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Solutions to some congruence equations via suborbital graphs
 Bahadır Özgür Güler^{1}Email author,
 Tuncay Kör^{1} and
 Zeynep Şanlı^{1}
 Received: 9 April 2016
 Accepted: 5 August 2016
 Published: 11 August 2016
Abstract
We relate the connection between the sizes of circuits in suborbital graph for the normalizer of \(\Gamma _0(m)\) in PSL(2,\(\mathbb {R}\)) and the congruence equations arising from related group action. We give a number theoretic result which says that all prime divisors of \(3u^2\mp 3u+1\) for any integer u must be congruent to \(1\pmod {3}\).
Keywords
 Normalizer
 Imprimitive action
 Suborbital graphs
Mathematics Subject Classification
 11F06
 20H10
 05C25
Background
It is known that the graph of a group provides a method by which a group can be visualized; in many cases it suggests an economical algebraic proof for a result and it gives same information but in a much more efficient way (Magnus et al. 1966). In this view, the idea of suborbital graph has been used mainly by finite group theorists.
After it was shown that this idea is also useful in the study of the modular group which is a finitely generated Fuchsian group (Jones et al. 1991), some other finitely generated groups have been studied by suborbital graphs (see Akbaş and Başkan 1996; Akbaş 2001; Akbaş et al. 2013; Beşenk et al. 2013; Deger et al. 2011; Güler et al. 2011, 2015; Kader et al. 2010; Kader and Güler 2013; Kesicioğlu et al. 2013; Keskin 2006; Kör et al. 2016). In most of them, it has been emphasized the connection between elliptic elements in group and circuits of the same order in graph closely related with the signature problem.

A shortest path in subgraphs can be expressed as a continued fraction (Jones et al. 1991);

A shortest path in trees of suborbital graphs is a special case of Pringsheim continued fraction (Deger et al. 2011);

The subgraph \(F_{1,2}\) can be defined as a new kind of continued fraction and any irrational numbers has a unique \(F_{1,2}\) expansion (Sarma et al. 2015);

The set of vertices of some suborbital graphs is strongly connected to the Fibonacci sequence (Akbaş et al. 2013).
The aim of this paper is to examine the action of the normalizer of \(\Gamma _0(m)\) which produce some congruence equations with solutions. Actually, the suborbital graphs of the normalizer were studied for some special cases (see Güler et al. 2011; Kader et al. 2010; Keskin 2006). In here, we take a different case that m will be of the form \(3p^2\) where p is prime number greater than or equal to 5.
Preliminaries
The action of Nor(m) on \(\hat{\mathbb {Q}}\)
Lemma 1
(Akbaş and Singerman 1992, Corollary 2) Let m has prime power decomposition \(2^{\alpha _1}\cdot 3^{\alpha _2}\cdot p_3^{\alpha _3}\cdots p_r^{\alpha _r}\) . Then Nor (m) acts transitively on \(\hat{\mathbb {Q}}\) if and only if \(\alpha _1\le 7\) , \(\alpha _2\le 3\) and \(\alpha _i\le 1\) for \(i=3,\dots ,r\).
Corollary 2
The action of the normalizer \(Nor(3p^2)\) is not transitive on \(\hat{\mathbb {Q}}\).
Proof
Since m is taken as the aforementioned case, the result is obvious by Lemma 1.
In this case, we will find a maximal subset of \(\hat{\mathbb {Q}}\) on which the normalizer acts transitively. Since \(\Gamma _0(m)\subset Nor(m)\), we give more special case before our desired result to understand the situation better. We now give a Lemma as follows.
Lemma 3
(Akbaş and Başkan 1996, Theorem 4.1) Given an arbitrary rational number k / s with \((k,s)=1\) , then there exists an element \(A\in \Gamma _0(m)\) such that \(A(k/s)=(k_1/s_1)\) with \(s_1m\).
The following known Theorem is also proved in the same paper. We will present a different proof for the sake of completeness.
Lemma 4
(Akbaş and Başkan 1996, Theorem 4.3) Let bm and let \((a_1,b)=(a_2,b)=1\) . Then \(\left( \begin{array}{cc} a_1 \\ b \end{array} \right)\) and \(\left( \begin{array}{cc} a_2 \\ b \end{array} \right)\) are conjugate under the action of \(\Gamma _0(m)\) if and only if \(a_1\equiv a_2\pmod {t}\) , where \(t=\left( b,\frac{m}{b}\right)\).
Proof
Lemma 5
(Güler et al. 2011, Corollary 2.4) Let bm. Then the orbit \(\left( {\begin{array}{c}a\\ b\end{array}}\right)\) of a/b under the action of \(\Gamma _0(m)\) is the set \(\Bigl \{x/y\in \hat{\mathbb {Q}}:(m,y)=b,a\equiv x\frac{y}{b}\pmod {\left( b,\frac{m}{b}\right) } \Bigr \}\) . Furthermore the number of orbits is \(\varphi \left( b,\frac{m}{b}\right)\) where \(\varphi (n)\) is Euler’s totient function which is the number of positive integers less than or equal to n that are coprime to n.
Proof
Lemma 3 and 4 complete the proof.
From the above we come to the following conclusion.
Corollary 6
Proof
Let us denote the representatives of the orbits by \(\left( \begin{matrix} a \\ b\end{matrix}\right)\) as above. The possible values of b are \(1, 3, p, 3p, p^{2}, 3p^{2}\) by Lemma 3. Hence, the number of nonconjugate classes of these orbits with Euler formula are 1 and \(p1\) for \(1, 3, p^{2}, 3p^{2}\) and p, 3p respectively. By Lemma 5, the result is obvious \(\square\)
Theorem 7
 (1)(a) If \(3\not \mid l\) and \(l\not \equiv p\pmod {3}\),(b) If \(3\not \mid l\) and \(l\equiv p\pmod {3}\),$$\begin{aligned} \left( {\begin{array}{c}l\\ p\end{array}}\right) \cup \left( {\begin{array}{c}pl\\ p\end{array}}\right) \cup \left( {\begin{array}{c}l\\ 3p\end{array}}\right) \cup \left( {\begin{array}{c}pl\\ 3p\end{array}}\right) \end{aligned}$$$$\begin{aligned} \left( {\begin{array}{c}l\\ p\end{array}}\right) \cup \left( {\begin{array}{c}pl\\ p\end{array}}\right) \cup \left( {\begin{array}{c}l\\ 3p\end{array}}\right) \cup \left( {\begin{array}{c}2pl\\ 3p\end{array}}\right) \end{aligned}$$
 (2)If \(3\mid l\) , then$$\begin{aligned} \left( {\begin{array}{c}l\\ p\end{array}}\right) \cup \left( {\begin{array}{c}pl\\ p\end{array}}\right) \cup \left( {\begin{array}{c}p+l\\ 3p\end{array}}\right) \cup \left( {\begin{array}{c}pl\\ 3p\end{array}}\right) \end{aligned}$$
 (3)$$\begin{aligned} \left( {\begin{array}{c}1\\ 1\end{array}}\right) \cup \left( {\begin{array}{c}1\\ 3\end{array}}\right) \cup \left( {\begin{array}{c}1\\ p^2\end{array}}\right) \cup \left( {\begin{array}{c}1\\ 3p^2\end{array}}\right) \end{aligned}$$
Proof
 (i)
If \(e=1\), then \(T\genfrac(){0.0pt}0{l}{p}=\genfrac(){0.0pt}0{l}{p}\).
 (ii)
If \(e=3\), then \(T\genfrac(){0.0pt}0{l}{p}=\left( \begin{matrix} 3a &\quad b \\ 3p^2c &\quad 3d\end{matrix}\right) \genfrac(){0.0pt}0{l}{p}=\frac{3al+bp}{3p^2cl+3dp}\). Since \(det\left( \begin{matrix} 3a &\quad b \\ p^2c &\quad d\end{matrix}\right) =1\), then \((3al+bp,p^2cl+3d)=1\). So \(\frac{3al+bp}{3p(pcl+d)}\in \left( {\begin{array}{c}x\\ 3p\end{array}}\right)\) and \(x\equiv (3al+bp)(pcl+d)\pmod {p}\). As \(detT=3\), then \(x\equiv l\pmod {p}\). Consequently \(\genfrac(){0.0pt}0{l}{p}\cup \genfrac(){0.0pt}0{l}{3p}\).
 (iii)
If \(e=p^2\), then \(T\genfrac(){0.0pt}0{l}{p}=\frac{apl+b}{p(3cl+dp)}\in \left( {\begin{array}{c}x\\ p\end{array}}\right)\). As \(detT=p^2\), then \(x\equiv pl\pmod {p}\). Consequently \(\genfrac(){0.0pt}0{l}{p}\cup \genfrac(){0.0pt}0{pl}{p}\).
 (iv)
If \(e=3p^2\), then \(T\genfrac(){0.0pt}0{l}{p}=\frac{3apl+b}{3p(cl+3dp)}\in \left( {\begin{array}{c}x\\ 3p\end{array}}\right)\). So \(x\equiv l\equiv pl\pmod {p}\). (i), (ii), (iii) and (iv) complete the proof
Corollary 8
\(\hat{\mathbb {Q}}(3p^2)=\left( {\begin{array}{c}1\\ 1\end{array}}\right) \cup \left( {\begin{array}{c}1\\ 3\end{array}}\right) \cup \left( {\begin{array}{c}1\\ p^2\end{array}}\right) \cup \left( {\begin{array}{c}1\\ 3p^2\end{array}}\right)\) is the maximal subset of \(\hat{\mathbb {Q}}\) on which the normalizer \(Nor(3p^2)\) acts transitively.
Lemma 9
The stabilizer of a point in \(\hat{\mathbb {Q}}(3p^2)\) is an infinite cyclic group.
Proof
Because of the transitive action, stabilizers of any two points are conjugate. So it is enough to look at just \(\infty =\frac{1}{0}\in \left( {\begin{array}{c}1\\ 3p^2\end{array}}\right)\). As \(T\genfrac(){0.0pt}0{1}{0}=\left( \begin{matrix} ae &\quad b \\ 3p^2c &\quad de\end{matrix}\right) \genfrac(){0.0pt}0{1}{0}=\frac{ae}{3p^2c}=\genfrac(){0.0pt}0{1}{0}\), then \(c=0\) and \(e=1\). From the determinant equality, \(T=\left( \begin{matrix} 1 &\quad b \\ 0 &\quad 1\end{matrix}\right)\). Consequently \((Nor(3p^2))_\infty =\left\langle \left( \begin{matrix} 1 &\quad 1 \\ 0 &\quad 1\end{matrix}\right) \right\rangle\).
Now we consider the imprimitivity of the action of \(Nor(3p^2)\) on \(\hat{\mathbb {Q}}(3p^2)\), beginning with a general discussion of primitivity of permutation groups.
Let \(\left( G,\Omega \right)\) be a transitive permutation group, consisting of a group G acting on a set \(\Omega\) transitively. An equivalence relation \(\approx\) on \(\Omega\) is called Ginvariant if, whenever \(\alpha ,\beta \in \Omega\) satisfy \(\alpha \approx \beta\), then \(g(\alpha )\approx g(\beta )\) for all \(g\in G.\) The equivalence classes are called blocks.
 (i)
the identity relation, \(\alpha \approx \beta\) if and only if \(\alpha =\beta\);
 (ii)
the universal relation, \(\alpha \approx \beta\) for all \(\alpha ,\beta \in \Omega\).
Lemma 10
(Biggs and White 1979, Theorem 1.6.5) Let \(\left( G,\Omega \right)\) be a transitive permutation group. Then \(\left( G,\Omega \right)\) is primitive if and only if \(G_{\alpha },\) the stabilizer of \(\alpha \in \Omega\) , is a maximal subgroup of G for each \(\alpha \in \Omega\).
Lemma 11
Theorem 12
Proof
The suborbital graph of \(Nor(3p^2)\) and \(\hat{\mathbb {Q}}(3p^2)\)
Sims (1967) introduced the idea of the suborbital graphs of a permutation group G acting on a set \(\Delta\) , these are graphs with vertexset \(\Delta\), on which G induces automorphisms. We summarize Sims’theory as follows: Let \((G,\Delta )\) be transitive permutation group. Then G acts on \(\Delta \times \Delta\) by \(g(\alpha ,\beta )=(g(\alpha ),g(\beta )) (g\in G,\alpha ,\beta \in \Delta )\). The orbits of this action are called suborbitals of G. The orbit containing \((\alpha ,\beta )\) is denoted by \(O(\alpha ,\beta )\). From \(O(\alpha ,\beta )\) we can form a suborbital graph \(G(\alpha ,\beta ):\) its vertices are the elements of \(\Delta\), and there is a directed edge from \(\gamma\) to \(\delta\) if \((\gamma ,\delta )\in O(\alpha ,\beta )\). A directed edge from \(\gamma\) to \(\delta\) is denoted by \((\gamma \rightarrow \delta )\). If \((\gamma ,\delta )\in O(\alpha ,\beta )\), then we will say that there exists an edge \((\gamma \rightarrow \delta )\) in \(G(\alpha ,\beta )\).
If \(\alpha =\beta\), the corresponding suborbital graph \(G(\alpha ,\alpha )\), called the trivial suborbital graph, is selfpaired: it consists of a loop based at each vertex \(\alpha \in \Delta\). By a circuit of length m (or a closed edge path), we mean a sequence \(\nu _1 \rightarrow \nu _2 \rightarrow \dots \rightarrow \nu _m \rightarrow \nu _1\) such that \(\nu _{i}\ne \nu _{j}\) for \(i\ne j\), where \(m\ge 3\). If \(m=3, 4\) and 6, then the circuit is called a triangle, a quadrilateral and a hexagon, respectively.
We now investigate the suborbital graphs for the action of \(Nor(3p^2)\) on \(\hat{\mathbb {Q}}(3p^2)\). Since the action of \(Nor(3p^2)\) on \(\hat{\mathbb {Q}}(3p^2)\) is transitive, \(Nor(3p^2)\) permutes the blocks transitively; so the subgraphs are all isomorphic. Hence it is sufficent to study with only one block. On the other hand, it is clear that each nontrivial suborbital graph contains a pair (\(\infty ,u/p^2\)) for some \(u/p^2\in \hat{\mathbb {Q}}(3p^2)\). We let \(F(\infty ,u/p^2)\) be the subgraph of \(G(\infty ,u/p^2)\) whose vertices form the block \([\infty ]=\left( \begin{matrix} 1\\ p^{2}\end{matrix}\right) \cup \left( \begin{matrix} 1\\ 3p^{2}\end{matrix}\right)\), so that \(G(\infty ,u/p^2)\) consists of two disjoint copies of \(F(\infty ,u/p^2)\).
Theorem 13
 (i)
If \(p^2s\) but \(3p^2\not \mid s\), then \(x\equiv \mp 3ur\pmod {p^2}, y\equiv \mp 3us\pmod {3p^2}\), \(rysx=\mp p^2\)
 (ii)
If \(3p^2s\) then \(x\equiv \mp ur\pmod {p^2}, y\equiv \mp us\pmod {p^2}, rysx=\mp p^2\)
Proof
In the opposite direction we do calculations only for (i)(a). The others are likewise done. So suppose \(x\equiv 3ur\pmod {p^2}\), \(y\equiv 3us\pmod {3p^2}\), \(rysx=p^2\), \(p^2s\) and \(3p^2\not \mid s\). Therefore there exist b, d in \(\mathbb {Z}\) such that \(x=3ur+p^2b\) and \(y=3su+3p^2d\). Since \(rysx=p^2\), we get \(3rdbs=1\), or \(9rd3bs=3\). Hence the element \(T:=\genfrac(){0.0pt}0{3r \quad b}{3s \quad 3d}\) is not only in the normalizer \(Nor(3p^2)\), but also in H. It is obvious that \(T(\infty )=\genfrac(){0.0pt}0{r}{s}\) and \(T\genfrac(){0.0pt}0{u}{p^2}=\genfrac(){0.0pt}0{x}{y}\). \(\square\)
Farey graph and subgraph \(F(\infty ,u/p^2)\)
Now, let us represent the edges of \(F(\infty ,u/p^2)\) as hyperbolic geodesics in the upper halfplane \(\mathbb {H}\), that is, as Euclidean semicircles or halflines perpendicular to real line as in Jones and Singerman (1987). To understand the situation better, we give the Farey graph and some its properties as follows:
Definition 14
The Farey graph, denoted by F, is defined as : the vertex \(\infty\) is joined to the integers, while two rational numbers r/s and x/y (in reduced form) are adjacent in F if and only if \(r/sx/y=\mp 1\), or equivalently if they are consecutive terms in some Farey sequence \(F_{m}\) (consisting of the rationals x/y with \(y\le m\), arranged in increasing order). See also Fig. 1.
Lemma 15
(Jones et al. 1991, Corollary 4.2) No edges of F cross in \(\mathbb {H}\).
Similar result can be given by both of following useful Lemma and Theorem 13 as in (Jones et al. 1991);
Lemma 16
Let r/s and x/y be rational numbers such that \(r/sx/y=1\) , where \(s\ge 1\), \(y\ge 1\) . Then there exist no integers between r/s and x/y.
Proof
Let k be an integer such that \(r/s<k<x/y\). Then \(r<sk\) and \(x>ky\). Thus \(1=sxry>sxsky=s(xky)\ge s\), which is a contradiction.
Theorem 17
No edges of the subgraph \(F(\infty ,u/p^2)\) of \(Nor(3p^2)\) cross in \(\mathbb {H}\).
Proof
Without loss of generality, because of the transitive action, we can take the edges \(\infty \rightarrow \frac{u}{p^2}\), \(\frac{x_1}{y_1p^2}\rightarrow \frac{x_2}{y_2p^2}\) and \(\frac{x_1}{y_1p^2}<\frac{u}{p^2}< \frac{x_2}{y_2p^2}\), where all letters are positive integers. It is easily seen that \(x_1y_2p^2x_2y_1p^2=p^2\) by Theorem 13. \(\frac{x_1}{y_1}<u< \frac{x_2}{y_2}\) and Lemma 16 complete the proof.
Results
Theorem 18
\(F(\infty ,u/p^2)\) has a selfpaired edge iff \(3u^2\equiv 1\pmod {p^2}\).
Proof
Because of the transitive action, the form of selfpaired edge can be taken of \(1/0\rightarrow u/p^2\rightarrow 1/0\). The condition follows immediately from the second edge by Theorem 13.
Theorem 19
\(F(\infty ,u/p^2)\) has no triangle or quadrilateral.
Proof
We suppose that it has a triangle. Because of the transitive action, it must be of the form \(1/0\rightarrow u/p^2\rightarrow x/3p^2y\rightarrow 1/0\). But this contradicts Theorem 13 which says that both denominators of vertices of an edge having the form \(\frac{r}{s}\rightarrow \frac{x}{y}\) are not divisible by 3 at the same time. We now suppose that it has a quadrilateral. It must be of the form \(1/0\rightarrow u/p^2\rightarrow x/3p^2y\rightarrow k/p^2\rightarrow 1/0\) by same reason. From second, third and fourth edges by Theorem 13, we have the equations; \(3uyx=1\), \(x3ky=1\) and \(1\equiv 3uk\pmod {p^2}\). Therefore we obtain a contradiction 32.
Theorem 20
If \(3u^2\mp 3u+1 \equiv 0\pmod {p^2}\), \(F(\infty ,u/p^2)\) has a hexagon.
Proof
Theorem 21
\(H_0(3p^2)\) contains an elliptic element \(\varphi\) of order 6 if and only if \(F(\infty ,u/p^2)\) contains a hexagon.
Proof
Taking into account (3), we suppose that \(\varphi =\left( \begin{array}{cc} {3a} &\quad {b} \\ {3p^2c} &\quad {3d} \end{array}\right)\) is an elliptic element of order 6. It is known that \(a+d=\pm 1\) for order 3, 4, 6. Since \(det=3\), we have \(3a(\pm 1a)\equiv 1\)(mod\(p^2\)), that is \(3a^2\mp 3a+1\equiv 1\)(mod\(p^2\)). As \((a,n)=1\), \(F(\infty ,u/p^2)\) contains a hexagon by above Theorem.
Lemma 22
(Akbaş and Singerman 1990, Theorem 2) The periods of elliptic elements of Nor (m) may be 2, 3, 4, 6. Nor( m) has at most one period of order 6. It has a period of order 6 iff \(3\Vert m/h^2\) and if p is an odd prime divisor of \(m/h^2\) then \(p\equiv 1\pmod {3}\).
Corollary 23
The prime divisors p of \(3u^2\mp 3u+1\) , for any \(u\in \mathbb {Z}\) , are of the form \(p\equiv 1\pmod {3}\).
Proof
Let p a prime number and a divisor of \(3u^2\mp 3u+1\) for any integer u. In this case, it is clear that Nor(3p) has the elliptic element \(\left( \begin{array}{cc} 3u &\quad (3u^2\mp 3u+1)/p \\ 3p &\quad 3u+3\end{array} \right)\) of order 6 as in \(Nor(3p^2)\). We get \(p\equiv 1\pmod {3}\) by above Lemma.
Conclusions
Because this work combine different fields of mathematics such as algebra, geometry, group theory and number theory, it can be seen as an example of multidisciplinary approach which offer a new understanding of some situations. We show that we can produce solutions for some number theoretic problems using finite group theory once again. Taking into account the conjecture (Güler et al. 2011) which is also confirmed for the simplest hexagonal case within nontransitive cases by this paper, the normalizer has a potential to suggest solutions for other congruence equations such as \(8u^2\mp 4u+1\equiv 0\pmod {p}\), \(9u^2\mp 3u+1\equiv 0\pmod {p}\), \(27u^2\mp 9u+1\equiv 0\pmod {p}\) etc.
Declarations
Authors’ contributions
BÖG, TK, ZŞ completed the paper together. All authors read and approved the final manuscript.
Acknowledgements
We would like to express our sincere gratitude to Professor M. Akbaş for his immense help during the preparation of this paper. The authors are also thankful to the anonymous referees for the valuable suggestions towards the improvement of this manuscript.
Competing interests
The authors declare that they have no competing interests.
Open AccessThis article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Authors’ Affiliations
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