# Number of Distinct Homomorphic Images in Coset Diagrams

Authors: Muhammad Aamir; Muhammad Awais Yousaf; Abdul Razaq
Journal: Journal of Mathematics
 (2021)

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## Abstract

The representation of the action ofPGL2,Z on Ft∪∞ in a graphical format is labeled as coset diagram. These finite graphs are acquired by the contraction of the circuits in infinite coset diagrams. A circuit in a coset diagram is a closed path of edges and triangles. If one vertex of the circuit is fixed bypqΔ1pq−1Δ2pqΔ3…pq−1Δm∈PSL2,Z, then this circuit is titled to be a length-m circuit, denoted byΔ1,Δ2,Δ3,…,Δm. In this manuscript, we consider a circuit Δ of length 6 as Δ1,Δ2,Δ3,Δ4,Δ5,Δ6 with vertical axis of symmetry, that is, Δ2=Δ6,Δ3=Δ5. Let Γ1 and Γ2 be the homomorphic images of Δ acquired by contracting the vertices a,u and b,v, respectively, then it is not necessary that Γ1 and Γ2 are different. In this study, we will find the total number of distinct homomorphic images of Δ by contracting its all pairs of vertices with the condition Δ1>Δ2>Δ3>Δ4. The homomorphic images are obtained in this way having versatile applications in coding theory and cryptography. One can attain maximum nonlinearity factor using this in the encryption process.


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## Body

## 1. Introduction

It is prominent that the finite presentationp,q;p2=q3=1 is known as the modular group PSL2,Z generated by the linear fractional transformations p:χ⟶−1/χandq:χ⟶χ−1/χ. In [1], Akbas discussed suborbitalgraphs for the modular group by showing that these graphs contains no circuit if and only if it contains no triangles. If we insert an extra generator r=1/χ with p andq, another group is emerged, denoted as PGL2,Z [2], an extension of PSL2,Z with the finite presentation as(1)p,q,r;p2=q3=r2=pr2=qr2=1.In 1978, Professor Graham Higman propounds an unfamiliar type of a graph, titled as coset diagram, which presents the action ofPGL2,Z on PLFt, where Ft is a finite field and t shows a prime power. In 1983, this foundation is laid by Qaisar Mushtaq [3]. Small triangles are proposed for the cycle q3, such that q permutes the vertices of triangles in the opposite direction of rotation of clock and an edge is attached to any two vertices that are interchanged by p. Heavy dots represent the fixed points of p and q. Note that pr2=1 equals rqr=q−1, which means r reverses the triangle orientation proposed for the cycle q3. For that reason, the diagram need not to be made more perplexing by interjecting r edges.A coset diagram (subdiagram)Γ1 is said to be a homomorphic image of the coset diagram (subdiagram) Γ2 if and only if VΓ1<VΓ2∀a∈VΓ2 withah=a, where h∈PSL2,Z, there exist a vertex u in VΓ1 such that uh=u.Coset diagrams obtained from the action ofPSL2,Z over Qϵ are infinite graphs [4], where Qε=a1+a2ε;a1,a2∈ℚandε∈Z+is a square free. These diagrams are not easy to study because they are infinite. Thence, coset diagrams are considered as important for the action of PSL2,Z on PLFq because this action presents finite graphs. The number x1+ε/x2 is an expression of the number a1+a2ε∈Qε, where x1,x2,x12−ε/x2=1. These finite coset diagrams are the homomorphic images of the coset diagrams for x1+ε/x2, where ε≡z2modp for some z∈N.To explain more, coset diagram in Figure1 illustrates the action on PLF17 by PGL2,Z with permutation representations p,q, and r by χp=−1/χ, χq=χ−1/χ, andχr=1/χ, respectively, as(2)p:0∞,116,28,311,4,510,614,712,915,13,q:0∞1,2916,3128,4511,61510,71314,r:0∞,1,29,36,413,57,815,1012,1114,16.Figure 1
Coset diagram for the action ofPGL2,Z on PLF17.Thus,13=82mod17 gives that the coset diagram PGL2,17 is a homomorphic image of the coset diagram for x1+13/x2.Coset diagrams obtained from the action ofPSL2,Z on ℚε have some attractive narrative. In [5], the quadratic irrational numbers are classified by taking prime modulus that proved helpful in investigating the modular group action on the real quadratic field. The number B¯=x1−ε/x2 is called the conjugate of B=x1+ε/x2, where x1 and x2 are integers and ε is a fixed number from Z+, which is not a perfect square. B is said to be an ambiguous number [6], if the sign of B is different from the sign of B¯. B is said to be a totally negative (positive) if B and B¯ both have the same signs. For a fixed ϵ, the number of ambiguous numbers of the form B=x1+ε/x2 is finite and that segment of the coset diagram attained by the ambiguous numbers forms a closed path (circuit) and it is the only closed path in B-orbit [4]. With the help of coset diagram, Anna Torstensson not only described the applications to study the finitely presented group but also discussed the one-relator quotients of the modular group [7].A closed path of triangles and edges in a coset diagram is called a circuit. In a coset diagram, a circuit is said to be a length-k circuit, denoted by Δ1,Δ2,Δ3,…,Δk, if its one vertex is fixed by(3)axΔ1ax−1Δ2axΔ3…ax−1Δk∈PSL2,Z.Alternatively, it means that one vertex of theΔ1 triangles lies outside of the circuit and one vertex of the Δ2 triangles lies inside of the circuit and likewise. Since Δ1,Δ2,Δ3,…,Δk is a cycle, so it does not matter if one vertex of the Δ1 triangles lies inside of the circuit and one vertex of the Δ2 triangles lies outside of the circuit and likewise. Note that k is always even.The circuit of the typeΔ1,Δ2,Δ3,…,Δl′,Δ1,Δ2,Δ3,…,Δl′,…,Δ1,Δ2,Δ3,…,Δl′ is termed as a periodic circuit with period of length l′.Note 1. By VΔ, we mean the collection of vertices lies on the circuit Δ=Δ1,Δ2,Δ3,…,Δk.Leta,u∈VΔ be any two vertices fixed by the words h1 and h2, that is, ah1=a and uh2=u. Suppose h3 is the word that maps a to u, then ah1−1h3=u. Note that h3 and h1−1h3 are the only two paths that assign a to u. Now, by contraction of the pair of vertices a and u, we mean that a and u melt together to become one node s=a=u such that sh3=sh1−1h3=s. As a result of this contraction, a closed path Γ is created that contains the vertex s fixed by h3 and h1−1h3. This closed path Γ is the homomorphic image of the circuit Δ. It is important to note that a and u is not the only pair of contraction in Δ that creates homomorphic image Γ. There are also many pairs of contraction other than a and u that create the same homomorphic image Γ. The following theorems proved in [8] will help us to find the total number of such contracted pairs that produce the same homomorphic image Γ of Δ.Theorem 1.
Let the verticesa and u in Δ are contracted and a homomorphic image Γof Δ is evolved, then Γ is also obtainable if the vertices ah and uh in Δ are contracted.Theorem 2.
Ifa and u are contracted to obtain Γ, then during this process E number of pairs are contracted all together, where E is the collection of words such that ∀h∈E,ah and uh are contained by Δ.Example 1.
Let us contract the verticesa and u from the circuit 3,4,2,4,2,4 (Figure 2) and acquire a homomorphic Γ (Figure 3) of the circuit 3,4,2,4,2,4. Thus, h1=qpq and h2=pqpq−14pq2pq−14pq2pq−14p are the two possible paths between a and u that are fixing the vertex s=a=u in Γ.
LetE be the family of words such that for all h∈E implies ah and uh lie on the circuit 3,4,2,4,2,4, then E=e,q,q−1,p,pq,pq−1,pqp,pq2,pqpq−1. Then by Theorem 2, the cardinality of E implies that there are 9 pairs of vertices contracted to generate the homomorphic image Γ.
Note that the cardinality ofE does not give the total number of contracted pairs to generate the homomorphic image Γ. In the following, we will discuss the process to find the total number of contracted pairs to generate Γ.
LetΓ∗ denote itself as the mirror image of Γ. Thus, the permutation r ensures that the coset diagram is symmetric along the vertical axis. This implies Γ∗ will assuredly occur.
Ifu=pqπ1pqπ2…pqπnπi=1or−1 is a word, then u∗=pq−π1pq−π2…pq−πn. If the word h fixes the vertex s, then the vertex s∗ is fixed by u∗.
A homomorphic imageΓ has a symmetry with respect to vertical axis if and only if by contracting a and u, the vertices a∗ and u∗ are also contracted.Figure 2
Circuit3,4,2,4,2,4.Figure 3
Homomorphic imageΓ.Remark 1.
In coset diagrams,r reverses the orientation of the triangles representing the three cycles of q (as reflection does). So corresponding to each vertex s fixed by the pair h1, h2, there is a vertex s∗ in Γ∗ (mirror image of Γ) such that s∗ is a fixed point of h1∗, h2∗. In other words, it is created by contracting a∗ and u∗. There are certain Γ’s which have a vertical symmetry and so have the same orientations as those of their mirror images. The homomorphic image Γ of a circuit Δ=Δ1,Δ2,Δ3,Δ4,Δ5,Δ6, which has a vertex s fixed by the pair h1, h2, has the same orientation as that of its mirror image if and only if there is a vertex s∗ in Γ such that s∗h1=s∗h2=s∗.


### 1.1. Counting the Number of Pairs of Contracting Vertices of a Homomorphic Image

LetΓ be a homomorphic image of the circuit Δ acquired by the contraction of pair of vertices a and u of Δ. Then by Theorem 2, Γ has E number of pairs of vertices. To find the total number of pairs of vertices, one should follow the following steps.To know how many total pairs of contracting vertices are there, special precaution must be taken.(1)
If by contractinga and u to create Γ, the pair of vertices a∗ and u∗ are not contracted, then Γ has different orientation from its mirror image Γ∗. So, there are E number of more pairs of vertices for the mirror image of Γ.(2)
If by contractinga and u to create Γ, the pair of vertices a∗ and u∗ are also contracted, then Γ has the same orientation as that of its mirror image Γ∗. So, in this case, Γ has E number of pairs of contracted vertices.Consider a circuit of length 6 asΔ=Δ1,Δ2,Δ3,Δ4,Δ5,Δ6 (Figure 4) with vertical axis of symmetry, that is, Δ2=Δ6,Δ3=Δ5. Suppose Δ1>Δ2>Δ3>Δ4. The coset diagrams are composed of circuits. The vertices of the circuits in infinite diagrams are contracted in a certain way, and a finite coset diagram evolves. It is therefore necessary to ask how many distinct homomorphic images are obtained if we contract all the pairs of vertices of the circuit Δ? We not only give the answer to this question for a circuit Δ but also mention those pairs of vertices which are “important”. There is no need to contract the pairs which are not mentioned as “important”. If we contract those, we obtain a homomorphic image, which we have already obtained by contracting “important” pairs.Figure 4
The circuitΔ=Δ1,Δ2,Δ3,Δ4,Δ5,Δ6.Note 2. It is clear from Figure 4 that(1)
The mirror image of the vertexelk is e3Δk−l−12−k¯, that is, elk∗=e3Δk−l−12−k¯(2)
The vertexa3m+1k is fixed by the word pq−1mpqΔk±5¯pq−1Δk±4¯pqΔk±3¯pq−1Δk±2¯pqΔk±1¯pq−1Δk−m(3)
The vertexx3mk is fixed by the word pqΔk−mpq−1Δk±1¯pqΔk±2¯pq−1Δk±3¯pqΔk±4¯pq−1Δk±5¯pqm, where k∈1,2,3,4,5,6;l=1,2,3,…,Δk;m=1,2,3,…,Δk−1,n¯=6,ifn=0,6nmod6otherwise and ±=+ifkis odd−otherwise⋅Lemma 1.
If we contract the verticese3l1+11:l1=0,1,2,…,Δ2−1, with the vertex e3Δ11 in the circuit Δ, then there arise Δ2 distinct homomorphic images of Δ and 3l1+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ2 and their mirror homomorphic images of Δ is 3Δ22+3Δ2−2.Proof.
LetH1=Ωl11;l1=0,1,2,…,Δ2−1 (Figure 5) be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l1+11:l1=0,1,2,…,Δ2−1 with the vertex e3Δ11 in Δ, where e3l1+11 and e3Δ11are fixed by the words pql1pq−1Δ6pqΔ5pq−1Δ4pqΔ3pq−1Δ2pqΔ1−l1 and pqΔ2pq−1Δ3pqΔ4pq−1Δ5pqΔ6pq−1Δ1. It is easy to verify that P1=q−1pq−1Δ1−l1−1 and P2=pqΔ2pq−1Δ3pqΔ4pq−1Δ5pqΔ6pq−1l1p are the possible paths between e3l1+11 and e3Δ11 (Figure 4). This implies that for each l1, the homomorphic image Ωl11 has a vertex s fixed by P1 and P2.
Thus,(4)E1=e,q,q−1,p,pq,pq−1,pqp,pq2,pqpq−1,…,pql1p,pql1+1,pql1pq−1is the family of elements in PSL2,Z such that ∀x∈E,Δ contains the vertices e3l1+11x and e3Δ11x. This gives that the cardinality of E1, that is, 3l1+2 is the number of contracted pairs of vertices to produce the homomorphic image Ωl11 (Theorem 2). For m≠n, let Ωm1 and Ωn1 be any two elements of H1, then the number of triangles in Ωm1 and Ωn1 are not equal (Figure 5). This implies that all the elements in H1 are different and no one is the mirror image of the other. This further forms the result as H1=Δ2. Thence, the number of contracted pairs of vertices of Δ to create all the elements of H1 is ∑l1=0Δ2−13l1+2.
From Figure5, it is also clear that no element of H1 except Ω01 has vertical axis of symmetry. So, Ω01 is the only homomorphic image whose orientation is not different from its mirror image Ω01∗ and all the remaining Δ2−1 elements of H1 have different orientations from their mirror images. Hence, there are(5)6∑l1=1Δ2−1l1+2+6=3Δ22+3Δ2−2pairs of contracted vertices to produce all the homomorphic images in H1.Figure 5
Homomorphic imagesΩl11.Lemma 2.
If we contract the verticese3l1+12:l1=0,1,2,…,Δ2−1, with the vertex e3Δ11 in the circuit Δ, then there arise Δ2 distinct homomorphic images of Δ and 3l1+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ2 and their mirror homomorphic images of Δ is 3Δ22+3Δ2.LetH2=Ωl12;l1=0,1,2,…,Δ2−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l1+12:l1=0,1,2,…,Δ2−1 with the vertex e3Δ11 in Δ. Suppose r shows itself as the remainder of l1/Δ2−l1. Then, graphically we make four partitions of H2 as follows:(i)
Ωl12:Δ2−2l1>1 (Figure 6(a))(ii)
Ωl12:Δ2−2l1=1 (Figure 6(b))(iii)
Ωl12:Δ2−2l1<1 and Δ2−l1>r+1 (Figure 6(c))(iv)
Ωl12:Δ2−2l1<1 and Δ2−l1=r+1 (Figure 6(d))From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma1.Letk1∈3,4,5.Figure 6
(a) Homomorphic imagesΩl12:Δ2−2l1>1. (b) Homomorphic images Ωl12:Δ2−2l1=1. (c) Homomorphic images Ωl12:Δ2−2l1<1 and Δ2−l1>r+1. (d) Homomorphic images Ωl12:Δ2−2l1<1 and Δ2−l1=r+1.
(a)(b)(c)(d)Lemma 3.
If we contract the verticese3l2,k1+1k1:l2,k1=0,1,2,…,Δk1−1, with the vertex e3Δ11 in the circuit Δ, then for each k1, there arise Δk1 distinct homomorphic images of Δ and 3l2,k1+2pairs of contracted vertices that create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δk1 and their mirror homomorphic images of Δ is 3Δk12+3Δk1.For a fix value ofk1, let H3k1=Ωl2,k13;l2,k1=0,1,2,…,Δk1−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l2,k1+1k1:l2,k1=0,1,2,…,Δk1−1, with the vertex e3Δ11 in Δ. Figures 7–9 present H3k1 graphically. From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself.This lemma can be proved by using the same procedure as that for Lemma1.Figure 7
Homomorphic imagesΩl2,33.Figure 8
Homomorphic imagesΩl2,53.Figure 9
Homomorphic imagesΩl2,43.Lemma 4.
If we contract the verticese3l3+16:l3=0,1,2,…,Δ6−1, with the vertex e3Δ11 in the circuit Δ, then there arise Δ6 distinct homomorphic images of Δ and 3l3+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ6 and their mirror homomorphic images of Δ is 3/2Δ62+3Δ6.LetH4=Ωl34;l3=0,1,2,…,Δ6−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l6+11:l3=0,1,2,…,Δ6−1 with the vertex e3Δ11 in Δ. Figure 10 presents H4 graphically. From all the homomorphic images presented in Figure 10, it is not intricated to check that every one is the mirror image of itself.This lemma can be proved by using the same procedure as that for Lemma1.Figure 10
Homomorphic imagesΩl34.Lemma 5.
If we contract the verticese3l4+11:l4=1,2,…,Δ1−1, with the vertex e3Δ22 in the circuit Δ, then there arise Δ1−1 distinct homomorphic images of Δ and 3l4+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ1−1 and their mirror homomorphic images of Δ is 3/2Δ12+3Δ1−4.
LetH5=Ωl45;l4=1,2,…,Δ1−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l4+11:l4=1,2,…,Δ1−1, with the vertex e3Δ22 in Δ. Suppose r shows itself as the remainder of l4/Δ1−l4, then graphically, we make four partitions of H5 as follows:(i)
Ωl45:Δ1−2l4>1 (Figure 11(a))(ii)
Ωl45:Δ1−2l4=1 (Figure 11(b))(iii)
Ωl45:Δ1−2l4<1 and Δ1−l4>r+1 (Figure 11(c))(iv)
Ωl45:Δ1−2l4<1 and Δ1−l4=r+1 (Figure 11(d))From all the homomorphic images presented in these figures, it is not intricated to check that every one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma1.Figure 11
(a) Homomorphic imagesΩl45:Δ1−2l4>1. (b) Homomorphic images Ωl45:Δ1−2l4=1. (c) Homomorphic images Ωl45:Δ1−2l4<1 and Δ1−l4>r+1. (d) Homomorphic images Ωl45:Δ1−2l4<1 and.Δ1−l4=r+1
(a)(b)(c)(d)Lemma 6.
If we contract the verticese3l5+12:l5=1,2,…,Δ2−1, with the vertex e3Δ22 in the circuit Δ, then there arise Δ2−1 distinct homomorphic images of Δ and 3l5+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ2−1 and their mirror homomorphic images of Δ is 3Δ22+3Δ2−4.
LetH6=Ωl56;l5=1,2,…,Δ2−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l5+12:l5=1,2,…,Δ2−1, with the vertex e3Δ22 in Δ. Graphically, we make two partitions of H5 as follows:(i)
Ωl56:l5<Δ2−1 (Figure 12(a))(ii)
Ωl56:l5=Δ2−1 (Figure 12(b))
From all the homomorphic images presented in Figures12(a) and 12(b), it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 1.
Letk1∈3,4,5.Figure 12
(a) Homomorphic imagesΩl56:l5<Δ2−1. (b) Homomorphic images Ωl56:l5=Δ2−1.
(a)(b)Lemma 7.
If we contract the verticese3l6,k1+1k1:l6,k1=1,2,…,Δk1−1, with the vertex e3Δ22 in the circuit Δ, then for each k1, there arise Δk1−1 distinct homomorphic images of Δ and 3l6,k1+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δk1−1 and their mirror homomorphic images of Δ is 3Δk12+3Δk1−4.
For a fix value ofk1, let H7k1=Ωl6,k17;l6,k1=1,2,…,Δk1−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l6,k1+1k1:l6,k1=1,2,…,Δk1−1, with the vertex e3Δ22 in Δ. Figures 13–15 present H7k1 graphically. From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 1.Figure 13
Homomorphic imagesΩl6,37.Figure 14
Homomorphic imagesΩl6,47.Figure 15
Homomorphic imagesΩl6,57.Lemma 8.
If we contract the verticese3l7+13:l7=1,2,…,Δ4−1, with the vertex e3Δ33 in the circuit Δ, then there arise Δ4−1 distinct homomorphic images of Δ and 3l7+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ4−1 and their mirror homomorphic images of Δ is 3Δ42+3Δ4−4.
LetH8=Ωl78;l7=1,2,…,Δ4−1be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l7+13:l7=1,2,…,Δ4−1, with the vertex e3Δ33 in Δ. Figure 16 presents H8 graphically. From all the homomorphic images presented in Figure 16, it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 1.Figure 16
Homomorphic imagesΩl78.Lemma 9.
If we contract the verticese3l7+14:l7=1,2,…,Δ4−1, with the vertex e3Δ33 in the circuit Δ, then there arise Δ4−1 distinct homomorphic images of Δ and 3l7+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ1−1 and their mirror homomorphic images of Δ is 3/2Δ42+3Δ4−4.
LetH9=Ωl79;l7=1,2,…,Δ4−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l7+14:l7=1,2,…,Δ4−1, with the vertex e3Δ33 in Δ. Suppose r shows itself as the remainder of l7/Δ4−l7, then graphically we make four partitions of H9 as follows:(i)
Ωl79:Δ4−2l7>1 (Figure 17(a))(ii)
Ωl79:Δ4−2l7=1 (Figure 17(b))(iii)
Ωl79:Δ4−2l7<1 and Δ4−l7>r+1 (Figure 17(c))(iv)
Ωl79:Δ4−2l7<1 and Δ4−l7=r+1 (Figure 17(d))
From all the homomorphic images presented in these figures, it is not intricated to check that every one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma1.Figure 17
(a) Homomorphic imagesΩl79:Δ4−2l7>1. (b) Homomorphic images Ωl79:Δ4−2l7=1. (c) Homomorphic images Ωl79:Δ4−2l7<1 and Δ4−l7>r+1. (d) Homomorphic images Ωl79:Δ4−2l7<1 and Δ4−l7=r+1.
(a)(b)(c)(d)Lemma 10.
If we contract the verticese3l8+14:l8=0,1,2,…,Δ4−1, with the vertex e3Δ44 in the circuit Δ, then there arise Δ4 distinct homomorphic images of Δ and 3l8+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ4 and their mirror homomorphic images of Δ is 3Δ42+3Δ4−2.
LetH10=Ωl810;l8=0,1,2,…,Δ4−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l8+14:l8=0,1,2,…,Δ4−1, with the vertex e3Δ44 in Δ. Graphically, we make two partitions of H10 as follows:(i)
Ωl810:l8<Δ4−1 (Figure 18(a))(ii)
Ωl810:l8=Δ4−1 (Figure 18(b))
From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself exceptΩ010. This lemma can be proved by using the same procedure as that for Lemma 1.Figure 18
(a) Homomorphic imagesΩl810:l8<Δ4−1. (b) Homomorphic images Ωl810:l8=Δ4−1.
(a)(b)Lemma 11.
If we contract the verticese3l9+15:l9=0,1,2,…,Δ5−1, with the vertex e3Δ44 in the circuit Δ, then there arise Δ5 distinct homomorphic images of Δ and 3l9+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ5 and their mirror homomorphic images of Δ is 3/2Δ52+3Δ5.
LetH11=Ωl911;l9=0,1,2,…,Δ5−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l9+15:l9=0,1,2,…,Δ5−1, with the vertex e3Δ44 in Δ. Figure 19 presents H11 graphically. From all the homomorphic images presented in Figure 19, it is not intricated to check that every one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 1.Figure 19
Homomorphic imagesΩl911.Lemma 12.
If we contract the verticese3l10+13:l10=0,1,2,…,Δ3−1, with the vertex e3Δ44 in the circuit Δ, then there arise Δ3 distinct homomorphic images of Δ and 3l10+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ3 and their mirror homomorphic images of Δ is 3Δ32+3Δ3.
LetH12=Ωl1012;l10=0,1,2,…,Δ3−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l10+13:l10=0,1,2,⋯,Δ3−1, with the vertex e3Δ44 in Δ. Suppose r shows itself as the reminder of l10/Δ3−l10, then graphically we make four partitions of H12 as follows:(i)
Ωl1012:Δ3−2l10>1 (Figure 20(a))(ii)
Ωl1012:Δ3−2l10=1 (Figure 20(b))(iii)
Ωl1012:Δ3−2l10<1 and Δ3−l10>r+1 (Figure 20(c))(iv)
Ωl1012:Δ3−2l10<1 and Δ3−l10=r+1 (Figure 20(d))
From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma1.Figure 20
(a) Homomorphic imagesΩl1012:Δ3−2l10>1. (b) Homomorphic images Ωl1012:Δ3−2l10=1. (c) Homomorphic images Ωl1012:Δ3−2l10<1 and Δ3−l10>r+1. (d) Homomorphic images Ωl1012:Δ3−2l10<1 and Δ3−l10=r+1.
(a)(b)(c)(d)Lemma 13.
If we contract the verticese3l1+12:l1=0,1,2,…,Δ2−1, with the vertex e3Δ55 in the circuit Δ, then there arise Δ2 distinct homomorphic images of Δ and 3l1+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ2 and their mirror homomorphic images of Δ is 3/2Δ22+3Δ2.
LetH13=Ωl113;l1=0,1,2,…,Δ2−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l1+12:l1=0,1,2,…,Δ2−1, with the vertex e3Δ55 in Δ. Figure 21 presents H13 graphically. From all the homomorphic images presented in Figure 21, it is not intricated to check that every one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 1.Figure 21
Homomorphic imagesΩl113.Lemma 14.
If we contract the verticese3l11+13:l11=0,1,2,…,Δ3−1, with the vertex e3Δ55 in the circuit Δ, then there arise Δ3 distinct homomorphic images of Δ and 3l11+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ3 and their mirror homomorphic images of Δ is 3Δ32+3Δ3−2.
LetH14=Ωl1114;l11=0,1,2,…,Δ3−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l11+13:l11=0,1,2,…,Δ3−1, with the vertex e3Δ55 in Δ. Figure 22 presents H14 graphically. From all the homomorphic images presented in Figure 22, it is not intricated to check that no one is the mirror image of itself except Ω014. This lemma can be proved by using the same procedure as that for Lemma 1.Figure 22
Homomorphic imagesΩl1114.Lemma 15.
If we contract the verticese3l7+14:l7=1,2,…,Δ4−1, with the vertex e3Δ55 in the circuit Δ, then there arise Δ4−1 distinct homomorphic images of Δ and 3l7+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ4−1 and their mirror homomorphic images of Δ is 3Δ42+3Δ4−4.
LetH15=Ωl715;l7=1,2,…,Δ4−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l7+14:l7=1,2,…,Δ4−1, with the vertex e3Δ55 in Δ. Figure 23 presents H15 graphically. From all the homomorphic images presented in Figure 23, it is not intricated to check that no one is the mirror image of itself.
This lemma can be proved by using the same procedure as that for Lemma1.Figure 23
Homomorphic imagesΩl715.Lemma 16.
If we contract the verticese3l12+15:l12=0,1,2,…,Δ5−1, with the vertex e3Δ55 in the circuit Δ, then there arise Δ5 distinct homomorphic images of Δ and 3l12+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ5 and their mirror homomorphic images of Δ is 3Δ52+3Δ5.
LetH16=Ωl1216;l12=0,1,2,…,Δ5−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l12+15:l12=0,1,2,…,Δ5−1, with the vertex e3Δ55 in Δ. Graphically, we make two partitions of H16 as follows:(i)
Ωl1216:l12<Δ5−1 (Figure 24(a))(ii)
Ωl1216:l12=Δ5−1 (Figure 24(b))
From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma1.Figure 24
(a) Homomorphic imagesΩl1216:l12<Δ5−1. (b) Homomorphic images Ωl1216:l12=Δ5−1.
(a)(b)Lemma 17.
If we contract the verticese3l13+13:l13=1,2,…,Δ3−1, with the vertex e3Δ66 in the circuit Δ, then there arise Δ3−1 distinct homomorphic images of Δ and 3l13+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ3−1 and their mirror homomorphic images of Δ is 3/2Δ32+3Δ3−4.
LetH17=Ωl1317;l13=1,2,…,Δ3−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l13+13:l13=1,2,…,Δ3−1, with the vertex e3Δ66 in Δ. Figure 25 presents H17 graphically. From all the homomorphic images presented in Figure 25, it is not intricated to check that every one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 1.Figure 25
Homomorphic imagesΩl1317.Lemma 18.
If we contract the verticese3l14+11:l14=Δ2+1,Δ2+2,…,Δ1−1, with the vertex e3Δ11 in the circuit Δ, then there arise Δ1−Δ2−1 distinct homomorphic images of Δ and 3Δ2+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ1−Δ2−1 and their mirror homomorphic images of Δ is 6Δ2+2Δ1−Δ2−1.Proof.
LetH18=Ωl1418;l14=Δ2+1,Δ2+2,…,Δ1−1 (Figure 26) be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l14+11:l14=Δ2+1,Δ2+2,…,Δ1−1, with the vertex e3Δ11 in Δ, where e3l14+11 and e3Δ11 are fixed by the words pqΔ1−l14pq−1Δ6pqΔ5pq−1Δ4pqΔ3pq−1Δ2pql14 and pqΔ2pq−1Δ3pqΔ4pq−1Δ5pqΔ2pq−1Δ3pqΔ4pq−1Δ5pqΔ6pq−1Δ1. It is easy to verify that P1=q−1pq−1Δ1−l14−1 and P2=pqΔ2pq−1Δ3pqΔ4pq−1Δ5pqΔ6pq−1l14p are the possible paths between e3l14+11 and e3Δ11 (Figure 4). This implies that, for each l14, the homomorphic image Ωl1418 has a vertex s fixed by P1 and P2. Thus,(6)E18=e,q,q−1,p,pq,pq−1,pqp,pq2,pqpq−1,…,pqΔ2p,,pqΔ2+1,pqΔ2pq−1is the family of elements in PSL2,Z such that ∀x∈E18,Δ contains the vertices e3l14+11x and e3Δ11x. This gives that the cardinality of E18, that is, 3Δ2+2 is the number of contracted pairs of vertices to produce the homomorphic image Ωl1418 (Theorem 2).
Now, we will prove that all the homomorphic images inH18 are distinct and no one is the mirror image of itself.
LetΩm18 and Ωn18 be any two elements of H18, then Ωm18 is produced by contraction of e3m+11 and e3Δ11 and Ωn18 is produced by contraction of e3n+11 and e3Δ11. Suppose Ωm18 and Ωn18 are the same. This concludes that Ωm18 is procurable also by contracting e3n+11 and e3Δ11, implying that e3n+11↔e3Δ11 is one of the pairs of contracted vertices for Ωm18. Then there must exist an element w∈E18 such that e3m+11w=e3n+11 and e3Δ11w=e3Δ11. e∈E18 is the only element such that e3Δ11e=e3Δ11but e3m+11e≠e3n+11. Thus, Ωm18 and Ωn18 are not the same, that is, by contracting e3m+11 and e3Δ11 to produce Ωm18, e3n+11 and e3Δ11 are not contracted. Now, if Ωm18 and Ωn18∗ are the same, then there must exist an element w∈E18 such that e3m+11w=e3n+11∗ and e3Δ11w=e3Δ11∗. But there does not such an element exist in E18. Thus, Ωm18 is not the same as the mirror image of Ωn18, that is, by contracting e3m+11 and e3Δ11 to create Ωm18, e3n+11∗ and e3Δ11∗ are not contracted. This implies that all the elements in H18 are distinct. Thus, H18=Ωl1418;l14=Δ2+1,Δ2+2,…,Δ1−1 gives H18=Δ1−Δ2−1.
To check does any element ofH18 is the mirror image of itself, we supposeΩm18 is the same to its mirror image Ωm18∗. Then by definition, there must exist an element w∈E18 such that e3m+11w=e3m+11∗ and e3Δ11w=e3Δ11∗. But there does not such an element exist in E18. Thus, Ωm18 is not the same as its mirror image, that is, by contracting e3m+11 and e3Δ11 to create Ωm18, e3m+11∗ and e3Δ11∗ are not contracted. Hence, there are(7)2×E18×H18=6Δ2+2Δ1−Δ2−1pairs of contracting vertices to produce all the homomorphic images in H18.
Letη1=0,ifΔ1+Δ4=0mod21,otherwise..Figure 26
Homomorphic imagesΩl1418.Lemma 19.
If we contract the verticese3l15+11:l15=Δ4+1,Δ4+2,…,Δ1+Δ4−η1/2, with the vertex e3Δ33 in the circuit Δ, then there arise Δ1−Δ4−η1/2 distinct homomorphic images of Δ and 3Δ4+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ1−Δ4−η1/2 and their mirror homomorphic images of Δ is 3Δ4+2Δ1−Δ4−1.Proof.
LetH19=Ωl1519;l15=Δ4+1,Δ4+2,…,Δ1+Δ4−η1/2 (Figure 27) be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l15+11:l15=Δ4+1,Δ4+2,…,Δ1+Δ4−η1/2, with the vertex e3Δ33 in Δ, where e3l15+11 and e3Δ33 are fixed by the words pqΔ1−l15pq−1Δ6pqΔ5pq−1Δ4pqΔ3pq−1Δ2pql15 and pqΔ4pq−1Δ5pqΔ6pq−1Δ1pqΔ2pq−1Δ3. It is easy to verify that P1=q−1pq−1Δ1−l15−1pqΔ2pq−1Δ3 and P2=pqΔ4pq−1Δ5pqΔ6pq−1l15p are the possible paths between e3l15+11 and e3Δ33 (Figure 4). This implies that, for each l15, the homomorphic image Ωl1519 has a vertex s fixed by P1 and P2. Thus, (8)E19=e,q,q−1,p,pq,pq−1,pqp,pq2,pqpq−1,…,pqΔ4p,pqΔ4+1,pqΔ4pq−1is the family of elements in PSL2,Z such that ∀x∈E19,Δ contains the vertices e3l15+11x and e3Δ33x. This gives that the cardinality of E19, that is, 3Δ4+2 is the number of contracted pairs of vertices to produce the homomorphic image Ωl1517 (Theorem 2).
Now, we will prove that (i) forΔ1+Δ4=1mod2, all the homomorphic images in H19 are distinct and no one is the mirror image of itself, and (ii) for Δ1+Δ4=0mod2, all the homomorphic images in H19 are distinct and only ΩΔ1+Δ4/219 is the mirror image of itself.
LetΩm19 and Ωn19 be any two elements of H19, then Ωm19 is produced by contraction of e3m+11 and e3Δ33 and Ωn19 is produced by contraction of e3n+11 and e3Δ33.
SupposeΩm19 and Ωn19 are the same. This concludes that Ωm19 is procurable also by contracting e3n+11 and e3Δ33, implying that e3n+11↔e3Δ33 is one of the pairs of contracted vertices for Ωm19. Then there must exist an element w∈E19 such that e3m+11w=e3n+11 and e3Δ33w=e3Δ33. e∈E19 is the only element such that e3Δ33e=e3Δ33 but e3m+11e≠e3n+11. Thus, Ωm19 and Ωn19 are not the same, that is, by contracting e3m+11 and e3Δ33 to produce Ωm19, e3n+11 and e3Δ33 are not contracted. Now, if Ωm19 and Ωn19∗ are the same, then there must exist an element w∈E19 such that e3m+11w=e3n+11∗ and e3Δ33w=e3Δ33∗. pqΔ4p∈E19 is the only element such that e3Δ33pqΔ4p=e3Δ33∗ and e3m+11pqΔ4p=e3Δ1+Δ4−m+11∗. This implies that Ωm19 and Ωn19 are the mirror images of each other if and only if n=Δ1+Δ4−m and Ωm19 is the mirror image of itself if and only if m=Δ1+Δ4−m, that is, m=Δ1+Δ4/2. Now,(i)
IfΔ1+Δ4=1mod2, then ∀m∈l15, Δ1+Δ4−m>Δ1+Δ4−1/2, implying that ΩΔ1+Δ4−m19∉H19. This states that no homomorphic image in H19 is the mirror image of other. Hence, all the homomorphic images in H19 are distinct. Thus, H19=Ωl1519;l15=Δ4+1,Δ4+2,…,Δ1+Δ4−1/2 gives H19=Δ1−Δ4−1/2.Also,Δ1+Δ4/2>Δ1+Δ4−1/2 gives that no homomorphic image in H19 is the mirror image of itself. Hence, there are(9)2×E19×H19=6Δ4+2Δ1−Δ4−12=3Δ4+2Δ3−Δ4−1pairs of contracting vertices to produce all the homomorphic images inH19.(ii)
IfΔ1+Δ4=0mod2, then ∀m∈l15/Δ1+Δ4/2, Δ1+Δ4−m>Δ1+Δ4/2, implying that ΩΔ1+Δ4−m19∉H19. This states that no homomorphic image in H19 is the mirror image of other. Hence, all the homomorphic images in H19 are distinct. Thus, H19=Ωl1519;l15=Δ4+1,Δ4+2,…,Δ1+Δ4/2 gives H19=Δ1−Δ4/2.
Form=Δ1+Δ4/2, we have Δ1+Δ4−m=Δ1+Δ4/2∈l15 gives ΩΔ1+Δ4/219∈H19 is the mirror image of itself. Hence, there are(10)2×E19×H19−1+E19=6Δ4+2Δ1−Δ42−1+3Δ4+2=3Δ4+2Δ3−Δ4−1pairs of contracting vertices to produce all the homomorphic images in H19.
Letk2∈2,3.Figure 27
Homomorphic imagesΩl1519.Lemma 20.
If we contract the verticese3l3,k2+1k2:l3,k2=Δ4+1,Δ4+2,…,Δk2−1, with the vertex e3Δ33 in the circuit Δ, then for each k2, there arise Δk2−Δ4−1 distinct homomorphic images of Δ and 3Δ4+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δk2−Δ4−1 and their mirror homomorphic images of Δ is 6Δ4+2Δk2−Δ4−1.
For a fix value ofk2, let H20k2=Ωl3,k220;l3,k2=Δ4+1,Δ4+2,…,Δk2−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l3,k2+1k2:l3,k2=Δ4+1,Δ4+2,…,Δk2−1, with the vertex e3Δ33 in Δ. Figures 28 and 29 present H3k2 graphically. From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself.
This lemma can be proved by using the same procedure as that for Lemma18.Figure 28
Homomorphic imagesΩl3,220.Figure 29
Homomorphic imagesΩl3,320.Lemma 21.
If we contract the verticese3l16+12:l16=Δ3+1,Δ3+2,…,Δ2−1, with the vertex e3Δ44 in the circuit Δ, then there arise Δ2−Δ3−1 distinct homomorphic images of Δ and 3Δ3+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ2−Δ3−1 and their mirror homomorphic images of Δ is 6Δ3+2Δ2−Δ3−1.
LetH21=Ωl1621;l16=Δ3+1,Δ3+2,…,Δ2−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l16+12:l16=Δ3+1,Δ3+2,…,Δ2−1, with the vertex e3Δ44 in Δ. Figure 30 presents H21 graphically. From all the homomorphic images presented in Figure 30, it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 18.
Letk3∈1,2.Figure 30
Homomorphic imagesΩl1621.Lemma 22.
If we contract the verticese3l4,k3+1k3:l4,k3=Δ5+1,Δ5+2,…,Δk3−1, with the vertex e3Δ66 in the circuit Δ, then for each k3, there arise Δk3−Δ5−1 distinct homomorphic images of Δ and 3Δ5+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δk3−Δ5−1 and their mirror homomorphic images of Δ is 6Δ5+2Δk3−Δ5−1.
For a fix value ofk3, let H22k3=Ωl4,k322;l4,k3=Δ5+1,Δ5+2,…,Δk3−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l4,k3+1k3:l4,k3=Δ5+1,Δ5+2,…,Δk3−1, with the vertex e3Δ66 in Δ. Figures 31 and 32 present H22k3 graphically. From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 18.
Letk2∈2,3.Figure 31
Homomorphic imagesΩl4,122.Figure 32
Homomorphic imagesΩl4,222.Lemma 23.
If we contract the verticese3l4+11:l4=1,2,…,Δ1−1, with the vertices e3l5,k2k2:l5,k2=1,2,…,Δk2−1, in the circuit Δ, then for each k2, there arise Δ1−1Δk2−1 distinct homomorphic images of Δ and 6 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ1−1Δk2−1 and their mirror homomorphic images of Δ is 12Δ1−1Δk2−1.Proof.
Fork2=2, let H232=Ωl4,l5,223;l4=1,2,…,Δ1−1;l5,2=1,2,…,Δ2−1(Figure 33) be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l4+11 with the vertex e3l5,22 in Δ, where e3l4+11 and e3l5,22 are fixed by the words pql4pq−1Δ6pqΔ5pq−1Δ4pqΔ3pq−1Δ2pqΔ1−l4 and pq−1Δ2−l5,2pqΔ1pq−1Δ6pqΔ5pq−1Δ4pqΔ3pq−1l5,2pqΔ5pq−1Δ4pqΔ3pq−1l5,2. It is easy to verify that P1=pq−1Δ2−l5,2pqΔ1−l4 and P2=pql4pq−1Δ6pqΔ5pq−1Δ4pqΔ3pq−1l5,2 are the possible paths between e3l4+11 and e3l5,22 (Figure 4). This implies that the homomorphic image Ωl4,l5,223 has a vertex s fixed by P1 and P2. Thus,(11)E232=e,q,q−1,p,pq,pq−1is the family of elements in PSL2,Z such that ∀x∈E232,Δ contains the vertices e3l4+11x and e3l5,22x. This gives that the cardinality of E232, that is, 6 is the number of contracted pairs of vertices to produce the homomorphic image Ωl4,l5,223 (Theorem 2).
Now, we will prove that all the homomorphic images inH232 are distinct and no one is the mirror image of itself.
LetΩm1,m223 and Ωn1,n223 be any two elements of H232, then Ωm1,m223 is produced by contraction of e3m1+11 and e3m22 and Ωn1,n223 is produced by contraction of e3n1+11 and e3n22.
SupposeΩm1,m223 and Ωn1,n223 are the same. This concludes that Ωm1,m223 is procurable also by contracting e3n1+11 and e3n22, implying that e3n1+11↔e3n22 is one of the pairs of contracted vertices for Ωm1,m223. Then there must exist an element w∈E232 such that e3m1+11w=e3n1+11 and e3m22w=e3n22. pq−1∈E52 is the only element such that e3m22pq−1=e3m2+12 but e3m1+11pq−1≠e3n1+11. Thus, Ωm1,m223 and Ωn1,n223 are not the same, that is, by contracting e3m1+11 and e3m22 to produce Ωm1,m223, e3n1+11 and e3n22 are not contracted. Now, if Ωm1,m223 and Ωn1,n223∗ are the same, then there must exist an element w∈E232 such that e3m1+11w=e3n1+11∗ and e3m22w=e3n22∗. p∈E232 is the only element such that e3m1+11p=e3Δ1−m1+11∗ but e3m22p≠e3n22∗. This implies that Ωm1,m223 and Ωn1,n223 are not the mirror images of each other, that is, by contracting e3m1+11 and e3m22 to produce Ωm1,m223, e3n1+11∗ and e3n22∗ are not contracted. Hence, all the homomorphic images in H232 are distinct. Thus, H232=Ωl4,l5,223;l4=1,2,…,Δ1−1;l5,2=1,2,…,Δ2−1 gives H232=Δ1−1Δ2−1.
SupposeΩm1,m223 is the mirror image of itself, then there must exist an element w∈E232 such that e3m1+11w=e3m1+11∗ and e3m22w=e3m22∗. p∈E232 is the only element such that e3m1+11p=e3Δ1−m1+11∗ but e3m22p≠e3m22∗. This implies that Ωm1,m223 is not the same as Ωm1,m223∗, that is, by contracting e3m1+11 and e3m22 to produce Ωm1,m223, e3m1+11∗ and e3m22∗ are not contracted.
Hence, there are(12)2×E232×H232=12Δ1−1Δ2−1pairs of contracting vertices to produce all the homomorphic images in H232.
We can prove this lemma fork2=3 (Figure 34) in similar way as that for k2=2.
Letη2=0ifΔ4=0mod21otherwise and η3=0ifΔ1=0mod21otherwise.Figure 33
Homomorphic imagesΩl4,l5,223.Figure 34
Homomorphic imagesΩl4,l5,323.Lemma 24.
If we contract the verticese3l17+11:l17=1,2,…,Δ1−η3/2,ifl18=Δ4/21,2,…,Δ1−1otherwise, with the vertices e3l184:l18=1,2,…,Δ4−η2/2, in the circuit Δ, then there arise 1/2Δ1−1Δ4−1ifΔ4=1mod21/2Δ1−1Δ4−1+1−η3ifΔ4=0mod2 distinct homomorphic images of Δ and 6 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all these homomorphic images and their mirror homomorphic images of Δ is 6Δ1−1Δ4−1.Proof.
LetH24=Ωl17,l1824;l17=1,2,…,Δ1−η3/2,ifl18=Δ4/2,l18=1,2,…,Δ4−η2/21,2,…,Δ1−1otherwise (Figure 35) be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l17+11 with the vertices e3l184 in Δ, where e3l17+11 and e3l184 are fixed by the words pql17pq−1Δ6pqΔ5pq−1Δ4pqΔ3pq−1Δ2pqΔ1−l18 and pq−1Δ4−l18pqΔ3pq−1Δ2pqΔ1pq−1Δ6pqΔ5pq−1l18. It is easy to verify that P1=pql17pq−1Δ6pqΔ5pq−1l18 and P2=pq−1Δ4−l18pqΔ3pq−1Δ2pqΔ1−l17 are the possible paths between e3l17+11 and e3l184 (Figure 4). This implies that the homomorphic image Ωl17,l1824 has a vertex s fixed by P1 and P2. Thus,(13)E24=e,q,q−1,p,pq,pq−1is the family of elements in PSL2,Z such that ∀x∈E24,Δ contains the vertices e3l17+11x and e3l184x. This gives that the cardinality of E24, that is, 6 is the number of contracted pairs of vertices to produce the homomorphic image Ωl17,l1824 (Theorem 2).
Now, we prove that (i) forΔ4=0mod2, all the elements in H24 are distinct and no one is the mirror image of itself, and (ii) for Δ4=1mod2, all the elements in H24 are distinct and the homomorphic image ΩΔ1/2,Δ4/224 is the mirror image of itself.
SupposeΩm1,m224 and Ωn1,n224 are the same. This concludes that Ωm1,m224 is procurable also by contracting e3n1+11 and e3n24, implying that e3n1+11↔e3n24 is one of the pairs of contracted vertices for Ωm1,m224. Then there must exist an element w∈E24 such that e3m1+11w=e3n1+11 and e3m24w=e3n24. pq−1∈E24 is the only element such that e3m24pq−1=e3m2+14 but e3m1+11pq−1≠e3n1+11. Thus, Ωm1,m224 and Ωn1,n224 are not the same, that is, by contracting e3m1+11 and e3m24 to produce Ωm1,m224, e3n1+11 and e3n24 are not contracted. Now, if Ωm1,m224 and Ωn1,n224 are the same, then there must exist an element w∈E24 such that e3m1+11w=e3n1+11∗ and e3m24w=e3n24∗. p∈E24 is the only element such that e3m1+11p=e3Δ1−m1+11∗ and e3m24p=e3Δ4−m24∗. This implies that Ωm1,m224 and Ωn1,n224 are the mirror images of each other if and only if n1=Δ1−m1 and n2=Δ4−m2, that is, by contracting e3m1+11 and e3m24 to produce Ωm1,m224, e3n1+11∗ and e3n24∗ are also contracted and the homomorphic image Ωm1,m224 is the mirror image of itself if and only if m1=Δ1−m1 and m2=Δ4−m2 that is m1=Δ1/2 and m2=Δ4/2. Now,(1
)IfΔ4=1mod2 then ∀m2∈l18, we get Δ4−m2>Δ4−1/2 implies ΩΔ1−m1,Δ4−m224∉H24. This indicates that no homomorphic image in H24 is the mirror image of other. Hence, all the elements in H24 are distinct. Thus, H24=Ωl17,l1824;l17=1,2,…,Δ1−η3/2ifl18=Δ4/2,l18=1,2,…,Δ4−η2/21,2,…,Δ1−1otherwise, implying that(14)H24=12Δ1−1Δ4−1⋅Also,Δ4/2>Δ4−1/2 implies no homomorphic image in H24 is the mirror image of itself. Hence, there are(15)2×E24×H24=6Δ1−1Δ4−1pairs of contracting vertices to produce all the homomorphic images inH24.(2)
IfΔ4=0mod2, then only for m2=Δ4/2, Δ4−m2=Δ4/2∈l18 and ∀m1∈l17Δ1/2, Δ1−m1∉l17, implying that ΩΔ1−m1,Δ4/224∉H24. Hence, all the elements in H24 are distinct. Thus, H24=Ωl17,l1824;l17=1,2,…,Δ1−η3/2ifl18=Δ4/2;l18=1,2,…,Δ4−η2/21,2,…,Δ1−1otherwise, implying that(16)H24=12Δ1−1Δ4−2+12Δ1−η3=12Δ1−1Δ4−1+1−η3⋅
Now,(a).
IfΔ1 is odd, then Δ1/2∉l17, implying that ΩΔ1/2,4/224∉H24. So, no homomorphic image in H24 is the mirror image of itself. Hence, there are(17)2×E24×H24=6Δ1−1Δ4−1pairs of contracting vertices to produce all the homomorphic images inH24.(b)
IfΔ1 is even, then Δ1/2∈l17, implying that ΩΔ1/2,Δ4/224∈H24 is the mirror image of itself. Hence, there are(18)2×E24×H24−1+E24=6Δ1−1Δ4−1pairs of contracting vertices to produce all the homomorphic images inH24.
Letk1∈3,4,5.Figure 35
Homomorphic imagesΩl17,l1824.Lemma 25.
If we contract the verticese3l5+12:l5=1,2,…,Δ2−1, with the vertices e3l6,k1k1:l6,k1=1,2,…,Δk1−1, in the circuit Δ, then for each k5, there arise Δ2−1Δk1−1 distinct homomorphic images of Δ and 6 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ2−1Δk1−1 and their mirror homomorphic images of Δ is 12Δ2−1Δk1−1.
For a fix value ofk1, let H25k1=Ωl5,l6,k125;l5=1,2,…,Δ2−1;l6,k1=1,2,…,Δk1−1be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l5+12 with the vertices e3l6,k1k1 in Δ. Figures 36, 37, and 38 present H25k1 graphically. From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself.
This lemma can be proved by using the same procedure as that for Lemma23.Figure 36
Homomorphic imagesΩl5,l6,325.Figure 37
Homomorphic imagesΩl5,l6,425.Figure 38
Homomorphic imagesΩl5,l6,525.Lemma 26.
If we contract the verticese3l13+13:l13=1,2,…,Δ3−1, with the vertices e3l74:l7=1,2,…,Δ4−1, in the circuit Δ, then there arise Δ3−1Δ4−1 distinct homomorphic images of Δ and 6 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ3−1Δ4−1 and their mirror homomorphic images of Δ is 12Δ3−1Δ4−1.
LetH26=Ωl13,l726;l13=1,2,…,Δ3−1;l7=1,2,…,Δ4−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l13+13 with the vertices e3l74 in Δ. Figure 39 presents H26 graphically. From all the homomorphic images presented in Figure 39, it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 23.
Letk2¯=8−k2.Figure 39
Homomorphic imagesΩl13,l726.Lemma 27.
If we contract the verticese3l7,k2+1k2:l7,k2=1,2,…,Δk2−1, with the vertices e3l8,k2k2¯:l8,k2=1,2,…,Δk2−l7,k2, in the circuit Δ, then for each k3, there arise 1/2Δk2Δk2−1 distinct homomorphic images of Δ and 6 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all 1/2Δk2Δk2−1 and their mirror homomorphic images of Δ is 6Δk2−12.Proof.
Fork2=2, let H272=Ωl7,2,l8,227;l7,2=1,2,…,Δ2−1;l8,2=1,2,…,Δ2−l7,2 (Figure 40) be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l7,2+12 with the vertex e3l8,26 in Δ, where e3l7,2+12 and e3l8,26 are fixed by the words pql7,2pq−1Δ3pqΔ4pq−1Δ5pqΔ6pq−1Δ1pqΔ2−l7,2 and pq−1Δ6−l8,2pqΔ5pq−1Δ4pqΔ3pq−1Δ2pqΔ1pq−1l8,2. It is easy to verify that P1=pql7,2pq−1Δ3pqΔ4pq−1Δ5pqΔ6−l8,2p and P2=qpql8,2−1pq−1Δ1pqΔ2−l7,2 are the possible paths between e3l7,2+12 and e3l8,26 (Figure 4). This implies that the homomorphic image Ωl7,2,l8,227 has a vertex s fixed by P1 and P2. Thus,(19)E272=e,q,q−1,p,pq,pq−1is the family of elements in PSL2,Z such that ∀w∈E272,Δ contains the vertices e3l7,2+12w and e3l8,26w. This gives that the cardinality of E272, that is, 6 is the number of contracted pairs of vertices to produce each homomorphic image Ωl7,2,l8,227 (Theorem 2).
SupposeΩm1,m227 and Ωn1,n227 are the same. This concludes that Ωm1,m227 is procurable also by contracting e3n1+12 and e3n26, implying that e3n1+12↔e3n26 is one of the pairs of contracted vertices for Ωm1,m227. Then there must exist an element w∈E272 such that e3m1+12w=e3n1+12 and e3m26w=e3n26. pq−1∈E272 is the only element such that e3m26pq−1=e3m2+16 but e3m1+12pq−1≠e3n1+12. Thus, Ωm1,m227 and Ωn1,n227 are not the same, that is, by contracting e3m1+12 and e3m26 to produce Ωm1,m227, e3n1+12 and e3n26 are not contracted. Now, if Ωm1,m227 and Ωn1,n227 are the same, then there must exist an element w∈E272 such that e3m1+12w=e3n1+12∗ and e3m26w=e3n26∗ or e3m1+12w=e3n1+16∗ and e3m26w=e3n22∗. p∈E272 is the only element such that e3m1+12p=e3m12=e3Δ2−m1+16∗ and e3m26p=e3m2+16=e3Δ2−m22∗. This implies that Ωm1,m227 and Ωn1,n227 are the mirror images of each other if and only if n1=Δ2−m2 and n2=Δ2−m1, that is, by contracting e3m1+12 and e3m26 to produce Ωm1,m227, e3n1+12∗ and e3n26∗ are also contracted. For a fix m1∈l7,2 and for all m2∈l8,2\Δ2−m1, ΩΔ2−m2,Δ2−m122∉H272, implying that Ωm1,m227 and Ωn1,n227 are not the mirror images of each other. This implies that all the homomorphic images in H272 are distinct. Thus, H272=Ωl7,2,l8,227;l7,2=1,2,…,Δ2−1;l8,2=1,2,…,Δ2−l7,2 implies(20)H272=12Δ2Δ2−1.
LetΩm1,m227 be the mirror image of itself, then from Figure 40, we have m1−1=Δ2−m2−1 and m2−1=Δ2−m1−1, implying m2=Δ2−m1. Now, ∀m1∈l7,2, Δ2−m1∈l8,2, implying that Ωm1,Δ2−m127∈H272 is the mirror image of itself. So, out of 1/2Δ2Δ2−1. homomorphic images in H272, Δ2−1 are the mirror images of itself. Hence, there are(21)2×E272×H272−Δ2−1+E272×Δ2−1=6Δ2−12pairs of contracting vertices to produce all the homomorphic images in H272. We can prove this lemma for k2=3 (Figure 41) in similar way as that for k2=2.
Letk4=1,4 and η4,k4=0ifΔk4=0mod21otherwise.Figure 40
Homomorphic imagesΩl7,2,l8,227.Figure 41
Homomorphic imagesΩl7,3,l8,327.Lemma 28.
If we contract the verticese3l7,k4+1k4:l9,k4=1,2,…,Δk4−2−η4,k4/2, with the vertices e3l10,k4k4:l10,k4=l9,k4+1,l9,k4+2,…,Δk4−l9,k4, in the circuit Δ, then for each k4, there arise 1/4Δk42−2Δk4+η4,k4 distinct homomorphic images of Δ and 6 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all 1/4Δk42−2Δk4+η4,k4 and their mirror homomorphic images of Δ is 3Δk42−3Δk4+2.Proof.
Fork4=1, let H281=Ωl9,1,l10,128;l9,1=1,2,…,Δ1−2−η4,1/2;l10,1=l9,1+1,l9,1+2,…,Δ1−l9,1 (Figure 42) be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l9,1+11 with the vertices e3l10,11 in Δ, where e3l9,1+11 and e3l10,11 are fixed by the words pql9,1pq−1Δ6pqΔ5pq−1Δ4pqΔ3pq−1Δ2pqΔ1−l9,1 and pq−1Δ1−l10,1pqΔ2pq−1Δ3pqΔ4pq−1Δ5pqΔ6pq−1l10,1. It is easy to verify that P1=pql9,1pq−1Δ6pqΔ5pq−1Δ4pqΔ3pq−1Δ2pqΔ1−l10,1p and P2=qpql10,1−l9,1−1 are the possible paths between e3l9,1+11 and e3l10,11 (Figure 4). This implies that the homomorphic image Ωl9,1,l10,128 has a vertex s fixed by P1 and P2. Thus,(22)E281=e,q,q−1,p,pq,pq−1is the family of elements in PSL2,Z such that ∀w∈E281,Δ contains the vertices e3l9,1+11w and e3l10,11w. This gives that the cardinality of E281, that is, 6 is the number of contracted pairs of vertices to produce each homomorphic image Ωl9,1,l10,128 (Theorem 2).
SupposeΩm1,m228 and Ωn1,n228 are the same. This concludes that Ωm1,m228 is procurable also by contracting e3n1+11 and e3n21, implying that e3n1+11↔e3n21 is one of the pairs of contracted vertices for Ωm1,m228. Then there must exist an element w∈E281 such that e3m1+11w=e3n1+11 and e3m21w=e3n21. pq,pq−1∈E281 are the only element such that e3m1+11pq=e3m1−1+11 and e3m21pq−1=e3m2+11 but e3m21pq≠e3n21 and e3m1+11pq−1≠e3n1+11. Ωm1,m228 and Ωn1,n228 are not the same, that is, by contracting e3m1+11 and e3m21 to produce Ωm1,m228, e3n1+11 and e3n21 are not contracted. Now, if Ωm1,m228 and Ωn1,n228∗ are same, then there must exist an element w∈E281 such that e3m1+11w=e3n1+11∗ and e3m21w=e3n21∗. p∈E281 is the only element such that e3m1+11p=e3m11=e3Δ1−m1+11∗ and e3m21p=e3m2+11=e3Δ1−m21∗. This implies that Ωm1,m228 and Ωn1,n228 are the mirror images of each other if and only if n1=Δ1−m1 and n2=Δ1−m2, that is, by contracting e3m1+11 and e3m21 to produce Ωm1,m228, e3n1+11∗ and e3n21∗ are also contracted. For all m1∈l9,1, Δ1−m1>Δ1−2−η4,1/2, implying that Ωn1,n228=ΩΔ2−m2,Δ2−m128∉H281 and that Ωm1,m228 and Ωn1,n228 are not the mirror images of each other. This implies that all the homomorphic images in H281 are distinct. Thus,(23)H281=Ωl9,1,l10,128;l9,1=1,2,…,Δ1−2−η4,12;l10,1=l9,1+1,l9,1+2,…,Δ1−l9,1implies(24)H281=Δ1−2+Δ2−4+Δ2−6+⋯+4−η4,1+2−η4,1,=14Δ12−2Δ1+η4,1.
LetΩm1,m228 be the mirror image of itself, then from Figure 42, we have m1−1=Δ1−m2−1, implying m2=Δ1−m1. Now, ∀m1∈l9,1, Δ1−m1∈l10,1, implying that Ωm1,Δ1−m128∈H281 is the mirror image of itself. So, out of 1/4Δ12−2Δ1+η4,1 homomorphic images in H281, Δ1−2−η4,1/2 are the mirror images of itself. Hence, there are(25)2×E282×H282−Δ1−2−η4,12+E282×Δ1−2−η4,12=3Δ12−3Δ1+2pairs of contracting vertices to produce all the homomorphic images in H281. We can prove this lemma for k4=4 (Figure 43) in similar way as that for k4=1.Figure 42
Homomorphic imagesΩl9,1,l10,128.Figure 43
Homomorphic imagesΩl9,4,l10,428.Lemma 29.
If we contract the verticese3l11,k2+1k2:l11,k2=1,2,…,Δk2−1, with the vertices e3l12,k2k2:l12,k2=1,2,…,Δk2−l11,k2, in the circuit Δ, then for each k2, there arise 1/6Δk2−1Δk2−2 distinct homomorphic images of Δ and 6 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all 1/2Δk2−1Δk2−2 and their mirror homomorphic images of Δ is 6Δk2−1Δk2−2.
For a fix value ofk2, let H29k2=Ωl11,k2,l12,k229;l11,k2=1,2,…,Δk2−1;l12,k2=1,2,…,Δk2−l11,k2 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l11,k2+1k2 with the vertices e3l12,k2k2 in Δ. Figures 44 and 45 present H29k2 graphically. From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 23.
Letk4=1,4 and η4,k4=0ifΔk4=0mod21otherwise.Figure 44
Homomorphic imagesΩl11,2,l12,229.Figure 45
Homomorphic imagesΩl11,3,l12,329.Lemma 30.
If we contract the verticese3l13,k4+1k4:l13,k4=1,2,…,Δk4−η4,k4/2, with the vertices e3l13,k4k4 in the circuit Δ, then for each k4, there arise 1/2Δk4−η4,k4 distinct homomorphic images of Δ and 3 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all 1/2Δk4−η4,k4 and their mirror homomorphic images of Δ is 3Δk4−1.
For a fix value ofk4, let H30k4=Ωl13,k4,l13,k430;l13,k4=1,2,…,Δk4−η4,k4/2 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l13,k4+1k4 with the vertices e3l13,k4k4 in Δ. Figures 46 and 47 present H30k4 graphically. From all the homomorphic images presented in these figures, it is not intricated to check that if Δk2=0mod2, then ΩΔk2/2,Δk2/230is the only homomorphic image which is the mirror image of itself; otherwise, no one is the mirror image of itself.
This lemma can be proved by using the same procedure as that for Lemma28.Figure 46
Homomorphic imagesΩl13,1,l13,130.Figure 47
Homomorphic imagesΩl13,4,l13,430.Lemma 31.
If we contract the verticese3l11,k2+1k2:l11,k2=1,2,…,Δk2−1, with the vertices e3l11,k2k2 in the circuit Δ, then for each k2, there arise Δk2−1 distinct homomorphic images of Δ and 3 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δk2−1 and their mirror homomorphic images of Δ is 6Δk2−1.
For a fix value ofk2, let H31k2=Ωl11,k2,l11,k231;l11,k2=1,2,…,Δk2−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l11,k2+1k2 with the vertices e3l11,k2k2 in Δ. Figures 48 and 49 present H31k2 graphically. From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using same procedure as that for Lemma 23.
Letk5=2,4.Figure 48
Homomorphic imagesΩl11,2,l11,231.Figure 49
Homomorphic imagesΩl11,3,l11,331.Lemma 32.
For eachk5, there are 6 Δk5+Δk5+1+2 pairs of contracting vertices to produce the homomorphic image of Δ by contracting the vertex e11 with the vertex e1k5.
Fork5=2, let Ω232 be the homomorphic image of Δ acquired by contracting the vertex e11 with the vertex e12 in Δ, where e11 and e12 are fixed by the words pq−1Δ6pqΔ5pq−1Δ4pqΔ3pq−1Δ2pqΔ1 and pq−1Δ3pqΔ4pq−1Δ5pqΔ6pq−1Δ1pqΔ2. It is easy to verify that P1=pq−1Δ6pqΔ5pq−1Δ4pqΔ3p and P2=q−1pq−1Δ2−1pqΔ1 are the possible paths between e11 and e12 (Figure 4). This implies that the homomorphic image Ω232 has a vertex s fixed by P1 and P2. Thus,(26)E232=e,q,q−1,p,pq,pq−11,pq−11p,pq−11pq,pq−12,…,pq−1Δ3p,pq−1Δ3pq,pq−1Δ3+1,q−1p1,q−1p1q,q−1p1q−1,…,q−1pΔ2,q−1pΔ2q,q−1pΔ2q−1is the family of elements in PSL2,Z such that ∀w∈E232,Δ contains the vertices e11w and e21w. This gives that the cardinality of E232, that is, 3Δ2+Δ3+2 is the number of contracted pairs of vertices to produce the homomorphic image Ω232 (Theorem 2). From Figure 50, it is not intricated to check that Ω232 and its mirror image Ω232∗ are not the same. Hence, there are(27)2×E322=3Δ2+Δ3+2,pairs of contracting vertices to produce the homomorphic images Ω232.
We can prove this lemma fork5=4 (Figure 51) in similar way as that for k5=2.Figure 50
Homomorphic imageΩ232.Figure 51
Homomorphic imageΩ432.Lemma 33.
For eachk3, there are 6 Δ4+Δ5+2 pairs of contracting vertices to produce the homomorphic image of Δ by contracting the vertex e15 with the vertex e1k3.
For a fix value ofk3, let Ωk333 be the homomorphic image of Δ acquired by the contraction of the vertex e15 with the vertex e1k3 in Δ. Figures 52 and 53 present Ωk333 graphically. From these figures, it is not intricated to check that Ωk333 is not the mirror image of itself.
This lemma can be proved by using the same procedure as that for Lemma32.
Letk6=1,2,5 and k6¯=6ifk6=1,3ifk6=2,4ifk6=5..Figure 52
Homomorphic imageΩ133.Figure 53
Homomorphic imageΩ233.Lemma 34.
For eachk6, there are 6 Δk6¯+1 pairs of contracting vertices to produce the homomorphic image of Δ by contracting the vertex e1k6 with the vertex e1k6¯.
For a fix value ofk6, let Ωk634 be the homomorphic image of Δ acquired by the contraction of the vertex e1k6 with the vertex e1k6¯ in Δ. Figures 54, 55, and 56 present Ωk634 graphically. From these figures, it is not intricated to check that Ωk634 is not the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 32.Figure 54
Homomorphic imageΩ134.Figure 55
Homomorphic imageΩ234.Figure 56
Homomorphic imageΩ534.Lemma 35.
For eachk3, there are 3 Δk3+1+2Δk3+2+2 pairs of contracting vertices to produce the homomorphic image of Δ by contracting the vertex e1k3 with the vertex e1k3+2.
For a fix value ofk3, let Ωk335 be the homomorphic image of Δ acquired by the contraction of the vertex e1k3 with the vertex e1k3+2 in Δ. Figures 57(a), 57(b), and 57(c) present Ω135 and Figure 58 presents Ω235 graphically. From these figures, it is not intricated to check that Ωk335 is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 32.
LetVΔ×Δ=a,b;a,b∈VΔanda≠b, then(28)VΔ×Δ=129Δ1+2Δ2+2Δ3+Δ4−3Δ1+2Δ2+2Δ3+Δ4⋅
We are now able to prove our primary outcome. Let(29)η=4,ifΔ1,Δ4=E,E,0,ifΔ1,Δ4=O,O,−1,otherwise.where E and O stand for even and odd, respectively.Figure 57
(a) Homomorphic imageΩ135: Δ1>Δ3+Δ5+1. (b) Homomorphic image Ω135: Δ1=Δ3+Δ5+1. (c) Homomorphic image Ω135: Δ1<Δ3+Δ5+1.
(a)(b)(c)Figure 58
Homomorphic imageΩ235.Theorem 5.
There are1/4Δ1+2Δ2+2Δ3+Δ42+4Δ1+2Δ2+2Δ3+Δ4+4Δ4+η number of distinct homomorphic images acquired by the contraction of all pairs of vertices of the circuit Δ.Proof.
Let us collect all the pairs of contracting vertices ofΔ mentioned in Lemma 1 to Lemma 35 in the form of set S as(30)S=e3l1+11,e3Δ11,e3l1+12,e3Δ11,e3l2,k1+1k1,e3Δ11,e3l3+16,e3Δ11,e3l4+11,e3Δ22,e3l5+12,e3Δ22,e3l6,k1+1k1,e3Δ22,e3l7+13,e3Δ33,e3l7+14,e3Δ33,e3l8+14,e3Δ44,e3l9+15,e3Δ44,e3l10+13,e3Δ44,e3l1+12,e3Δ55,e3l11+13,e3Δ55,e3l7+14,e3Δ55,e3l12+15,e3Δ55,e3l13+13,e3Δ66,e3l14+11,e3Δ11,e3l15+11,e3Δ33,e3l3,k2+1k2,e3Δ33,e3l16+12,e3Δ44,e3l4,k3+1k3,e3Δ66,e3l4+11,e3l5,k2k2,e3l4+11,e3l174,e3l5+12,e3l6,k1k1,e3l13+13,e3l74,e3l7,k2+1k2,e3l8,k2k¯2,e3l7,k4+1k4,e3l10,k4k4,e3l11,k2+1k2,e3l12,k2k2,e3l13,k4+1k4,e3l13,k4k4,e3l11,k2+1k2,e3l11,k2k2,e1k5,e11,e1k3,e15,e1k6,e1k¯6,e1k3,e1k3+2.
LetH be the set of homomorphic images obtained by contracting each element of S, then(31)H=Ωl11,Ωl12,Ωl2,k13,Ωl34,Ωl45,Ωl56,Ωl6,k17,Ωl78,Ωl79,Ωl810,Ωl911,Ωl1012,Ωl113,Ωl1114,Ωl715,Ωl1216,Ωl1317,Ωl1418,Ωl1519,Ωl3,k220,Ωl1621,Ωl4,k322,Ωl4,l5,k223,Ωl4,l1724,Ωl5,l6,k125,Ωl13,l726,Ωl7,k2,l8,k227,Ωl9,k4,l10,k428,Ωl11,k2,l12,k229,Ωl13,k4,l13,k430,Ωl11,k2,l11,k231,Ωk532,Ωk333,Ωk634,Ωk335⋅
Let S′ be the sum of contracted pairs obtained by contracting each element of S, then(32)S′=3Δ22+3Δ2−2+3Δ22+3Δ2+32Δ62+3Δ6+32Δ12+3Δ1−4+3Δ22+3Δ2−4+3Δ42+3Δ4−4+32Δ42+3Δ4−4+3Δ42+3Δ4−2+32Δ52+3Δ5+3Δ32+3Δ3+32Δ22+3Δ2+3Δ32+3Δ3−2+3Δ42+3Δ4−4+3Δ52+3Δ5+32Δ32+3Δ3−4+6Δ2+2Δ1−Δ2−1+3Δ4+2Δ1−Δ4−1+6Δ3+2Δ2−Δ3−1+6Δ1−1Δ4−1+12Δ3−1Δ4−1+∑k1=353Δk12+3Δk1+3Δk12+3Δk1−4+12Δ2−1Δk1−1+∑k2=236Δ4+2Δk2−Δ4−1+12Δ1−1Δk2−1+6Δk2−12+6Δk2−1Δk2−2+6Δk2−1+∑k3=126Δ5+2Δk3−Δ5−1+6Δ4+Δ5+2+3Δk3+1+2Δk3+2+2+∑k4=143Δk42−3Δk4+2+3Δk4−1+∑k5=2,46Δk5+Δk5+1+2+∑k6=1,2,56Δk6+1=129Δ1+2Δ2+2Δ3+Δ4−3Δ1+2Δ2+2Δ3+Δ4⋅shows that each element of VΔ×Δ is contracted.
Now,(33)H=Δ2+Δ2+Δ6+Δ1−1+Δ2−1+Δ4−1+Δ1−1+Δ4+Δ5+Δ3+Δ2+Δ3+Δ4−1+Δ5+Δ3−1+Δ1−Δ2−1+Δ1−Δ4−η12+Δ2−Δ3−1+12Δ1−1Δ4−1ifΔ4=1mod212Δ1−1Δ4−1+1−η3ifΔ4=0mod2+Δ3−1Δ4−1+∑k1=35Δk1+Δk1−1+Δ2−1Δk1−1+∑k1=35Δk2−Δ4−1+Δ1−1Δk2−1+12Δk2Δk2−1+Δk2−1+12Δk2−1Δk2−2+∑k3=12Δk3−Δ5−1+1+1+∑k4=1,414Δk42−2Δk4+Δ4,k4+12Δk4−η4,k4+∑k5=2,41+∑k6=1,2,51=14Δ1+2Δ2+2Δ3+Δ42+4Δ1+2Δ2+2Δ3+Δ4+4Δ4+η.
The value ofH gives the total number of homomorphic images produced in the contraction of all pairs of vertices of Δ. The value of S′ guaranteed that the set H contains all homomorphic images.


## 1.1. Counting the Number of Pairs of Contracting Vertices of a Homomorphic Image

LetΓ be a homomorphic image of the circuit Δ acquired by the contraction of pair of vertices a and u of Δ. Then by Theorem 2, Γ has E number of pairs of vertices. To find the total number of pairs of vertices, one should follow the following steps.To know how many total pairs of contracting vertices are there, special precaution must be taken.(1)
If by contractinga and u to create Γ, the pair of vertices a∗ and u∗ are not contracted, then Γ has different orientation from its mirror image Γ∗. So, there are E number of more pairs of vertices for the mirror image of Γ.(2)
If by contractinga and u to create Γ, the pair of vertices a∗ and u∗ are also contracted, then Γ has the same orientation as that of its mirror image Γ∗. So, in this case, Γ has E number of pairs of contracted vertices.Consider a circuit of length 6 asΔ=Δ1,Δ2,Δ3,Δ4,Δ5,Δ6 (Figure 4) with vertical axis of symmetry, that is, Δ2=Δ6,Δ3=Δ5. Suppose Δ1>Δ2>Δ3>Δ4. The coset diagrams are composed of circuits. The vertices of the circuits in infinite diagrams are contracted in a certain way, and a finite coset diagram evolves. It is therefore necessary to ask how many distinct homomorphic images are obtained if we contract all the pairs of vertices of the circuit Δ? We not only give the answer to this question for a circuit Δ but also mention those pairs of vertices which are “important”. There is no need to contract the pairs which are not mentioned as “important”. If we contract those, we obtain a homomorphic image, which we have already obtained by contracting “important” pairs.Figure 4
The circuitΔ=Δ1,Δ2,Δ3,Δ4,Δ5,Δ6.Note 2. It is clear from Figure 4 that(1)
The mirror image of the vertexelk is e3Δk−l−12−k¯, that is, elk∗=e3Δk−l−12−k¯(2)
The vertexa3m+1k is fixed by the word pq−1mpqΔk±5¯pq−1Δk±4¯pqΔk±3¯pq−1Δk±2¯pqΔk±1¯pq−1Δk−m(3)
The vertexx3mk is fixed by the word pqΔk−mpq−1Δk±1¯pqΔk±2¯pq−1Δk±3¯pqΔk±4¯pq−1Δk±5¯pqm, where k∈1,2,3,4,5,6;l=1,2,3,…,Δk;m=1,2,3,…,Δk−1,n¯=6,ifn=0,6nmod6otherwise and ±=+ifkis odd−otherwise⋅Lemma 1.
If we contract the verticese3l1+11:l1=0,1,2,…,Δ2−1, with the vertex e3Δ11 in the circuit Δ, then there arise Δ2 distinct homomorphic images of Δ and 3l1+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ2 and their mirror homomorphic images of Δ is 3Δ22+3Δ2−2.Proof.
LetH1=Ωl11;l1=0,1,2,…,Δ2−1 (Figure 5) be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l1+11:l1=0,1,2,…,Δ2−1 with the vertex e3Δ11 in Δ, where e3l1+11 and e3Δ11are fixed by the words pql1pq−1Δ6pqΔ5pq−1Δ4pqΔ3pq−1Δ2pqΔ1−l1 and pqΔ2pq−1Δ3pqΔ4pq−1Δ5pqΔ6pq−1Δ1. It is easy to verify that P1=q−1pq−1Δ1−l1−1 and P2=pqΔ2pq−1Δ3pqΔ4pq−1Δ5pqΔ6pq−1l1p are the possible paths between e3l1+11 and e3Δ11 (Figure 4). This implies that for each l1, the homomorphic image Ωl11 has a vertex s fixed by P1 and P2.
Thus,(4)E1=e,q,q−1,p,pq,pq−1,pqp,pq2,pqpq−1,…,pql1p,pql1+1,pql1pq−1is the family of elements in PSL2,Z such that ∀x∈E,Δ contains the vertices e3l1+11x and e3Δ11x. This gives that the cardinality of E1, that is, 3l1+2 is the number of contracted pairs of vertices to produce the homomorphic image Ωl11 (Theorem 2). For m≠n, let Ωm1 and Ωn1 be any two elements of H1, then the number of triangles in Ωm1 and Ωn1 are not equal (Figure 5). This implies that all the elements in H1 are different and no one is the mirror image of the other. This further forms the result as H1=Δ2. Thence, the number of contracted pairs of vertices of Δ to create all the elements of H1 is ∑l1=0Δ2−13l1+2.
From Figure5, it is also clear that no element of H1 except Ω01 has vertical axis of symmetry. So, Ω01 is the only homomorphic image whose orientation is not different from its mirror image Ω01∗ and all the remaining Δ2−1 elements of H1 have different orientations from their mirror images. Hence, there are(5)6∑l1=1Δ2−1l1+2+6=3Δ22+3Δ2−2pairs of contracted vertices to produce all the homomorphic images in H1.Figure 5
Homomorphic imagesΩl11.Lemma 2.
If we contract the verticese3l1+12:l1=0,1,2,…,Δ2−1, with the vertex e3Δ11 in the circuit Δ, then there arise Δ2 distinct homomorphic images of Δ and 3l1+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ2 and their mirror homomorphic images of Δ is 3Δ22+3Δ2.LetH2=Ωl12;l1=0,1,2,…,Δ2−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l1+12:l1=0,1,2,…,Δ2−1 with the vertex e3Δ11 in Δ. Suppose r shows itself as the remainder of l1/Δ2−l1. Then, graphically we make four partitions of H2 as follows:(i)
Ωl12:Δ2−2l1>1 (Figure 6(a))(ii)
Ωl12:Δ2−2l1=1 (Figure 6(b))(iii)
Ωl12:Δ2−2l1<1 and Δ2−l1>r+1 (Figure 6(c))(iv)
Ωl12:Δ2−2l1<1 and Δ2−l1=r+1 (Figure 6(d))From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma1.Letk1∈3,4,5.Figure 6
(a) Homomorphic imagesΩl12:Δ2−2l1>1. (b) Homomorphic images Ωl12:Δ2−2l1=1. (c) Homomorphic images Ωl12:Δ2−2l1<1 and Δ2−l1>r+1. (d) Homomorphic images Ωl12:Δ2−2l1<1 and Δ2−l1=r+1.
(a)(b)(c)(d)Lemma 3.
If we contract the verticese3l2,k1+1k1:l2,k1=0,1,2,…,Δk1−1, with the vertex e3Δ11 in the circuit Δ, then for each k1, there arise Δk1 distinct homomorphic images of Δ and 3l2,k1+2pairs of contracted vertices that create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δk1 and their mirror homomorphic images of Δ is 3Δk12+3Δk1.For a fix value ofk1, let H3k1=Ωl2,k13;l2,k1=0,1,2,…,Δk1−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l2,k1+1k1:l2,k1=0,1,2,…,Δk1−1, with the vertex e3Δ11 in Δ. Figures 7–9 present H3k1 graphically. From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself.This lemma can be proved by using the same procedure as that for Lemma1.Figure 7
Homomorphic imagesΩl2,33.Figure 8
Homomorphic imagesΩl2,53.Figure 9
Homomorphic imagesΩl2,43.Lemma 4.
If we contract the verticese3l3+16:l3=0,1,2,…,Δ6−1, with the vertex e3Δ11 in the circuit Δ, then there arise Δ6 distinct homomorphic images of Δ and 3l3+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ6 and their mirror homomorphic images of Δ is 3/2Δ62+3Δ6.LetH4=Ωl34;l3=0,1,2,…,Δ6−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l6+11:l3=0,1,2,…,Δ6−1 with the vertex e3Δ11 in Δ. Figure 10 presents H4 graphically. From all the homomorphic images presented in Figure 10, it is not intricated to check that every one is the mirror image of itself.This lemma can be proved by using the same procedure as that for Lemma1.Figure 10
Homomorphic imagesΩl34.Lemma 5.
If we contract the verticese3l4+11:l4=1,2,…,Δ1−1, with the vertex e3Δ22 in the circuit Δ, then there arise Δ1−1 distinct homomorphic images of Δ and 3l4+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ1−1 and their mirror homomorphic images of Δ is 3/2Δ12+3Δ1−4.
LetH5=Ωl45;l4=1,2,…,Δ1−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l4+11:l4=1,2,…,Δ1−1, with the vertex e3Δ22 in Δ. Suppose r shows itself as the remainder of l4/Δ1−l4, then graphically, we make four partitions of H5 as follows:(i)
Ωl45:Δ1−2l4>1 (Figure 11(a))(ii)
Ωl45:Δ1−2l4=1 (Figure 11(b))(iii)
Ωl45:Δ1−2l4<1 and Δ1−l4>r+1 (Figure 11(c))(iv)
Ωl45:Δ1−2l4<1 and Δ1−l4=r+1 (Figure 11(d))From all the homomorphic images presented in these figures, it is not intricated to check that every one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma1.Figure 11
(a) Homomorphic imagesΩl45:Δ1−2l4>1. (b) Homomorphic images Ωl45:Δ1−2l4=1. (c) Homomorphic images Ωl45:Δ1−2l4<1 and Δ1−l4>r+1. (d) Homomorphic images Ωl45:Δ1−2l4<1 and.Δ1−l4=r+1
(a)(b)(c)(d)Lemma 6.
If we contract the verticese3l5+12:l5=1,2,…,Δ2−1, with the vertex e3Δ22 in the circuit Δ, then there arise Δ2−1 distinct homomorphic images of Δ and 3l5+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ2−1 and their mirror homomorphic images of Δ is 3Δ22+3Δ2−4.
LetH6=Ωl56;l5=1,2,…,Δ2−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l5+12:l5=1,2,…,Δ2−1, with the vertex e3Δ22 in Δ. Graphically, we make two partitions of H5 as follows:(i)
Ωl56:l5<Δ2−1 (Figure 12(a))(ii)
Ωl56:l5=Δ2−1 (Figure 12(b))
From all the homomorphic images presented in Figures12(a) and 12(b), it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 1.
Letk1∈3,4,5.Figure 12
(a) Homomorphic imagesΩl56:l5<Δ2−1. (b) Homomorphic images Ωl56:l5=Δ2−1.
(a)(b)Lemma 7.
If we contract the verticese3l6,k1+1k1:l6,k1=1,2,…,Δk1−1, with the vertex e3Δ22 in the circuit Δ, then for each k1, there arise Δk1−1 distinct homomorphic images of Δ and 3l6,k1+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δk1−1 and their mirror homomorphic images of Δ is 3Δk12+3Δk1−4.
For a fix value ofk1, let H7k1=Ωl6,k17;l6,k1=1,2,…,Δk1−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l6,k1+1k1:l6,k1=1,2,…,Δk1−1, with the vertex e3Δ22 in Δ. Figures 13–15 present H7k1 graphically. From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 1.Figure 13
Homomorphic imagesΩl6,37.Figure 14
Homomorphic imagesΩl6,47.Figure 15
Homomorphic imagesΩl6,57.Lemma 8.
If we contract the verticese3l7+13:l7=1,2,…,Δ4−1, with the vertex e3Δ33 in the circuit Δ, then there arise Δ4−1 distinct homomorphic images of Δ and 3l7+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ4−1 and their mirror homomorphic images of Δ is 3Δ42+3Δ4−4.
LetH8=Ωl78;l7=1,2,…,Δ4−1be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l7+13:l7=1,2,…,Δ4−1, with the vertex e3Δ33 in Δ. Figure 16 presents H8 graphically. From all the homomorphic images presented in Figure 16, it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 1.Figure 16
Homomorphic imagesΩl78.Lemma 9.
If we contract the verticese3l7+14:l7=1,2,…,Δ4−1, with the vertex e3Δ33 in the circuit Δ, then there arise Δ4−1 distinct homomorphic images of Δ and 3l7+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ1−1 and their mirror homomorphic images of Δ is 3/2Δ42+3Δ4−4.
LetH9=Ωl79;l7=1,2,…,Δ4−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l7+14:l7=1,2,…,Δ4−1, with the vertex e3Δ33 in Δ. Suppose r shows itself as the remainder of l7/Δ4−l7, then graphically we make four partitions of H9 as follows:(i)
Ωl79:Δ4−2l7>1 (Figure 17(a))(ii)
Ωl79:Δ4−2l7=1 (Figure 17(b))(iii)
Ωl79:Δ4−2l7<1 and Δ4−l7>r+1 (Figure 17(c))(iv)
Ωl79:Δ4−2l7<1 and Δ4−l7=r+1 (Figure 17(d))
From all the homomorphic images presented in these figures, it is not intricated to check that every one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma1.Figure 17
(a) Homomorphic imagesΩl79:Δ4−2l7>1. (b) Homomorphic images Ωl79:Δ4−2l7=1. (c) Homomorphic images Ωl79:Δ4−2l7<1 and Δ4−l7>r+1. (d) Homomorphic images Ωl79:Δ4−2l7<1 and Δ4−l7=r+1.
(a)(b)(c)(d)Lemma 10.
If we contract the verticese3l8+14:l8=0,1,2,…,Δ4−1, with the vertex e3Δ44 in the circuit Δ, then there arise Δ4 distinct homomorphic images of Δ and 3l8+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ4 and their mirror homomorphic images of Δ is 3Δ42+3Δ4−2.
LetH10=Ωl810;l8=0,1,2,…,Δ4−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l8+14:l8=0,1,2,…,Δ4−1, with the vertex e3Δ44 in Δ. Graphically, we make two partitions of H10 as follows:(i)
Ωl810:l8<Δ4−1 (Figure 18(a))(ii)
Ωl810:l8=Δ4−1 (Figure 18(b))
From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself exceptΩ010. This lemma can be proved by using the same procedure as that for Lemma 1.Figure 18
(a) Homomorphic imagesΩl810:l8<Δ4−1. (b) Homomorphic images Ωl810:l8=Δ4−1.
(a)(b)Lemma 11.
If we contract the verticese3l9+15:l9=0,1,2,…,Δ5−1, with the vertex e3Δ44 in the circuit Δ, then there arise Δ5 distinct homomorphic images of Δ and 3l9+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ5 and their mirror homomorphic images of Δ is 3/2Δ52+3Δ5.
LetH11=Ωl911;l9=0,1,2,…,Δ5−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l9+15:l9=0,1,2,…,Δ5−1, with the vertex e3Δ44 in Δ. Figure 19 presents H11 graphically. From all the homomorphic images presented in Figure 19, it is not intricated to check that every one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 1.Figure 19
Homomorphic imagesΩl911.Lemma 12.
If we contract the verticese3l10+13:l10=0,1,2,…,Δ3−1, with the vertex e3Δ44 in the circuit Δ, then there arise Δ3 distinct homomorphic images of Δ and 3l10+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ3 and their mirror homomorphic images of Δ is 3Δ32+3Δ3.
LetH12=Ωl1012;l10=0,1,2,…,Δ3−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l10+13:l10=0,1,2,⋯,Δ3−1, with the vertex e3Δ44 in Δ. Suppose r shows itself as the reminder of l10/Δ3−l10, then graphically we make four partitions of H12 as follows:(i)
Ωl1012:Δ3−2l10>1 (Figure 20(a))(ii)
Ωl1012:Δ3−2l10=1 (Figure 20(b))(iii)
Ωl1012:Δ3−2l10<1 and Δ3−l10>r+1 (Figure 20(c))(iv)
Ωl1012:Δ3−2l10<1 and Δ3−l10=r+1 (Figure 20(d))
From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma1.Figure 20
(a) Homomorphic imagesΩl1012:Δ3−2l10>1. (b) Homomorphic images Ωl1012:Δ3−2l10=1. (c) Homomorphic images Ωl1012:Δ3−2l10<1 and Δ3−l10>r+1. (d) Homomorphic images Ωl1012:Δ3−2l10<1 and Δ3−l10=r+1.
(a)(b)(c)(d)Lemma 13.
If we contract the verticese3l1+12:l1=0,1,2,…,Δ2−1, with the vertex e3Δ55 in the circuit Δ, then there arise Δ2 distinct homomorphic images of Δ and 3l1+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ2 and their mirror homomorphic images of Δ is 3/2Δ22+3Δ2.
LetH13=Ωl113;l1=0,1,2,…,Δ2−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l1+12:l1=0,1,2,…,Δ2−1, with the vertex e3Δ55 in Δ. Figure 21 presents H13 graphically. From all the homomorphic images presented in Figure 21, it is not intricated to check that every one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 1.Figure 21
Homomorphic imagesΩl113.Lemma 14.
If we contract the verticese3l11+13:l11=0,1,2,…,Δ3−1, with the vertex e3Δ55 in the circuit Δ, then there arise Δ3 distinct homomorphic images of Δ and 3l11+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ3 and their mirror homomorphic images of Δ is 3Δ32+3Δ3−2.
LetH14=Ωl1114;l11=0,1,2,…,Δ3−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l11+13:l11=0,1,2,…,Δ3−1, with the vertex e3Δ55 in Δ. Figure 22 presents H14 graphically. From all the homomorphic images presented in Figure 22, it is not intricated to check that no one is the mirror image of itself except Ω014. This lemma can be proved by using the same procedure as that for Lemma 1.Figure 22
Homomorphic imagesΩl1114.Lemma 15.
If we contract the verticese3l7+14:l7=1,2,…,Δ4−1, with the vertex e3Δ55 in the circuit Δ, then there arise Δ4−1 distinct homomorphic images of Δ and 3l7+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ4−1 and their mirror homomorphic images of Δ is 3Δ42+3Δ4−4.
LetH15=Ωl715;l7=1,2,…,Δ4−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l7+14:l7=1,2,…,Δ4−1, with the vertex e3Δ55 in Δ. Figure 23 presents H15 graphically. From all the homomorphic images presented in Figure 23, it is not intricated to check that no one is the mirror image of itself.
This lemma can be proved by using the same procedure as that for Lemma1.Figure 23
Homomorphic imagesΩl715.Lemma 16.
If we contract the verticese3l12+15:l12=0,1,2,…,Δ5−1, with the vertex e3Δ55 in the circuit Δ, then there arise Δ5 distinct homomorphic images of Δ and 3l12+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ5 and their mirror homomorphic images of Δ is 3Δ52+3Δ5.
LetH16=Ωl1216;l12=0,1,2,…,Δ5−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l12+15:l12=0,1,2,…,Δ5−1, with the vertex e3Δ55 in Δ. Graphically, we make two partitions of H16 as follows:(i)
Ωl1216:l12<Δ5−1 (Figure 24(a))(ii)
Ωl1216:l12=Δ5−1 (Figure 24(b))
From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma1.Figure 24
(a) Homomorphic imagesΩl1216:l12<Δ5−1. (b) Homomorphic images Ωl1216:l12=Δ5−1.
(a)(b)Lemma 17.
If we contract the verticese3l13+13:l13=1,2,…,Δ3−1, with the vertex e3Δ66 in the circuit Δ, then there arise Δ3−1 distinct homomorphic images of Δ and 3l13+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ3−1 and their mirror homomorphic images of Δ is 3/2Δ32+3Δ3−4.
LetH17=Ωl1317;l13=1,2,…,Δ3−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l13+13:l13=1,2,…,Δ3−1, with the vertex e3Δ66 in Δ. Figure 25 presents H17 graphically. From all the homomorphic images presented in Figure 25, it is not intricated to check that every one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 1.Figure 25
Homomorphic imagesΩl1317.Lemma 18.
If we contract the verticese3l14+11:l14=Δ2+1,Δ2+2,…,Δ1−1, with the vertex e3Δ11 in the circuit Δ, then there arise Δ1−Δ2−1 distinct homomorphic images of Δ and 3Δ2+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ1−Δ2−1 and their mirror homomorphic images of Δ is 6Δ2+2Δ1−Δ2−1.Proof.
LetH18=Ωl1418;l14=Δ2+1,Δ2+2,…,Δ1−1 (Figure 26) be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l14+11:l14=Δ2+1,Δ2+2,…,Δ1−1, with the vertex e3Δ11 in Δ, where e3l14+11 and e3Δ11 are fixed by the words pqΔ1−l14pq−1Δ6pqΔ5pq−1Δ4pqΔ3pq−1Δ2pql14 and pqΔ2pq−1Δ3pqΔ4pq−1Δ5pqΔ2pq−1Δ3pqΔ4pq−1Δ5pqΔ6pq−1Δ1. It is easy to verify that P1=q−1pq−1Δ1−l14−1 and P2=pqΔ2pq−1Δ3pqΔ4pq−1Δ5pqΔ6pq−1l14p are the possible paths between e3l14+11 and e3Δ11 (Figure 4). This implies that, for each l14, the homomorphic image Ωl1418 has a vertex s fixed by P1 and P2. Thus,(6)E18=e,q,q−1,p,pq,pq−1,pqp,pq2,pqpq−1,…,pqΔ2p,,pqΔ2+1,pqΔ2pq−1is the family of elements in PSL2,Z such that ∀x∈E18,Δ contains the vertices e3l14+11x and e3Δ11x. This gives that the cardinality of E18, that is, 3Δ2+2 is the number of contracted pairs of vertices to produce the homomorphic image Ωl1418 (Theorem 2).
Now, we will prove that all the homomorphic images inH18 are distinct and no one is the mirror image of itself.
LetΩm18 and Ωn18 be any two elements of H18, then Ωm18 is produced by contraction of e3m+11 and e3Δ11 and Ωn18 is produced by contraction of e3n+11 and e3Δ11. Suppose Ωm18 and Ωn18 are the same. This concludes that Ωm18 is procurable also by contracting e3n+11 and e3Δ11, implying that e3n+11↔e3Δ11 is one of the pairs of contracted vertices for Ωm18. Then there must exist an element w∈E18 such that e3m+11w=e3n+11 and e3Δ11w=e3Δ11. e∈E18 is the only element such that e3Δ11e=e3Δ11but e3m+11e≠e3n+11. Thus, Ωm18 and Ωn18 are not the same, that is, by contracting e3m+11 and e3Δ11 to produce Ωm18, e3n+11 and e3Δ11 are not contracted. Now, if Ωm18 and Ωn18∗ are the same, then there must exist an element w∈E18 such that e3m+11w=e3n+11∗ and e3Δ11w=e3Δ11∗. But there does not such an element exist in E18. Thus, Ωm18 is not the same as the mirror image of Ωn18, that is, by contracting e3m+11 and e3Δ11 to create Ωm18, e3n+11∗ and e3Δ11∗ are not contracted. This implies that all the elements in H18 are distinct. Thus, H18=Ωl1418;l14=Δ2+1,Δ2+2,…,Δ1−1 gives H18=Δ1−Δ2−1.
To check does any element ofH18 is the mirror image of itself, we supposeΩm18 is the same to its mirror image Ωm18∗. Then by definition, there must exist an element w∈E18 such that e3m+11w=e3m+11∗ and e3Δ11w=e3Δ11∗. But there does not such an element exist in E18. Thus, Ωm18 is not the same as its mirror image, that is, by contracting e3m+11 and e3Δ11 to create Ωm18, e3m+11∗ and e3Δ11∗ are not contracted. Hence, there are(7)2×E18×H18=6Δ2+2Δ1−Δ2−1pairs of contracting vertices to produce all the homomorphic images in H18.
Letη1=0,ifΔ1+Δ4=0mod21,otherwise..Figure 26
Homomorphic imagesΩl1418.Lemma 19.
If we contract the verticese3l15+11:l15=Δ4+1,Δ4+2,…,Δ1+Δ4−η1/2, with the vertex e3Δ33 in the circuit Δ, then there arise Δ1−Δ4−η1/2 distinct homomorphic images of Δ and 3Δ4+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ1−Δ4−η1/2 and their mirror homomorphic images of Δ is 3Δ4+2Δ1−Δ4−1.Proof.
LetH19=Ωl1519;l15=Δ4+1,Δ4+2,…,Δ1+Δ4−η1/2 (Figure 27) be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l15+11:l15=Δ4+1,Δ4+2,…,Δ1+Δ4−η1/2, with the vertex e3Δ33 in Δ, where e3l15+11 and e3Δ33 are fixed by the words pqΔ1−l15pq−1Δ6pqΔ5pq−1Δ4pqΔ3pq−1Δ2pql15 and pqΔ4pq−1Δ5pqΔ6pq−1Δ1pqΔ2pq−1Δ3. It is easy to verify that P1=q−1pq−1Δ1−l15−1pqΔ2pq−1Δ3 and P2=pqΔ4pq−1Δ5pqΔ6pq−1l15p are the possible paths between e3l15+11 and e3Δ33 (Figure 4). This implies that, for each l15, the homomorphic image Ωl1519 has a vertex s fixed by P1 and P2. Thus, (8)E19=e,q,q−1,p,pq,pq−1,pqp,pq2,pqpq−1,…,pqΔ4p,pqΔ4+1,pqΔ4pq−1is the family of elements in PSL2,Z such that ∀x∈E19,Δ contains the vertices e3l15+11x and e3Δ33x. This gives that the cardinality of E19, that is, 3Δ4+2 is the number of contracted pairs of vertices to produce the homomorphic image Ωl1517 (Theorem 2).
Now, we will prove that (i) forΔ1+Δ4=1mod2, all the homomorphic images in H19 are distinct and no one is the mirror image of itself, and (ii) for Δ1+Δ4=0mod2, all the homomorphic images in H19 are distinct and only ΩΔ1+Δ4/219 is the mirror image of itself.
LetΩm19 and Ωn19 be any two elements of H19, then Ωm19 is produced by contraction of e3m+11 and e3Δ33 and Ωn19 is produced by contraction of e3n+11 and e3Δ33.
SupposeΩm19 and Ωn19 are the same. This concludes that Ωm19 is procurable also by contracting e3n+11 and e3Δ33, implying that e3n+11↔e3Δ33 is one of the pairs of contracted vertices for Ωm19. Then there must exist an element w∈E19 such that e3m+11w=e3n+11 and e3Δ33w=e3Δ33. e∈E19 is the only element such that e3Δ33e=e3Δ33 but e3m+11e≠e3n+11. Thus, Ωm19 and Ωn19 are not the same, that is, by contracting e3m+11 and e3Δ33 to produce Ωm19, e3n+11 and e3Δ33 are not contracted. Now, if Ωm19 and Ωn19∗ are the same, then there must exist an element w∈E19 such that e3m+11w=e3n+11∗ and e3Δ33w=e3Δ33∗. pqΔ4p∈E19 is the only element such that e3Δ33pqΔ4p=e3Δ33∗ and e3m+11pqΔ4p=e3Δ1+Δ4−m+11∗. This implies that Ωm19 and Ωn19 are the mirror images of each other if and only if n=Δ1+Δ4−m and Ωm19 is the mirror image of itself if and only if m=Δ1+Δ4−m, that is, m=Δ1+Δ4/2. Now,(i)
IfΔ1+Δ4=1mod2, then ∀m∈l15, Δ1+Δ4−m>Δ1+Δ4−1/2, implying that ΩΔ1+Δ4−m19∉H19. This states that no homomorphic image in H19 is the mirror image of other. Hence, all the homomorphic images in H19 are distinct. Thus, H19=Ωl1519;l15=Δ4+1,Δ4+2,…,Δ1+Δ4−1/2 gives H19=Δ1−Δ4−1/2.Also,Δ1+Δ4/2>Δ1+Δ4−1/2 gives that no homomorphic image in H19 is the mirror image of itself. Hence, there are(9)2×E19×H19=6Δ4+2Δ1−Δ4−12=3Δ4+2Δ3−Δ4−1pairs of contracting vertices to produce all the homomorphic images inH19.(ii)
IfΔ1+Δ4=0mod2, then ∀m∈l15/Δ1+Δ4/2, Δ1+Δ4−m>Δ1+Δ4/2, implying that ΩΔ1+Δ4−m19∉H19. This states that no homomorphic image in H19 is the mirror image of other. Hence, all the homomorphic images in H19 are distinct. Thus, H19=Ωl1519;l15=Δ4+1,Δ4+2,…,Δ1+Δ4/2 gives H19=Δ1−Δ4/2.
Form=Δ1+Δ4/2, we have Δ1+Δ4−m=Δ1+Δ4/2∈l15 gives ΩΔ1+Δ4/219∈H19 is the mirror image of itself. Hence, there are(10)2×E19×H19−1+E19=6Δ4+2Δ1−Δ42−1+3Δ4+2=3Δ4+2Δ3−Δ4−1pairs of contracting vertices to produce all the homomorphic images in H19.
Letk2∈2,3.Figure 27
Homomorphic imagesΩl1519.Lemma 20.
If we contract the verticese3l3,k2+1k2:l3,k2=Δ4+1,Δ4+2,…,Δk2−1, with the vertex e3Δ33 in the circuit Δ, then for each k2, there arise Δk2−Δ4−1 distinct homomorphic images of Δ and 3Δ4+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δk2−Δ4−1 and their mirror homomorphic images of Δ is 6Δ4+2Δk2−Δ4−1.
For a fix value ofk2, let H20k2=Ωl3,k220;l3,k2=Δ4+1,Δ4+2,…,Δk2−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l3,k2+1k2:l3,k2=Δ4+1,Δ4+2,…,Δk2−1, with the vertex e3Δ33 in Δ. Figures 28 and 29 present H3k2 graphically. From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself.
This lemma can be proved by using the same procedure as that for Lemma18.Figure 28
Homomorphic imagesΩl3,220.Figure 29
Homomorphic imagesΩl3,320.Lemma 21.
If we contract the verticese3l16+12:l16=Δ3+1,Δ3+2,…,Δ2−1, with the vertex e3Δ44 in the circuit Δ, then there arise Δ2−Δ3−1 distinct homomorphic images of Δ and 3Δ3+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ2−Δ3−1 and their mirror homomorphic images of Δ is 6Δ3+2Δ2−Δ3−1.
LetH21=Ωl1621;l16=Δ3+1,Δ3+2,…,Δ2−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l16+12:l16=Δ3+1,Δ3+2,…,Δ2−1, with the vertex e3Δ44 in Δ. Figure 30 presents H21 graphically. From all the homomorphic images presented in Figure 30, it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 18.
Letk3∈1,2.Figure 30
Homomorphic imagesΩl1621.Lemma 22.
If we contract the verticese3l4,k3+1k3:l4,k3=Δ5+1,Δ5+2,…,Δk3−1, with the vertex e3Δ66 in the circuit Δ, then for each k3, there arise Δk3−Δ5−1 distinct homomorphic images of Δ and 3Δ5+2 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δk3−Δ5−1 and their mirror homomorphic images of Δ is 6Δ5+2Δk3−Δ5−1.
For a fix value ofk3, let H22k3=Ωl4,k322;l4,k3=Δ5+1,Δ5+2,…,Δk3−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l4,k3+1k3:l4,k3=Δ5+1,Δ5+2,…,Δk3−1, with the vertex e3Δ66 in Δ. Figures 31 and 32 present H22k3 graphically. From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 18.
Letk2∈2,3.Figure 31
Homomorphic imagesΩl4,122.Figure 32
Homomorphic imagesΩl4,222.Lemma 23.
If we contract the verticese3l4+11:l4=1,2,…,Δ1−1, with the vertices e3l5,k2k2:l5,k2=1,2,…,Δk2−1, in the circuit Δ, then for each k2, there arise Δ1−1Δk2−1 distinct homomorphic images of Δ and 6 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ1−1Δk2−1 and their mirror homomorphic images of Δ is 12Δ1−1Δk2−1.Proof.
Fork2=2, let H232=Ωl4,l5,223;l4=1,2,…,Δ1−1;l5,2=1,2,…,Δ2−1(Figure 33) be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l4+11 with the vertex e3l5,22 in Δ, where e3l4+11 and e3l5,22 are fixed by the words pql4pq−1Δ6pqΔ5pq−1Δ4pqΔ3pq−1Δ2pqΔ1−l4 and pq−1Δ2−l5,2pqΔ1pq−1Δ6pqΔ5pq−1Δ4pqΔ3pq−1l5,2pqΔ5pq−1Δ4pqΔ3pq−1l5,2. It is easy to verify that P1=pq−1Δ2−l5,2pqΔ1−l4 and P2=pql4pq−1Δ6pqΔ5pq−1Δ4pqΔ3pq−1l5,2 are the possible paths between e3l4+11 and e3l5,22 (Figure 4). This implies that the homomorphic image Ωl4,l5,223 has a vertex s fixed by P1 and P2. Thus,(11)E232=e,q,q−1,p,pq,pq−1is the family of elements in PSL2,Z such that ∀x∈E232,Δ contains the vertices e3l4+11x and e3l5,22x. This gives that the cardinality of E232, that is, 6 is the number of contracted pairs of vertices to produce the homomorphic image Ωl4,l5,223 (Theorem 2).
Now, we will prove that all the homomorphic images inH232 are distinct and no one is the mirror image of itself.
LetΩm1,m223 and Ωn1,n223 be any two elements of H232, then Ωm1,m223 is produced by contraction of e3m1+11 and e3m22 and Ωn1,n223 is produced by contraction of e3n1+11 and e3n22.
SupposeΩm1,m223 and Ωn1,n223 are the same. This concludes that Ωm1,m223 is procurable also by contracting e3n1+11 and e3n22, implying that e3n1+11↔e3n22 is one of the pairs of contracted vertices for Ωm1,m223. Then there must exist an element w∈E232 such that e3m1+11w=e3n1+11 and e3m22w=e3n22. pq−1∈E52 is the only element such that e3m22pq−1=e3m2+12 but e3m1+11pq−1≠e3n1+11. Thus, Ωm1,m223 and Ωn1,n223 are not the same, that is, by contracting e3m1+11 and e3m22 to produce Ωm1,m223, e3n1+11 and e3n22 are not contracted. Now, if Ωm1,m223 and Ωn1,n223∗ are the same, then there must exist an element w∈E232 such that e3m1+11w=e3n1+11∗ and e3m22w=e3n22∗. p∈E232 is the only element such that e3m1+11p=e3Δ1−m1+11∗ but e3m22p≠e3n22∗. This implies that Ωm1,m223 and Ωn1,n223 are not the mirror images of each other, that is, by contracting e3m1+11 and e3m22 to produce Ωm1,m223, e3n1+11∗ and e3n22∗ are not contracted. Hence, all the homomorphic images in H232 are distinct. Thus, H232=Ωl4,l5,223;l4=1,2,…,Δ1−1;l5,2=1,2,…,Δ2−1 gives H232=Δ1−1Δ2−1.
SupposeΩm1,m223 is the mirror image of itself, then there must exist an element w∈E232 such that e3m1+11w=e3m1+11∗ and e3m22w=e3m22∗. p∈E232 is the only element such that e3m1+11p=e3Δ1−m1+11∗ but e3m22p≠e3m22∗. This implies that Ωm1,m223 is not the same as Ωm1,m223∗, that is, by contracting e3m1+11 and e3m22 to produce Ωm1,m223, e3m1+11∗ and e3m22∗ are not contracted.
Hence, there are(12)2×E232×H232=12Δ1−1Δ2−1pairs of contracting vertices to produce all the homomorphic images in H232.
We can prove this lemma fork2=3 (Figure 34) in similar way as that for k2=2.
Letη2=0ifΔ4=0mod21otherwise and η3=0ifΔ1=0mod21otherwise.Figure 33
Homomorphic imagesΩl4,l5,223.Figure 34
Homomorphic imagesΩl4,l5,323.Lemma 24.
If we contract the verticese3l17+11:l17=1,2,…,Δ1−η3/2,ifl18=Δ4/21,2,…,Δ1−1otherwise, with the vertices e3l184:l18=1,2,…,Δ4−η2/2, in the circuit Δ, then there arise 1/2Δ1−1Δ4−1ifΔ4=1mod21/2Δ1−1Δ4−1+1−η3ifΔ4=0mod2 distinct homomorphic images of Δ and 6 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all these homomorphic images and their mirror homomorphic images of Δ is 6Δ1−1Δ4−1.Proof.
LetH24=Ωl17,l1824;l17=1,2,…,Δ1−η3/2,ifl18=Δ4/2,l18=1,2,…,Δ4−η2/21,2,…,Δ1−1otherwise (Figure 35) be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l17+11 with the vertices e3l184 in Δ, where e3l17+11 and e3l184 are fixed by the words pql17pq−1Δ6pqΔ5pq−1Δ4pqΔ3pq−1Δ2pqΔ1−l18 and pq−1Δ4−l18pqΔ3pq−1Δ2pqΔ1pq−1Δ6pqΔ5pq−1l18. It is easy to verify that P1=pql17pq−1Δ6pqΔ5pq−1l18 and P2=pq−1Δ4−l18pqΔ3pq−1Δ2pqΔ1−l17 are the possible paths between e3l17+11 and e3l184 (Figure 4). This implies that the homomorphic image Ωl17,l1824 has a vertex s fixed by P1 and P2. Thus,(13)E24=e,q,q−1,p,pq,pq−1is the family of elements in PSL2,Z such that ∀x∈E24,Δ contains the vertices e3l17+11x and e3l184x. This gives that the cardinality of E24, that is, 6 is the number of contracted pairs of vertices to produce the homomorphic image Ωl17,l1824 (Theorem 2).
Now, we prove that (i) forΔ4=0mod2, all the elements in H24 are distinct and no one is the mirror image of itself, and (ii) for Δ4=1mod2, all the elements in H24 are distinct and the homomorphic image ΩΔ1/2,Δ4/224 is the mirror image of itself.
SupposeΩm1,m224 and Ωn1,n224 are the same. This concludes that Ωm1,m224 is procurable also by contracting e3n1+11 and e3n24, implying that e3n1+11↔e3n24 is one of the pairs of contracted vertices for Ωm1,m224. Then there must exist an element w∈E24 such that e3m1+11w=e3n1+11 and e3m24w=e3n24. pq−1∈E24 is the only element such that e3m24pq−1=e3m2+14 but e3m1+11pq−1≠e3n1+11. Thus, Ωm1,m224 and Ωn1,n224 are not the same, that is, by contracting e3m1+11 and e3m24 to produce Ωm1,m224, e3n1+11 and e3n24 are not contracted. Now, if Ωm1,m224 and Ωn1,n224 are the same, then there must exist an element w∈E24 such that e3m1+11w=e3n1+11∗ and e3m24w=e3n24∗. p∈E24 is the only element such that e3m1+11p=e3Δ1−m1+11∗ and e3m24p=e3Δ4−m24∗. This implies that Ωm1,m224 and Ωn1,n224 are the mirror images of each other if and only if n1=Δ1−m1 and n2=Δ4−m2, that is, by contracting e3m1+11 and e3m24 to produce Ωm1,m224, e3n1+11∗ and e3n24∗ are also contracted and the homomorphic image Ωm1,m224 is the mirror image of itself if and only if m1=Δ1−m1 and m2=Δ4−m2 that is m1=Δ1/2 and m2=Δ4/2. Now,(1
)IfΔ4=1mod2 then ∀m2∈l18, we get Δ4−m2>Δ4−1/2 implies ΩΔ1−m1,Δ4−m224∉H24. This indicates that no homomorphic image in H24 is the mirror image of other. Hence, all the elements in H24 are distinct. Thus, H24=Ωl17,l1824;l17=1,2,…,Δ1−η3/2ifl18=Δ4/2,l18=1,2,…,Δ4−η2/21,2,…,Δ1−1otherwise, implying that(14)H24=12Δ1−1Δ4−1⋅Also,Δ4/2>Δ4−1/2 implies no homomorphic image in H24 is the mirror image of itself. Hence, there are(15)2×E24×H24=6Δ1−1Δ4−1pairs of contracting vertices to produce all the homomorphic images inH24.(2)
IfΔ4=0mod2, then only for m2=Δ4/2, Δ4−m2=Δ4/2∈l18 and ∀m1∈l17Δ1/2, Δ1−m1∉l17, implying that ΩΔ1−m1,Δ4/224∉H24. Hence, all the elements in H24 are distinct. Thus, H24=Ωl17,l1824;l17=1,2,…,Δ1−η3/2ifl18=Δ4/2;l18=1,2,…,Δ4−η2/21,2,…,Δ1−1otherwise, implying that(16)H24=12Δ1−1Δ4−2+12Δ1−η3=12Δ1−1Δ4−1+1−η3⋅
Now,(a).
IfΔ1 is odd, then Δ1/2∉l17, implying that ΩΔ1/2,4/224∉H24. So, no homomorphic image in H24 is the mirror image of itself. Hence, there are(17)2×E24×H24=6Δ1−1Δ4−1pairs of contracting vertices to produce all the homomorphic images inH24.(b)
IfΔ1 is even, then Δ1/2∈l17, implying that ΩΔ1/2,Δ4/224∈H24 is the mirror image of itself. Hence, there are(18)2×E24×H24−1+E24=6Δ1−1Δ4−1pairs of contracting vertices to produce all the homomorphic images inH24.
Letk1∈3,4,5.Figure 35
Homomorphic imagesΩl17,l1824.Lemma 25.
If we contract the verticese3l5+12:l5=1,2,…,Δ2−1, with the vertices e3l6,k1k1:l6,k1=1,2,…,Δk1−1, in the circuit Δ, then for each k5, there arise Δ2−1Δk1−1 distinct homomorphic images of Δ and 6 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ2−1Δk1−1 and their mirror homomorphic images of Δ is 12Δ2−1Δk1−1.
For a fix value ofk1, let H25k1=Ωl5,l6,k125;l5=1,2,…,Δ2−1;l6,k1=1,2,…,Δk1−1be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l5+12 with the vertices e3l6,k1k1 in Δ. Figures 36, 37, and 38 present H25k1 graphically. From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself.
This lemma can be proved by using the same procedure as that for Lemma23.Figure 36
Homomorphic imagesΩl5,l6,325.Figure 37
Homomorphic imagesΩl5,l6,425.Figure 38
Homomorphic imagesΩl5,l6,525.Lemma 26.
If we contract the verticese3l13+13:l13=1,2,…,Δ3−1, with the vertices e3l74:l7=1,2,…,Δ4−1, in the circuit Δ, then there arise Δ3−1Δ4−1 distinct homomorphic images of Δ and 6 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δ3−1Δ4−1 and their mirror homomorphic images of Δ is 12Δ3−1Δ4−1.
LetH26=Ωl13,l726;l13=1,2,…,Δ3−1;l7=1,2,…,Δ4−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l13+13 with the vertices e3l74 in Δ. Figure 39 presents H26 graphically. From all the homomorphic images presented in Figure 39, it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 23.
Letk2¯=8−k2.Figure 39
Homomorphic imagesΩl13,l726.Lemma 27.
If we contract the verticese3l7,k2+1k2:l7,k2=1,2,…,Δk2−1, with the vertices e3l8,k2k2¯:l8,k2=1,2,…,Δk2−l7,k2, in the circuit Δ, then for each k3, there arise 1/2Δk2Δk2−1 distinct homomorphic images of Δ and 6 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all 1/2Δk2Δk2−1 and their mirror homomorphic images of Δ is 6Δk2−12.Proof.
Fork2=2, let H272=Ωl7,2,l8,227;l7,2=1,2,…,Δ2−1;l8,2=1,2,…,Δ2−l7,2 (Figure 40) be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l7,2+12 with the vertex e3l8,26 in Δ, where e3l7,2+12 and e3l8,26 are fixed by the words pql7,2pq−1Δ3pqΔ4pq−1Δ5pqΔ6pq−1Δ1pqΔ2−l7,2 and pq−1Δ6−l8,2pqΔ5pq−1Δ4pqΔ3pq−1Δ2pqΔ1pq−1l8,2. It is easy to verify that P1=pql7,2pq−1Δ3pqΔ4pq−1Δ5pqΔ6−l8,2p and P2=qpql8,2−1pq−1Δ1pqΔ2−l7,2 are the possible paths between e3l7,2+12 and e3l8,26 (Figure 4). This implies that the homomorphic image Ωl7,2,l8,227 has a vertex s fixed by P1 and P2. Thus,(19)E272=e,q,q−1,p,pq,pq−1is the family of elements in PSL2,Z such that ∀w∈E272,Δ contains the vertices e3l7,2+12w and e3l8,26w. This gives that the cardinality of E272, that is, 6 is the number of contracted pairs of vertices to produce each homomorphic image Ωl7,2,l8,227 (Theorem 2).
SupposeΩm1,m227 and Ωn1,n227 are the same. This concludes that Ωm1,m227 is procurable also by contracting e3n1+12 and e3n26, implying that e3n1+12↔e3n26 is one of the pairs of contracted vertices for Ωm1,m227. Then there must exist an element w∈E272 such that e3m1+12w=e3n1+12 and e3m26w=e3n26. pq−1∈E272 is the only element such that e3m26pq−1=e3m2+16 but e3m1+12pq−1≠e3n1+12. Thus, Ωm1,m227 and Ωn1,n227 are not the same, that is, by contracting e3m1+12 and e3m26 to produce Ωm1,m227, e3n1+12 and e3n26 are not contracted. Now, if Ωm1,m227 and Ωn1,n227 are the same, then there must exist an element w∈E272 such that e3m1+12w=e3n1+12∗ and e3m26w=e3n26∗ or e3m1+12w=e3n1+16∗ and e3m26w=e3n22∗. p∈E272 is the only element such that e3m1+12p=e3m12=e3Δ2−m1+16∗ and e3m26p=e3m2+16=e3Δ2−m22∗. This implies that Ωm1,m227 and Ωn1,n227 are the mirror images of each other if and only if n1=Δ2−m2 and n2=Δ2−m1, that is, by contracting e3m1+12 and e3m26 to produce Ωm1,m227, e3n1+12∗ and e3n26∗ are also contracted. For a fix m1∈l7,2 and for all m2∈l8,2\Δ2−m1, ΩΔ2−m2,Δ2−m122∉H272, implying that Ωm1,m227 and Ωn1,n227 are not the mirror images of each other. This implies that all the homomorphic images in H272 are distinct. Thus, H272=Ωl7,2,l8,227;l7,2=1,2,…,Δ2−1;l8,2=1,2,…,Δ2−l7,2 implies(20)H272=12Δ2Δ2−1.
LetΩm1,m227 be the mirror image of itself, then from Figure 40, we have m1−1=Δ2−m2−1 and m2−1=Δ2−m1−1, implying m2=Δ2−m1. Now, ∀m1∈l7,2, Δ2−m1∈l8,2, implying that Ωm1,Δ2−m127∈H272 is the mirror image of itself. So, out of 1/2Δ2Δ2−1. homomorphic images in H272, Δ2−1 are the mirror images of itself. Hence, there are(21)2×E272×H272−Δ2−1+E272×Δ2−1=6Δ2−12pairs of contracting vertices to produce all the homomorphic images in H272. We can prove this lemma for k2=3 (Figure 41) in similar way as that for k2=2.
Letk4=1,4 and η4,k4=0ifΔk4=0mod21otherwise.Figure 40
Homomorphic imagesΩl7,2,l8,227.Figure 41
Homomorphic imagesΩl7,3,l8,327.Lemma 28.
If we contract the verticese3l7,k4+1k4:l9,k4=1,2,…,Δk4−2−η4,k4/2, with the vertices e3l10,k4k4:l10,k4=l9,k4+1,l9,k4+2,…,Δk4−l9,k4, in the circuit Δ, then for each k4, there arise 1/4Δk42−2Δk4+η4,k4 distinct homomorphic images of Δ and 6 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all 1/4Δk42−2Δk4+η4,k4 and their mirror homomorphic images of Δ is 3Δk42−3Δk4+2.Proof.
Fork4=1, let H281=Ωl9,1,l10,128;l9,1=1,2,…,Δ1−2−η4,1/2;l10,1=l9,1+1,l9,1+2,…,Δ1−l9,1 (Figure 42) be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l9,1+11 with the vertices e3l10,11 in Δ, where e3l9,1+11 and e3l10,11 are fixed by the words pql9,1pq−1Δ6pqΔ5pq−1Δ4pqΔ3pq−1Δ2pqΔ1−l9,1 and pq−1Δ1−l10,1pqΔ2pq−1Δ3pqΔ4pq−1Δ5pqΔ6pq−1l10,1. It is easy to verify that P1=pql9,1pq−1Δ6pqΔ5pq−1Δ4pqΔ3pq−1Δ2pqΔ1−l10,1p and P2=qpql10,1−l9,1−1 are the possible paths between e3l9,1+11 and e3l10,11 (Figure 4). This implies that the homomorphic image Ωl9,1,l10,128 has a vertex s fixed by P1 and P2. Thus,(22)E281=e,q,q−1,p,pq,pq−1is the family of elements in PSL2,Z such that ∀w∈E281,Δ contains the vertices e3l9,1+11w and e3l10,11w. This gives that the cardinality of E281, that is, 6 is the number of contracted pairs of vertices to produce each homomorphic image Ωl9,1,l10,128 (Theorem 2).
SupposeΩm1,m228 and Ωn1,n228 are the same. This concludes that Ωm1,m228 is procurable also by contracting e3n1+11 and e3n21, implying that e3n1+11↔e3n21 is one of the pairs of contracted vertices for Ωm1,m228. Then there must exist an element w∈E281 such that e3m1+11w=e3n1+11 and e3m21w=e3n21. pq,pq−1∈E281 are the only element such that e3m1+11pq=e3m1−1+11 and e3m21pq−1=e3m2+11 but e3m21pq≠e3n21 and e3m1+11pq−1≠e3n1+11. Ωm1,m228 and Ωn1,n228 are not the same, that is, by contracting e3m1+11 and e3m21 to produce Ωm1,m228, e3n1+11 and e3n21 are not contracted. Now, if Ωm1,m228 and Ωn1,n228∗ are same, then there must exist an element w∈E281 such that e3m1+11w=e3n1+11∗ and e3m21w=e3n21∗. p∈E281 is the only element such that e3m1+11p=e3m11=e3Δ1−m1+11∗ and e3m21p=e3m2+11=e3Δ1−m21∗. This implies that Ωm1,m228 and Ωn1,n228 are the mirror images of each other if and only if n1=Δ1−m1 and n2=Δ1−m2, that is, by contracting e3m1+11 and e3m21 to produce Ωm1,m228, e3n1+11∗ and e3n21∗ are also contracted. For all m1∈l9,1, Δ1−m1>Δ1−2−η4,1/2, implying that Ωn1,n228=ΩΔ2−m2,Δ2−m128∉H281 and that Ωm1,m228 and Ωn1,n228 are not the mirror images of each other. This implies that all the homomorphic images in H281 are distinct. Thus,(23)H281=Ωl9,1,l10,128;l9,1=1,2,…,Δ1−2−η4,12;l10,1=l9,1+1,l9,1+2,…,Δ1−l9,1implies(24)H281=Δ1−2+Δ2−4+Δ2−6+⋯+4−η4,1+2−η4,1,=14Δ12−2Δ1+η4,1.
LetΩm1,m228 be the mirror image of itself, then from Figure 42, we have m1−1=Δ1−m2−1, implying m2=Δ1−m1. Now, ∀m1∈l9,1, Δ1−m1∈l10,1, implying that Ωm1,Δ1−m128∈H281 is the mirror image of itself. So, out of 1/4Δ12−2Δ1+η4,1 homomorphic images in H281, Δ1−2−η4,1/2 are the mirror images of itself. Hence, there are(25)2×E282×H282−Δ1−2−η4,12+E282×Δ1−2−η4,12=3Δ12−3Δ1+2pairs of contracting vertices to produce all the homomorphic images in H281. We can prove this lemma for k4=4 (Figure 43) in similar way as that for k4=1.Figure 42
Homomorphic imagesΩl9,1,l10,128.Figure 43
Homomorphic imagesΩl9,4,l10,428.Lemma 29.
If we contract the verticese3l11,k2+1k2:l11,k2=1,2,…,Δk2−1, with the vertices e3l12,k2k2:l12,k2=1,2,…,Δk2−l11,k2, in the circuit Δ, then for each k2, there arise 1/6Δk2−1Δk2−2 distinct homomorphic images of Δ and 6 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all 1/2Δk2−1Δk2−2 and their mirror homomorphic images of Δ is 6Δk2−1Δk2−2.
For a fix value ofk2, let H29k2=Ωl11,k2,l12,k229;l11,k2=1,2,…,Δk2−1;l12,k2=1,2,…,Δk2−l11,k2 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l11,k2+1k2 with the vertices e3l12,k2k2 in Δ. Figures 44 and 45 present H29k2 graphically. From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 23.
Letk4=1,4 and η4,k4=0ifΔk4=0mod21otherwise.Figure 44
Homomorphic imagesΩl11,2,l12,229.Figure 45
Homomorphic imagesΩl11,3,l12,329.Lemma 30.
If we contract the verticese3l13,k4+1k4:l13,k4=1,2,…,Δk4−η4,k4/2, with the vertices e3l13,k4k4 in the circuit Δ, then for each k4, there arise 1/2Δk4−η4,k4 distinct homomorphic images of Δ and 3 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all 1/2Δk4−η4,k4 and their mirror homomorphic images of Δ is 3Δk4−1.
For a fix value ofk4, let H30k4=Ωl13,k4,l13,k430;l13,k4=1,2,…,Δk4−η4,k4/2 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l13,k4+1k4 with the vertices e3l13,k4k4 in Δ. Figures 46 and 47 present H30k4 graphically. From all the homomorphic images presented in these figures, it is not intricated to check that if Δk2=0mod2, then ΩΔk2/2,Δk2/230is the only homomorphic image which is the mirror image of itself; otherwise, no one is the mirror image of itself.
This lemma can be proved by using the same procedure as that for Lemma28.Figure 46
Homomorphic imagesΩl13,1,l13,130.Figure 47
Homomorphic imagesΩl13,4,l13,430.Lemma 31.
If we contract the verticese3l11,k2+1k2:l11,k2=1,2,…,Δk2−1, with the vertices e3l11,k2k2 in the circuit Δ, then for each k2, there arise Δk2−1 distinct homomorphic images of Δ and 3 pairs of contracted vertices create each homomorphic image. Moreover, the number of total pairs of contracted vertices to generate all Δk2−1 and their mirror homomorphic images of Δ is 6Δk2−1.
For a fix value ofk2, let H31k2=Ωl11,k2,l11,k231;l11,k2=1,2,…,Δk2−1 be the collection of homomorphic images of Δ acquired by the contraction of the vertices e3l11,k2+1k2 with the vertices e3l11,k2k2 in Δ. Figures 48 and 49 present H31k2 graphically. From all the homomorphic images presented in these figures, it is not intricated to check that no one is the mirror image of itself. This lemma can be proved by using same procedure as that for Lemma 23.
Letk5=2,4.Figure 48
Homomorphic imagesΩl11,2,l11,231.Figure 49
Homomorphic imagesΩl11,3,l11,331.Lemma 32.
For eachk5, there are 6 Δk5+Δk5+1+2 pairs of contracting vertices to produce the homomorphic image of Δ by contracting the vertex e11 with the vertex e1k5.
Fork5=2, let Ω232 be the homomorphic image of Δ acquired by contracting the vertex e11 with the vertex e12 in Δ, where e11 and e12 are fixed by the words pq−1Δ6pqΔ5pq−1Δ4pqΔ3pq−1Δ2pqΔ1 and pq−1Δ3pqΔ4pq−1Δ5pqΔ6pq−1Δ1pqΔ2. It is easy to verify that P1=pq−1Δ6pqΔ5pq−1Δ4pqΔ3p and P2=q−1pq−1Δ2−1pqΔ1 are the possible paths between e11 and e12 (Figure 4). This implies that the homomorphic image Ω232 has a vertex s fixed by P1 and P2. Thus,(26)E232=e,q,q−1,p,pq,pq−11,pq−11p,pq−11pq,pq−12,…,pq−1Δ3p,pq−1Δ3pq,pq−1Δ3+1,q−1p1,q−1p1q,q−1p1q−1,…,q−1pΔ2,q−1pΔ2q,q−1pΔ2q−1is the family of elements in PSL2,Z such that ∀w∈E232,Δ contains the vertices e11w and e21w. This gives that the cardinality of E232, that is, 3Δ2+Δ3+2 is the number of contracted pairs of vertices to produce the homomorphic image Ω232 (Theorem 2). From Figure 50, it is not intricated to check that Ω232 and its mirror image Ω232∗ are not the same. Hence, there are(27)2×E322=3Δ2+Δ3+2,pairs of contracting vertices to produce the homomorphic images Ω232.
We can prove this lemma fork5=4 (Figure 51) in similar way as that for k5=2.Figure 50
Homomorphic imageΩ232.Figure 51
Homomorphic imageΩ432.Lemma 33.
For eachk3, there are 6 Δ4+Δ5+2 pairs of contracting vertices to produce the homomorphic image of Δ by contracting the vertex e15 with the vertex e1k3.
For a fix value ofk3, let Ωk333 be the homomorphic image of Δ acquired by the contraction of the vertex e15 with the vertex e1k3 in Δ. Figures 52 and 53 present Ωk333 graphically. From these figures, it is not intricated to check that Ωk333 is not the mirror image of itself.
This lemma can be proved by using the same procedure as that for Lemma32.
Letk6=1,2,5 and k6¯=6ifk6=1,3ifk6=2,4ifk6=5..Figure 52
Homomorphic imageΩ133.Figure 53
Homomorphic imageΩ233.Lemma 34.
For eachk6, there are 6 Δk6¯+1 pairs of contracting vertices to produce the homomorphic image of Δ by contracting the vertex e1k6 with the vertex e1k6¯.
For a fix value ofk6, let Ωk634 be the homomorphic image of Δ acquired by the contraction of the vertex e1k6 with the vertex e1k6¯ in Δ. Figures 54, 55, and 56 present Ωk634 graphically. From these figures, it is not intricated to check that Ωk634 is not the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 32.Figure 54
Homomorphic imageΩ134.Figure 55
Homomorphic imageΩ234.Figure 56
Homomorphic imageΩ534.Lemma 35.
For eachk3, there are 3 Δk3+1+2Δk3+2+2 pairs of contracting vertices to produce the homomorphic image of Δ by contracting the vertex e1k3 with the vertex e1k3+2.
For a fix value ofk3, let Ωk335 be the homomorphic image of Δ acquired by the contraction of the vertex e1k3 with the vertex e1k3+2 in Δ. Figures 57(a), 57(b), and 57(c) present Ω135 and Figure 58 presents Ω235 graphically. From these figures, it is not intricated to check that Ωk335 is the mirror image of itself. This lemma can be proved by using the same procedure as that for Lemma 32.
LetVΔ×Δ=a,b;a,b∈VΔanda≠b, then(28)VΔ×Δ=129Δ1+2Δ2+2Δ3+Δ4−3Δ1+2Δ2+2Δ3+Δ4⋅
We are now able to prove our primary outcome. Let(29)η=4,ifΔ1,Δ4=E,E,0,ifΔ1,Δ4=O,O,−1,otherwise.where E and O stand for even and odd, respectively.Figure 57
(a) Homomorphic imageΩ135: Δ1>Δ3+Δ5+1. (b) Homomorphic image Ω135: Δ1=Δ3+Δ5+1. (c) Homomorphic image Ω135: Δ1<Δ3+Δ5+1.
(a)(b)(c)Figure 58
Homomorphic imageΩ235.Theorem 5.
There are1/4Δ1+2Δ2+2Δ3+Δ42+4Δ1+2Δ2+2Δ3+Δ4+4Δ4+η number of distinct homomorphic images acquired by the contraction of all pairs of vertices of the circuit Δ.Proof.
Let us collect all the pairs of contracting vertices ofΔ mentioned in Lemma 1 to Lemma 35 in the form of set S as(30)S=e3l1+11,e3Δ11,e3l1+12,e3Δ11,e3l2,k1+1k1,e3Δ11,e3l3+16,e3Δ11,e3l4+11,e3Δ22,e3l5+12,e3Δ22,e3l6,k1+1k1,e3Δ22,e3l7+13,e3Δ33,e3l7+14,e3Δ33,e3l8+14,e3Δ44,e3l9+15,e3Δ44,e3l10+13,e3Δ44,e3l1+12,e3Δ55,e3l11+13,e3Δ55,e3l7+14,e3Δ55,e3l12+15,e3Δ55,e3l13+13,e3Δ66,e3l14+11,e3Δ11,e3l15+11,e3Δ33,e3l3,k2+1k2,e3Δ33,e3l16+12,e3Δ44,e3l4,k3+1k3,e3Δ66,e3l4+11,e3l5,k2k2,e3l4+11,e3l174,e3l5+12,e3l6,k1k1,e3l13+13,e3l74,e3l7,k2+1k2,e3l8,k2k¯2,e3l7,k4+1k4,e3l10,k4k4,e3l11,k2+1k2,e3l12,k2k2,e3l13,k4+1k4,e3l13,k4k4,e3l11,k2+1k2,e3l11,k2k2,e1k5,e11,e1k3,e15,e1k6,e1k¯6,e1k3,e1k3+2.
LetH be the set of homomorphic images obtained by contracting each element of S, then(31)H=Ωl11,Ωl12,Ωl2,k13,Ωl34,Ωl45,Ωl56,Ωl6,k17,Ωl78,Ωl79,Ωl810,Ωl911,Ωl1012,Ωl113,Ωl1114,Ωl715,Ωl1216,Ωl1317,Ωl1418,Ωl1519,Ωl3,k220,Ωl1621,Ωl4,k322,Ωl4,l5,k223,Ωl4,l1724,Ωl5,l6,k125,Ωl13,l726,Ωl7,k2,l8,k227,Ωl9,k4,l10,k428,Ωl11,k2,l12,k229,Ωl13,k4,l13,k430,Ωl11,k2,l11,k231,Ωk532,Ωk333,Ωk634,Ωk335⋅
Let S′ be the sum of contracted pairs obtained by contracting each element of S, then(32)S′=3Δ22+3Δ2−2+3Δ22+3Δ2+32Δ62+3Δ6+32Δ12+3Δ1−4+3Δ22+3Δ2−4+3Δ42+3Δ4−4+32Δ42+3Δ4−4+3Δ42+3Δ4−2+32Δ52+3Δ5+3Δ32+3Δ3+32Δ22+3Δ2+3Δ32+3Δ3−2+3Δ42+3Δ4−4+3Δ52+3Δ5+32Δ32+3Δ3−4+6Δ2+2Δ1−Δ2−1+3Δ4+2Δ1−Δ4−1+6Δ3+2Δ2−Δ3−1+6Δ1−1Δ4−1+12Δ3−1Δ4−1+∑k1=353Δk12+3Δk1+3Δk12+3Δk1−4+12Δ2−1Δk1−1+∑k2=236Δ4+2Δk2−Δ4−1+12Δ1−1Δk2−1+6Δk2−12+6Δk2−1Δk2−2+6Δk2−1+∑k3=126Δ5+2Δk3−Δ5−1+6Δ4+Δ5+2+3Δk3+1+2Δk3+2+2+∑k4=143Δk42−3Δk4+2+3Δk4−1+∑k5=2,46Δk5+Δk5+1+2+∑k6=1,2,56Δk6+1=129Δ1+2Δ2+2Δ3+Δ4−3Δ1+2Δ2+2Δ3+Δ4⋅shows that each element of VΔ×Δ is contracted.
Now,(33)H=Δ2+Δ2+Δ6+Δ1−1+Δ2−1+Δ4−1+Δ1−1+Δ4+Δ5+Δ3+Δ2+Δ3+Δ4−1+Δ5+Δ3−1+Δ1−Δ2−1+Δ1−Δ4−η12+Δ2−Δ3−1+12Δ1−1Δ4−1ifΔ4=1mod212Δ1−1Δ4−1+1−η3ifΔ4=0mod2+Δ3−1Δ4−1+∑k1=35Δk1+Δk1−1+Δ2−1Δk1−1+∑k1=35Δk2−Δ4−1+Δ1−1Δk2−1+12Δk2Δk2−1+Δk2−1+12Δk2−1Δk2−2+∑k3=12Δk3−Δ5−1+1+1+∑k4=1,414Δk42−2Δk4+Δ4,k4+12Δk4−η4,k4+∑k5=2,41+∑k6=1,2,51=14Δ1+2Δ2+2Δ3+Δ42+4Δ1+2Δ2+2Δ3+Δ4+4Δ4+η.
The value ofH gives the total number of homomorphic images produced in the contraction of all pairs of vertices of Δ. The value of S′ guaranteed that the set H contains all homomorphic images.


## 2. Conclusion

Thus, there are total1/29Δ1+2Δ2+2Δ3+Δ4−3Δ1+2Δ2+2Δ3+Δ4 number of elements in VΔ×Δ. To find all distinct homomorphic images, we do not need to contract each pair of vertices of VΔ×Δ. We have to contract only those pairs of vertices, which are in set S and they are 1/4Δ1+2Δ2+2Δ3+Δ42+4Δ1+2Δ2+2Δ3+Δ4+4Δ4+η in numbers because if we contract the pair which is not belong to set S, we attain the homomorphic image, which we already acquired by contracting the element of set S.


## 3. Applications of Homomorphic Copies in Lightweight Cryptography and Chemistry

The construction of any product by using minimum resources without compromising on quality is primary objective of scientists.The coset diagrams with 256 vertices are used in the construction of strong 8 × 8 S-boxes [9]. In present-day block ciphers, cryptographically secure S-boxes are designed to attain the requirements of Shannon's necessity for perplexity. Substitution boxes are the fundamental segments in numerous Feistel network-based block cryptosystems or substitution-permutation (S-P) networks. The use of 8 × 8 S-boxes in block cipher is excessively costly. So, it is not surprising that we are seeing strong progress in the field of lightweight cryptography in recent years, for instance PRESENT. Lightweight cryptographic calculations are utilized in business items, including DESL, PRINTCIPHER, SEA, HIGHT, PRESENT, LED, and KATAN/KTANTAN. For instance, Keeloq is a 32-bit block cipher frequently utilized in the automobile business. Digital signature transponder is a 40-bit block cipher, implemented in wireless authentication systems. Since a homomorphic copy of a graph is smaller graph with the same algebraic structure, it can be quite handy to generate small sized S-boxes by using homomorphic copies of the coset diagram having 256 vertices.Interrelation of certain types of coset graphs and structure of carbon allotropes not only highlights the connection but also improvises applications in many fields.Several forms of carbon can be found in nature. One of the most important allotropes of carbon is fullerene which was discovered in1985. Fullerenes are carbon-cage-like polyhedral molecules in which a large number of carbon atoms are bonded in a nearly spherical symmetric configuration. Fullerenes Cn can be drawn for n=20 and for all even n≥24. They have n carbon atoms, 3n/2 bonds, 12 pentagonal faces, and n/2−10 hexagonal faces. The most important member of the family of fullerenes is C60. Fullerenes can be classified in terms of their groups of symmetries. These groups are also known as point groups. Every element of a point group is an isometry of the Euclidian space and so it is either a rotation around an axis or it is a reflection in a plane. The list of all 28 fullerene point groups is: Ih, I, Th, Td, T, D6h, D6d, D6, D5h, D5d, D5, D3h, D3d, D3, D2h, D2d, D2, S6, S4, C3h, C2h, C3v, C3, C2v, C2, Cs, Ci, and C1 [10]. Fullerene C60 has icosahedral symmetry, that is, the symmetry group is isomorphic to A5. A constructive enumeration of fullerenes has been dealt in detail in [11]. The structure of coset graphs of fullerene C60, symmetry group PSL2,5 and adjacency matrix has been explored through the action of modular group in [12]. Another carbon allotrope with high permutational symmetry is allotrope D168 Schwarzite, proposed by Vanderbilt and Tersoff, which has an automorphism group of order 168. In [13], it has been shown that coset diagram for PSL2,7 points has interesting relation with carbon allotrope with negative curvature D168 Schwarzite. The future studies may extend the present study by investigating the homomorphic copies of the coset graphs for PSL2,5 and PSL2,7 and their related chemical structures.



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