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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 16, Issue 1

# Involutory biquandles and singular knots and links

Khaled Bataineh
Published Online: 2018-05-10 | DOI: https://doi.org/10.1515/math-2018-0031

## Abstract

We define a new algebraic structure for singular knots and links. It extends the notion of a bikei (or involutory biquandle) from regular knots and links to singular knots and links. We call this structure a singbikei. This structure results from the generalized Reidemeister moves representing singular isotopy. We give several examples on singbikei and we use singbikei to distinguish several singular knots and links.

Keywords: Knots and links; Quandles; Groups

MSC 2010: 57M27

## 1 Introduction

Singular knots and links are viewed diagrammatically as knots and links with some of the crossings being 4-valent rigid vertices. The theory of Vassiliev Invariants of knots and links shed the light on studying singular knots and links as a larger space that involves usual knots and links as a subspace. See [1, 2, 3]. Since then, many knot and link invariants have been generalized to singular knots and links. For example see [4, 5, 6, 7, 8].

Kei and quandles are algebraic structures, which were constructed to describe knots and links via generators and relations resulting from the arcs and the crossings of a knot or link diagram, and respecting the invariance of these diagrams under the Reidemeister moves. See [9, 10, 11, 12, 13, 14, 15, 16, 17].

Kei and quandles for singular knots were constructed in [18, 19], respectively. This paper introduces a new algebraic structure that generalizes involutory biquandles to singular knots. We call this structure singbikei. We give a plethora of non-trivial natural algebraic objects in examples that satisfy the axioms of this new algebraic structure. As a byproduct of this structure, we show how to apply the theory for distinguishing singular knots by giving several examples.

This paper is organized as follow. In Section 2 we give the basic concepts and terminology for kei and bikei. We also define singular knots and links and their isotopy invariance. In Section 3 we introduce the structure of a bikei for singular knots and links and we give several examples on this new structure with some related results. In Section 4 we give some examples of singular knots and links and distinguish them using several singular bikei colorings.

## 2 Basic concepts and terminology

Most of the basic concepts and terminology in this research can be found in [8].

We begin this section with the definition of a kei, and after the definition we will see how the axioms of a kei result from the three Reidemeister moves. The coloring of a regular crossing is drawn as in the following figure.

Fig. 1

The coloring of a regular crossing with one operation

#### Definition 2.1

A kei (involutory quandle) is a set X with a binary operation ▹ : X × XX satifying the following three axioms:

1. xx = x, for all xX.

2. (xy)▹ y = x, for all x, yX.

3. (xy)▹ z = (xz) ▹ (yz), for all x, y, zX.

The three axioms of a kei result from the invariance of the three Reidemeister moves as in the following figures.

Fig. 2

The coloring of Reidemeister moves

Now, instead of one operation at a crossing, two operations are defined, and instead of coloring the arcs from the top, the arcs are colored from left to right, so we get:

Fig. 3

The coloring of a regular crossing with two operations

Next we give the definition of a bikei. After this we will see how the axioms of a bikei result from the three Reidemeister moves.

#### Definition 2.2

A bikei (involutory biquandle) is a set X with two binary operations *, * : X × XX such that for all x, y, zX, we have

$x∗_x=x∗¯x(i)x∗_(y∗¯x)=x∗_y(ii.i)x∗¯(y∗_x)=x∗¯y(ii.ii)(x∗_y)∗_y=x(ii.iii)(x∗¯y)∗¯y=x(ii.iv)(x∗_y)∗_(z∗_y)=(x∗_z)∗_(y∗¯z)(iii.i)(z∗_y)∗¯(x∗_y)=(z∗¯x)∗_(y∗¯x)(iii.ii)(y∗¯x)∗¯(z∗¯x)=(y∗¯z)∗¯(x∗_z)(iii.iii)$

The axioms of a bikei result from the invariance of the three Reidemeister moves as in figures 4, 6 and 7.

Fig. 4

Reidemeister move RI for bikei

Fig. 5

Reidemeister move RII for bikei

Fig. 6

Rotated coloring crossing

Fig. 7

Reidemeister move RIII for bikei

The Reidemeister move RIII gives us what is called the exchange laws:

The Reidemeister move RII means that the over crossing and under crossing operations do not depend on which way the crossing is rotated.

Let X and Y be bikei with operations *X, *X and *Y, *Y respectively. Then a map f : XY is a bikei homomorphism if for all x, x′ ∈ X we have

$f(x∗_Xx′)=f(x)∗_Yf(x′) andf(x∗¯Xx′)=f(x)∗¯Yf(x′).$

A bikei isomorphism is a bijective bikei homomorphism, and two bikei are isomorphic if there is a bikei isomorphism between them.

#### Remark 2.3

Every Kei is a bikei with the operations x * y = xy and x * y = x.

Typical examples of bikei include the following :

• A non-empty set X with operation x * y = x * y = σ(x), where σ is any involution from X to X, is a bikei. It is called a constant action bikei.

• Let Λ = Z[t, s]/(t2 − 1, s2 − 1, (st)(1 − s)), then any Λ - module X with x * y = tx + (st)y, and x * y = sx, is called an Alexander bikei.

• A group X = G with x * y = yx−1y and x * y = x is a bikei. It is called the core bikei of the group G.

A singular link in S3 is the image of a smooth immersion of n circles in S3 that has finitely many double points, called singular points.

Two singular knots K1 and K2 are isotopy equivalent if we can get one of them from the other by a finite sequence of the generalized Reidemeister moves RI, RII, RIII, RIVa, RIVb and RV in the following figure.

Fig. 8

Generalized Reidemeister moves RI, RII, RIII, RIVa, RIVb and RV

## 3 Construction of singbikei

We will define the notion of a singbikei and give some examples and use them to construct an invariant of singular knots and links. The invariant is the set of colorings of a given singular knot or link by a singbikei. We draw the colorings of the regular and singular crossings as in the following figure.

Since our singular crossings are unoriented, we need the operations to be symmetric in the sense that if we rotate the crossing in the right diagram of the above figure by 90, 180 or 270 degrees, the operations should stay the same in order for colorings to be well-defined. Therefore we get the following three axioms:

$x=R1(y,R2(x,y))=R2(R2(x,y),R1(x,y))$(1)

$y=R2(R1(x,y),x)=R1(R2(x,y),R1(x,y))$(2)

$R(x,y)=(R2(y,R2(x,y)),R1(R1(x,y),x))$(3)

We have 5 generalized Reidemeister moves; I, II, III are for regular crossings; IV, V are for singular crossings. Next we show how the generalized Reidemeister moves induce relations considering the colorings of the singular crossings.

The following definition is coming from the generalized Reidemeister moves and the axioms are justified from the figures [9-11].

Fig. 9

Reidemeister move RIVa

Fig. 10

Reidemeister move RIVb

Fig. 11

Reidemeister move RV

#### Definition 3.1

Let (X, *, *) be a bikei. Let R1 and R2 be two maps from X × X to X. Then (X, *, *, R1, R2) is called a singbikei if, in addition to the three axioms 3.1, 3.2 and 3.3, the following axioms are satisfied

$R1(x∗¯y,z∗¯y)=R1(x,z)∗¯(y∗_x)$(4)

$R2(x∗¯y,z∗¯y)=R2(x,z)∗¯(y∗_z)$(5)

$(y∗_x)∗_R1(x,z)=(y∗_z)∗_R2(x,z)$(6)

$R1(x∗_y,z∗_y)=R1(x,z)∗_(y∗¯x)$(7)

$R2(x∗_y,z∗_y)=R2(x,z)∗_(y∗¯z)$(8)

$(y∗¯x)∗¯R1(x,z)=(y∗¯z)∗¯R2(x,z)$(9)

$R1(x,y)=R2(y∗¯x,x∗_y)∗_R1(y∗¯x,x∗_y)$(10)

$R2(x,y)=R1(y∗¯x,x∗_y)∗¯R2(y∗¯x,x∗_y)$(11)

The following straightforward lemma makes the set of colorings of a singular knot or link by a singbikei an invariant of singular knots and links.

#### Lemma 3.2

The set of colorings of a singular knot by a singbikei does not change by the Reidemeister moves RI, RII, RIII, RIVa, RIVb and RV.

#### Example 3.3

It is known that every kei is a bikei with the operations x * y = xy and x * y = x. Then X is a bikei with these operations so (X, *,*,R1, R2) is a singbikei if R1, R2 : X × XX satisfy the following equations:

$x=R1(y,R2(x,y))=R2(R2(x,y),R1(x,y))y=R2(R1(x,y),x)=R1(R2(x,y),R1(x,y))R(x,y)=(R2(y,R2(x,y)),R1(R1(x,y),x))(y▹x)▹R1(x,z)=(y▹z)▹R2(x,z)R1(x▹y,z▹y)=R1(x,z)▹yR2(x▹y,z▹y)=R2(x,z)▹yR1(x,y)=R2(y,x▹y)▹R1(y,x▹y)R2(x,y)=R1(y,x▹y)$

#### Example 3.4

Let X be a set and σ : XX be any involution on X, (i.e any map such that σ2 = IdX) with x * y = x * y = σ(x).

Let R1, R2 : X × XX be two maps, then (X, *, *, R1, R2) is a singbikei if R1 and R2 satisfy the following equations:

$x=R1(y,R2(x,y))=R2(R2(x,y),R1(x,y))y=R2(R1(x,y),x)=R1(R2(x,y),R1(x,y))R(x,y)=(R2(y,R2(x,y)),R1(R1(x,y),x))R1(σ(x),σ(z))=σ(R1(x,z))R2(σ(x),σ(z))=σ(R2(x,z))R1(x,y)=σ(R2(σ(y),σ(x))R2(x,y)=σ(R1(σ(y),σ(x))$

#### Proposition 3.5

Let X = Zn and σ : ZnZn be given by one of the following rules

$σ(x)=(n−1)x+d,wheredisarbitraryinZn$

or

$σ(x)=x$

or

$σ(x)=x+n2,whereniseven$

then σ is an involution.

#### Proof

1. If σ is given by σ (x)= (n − 1)x + d, then

$σ(σ(x))=σ((n−1)x+d)=(n−1)[(n−1)x+d]+d=(n−1)2x+(n−1)d+d=(n−1)2x+nd=x$

2. If σ is given by σ (x)= x, then

$σ(σ(x))=σ(x)=x$

3. If σ is given by $\begin{array}{}\sigma \left(x\right)=x+\frac{n}{2},\end{array}$ is even, then

$σ(σ(x))=σ(x+n2)=[x+n2]+n2=x+n=x$

This completes the proof.  □

Sometimes these are the only linear (i.e functions of the form f(x)= ax + b) involutions σ on Zn and sometimes Zn has other linear involutions. For example if X = Z8, in addition to the previous solutions,

$σ(x)=5x+4σ(x)=3x+2σ(x)=3x+6$

are also linear involutions in Z8.

#### Lemma 3.6

If n is prime, then the only linear formulas for an involution σ on Zn are:

$σ(x)=(n−1)x+d,d∈Znorσ(x)=x.$

#### Proof

The general linear formula of σ is σ (x)= cx + d, where c, dZn.

Since σ is an involution, we have

$σ(σ(x))=c[cx+d]+d=c2x+cd+d=c2x+(c+1)d=x$

Since c2 = 1 in Zn and n is prime, we have

$c2≡1(modn)(c2−1)≡0(modn)(c−1)(c+1)≡0(modn)n/(c−1)(c+1)⟹n/(c−1)orn/(c+1)c−1≡0(modn)orc+1≡0(modn)c≡1(modn)orc≡−1(modn)c≡1(modn)orc≡n−1(modn)c=1orc=n−1inZn$

So, we have two cases:

If c = n − 1, then

$σ(σ(x))=(n−1)2x+nd=n2x−2nx+x=x.$

Therefore σ (x)= (n − 1)x + d, where dZn.

If c = 1, then

$σ(σ(x))=x+2d=x,when2d=0wegetd=0or2d=n.$

• If d = 0, then σ (x)= x.

• If $\begin{array}{}d=\frac{n}{2}\end{array}$, then n is even and prime. Therefore n = 2 so d = 1, then σ (x) = x + 1, which is included in the case when c = n − 1.

So σ must be σ (x)= (n − 1)x + d, dZn   or    σ (x) = x.

#### Theorem 3.7

Let X = Zn and σ : ZnZn be an involution and x * y = x * y = σ(x). Then R1, R2 : Zn × ZnZn given below make (Zn, *, *, R1, R2) a singbikei.

1. If σ (x) = (n − 1)x + d, dZn then

R1(x, y) = (n − 1)y + c when (n − 2)d = (n − 2)c and

R2(x, y)= (n − 1)x + c.

2. If σ (x) = x, then

R1(x, y)= (n − 1)y + c and

R2(x, y) = (n − 1)x + c.

3. If σ (x) = $\begin{array}{}x+\frac{n}{2},\end{array}$ where n is even, then

R1(x, y) = (n − 1)y + c when $\begin{array}{}\left(n-1\right)\phantom{\rule{1mm}{0ex}}\frac{n}{2}=\frac{n}{2}\end{array}$ and

R2(x, y) = (n − 1)x + c.

#### Proof

We show that R1 and R2 satisfy all the equations in Example 3.4,

1. If σ (x)= (n − 1)x + d, where dZn, then

$R1(y,R2(x,y))=(n−1)2x+(n−1)c+c=n2x−2nx+x+nc=xR2(R2(x,y),R1(x,y))=(n−1)2x+(n−1)c+c=n2x−2nx+x+nc=xR2(R1(x,y),x)=(n−1)2y+(n−1)c+c=n2y−2ny+y+nc=yR1(R2(x,y),R1(x,y))=(n−1)2y+(n−1)c+c=n2y−2ny+y+nc=y(R2(y,R2(x,y)),R1(R1(x,y),x))=((n−1)y+c,(n−1)x+c)=R(x,y)R1(σ(x),σ(z))=R1((n−1)x+d,(n−1)z+d)=(n−1)[(n−1)z+d]+c=(n−1)2z+(n−1)d+c=(n−1)2z+(n−2)d+d+c=(n−1)2z+(n−2)c+d+c=(n−1)2z+(n−1)c+d=(n−1)[(n−1)z+c]+d=(n−1)R1(x,z)+d=σ(R1(x,z))R2(σ(x),σ(z))=R2((n−1)x+d,(n−1)z+d)=(n−1)[(n−1)x+d]+c=(n−1)2x+(n−1)d+c=(n−1)2x+(n−2)d+d+c=(n−1)2x+(n−2)c+d+c=(n−1)2x+(n−1)c+d=(n−1)[(n−1)x+c]+d=(n−1)R2(x,z)+d=σ(R2(x,z))$

Let I = σ(R2(σ(y), σ(x)).

$I=σ(R2((n−1)y+d,(n−1)x+d))=(n−1)[(n−1)[(n−1)y+d]+c]+d=(n−1)[(n−1)2y+(n−1)d+c]+d=(n−1)3y+(n−1)2d+(n−1)c+d=−y+2d+(n−1)c=(n−1)y+2d+c+(n−2)c=(n−1)y+2d+c+(n−2)d=(n−1)y+c=R1(x,y)$

Let H = σ(R1(σ(y), σ(x)).

$H=σ(R1((n−1)y+d,(n−1)x+d))=(n−1)[(n−1)[(n−1)x+d]+c]+d=(n−1)[(n−1)2x+(n−1)d+c]+d=(n−1)3x+(n−1)2d+(n−1)c+d=−x+2d+(n−1)c=(n−1)x+2d+c+(n−2)c=(n−1)x+2d+c+(n−2)d=(n−1)x+c=R2(x,y)$

2. If σ(x) = x, then

$R1(y,R2(x,y))=R1(y,(n−1)x+c)=(n−1)[(n−1)x+c]+c=(n−1)2x+(n−1)c+c=n2x−2nx+x+nc=x$

$R2(R2(x,y),R1(x,y))=R2((n−1)x+c,(n−1)y+c)=(n−1)[(n−1)x+c]+c=(n−1)2x+(n−1)c+c=n2x−2nx+x+nc=x$

$R2(R1(x,y),x)=R2((n−1)y+c,x)=(n−1)[(n−1)y+c]+c=(n−1)2y+(n−1)c+c=n2y−2ny+y+nc=y$

$R1(R2(x,y),R1(x,y))=R1((n−1)x+c,(n−1)y+c)=(n−1)[(n−1)y+c]+c=(n−1)2y+(n−1)c+c=n2y−2ny+y+nc=y$

$(R2(y,R2(x,y)),R1(R1(x,y),x))=(R2(y,(n−1)x+c),R1((n−1)y+c,x))=((n−1)y+c,(n−1)x+c)=R(x,y)$

$R1(σ(x),σ(z))=R1(x,z)=σ(R1(x,z))$

$R2(σ(x),σ(z))=R2(x,z)=σ(R2(x,z))$

$σ(R2(σ(y),σ(x))=σ(R2(y,x))=R2(y,x)=(n−1)y+c=R1(x,y)$

$σ(R1(σ(y),σ(x))=σ(R1(y,x))=R1(y,x)=(n−1)x+c=R2(x,y)$

3. If σ(x) = x + $\begin{array}{}\frac{n}{2}\end{array}$, where n is even, then

$R1(y,R2(x,y))=R1(y,(n−1)x+c)=(n−1)[(n−1)x+c]+c=(n−1)2x+(n−1)c+c=n2x−2nx+x+nc=x$

$R2(R2(x,y),R1(x,y))=R2((n−1)x+c,(n−1)y+c)=(n−1)[(n−1)x+c]+c=(n−1)2x+(n−1)c+c=n2x−2nx+x+nc=x$

$R2(R1(x,y),x)=R2((n−1)y+c,x)=(n−1)[(n−1)y+c]+c=(n−1)2y+(n−1)c+c=n2y−2ny+y+nc=y$

$R1(R2(x,y),R1(x,y))=R1((n−1)x+c,(n−1)y+c)=(n−1)[(n−1)y+c]+c=(n−1)2y+(n−1)c+c=n2y−2ny+y+nc=y$

$(R2(y,R2(x,y)),R1(R1(x,y),x))=(R2(y,(n−1)x+c),R1((n−1)y+c,x))=((n−1)y+c,(n−1)x+c)=R(x,y)$

$R1(σ(x),σ(z))=R1(x+n2,z+n2)=(n−1)[z+n2]+c=(n−1)z+(n−1)n2+c=R1(x,z)+n2=σ(R1(x,z))$

$R2(σ(x),σ(z))=R2(x+n2,z+n2)=(n−1)[x+n2]+c=(n−1)x+(n−1)n2+c=R2(x,z)+n2=σ(R2(x,z))$

$σ(R2(σ(y),σ(x))=σ(R2(y+n2,x+n2))=(n−1)[y+n2]+c+n2=(n−1)y+(n−1)n2+c+n2=(n−1)y+c=R1(x,y)$

$σ(R1(σ(y),σ(x))=σ(R1(y+n2,x+n2))=(n−1)[x+n2]+c+n2=(n−1)x+(n−1)n2+c+n2=(n−1)x+c=R2(x,y)$

So (Zn,*, *, R1, R2) is a singbikei. □

#### Example 3.8

As examples on the last theorem, we give

1. let X = Z10, with

$σ(x)=9x+3R1(x,y)=9y+8R2(x,y)=9x+8$

then (Z10,*, *, R1, R2) is a singbikei.

2. let X = Z8, with

$σ(x)=x+4R1(x,y)=7y+5R2(x,y)=7x+5$

then (Z8, *, *, R1, R2) is a singbikei.

#### Example 3.9

Let Λ = Z[t, s]/(t2–1 , s2–1, (st)(1–s)) be the quotient of the ring of two-variable polynomials with integer coefficients such that s2 = t2 = 1 by the ideal generated by (st)(1–s). Let R1, R2 : X × XX be two maps. Then any Λ - module X with x*y = tx + (st)y, and x*y = sx, is a singbikei if R1 and R2 satisfy the following equations:

$x=R1(y,R2(x,y))=R2(R2(x,y),R1(x,y))y=R2(R1(x,y),x)=R1(R2(x,y),R1(x,y))R(x,y)=(R2(y,R2(x,y)),R1(R1(x,y),x))R1(sx,sz)=sR1(x,z)R2(sx,sz)=sR2(x,z)t(ty+(s−t)x)+(s−t)R1(x,z)=t(ty+(s−t)z)+(s−t)R2(x,z)R1(tx+(s−t)y,tz+(s−t)y)=tR1(x,z)+(s−t)(sy)R2(tx+(s−t)y,tz+(s−t)y)=tR2(x,z)+(s−t)(sy)R1(x,y)=tR2(sy,tx+(s−t)y)+(s−t)R1(sy,tx+(s−t)y)R2(x,y)=sR1(sy,tx+(s−t)y)$

We call such a singbikei an Alexander singbikeis. For example,

1. Z10 with x*y = 9x + 2y and x*y = x, R1(x, y) = 8x + 3y and R2(x, y) = 7x + 4y satisfy all the equations in Example 3.9. So (Z10, * ,*, R1,R2) is an Alexander singbikei.

2. Z13 with x*y = 12x + 2y and x*y = x, R1(x, y) = 9x + 5y and R2(x, y) = 8x + 6y satisfy all the equations in Example 3.9. So (Z13, * ,*, R1,R2) is an Alexander singbikei.

#### Example 3.10

Let X = G be a group with x*y = yx−1y and x*y = x.

Let R1 , R2 : X × XX be two maps, then (X, *, *, R1, R2) is a singbikei if R1 and R2 satisfy the following equations:

$x=R1(y,R2(x,y))=R2(R2(x,y),R1(x,y))y=R2(R1(x,y),x)=R1(R2(x,y),R1(x,y))R(x,y)=(R2(y,R2(x,y)),R1(R1(x,y),x))R1(x,z)(x−1yx−1)R1(x,z)=R2(x,z)(z−1yz−1)R2(x,z)R1(yx−1y,yz−1y)=yR1−1(x,z)yR2(yx−1y,yz−1y)=yR2−1(x,z)yR1(x,y)=R1(y,yx−1y)R2−1(y,yx−1y)R1(y,yx−1y)R2(x,y)=R1(y,yx−1y)$

#### Proposition 3.11

The following maps

$R1(x,y)=(xy−1)nxR2(x,y)=(xy−1)n−1x,n∈N$

with the condition

$(yx−1)n2+1=1$

are solutions for the system in Example 3.10.

#### Proof

We show that R1 and R2 satisfy the equations in Example 3.10.

$R1(y,R2(x,y))=R1(y,(xy−1)n−1x)=[yx−1(yx−1)n−1]ny=(yx−1)n2y=xy−1y=x$

$R2(R2(x,y),R1(x,y))=R2((xy−1)n−1x,(xy−1)nx)=[(xy−1)n−1xx−1(yx−1)n]n−1(xy−1)n−1x=[yx−1]n−1(xy−1)n−1x=x$

$R2(R1(x,y),x)=R2((xy−1)nx,x)=[(xy−1)nxx−1]n−1(xy−1)nx=[xy−1]n2−n(xy−1)nx=[xy−1]n2x=yx−1x=y$

$R1(R2(x,y),R1(x,y))=R1((xy−1)n−1x,(xy−1)nx)=[(xy−1)n−1xx−1(yx−1)n]n(xy−1)n−1x=[yx−1]n(xy−1)n−1x=yx−1x=y$

$(R2(y,R2(x,y)),R1(R1(x,y),x))=(R2(y,(xy−1)n−1x),R1((xy−1)nx,x))=([yx−1(yx−1)n−1]n−1y,[(xy−1)nxx−1]n(xy−1)nx)=([(yx−1)n]n−1y,[(xy−1)n]n(xy−1)nx)=((yx−1)n2−ny,(xy−1)n2(xy−1)nx)=(xy−1(xy−1)ny,(xy−1)n2+1(xy−1)n−1x)=((xy−1)n+1y,(xy−1)n−1x)=((xy−1)nxy−1y,(xy−1)n−1x)=((xy−1)nx,(xy−1)n−1x)=R(x,y)$

$R1(x,z)(x−1yx−1)R1(x,z)=R2(x,z)(z−1yz−1)R2(x,z)(xz−1)nx[x−1yx−1](xz−1)nx=(xz−1)n−1x[z−1yz−1](xz−1)n−1x(xz−1)nyx−1(xz−1)nx=(xz−1)nyz−1(xz−1)n−1x=(xz−1)nyx−1xz−1(xz−1)n−1x=(xz−1)nyx−1(xz−1)nx$

$R1(yx−1y,yz−1y)=yR1−1(x,z)y(yx−1yy−1zy−1)nyx−1y=y[(xz−1)nx]−1yy(x−1z)nx−1y=y(xz−1)nx−1y$

$R2(yx−1y,yz−1y)=yR2−1(x,z)y(yx−1yy−1zy−1)n−1yx−1y=y[(xz−1)n−1x]−1yyx−1(zx−1)n−1y=yx−1(zx−1)n−1y$

$R1(x,y)=R1(y,yx−1y)R2−1(y,yx−1y)R1(y,yx−1y)(xy−1)nx=(yy−1xy−1)ny[(yy−1xy−1)n−1y]−1(yy−1xy−1)ny=(xy−1)nyy−1(yx−1)n−1(xy−1)ny=(xy−1)nxy−1y=(xy−1)nx$

$R2(x,y)=R1(y,yx−1y)(xy−1)n−1x=(yy−1xy−1)ny=(xy−1)ny=(xy−1)n−1xy−1y=(xy−1)n−1x$

This completes the proof. □

#### Remark 3.12

From the proposition above, note that:

1. If n = 1 , then R1(x, y) = xy–1x and R2(x, y) = x with the condition (yx–1)2 = 1 is a solution for Example 3.10.

2. If n = 2 , then R1(x, y) = xy–1xy–1x and R2(x, y) = xy–1x with the condition (yx–1)5 = 1 is a solution for Example 3.10.

## 4 Applications

In this section we use singbikei and coloring invariants to distinguish singular knots and links.

#### Example 4.1

Consider the two singular knots in the graph below.

(a) Let X = G be a group generated by (yx–1)2 = 1 with x*y = yx–1y and x*y = x, R1(x, y) = xy–1x and R2 (x, y) = x.

In Figure 12(a), the relations at the crossings give

Fig. 12

The singular knots in Example 4.1(a).

$x=(xy−1)3xx=yz−1yz=xy−1x$

Thus the set of colorings is {(x, y, z) ∈ G × G × G : x = y = z}.

In Figure 12(b), the relations at the crossings give

$x=yx=yz−1yz=(yx−1)2y$

Thus the set of colorings is {(x, y, z) ∈ G × G × G : x = y = z}.

The solution set is the same for both of the sets of colorings above. Therefore, this coloring invariant fails to distinguish these two singular knots.

(b) Let X = G be a group generated by (yx–1)5 = 1 with x*y = yx–1y and x*y = x, R1(x, y) = (xy–1)2x and R2 (x, y) = xy–1x.

In Figure 13(a), the relations at the crossings give

Fig. 13

The singular knots in Example 4.1(b).

$x=(xy−1)3zy−1(xy−1)2x(xy−1)2=(yz−1)2z=(xy−1)2x$

Thus the set of colorings is {(x, y, z) ∈ G × G × G: 1 = (xy–1)3, (xy–1)2 = (yz–1)2}.

In Figure 13(b), the relations at the crossings give

$x=yz−1y(xy−1)2=(yz−1)2z=yz−1(yx−1)3yz−1y.$

Thus the set of colorings is {(x, y, z) ∈ G × G × G: xy–1 = yz–1}.

One can always choose a group G such that these two coloring sets are distinct.

#### Example 4.2

Consider the two singular links in the graph below, let X = G be a group generated by (yx–1)5 = 1 with x*y = yx–1y and x*y = x, R1(x, y) = (xy–1)2x and R2 (x, y) = xy–1x.

In Figure 14(a), the relations at the crossings give

Fig. 14

The singular links in Example 4.2.

$x=xy−1xx=yz−1yz=y$

Thus the set of colorings is {(x, y, z) ∈ G × G × G : x = y = z}.

In Figure 14(b), the relations at the crossings give

$x=xy−1xx=(yz−1)3yz=(yz−1)2y$

Thus the set of colorings is {(x, y, z) ∈ G × G × G : x = y, 1 = (yz–1)3}.

One can always choose a group G such that these two coloring sets are distinct.

#### Example 4.3

Consider the two singular links in the graph below. Each of them has one singular crossing followed by (n + 1) regular crossings, let X = G be a group generated by (yx–1)5 = 1 with x*y = yx–1y and x*y = x, R1(x, y) = (xy–1)2x and R2 (x, y) = xy–1x.

In Figure 15(a), the relations at the crossings give

Fig. 15

The singular links in Example 4.3.

$x=(xy−1)n+3xy=(xy−1)n+2x$

Thus the set of colorings is {(x, y) ∈ G × G : 1 = (xy–1)n+3}.

In Figure 15(b), the relations at the crossings give

$x=(xy−1)2−nyy=(xy−1)1−ny$

Thus the set of colorings is {(x, y) ∈ G × G : 1 = (xy–1)1–n}.

One can always choose a group G such that these two coloring sets are distinct.

#### Example 4.4

Consider the two singular knots in the graph below, let X = G be a group generated by (yx–1)5 = 1 with x*y = yx–1y and x*y = x, R1(x, y) = (xy–1)2x and R2 (x, y) = xy–1x.

In Figure 16(a), the relations at the crossings give

Fig. 16

The singular knots in Example 4.4.

$x=zy=zy−1(xy−1)2zz=zy−1(xy−1)3z$

Thus the set of colorings is {(x, y, z) ∈ G × G × G : x = z, 1 = (xy–1)4}.

In Figure 16(b), the relations at the crossings give

$x=xy−1zy−1xy=yz=yx−1y$

Thus the set of colorings is {(x, y, z) ∈ G × G × G : z = yx–1y}.

Therefore, this coloring invariant distinguishes these two singular knots.

#### Example 4.5

Consider the two singular knots in the graph below, let X = G be a group generated by (yx–1)5 = 1 with x*y = yx–1y and x*y = x, R1(x, y) = (xy–1)2x and R2(x, y) = xy–1x.

In Figure 17(a), the relations at the crossings give

Fig. 17

The singular knots in Example 4.5.

$x=(xy−1)5xy=(xy−1)4x$

Thus the set of colorings is G × G.

In Figure 17(b), the set of colorings is G. Therefore, this coloring invariant distinguishes these two singular knots.

## Acknowledgement

The second author would like to thank her masters thesis advisor, the first author, for his guidance and big efforts to accomplish this work.

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## About the article

Accepted: 2018-02-14

Published Online: 2018-05-10

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 469–489, ISSN (Online) 2391-5455,

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