# Representations of virtual braids by automorphisms and virtual knot groups

V. G. Bardakov, Yu. A. Mikhalchishina, M. V. Neshchadim

###### Abstract

In the present paper the representation of the virtual braid group $VB_{n}$
into the automorphism group of free product of the free group and free abelian
group is constructed. This representation generalizes the previously
constructed ones. The fact that these already known representations are not
faithful for $n\geq 4$ is verified. Using representations of $VB_{n}$, the
virtual link group is defined. Also representations of welded braid group
$WB_{n}$ are constructed and the welded link group is defined.

Introduction

The Artin representation [5, Corollary 1.8.3] of the braid group by
automorphisms of the free group allows to solve the word problem in the braid
group, to construct the linear representations of the braid group (the Burau
representation) as well as the linear representation of pure braid group (the
Gassner representation) etc. The virtual braid group (see [23], [11]) is one
of generalizations of the classical braid group. A natural question about the
representation of the virtual braid group by automorphisms of some group
arises.

For each virtual braid the virtual link could be constructed. Contrary to the
classical link case there is no natural definition of a virtual link group.
The following question arises: to define the virtual link group.

There exist several approaches to determine the group of virtual knots and
links. The first definition was given by L. Kauffman [11] in 1996. Next there
was the definition by V. Manturov [16] (see also [4], where the same groups
were defined using the virtual braid group representation by automorphisms of
a free group). Further there was the definition by D. Silver and S. Williams
[21]-[22] and lately another version by H. Boden, E. Dies, A. Gaudreau, A.
Gerlings, E. Harper, A. Nicas [6].

The fact that all representations known before are not faithful for $n\geq 4$
will be shown in the present work. A new representation which generalizes all
previous ones will be suggested. Moreover the fact that the new representation
is equivalent to the extension of the Artin representation of the braid group
will be shown. Using the new representation the virtual link group is defined
and the fact that it is the virtual link invariant is proved. There will be
shown the fact that for the classical link the constructed group is isomorphic
to the free product of the fundamental group of the link complement in the
tree dimensional sphere and some free abelian group.

The principally new way to the definition of the virtual link group was
suggested in the paper of J. Carter, D. Silver and S. Williams [7] where the
virtual link group was defined as the fundamental group of the complement of
some classical link in a thickened surface. In the present work there will be
shown that the group of virtual trefoil defined in [7] is not isomorphic to
the group of virtual trefoil defined using the representation of virtual braid
group .

Authors are grateful to Oleg Chterental who pointed out an element in the
kernel of the Artin representation of the virtual braid group (see [8]).

§ 1. The braid group, the virtual braid group and the welded braid group

In the paper the following notations are used. If $a$ and $b$ are elements of
some group $G$, then $a^{b}=b^{-1}ab$ is conjugacy of $a$ by $b$;
$[a,b]=a^{-1}b^{-1}ab$ is the commutator of elements $a$ and $b$.

The braid group. Recall that the braid group $B_{n}$ on $n$ strands, where
$n\geq 2$, is defined by generators
$\sigma_{1},\sigma_{2},\ldots,\sigma_{n-1}$ and defining relations

$\displaystyle\sigma_{i}\sigma_{i+1}\sigma_{i}=~{}$
$\displaystyle\sigma_{i+1}\sigma_{i}\sigma_{i+1}\hskip
56.9055pt\mbox{for}~{}i=1,2,\ldots,n-2,$ (1)
$\displaystyle\sigma_{i}\sigma_{j}=~{}$
$\displaystyle\sigma_{j}\sigma_{i}\hskip 85.35826pt~{}\mbox{for}~{}|i-j|\geq
2.$ (2)

There exists a homomorphism of the group $B_{n}$ on the permutation group
$S_{n}$ that maps the generator $\sigma_{i}$ into the transposition
$(i,{i+1}),$ $i=1,2,\ldots,n-1.$ The kernel of this homomorphism is the pure
braid group and is denoted by the symbol $P_{n}$. The group $P_{n}$ is
generated by elements $a_{ij},1\leq i<j\leq n$, which can be expressed through
the generators of $B_{n}$ in the following way

$a_{i,i+1}=\sigma_{i}^{2},$
$a_{ij}=\sigma_{j-1}\sigma_{j-2}\ldots\sigma_{i+1}\sigma_{i}^{2}\sigma_{i+1}^{-1}\ldots\sigma_{j}^{-1}\sigma_{j-1}^{-1}.$

The group $P_{n}$ composes in the semi direct product of free groups

$P_{n}=U_{n}\leftthreetimes(U_{n-1}\leftthreetimes(\ldots\leftthreetimes(U_{3}\leftthreetimes
U_{2}))\ldots),$

where $U_{i}=\langle a_{1i},a_{2i},\ldots,a_{i-1,i}\rangle\simeq F_{i-1}$ is
the free group of the rank $n-1$, $i=2,3,\ldots,n$ (see [5, Section 1.4],
[17]).

The virtual braid group and the welded braid group. The virtual braid group
$VB_{n}$ is presented in the paper [11], where a system of its generators and
defining relations is written. In the paper [23] the more compact system of
defining relations is constructed (see relations below). The virtual braid
group $VB_{n}$ is generated by the classical braid group
$B_{n}=\langle\sigma_{1},\ldots,\sigma_{n-1}\rangle$ and the permutation group
$S_{n}=\langle\rho_{1},\ldots,\rho_{n-1}\rangle$. Generators
$\sigma_{i},i=1,\ldots,n-1$ satisfy the relations (1), (2) and generators
$\rho_{i},i=1,\ldots,n-1,$ satisfy the following relations (relations in the
permutation group $S_{n}$):

$\displaystyle\rho_{i}^{2}$ $\displaystyle=1~{}~{}\hskip
99.58464pt\mbox{for}~{}~{}i=1,2,\ldots,n-1,$ (3)
$\displaystyle\rho_{i}\rho_{j}$ $\displaystyle=\rho_{j}\rho_{i}~{}~{}\hskip
85.35826pt\mbox{for}~{}~{}|i-j|\geq 2,$ (4)
$\displaystyle\rho_{i}\rho_{i+1}\rho_{i}$
$\displaystyle=\rho_{i+1}\rho_{i}\rho_{i+1}~{}~{}\hskip
56.9055pt\mbox{for}~{}~{}i=1,2\ldots,n-2.$ (5) Other defining relations of the
group $VB_{n}$ are mixed and they are as follows
$\displaystyle\sigma_{i}\rho_{j}$
$\displaystyle=\rho_{j}\sigma_{i}~{}~{}\hskip
85.35826pt\mbox{for}~{}~{}|i-j|\geq 2,$ (6)
$\displaystyle\rho_{i}\rho_{i+1}\sigma_{i}$
$\displaystyle=\sigma_{i+1}\rho_{i}\rho_{i+1}~{}~{}\hskip
56.9055pt\mbox{for}~{}~{}i=1,2,\ldots,n-2.$ (7) Note that the last relation is
equivalent to: $\displaystyle\rho_{i+1}\rho_{i}\sigma_{i+1}$
$\displaystyle=\sigma_{i}\rho_{i+1}\rho_{i}~{}~{}\hskip
71.13188pt\mbox{for}~{}~{}i=1,2,\ldots,n-2.$ (8)

As it was pointed out in the paper [11], in the group $VB_{n}$ the following
relations do not hold

$\displaystyle\mathcal{F}_{1}:~{}~{}~{}~{}\rho_{i}\sigma_{i+1}\sigma_{i}$
$\displaystyle=\sigma_{i+1}\sigma_{i}\rho_{i+1},$
$\displaystyle\mbox{for}~{}~{}i=1,2,\ldots,n-2,$
$\displaystyle\mathcal{F}_{2}:~{}\rho_{i+1}\sigma_{i}\sigma_{i+1}$
$\displaystyle=\sigma_{i}\sigma_{i+1}\rho_{i},$
$\displaystyle\mbox{for}~{}~{}i=1,2,\ldots,n-2,$

which look like relations (8). These relations are referred to as forbidden
relations.

In the work [9] the welded braid group denoted by $WB_{n}$ is presented. This
group is generated by elements $\sigma_{i}$ and
$\alpha_{i},~{}i=1,2,\ldots,n-1$. The group generated by elements $\sigma_{i}$
is the classical braid group $B_{n}$, the group generated by elements
$\alpha_{i}$ is the permutation group $S_{n}$ also the mixed relations hold

$\displaystyle\alpha_{i}\sigma_{j}$ $\displaystyle=\sigma_{j}\alpha_{i}\hskip
99.58464pt~{}\mbox{for}~{}~{}|i-j|\geq 2,$ (9)
$\displaystyle\alpha_{i}\alpha_{i+1}\sigma_{i}$
$\displaystyle=\sigma_{i+1}\alpha_{i}\alpha_{i+1}\hskip
71.13188pt\mbox{for}~{}~{}i=1,2,\ldots,n-2,$ (10)
$\displaystyle\sigma_{i}\sigma_{i+1}\alpha_{i}$
$\displaystyle=\alpha_{i+1}\sigma_{i}\sigma_{i+1}\hskip
71.13188pt\mbox{for}~{}~{}i=1,2,\ldots,n-2.$ (11)

Comparing the relations of groups $VB_{n}$ and $WB_{n}$ it is clear that
$WB_{n}$ is obtained from $VB_{n}$ by adding the additional relation (11)
which coincide with $\mathcal{F}_{2}$. Therefore, there exists the
homomorphism

$\varphi_{VW}:VB_{n}\rightarrow WB_{n},$

putting $\sigma_{i}$ into $\sigma_{i}$ and $\rho_{i}$ into $\alpha_{i}$ for
all $i=1,2,\ldots,n-1$. Hence $WB_{n}$ is the homomorphic image of the group
$VB_{n}$.

Note that $WB_{n}$ is isomorphic to the group of conjugating automorphisms
$C_{n}$, which is a subgroup of $\mathrm{Aut}(F_{n})$ and is studied in [1, 2,
19, 20].

§ 2. Representations of virtual braids by automorphisms

The present section we begin by listing some known representations of the
virtual braid group $VB_{n}$ by automorphisms of the free group. Next we
define the representation that generalizes all previous ones and in the
conclusion we show that all previously written representations are not
faithful for $n\geq 4$.

The extension of the Artin representation. The representation
$VB_{n}\longrightarrow\mathrm{Aut}(F_{n+1})$ (see [3], [16]) into the
automorphism group of the free group $F_{n+1}=\langle
x_{1},x_{2},\ldots,x_{n},y\rangle$ of the rank $n+1$:

$\varphi_{A}(\sigma_{i}):\left\\{\begin{array}[]{ll}x_{i}\longmapsto
x_{i}x_{i+1}x_{i}^{-1},&\\\ x_{i+1}\longmapsto x_{i},&\\\
\end{array}\right.~{}~{}~{}~{}~{}\varphi_{A}(\rho_{i}):\left\\{\begin{array}[]{ll}x_{i}\longmapsto
x_{i+1}^{y^{-1}},&\\\ x_{i+1}\longmapsto x_{i}^{y},&\\\ \end{array}\right.$

(here and hereafter we write only non-trivial actions on the generators,
assume that all other generators are fixed) is an extension of the Artin
representation $\varphi_{A}:B_{n}\longrightarrow\mathrm{Aut}(F_{n})$ (see
[5]). Hence we will hereafter denote it with the same symbol $\varphi_{A}$.

The Silver-Williams representation. For the pair of natural numbers $n$ and
$k$ we define the group $F_{n,k+1}=F_{n}*\mathbb{Z}^{k+1}$ as the free product
of the free group of the rank $n$ and the free abelian group of the rank
$k+1$. We assume that $F_{n}$ is freely generated by elements
$x_{1},x_{2},\ldots,x_{n}$ and $\mathbb{Z}^{k+1}$ is freely generated by
elements $v,u_{1},u_{2},\ldots,u_{k}$. Note that $F_{n,1}=F_{n+1}$.

Using the definition of the generalized Alexander group for virtual links
given by D. Silver and S. Williams [21], we construct the representation
$\varphi_{SW}:VB_{n}\longrightarrow\mathrm{Aut}(F_{n,n+1})$ of the virtual
braid group $VB_{n}$ into the automorphism group of the group $F_{n,n+1}$
defined by the action on the generators

$\varphi_{SW}(\sigma_{i}):\left\\{\begin{array}[]{ll}x_{i}\longmapsto
x_{i}x_{i+1}^{u_{i}}x_{i}^{-vu_{i+1}},&\\\ x_{i+1}\longmapsto x_{i}^{v},&\\\
\end{array}\right.~{}~{}~{}\varphi_{SW}(\sigma_{i}):\left\\{\begin{array}[]{ll}u_{i}\longmapsto
u_{i+1},&\\\ u_{i+1}\longmapsto u_{i},&\\\ \end{array}\right.~{}~{}~{}$
$\varphi_{SW}(\rho_{i}):\left\\{\begin{array}[]{ll}x_{i}\longmapsto
x_{i+1},&\\\ x_{i+1}\longmapsto x_{i},&\\\
\end{array}\right.~{}~{}~{}\varphi_{SW}(\rho_{i}):\left\\{\begin{array}[]{ll}u_{i}\longmapsto
u_{i+1},&\\\ u_{i+1}\longmapsto u_{i},&\\\ \end{array}\right.~{}~{}~{}$

It is easy to verify that the following automorphism corresponds to the
element $\sigma_{i}^{-1}$

$\varphi_{SW}(\sigma_{i}^{-1}):\left\\{\begin{array}[]{ll}x_{i}\longmapsto
x_{i+1}^{v^{-1}},&\\\
x_{i+1}\longmapsto\left(x_{i+1}^{-v^{-1}}x_{i}x_{i+1}^{u_{i}}\right)^{u_{i+1}^{-1}},&\\\
\end{array}\right.~{}~{}~{}\varphi_{SW}(\sigma_{i}^{-1}):\left\\{\begin{array}[]{ll}u_{i}\longmapsto
u_{i+1},&\\\ u_{i+1}\longmapsto u_{i}.&\\\ \end{array}\right.~{}~{}~{}$

We shall prove that this representation generalizes the representation
$\varphi_{A}:VB_{n}\longrightarrow\mathrm{Aut}(F_{n+1})$. More precisely the
following proposition is true (The idea of this proposition was suggested by
D. Silver).

Proposition 1. If we put

$u_{1}=u_{2}=\ldots=u_{n}=y,v=y^{-1},$

and instead of generators $x_{1},x_{2},\ldots,x_{n}$ we put new generators

$z_{1}=x_{1},z_{2}=x_{2}^{y},z_{3}=x_{3}^{y^{2}},\ldots,z_{n}=x_{n}^{y^{n-1}},$

then the obtained group is isomorphic to $F_{n+1}$ and the representation
$\varphi_{SW}$ is coincide with the representation $\varphi_{A}$ on this
group.

Proof. The Silver-Williams representation acts on generators
$z_{1},z_{2},\ldots,z_{n}$ by the rule

$\varphi_{SW}(\sigma_{i}):\left\\{\begin{array}[]{ll}z_{i}\longmapsto
z_{i}z_{i+1}z_{i}^{-1},&\\\ z_{i+1}\longmapsto z_{i},&\\\
\end{array}\right.\varphi_{SW}(\rho_{i}):\left\\{\begin{array}[]{ll}z_{i}\longmapsto
z_{i+1}^{y^{-1}},&\\\ z_{i+1}\longmapsto z_{i}^{y},&\\\ \end{array}\right.$

but it is the representation $\varphi_{A}$.

The Boden - Dies - Gaudreau - Gerlings - Harper - Nicas representation [6].
For brevity sake we denote this representation by the symbol $\varphi_{BD}$.
Let $F_{n,2}=F_{n}*\mathbb{Z}^{2}$, where $F_{n}$ is freely generated by
elements $x_{1},x_{2},\ldots,x_{n}$ and $\mathbb{Z}^{2}$ is freely generated
by elements $v,u$. The representation
$\varphi_{BD}:VB_{n}\longrightarrow\mathrm{Aut}(F_{n,2})$ of the virtual braid
group $VB_{n}$ into the automorphism group of the group $F_{n,2}$ is defined
by the action on the generators:

$\varphi_{BD}(\sigma_{i}):\left\\{\begin{array}[]{ll}x_{i}\longmapsto
x_{i}x_{i+1}x_{i}^{-u},&\\\ x_{i+1}\longmapsto x_{i}^{u},&\\\
\end{array}\right.~{}~{}~{}\varphi_{BD}(\rho_{i}):\left\\{\begin{array}[]{ll}x_{i}\longmapsto
x_{i+1}^{v^{-1}},&\\\ x_{i+1}\longmapsto x_{i}^{v}.&\\\ \end{array}\right.$

A new representation of the virtual braid group. Now we define a new
representation of $VB_{n}$ which generalizes the representations listed above.
We consider the free product $F_{n,2n+1}=F_{n}*\mathbb{Z}^{2n+1}$, where
$F_{n}$ is a free group of the rank $n$ generated by elements
$x_{1},x_{2},\ldots,x_{n}$ and $\mathbb{Z}^{2n+1}$ is a free abelian group of
the rank $2n+1$ freely generated by elements $u_{1},u_{2},\ldots,u_{n}$,
$v_{0},v_{1},v_{2},\ldots,v_{n}$. The following statement is true

Theorem 1. The following mapping
$\varphi_{M}:VB_{n}\longrightarrow\mathrm{Aut}(F_{n,2n+1})$ defined by the
action on the generators:

$\varphi_{M}(\sigma_{i}):\left\\{\begin{array}[]{l}x_{i}\longmapsto
x_{i}x_{i+1}^{u_{i}}x_{i}^{-v_{0}u_{i+1}},\\\ x_{i+1}\longmapsto
x_{i}^{v_{0}},\\\
\end{array}\right.~{}~{}~{}\varphi_{M}(\sigma_{i}):\left\\{\begin{array}[]{l}u_{i}\longmapsto
u_{i+1},\\\ u_{i+1}\longmapsto u_{i},\\\ \end{array}\right.$
$\varphi_{M}(\sigma_{i}):\left\\{\begin{array}[]{l}v_{i}\longmapsto
v_{i+1},\\\ v_{i+1}\longmapsto v_{i},\\\ \end{array}\right.$
$\varphi_{M}(\rho_{i}):\left\\{\begin{array}[]{l}x_{i}\longmapsto
x_{i+1}^{v_{i}^{-1}},\\\ x_{i+1}\longmapsto x_{i}^{v_{i+1}},\\\
\end{array}\right.~{}~{}~{}\varphi_{M}(\rho_{i}):\left\\{\begin{array}[]{l}u_{i}\longmapsto
u_{i+1},\\\ u_{i+1}\longmapsto u_{i},\\\ \end{array}\right.$
$\varphi_{M}(\rho_{i}):\left\\{\begin{array}[]{l}v_{i}\longmapsto v_{i+1},\\\
v_{i+1}\longmapsto v_{i},\\\ \end{array}\right.$

is provided a representation of $VB_{n}$ into  $\mathrm{Aut}(F_{n,2n+1})$.

Proof. It is need to verify that the relations of the group $VB_{n}$ go to the
relations of $VB_{n}$. It is easy to check that it is so for the commutativity
relations since the representation $\varphi_{M}$ is the local representation.

Let us verify the correctness of the relation

$\varphi_{M}(\sigma_{i})\varphi_{M}(\sigma_{i+1})\varphi_{M}(\sigma_{i})=\varphi_{M}(\sigma_{i+1})\varphi_{M}(\sigma_{i})\varphi_{M}(\sigma_{i+1}).$

For that purpose we find the action of
$\varphi_{M}(\sigma_{i})\varphi_{M}(\sigma_{i+1})\varphi_{M}(\sigma_{i})$ on
the generators $x_{1}$, $\ldots$, $x_{n}$. We have

$\varphi_{M}(\sigma_{i})\varphi_{M}(\sigma_{i+1})\varphi_{M}(\sigma_{i}):\left\\{\begin{array}[]{ll}x_{i}\longmapsto
x_{i}x_{i+1}^{u_{i}}x_{i+2}^{u_{i}u_{i+1}}x_{i+1}^{-u_{i}u_{i+2}v_{0}}x_{i}^{-u_{i+2}v_{0}},&\\\
x_{i+1}\longmapsto{(x_{i}x_{i+1}^{u_{i}}x_{i}^{-u_{i+1}v_{0}})}^{v_{0}},&\\\
x_{i+2}\longmapsto x_{i}^{v_{0}^{2}}.&\\\ \end{array}\right.$

Analogously we find the action of
$\varphi_{M}(\sigma_{i+1})\varphi_{M}(\sigma_{i})\varphi_{M}(\sigma_{i+1})$.
We obtain

$\varphi_{M}(\sigma_{i+1})\varphi_{M}(\sigma_{i})\varphi_{M}(\sigma_{i+1}):\left\\{\begin{array}[]{ll}x_{i}\longmapsto
x_{i}\left(x_{i+1}x_{i+2}^{u_{i+1}}x_{i+1}^{-u_{i+2}v_{0}}\right)^{u_{i}}x_{i}^{-u_{i+2}v_{0}},&\\\
x_{i+1}\longmapsto
x_{i}^{v_{0}}x_{i+1}^{v_{0}u_{i}}x_{i}^{-{v_{0}}^{2}u_{i+1}},&\\\
x_{i+2}\longmapsto x_{i}^{v_{0}^{2}}.&\\\ \end{array}\right.$

Comparing the obtained automorphisms it can be seen that the long relation of
the braid group holds when it acts on the generators of $F_{n}$. The fact that
the relation holds with the action on subgroups generated by elements $u_{j}$
and $v_{j}$ follows from the fact that the automorphisms
$\varphi_{M}(\sigma_{i})$ act as permutations.

We verify the correctness of the mixed relation

$\varphi_{M}(\rho_{i})\varphi_{M}(\rho_{i+1})\varphi_{M}(\sigma_{i})=\varphi_{M}(\sigma_{i+1})\varphi_{M}(\rho_{i})\varphi_{M}(\rho_{i+1}).$

For that purpose we find the action of
$\varphi_{M}(\rho_{i})\varphi_{M}(\rho_{i+1})\varphi_{M}(\sigma_{i})$ on the
generators $x_{1}$, $\ldots$, $x_{n}$. We have

$\varphi_{M}(\rho_{i})\varphi_{M}(\rho_{i+1})\varphi_{M}(\sigma_{i}):\left\\{\begin{array}[]{ll}x_{i}\longmapsto
x_{i+2}^{v_{i}^{-1}v_{i+1}^{-1}},&\\\
x_{i+1}\longmapsto\left(x_{i}x_{i+1}^{u_{i}}x_{i}^{-u_{i+1}v_{0}}\right)^{v_{i+2}},&\\\
x_{i+2}\longmapsto x_{i}^{v_{0}v_{i+2}}.&\\\ \end{array}\right.$

Analogously we find the action of
$\varphi_{M}(\sigma_{i+1})\varphi_{M}(\rho_{i})\varphi_{M}(\rho_{i+1})$. We
obtain

$\varphi_{M}(\sigma_{i+1})\varphi_{M}(\rho_{i})\varphi_{M}(\rho_{i+1}):\left\\{\begin{array}[]{ll}x_{i}\longmapsto
x_{i+2}^{v_{i+1}^{-1}v_{i}^{-1}},&\\\ x_{i+1}\longmapsto
x_{i}^{v_{i+2}}x_{i+1}^{v_{i+2}u_{i}}x_{i}^{-v_{i+2}u_{i+1}v_{0}},&\\\
x_{i+2}\longmapsto x_{i}^{v_{i+2}v_{0}}.&\\\ \end{array}\right.$

Comparing the obtained automorphisms it is clear that the mixed relation of
the braid group holds on the generators of $F_{n}$. The fact that the relation
holds with the action on subgroups generated by elements $u_{j}$ and $v_{j}$
follows because the automorphisms $\varphi_{M}(\sigma_{i})$ and
$\varphi_{M}(\rho_{i})$ act as permutations.

The remaining relations are verified analogously. The theorem is proved.

It is easy to check that the element $\varphi_{M}(\sigma_{i}^{-1})$ acts on
the generators of the group $F_{n}$ by the rule

$\varphi_{M}(\sigma_{i}^{-1}):\left\\{\begin{array}[]{ll}x_{i}\longmapsto{x_{i+1}}^{v_{0}^{-1}},&\\\
x_{i+1}\longmapsto\left(x_{i+1}^{-{v_{0}^{-1}}}x_{i}x_{i+1}^{u_{i}}\right)^{u_{i+1}^{-1}}.&\\\
\end{array}\right.$

We show that this representation generalizes the representations
$\varphi_{SW}$ and $\varphi_{BD}$. It is true that

Proposition 2. 1) If we put

$v_{1}=v_{2}=\ldots=v_{n}=v$

in the group $F_{n,2n+1}$ and insert new generators $z_{1},\ldots z_{n}$ such
that

$z_{1}=x_{1},z_{2}=x_{2}^{v},z_{3}=x_{3}^{v^{2}},\ldots,z_{n}=x_{n}^{v^{n-1}},~{}~{}~{}w_{i}=vu_{i},~{}~{}~{}i=1,2,\ldots,n,$

then we obtain the group $F_{n,n+1}$ on which the representation $\varphi_{M}$
induces the representation $\varphi_{SW}$.

2) If we put

$u_{1}=u_{2}=\ldots=u_{n}=1,v_{0}=u,v_{1}=v_{2}=\ldots=v_{n}=v$

in the group $F_{n,2n+1}$, then we obtain the group $F_{n,2}$ on which the
representation $\varphi_{M}$ induces the representation $\varphi_{BD}$.

Proof. 1) Note that on the new generators $z_{1},\ldots,z_{n}$ the
automorphisms $\varphi_{M}(\sigma_{i})$ and $\varphi_{M}(\rho_{i})$ act by the
rule

$\varphi_{M}(\sigma_{i}):\left\\{\begin{array}[]{ll}z_{i}\longmapsto
z_{i}z_{i+1}^{vu_{i}}z_{i}^{-u_{i+1}},&\\\ z_{i+1}\longmapsto
z_{i}^{v^{-1}},&\\\
\end{array}\right.~{}~{}~{}\varphi_{M}(\rho_{i}):\left\\{\begin{array}[]{ll}z_{i}\longmapsto
z_{i+1},&\\\ z_{i+1}\longmapsto z_{i}.&\\\ \end{array}\right.$

Assuming $w_{i}=vu_{i}$ for $i=1,2,\ldots,n$, we obtain the Silver-Williams
representation.

The second part of the proposition is verified analogously. The proposition is
proved.

The representation of the virtual pure braid group. The virtual pure braid
group $VP_{n}$ is the kernel of the homomorphism $VB_{n}\longrightarrow S_{n}$
which acts on the generators:

$\sigma_{i}\longmapsto\rho_{i},~{}~{}\rho_{i}\longmapsto\rho_{i},~{}~{}i=1,2,\ldots,n-1.$

It was found out in [1] that the group $VP_{n}$ is generated by elements

$\lambda_{i,i+1}=\rho_{i}\,\sigma_{i}^{-1},~{}~{}~{}\lambda_{i+1,i}=\rho_{i}\,\lambda_{i,i+1}\,\rho_{i}=\sigma_{i}^{-1}\,\rho_{i},~{}~{}~{}i=1,2,\ldots,n-1,$
$\lambda_{ij}=\rho_{j-1}\,\rho_{j-2}\ldots\rho_{i+1}\,\lambda_{i,i+1}\,\rho_{i+1}\ldots\rho_{j-2}\,\rho_{j-1},$
$\lambda_{ji}=\rho_{j-1}\,\rho_{j-2}\ldots\rho_{i+1}\,\lambda_{i+1,i}\,\rho_{i+1}\ldots\rho_{j-2}\,\rho_{j-1},~{}~{}~{}1\leq
i<j-1\leq n-1.$

We find the images of the generators $\lambda_{ij}$, $\lambda_{ji}$ under the
representation $\varphi_{M}$. It is true

Proposition 3. The representation $\varphi_{M}$ acts on the generators of
$VP_{n}$ by the formulae

$\varphi_{M}(\lambda_{i,i+1}):\left\\{\begin{array}[]{l}x_{i}\longmapsto\left(x_{i+1}^{-1}x_{i}x_{i+1}^{u_{i+1}}\right)^{u_{i}^{-1}v_{i+1}^{-1}},\\\
x_{i+1}\longmapsto x_{i+1}^{v_{i}},\\\
\end{array}\right.~{}~{}\varphi_{M}(\lambda_{i+1,i}):\left\\{\begin{array}[]{l}x_{i}\longmapsto
x_{i}^{v_{i+1}},\\\
x_{i+1}\longmapsto\left(x_{i}^{-v_{i+1}}x_{i+1}^{v_{i}^{-1}}x_{i}^{v_{i+1}u_{i}}\right)^{u_{i+1}^{-1}},\\\
\end{array}\right.$
$\varphi_{M}(\lambda_{ij}):\left\\{\begin{array}[]{l}x_{i}\longmapsto\left(x_{j}^{-v_{i+1}^{-1}\ldots
v_{j-1}^{-1}}x_{i}x_{j}^{-v_{i+1}^{-1}\ldots
v_{j-1}^{-1}u_{j}}\right)^{u_{i}^{-1}v_{j}^{-1}},\\\ x_{j}\longmapsto
x_{j}^{v_{i}},\\\ \end{array}\right.$
$\varphi_{M}(\lambda_{ji}):\left\\{\begin{array}[]{l}x_{i}\longmapsto
x_{i}^{v_{j}},\\\ x_{j}\longmapsto\left(x_{i}^{-v_{j}}x_{j}^{v_{i}^{-1}\ldots
v_{j-1}^{-1}}x_{i}^{v_{j}u_{i}}\right)^{u_{j}^{-1}v_{i+1}\ldots v_{j-1}},\\\
\end{array}\right.$

where $1\leq i<j-1\leq n-1$.

The representation of the welded braid group. We consider $F_{n,n+1}$ which
equals to the free product $F_{n}*\mathbb{Z}^{n+1}$, where $\mathbb{Z}^{n+1}$
is a free abelian group of the rank $n+1$ freely generated by elements
$v,u_{1},u_{2},\ldots,u_{n}$. The following statement is true

Proposition 4. The mapping
$\psi_{M}:WB_{n}\longrightarrow\mathrm{Aut}(F_{n,n+1})$ defined by the action
on generators

$\psi_{M}(\sigma_{i}):\left\\{\begin{array}[]{ll}x_{i}\longmapsto
x_{i}x_{i+1}^{u_{i}}x_{i}^{-vu_{i+1}},&\\\ x_{i+1}\longmapsto x_{i}^{v},&\\\
\end{array}\right.~{}~{}~{}\psi_{M}(\sigma_{i}):\left\\{\begin{array}[]{ll}u_{i}\longmapsto
u_{i+1},&\\\ u_{i+1}\longmapsto u_{i},&\\\ \end{array}\right.$
$\psi_{M}(\alpha_{i}):\left\\{\begin{array}[]{ll}x_{i}\longmapsto x_{i+1},&\\\
x_{i+1}\longmapsto x_{i},&\\\
\end{array}\right.~{}~{}~{}\psi_{M}(\alpha_{i}):\left\\{\begin{array}[]{ll}u_{i}\longmapsto
u_{i+1},&\\\ u_{i+1}\longmapsto u_{i}&\\\ \end{array}\right.$

determines the representation of the group $WB_{n}$ into the group
$\mathrm{Aut}(F_{n,n+1})$.

It could be seen that the representation of the group $WB_{n}$ is obtained
from the one of the group $VB_{n}$ providing all generators $v_{i}$ are sent
to 1, where $1\leq i\leq n$.

Proof. It is sufficient to check that the relation $\mathcal{F}_{1}$ holds

$\psi_{M}(\rho_{i})\psi_{M}(\sigma_{i+1})\psi_{M}(\sigma_{i})=\psi_{M}(\sigma_{i+1})\psi_{M}(\sigma_{i})\psi_{M}(\rho_{i+1})$

which is the forbidden relation in $VB_{n}$. We find the action of the
automorphism $\psi_{M}(\rho_{i})\psi_{M}(\sigma_{i+1})\psi_{M}(\sigma_{i})$ on
the generators of $F_{n}$. We obtain

$\psi_{M}(\rho_{i})\psi_{M}(\sigma_{i+1})\psi_{M}(\sigma_{i}):\left\\{\begin{array}[]{ll}x_{i}\longmapsto
x_{i}x_{i+2}^{u_{i}}x_{i}^{-u_{i+2}},&\\\ x_{i+1}\longmapsto
x_{i}x_{i+1}^{u_{i}}x_{i}^{-u_{i+1}},&\\\ x_{i+2}\longmapsto x_{i}.&\\\
\end{array}\right.$

Analogously we find the action of the automorphism
$\psi_{M}(\sigma_{i+1})\psi_{M}(\sigma_{i})\psi_{M}(\rho_{i+1})$. We have

$\psi_{M}(\sigma_{i+1})\psi_{M}(\sigma_{i})\psi_{M}(\rho_{i+1}):\left\\{\begin{array}[]{ll}x_{i}\longmapsto
x_{i}x_{i+2}^{u_{i}}x_{i}^{-u_{i+2}},&\\\ x_{i+1}\longmapsto
x_{i}x_{i+1}^{u_{i}}x_{i}^{-u_{i+1}},&\\\ x_{i+2}\longmapsto x_{i}.&\\\
\end{array}\right.$

Comparing the obtained automorphisms we see that the required relation holds.

The proposition is proved.

As it is known the relation $\mathcal{F}_{1}$ holds in $WB_{n}$ but at the
same time the relation $\mathcal{F}_{2}$ does not hold. It would be
interesting to construct representations of other factor-groups of the group
$VB_{n}$. For example, the representation of the group $UVB_{n}$ in which both
forbidden relations hold. We verify the conditions for the second forbidden
relation $\mathcal{F}_{2}$ to hold:

$\psi_{M}(\rho_{i+1})\psi_{M}(\sigma_{i})\psi_{M}(\sigma_{i+1})=\psi_{M}(\sigma_{i})\psi_{M}(\sigma_{i+1})\psi_{M}(\rho_{i}).$

We find the action of the automorphism
$\psi_{M}(\rho_{i+1})\psi_{M}(\sigma_{i})\psi_{M}(\sigma_{i+1})$ on the
generators of $F_{n}$. We obtain

$\psi_{M}(\rho_{i+1})\psi_{M}(\sigma_{i})\psi_{M}(\sigma_{i+1}):\left\\{\begin{array}[]{ll}x_{i}\longmapsto
x_{i}\left(x_{i+1}x_{i+2}^{u_{i+1}}x_{i+1}^{-u_{i+2}}\right)^{u_{i}}x_{i}^{-u_{i+2}},&\\\
x_{i+1}\longmapsto x_{i+1},&\\\ x_{i+2}\longmapsto x_{i}.&\\\
\end{array}\right.$

Analogously we find the action of the automorphism
$\psi_{M}(\sigma_{i})\psi_{M}(\sigma_{i+1})\psi_{M}(\rho_{i})$. We have

$\psi_{M}(\sigma_{i})\psi_{M}(\sigma_{i+1})\psi_{M}(\rho_{i}):\left\\{\begin{array}[]{ll}x_{i}\longmapsto
x_{i+1}\left(x_{i}x_{i+2}^{u_{i}}x_{i}^{-u_{i+2}}\right)^{u_{i+1}}x_{i+1}^{-u_{i+2}},&\\\
x_{i+1}\longmapsto x_{i+1},&\\\ x_{i+2}\longmapsto x_{i}.&\\\
\end{array}\right.$

Comparing the obtained automorphisms it can be seen that the required relation
holds providing the following restriction

$\left(x_{i+1}x_{i+2}^{u_{i+1}}x_{i+1}^{-u_{i+2}}\right)^{u_{i}}x_{i}^{-u_{i+2}}=x_{i+1}\left(x_{i}x_{i+2}^{u_{i}}x_{i}^{-u_{i+2}}\right)^{u_{i+1}}x_{i+1}^{-u_{i+2}},$

i. e.

$\left(x_{i+1}^{-u_{i}}x_{i}^{-1}x_{i}x_{i}^{u_{i+1}}\right)\cdot
x_{i+2}^{u_{i}u_{i+1}}=x_{i+2}^{u_{i}u_{i+1}}\cdot\left(x_{i+1}^{-u_{i}}x_{i}^{-1}x_{i}x_{i}^{u_{i+1}}\right)^{u_{i+2}}.$

These are the relations that are of the form $ab=ba^{c}$ or equally
$[ac,b]=1$.

Question 1. What is the group where these relations hold?

§ 3. Representation that is equivalent to the representation $\varphi_{M}$

The constructed representation $\varphi_{M}$ is not an extension of the Artin
representation which is known to be faithful on the subgroup $B_{n}$. We show
that the representation $\varphi_{M}$ is equivalent to the simpler one which
is an extension of the Artin representation. Let
$F_{n,n}=F_{n}*\mathbb{Z}^{n}$ where $F_{n}=\langle
y_{1},y_{2},\ldots,y_{n}\rangle$ is the free group and $\mathbb{Z}^{n}=\langle
v_{1},v_{2},\ldots,v_{n}\rangle$ is the free abelian group of the rank $n$. It
is true

Theorem 2. The representation
$\widetilde{\varphi}_{M}:VB_{n}\longrightarrow\mathrm{Aut}(F_{n,n})$ defined
by the action on the generators

$\widetilde{\varphi}_{M}(\sigma_{i}):\left\\{\begin{array}[]{l}y_{i}\longmapsto
y_{i}y_{i+1}y_{i}^{-1},\\\ y_{i+1}\longmapsto y_{i},\\\
\end{array}\right.~{}~{}~{}\widetilde{\varphi}_{M}(\sigma_{i}):\left\\{\begin{array}[]{l}v_{i}\longmapsto
v_{i+1},\\\ v_{i+1}\longmapsto v_{i},\\\ \end{array}\right.$
$\widetilde{\varphi}_{M}(\rho_{i}):\left\\{\begin{array}[]{l}y_{i}\longmapsto
y_{i+1}^{v_{i}^{-1}},\\\ y_{i+1}\longmapsto y_{i}^{v_{i+1}},\\\
\end{array}\right.~{}~{}~{}\widetilde{\varphi}_{M}(\rho_{i}):\left\\{\begin{array}[]{l}v_{i}\longmapsto
v_{i+1},\\\ v_{i+1}\longmapsto v_{i}\\\ \end{array}\right.$

is equivalent to the representation $\varphi_{M}$.

The theorem proof is due to the following lemmas.

Lemma 1.  If we take new generators $y_{i}$ and $w_{i}$ in the group
$F_{n,2n+1}$ such that $y_{i}=(x_{i}u_{i}^{-1}v_{0}^{-1})^{v_{0}^{-(i-1)}}$,
$w_{i}=v_{i}v_{0}^{-1}$, $i=1,\ldots,n$, instead of generators $x_{i}$,
$v_{i}$, $i=1,\ldots,n$, than the representation

$\varphi_{M}:VB_{n}\longrightarrow\mathrm{Aut}(F_{n,2n+1})$

acts on new generators $y_{1},\ldots,y_{n}$, $u_{1},\ldots,u_{n}$,
$v_{0},w_{1},\ldots,w_{n}$ by the following way

$\varphi_{M}(\sigma_{i}):\left\\{\begin{array}[]{l}y_{i}\longmapsto
y_{i}y_{i+1}y_{i}^{-1},\\\ y_{i+1}\longmapsto y_{i},\\\
\end{array}\right.~{}~{}~{}\varphi_{M}(\sigma_{i}):\left\\{\begin{array}[]{l}u_{i}\longmapsto
u_{i+1},\\\ u_{i+1}\longmapsto u_{i},\\\
\end{array}\right.~{}~{}~{}\varphi_{M}(\sigma_{i}):\left\\{\begin{array}[]{l}w_{i}\longmapsto
w_{i+1},\\\ w_{i+1}\longmapsto w_{i},\\\ \end{array}\right.$
$\varphi_{M}(\rho_{i}):\left\\{\begin{array}[]{l}y_{i}\longmapsto
y_{i+1}^{w_{i}^{-1}},\\\ y_{i+1}\longmapsto y_{i}^{w_{i+1}},\\\
\end{array}\right.~{}~{}~{}\varphi_{M}(\rho_{i}):\left\\{\begin{array}[]{l}u_{i}\longmapsto
u_{i+1},\\\ u_{i+1}\longmapsto u_{i},\\\
\end{array}\right.~{}~{}~{}\varphi_{M}(\rho_{i}):\left\\{\begin{array}[]{l}w_{i}\longmapsto
w_{i+1},\\\ w_{i+1}\longmapsto w_{i}.\\\ \end{array}\right.$

Proof. It is sufficient to find the action $\varphi_{M}(\sigma_{i})$ and
$\varphi_{M}(\rho_{i})$ on $y_{i}$, $y_{i+1}$, $w_{i}$, $w_{i+1}$,
$i=1,\ldots,n-1$. We obtain

$\begin{array}[]{rl}\varphi_{M}(\sigma_{i})(y_{i})&=(x_{i}u_{i}^{-1}v_{0}^{-1})^{v_{0}^{-(i-1)}}\longmapsto(x_{i}x_{i+1}^{u_{i}}x_{i}^{-v_{0}u_{i+1}}u_{i+1}^{-1}v_{0}^{-1})^{v_{0}^{-(i-1)}}\\\
&=(x_{i}u_{i}^{-1}x_{i+1}u_{i}v_{0}^{-1}u_{i+1}^{-1}x_{i}^{-1}v_{0}u_{i+1}u_{i+1}^{-1}v_{0}^{-1})^{v_{0}^{-(i-1)}}\\\
&=(x_{i}u_{i}^{-1}x_{i+1}u_{i+1}^{-1}u_{i}v_{0}^{-1}x_{i}^{-1})^{v_{0}^{-(i-1)}}\\\
&=(x_{i}u_{i}^{-1}v_{0}^{-1}v_{0}x_{i+1}u_{i+1}^{-1}v_{0}^{-1}v_{0}^{-1}v_{0}u_{i}x_{i}^{-1})^{v_{0}^{-(i-1)}}\\\
&=(x_{i}u_{i}^{-1}v_{0}^{-1})^{v_{0}^{-(i-1)}}(x_{i+1}u_{i+1}^{-1})^{v_{0}^{-i}}(x_{i}u_{i}^{-1}v_{0}^{-1})^{-v_{0}^{-(i-1)}}\\\
&=y_{i}y_{i+1}y_{i}^{-1},\\\
\varphi_{M}(\sigma_{i})(y_{i+1})&=(x_{i+1}u_{i+1}^{-1}v_{0}^{-1})^{v_{0}^{-i}}\mapsto(x_{i}^{v_{0}}u_{i}^{-1}v_{0}^{-1})^{v_{0}^{-i}}=(x_{i}u_{i}^{-1}v_{0}^{-1})^{v_{0}^{-(i-1)}}=y_{i},\\\
\varphi_{M}(\sigma_{i})(w_{i})&=v_{i}v_{0}^{-1}\mapsto
v_{i+1}v_{0}^{-1}=w_{i+1},\\\
\varphi_{M}(\sigma_{i})(w_{i+1})&=v_{i+1}v_{0}^{-1}\mapsto
v_{i}v_{0}^{-1}=w_{i},\\\ \end{array}$
$\begin{array}[]{ll}\varphi_{M}(\rho_{i})(y_{i})&=(x_{i}u_{i}^{-1}v_{0}^{-1})^{v_{0}^{-(i-1)}}\longmapsto(x_{i+1}^{v_{i}^{-1}}u_{i+1}^{-1}v_{0}^{-1})^{v_{0}^{-(i-1)}}\\\
&=(x_{i+1}u_{i+1}^{-1}v_{0}^{-1})^{v_{0}^{-i}v_{i}^{-1}v_{0}}=y_{i+1}^{w_{i}^{-1}},\\\
\varphi_{M}(\rho_{i})(w_{i})&=v_{i}v_{0}^{-1}\mapsto
v_{i+1}v_{0}^{-1}=w_{i+1},\\\
\varphi_{M}(\rho_{i})(w_{i+1})&=v_{i+1}v_{0}^{-1}\mapsto
v_{i}v_{0}^{-1}=w_{i}.\\\ \end{array}$

The lemma is proved.

On account of the Lemma 1 the subgroups

$F_{n,n}=\left\langle y_{1},\ldots,y_{n},w_{1},\ldots,w_{n}\right\rangle,\quad
W=\left\langle w_{1},\ldots,w_{n}\right\rangle,\quad U=\left\langle
u_{1},\ldots,u_{n}\right\rangle,\quad V_{0}=\left\langle v_{0}\right\rangle$

of $F_{n,2n+1}$ are invariants under the action of $\varphi_{M}(VB_{n})$. In
particular, the representation $\varphi_{M}$ induces the representation
$\widetilde{\varphi}_{M}$ of the group $VB_{n}$ into the group $Aut(F_{n,n})$.
Also note that the restrictions $\varphi_{M}(VB_{n})$ on the subgroups $W$ and
$U$ are equivalent with respect to the isomorphisms $u_{i}\leftrightarrows
w_{i}$ where $i=1,\ldots,n$.

Lemma 2.  The representations $\varphi_{M}$ and $\widetilde{\varphi}_{M}$ are
equivalent, i. e. $\varphi_{M}(\beta)=1$ if and only if
$\widetilde{\varphi}_{M}(\beta)=1$ for some $\beta\in VB_{n}$.

Proof follows directly from the fact that the restrictions of
$\varphi_{M}(VB_{n})$ on the subgroups $W$ and $U$ are equivalent. The lemma
is proved.

Theorem 2 follows from the proved lemmas.

Hence the most general representation $\widetilde{\varphi}_{M}$ is in fact the
extension of the Artin representation of the braid group $B_{n}$.

Using the result of O. Chterental [8] who proved that the representation
$\varphi_{A}$ of the group $VB_{n}$ is not faithful for $n\geq 4$, it could be
shown that the representations $\varphi_{SW}$ and $\varphi_{BD}$ are not
faithful for $n\geq 4$ as well. It is true

Proposition 5. The element
$\beta=(\sigma_{2}^{-1}\rho_{1}\sigma_{2}\rho_{3})^{3}\in VP_{4}$ lies in the
kernel of the representations $\varphi_{SW}$ as well as $\varphi_{BD}$ and it
does not lie in the kernel of the representation $\widetilde{\varphi}_{M}$.

Proof. The straightforward verification that $\varphi_{SW}(\beta)=1$ and
$\varphi_{BD}(\beta)=1$ gives us the first part of the proof.

Further we denote $\sigma_{2}^{-1}\rho_{1}\sigma_{2}\rho_{3}$ by $\alpha$,
then $\beta=\alpha^{3}$. To check the remaining part of the proof we find the
automorphism $\widetilde{\varphi}_{M}(\alpha)$. We obtain

$\widetilde{\varphi}_{M}(\alpha):\left\\{\begin{array}[]{l}x_{1}\longmapsto(x_{2}x_{4}^{v_{3}^{-1}}x_{2}^{-1})^{v_{1}^{-1}},\\\
x_{2}\longmapsto x_{2},\\\ x_{3}\longmapsto x_{2}^{-1}x_{1}^{v_{4}}x_{2},\\\
x_{4}\longmapsto x_{3}^{v_{4}},{}{}\\\
\end{array}\right.~{}~{}~{}\widetilde{\varphi}_{M}(\alpha):\left\\{\begin{array}[]{l}u_{1}\longmapsto
u_{4},\\\ u_{2}\longmapsto u_{2},\\\ u_{3}\longmapsto u_{1},\\\
u_{4}\longmapsto u_{3}.\\\ \end{array}\right.$

Next we find

$\widetilde{\varphi}_{M}(\beta):\left\\{\begin{array}[]{l}x_{1}\longmapsto
x_{2}^{v_{3}^{-1}}\left(x_{2}^{-1}x_{1}^{v_{4}}x_{2}\right)^{v_{4}^{-1}}x_{2}^{-v_{3}^{-1}},\\\
x_{2}\longmapsto x_{2},\\\ x_{3}\longmapsto
x_{2}^{-1}x_{2}^{v_{1}}x_{3}x_{2}^{-v_{4}^{-1}}x_{2},\\\ x_{4}\longmapsto
x_{2}^{-v_{1}}\left(x_{2}x_{4}^{v_{3}^{-1}}x_{2}^{-1}\right)^{v_{3}}x_{2}^{v_{1}},{}{}\\\
\end{array}\right.$

i. e. $\widetilde{\varphi}_{M}(\beta)\not=1$. The proposition is proved.

Nevertheless the following question remains open

Question 2. Is the representation
$\varphi_{A}:VB_{n}\longrightarrow\mathrm{Aut}(F_{n+1})$ faithful for $n=3$?

§ 4. The link groups

To find a presentation of the link group two approaches could be used. The
first one is the Wirtinger approach when the link diagram on the plane is used
to find the generators and defining relations of the link group. The second
one is to consider the link as a closed braid and use the braid group
representation into the automorphism group of some group. Both these
approaches are described in the book [5] for classical links and in the paper
[4] for virtual ones. In particular, for each virtual braid $\beta$ the groups
$G_{A}(\beta)$, $G_{SW}(\beta)$, $G_{BD}(\beta)$ and $G_{M}(\beta)$ could be
defined using the representations $\varphi_{A}$, $\varphi_{SW}$,
$\varphi_{BD}$ and $\varphi_{M}$ correspondingly. In addition it should be
proved that the corresponding group depends only on the link $\widehat{\beta}$
(the link $\widehat{\beta}$ is the closure of the braid $\beta$) if we want to
construct the link invariant.

The classical link group. Recall that for a classical link $L$ in
$\mathbb{S}^{3}$ its group $G(L)$ is the fundamental group
$\pi_{1}(\mathbb{S}^{3}\setminus L)$ of the complement of $L$ in
$\mathbb{S}^{3}$. To find generators and relations of the group $G(L)$ either
the link diagram or the consideration of the link as the closed braid and the
Artin representation could be used. In the case we use the link diagram the
generator is assigned to each connected component and the following relation
is assigned to each classical crossing: $ab=bc$ or its equivalent $c=b^{-1}ab$
for the positive crossing (see Fig. 1 on the left) and $ab=ca$ or its
equivalent $c=aba^{-1}$ for the negative crossing (see Fig. 1 on the right).

Figure 1: Relations in classical crossings

The generalized Alexander group. D. Silver and S. Williams [21] defined the
generalized Alexander group $\widetilde{A}_{L}$ for the virtual $d$-component
link $L=K_{1}\cup K_{2}\cup\ldots\cup K_{d}$. For that purpose the free
abelian group $\Pi_{d}=\langle u_{1},u_{2},\ldots,u_{d},v\rangle$ of the rank
$d+1$ was used. In particular provided $L$ a knot, the free abelian group
$\Pi_{1}=\langle u,v\rangle$ of the rank $2$ is used.

Figure 2: Relation in the Alexander generalized group

The particular generator is assigned to each arc of the diagram $L$ from one
classical crossing to another one. Two relations are assigned to each positive
crossing: $ab^{u_{i}}=cd^{u_{j}}$, $b=c^{v}$ (see Fig. 2 on the left) which
could be written in the form

$c=b^{v^{-1}},~{}~{}d=b^{-v^{-1}u_{j}^{-1}}a^{u_{j}^{-1}}b^{u_{i}u_{j}^{-1}}.$

Two relations are assigned to each negative crossing: $ab^{u_{i}}=cd^{u_{j}}$,
$d=a^{v}$ (see Fig. 2 on the right) which could be written in the form

$c=ab^{u_{i}}a^{-vu_{j}},~{}~{}d=a^{v}.$

No relation is assigned to the virtual crossing.

The welded braid group. For the welded $d$ component link $L$ the generalized
Alexander group $A_{L}$ is constructed (see [21]) which is obtained from the
corresponding virtual link by putting $v=1$, i. e. the following relations
correspond to each classical crossing

$ab^{u_{i}}=cd^{u_{j}},~{}~{}b=c$

for the positive case and the relations

$ab^{u_{i}}=cd^{u_{j}},~{}~{}d=a$

for the negative one.

The generalized Alexander group of the trivial $d$ component link $U_{d}$ is a
free product $F_{d}*\mathbb{Z}^{d}$ of the free group of the rank $d$ and the
free abelian group of the same rank.

§ 5. The virtual link groups

In the present section the new definition of the virtual link group is given.
For this two approaches are used: the group will be defined at first with the
help of the representation $\varphi_{M}$ (the braid approach) then through the
virtual diagram. Next the equivalence of these two definitions is shown.

The braid approach. We describe the general way which allows to construct the
link invariants through the representation of the virtual braid group by
automorphisms of some group. Assume that we have a representation
$\varphi:VB_{n}\longrightarrow\mathrm{Aut}(H)$ of the virtual braid group into
the automorphism group of some group $H=\langle
h_{1},h_{2},\ldots,h_{m}~{}\|~{}\mathcal{R}\rangle$, where $\mathcal{R}$ is
the set of defining relations. The following group is assigned to the virtual
braid $\beta\in VB_{n}$:

$G_{\varphi}(\beta)=\langle
h_{1},h_{2},\ldots,h_{m}~{}\|~{}\mathcal{R},h_{i}=\varphi(\beta)(h_{i}),~{}~{}i=1,2,\ldots,m\rangle.$

The group $G_{\varphi}$ is the invariant of the virtual link providing the
verification that the group $G_{\varphi}(\beta)$ is isomorphic to
$G_{\varphi}(\beta^{\prime})$ for each braid $\beta^{\prime}$ such that the
links $\widehat{\beta}$ and $\widehat{\beta^{\prime}}$ are equivalent.

This approach is used for the previously defined representation $\varphi_{M}$.
Given $\beta\in VB_{n}$, the  group of the braid $\beta$ is the following
group

$\displaystyle\begin{split}G_{M}(\beta)&=\langle
x_{1},x_{2},\ldots,x_{n},u_{1},u_{2},\ldots,u_{n},v_{0},v_{1},\ldots,v_{n}~{}||~{}[u_{i},u_{j}]=[v_{k},v_{l}]=[u_{i},v_{k}]=1,\\\
x_{i}=\varphi_{M}(\beta)&(x_{i}),~{}~{}u_{i}=\varphi_{M}(\beta)(u_{i}),~{}~{}v_{i}=\varphi_{M}(\beta)(v_{i}),~{}~{}i,j=1,2,\ldots,n,~{}~{}k,l=0,1,\ldots,n\rangle.\end{split}$

Let show that the group defined in this manner is indeed the link invariant.
More precisely it is true

Theorem 3. Given $\beta\in VB_{n}$ and $\beta^{\prime}\in VB_{m}$ the two
virtual braids such that theirs closures define the same link $L$, then
$G_{M}(\beta)\cong G_{M}(\beta^{\prime})$, i. e. the group $G_{M}(\beta)$ is
the link $L$ invariant and it could be denoted by the symbol
$G_{M}(\widehat{\beta})=G_{M}(L)$.

The next paragraph is dedicated to the proof of the theorem.

The diagram approach. Now we show the way of how to find the virtual link
group using its diagram. Let $D_{L}$ be a diagram of a virtual $d$-component
link $L$. This diagram is divided into arcs from one crossing (classical or
virtual) to other and for every obtained arc a generator is assigned. Thus we
have some set of generators $a,b,c,\ldots$. We numerate all components of
$D_{L}$ by numbers from 1 to $d$ and assign to the $i$ th component two
generators: $u_{i}$ and $v_{i}$.

A group of the diagram $D_{L}$ is the group generated by elements
$a,b,c,\ldots$, $u_{i}$, $v_{0}$, $v_{i}$, $i=1,\ldots,d$, and defined by the
following system of relations: two relations correspond to each crossing(see
Fig. 3, 4).

Figure 3: The relations in classical crossings of the group $G_{M}$ Figure 4:
The relations in the virtual crossing of the group $G_{M}$

Furthermore, all elements $u_{i}$ and $v_{j}$ commute pairwise , i. e.
$\langle u_{1},\ldots,u_{d},v_{0},v_{1},\ldots,v_{d}\rangle$ is a free abelian
group of the rank $2d+1$.

This group is denoted by $G_{D}(D_{L})$. If $D_{L}^{\prime}$ is another
diagram of $L$, then $D_{L}$ can be transformed to $D_{L}^{\prime}$ with the
help of a finite number of generalized Reidemeister moves and a flat isotopy.
Analogously to the case of classical link diagrams it is not difficult to
prove

Proposition 6. If $D_{L}$ and $D_{L}^{\prime}$ are two diagrams of a virtual
link $L$, then groups $G_{D}(D_{L})$ and $G_{D}(D_{L}^{\prime})$ are
isomorphic. Hence $G_{D}(D_{L})$ is an invariant of $L$ and we denote it by
$G_{D}(L)$.

Now we show that the group of a virtual link $L$ constructed by the braid is
isomorphic to the group constructed by the diagram. More precisely, it is true

Proposition 7. Let $L$ be a virtual link, $D_{L}$ be its diagram, $\beta$ be a
braid such that its closure $\widehat{\beta}$ is equivalent to $L$, then
$G_{D}(D_{L})$ is isomorphic to $G_{M}(\beta)$.

Proof. Let $\beta$ be a virtual braid from the group $VB_{n}$ for some $n$. We
consider the braid $\beta$ as a word in the alphabet $\\{\sigma_{1}^{\pm
1},\sigma_{2}^{\pm 1},\ldots,\sigma_{n-1}^{\pm
1},\rho_{1},\rho_{2},\ldots,\rho_{n-1}\\}$. Every generator
$\sigma_{i},\sigma_{i}^{-1}$ and $\rho_{i}$ has a geometric interpretation and
can be presented by its diagram. Hence the braid $\beta$ can be presented by a
diagram which is divided in the layers so that each layer contains only one
crossing: classical or virtual. The generators $\sigma_{i}$ or
$\sigma_{i}^{-1}$ correspond to the classical crossing and the generator
$\rho_{i}$ corresponds to the virtual one. If the braid $\beta$ is a word of
the length $m$ in the group $VB_{n}$, then its diagram contains $m$ layers. We
consider the first layer. It contains $n$ arcs. The middle line of the first
layer divides these arcs in half. We assign to starting parts of these arcs
the symbols $x_{1,j},u_{1,j},v_{1,j}$, $j=1,2,\ldots,n$. To the next parts of
the arcs which end at the middle of the second layer we assign the symbols
$x_{2,j},u_{2,j},v_{2,j}$, $j=1,2,\ldots,n$. By continuing this process we
divide the diagram $\beta$ into $n(m+1)$ arcs.

We are moving alongside the braid diagram from the bottom upwards, i.e. we are
reading the word $\beta$ from right to left. Let the $k$-th layer contain the
classical crossing, which corresponds to the elementary braid $\sigma_{i}$. In
this layer we have two sets of arcs with the corresponding sets of symbols
$x_{kj}$, $u_{kj}$, $v_{kj}$ and $x_{k+1,j}$, $u_{k+1,j}$, $v_{k+1,j}$,
$j=1,\ldots,n$.

Figure 5: Classical positive crossing

Due to the formulae of the Theorem 1, the following automorphism corresponds
to the braid $\sigma_{i}$

$\varphi_{M}(\sigma_{i}):\left\\{\begin{array}[]{l}x_{k+1,i}\longmapsto
x_{k,i}x_{k,i+1}^{u_{ki}}x_{k,i}^{-v_{0}u_{k,i+1}},\\\ x_{k+1,i+1}\longmapsto
x_{k,i}^{v_{0}},\\\ x_{k+1,j}\longmapsto x_{k,j},\\\
\end{array}\right.\left\\{\begin{array}[]{l}u_{k+1,i}\longmapsto u_{k,i+1},\\\
u_{k+1,i+1}\longmapsto u_{k,i},\\\ u_{k+1,j}\longmapsto u_{k,j},\\\
\end{array}\right.\left\\{\begin{array}[]{l}v_{k+1,i}\longmapsto v_{k,i+1},\\\
v_{k+1,i+1}\longmapsto v_{k,i},\\\ v_{k+1,j}\longmapsto v_{k,j},\\\
\end{array}\right.$

where $j\neq i,i+1$. The case when the $k$-th layer contains $\sigma_{i}^{-1}$
is considered by analogy.

If the $k$-th layer contains $\rho_{i}$ (see Fig. 6), then we have the
following automorphism

Figure 6: Virtual crossing

$\varphi_{M}(\rho_{i}):\left\\{\begin{array}[]{l}x_{k+1,i}\longmapsto
x_{k,i+1}^{v_{ki}^{-1}},\\\ x_{k+1,i+1}\longmapsto x_{k,i}^{v_{k,i+1}},\\\
x_{k+1,j}\longmapsto x_{k,j},\\\
\end{array}\right.\left\\{\begin{array}[]{l}u_{k+1,i}\longmapsto u_{k,i+1},\\\
u_{k+1,i+1}\longmapsto u_{k,i},\\\ u_{k+1,j}\longmapsto u_{k,j},\\\
\end{array}\right.\left\\{\begin{array}[]{l}v_{k+1,i}\longmapsto v_{k,i+1},\\\
v_{k+1,i+1}\longmapsto v_{k,i},\\\ v_{k+1,j}\longmapsto v_{k,j},\\\
\end{array}\right.$

where $j\neq i,i+1$.

Since in every layer top and bottom arcs are identified we can write the set
of relations

1)

$\left\\{\begin{array}[]{l}x_{k+1,i}=x_{k,i}x_{k,i+1}^{u_{ki}}x_{k,i}^{-v_{0}u_{k,i+1}},\\\
x_{k+1,i+1}=x_{k,i}^{v_{0}},\\\ x_{k+1,j}=x_{k,j},\\\
\end{array}\right.\left\\{\begin{array}[]{l}u_{k+1,i}=u_{k,i+1},\\\
u_{k+1,i+1}=u_{k,i},\\\ u_{k+1,j}=u_{k,j},\\\
\end{array}\right.\left\\{\begin{array}[]{l}v_{k+1,i}=v_{k,i+1},\\\
v_{k+1,i+1}=v_{k,i},\\\ v_{k+1,j}=v_{k,j},\\\ \end{array}\right.$

where $j\neq i,i+1$, for the $k$-th layer corresponded to $\sigma_{i}$;

2)

$\left\\{\begin{array}[]{l}x_{k+1,i}=x_{k,i+1}^{v_{ki}^{-1}},\\\
x_{k+1,i+1}=x_{k,i}^{v_{k,i+1}},\\\ x_{k+1,j}=x_{k,j},\\\
\end{array}\right.\left\\{\begin{array}[]{l}u_{k+1,i}=u_{k,i+1},\\\
u_{k+1,i+1}=u_{k,i},\\\ u_{k+1,j}=u_{k,j},\\\
\end{array}\right.\left\\{\begin{array}[]{l}v_{k+1,i}=v_{k,i+1},\\\
v_{k+1,i+1}=v_{k,i},\\\ v_{k+1,j}=v_{k,j},\\\ \end{array}\right.$

where $j\neq i,i+1$, for $k$-th layer which corresponds to $\rho_{i}$.

Also we have to add

3) commutation relations:

$[u_{kl},u_{pq}]=[u_{kl},v_{pq}]=[v_{kl},v_{pq}]=[u_{kl},v_{0}]=[v_{kl},v_{0}]=1,~{}~{}~{}k,p=1,\ldots,m+1,l,q=1,\ldots,n.$

Using these relations, we move alongside the braid $\beta$ from the bottom
upwards and we express the generators $x_{m+1,j}$, $u_{m+1,j}$, $v_{m+1,j}$
from the $m$-th layer through the generators $x_{m,j}$, $u_{m,j}$, $v_{m,j}$,
$j=1,2,\ldots,n$. Next we consider the $m-1$-th layer and expressing the
generators $x_{m+1,j}$, $u_{m+1,j}$, $v_{m+1,j}$ through the generators
$x_{m-1,j}$, $u_{m-1,j}$, $v_{m-1,j}$, $j=1,2,\ldots,n$, and so on. And we
express the generators $x_{m+1,j}$, $u_{m+1,j}$, $v_{m+1,j}$, through the
generators $x_{1,j}$, $u_{1,j}$, $v_{1,j}$, $j=1,2,\ldots,n$ when we arrive to
the first layer.

To come to the closed braid $\widehat{\beta}$ we must connect initial arcs of
the diagram $\beta$ with the terminal arcs of this diagram, i.e. we must add
the relations

4)

$x_{m+1,j}=x_{1,j},\quad u_{m+1,j}=u_{1,j},\quad v_{m+1,j}=v_{1,j},\quad
j=1,\ldots,n.$

It follows from relations 1), 2), 4) that the generators $u_{k,l}$, $v_{k,l}$
that correspond to one component of the link are equal. This corresponds to
the fact that under the homomorphism of $VB_{n}$ onto $S_{n}$ the image
$\beta$ in $S_{n}$ is a product of $d$ independent cycles. Hence among the
generators $u_{i,l}$ (also as among the generators $v_{i,l}$) there are only
$d$ different ones.

Excluding transitional letters $x_{k,l}$, $u_{k,l}$, $v_{k,l}$, $k\geq 2$, we
obtain relations

$\varphi_{M}(\beta)(x_{1})=x_{1},\ldots,\varphi_{M}(\beta)(x_{n})=x_{n},$
$\varphi_{M}(\beta)(u_{1})=u_{1},\ldots,\varphi_{M}(\beta)(u_{n})=u_{n},$
$\varphi_{M}(\beta)(v_{1})=v_{1},\ldots,\varphi_{M}(\beta)(v_{n})=v_{n},$

of the virtual link group constructed using the braid $\beta$. Here
$x_{1,1}=x_{1}$, $\ldots$, $x_{1,n}=x_{n}$, $u_{1,1}=u_{1}$, $\ldots$,
$u_{1,n}=u_{n}$, $v_{1,1}=v_{1}$, $\ldots$, $v_{1,n}=v_{n}$. Hence we get the
representation of the group $G_{M}(\beta)$. The proposition is proved.

Welded link group. A welded ling group could be defined analogously using the
representation $\psi_{M}$. Let a welded link $L_{w}$ be equivalent to a
closure of the braid $\beta_{w}\in WB_{n}$ for some $n$. The group of the
braid $\beta_{w}$ is the following group

$G(\beta_{w})=\langle
x_{1},x_{2},\ldots,x_{n},u_{1},u_{2},\ldots,u_{n}~{}||~{}[u_{i},u_{j}]=1,$
$x_{i}=\psi_{M}(\beta_{w})(x_{i}),~{}~{}u_{i}=\psi_{M}(\beta_{w})(u_{i}),~{}~{}i,j=1,2,\ldots,n\rangle.$

Taking into account the fact that the group $WB_{n}$ is a homomorphic image of
the group $VB_{n}$, it could be concluded from Theorem 3 that the group
$G(\beta_{w})$ is an invariant of the link $L_{w}$.

§ 6. Proof of Theorem 3

The proof of this theorem is divided into several steps. At first it will be
shown that the group $G_{\widetilde{M}}(\beta)$ which is constructed using the
representation $\widetilde{\varphi_{M}}$ is the invariant of closed braid
$\widehat{\beta}$. It is true

Theorem 4. Let $\beta\in VB_{n}$ and $\beta^{\prime}\in VB_{m}$ be two virtual
braids such that theirs closures define the same link $L$, then
$G_{\widetilde{M}}(\beta)\cong G_{\widetilde{M}}(\beta^{\prime})$, i. e. the
group $G_{\widetilde{M}}(\beta)$ is the invariant of the link
$L=\widehat{\beta}$ and it will be denoted by the symbol
$G_{\widetilde{M}}(L)$.

Proof. Kamada proved the analogue of Markov’s theorem in the case of virtual
braids [10]. This theorem formulated in the algebraic terms by Kauffman and
Lambropoulou [12] is more convenient for the verification. According to this
theorem two oriented virtual links are equivalent if and only if two
corresponding virtual braids differ by braid relations in $VB_{n}$ and a
finite sequence of the following moves or their inverses:

1) Virtual and real conjugation:

$\beta\to\rho_{k}\beta\rho_{k},~{}~{}~{}\beta\to\sigma_{k}\beta\sigma_{k}^{-1};$

2) Right virtual and real stabilization:

$\beta\to\beta\rho_{n},~{}~{}~{}\beta\to\beta\sigma_{n}^{\pm 1},$

3) Algebraic right over/under threading:

$\beta\to\beta\sigma_{n}^{\pm 1}\rho_{n-1}\sigma_{n}^{\mp 1},$

4) Algebraic left over/under threading:

$\beta\to\beta\rho_{n}\rho_{n-1}\sigma_{n-1}^{\mp 1}\rho_{n}\sigma_{n-1}^{\pm
1}\rho_{n-1}\rho_{n},$

where $\beta,\rho_{k},\sigma_{k}\in VB_{n},k=1,\ldots,n-1$ and
$\rho_{n},\sigma_{n}\in VB_{n+1}.$

It is needed to verify that groups obtained by moves 1) – 4) are isomorphic to
the group $G_{\widetilde{M}}(\beta)$.

To simplify notations we will write $\beta$ instead of
$\widetilde{\varphi}_{M}(\beta)$ and write automorphisms to the right of the
arguments. We denote by symbol $G_{1}$ the group constructed using the braid
$\beta$:

$G_{1}=G_{\widetilde{M}}(\beta)=\langle
y_{1},y_{2},\ldots,y_{n},v_{1},v_{2},\ldots,v_{n}~{}\|~{}y_{i}=y_{i}\beta,~{}~{}v_{i}=v_{i}\beta,~{}i=1,2,\ldots,n\rangle.$

From now on we will not write commutativity relations of the elements $v_{i}$.

1) We consider the conjugacy of the braid $\beta$ in the group $VB_{n}$.
Conjugating by the generator $\sigma_{k}^{-1}$ the following group is obtained

$G_{2}=G_{\widetilde{M}}(\sigma_{k}\beta\sigma_{k}^{-1})=\langle
y_{1},\ldots,y_{n},v_{1},\ldots,v_{n}~{}\|~{}y_{i}=y_{i}(\sigma_{k}\beta\sigma_{k}^{-1}),~{}v_{i}=v_{i}(\sigma_{k}\beta\sigma_{k}^{-1}),~{}i=1,2,\ldots,n\rangle,$

where $k\in{1,2,\ldots,n-1}$. To prove that $G_{2}\cong G_{1}$ defining
relations of $G_{2}$ are rewritten in the form

$y_{i}\sigma_{k}=y_{i}(\sigma_{k}\beta),~{}i=1,2,\ldots,n.$

If $i\neq k,k+1$, due to the equality $y_{i}\sigma_{k}=y_{i}$, this relation
is equivalent consequently

$y_{i}=y_{i}\beta$

which is the relation in $G_{1}$. Thus only four relations are needed to
consider:

$\begin{split}y_{k}\sigma_{k}=y_{k}(\sigma_{k}\beta),\hskip
28.45274pty_{k+1}\sigma_{k}=y_{k+1}(\sigma_{k}\beta),\\\
v_{k}\sigma_{k}=v_{k}(\sigma_{k}\beta),\hskip
28.45274ptv_{k+1}\sigma_{k}=v_{k+1}(\sigma_{k}\beta).\end{split}$ (12)

By definition of $\widetilde{\varphi}_{M}$ relations (12) are equivalent to

$\displaystyle\begin{split}y_{k}y_{k+1}y_{k}^{-1}&=(y_{k}y_{k+1}y_{k}^{-1})\beta,\hskip
19.91684pty_{k}=y_{k}\beta,\\\ v_{k+1}&=v_{k+1}\beta,\hskip
56.9055ptv_{k}=v_{k}\beta.\end{split}$

It is clear that the second, the third and the fourth relations are relations
in $G_{1}$. We rewrite the first relation:

$y_{k}y_{k+1}y_{k}^{-1}=(y_{k}\beta)(y_{k+1}\beta)(y_{k}^{-1}\beta)$

Using the second relation we obtain

$y_{k+1}=y_{k+1}\beta,$

i. e. it is a relation in $G_{1}$. Hence it has been proved that it is
possible to go from the group $G_{2}$ to the group $G_{1}$ using the Tietze
transformations.

We are considering the conjugation by the element $\rho_{k}$. In that case we
have

$G_{2}=G_{\widetilde{M}}(\rho_{k}\beta\rho_{k})=\langle
y_{1},\ldots,y_{n},v_{1},\ldots,v_{n}~{}\|~{}y_{i}=y_{i}(\rho_{k}\beta\rho_{k}),~{}v_{i}=v_{i}(\rho_{k}\beta\rho_{k}),~{}i=1,2,\ldots,n\rangle.$

We rewrite defining relations of $G_{2}$ in the form

$y_{i}\rho_{k}=y_{i}(\rho_{k}\beta),~{}i=1,2,\ldots,n.$

If $i\neq k,k+1$, then we obtain

$y_{i}=y_{i}\beta,$

due to $y_{i}\rho_{k}=y_{i}$. But it is the relation in $G_{1}$. Hence only
four relations have to be considered:

$\displaystyle\begin{split}y_{k}\rho_{k}=y_{k}(\rho_{k}\beta),\hskip
28.45274pty_{k+1}\rho_{k}=y_{k+1}(\rho_{k}\beta),\\\
v_{k}\rho_{k}=v_{k}(\rho_{k}\beta),\hskip
28.45274ptv_{k+1}\rho_{k}=v_{k+1}(\rho_{k}\beta).\end{split}$

These relations are equivalent to the following ones

$\displaystyle\begin{split}y_{k+1}^{v_{k}^{-1}}&=y_{k+1}^{v_{k}^{-1}}\beta,\hskip
28.45274pty_{k}^{v_{k+1}}=y_{k}^{v_{k+1}}\beta,\\\
v_{k+1}&=v_{k+1}\beta,\hskip 45.52458ptv_{k}=v_{k}\beta.\end{split}$

It is easy to see that the relation of $G_{1}$ is obtained. Thus the fact has
been proved that the set of relations in $G_{2}$ is equivalent to the set of
relations in $G_{1}$.

2) We are considering the move of the braid $\beta\in VB_{n}$ into the braid
$\beta\sigma_{n}^{-1}\in VB_{n+1}$. We are dealing with two groups $G_{1}$ and

$G_{2}=G_{\widetilde{M}}(\beta\sigma_{n}^{-1})=\langle
y_{1},\ldots,y_{n},y_{n+1},v_{1},\ldots,v_{n},v_{n+1}~{}\|~{}y_{i}=y_{i}(\beta\sigma_{n}^{-1}),~{}v_{i}=v_{i}(\beta\sigma_{n}^{-1}),~{}i=1,2,\ldots,n+1\rangle$

and it is needed to prove that they are isomorphic.

We rewrite relations of $G_{2}$ in the form

$y_{i}\sigma_{n}=y_{i}\beta,~{}i=1,2,\ldots,n+1.$

Using the same reasoning as in 1) we have to consider only four relations:

$\begin{split}y_{n}\sigma_{n}=y_{n}\beta,\hskip
28.45274pty_{n+1}\sigma_{n}=y_{n+1}\beta\\\ v_{n}\sigma_{n}=v_{n}\beta,\hskip
28.45274ptv_{n+1}\sigma_{n}=v_{n+1}\beta.\end{split}$ (13)

By definition $\varphi_{\widetilde{M}}$ relations (13) are equivalent to

$\displaystyle\begin{split}y_{n}y_{n+1}y_{n}^{-1}&=y_{n}\beta,\hskip
28.45274pty_{n}=y_{n+1},\\\ v_{n+1}&=v_{n}\beta,\hskip
28.45274ptv_{n}=v_{n+1}.\end{split}$

Using the second relation the first one is rewritten in the form

$y_{n}=y_{n}\beta$

which is the relation in $G_{1}$. Using relations $y_{n+1}=y_{n}$ and
$v_{n+1}=v_{n}$ we can exclude $y_{n+1}$ and $v_{n+1}$ from the generator set
of the group $G_{2}$. Thus it has been proved that the group $G_{2}$ is
isomorphic to the group $G_{1}$.

The move of the braid $\beta\in VB_{n}$ into the braid $\beta\sigma_{n}\in
VB_{n+1}$ are considered analogously.

Now we consider the move of the braid $\beta\in VB_{n}$ into the braid
$\beta\rho_{n}\in VB_{n+1}$. Two groups are obtained, $G_{1}$ and

$G_{2}=G_{\widehat{M}}(\beta\rho_{n})=\langle
y_{1},\ldots,y_{n},y_{n+1},v_{1},\ldots,v_{n},v_{n+1}~{}\|~{}y_{i}=y_{i}(\beta\rho_{n}),~{}v_{i}=v_{i}(\beta\rho_{n}),~{}i=1,2,\ldots,n+1\rangle.$

For $i=n$ and $i=n+1$ we have the following relations in $G_{2}$:

$\displaystyle\begin{split}y_{n}\rho_{n}=y_{n}\beta,\hskip
28.45274pty_{n+1}\rho_{n}=y_{n+1}\beta\\\ v_{n}\rho_{n}=v_{n}\beta,\hskip
28.45274ptv_{n+1}\rho_{n}=v_{n+1}\beta\end{split}$

which are equivalent to relations

$\displaystyle\begin{split}y_{n+1}^{v_{n}^{-1}}&=y_{n}\beta,\hskip
28.45274pty_{n}^{v_{n+1}}=y_{n+1}\beta=y_{n+1},\\\ v_{n+1}&=v_{n}\beta,\hskip
28.45274ptv_{n}=v_{n+1}.\end{split}$

Substituting $y_{n+1}$ from the second equation into the first relation, we
obtain

$y_{n}=y_{n}\beta$

and that is the relation in $G_{1}$. Using the second and the fourth relations
variables $y_{n+1}$ and $v_{n+1}$ could be excluded from the set of generators
of the group $G_{2}$. Hence we have proved that the group $G_{2}$ is
isomorphic to the group $G_{1}$.

3) We are considering the algebraic right over threading, i. e. the move of
the braid $\beta\in VB_{n}$ into the braid
$\beta\sigma_{n}\rho_{n-1}\sigma_{n}^{-1}\in VB_{n+1}$. In this case we have
$G_{1}$ and

$\displaystyle\begin{split}G_{2}=G_{\widetilde{M}}(\beta\sigma_{n}\rho_{n-1}\sigma_{n}^{-1})=&\langle
y_{1},y_{2},\ldots,y_{n},y_{n+1},v_{1},\ldots,v_{n},v_{n+1}~{}\|~{}\\\
y_{i}=y_{i}(\beta\sigma_{n}&\rho_{n-1}\sigma_{n}^{-1}),~{}~{}v_{i}=v_{i}(\beta\sigma_{n}\rho_{n-1}\sigma_{n}^{-1}),~{}~{}i=1,2,\ldots,{n+1}\rangle.\end{split}$

To prove that $G_{1}\cong G_{2}$ we rewrite relations of $G_{2}$ in the form

$y_{i}\sigma_{n}\rho_{n-1}\sigma_{n}^{-1}=y_{i}\beta,~{}~{}v_{i}\sigma_{n}\rho_{n-1}\sigma_{n}^{-1}=v_{i}\beta,~{}i=1,2,\ldots,n+1.$

If $i=1,2,\ldots,n-2$ then we have

$y_{i}=y_{i}\beta,~{}~{}v_{i}=v_{i}\beta.$

These are relations in $G_{1}$. Hence we have to consider only six relations:

$\begin{split}y_{n-1}(\sigma_{n}\rho_{n-1}\sigma_{n}^{-1})=y_{n-1}\beta,~{}~{}y_{n}(\sigma_{n}\rho_{n-1}\sigma_{n}^{-1})=y_{n}\beta,~{}~{}y_{n+1}(\sigma_{n}\rho_{n-1}\sigma_{n}^{-1})=y_{n+1}\beta,\\\
v_{n-1}(\sigma_{n}\rho_{n-1}\sigma_{n}^{-1})=v_{n-1}\beta,~{}~{}v_{n}(\sigma_{n}\rho_{n-1}\sigma_{n}^{-1})=v_{n}\beta,~{}~{}v_{n+1}(\sigma_{n}\rho_{n-1}\sigma_{n}^{-1})=v_{n+1}\beta.\end{split}$
(14)

By definition $\widetilde{\varphi}_{M}$ the braid
$\sigma_{n}\rho_{n-1}\sigma_{n}^{-1}$ acts on generators
$y_{n-1},y_{n},y_{n+1},v_{n-1},v_{n},v_{n+1}$ in the following way:

$\displaystyle\begin{split}\sigma_{n}\rho_{n-1}\sigma_{n}^{-1}:\left\\{\begin{array}[]{ll}y_{n-1}\xrightarrow{\sigma_{n}}y_{n-1}\xrightarrow{\rho_{n-1}}y_{n}^{v_{n-1}^{-1}}\xrightarrow{\sigma_{n}^{-1}}y_{n+1}^{v_{n-1}^{-1}},\\\
y_{n}\xrightarrow{\sigma_{n}}y_{n}y_{n+1}y_{n}^{-1}\xrightarrow{\rho_{n-1}}y_{n-1}^{v_{n}}y_{n+1}y_{n-1}^{-v_{n}}\xrightarrow{\sigma_{n}^{-1}}y_{n-1}^{v_{n+1}}y_{n+1}^{-1}y_{n}y_{n+1}y_{n-1}^{-v_{n+1}},\\\
y_{n+1}\xrightarrow{\sigma_{n}}y_{n}\xrightarrow{\rho_{n-1}}y_{n-1}^{v_{n}}\xrightarrow{\sigma_{n}^{-1}}y_{n-1}^{v_{n+1}}.\\\
\end{array}\right.\end{split}$

Analogously

$\displaystyle\sigma_{n}\rho_{n-1}\sigma_{n}^{-1}:\left\\{\begin{array}[]{ll}v_{n-1}\xrightarrow{\sigma_{n}\rho_{n-1}\sigma_{n}^{-1}}v_{n+1},\\\
v_{n}\xrightarrow{\sigma_{n}\rho_{n-1}\sigma_{n}^{-1}}v_{n},\\\
v_{n+1}\xrightarrow{\sigma_{n}\rho_{n-1}\sigma_{n}^{-1}}v_{n-1}.\\\
\end{array}\right.$

Therefore, relations (14) for the representation $\widetilde{\varphi}_{M}$ are
equivalent to

$y_{n+1}^{v_{n-1}^{-1}}=y_{n-1}\beta,~{}~{}y_{n-1}^{v_{n+1}}y_{n+1}^{-1}y_{n}y_{n+1}y_{n-1}^{-v_{n+1}}=y_{n}\beta,~{}~{}y_{n-1}^{v_{n+1}}=y_{n+1}\beta,$
$v_{n+1}=v_{n-1}\beta,~{}~{}v_{n}=v_{n}\beta,~{}~{}v_{n-1}=v_{n+1}\beta.$

The braid $\beta$ acts on generators $y_{n+1}$ and $v_{n+1}$ trivially since
$\beta\in VB_{n}$, i. e. $y_{n+1}\beta=y_{n+1},~{}v_{n+1}\beta=v_{n+1}$.
Consequently, the third and the sixth relations are equivalent to

$y_{n-1}^{v_{n+1}}=y_{n+1},~{}~{}v_{n-1}=v_{n+1},$

correspondingly. Substituting the third relation into the first one and taking
into account the sixth relation, we obtain

$y_{n-1}=y_{n-1}\beta,$

which is the relation in $G_{1}$. Using the third relation we obtain that the
second relation is equivalent to

$y_{n}=y_{n}\beta,$

which is the relation in $G_{1}$. Besides $y_{n+1}$ and $v_{n+1}$ could be
excluded from the generator set of $G_{2}$. Therefore, it has been proved that
the set of relations of $G_{2}$ is equivalent to the set of relations of
$G_{1}$.

4) We are considering the algebraic left over threading, i. e. the move of the
braid $\beta\in VB_{n}$ into the braid
$\beta\rho_{n}\rho_{n-1}\sigma_{n-1}^{-1}\rho_{n}\sigma_{n-1}\rho_{n-1}\rho_{n}\in
VB_{n+1}$. In this case

$\displaystyle\begin{split}G_{2}=G(\beta\sigma_{n}\rho_{n-1}\sigma_{n}^{-1})&=\langle
y_{1},y_{2},\ldots,y_{n},y_{n+1},v_{1},\ldots,v_{n},v_{n+1}~{}\|\\\
~{}y_{i}=y_{i}(\beta\rho_{n}\rho_{n-1}\sigma_{n-1}^{-1}\rho_{n}\sigma_{n-1}\rho_{n-1}\rho_{n})&,~{}v_{i}=v_{i}(\beta\rho_{n}\rho_{n-1}\sigma_{n-1}^{-1}\rho_{n}\sigma_{n-1}\rho_{n-1}\rho_{n}),~{}i=1,2,\ldots,{n+1}\rangle.\end{split}$

To prove that $G_{1}\cong G_{2}$ we rewrite relations of $G_{2}$ in the form

$y_{i}\rho_{n}\rho_{n-1}\sigma_{n-1}^{-1}\rho_{n}\sigma_{n-1}\rho_{n-1}\rho_{n}=y_{i}\beta,~{}i=1,2,\ldots,n+1.$

If $i=1,2,\ldots,n-2$ we have

$y_{i}=y_{i}\beta$

which is the relation in $G_{1}$. Hence it has to be considered only six
relations:

$\begin{split}y_{n-1}(\rho_{n}\rho_{n-1}\sigma_{n-1}^{-1}\rho_{n}\sigma_{n-1}\rho_{n-1}\rho_{n})&=y_{n-1}\beta,\\\
y_{n}(\rho_{n}\rho_{n-1}\sigma_{n-1}^{-1}\rho_{n}\sigma_{n-1}\rho_{n-1}\rho_{n})&=y_{n}\beta,\\\
y_{n+1}(\rho_{n}\rho_{n-1}\sigma_{n-1}^{-1}\rho_{n}\sigma_{n-1}\rho_{n-1}\rho_{n})&=y_{n+1}\beta,\\\
v_{n-1}(\rho_{n}\rho_{n-1}\sigma_{n-1}^{-1}\rho_{n}\sigma_{n-1}\rho_{n-1}\rho_{n})&=v_{n-1}\beta,\\\
v_{n}(\rho_{n}\rho_{n-1}\sigma_{n-1}^{-1}\rho_{n}\sigma_{n-1}\rho_{n-1}\rho_{n})&=v_{n}\beta,\\\
v_{n+1}(\rho_{n}\rho_{n-1}\sigma_{n-1}^{-1}\rho_{n}\sigma_{n-1}\rho_{n-1}\rho_{n})&=v_{n+1}\beta.\end{split}$
(15)

Denote the braid
$\rho_{n}\rho_{n-1}\sigma_{n-1}^{-1}\rho_{n}\sigma_{n-1}\rho_{n-1}\rho_{n}$ by
$b$. It is easy to see that $b$ acts on generators $y_{n-1},y_{n},y_{n+1}$,
$v_{n-1},v_{n},v_{n+1}$ in the following way:

$\displaystyle
b:\left\\{\begin{array}[]{ll}y_{n-1}\longrightarrow(y_{n}^{-1}y_{n+1}^{v_{n}^{-1}v_{n-1}^{-1}}y_{n-1}^{v_{n+1}}y_{n+1}^{-v_{n}^{-1}v_{n-1}^{-1}}y_{n})^{v_{n}^{-1}},\\\
y_{n}\longrightarrow y_{n+1}^{v_{n}^{-1}v_{n-1}^{-1}},\\\
y_{n+1}\longrightarrow y_{n}^{v_{n+1}v_{n-1}}.\\\ \end{array}\right.$
$\displaystyle b:\left\\{\begin{array}[]{ll}v_{n-1}\longrightarrow v_{n-1},\\\
v_{n}\longrightarrow v_{n+1},\\\ v_{n+1}\longrightarrow v_{n}.\\\
\end{array}\right.$

Thus relations (15) are equivalent to

$(y_{n}^{-1}y_{n+1}^{v_{n}^{-1}v_{n-1}^{-1}}y_{n-1}^{v_{n+1}}y_{n+1}^{-v_{n}^{-1}v_{n-1}^{-1}}y_{n})^{v_{n}^{-1}}=y_{n-1}\beta,~{}~{}y_{n+1}^{v_{n}^{-1}v_{n-1}^{-1}}=y_{n}\beta,~{}~{}y_{n}^{v_{n+1}v_{n-1}}=y_{n+1}\beta,$
$v_{n-1}=v_{n-1}\beta,~{}~{}v_{n+1}=v_{n}\beta,~{}~{}v_{n}=v_{n+1}\beta.$

Since $\beta\in VB_{n}$ it acts on generators $y_{n+1}$ and $v_{n+1}$
trivially. Therefore, the third relation is equivalent to

$y_{n}^{v_{n+1}v_{n-1}}=y_{n+1}$

and the sixth relation is equivalent to

$v_{n}=v_{n+1}.$

Obtained equations are substituted in the first and the second relations and
it results in

$y_{n-1}=y_{n-1}\beta,~{}~{}y_{n}=y_{n}\beta,$

which are the relations in $G_{1}$. The generator $y_{n+1}$ and $v_{n+1}$
could be excluded from the generating set of $G_{2}$. Therefore, the fact has
been proved that the set of relations of $G_{2}$ is equivalent to the set of
relations of $G_{1}$.

The theorem is proved.

We proved that the representation $\varphi_{M}$ is equivalent to the
representation $\widetilde{\varphi}_{M}$. Now we show that the group
$G_{M}(\beta)$ is isomorphic to the group $G_{\widehat{M}}(\beta)$. According
to the proved theorem, the group $G_{M}(\beta)$ is an invariant of the link
$\widehat{\beta}$ and this fact leads to the proof of Theorem 3.

Let a group $G$ have the presentation $G=\langle X\,\|\,\mathcal{R}\rangle$,
where $X=\left\\{x_{1},\ldots,x_{n}\right\\}$ is the generating set and
$\mathcal{R}$ is the set of defining relations. If in the group $G$ another
generating set $Y=\left\\{y_{1},\ldots,y_{m}\right\\}$ is chosen, then the
group $G$ has the presentation $G=\langle
Y\,\|\,\widetilde{\mathcal{R}}\rangle$ which connected with the initial
presentation by the sequence of Tietze transformations. From now on we assume
the generating sets $X$, $Y$ and their corresponding sets of defining
relations are fixed.

If $\varphi$ is some automorphism of the group $G$, then a group could be
defined using the generating set $X$

$G(X,\varphi)=\langle
X\,\|\,\mathcal{R},\,\,\varphi(x_{i})=x_{i},\,i=1,\ldots,n\rangle.$

Analogously using the generating set $Y$ and the automorphism $\varphi$, a
group could be defined

$G(Y,\varphi)=\langle
Y\,\|\,\widetilde{\mathcal{R}},\,\,\varphi(y_{j})=y_{j},\,j=1,\ldots,m\rangle.$

It is true

Theorem 5. Groups $G(X,\varphi)$ and $G(Y,\varphi)$ are isomorphic.

Proof. Let the automorphism $\varphi$ act on generating sets $X$ and $Y$ in
the following way

$\varphi(x_{i})=w_{i}(X),\quad\varphi(y_{j})=u_{j}(Y),$

where $w_{i}(X)$ and $u_{j}(Y)$ are words in the terms of generators $X$ and
$Y$ as well as their inverses consequently for $i=1,\ldots,n$, $j=1,\ldots,m$.
We fix expressions

$x_{i}=b_{i}(Y),\quad y_{j}=a_{j}(X)$

to express generators of $X$ through $Y$ and, visa versa, generators $Y$
through $X$. To simplify the notation indices $i$ and $j$ are dropped out so
there are obtained systems of equations

$\varphi(X)=W(X),\quad\varphi(Y)=U(Y),\quad X=B(Y),\quad Y=A(X).$

Since the automorphism $\varphi$ of the group $G$ that acts on generators of
$X$ and $Y$ is the same, the following systems of equations

$A(W(B(Y)))=U(Y)\Leftrightarrow A(W(X))=U(A(X))\Leftrightarrow
W(B(Y))=B(U(Y))$

are equivalent in the group $G$.

The proof of the theorem statement is to apply sequentially the Tietze
transformations to the presentation of the group $G(X,\varphi)$. Thus we
consider the group $G(X,\varphi)$. By definition we have

$G(X,\varphi)=\langle X\,\|\,\mathcal{R},\,\,X=W(X)\rangle.$

We add elements $Y$ into the generating set and the equations $Y=A(X)$ into
the set of relations

$G(X,\varphi)=\langle X,\,Y\,\|\,\mathcal{R},\,\,X=W(X),\,\,Y=A(X)\rangle.$

Relations $Y=A(X)$ are equivalent to $X=B(Y)$ with the module of relations
$\mathcal{R}$. Thus

$\langle X,\,Y\,\|\,\mathcal{R},\,\,X=W(X),\,\,Y=A(X)\rangle\simeq\langle
X,\,Y\,\|\,\mathcal{R},\,\,X=W(X),\,\,X=B(Y)\rangle.$

Excluding generators $X$ and relations $X=B(Y)$ we obtain

$\langle X,\,Y\,\|\,\mathcal{R},\,\,X=W(X),\,\,X=B(Y)\rangle\simeq\langle
Y\,\|\,\widetilde{\mathcal{R}},\,\,B(Y)=W(B(Y))\rangle.$

Since $W(B(Y))=B(U(Y))$ we have

$\langle Y\,\|\,\widetilde{\mathcal{R}},\,\,B(Y)=W(B(Y))\rangle\simeq\langle
Y\,\|\,\widetilde{\mathcal{R}},\,\,B(Y)=B(U(Y))\rangle.$

Taking into account that in the group $G$ the equations $A(B(Y))=Y$ and
$B(A(X))=X$ hold, it could be seen that systems of relations $B(Y)=B(Z)$ and
$Y=Z$ are equivalent in the group $G$ (i. e. they are equivalent with the
module of relations $\widetilde{\mathcal{R}}$). Consequently

$\langle Y\,\|\,\widetilde{\mathcal{R}},\,\,B(Y)=B(U(Y))\rangle\simeq\langle
Y\,\|\,\widetilde{\mathcal{R}},\,\,Y=U(Y)\rangle=G(Y,\varphi).$

Therefore,

$G(X,\varphi)\simeq G(Y,\varphi).$

The theorem is proved.

As a corollary the fact is obtained that groups $G_{M}(L)$ and
$G_{\widetilde{M}}(L)$ are isomorphic. Consequently Theorem 3 is proved.

§ 7. Properties of link groups

Directly from the definition of the group $G_{M}$ it could be derived

Proposition 6. If $L$ is $d$-component link which is presented by a closure of
a $n$-strand braid, then in the group $G_{M}(L)$ there are $d$ different
generators among $u_{1},u_{2},\ldots,u_{n}$. In particular, provided $L$ a
knot, $u_{1}=u_{2}=\ldots=u_{n}$.

As shown earlier the representation $\widetilde{\varphi}_{M}$ is an extension
of the Artin representation. As a corollary one can prove

Proposition 7. Given a classical $d$-component link $L$, there is the
following isomorphism of groups

$G_{\widetilde{M}}(L)=G(L)*\mathbb{Z}^{d},~{}\mbox{where}~{}G(L)=\pi_{1}(\mathbb{S}^{3}\setminus
L).$

Figure 7: Virtual trefoil

Example. Given $T_{v}$ the virtual trefoil (see Fig. 7), the generalized
Alexander group is generated by elements $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$,
$u$, $v$ and it has in the first crossing the relations

$x_{1}x_{2}^{u}=x_{3}x_{4}^{u},~{}~{}x_{2}=x_{3}^{v},$

and in the second crossing the relations

$x_{3}x_{4}^{u}=x_{2}x_{1}^{u},~{}~{}x_{4}=x_{2}^{v}.$

Excluding generators $x_{3}$ and $x_{4}$, we obtain the presentation

$\widetilde{A}_{T_{v}}=\langle
x_{1},x_{2},u,v~{}||~{}[u,v]=1,~{}~{}x_{1}x_{2}^{u}=x_{2}^{v^{-1}}x_{2}^{vu},~{}~{}x_{2}^{v^{-1}}x_{2}^{vu}=x_{2}x_{1}^{u}\rangle.$

Excluding the generator $x_{1}$, we have

$\widetilde{A}_{T_{v}}=\langle
x_{2},u,v~{}||~{}[u,v]=1,~{}~{}x_{2}x_{2}^{v^{-1}u}x_{2}^{vu^{2}}=x_{2}^{v^{-1}}x_{2}^{vu}x_{2}^{u^{2}}\rangle.$

If a generator $x_{2}$ replaced by a new one $b$ which equals to
$x_{2}u^{-1}$, then the presentation is of the form

$\widetilde{A}_{T_{v}}=\langle
b,u,v~{}||~{}[u,v]=1,~{}~{}b(b^{v^{-1}}b^{v})=(b^{v^{-1}}b^{v})b\rangle=\langle
b,u,v~{}||~{}[u,v]=1,~{}~{}[b^{v^{-1}}b^{v},b]=1\rangle.$

On the other hand, the virtual trefoil could be presented as a classical knot
in the thickened torus. The fundamental group of the knot complement in the
thickened torus was found in [7] and it has the following presentation

$G=\langle
a,x,y~{}||~{}[x,y]=1,~{}~{}(a^{x}a^{y})a^{xy}=a(a^{x}a^{y})\rangle.$

A natural question arises about the isomorphism of groups
$\widetilde{A}_{T_{v}}$ and $G$. The answer on this question is given by

Proposition 8.  Groups $\widetilde{A}_{T_{v}}$ and $G$ are not isomorphic.

Proof. For brevity we denote $H=\widetilde{A}_{T_{v}}$. We show that the
factor-groups $H/\gamma_{3}H$ and $G/\gamma_{3}G$ of these groups by the third
term of lower central series are not isomorphic to each other. We obtain

$bb^{v^{-1}}b^{v}=b^{2}[b,v^{-1}]b[b,v]\equiv
b^{3}\,(\mathrm{mod}\,\gamma_{3}H),$
$b^{v^{-1}}b^{v}b=b[b,v^{-1}]b[b,v]b\equiv
b^{3}\,(\mathrm{mod}\,\gamma_{3}H).$

Hence

$H/\gamma_{3}H=\langle b,u,v~{}||~{}[u,v]=1,~{}~{}\gamma_{3}H=1\rangle.$

In particular, $\gamma_{2}H/\gamma_{3}H\cong\mathbb{Z}\times\mathbb{Z}$ (it is
generated by images of the commutators $[b,u]$, $[b,v]$).

On the other hand

$a^{x}a^{y}a^{xy}=a[a,x]a[a,y]a[a,xy]\equiv
a^{3}[a,xy]^{2}\,(\mathrm{mod}\,\gamma_{3}G),$
$aa^{x}a^{y}=a^{2}[a,x]a[a,y]\equiv a^{3}[a,xy]\,(\mathrm{mod}\,\gamma_{3}G).$

Therefore,

$G/\gamma_{3}G=\langle
a,x,y~{}||~{}[x,y]=1,~{}~{}[a,xy]=1,~{}~{}\gamma_{3}G=1\rangle.$

In particular, $\gamma_{2}G/\gamma_{3}G\cong\mathbb{Z}$ (it is generated by
images of the commutator $[a,x]$).

As factor-groups $\gamma_{2}H/\gamma_{3}H$ and $\gamma_{2}G/\gamma_{3}G$ are
not isomorphic so the correspondent groups $H$ and $G$ are not isomorphic. The
proposition is proved.

We show that the group $\widetilde{A}_{T_{v}}$ is a HNN-extension of the one-
relator group.

Proposition 9.  The group

$\widetilde{A}_{T_{v}}=\langle
b,u,v~{}||~{}[u,v]=1,~{}~{}bb^{v^{-1}}b^{v}=b^{v^{-1}}b^{v}b\rangle$

is a $HNN$-extension.

Proof. Using Tietze transformations for the presentation of
$\widetilde{A}_{T_{v}}$, we obtain

$\widetilde{A}_{T_{v}}=\langle
b,u,v~{}||~{}[u,v]=1,~{}~{}bb^{v^{-1}}b^{v}=b^{v^{-1}}b^{v}b\rangle=$
$=\langle b,u,v~{}||~{}u^{v}=u,~{}~{}[b^{v^{-1}}b^{v},b]=1\rangle=$ $=\langle
b,u,v,z,t~{}||~{}u^{v}=u,~{}~{}z=b^{v^{-1}},~{}~{}t=b^{v},~{}~{}[zt,b]=1\rangle=$
$=\langle
b,u,v,z,t~{}||~{}u^{v}=u,~{}~{}z^{v}=b,~{}~{}b^{v}=t,~{}~{}[zt,b]=1\rangle.$

We consider the group

$G=\langle b,u,z,t~{}||~{}[zt,b]=1\rangle$

and its subgroups

$A=\langle b,u,z\rangle,\quad B=\langle b,u,t\rangle.$

It is sufficient to show that the mapping

$u\mapsto u,\quad z\mapsto b,\quad b\mapsto t$

is an isomorphism of the subgroup $A$ onto the subgroup $B$. For that purpose
it is sufficient to show that $A$ and $B$ are free subgroups of the rank 3
with free generators $u,b,z$ and $u,b,t$ respectively. Taking into account
that

$G=\langle b,u,z,t~{}||~{}[zt,b]=1\rangle=\langle u\rangle\ast\widetilde{G},$

where

$\widetilde{G}=\langle b,z,t~{}||~{}[zt,b]=1\rangle,$

it is sufficient to show that subgroups

$\widetilde{A}=\langle b,z\rangle,\quad\widetilde{B}=\langle b,t\rangle$

are free groups of the rank 2 in the group $\widetilde{G}$. We consider the
homomorphic image of the group $\widetilde{G}$ which is obtained by adding
relations $t=z^{-1}$:

$\langle b,z,t~{}||~{}[zt,b]=1,~{}~{}t=z^{-1}\rangle.$

Excluding the generator $t$, the free group $\langle b,z\rangle$ is obtained
which is covered by the image of the subgroup $\widetilde{A}$. And excluding
the generator $z$, the free group $\langle t,z\rangle$ is obtained which is
covered by the image of the subgroup $\widetilde{B}$. Therefore,
$\widetilde{A}$ and $\widetilde{B}$ are free subgroups of the rank 2 with free
generators $b,z$ and $b,t$ respectively. It leads to the fact that $A$ and $B$
are free subgroups of the rank 3 with free generators $u,b,z$ and $u,b,t$
respectively.

Hence

$\widetilde{A}_{T_{v}}=\langle
b,u,v,z,t~{}||~{}u^{v}=u,~{}~{}z^{v}=b,~{}~{}b^{v}=t,~{}~{}[zt,b]=1\rangle$

is a HNN-extension of the group

$G=\langle b,u,z,t~{}||~{}[zt,b]=1\rangle$

with the stable letter $v$ and associated subgroups

$A=\langle b,u,z\rangle,\quad B=\langle b,u,t\rangle.$

The proposition is proved.

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