Abstract
Coset diagrams [1, 2] are used to demonstrate the graphical representation of the action of the extended modular group over . In these sorts of graphs, a closed path of edges and triangles is known as a circuit, and a fragment is emerged by the connection of two or more circuits. The coset diagram evolves through the joining of these fragments. If one vertex of the circuit is fixed by , then this circuit is termed to be a length – circuit, denoted by . In this study, we consider two circuits of length as and with the vertical axis of symmetry that is and . It is supposed that is a fragment formed by joining and at a certain point. The condition for existence of a fragment is given in [3] in the form of a polynomial in . If we change the pair of vertices and connect them, then the resulting fragment and the fragment may coincide. In this article, we find the total number of distinct fragments by joining all the vertices of with the vertices of provided the condition .
1. Introduction
It is considered that is known as the Lobachevski plane model, the model of the upper-half plane of hyperbolic plane geometry. Then, the group of Mobius Transformations [4], with where and is the group of isometries that preserve the orientation in . It is isomorphic to the quotient group , which is called the projective special linear group. Geometrically, the group of isometries of is the action of on [5] with left action (faithful) as
The Mobius transformations of with coefficients from the set of integers form a group known as a discrete group [6], a subgroup of , symbolically written as , it is a quotient group of special linear group by its center .
It is eminent that the transformations of linear fractions and are used to generate , so-called the modular group, with presentation
By introducing a new generator with and , we obtain a group , the extension of , usingas a relation.
The coset diagrams present the action of on , where is a finite field and shows a prime power. These graphs have a long and rich history [7, 8]. Small triangles are proposed for the cycle , such that 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 . Heavy dots represent the fixed point of and . Note that, equals , that means reverses the triangle orientation proposed for the cycle . For that reason, the diagram need not be made more intricate by inserting – edges.
Definition 1. A coset diagram (subdiagram) is said to be a homomorphic image of the coset diagram (subdiagram) if and only if(i)(ii) with , where , there exist a vertex in such that (iii)-edges map to -edges(iv)-edges mapped to -edgesCoset diagrams obtained from the action of over are infinite graphs, where , whereas coset diagrams for the action of on presents finite graphs. The number is an expression of the number , where The finite coset diagrams are homomorphic images of the coset diagrams for , where for some .
To explain more, the coset diagram in the following (Figure 1), illustrate the action on by with permutation representations , and by , , and , respectively, asFor a comprehensive understanding of coset diagrams, we propose [9–13].
In the study of [3], it has been shown that for a fixed value of there are only a finite number of real quadratic irrational ambiguous numbers of the form and that part of the coset diagram containing ambiguous numbers forms a circuit and it is the only circuit in the orbit of .
In a coset diagram, a closed path of triangles and edges is called a circuit. A circuit is said to be of length –, denoted by , if its one vertex is fixed byAlternatively, it means that one vertex of the triangles lie outside of the circuit and one vertex of the triangles lies inside of the circuit and likewise. Since is a cycle, so it does not matter if one vertex of the triangles lie inner of the circuit and one vertex of the triangles lies outer of the circuit and likewise. Note that, is always even.
The circuit of the type is termed as a periodic circuit with the period of length .
For more on circuits in coset diagrams, we refer [14].
Consider two nonperiodic and simple circuits and . Let and be any vertices from and fixed by the words and from , respectively. To connect and at and , we arbitrarily choose the circuit and apply on in such a way that ends at Appropriately, a fragment (say) emerge, consisting a vertex fixed by the pair , .
Let shows itself as the mirror image of . Since the permutation ensures that the coset diagram is symmetric along the vertical axis. This implies will assuredly occur.
If is a word then If the word fixes the vertex , then the vertex is fixed by .
There are two components involves in the action of on and they are and . Let denote itself as the complement . In what follows, by , we shall mean a nonsimple fragment composed by connecting two nontrivial and nonperiodic circuits. Coset diagrams corresponding to the actions of on via a homomorphism with parameter are denoted by [3]. These diagrams are composed of fragments. There is a question that must revolve in minds when a fragment exists in . In [3], the response is found in the following way.
Theorem 1. Given a fragment , there is a polynomial in such that(i)if occurs in , then (ii)if then or occurs in or in How to calculate a polynomial from ? The answer is given in [3].
Let be a polynomial acquired from the fragment , which is emerged in the connection of two nonperiodic circuits. Then, there exists a homomorphic image of other than corresponding to each zero of in the appropriate coset diagrams. Thus, we are compromising with the fragments, which are set up by connecting a pair of nonperiodic circuits.
Remark 1. The direction of the triangles describing the three-cycles of completely changed by the action of (like as reflection). So if is a vertex of fixed by the pair , then obviously the pair fixed the vertex of . Since a vertical axis of symmetry possesses by , therefore if founds in , then also contains . So and have the same existence condition in implying that, they give a unique polynomial. There are specific fragments that admit a vertical axis of symmetry. In this case, the orientation of the mirror image is the same as that of the fragment. These types of fragments may have fixed points of . If is a couple of words fixing the vertex of the fragment , then the orientation of is same as that of if and only if there exists a vertex in which is fixed by a couple of words .
2. Pairs of Connecting Vertices
If a fragment emerged by connecting vertices and from and , respectively; then and are not the single pair of joined vertices but there are many (depends upon and ) pairs of vertices in and , that are connected. That is, a finite number of pairs of connected vertices results the same fragment.
Definition 2. Let , and , be the vertices in and such that , , , and are fixed by , , , and , respectively. Let be the fragment set up in the connection of with . Then, the pair of vertices is identical to the pair of vertices if and only if in the connection of with to produce , , and also, get connected with each other. If two pairs of vertices and are identical, then we write ∼ .
Let be a fragment formed by connecting the vertex fixed by in with the vertex fixed by in and denotes itself as the set of pairs of joining vertices that are identical to . Let be the collection of words such that for all , the vertices and lies on and , respectively.
The following theorems proved in [15] will help us to find all the pairs of joining vertices for the fragment acquired by the connection of and .
Theorem 2. For any , there is a pair of joining vertices in .
Theorem 3. Corresponding to each pair of joining vertices , there is a unique word such that , .
Theorem 4. There is a one-to-one correspondence between and .
3. Counting the Number of Pairs of Connecting Vertices for a Fragment
Each joining point gives a couple of words, which further ensures a polynomial. Since a polynomial acquired from a fragment is unique. Thus, for all pairs of joining vertices for a fragment, a unique polynomial is evolved. Therefore, all the pairs of joining vertices for a fragment must be identified.
Let us join the vertices and of and fixed by and , respectively, to acquire the fragment . Let , then there are at least pairs of joining vertices in and to obtain . Note that, is not the total number of pairs of joining vertices in and to compose .
Let be any two vertices from the circuit fixed by the words and that is and . Suppose is the word that maps to that is Note that, and are the only paths that assign to . By contraction of vertices and , we mean that and melt together to become one node such that . As a result of this contraction, a closed path is created that containing the vertex fixed by and . This closed path is the homomorphic image of the circuit . Note that, and is not the only pair of contraction in that creates homomorphic image . There are also many pairs of contraction other than and that create the same homomorphic image . It is therefore necessary to ask how many distinct homomorphic images are obtained if we contract all pairs of vertices of the circuit ? The answer to this question is given in [15]. The process to find the number of pairs of connecting vertices for a fragment is same as that the number of pairs of contracting vertices for a homomorphic image. To know, how many total pairs of joining vertices for a fragment, one has to be extra careful.(i)If the orientation of is different from , then the pair of words , does not fix any vertex in (Remark 1). That is, to produce , with the joining of and , the pair of vertices , is not connected. So, there are total pairs of joining vertices for and .(ii)If the orientation of is same as , then there exists a vertex in fixed by the pair of words , (Remark 1). That is, to produce , with the joining of and , the pair of vertices , is also connected. So, there are total pairs of joining vertices for and .
In the literature, the question that how many pairs of connecting vertices form the fragment is answered for the pairs of circuits of length-2 [15] and contracting vertices produce the homomorphic image is responded for the circuit of length-4 [11] under certain conditions. We have solved this problem for the pair of circuits of length 6.
4. Connection of Circuits
Consider two circuits of length-6 as (Figure 2) and (Figure 3) with the vertical axis of symmetry that is and and impose a condition
Let us connect and at a certain point and obtained a fragment . Since there are finitely many pairs of connecting vertices that build same fragment; therefore, it is not necessary that if we altered the pair of connecting vertices of and , we obtain a fragment other than . But it is necessary to inquire that how many distinct fragments are obtained by joining the circuits and at all pairs of connecting vertices. In this article, we will not only answer this question but also identify the pairs of connecting vertices of and that are important. At those pairs of connecting vertices that are not stated as important, and need not to connect because if we join and at such pairs, we attain fragments that we already obtained by connecting important pairs. Since and , therefore and have and number of vertices, respectively, implies that there are total pairs of connecting vertices of and . We join a vertex of with the vertex of and compose a fragment.
From Figures 2 and 3, for each ; and , we have(i) is mirror image of the vertex (ii) is the mirror image of the vertex (iii)The vertex is fixed by the word(iv)The vertex is fixed by the word
We take + at the place of if is odd and–if is even.
Before proving the results, we define some symbolic notations in the following:
Note. All lemma’s presented in the following, we take sign at the place of , if and are even, and otherwise.
Lemma 1. If we join , the vertex from , with the vertices from , then for a fix , there occurs number of distinct fragments and pairs of connecting vertices generate each (same) fragment. Furthermore, a total of pairs of connecting vertices produces all fragments and their mirror images.
Proof. For a fix , let us join , the vertex from , with the vertices from and attain a set of fragments (Figure 4), where the wordfixing the vertex and the vertex is fixed by the wordThen, the setcontains words such that if is any word from implies and lies on and , respectively. Thus, the fragment has pairs of connecting vertices in and . In other words, for each , the connection of and give the same fragment .
Now, we prove that all the elements in are distinct and no one is the mirror image of itself. For this, let and be any two fragments from . Then, is set up by connecting and and is set up by and . If , then there exists an element such that and . is the only word such that but . Thus, is disequivalent to , that is by joining with to produce , is not connected with . Now, if , then there exists an element such that and . But there does not such an element exist in . Thus, is disequivalent to , that is by joining with to produce , is not connected with . Hence, all the elements in are distinct. Since, implying that .
If , then there exists an element such that and . But there does not such an element exist in . Thus, is disequivalent to , that is by joining with to produce , is not connected with . Therefore, is orientally different from its mirror image . Alternatively, the vertical axis of symmetry does not possess by any of the elements in . Hence, there are totalpairs of connecting vertices to produce fragments in and their mirror images.
Lemma 2. If we join , the vertex from , with the vertices from , then for a fix , there occurs number of distinct fragments and pairs of connecting vertices generate each (same) fragment. Furthermore, a total of pairs of connecting vertices produces all fragments and their mirror images. (Figure 5).
We can prove Lemma 2 in a similar way as Lemma 1 by replacing , , , , , and by , , , , , and , respectively.
Lemma 3. If we join , the vertex from , with the vertices from , then for a fix , there occurs number of distinct fragments and pairs of connecting vertices generate each (same) fragment. Furthermore, a total of pairs of connecting vertices produces all fragments and their mirror images (Figure 6).
We can prove Lemma 3 in a similar way as Lemma 1 by replacing , , , , , and by , , , , , and , respectively.
Lemma 4. If we join , the vertex from , with the vertices from , then for a fix , there occurs number of distinct fragments and pairs of connecting vertices generate each (same) fragment. Furthermore, a total of pairs of connecting vertices produces all fragment and their mirror images (Figure 7).
We can prove Lemma 4 in a similar way as Lemma 1 by replacing , , , , , and by , , , , , and , respectively.
Lemma 5. If we join , the vertex from , with the vertices from , then for a fix , there occurs number of distinct fragments and pairs of connecting vertices generate each (same) fragment. Furthermore, a total of pairs of connecting vertices produces all fragments and their mirror images (Figure 8).
We can prove Lemma 5 in a similar way as Lemma 1 by replacing , , , , , and by , , , , , and , respectively.
Lemma 6. If we join , the vertex from , with the vertices from , then for a fix , there occurs number of distinct fragments and pairs of connecting vertices generate each (same) fragment. Furthermore, a total of pairs of connecting vertices produces all fragments and their mirror images (Figure 9).
We can prove Lemma 6 in a similar way as Lemma 1 by replacing , , , , , and by , , , , , and , respectively.
Lemma 7. If we join , the vertex from , with the vertices from , then for a fix , there occurs number of distinct fragments and pairs of connecting vertices generate each (same) fragment. Furthermore, a total of pairs of connecting vertices produces all fragments and their mirror images.
Proof. For a fix , let us join , the vertex from , with the vertices from and attain a set of fragments (Figure 10), where the wordfixing the vertex and the vertex is fixed by the wordThen, the setcontains words such that if is any word from implies and lies on and , respectively. Thus, the fragment has pairs of connecting vertices in and . In other words, for each , the connection of and give the same fragment .
Now, we prove that all the elements in are distinct and no one is the mirror image of itself. For this, let and be any two fragments from . Then, is set up by connecting and and is set up by and . If , then there exists an element such that and . is the only word such that but . Thus, is disequivalent to , that is by joining with to produce , is not connected with . Now, if , then there exists an element such that and . But there does not such an element exist in . Thus, is disequivalent to , that is by joining with to produce , is not connected with . Hence, all elements in are distinct. Since, implying that .
If , then there exists an element such that and . But there does not such an element exist in . Thus, is disequivalent to , that is by joining with to produce , is not connected with . Therefore, is orientally different from its mirror image . Alternatively, the vertical axis of symmetry does not possess by any of the elements in . Hence, there are totalpairs of connecting vertices to produce fragments in and their mirror images.
Lemma 8. If we join , the vertex from , with the vertices from , then for a fix , there occurs number of distinct fragments and pairs of connecting vertices generate each (same) fragment. Furthermore, a total of pairs of connecting vertices produces all fragments and their mirror images (Figure 11).
We can prove Lemma 8 in a similar way as Lemma 7 by replacing , , , , , and by , , , , , and , respectively.
Lemma 9. If we join , the vertex from , with the vertices from , then for a fix , there occurs number of distinct fragments and pairs of connecting vertices generate each (same) fragment. Furthermore, a total of pairs of connecting vertices produces all fragments and their mirror images (Figure 12).
We can prove Lemma 9 in a similar way as Lemma 7 by replacing , , , , , and by , , , , , and , respectively.
Lemma 10. If we join , the vertex from , with the vertices from , then for a fix , there occurs number of distinct fragments and pairs of connecting vertices generate each (same) fragment. Furthermore, a total of pairs of connecting vertices produces all fragments and their mirror images (Figure 13).
We can prove Lemma 10 in a similar way as Lemma 7 by replacing , , , , , and by , , , , , and , respectively.
Lemma 11. If we join , the vertex from , with the vertices from , then for a fix , there occurs number of distinct fragments and pairs of connecting vertices generate each (same) fragment. Furthermore, a total pairs of connecting vertices produces all fragments and their mirror images.
Proof. For a fix , let us join , the vertex from , with the vertices from and attain a set of fragments (Figure 14), where the wordfixing the vertex and the vertex is fixed by the wordThen, the setcontains words such that if is any word from implies and lies on and respectively. Thus, the fragment has pairs of connecting vertices in and . In other words, for each , the connection of and give the same fragment .
Now, we prove that (i) for , all the elements in are distinct and no one is the mirror image of itself and (ii) for , all the elements in are distinct and is the only fragment which is orientally the same as its mirror image. For this, let and be any two fragments from . Then, is set up by connecting and and is set up by and . If , then there exists an element such that and . is the only word such that but . Thus, is disequivalent to , that is by joining with to produce , is not connected with . Now, if then there exist an element such that and . is the only word such that but . This implies that for , is equivalent to , that is by joining with to produce , is connected with . So, the fragments and are the mirror images of each other if and only if and the fragment is the mirror image of itself if and only if , that is . Now,(1)if is odd, then implying that . This indicates that no fragment in is the mirror image of others. Hence, all the elements in are distinct. Since implying that Also, implies no fragment in is the mirror image of itself. Hence, there are pairs of connecting vertices to produce fragments in and their mirror images.(2)If is even, then , implying that . This indicates that no fragment in is the mirror image of others. Hence, all the elements in are distinct. Sinceimplying thatFor , we have implies is the mirror image of itself. Hence, there arepairs of connecting vertices to produce fragments in and their mirror images.
Lemma 12. If we join , the vertex from , with the vertices from , then for a fix , there occurs number of distinct fragments and pairs of connecting vertices generate each (same) fragment. Furthermore, a total of pairs of connecting vertices produces all fragments and their mirror images (Figure 15).
We can prove Lemma 12 in a similar way as Lemma 11 by replacing , , , , , and by , , , , , and , respectively
Lemma 13. If we join , the vertex from , with the vertices from , then for a fix , there occurs , number of distinct fragments and 6 pairs of connecting vertices generates each (same) fragment. Furthermore, a total of pairs of connecting vertices produces all these fragments and their mirror images.
Proof. For a fix , let us join , the vertex from , with the vertices from and attain a set of fragments (Figure 16), where the wordfixing the vertex and the vertex is fixed by the wordThen, the setcontains words such that if is any word from implies and lies on and , respectively. Thus, the fragment has pairs of connecting vertices in and . In other words, for each , the connection of and gives the same fragment .
Now, we prove that (i) for , all the elements in are distinct and no one is the mirror image of itself and (ii) for , is the mirror image of for all and is the mirror image of itself. Let, and be any two fragments from . Then, is set up by connecting and and is set up by and . If , , , then, there exists an element such that and . is the only word such that but . Thus, , is disequivalent to , , that is by joining with to produce , is not connected with .
If , , , then there exists an element such that and . is the only word such that and . This implies that for and , , , , that is by joining with to produce , get also connected with . So, the fragments and are the mirror image of each other if and only if and and the fragment is the mirror image of itself if and only if and , that is and . Now,(1)If is odd, then for all , gives . This indicates, no fragment in is the mirror image of others. Hence, all the elements in are distinct. Since, implying that Also, implies no fragment in is the mirror image of itself. Hence, there are pairs of connecting vertices to produce fragments in and their mirror images.(2)If is even, then only for , gives, for all , implying that . Hence, all the elements in are distinct. Since,implying thatNow,(a)If is odd, then implying that . So, no one is the mirror image of itself. Hence, there are pairs of connecting vertices to produce fragments in and their mirror images.(b)If is even, then implying that is orientally the same as its mirror image. Hence, there arepairs of connecting vertices to produce fragments in and their mirror images.
Lemma 14. If we join , the vertex from , with the vertices from , then for a fix , there occurs , number of distinct fragments and 6 pairs of connecting vertices generates each (same) fragment. Furthermore, a total of pairs of connecting vertices produces all these fragments and their mirror images (Figure 17).
We can prove Lemma 14 in the same way as Lemma 13
Lemma 15. If we join , the vertex from , with the vertices from , then for a fix , there occurs number of distinct fragments and pairs of connecting vertices generates each (same) fragment. Furthermore, a total of pairs of connecting vertices produces all these fragments and their mirror images.
Proof. For a fix , let us join , the vertex from , with the vertices from , and attain a set of fragments (Figure 18), where the wordfixing the vertex and the vertex is fixed by the wordThen, the setcontains words such that if is any word from implies and lies on and , respectively. Thus, the fragment has pairs of connecting vertices in and . In other words, for each , the connection of and gives the same fragment .
Now, we prove that all the elements in are distinct and no one is the mirror image of itself. Let and be any two fragments from . Then, is set up by connecting and and is set up by and . If , , , then there exists an element such that and . is the only word such that but . Thus, , is disequivalent to , that is by joining with to produce , is not connected with . Now, If , , , then there exists an element such that and . is the only word such that but . Thus, , is dis-equivalent to , that is by joining with to produce , is not connected with . Hence, all the fragments in are distinct. SinceImplying that .
Now, if , , , then, there exists an element such that and . is the only word such that but . Thus, , is dis-equivalent to , that is by joining with to produce , is not connected with . Therefore, is orientally different from its mirror image . Alternatively, the vertical axis of symmetry does not possess by any of the elements in . Hence there are totalpairs of connecting vertices to produce fragments in and their mirror images.
Lemma 16. If we join , the vertex from , with the vertices from , then for a fix , there occurs number of distinct fragments and pairs of connecting vertices generates each (same) fragment. Furthermore, a total of pairs of connecting vertices produces all these fragments and their mirror images (Figure 19).
We can prove Lemma 16 in a similar way as Lemma 15 by replacing , , , , and by , , , , and , respectively.
Lemma 17. If we join , the vertex from , with the vertices from , then for a fix , there occurs number of distinct fragments and pairs of connecting vertices generates each (same) fragment. Furthermore, a total of pairs of connecting vertices produces all these fragments and their mirror images (Figure 20).
We can prove Lemma 17 in a similar way as Lemma 15 by replacing , , , , , , and by , , , , , , and , respectively.
Lemma 18. If we join , the vertex from , with the vertices from , then for a fix , there occurs number of distinct fragments and pairs of connecting vertices generates each (same) fragment. Furthermore, a total of pairs of connecting vertices produces all these fragments and their mirror images (Figure 21).
We can prove Lemma 18 in a similar way as Lemma 15 by replacing , , , , , , and by , , , , , , and , respectively.
Lemma 19. For a fix , there are pairs of connecting vertices for the fragment and its mirror image by joining , the vertex from , with the vertex from .
Proof. For a fix , let us join , the vertex from , with the vertices from and attain a fragments (Figure 22), where the wordfixing the vertex and the vertex is fixed by the wordThen, the setcontains words such that if is any word from implies and lies on and , respectively. Thus, the fragment has pairs of connecting vertices in and . In other words, for each , the connection of and give the same fragment .
Now, we show that the fragment is orientally different from its mirror image . If , then there exists an element such that and . But there does not such an element exist in . Thus, is disequivalent to , that is by joining with to produce , is not connected with . Therefore, is orientally different from its mirror image . Hence, there are totalpairs of connecting vertices to produce fragment and its mirror images.
Lemma 20. For a fix , there are pairs of connecting vertices for the fragment and its mirror image by joining , the vertex from , with the vertex from (Figure 23).
We can prove Lemma 20 in a similar way as Lemma 19 by replacing , , , and by , , , and , respectively.
Lemma 21. For a fix , there are pairs of connecting vertices for the fragment and its mirror image by joining , the vertex from , with the vertex from (Figure 24).
We can prove Lemma 21 in a similar way as Lemma 19 by replacing , , , and by , , , and , respectively.
Now, to prove our main result, we define some symbolic notations as follows:where and stand for even and odd positive integers, respectively.
Let
Theorem 5. If we connect the circuit with the circuit at all pairs of vertices then the total number of distinct fragments are .
Proof. Let us collect all the pairs of connecting vertices of and mentioned in Lemma 1 to Lemma 21 in the form of set asLet be the set of fragments obtained by joining each element of set , thenThis impliesshows that each pair of vertices in is connected, where is the Cartesian product of the vertices of and .
Now,The value of gives the total number of distinct fragments produced in the connection of and at all pairs of vertices. The value of promised that the set contains all fragments obtained by the connection of and . A unique polynomial is obtained by a fragment, so there are polynomials obtained in the connection of and .
5. Conclusion
Since there are total pairs of vertices in . To find all the fragments, we do not need to connect each pair of . We have to join only those pairs of vertices, which are in the set and they arein numbers because if we connect the pair which do not belong to set , we will attain a fragment, which has already been acquired earlier by connecting the elements of set 𝑆.
Data Availability
No data were used to support this study.
Conflicts of Interest
The authors declare that they have no conflicts of interest.