows that z = x ⁢ z 1 ⁢ y , x ∈ K n ⁢ [ X 1 ∩ X ] , y ∈ K n ⁢ [ X 2 ∩ X ] . Hence, we are done for the case a l ⁢ ρ 2 ⁢ a l − 1 ⁢ ρ 2 ∈ H . The case a l − 1 ⁢ ρ 1 ⁢ a l ⁢ ρ 1 ∈ H also follows similarly, and the proof is complete. ∎

Now, we set V 3 = U n and

V k = { α i , j ∈ U n ∣ ( i , j ) ∉ { 4 , 5 , … , k } × { 1 , 2 , 3 } }

for 4 ≤ k ≤ n . Note that

V n ⊆ V n − 1 ⊆ ⋯ ⊆ V k ⊆ V k − 1 ⊆ ⋯ ⊆ V 3 = U n

and each V k is invariant under the conjugation action of ρ 1 and ρ 2 .

Lemma 5.17

Let z ∈ K n ⁢ [ V k − 1 ] such that 𝑧 satisfies condition (C). Then there exist x ∈ K n ⁢ [ X 1 ∩ X ] , y ∈ K n ⁢ [ X 2 ∩ X ] and z 1 ∈ K n ⁢ [ V k ] such that z = x ⁢ z 1 ⁢ y and z 1 satisfies condition (C).

Proof

We proceed by induction on 𝑘. The case k = 3 holds by Lemma 5.16. We begin by noting that if z ∈ K n ⁢ [ V k ] , then we are done by taking x = 1 , y = 1 and z 1 = z . So we assume that z ∈ K n ⁢ [ V k − 1 ] ∖ K n ⁢ [ V k ] . Since both V k − 1 and V k are invariant under the conjugation action of ρ 1 and ρ 2 , we have

V k − 1 = V k ∪ { α k , 1 , α k , 2 , α k , 3 } .

We set

G j = K n ⁢ [ V k ∪ { α k , j } ] , 1 ≤ j ≤ 3 , H = G 1 ∩ G 2 ∩ G 3 = K n ⁢ [ V k ] , G = G 1 * H G 2 * H G 3 = K n ⁢ [ V k − 1 ] .

It is not difficult to check that

ρ 1 ⁢ G 1 ⁢ ρ 1 = G 2 , ρ 1 ⁢ G 2 ⁢ ρ 1 = G 1 , ρ 1 ⁢ G 3 ⁢ ρ 1 = G 3 , ρ 2 ⁢ G 1 ⁢ ρ 2 = G 1 , ρ 2 ⁢ G 2 ⁢ ρ 2 = G 3 , ρ 2 ⁢ G 3 ⁢ ρ 2 = G 2 .

By Lemma 5.4, we can write z = a 1 ⁢ a 2 ⁢ … ⁢ a l for some integer l ≥ 1 such that

  1. for each 1 ≤ i ≤ l , there exists j = j ⁢ ( i ) ∈ { 1 , 2 , 3 } such that a i ∈ G j ∖ H , and

  2. j ⁢ ( i ) ≠ j ⁢ ( i + 1 ) for all 1 ≤ i ≤ l − 1 .

As in the preceding lemma, we now argue by induction on 𝑙, the length of z = a 1 ⁢ a 2 ⁢ … ⁢ a l . Suppose that z = a 1 . Then either a 1 ∈ G 1 ∖ H or a 1 ∈ G 2 ∖ H or a 1 ∈ G 3 ∖ H . We first suppose that z = a 1 ∈ G 1 ∖ H and set

b 1 = a 1 ∈ G 1 ∖ H , b 2 = ρ 2 ⁢ a 1 − 1 ⁢ ρ 2 ∈ G 1 ∖ H , b 3 = ρ 2 ⁢ ρ 1 ⁢ a 1 ⁢ ρ 1 ⁢ ρ 2 ∈ G 3 ∖ H , b 4 = ρ 2 ⁢ ρ 1 ⁢ ρ 2 ⁢ a 1 − 1 ⁢ ρ 2 ⁢ ρ 1 ⁢ ρ 2 ∈ G 3 ∖ H , b 5 = ρ 1 ⁢ ρ 2 ⁢ a 1 ⁢ ρ 2 ⁢ ρ 1 ∈ G 2 ∖ H , b 6 = ρ 1 ⁢ a 1 − 1 ⁢ ρ 1 ∈ G 2 ∖ H .

Since z = a 1 satisfies condition (C), we have b 1 ⁢ b 2 ⁢ b 3 ⁢ b 4 ⁢ b 5 ⁢ b 6 = 1 . In view of Lemma 5.4 and the fact that 𝐻 is invariant under the conjugation action of ρ 1 and ρ 2 , this is possible only if b 1 ⁢ b 2 , b 3 ⁢ b 4 and b 5 ⁢ b 6 all lie in 𝐻. Set

c = a 1 ⁢ ρ 2 ⁢ a 1 − 1 ⁢ ρ 2 ∈ H .

Note that

ρ 2 ⁢ c ⁢ ρ 2 = ρ 2 ⁢ a 1 ⁢ ρ 2 ⁢ a 1 − 1 ⁢ ρ 2 ⁢ ρ 2 = ρ 2 ⁢ a 1 ⁢ ρ 2 ⁢ a 1 − 1 = c − 1 .

Hence, by Lemma 5.11, there exist α ∈ H = K n ⁢ [ V k ] and β ∈ K n ⁢ [ V k ∩ X 2 ] such that c = α ⁢ β ⁢ ρ 2 ⁢ α − 1 ⁢ ρ 2 with β 2 = 1 . This gives us a 1 ⁢ ρ 2 ⁢ a 1 − 1 ⁢ ρ 2 = c = α ⁢ β ⁢ ρ 2 ⁢ α − 1 ⁢ ρ 2 and consequently,

ρ 2 = a 1 − 1 ⁢ α ⁢ β ⁢ ρ 2 ⁢ α − 1 ⁢ a 1 ,

where α − 1 ⁢ a 1 ∈ K n . By Corollary 5.15, we have β = 1 , that is, c = α ⁢ ρ 2 ⁢ α − 1 ⁢ ρ 2 . Now, set

y = α − 1 ⁢ a 1 , x = 1 and z 1 = α ∈ H = K n ⁢ [ V k ] .

This gives us z = a 1 = x ⁢ z 1 ⁢ y , x ∈ K n ⁢ [ X 1 ∩ X ] and

ρ 2 ⁢ y ⁢ ρ 2 = ρ 2 ⁢ α − 1 ⁢ a 1 ⁢ ρ 2 = ( ρ 2 ⁢ α − 1 ⁢ ρ 2 ) ⁢ ( ρ 2 ⁢ a 1 ⁢ ρ 2 ) = ( α − 1 ⁢ c ) ⁢ ( c − 1 ⁢ a 1 ) = y ,

i.e., y ∈ K n ⁢ [ X 2 ∩ X ] . Thus, we are done for the case a 1 ∈ G 1 ∖ H .

Next suppose that a 1 ∈ G 2 ∖ H . We again set

b 1 = a 1 ∈ G 2 ∖ H , b 2 = ρ 2 ⁢ a 1 − 1 ⁢ ρ 2 ∈ G 3 ∖ H , b 3 = ρ 2 ⁢ ρ 1 ⁢ a 1 ⁢ ρ 1 ⁢ ρ 2 ∈ G 1 ∖ H , b 4 = ρ 2 ⁢ ρ 1 ⁢ ρ 2 ⁢ a 1 − 1 ⁢ ρ 2 ⁢ ρ 1 ⁢ ρ 2 ∈ G 2 ∖ H , b 5 = ρ 1 ⁢ ρ 2 ⁢ a 1 ⁢ ρ 2 ⁢ ρ 1 ∈ G 3 ∖ H , b 6 = ρ 1 ⁢ a 1 − 1 ⁢ ρ 1 ∈ G 1 ∖ H .

Since z = a 1 satisfies condition (C), we have b 1 ⁢ b 2 ⁢ b 3 ⁢ b 4 ⁢ b 5 ⁢ b 6 = 1 . This leads to a contradiction due to Lemma 5.4. Finally, suppose that a 1 ∈ G 3 ∖ H . We again set

b 1 = a 1 ∈ G 3 ∖ H , b 2 = ρ 2 ⁢ a 1 − 1 ⁢ ρ 2 ∈ G 2 ∖ H , b 3 = ρ 2 ⁢ ρ 1 ⁢ a 1 ⁢ ρ 1 ⁢ ρ 2 ∈ G 2 ∖ H , b 4 = ρ 2 ⁢ ρ 1 ⁢ ρ 2 ⁢ a 1 − 1 ⁢ ρ 2 ⁢ ρ 1 ⁢ ρ 2 ∈ G 1 ∖ H , b 5 = ρ 1 ⁢ ρ 2 ⁢ a 1 ⁢ ρ 2 ⁢ ρ 1 ∈ G 1 ∖ H , b 6 = ρ 1 ⁢ a 1 − 1 ⁢ ρ 1 ∈ G 3 ∖ H .

Since a 1 satisfies condition (C), again we have b 1 ⁢ b 2 ⁢ b 3 ⁢ b 4 ⁢ b 5 ⁢ b 6 = 1 . It follows that both b 2 ⁢ b 3 and b 4 ⁢ b 5 is in 𝐻. Hence, we have a 1 − 1 ⁢ ρ 1 ⁢ a 1 ⁢ ρ 1 = c (say) ∈ H . Note that ρ 1 ⁢ c ⁢ ρ 1 = c − 1 . Hence, by Lemma 5.11, there exist d ∈ H = K n ⁢ [ V k ] and t ∈ K n ⁢ [ V k ∩ X 1 ] such that c = d ⁢ t ⁢ ρ 1 ⁢ d − 1 ⁢ ρ 1 with t 2 = 1 . Thus, we have

a 1 − 1 ⁢ ρ 1 ⁢ a 1 ⁢ ρ 1 = c = d ⁢ t ⁢ ρ 1 ⁢ d − 1 ⁢ ρ 1 ,

and consequently ρ 1 = a 1 ⁢ d ⁢ t ⁢ ρ 1 ⁢ d − 1 ⁢ a 1 − 1 , where a 1 ⁢ d ∈ K n . By Corollary 5.15, we have t = 1 . Now, set x = a 1 ⁢ d , y = 1 and z 1 = d − 1 ∈ K n ⁢ [ V k ] . It follows that z = x ⁢ z 1 ⁢ y and

ρ 1 ⁢ x ⁢ ρ 1 = ρ 1 ⁢ a 1 ⁢ d ⁢ ρ 1 = ( ρ 1 ⁢ a 1 ⁢ ρ 1 ) ⁢ ( ρ 1 ⁢ d ⁢ ρ 1 ) = ( a 1 ⁢ c ) ⁢ ( c − 1 ⁢ d ) = a 1 ⁢ d = x ,

i.e., x ∈ K n ⁢ [ X 1 ∩ X ] . Hence, we are done for the case l = 1 .

Let us now suppose that l ≥ 2 and that the induction hypothesis holds. Since z = a 1 ⁢ a 2 ⁢ … ⁢ a l satisfy condition (C), we have

( a 1 ⁢ … ⁢ a l ) ⁢ ρ 2 ̂ ⁢ ( a l − 1 ⁢ … ⁢ a 1 − 1 ) ⁢ ρ 2 ⁢ ρ 1 ̂ ⁢ ( a 1 ⁢ … ⁢ a l ) ⁢ ρ 2 ⁢ ρ 1 ⁢ ρ 2 ̂ ( a l − 1 ⁢ … ⁢ a 1 − 1 ) ⁢ ρ 1 ⁢ ρ 2 ̂ ⁢ ( a 1 ⁢ … ⁢ a l ) ⁢ ρ 1 ̂ ⁢ ( a l − 1 ⁢ … ⁢ a 1 − 1 ) = 1 .

Then, by Lemma 5.4, we have either of the following:

  1. a l ⁢ ρ 2 ̂ ⁢ ( a l − 1 ) ∈ H ,

  2. ρ 2 ̂ ⁢ ( a 1 − 1 ) ⁢ ρ 2 ⁢ ρ 1 ̂ ⁢ ( a 1 ) ∈ H ,

  3. ρ 2 ⁢ ρ 1 ̂ ⁢ ( a l ) ⁢ ρ 2 ⁢ ρ 1 ⁢ ρ 2 ̂ ⁢ ( a l − 1 ) = ρ 2 ⁢ ρ 1 ̂ ⁢ ( a l ) ⁢ ρ 1 ⁢ ρ 2 ⁢ ρ 1 ̂ ⁢ ( a l − 1 ) ∈ H ,

  4. ρ 2 ⁢ ρ 1 ⁢ ρ 2 ̂ ⁢ ( a 1 − 1 ) ⁢ ρ 1 ⁢ ρ 2 ̂ ⁢ ( a 1 ) = ρ 1 ⁢ ρ 2 ⁢ ρ 1 ̂ ⁢ ( a 1 − 1 ) ⁢ ρ 1 ⁢ ρ 2 ̂ ⁢ ( a 1 ) ∈ H ,

  5. ρ 1 ⁢ ρ 2 ̂ ⁢ ( a l ) ⁢ ρ 1 ̂ ⁢ ( a l − 1 ) ∈ H .

Since 𝐻 is invariant under the conjugation action of ρ 1 and ρ 2 , either of the above possibilities can be reduced to either a l ⁢ ρ 2 ̂ ⁢ ( a l − 1 ) ∈ H or a 1 − 1 ⁢ ρ 1 ̂ ⁢ ( a 1 ) ∈ H .

We first suppose that c = a l ⁢ ρ 2 ⁢ a l − 1 ⁢ ρ 2 ∈ H . So we get ρ 2 ⁢ c ⁢ ρ 2 = c − 1 . Repeating the above arguments, we get c = d ⁢ ρ 2 ⁢ d − 1 ⁢ ρ 2 for some d ∈ H = K n ⁢ [ V k ] . Set

x l = 1 , w l = a l − 1 ⁢ d , y l = d − 1 ⁢ a l and z l = a 1 ⁢ a 2 ⁢ … ⁢ a l − 2 ⁢ w l .

Note that z = x l ⁢ z l ⁢ y l . It is easy to check that y l ∈ K n ⁢ [ X 2 ∩ X ] , and hence z l satisfies condition (C). Note that the length of z l is l − 1 , whereas the length of 𝑧 is 𝑙. Hence, by the induction hypothesis, there exist x ′ , y ′ ∈ K n and z ′ ∈ H = K n ⁢ [ V k ] such that z l = x ′ ⁢ z ′ ⁢ y ′ with ρ 1 ⁢ x ′ ⁢ ρ 1 = x ′ and ρ 2 ⁢ y ′ ⁢ ρ 2 = y ′ . Now, set x = x l ⁢ x ′ , y = y ′ ⁢ y l and z = z ′ ∈ H = K n ⁢ [ V k ] . It follows that z = x ⁢ z 1 ⁢ y , x ∈ K n ⁢ [ X 1 ∩ X ] , y ∈ K n ⁢ [ X 2 ∩ X ] . Hence, we are done for the case a l ⁢ ρ 2 ⁢ a l − 1 ⁢ ρ 2 ∈ H . The other cases follow in a similar manner. ∎

Now, we set

(5.2) W 1 , 1 = { α i , j ∈ X ∣ ( i , j ) ∈ { ( 1 , 2 ) , ( 2 , 3 ) , ( 3 , 1 ) } } ,
(5.3) W 1 , 2 = { α i , j ∈ X ∣ ( i , j ) ∈ { ( 2 , 1 ) , ( 3 , 2 ) , ( 1 , 3 ) } } .
It is easy to check that

ρ 1 ⁢ ( W 1 , 1 ) ⁢ ρ 1 = ρ 2 ⁢ ( W 1 , 1 ) ⁢ ρ 2 = W 1 , 2 and ρ 1 ⁢ ( W 1 , 2 ) ⁢ ρ 1 = ρ 2 ⁢ ( W 1 , 2 ) ⁢ ρ 2 = W 1 , 1 .

Lemma 5.18

If z ∈ K n ⁢ [ W 1 , 1 ] ∗ K n ⁢ [ W 1 , 2 ] satisfies condition (C), then z = 1 .

Proof

Suppose that z ≠ 1 . Set G 1 = K n ⁢ [ W 1 , 1 ] , G 2 = K n ⁢ [ W 1 , 2 ] , G = G 1 ∗ G 2 . Due to Lemma 5.4, we can write z = a 1 ⁢ a 2 ⁢ … ⁢ a l for some integer l ≥ 1 such that

  1. for each 1 ≤ i ≤ l , there exists j = j ⁢ ( i ) ∈ { 1 , 2 } such that a i ∈ G j ∖ { 1 } , and

  2. j ⁢ ( i ) ≠ j ⁢ ( i + 1 ) for all 1 ≤ i ≤ l − 1 .

We now argue by induction on 𝑙 to arrive at a contradiction. Suppose that l = 1 . Then either a 1 ∈ G 1 ∖ { 1 } or a 1 ∈ G 2 ∖ { 1 } . We can assume that a 1 ∈ G 1 ∖ { 1 } since the case a 1 ∈ G 2 ∖ { 1 } goes along parallel lines. We now set

b 1 = a 1 ∈ G 1 ∖ { 1 } , b 2 = ρ 2 ⁢ a 1 − 1 ⁢ ρ 2 ∈ G 2 ∖ { 1 } , b 3 = ρ 2 ⁢ ρ 1 ⁢ a 1 ⁢ ρ 1 ⁢ ρ 2 ∈ G 1 ∖ { 1 } , b 4 = ρ 2 ⁢ ρ 1 ⁢ ρ 2 ⁢ a 1 − 1 ⁢ ρ 2 ⁢ ρ 1 ⁢ ρ 2 ∈ G 2 ∖ { 1 } , b 5 = ρ 1 ⁢ ρ 2 ⁢ a 1 ⁢ ρ 2 ⁢ ρ 1 ∈ G 1 ∖ { 1 } , b 6 = ρ 1 ⁢ a 1 − 1 ⁢ ρ 1 ∈ G 2 ∖ { 1 } .

Since z = a 1 satisfies condition (C), we have b 1 ⁢ b 2 ⁢ b 3 ⁢ b 4 ⁢ b 5 ⁢ b 6 = 1 . But this leads to a contradiction due to Lemma 5.4.

Now, suppose that l ≥ 2 and the induction hypothesis holds. Due to Lemma 5.4, either a l − 1 ⁢ ρ 1 ⁢ a l ⁢ ρ 1 = 1 or a l ⁢ ρ 2 ⁢ a l − 1 ⁢ ρ 2 = 1 . We suppose that a l ⁢ ρ 2 ⁢ a l − 1 ⁢ ρ 2 = 1 since the case a l − 1 ⁢ ρ 1 ⁢ a l ⁢ ρ 1 = 1 also goes along parallel lines. Thus, we have ρ 2 ⁢ a l ⁢ ρ 2 = a l . Set x = 1 , w = a 1 ⁢ a 2 ⁢ … ⁢ a l − 1 ∈ G and y = a l . Note that z = x ⁢ w ⁢ y and w ≠ 1 (since the length of 𝑤 is non-zero). Since ρ 2 ⁢ a l ⁢ ρ 2 = a l , it follows that

y = a l ∈ K n ⁢ [ X 2 ] ,

and hence 𝑤 satisfies condition (C). Note that the length of 𝑤 is l − 1 , whereas the length of 𝑧 is 𝑙. Hence, by the induction hypothesis, w = 1 , a contradiction. ∎

Lemma 5.19

Let n ≥ 3 and let 𝑋 be a subset of 𝒮 that is invariant under the conjugation action of ρ 1 and ρ 2 . Suppose ϕ : S n → VT n is a homomorphism such that θ ⁢ ϕ is the identity on S n with ϕ ⁢ ( τ 1 ) = ρ 1 and ϕ ⁢ ( τ 2 ) = α ⁢ ρ 2 ⁢ α − 1 for some α ∈ K n ⁢ [ X ] . Then there exist α 1 ∈ K n ⁢ [ X 1 ] and α 2 ∈ K n ⁢ [ X 2 ] such that α = α 1 ⁢ α 2 .

Proof

Since ( τ 2 ⁢ τ 1 ) 3 = 1 , we have ( α ⁢ ρ 2 ⁢ α − 1 ⁢ ρ 1 ) 3 = 1 , which shows upon expansion that 𝛼 satisfies condition (C). We proceed along the following three steps.

Step (1): We first show by induction on 3 ≤ k ≤ n that there exist

x ∈ K n ⁢ [ X 1 ∩ X ] , y ∈ K n ⁢ [ X 2 ∩ X ] and z ∈ K n ⁢ [ U k ]

such that α = x ⁢ z ⁢ y and 𝑧 satisfies condition (C). The case k = 3 is trivial since we can take x = y = 1 and z = α . Suppose that 4 ≤ k ≤ n and the induction hypothesis holds, i.e., there exist x 1 ∈ K n ⁢ [ X 1 ∩ X ] , y 1 ∈ K n ⁢ [ X 2 ∩ X ] and z 1 ∈ K n ⁢ [ U k − 1 ] such that α = x 1 ⁢ z 1 ⁢ y 1 and z 1 satisfies condition (C). By Lemma 5.16, we get x 2 ∈ K n ⁢ [ X 1 ∩ X ] , y 2 ∈ K n ⁢ [ X 2 ∩ X ] and z 2 ∈ K n ⁢ [ U k ] such that z 2 satisfies condition (C) and z 1 = x 2 ⁢ z 2 ⁢ y 2 . Thus, we have α = x 1 ⁢ x 2 ⁢ z 2 ⁢ y 2 ⁢ y 1 . Taking x = x 1 ⁢ x 2 , z = z 2 and y = y 2 ⁢ y 1 completes step (1).

Step (2): We now show by induction on 3 ≤ k ≤ n that there exist

x ∈ K n ⁢ [ X 1 ∩ X ] , y ∈ K n ⁢ [ X 2 ∩ X ] and z ∈ K n ⁢ [ V k ]

such that α = x ⁢ z ⁢ y and 𝑧 satisfies condition (C). Since V 3 = U n , we are done for the case k = 3 from step (1). Now, suppose that 4 ≤ k ≤ n and the induction hypothesis holds, i.e., there exist x 1 ∈ K n ⁢ [ X 1 ∩ X ] , y 1 ∈ K n ⁢ [ X 2 ∩ X ] , z 1 ∈ K n ⁢ [ V k − 1 ] such that α = x 1 ⁢ z 1 ⁢ y 1 and z 1 satisfies condition (C). By Lemma 5.17, we get x 2 ∈ K n ⁢ [ X 1 ∩ X ] , y 2 ∈ K n ⁢ [ X 2 ∩ X ] and z 2 ∈ K n ⁢ [ V k ] such that z 2 satisfies condition (C) and z 1 = x 2 ⁢ z 2 ⁢ y 2 . Thus, we have α = x 1 ⁢ x 2 ⁢ z 2 ⁢ y 2 ⁢ y 1 . Taking x = x 1 ⁢ x 2 , z = z 2 and y = y 2 ⁢ y 1 completes step (2).

Step (3): We now set

W 1 = { α i , j ∈ X ∣ ( i , j ) ∈ { 1 , 2 , 3 } × { 1 , 2 , 3 } } , W 2 = { α i , j ∈ X ∣ ( i , j ) ∈ { 4 , 5 , … , k } × { 4 , 5 , … , k } } .

It follows that V n = W 1 ⊔ W 2 and hence

K n ⁢ [ V n ] = K n ⁢ [ W 1 ] × K n ⁢ [ W 2 ] .

Also, note that

K n ⁢ [ W 1 ] = K n ⁢ [ W 1 , 1 ] ∗ K n ⁢ [ W 1 , 2 ] ⁢ and K n ⁢ [ W 2 ] = K n ⁢ [ X 1 ∩ X ] ∩ K n ⁢ [ X 2 ∩ X ] ,

where W 1 , 1 and W 1 , 2 are given by (5.2) and (5.3). By step (2), we have

x ∈ K n ⁢ [ X 1 ∩ X ] , y ∈ K n ⁢ [ X 2 ∩ X ] and z ∈ K n ⁢ [ V n ] = K n ⁢ [ W 1 ] × K n ⁢ [ W 2 ]

such that α = x ⁢ z ⁢ y and 𝑧 satisfies condition (C). Let z 1 ∈ K n ⁢ [ W 1 ] and z 2 ∈ K n ⁢ [ W 2 ] such that z = z 1 ⁢ z 2 = 1 ⁢ z 1 ⁢ z 2 . It is easy to check that z 2 ∈ K n ⁢ [ X 2 ] , and hence z 1 satisfies condition (C). Now, Lemma 5.18 gives z 1 = 1 . Thus, we get α = x ⁢ z 2 ⁢ y , where

x ∈ K n ⁢ [ X 1 ∩ X ] , y ∈ K n ⁢ [ X 2 ∩ X ] , z 2 ∈ K n ⁢ [ W 2 ] = K n ⁢ [ X 1 ∩ X ] ∩ K n ⁢ [ X 2 ∩ X ] .

Taking α 1 = x and α 2 = z 2 ⁢ y completes the proof of the lemma. ∎

Lemma 5.19 can be generalised as follows.

Lemma 5.20

Let n ≥ 3 and let 𝑋 be a subset of 𝒮 that is invariant under the conjugation action of both ρ k and ρ k + 1 for some fixed 1 ≤ k ≤ n − 2 . Suppose that ϕ : S n → VT n is a homomorphism such that θ ⁢ ϕ is the identity on S n , ϕ ⁢ ( τ k ) = ρ k and ϕ ⁢ ( τ k + 1 ) = α ⁢ ρ k + 1 ⁢ α − 1 for some α ∈ K n ⁢ [ X ] . Then there exist α 1 ∈ K n ⁢ [ X k ∩ X ] and α 2 ∈ K n ⁢ [ X k + 1 ∩ X ] such that α = α 1 ⁢ α 2 .

Proof

The case k = 1 is considered in Lemma 5.19. So we assume that k ≥ 2 . Choose an element

w ∈ ⟨ ρ 1 , … , ρ n − 1 ⟩ such that w − 1 ⁢ ρ k ⁢ w = ρ 1 ⁢ and ⁢ w − 1 ⁢ ρ k + 1 ⁢ w = ρ 2 .

It is not difficult to see that such 𝑤 exists. In fact, if w 0 is the cycle ( 1 , 2 , … , n ) , then taking w = w 0 k − 1 , we see that

w − 1 ⁢ ρ k ⁢ w = ρ 1 and w − 1 ⁢ ρ k + 1 ⁢ w = ρ 2 .

Set g = θ ⁢ ( w ) . It follows that g − 1 ⁢ τ k ⁢ g = τ 1 and g − 1 ⁢ τ k + 1 ⁢ g = τ 2 .

Now, set ϕ ̄ = w ̂ − 1 ⁢ ϕ ⁢ g ̂ , where w ̂ and g ̂ are inner automorphisms induced by 𝑤 and 𝑔 in VT n and S n , respectively. Note that

θ ⁢ w ̂ − 1 ⁢ ϕ ⁢ g ̂ ⁢ ( τ ) = θ ⁢ w ̂ − 1 ⁢ ϕ ⁢ ( g ⁢ τ ⁢ g − 1 ) = θ ⁢ w ̂ − 1 ⁢ ( ϕ ⁢ ( g ) ⁢ ϕ ⁢ ( τ ) ⁢ ϕ ⁢ ( g − 1 ) ) = θ ⁢ ( w − 1 ⁢ ϕ ⁢ ( g ) ⁢ ϕ ⁢ ( τ ) ⁢ ϕ ⁢ ( g − 1 ) ⁢ w ) = θ ⁢ ( w − 1 ) ⁢ θ ⁢ ( ϕ ⁢ ( g ) ) ⁢ θ ⁢ ( ϕ ⁢ ( τ ) ) ⁢ θ ⁢ ( ϕ ⁢ ( g − 1 ) ) ⁢ θ ⁢ ( w ) = g − 1 ⁢ g ⁢ τ ⁢ g − 1 ⁢ g = τ

for all τ ∈ S n , and hence θ ⁢ w ̂ − 1 ⁢ ϕ ⁢ g ̂ ⁢ ( τ ) is the identity on S n . Also, note that

w ̂ − 1 ⁢ ϕ ⁢ g ̂ ⁢ ( τ 1 ) = w ̂ − 1 ⁢ ϕ ⁢ ( g ⁢ τ 1 ⁢ g − 1 ) = w ̂ − 1 ⁢ ϕ ⁢ ( τ k ) = w ̂ − 1 ⁢ ( ρ k ) = w − 1 ⁢ ρ k ⁢ w = ρ 1 ,
w ̂ ⁢ ϕ ⁢ g ̂ − 1 ⁢ ( τ 2 ) = w ̂ ⁢ ϕ ⁢ ( g ⁢ τ 2 ⁢ g − 1 ) = w ̂ ⁢ ϕ ⁢ ( τ k + 1 ) = w ̂ ⁢ ( α ⁢ ρ k + 1 ⁢ α − 1 ) = ( w − 1 ⁢ α ⁢ w ) ⁢ ρ 2 ⁢ ( w − 1 ⁢ α − 1 ⁢ w ) .
The result now follows from Lemma 5.19 and Lemma 5.7. ∎

5.4 Main results

We now prove the main results of this section. For each 1 ≤ k ≤ n , we set

Y k = { α i , j ∈ S ∣ k ≤ i ≠ j ≤ n } .

We note that

∅ = Y n ⊆ Y n − 1 ⊆ ⋯ ⊆ Y k ⊆ Y k − 1 ⊆ ⋯ ⊆ Y 1 = S .

Proposition 5.21

Let n ≥ 3 and let ϕ : S n → VT n be a homomorphism such that θ ⁢ ϕ is the identity on S n . Then 𝜙 is conjugate to 𝜆.

Proof

We claim that, for each 1 ≤ m ≤ n − 1 , there exists an inner automorphism w ̂ of VT n induced by some w ∈ K n such that w ̂ ⁢ ϕ ⁢ ( τ i ) = ρ i for all 1 ≤ i ≤ m . The case m = n − 1 would establish the proposition.

We prove the claim by induction on 𝑚. By Corollary 5.14, there exists x 1 ∈ K n such that ϕ ⁢ ( τ 1 ) = x 1 ⁢ ρ 1 ⁢ x 1 − 1 . This gives us x 1 ̂ − 1 ⁢ ϕ ⁢ ( τ 1 ) = ρ 1 . We now consider the case m = 2 . Set ϕ 1 = x 1 ̂ − 1 ⁢ ϕ , and note that θ ⁢ ϕ 1 is also the identity on S n . Again, by Corollary 5.14, there exists x 2 ∈ K n such that ϕ 1 ⁢ ( τ 2 ) = x 2 ⁢ ρ 2 ⁢ x 2 − 1 . We also have ϕ 1 ⁢ ( τ 1 ) = ρ 1 . Thus, by Lemma 5.19, there exist α 1 ∈ K n ⁢ [ X 1 ] , α 2 ∈ K n ⁢ [ X 2 ] such that x 2 = α 1 ⁢ α 2 . This gives

ϕ 1 ⁢ ( τ 2 ) = x 2 ⁢ ρ 2 ⁢ x 2 − 1 = α 1 ⁢ α 2 ⁢ ρ 2 ⁢ α 2 − 1 ⁢ α 1 − 1 = α 1 ⁢ ρ 2 ⁢ α 1 − 1 .

We can check that x 1 ⁢ α 1 ̂ − 1 ⁢ ϕ ⁢ ( τ i ) = ρ i for i = 1 , 2 , and the claim holds for m = 2 .

Now, suppose that 3 ≤ m ≤ n − 1 and that the claim holds for m − 1 , i.e., there exists some w ∈ K n such that w ̂ ⁢ ϕ ⁢ ( τ i ) = ρ i for all 1 ≤ i ≤ m − 1 . Set ϕ m − 1 = w ̂ ⁢ ϕ . Since θ ⁢ ϕ m − 1 is the identity on S n , there exists y ∈ K n such that ϕ m − 1 ⁢ ( τ m ) = y ⁢ ρ m . Since τ m ⁢ τ i = τ i ⁢ τ m , we have

y ⁢ ρ m ⁢ ρ i = ρ i ⁢ y ⁢ ρ m for all ⁢ 1 ≤ i ≤ m − 2 .

Thus, ρ i ⁢ y ⁢ ρ i = y , i.e., 𝑦 is a fixed-point under the conjugation action of ρ i for each 1 ≤ i ≤ m − 2 . By Corollary 5.8, we have

y ∈ K n ρ i ̂ = K n ⁢ [ X i ] for each ⁢ 1 ≤ i ≤ m − 2 .

Since

⋂ i = 1 m − 2 X i = Y m ,

it follows that y ∈ K n ⁢ [ Y m ] . Since τ m 2 = 1 , we have 1 = ϕ m − 1 ⁢ ( τ m ) 2 = y ⁢ ρ m ⁢ y ⁢ ρ m , i.e., ρ m ⁢ y ⁢ ρ m = y − 1 . Note that Y m is invariant under conjugation by ρ m . Thus, by Lemma 5.11, there exist

x m ∈ K n ⁢ [ Y m ] ≤ K n ⁢ [ Y m − 1 ] and β ∈ K n ⁢ [ Y m ∩ X m ] ≤ K n ⁢ [ X m ]

such that y = x m ⁢ β ⁢ ρ m ⁢ x m − 1 ⁢ ρ m and β 2 = 1 . This gives

ϕ m − 1 ⁢ ( τ m ) = y ⁢ ρ m = x m ⁢ β ⁢ ρ m ⁢ x m − 1 .

Now, due to Corollary 5.13, we have β = 1 , and consequently

ϕ m − 1 ⁢ ( τ m ) = x m ⁢ ρ m ⁢ x m − 1 .

Note that Y m − 1 is invariant under the conjugation action of both ρ m − 1 and ρ m . Due to Lemma 5.20, there exist

u ∈ K n ⁢ [ Y m − 1 ∩ X m − 1 ] = K n ⁢ [ Y m + 1 ] and v ∈ K n ⁢ [ Y m − 1 ∩ X m ]

such that x m = u ⁢ v . Thus, we have ϕ m − 1 ⁢ ( τ m ) = x m ⁢ ρ m ⁢ x m − 1 = u ⁢ ρ m ⁢ u − 1 . It follows from the choice of 𝑢 that u ̂ − 1 ⁢ ϕ m − 1 ⁢ ( τ i ) = ρ i for all 1 ≤ i ≤ m . This is equivalent to u ⁢ w ̂ − 1 ⁢ ϕ ⁢ ( τ i ) = ρ i for all 1 ≤ i ≤ m , and the proof is complete. ∎

Finally, we present the main result of this section.

Theorem 5.22

Let 𝑛 and 𝑚 be integers such that n ≥ m , n ≥ 5 and m ≥ 2 . Let ϕ : S n → VT m be a homomorphism. Then, up to conjugation of homomorphisms, one of the following assertions holds:

  1. 𝜙 is abelian,

  2. n = m and ϕ = λ ,

  3. n = m = 6 and ϕ = λ ⁢ ν .

Proof

Consider the composition S n → ϕ VT m → θ S m . By Proposition 4.1, one of the following assertions holds for θ ⁢ ϕ :

  1. θ ⁢ ϕ is abelian,

  2. n = m and θ ⁢ ϕ = id ,

  3. n = m = 6 and θ ⁢ ϕ = ν .

Case (1): Let θ ⁢ ϕ be abelian. We claim that there exists

w ∈ S m such that θ ⁢ ϕ ⁢ ( τ i ) = w ⁢ for all ⁢ i .

Suppose on the contrary that there exist 𝑖 and elements w 1 ≠ w 2 in S m such that θ ⁢ ϕ ⁢ ( τ i ) = w 1 and θ ⁢ ϕ ⁢ ( τ i + 1 ) = w 2 . The braid relation τ i ⁢ τ i + 1 ⁢ τ i = τ i + 1 ⁢ τ i ⁢ τ i + 1 gives w 1 ⁢ w 2 ⁢ w 1 = w 2 ⁢ w 1 ⁢ w 2 . Now, θ ⁢ ϕ being abelian implies that w 1 = w 2 , and the claim holds.

Set λ ⁢ ( w ) = g . It follows that, for each 1 ≤ i ≤ n − 1 , there exists α i ∈ K n such that ϕ ⁢ ( τ i ) = α i ⁢ g . We claim that α i = α i + 1 for all 𝑖. Since τ i 2 = 1 , it follows that g 2 = 1 and α i ⁢ g = g ⁢ α i − 1 for all 𝑖. Since ( τ i ⁢ τ i + 1 ) 3 = 1 , it follows that 1 = ( α i ⁢ g ⁢ α i + 1 ⁢ g ) 3 = ( α i ⁢ g ⁢ g ⁢ α i + 1 − 1 ) 3 = ( α i ⁢ α i + 1 − 1 ) 3 . Thus, the element α i ⁢ α i + 1 − 1 ∈ K n has order dividing three. Since K n is a right-angled Coxeter group, a non-trivial finite order element must have order two [10, Proposition 1.2]. This implies that α i = α i + 1 . Hence, the homomorphism 𝜙 is abelian.

Case (2): Let n = m and θ ⁢ ϕ = id . Then, by Proposition 5.21, the homomorphism 𝜙 is conjugate to 𝜆.

Case (3): Lastly, suppose that n = m = 6 and θ ⁢ ϕ = ν . Then θ ⁢ ϕ ⁢ ν − 1 is the identity on S n . By Proposition 5.21, the homomorphism ϕ ⁢ ν − 1 is conjugate to 𝜆, and equivalently, 𝜙 is conjugate to λ ⁢ ν . ∎

6 Homomorphisms from VT n to VT m

Recall that the non-inner automorphism ν : S 6 → S 6 is defined on generators as

ν ⁢ ( τ 1 ) = ( 1 , 2 ) ⁢ ( 3 , 4 ) ⁢ ( 5 , 6 ) , ν ⁢ ( τ 2 ) = ( 2 , 3 ) ⁢ ( 1 , 5 ) ⁢ ( 4 , 6 ) , ν ⁢ ( τ 3 ) = ( 1 , 3 ) ⁢ ( 2 , 4 ) ⁢ ( 5 , 6 ) , ν ⁢ ( τ 4 ) = ( 1 , 2 ) ⁢ ( 3 , 5 ) ⁢ ( 4 , 6 ) , ν ⁢ ( τ 5 ) = ( 2 , 3 ) ⁢ ( 1 , 4 ) ⁢ ( 5 , 6 ) .

We set v i = λ ⁢ ν ⁢ ( τ i ) for each 1 ≤ i ≤ 5 .

Lemma 6.1

If H = ⟨ v 3 , v 4 , v 5 ⟩ , then

KT 6 H := { x ∈ KT 6 ∣ w ⁢ x ⁢ w − 1 = x ⁢ for all ⁢ w ∈ H } = { 1 } .

Proof

Consider the subset H ′ = { v 3 , v 4 , v 5 , v 3 ⁢ v 4 ⁢ v 3 , v 4 ⁢ v 5 ⁢ v 4 , v 3 ⁢ v 4 ⁢ v 5 ⁢ v 4 ⁢ v 3 } of 𝐻. Since 𝐻 is a subgroup of ⟨ ρ 1 , … , ρ 5 ⟩ ≅ S 6 , we can view elements of H ′ in terms of permutations as

v 3 = ( 1 , 3 ) ⁢ ( 2 , 4 ) ⁢ ( 5 , 6 ) , v 3 ⁢ v 4 ⁢ v 3 = ( 1 , 6 ) ⁢ ( 2 , 5 ) ⁢ ( 3 , 4 ) , v 4 = ( 1 , 2 ) ⁢ ( 3 , 5 ) ⁢ ( 4 , 6 ) , v 4 ⁢ v 5 ⁢ v 4 = ( 1 , 5 ) ⁢ ( 2 , 6 ) ⁢ ( 3 , 4 ) , v 5 = ( 2 , 3 ) ⁢ ( 1 , 4 ) ⁢ ( 5 , 6 ) , v 3 ⁢ v 4 ⁢ v 5 ⁢ v 4 ⁢ v 3 = ( 1 , 2 ) ⁢ ( 3 , 6 ) ⁢ ( 4 , 5 ) .

For fixed 1 ≤ i < j ≤ 6 , set U i ⁢ j = S ∖ { α i , j , α j , i } , U i ⁢ j ′ = U i ⁢ j ∪ { α i , j } and U i ⁢ j ′′ = U i ⁢ j ∪ { α j , i } . Then, by Lemma 5.2, we have

KT 6 = KT 6 ⁢ [ U i ⁢ j ′ ] ∗ KT 6 ⁢ [ U i ⁢ j ] KT 6 ⁢ [ U i ⁢ j ′′ ] .

Note that the set H ′ is taken in such a way that each transposition ( i , j ) appears in the decomposition of some element of H ′ . Let us choose an element w ∈ H ′ containing the transposition ( i , j ) in its decomposition. We note that w ⁢ ( KT 6 ⁢ [ U i ⁢ j ′ ] ) ⁢ w = KT 6 ⁢ [ U i ⁢ j ′′ ] and w ⁢ ( K 6 ⁢ [ U i ⁢ j ′′ ] ) ⁢ w = K 6 ⁢ [ U i ⁢ j ′ ] . Thus, by Lemma 5.5, we have

KT 6 H ⊆ KT 6 w ̂ ⊆ KT 6 ⁢ [ U i ⁢ j ] for all ⁢ 1 ≤ i < j ≤ 6 .

Since ⋂ 1 ≤ i < j ≤ 6 U i ⁢ j = ∅ , by Lemma 5.1, we have KT 6 H = { 1 } . ∎

For each m ∈ Z , consider the homomorphism ϕ m : VT n → VT n given on generators by

ϕ m ⁢ ( s i ) = ( s i ⁢ ρ i ) m ⁢ ρ i and ϕ m ⁢ ( ρ i ) = ρ i .

Setting ϕ − 1 = ζ , the main result of this section is as follows.

Theorem 6.2

Let 𝑛 and 𝑚 be integers such that n ≥ m , n ≥ 5 and m ≥ 2 , and let ϕ : VT n → VT m be a homomorphism. Then, up to conjugation of homomorphisms, one of the following assertions holds:

  1. 𝜙 is abelian,

  2. n = m and ϕ ∈ { λ ⁢ π , λ ⁢ θ , ϕ m , ζ ⁢ ϕ m , where ⁢ m ∈ Z } ,

  3. n = m = 6 and ϕ ∈ { λ ⁢ ν ⁢ θ , λ ⁢ ν ⁢ π } .

Proof

Consider the composition S n → λ VT n → ϕ VT m . By Proposition 5.22, one of the following assertions holds for ϕ ⁢ λ :

  1. ϕ ⁢ λ is abelian,

  2. n = m and ϕ ⁢ λ = λ ,

  3. n = m = 6 and ϕ ⁢ λ = λ ⁢ ν .

Case (1): This case is similar to case (1) of Theorem 4.2. If ϕ ⁢ λ is abelian, then there exists w ∈ VT m such that ϕ ⁢ λ ⁢ ( τ i ) = w for all 𝑖. Equivalently, ϕ ⁢ ( ρ i ) = w for all 𝑖. Let ϕ ⁢ ( s 1 ) = z . The relation ρ i ⁢ s i + 1 ⁢ ρ i = ρ i + 1 ⁢ s i ⁢ ρ i + 1 gives

ϕ ⁢ ( s i ) = ϕ ⁢ ( s i + 1 ) = z for all ⁢ i .

Finally, the relation s 1 ⁢ ρ 3 = ρ 3 ⁢ s 1 gives z ⁢ w = w ⁢ z , and hence 𝜙 is abelian.

Case (2): Let m = n and ϕ ⁢ λ = λ . This implies that ϕ ⁢ ( ρ i ) = ρ i for all 𝑖. We now determine ϕ ⁢ ( s i ) for all 𝑖. Since VT n = K n ⋊ S n , we have ϕ ⁢ ( s i ) = a i ⁢ λ ⁢ ( w i ) for some a i ∈ K n and w i ∈ S n . For each 3 ≤ j ≤ n − 1 , we have

τ j ⁢ w 1 = θ ⁢ ϕ ⁢ ( ρ j ⁢ s 1 ) = θ ⁢ ϕ ⁢ ( s 1 ⁢ ρ j ) = w 1 ⁢ τ j .

This implies that w 1 lies in centraliser of ⟨ τ 3 , τ 4 , … , τ n − 1 ⟩ in S n , which is ⟨ τ 1 ⟩ . Thus, either w 1 = 1 or w 1 = τ 1 .

Case (2a): Suppose that w 1 = 1 . For each 3 ≤ k ≤ n − 1 , we have s 1 ⁢ ρ k = ρ k ⁢ s 1 , and hence a 1 = ρ k ⁢ a 1 ⁢ ρ k . Thus, if we set X k = { α i , j ∈ S ∣ i , j ∉ { k , k + 1 } } , then a 1 ∈ K n ⁢ [ X k ] for all 3 ≤ k ≤ n − 1 . We have

⋂ 3 ≤ k ≤ n − 1 X k = { α 1 , 2 , α 2 , 1 } ,

and hence a 1 ∈ K n ⁢ [ { α 1 , 2 , α 2 , 1 } ] ≅ Z 2 ∗ Z 2 . Elements in K n ⁢ [ { α 1 , 2 , α 2 , 1 } ] are of the form ( α 1 , 2 ⁢ α 2 , 1 ) m or α 1 , 2 ⁢ ( α 2 , 1 ⁢ α 1 , 2 ) m or α 2 , 1 ⁢ ( α 1 , 2 ⁢ α 2 , 1 ) m for some integer 𝑚. The only order two elements are α 1 , 2 , α 2 , 1 , α 1 , 2 ⁢ ( α 2 , 1 ⁢ α 1 , 2 ) m and α 2 , 1 ⁢ ( α 1 , 2 ⁢ α 2 , 1 ) m . Since a 1 2 = 1 , it follows that a 1 is either 1 or any of the order two elements mentioned beforehand. We use the relation s i + 1 = ρ i ⁢ ρ i + 1 ⁢ s i ⁢ ρ i + 1 ⁢ ρ i to determine a i as follows.

  • If a 1 = 1 , then ϕ ⁢ ( s i ) = 1 for all 𝑖. Thus, we obtain ϕ = λ ⁢ θ .

  • If ϕ ⁢ ( s 1 ) = α 1 , 2 = s 1 , then ϕ ⁢ ( s i ) = α i , i + 1 = s i for all 𝑖. Consequently, ϕ = id .

  • If a 1 = α 2 , 1 = ρ 1 ⁢ s 1 ⁢ ρ 1 , then ϕ ⁢ ( s i ) = α i + 1 , i = ρ i ⁢ s i ⁢ ρ i for all 𝑖. Thus, ϕ = ζ .

  • Let

    ϕ ⁢ ( s 1 ) = a 1 = α 1 , 2 ⁢ ( α 2 , 1 ⁢ α 1 , 2 ) m = s 1 ⁢ ( ρ 1 ⁢ s 1 ⁢ ρ 1 ⁢ s 1 ) m = s 1 ⁢ ( ρ 1 ⁢ s 1 ) 2 ⁢ m = ( s 1 ⁢ ρ 1 ) 2 ⁢ m ⁢ s 1 = ( s 1 ⁢ ρ 1 ) 2 ⁢ m + 1 ⁢ ρ 1 .

    Then we get ϕ ⁢ ( s i ) = ( s i ⁢ ρ i ) 2 ⁢ m + 1 ⁢ ρ i for all 𝑖, and hence ϕ = ϕ 2 ⁢ m + 1 .

  • Lastly, let

    ϕ ⁢ ( s 1 ) = a 1 = α 2 , 1 ⁢ ( α 1 , 2 ⁢ α 2 , 1 ) m = ( ρ 1 ⁢ s 1 ⁢ ρ 1 ) ⁢ ( s 1 ⁢ ρ 1 ⁢ s 1 ⁢ ρ 1 ) m = ρ 1 ⁢ ( s 1 ⁢ ρ 1 ) 2 ⁢ m + 1 .

    Then we have ϕ ⁢ ( s i ) = ρ i ⁢ ( s i ⁢ ρ i ) 2 ⁢ m + 1 for all 𝑖, and consequently ϕ = ζ ⁢ ϕ 2 ⁢ m + 1 .

Case (2b): Suppose that w 1 = τ 1 . Then ϕ ⁢ ( s 1 ) = a 1 ⁢ ρ 1 , and hence ρ 1 ⁢ a 1 ⁢ ρ 1 = a 1 − 1 . As in case (2a), the commuting relation s 1 ⁢ ρ k = ρ k ⁢ s 1 for 3 ≤ k ≤ n − 1 shows that a 1 ∈ K n ⁢ [ { α 1 , 2 , α 2 , 1 } ] . A direct check shows that an element of K n ⁢ [ { α 1 , 2 , α 2 , 1 } ] satisfying ρ 1 ⁢ a 1 ⁢ ρ 1 = a 1 − 1 must be of the form ( α 1 , 2 ⁢ α 2 , 1 ) m , where m ∈ Z .

  • If a 1 = 1 , then ϕ ⁢ ( s i ) = ρ i for all 𝑖, and hence ϕ = λ ⁢ π .

  • Let a 1 = ( α 1 , 2 ⁢ α 2 , 1 ) m for some non-zero integer 𝑚. Then

    ϕ ⁢ ( s 1 ) = ( s 1 ⁢ ρ 1 ⁢ s 1 ⁢ ρ 1 ) m ⁢ ρ 1 = ( s 1 ⁢ ρ 1 ) 2 ⁢ m ⁢ ρ 1 .

    Using the relation s 2 = ρ 1 ⁢ ρ 2 ⁢ s 1 ⁢ ρ 2 ⁢ ρ 1 gives

    ϕ ⁢ ( s 2 ) = ρ 1 ⁢ ρ 2 ⁢ ( α 1 , 2 ⁢ α 2 , 1 ) m ⁢ ρ 1 ⁢ ρ 2 ⁢ ρ 1 = ρ 1 ⁢ ρ 2 ⁢ ( α 1 , 2 ⁢ α 2 , 1 ) m ⁢ ρ 2 ⁢ ρ 1 ⁢ ρ 2 = ( α 2 , 3 ⁢ α 3 , 2 ) m ⁢ ρ 2 = ( s 2 ⁢ ρ 2 ) 2 ⁢ m ⁢ ρ 2 .

    Iterating the process gives ϕ ⁢ ( s i ) = ( s i ⁢ ρ i ) 2 ⁢ m ⁢ ρ i for all 𝑖. Thus, ϕ = ϕ 2 ⁢ m for some m ∈ Z .

Case (3): Let n = m = 6 and ϕ ⁢ λ = λ ⁢ ν . Let u i = ν ⁢ ( τ i ) and v i = λ ⁢ ( u i ) for all 1 ≤ i ≤ 5 . Then we have ϕ ⁢ ( ρ i ) = v i for all 𝑖. We set ϕ ⁢ ( s i ) = a i ⁢ λ ⁢ ( w i ) , where a i ∈ K 6 and w i ∈ S 6 . We note that, for i = 3 , 4 , 5 , we have

u i ⁢ w 1 = ( θ ⁢ ϕ ) ⁢ ( s 1 ⁢ ρ i ) = ( θ ⁢ ϕ ) ⁢ ( ρ i ⁢ s 1 ) = w 1 ⁢ u i .

This implies that w 1 belongs to the centraliser of ⟨ u 3 , u 4 , u 5 ⟩ in S 6 , and hence w 1 ∈ ⟨ u 1 ⟩ . Further, we note that λ ⁢ ( w 1 ) ⁢ v i = v i ⁢ λ ⁢ ( w 1 ) for i = 3 , 4 , 5 . The relation s 1 ⁢ ρ i = ρ i ⁢ s 1 gives a 1 = v i ⁢ a 1 ⁢ v i − 1 for i = 3 , 4 , 5 . It follows from Lemma 6.1 that a 1 = 1 , and hence ϕ ⁢ ( s 1 ) = λ ⁢ ( w 1 ) ∈ ⟨ v 1 ⟩ . We now have two possibilities: either ϕ ⁢ ( s 1 ) = 1 or ϕ ⁢ ( s 1 ) = v 1 . If ϕ ⁢ ( s 1 ) = 1 , then ϕ ⁢ ( s i ) = 1 for all 𝑖, and hence ϕ = λ ⁢ ν ⁢ θ . If ϕ ⁢ ( s 1 ) = v 1 , then ϕ ⁢ ( s i ) = v i for all 𝑖, and hence ϕ = λ ⁢ ν ⁢ π . This proves the theorem.∎

We now build the set-up for determining Aut ⁡ ( VT n ) .

Proposition 6.3

The following statements hold for each n ≥ 2 .

  1. ϕ m is not surjective for each even 𝑚.

  2. ϕ m is an automorphism of VT n for m = 1 , − 1 .

  3. ϕ m is injective but not surjective for each odd m ≠ 1 , − 1 .

Proof

Assertion (1) is immediate since the induced map on the abelianisation is not surjective for each even 𝑚. Assertion (2) is also clear since ϕ 1 is the identity automorphism and ϕ − 1 = ζ is an order two automorphism of VT n .

For assertion (3), we first observe that ϕ − m = ζ ⁢ ϕ m for each integer 𝑚. For,

ζ ⁢ ( ϕ m ⁢ ( s i ) ) = ζ ⁢ ( ( s i ⁢ ρ i ) m ⁢ ρ i ) = ( ρ i ⁢ s i ⁢ ρ i ⁢ ρ i ) m ⁢ ρ i = ( ρ i ⁢ s i ) m ⁢ ρ i = ( s i ⁢ ρ i ) − m ⁢ ρ i = ϕ − m ⁢ ( s i ) .

Thus, it is enough to take m = 2 ⁢ t + 1 , where t ≥ 1 . It is easy to see that

ϕ m ⁢ ( α i , i + 1 ) = α i , i + 1 ⁢ ( α i + 1 , i ⁢ α i , i + 1 ) t and ϕ m ⁢ ( α i , j ) = α i , j ⁢ ( α j , i ⁢ α i , j ) t

for all 1 ≤ i ≠ j ≤ n . This implies that ϕ m ⁢ ( K n ) ⊆ K n . We observe that, if

w = α i 1 , j 1 ⁢ ⋯ ⁢ α i k , j k

is a reduced word of length 𝑘, then

ϕ m ⁢ ( w ) = α i 1 , j 1 ⁢ ( α j 1 , i 1 ⁢ α i 1 , j 1 ) t ⁢ ⋯ ⁢ α i k , j k ⁢ ( α j k , i k ⁢ α i k , j k ) t

is a reduced word of length ( 2 ⁢ t + 1 ) ⁢ k .

Let x ⁢ y ∈ VT n with x ∈ K n and y ∈ S n such that

1 = ϕ m ⁢ ( x ⁢ y ) = ϕ m ⁢ ( x ) ⁢ y .

It follows that y = 1 and ϕ m ⁢ ( x ) = 1 . The preceding observation implies that x = 1 , and hence ϕ m is injective. Let x ⁢ y ∈ VT n with x ∈ K n and y ∈ S n such that α 1 , 2 = ϕ m ⁢ ( x ⁢ y ) = ϕ m ⁢ ( x ) ⁢ y . It follows that y = 1 and ϕ m ⁢ ( x ) = α 1 , 2 . Comparing the lengths of 𝑥 and ϕ m ⁢ ( x ) = α 1 , 2 , the preceding observation leads to a contradiction. Hence, ϕ m is not surjective, which proves assertion (3). ∎

Recall that a group is called co-Hopfian if every injective endomorphism is surjective. Proposition 6.3 (3) yields the following result.

Corollary 6.4

The groups VT n and KT n are not co-Hopfian for each n ≥ 2 .

Table 1
𝐺 B n P n VB n VP n
Braid group Pure braid group Virtual braid group Pure virtual braid group
Hopfian Yes for n ≥ 2 Yes for n ≥ 2 Yes for n ≥ 5 Yes for n = 2
By linearity By linearity [8] VP 2 is free group of rank two
Unknown for n ≥ 3
Co-Hopfian No for n ≥ 2 No for n ≥ 2 Yes for n ≥ 5 No for n = 2
Easy to see [6] Easy to see [6] [8] Unknown for n ≥ 3
Aut ⁡ ( G ) Known for n ≥ 2 Known for n ≥ 2 Known for n ≥ 5 Known for n = 2
[12] [7] [8] Unknown for n ≥ 3
Table 2
𝐺 T n PT n VT n PVT n
Twin group Pure twin group Virtual twin group Pure virtual twin group
Hopfian Yes for n ≥ 2 Yes for n ≥ 2 Yes for n ≥ 2 Yes for n ≥ 2
By linearity By linearity [32] PVT n is a RAAG [32]
Co-Hopfian No for n ≥ 3 No for 3 ≤ n ≤ 6 No for n ≥ 2 No for n ≥ 2
[31] PT n is free for 3 ≤ n ≤ 5 This work [32]
and PT 6 is RAAG [2, 17, 29]
Unknown for n ≥ 7
Aut ⁡ ( G ) Known for n ≥ 2 Known for 2 ≤ n ≤ 5 Known for n ≥ 5 Known for n ≥ 2
[30] Unknown for n ≥ 6 This work [32]
Theorem 6.5

For n ≥ 5 , we have Aut ⁡ ( VT n ) = Inn ⁡ ( VT n ) ⋊ ⟨ ζ ⟩ ≅ VT n ⋊ Z 2 and Out ⁡ ( VT n ) ≅ ⟨ ζ ⟩ ≅ Z 2 .

Proof

We first claim that the map 𝜁 is a non-inner automorphism of VT n for each n ≥ 3 . Suppose that 𝜁 is an inner automorphism of VT n , say, ζ = z ̂ for some z ∈ VT n . Since z ̂ ⁢ ( ρ i ) = ζ ⁢ ( ρ i ) = ρ i for all 𝑖, we have z ∈ C VT n ⁡ ( S n ) . By Corollary 5.9, we have z = 1 . But this is a contradiction since 𝜁 is not the identity map.

It follows from Theorem 6.2 (2) and Proposition 6.3 that any automorphism 𝜙 of VT n is of the form ϕ = w ̂ ⁢ ζ for some w ∈ VT n . We already showed above that 𝜁 is a non-inner automorphism. Further, it is known from [32, Corollary 4.2] that Z ⁡ ( VT n ) = 1 . Hence,

Aut ⁡ ( VT n ) = Inn ⁡ ( VT n ) ⋊ ⟨ ζ ⟩ ≅ VT n ⋊ Z 2 and Out ⁡ ( VT n ) = ⟨ ζ ⟩ ≅ Z 2 . ∎

Note that VT 2 ≅ T 3 ≅ Z 2 * Z 2 and Aut ⁡ ( VT 2 ) ≅ Inn ⁡ ( VT 2 ) ⋊ Z 2 by [30, Theorem 6.1 (1)]. The groups VT 3 and VT 4 need to be dealt with separately with the latter appearing to be more challenging. We leave these cases for interested readers.

In Tables 1 and 2, we list the status of some properties of braid groups, virtual braid groups, twin groups, virtual twin groups and their pure subgroups.

Funding source: National Board for Higher Mathematics

Award Identifier / Grant number: 0204/3/2020/R&D-II/2475

Funding source: Department of Science and Technology, Ministry of Science and Technology, India

Award Identifier / Grant number: DST/SJF/MSA-02/2018-19

Award Identifier / Grant number: SB/SJF/2019-20/04

Funding statement: Tushar Kanta Naik would like to acknowledge support from NBHM via grant 0204/3/2020/R&D-II/2475. Mahender Singh is supported by the Swarna Jayanti Fellowship grants DST/SJF/MSA-02/2018-19 and SB/SJF/2019-20/04.

Acknowledgements

Neha Nanda thanks IISER Bhopal for the Institute Post-Doctoral Fellowship.

  1. Communicated by: Evgenii I. Khukhro

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Received: 2023-01-28
Revised: 2023-10-19
Published Online: 2023-11-30
Published in Print: 2024-05-01

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