National University of Computer and EmergingSciences Lahore Campus

Department of Sciences and Humanities

Power Digraphs in Number Theory

Submitted byUzma Ahmad, MPhil (Mathematics)

In partial fulfillment of the requirements for the degree of

Doctor of Philosophy(2013)

Supervised by: Dr. S. M. Husnine

Author’s Declaration

I, Uzma Ahmad, declare that the work in this dissertation has been carriedout in accordance with the Regulations of the National University of Computerand Emerging Sciences. The work is original except where indicated by specialreferences in the text and no part of the dissertation has been submitted for anyother degree. The dissertation has not been presented to any other University forexamination.

Dated: March 26, 2013

Author:Uzma Ahmad

Plagiarism Undertaking

I, Uzma Ahmad, solemnly declare that the research work presented in thePhD thesis titled “Power Digraphs in Number Theory” has been carried out solelyby myself with no significant help from any other person except few of those whichare duly acknowledged. I confirm that no portion of my thesis has been plagiarizedand any material used in the thesis from other sources is properly referenced.

Dated: March 26, 2013

Author:Uzma Ahmad

To my husband and parents

Abstract

The modular exponentiation is considered to be one of the renowned problems innumber theory and is of paramount importance in the field of cryptography. Nowa days many security systems are based on powerful cryptographic algorithms.Most of them are designed by using the exponentiation xk ≡ y (mod n) as in RSA,Diffie- Hellman key exchange, Pseudo-random number generators etc. For the lasttwo decades, this problem is being studied by associating the power digraphs withmodular exponentiation. For the fixed values of n and k, a power digraph G(n, k) isformed by taking Zn as the set of vertices and the directed edges (x, y) from x to yif xk ≡ y (mod n) for the vertices x and y. These digraphs make a novel connectionbetween three disciplines of discrete mathematics namely number theory, graphtheory and cryptography. The objective of this dissertation is to generalize theresults on symmetry, heights, isolated fixed points, the number of componentsof a power digraph and the primality of Fermat numbers. To obtain the desiredgoal, a power digraph is decomposed into the direct product of smaller powerdigraphs by using the Chinese Remainder Theorem. The method of eliminationis adopted to discard those values of n and k which do not provide desired results.During the entire course of research, the Carmichael lambda-function λ(n) is usedfor developing the relations between the properties of a power digraph and theparameters n, k. For any prime divisor p of n, the concept of equivalence classeshas been used to discuss the symmetry of order p of G(n, k). The general rules todetermine the heights are formulated by comparing the prime factorizations of k,λ(n) and the orders of vertices. Some necessary and sufficient conditions for theexistence of symmetric power digraphs G(n, k), where n = pαq1q2 · · · qm such thatp, qi are distinct primes and α > 1, of order p are established. Explicit formulae forthe determination of the heights of the vertices and components of a power digraphin terms of n, k, λ(n) and the orders of vertices are formulated. An expression for thenumber of vertices at a specific height is established. The power digraphs in whicheach vertex of indegree 0 of a certain subdigraph is at height q ≥ 1 are characterized.The necessary and sufficient conditions on n and k for a digraph to have at least oneisolated fixed point are obtained. The work ends with the complete classification ofthe power digraphs with exactly two components.

Acknowledgements

My deepest gratitude is for the Almighty Allah for His blessings at every level ofmy life. I feel immense pleasure for conveying my heartiest thanks to my PhDSupervisor, Dr. S. M. Husnine who always placed me on the right track. It isneedless to say that without his utmost support and encouragement it would nothave been possible for me to reach at the final stages of my PhD.

I am also very thankful to the worthy Director Dr. Arshad Hussain at FAST,NUCES for providing an excellent research environment and exceptionally learnedteachers, especially Dr. Abdul Majeed and Dr. Abdul Khaliq. I am also indebted toour Head of the Department Ms. Sumaira Sarfraz who has been always very kindand supportive.

I feel obliged to thank Dr. Shahid H. Siddique, Chairman, Departmentof Mathematics, University of the Punjab for providing me with all possible helpduring the entire course of my study. I also like to pay my gratitude to my teachersat the secondary school, especially the mathematics teachers Mr. Jamsheed andMr. Imran Naseem who always played a significant role in the development of mymathematical skills.

It is pertinent to mention that the scholarship offering agency, the HigherEducation Commission of Pakistan for providing me with the financial support formy PhD studies. Lastly, I owe a lot to my husband Ahmad Islam, my parents,my friends (especially Farheen Ibraheem), colleagues and my loving kids for theirpatience and support.

Uzma Ahmad

Table of Contents

Table of Contents ix

List of Figures xi

Nomenclature xiii

List of Publications xv

1 Introduction 11.1 History . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Problem Statement and Motivations . . . . . . . . . . . . . . . . . 21.3 Organization of Thesis . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Literature Review 9

3 Preliminaries 153.1 Carmichael Lambda-Function λ(n) . . . . . . . . . . . . . . . . . . 153.2 Partition of a Power Digraph . . . . . . . . . . . . . . . . . . . . . 163.3 Product of Digraphs . . . . . . . . . . . . . . . . . . . . . . . . . . 163.4 Cycles and Components . . . . . . . . . . . . . . . . . . . . . . . 173.5 Indegree of the Vertices . . . . . . . . . . . . . . . . . . . . . . . . 22

4 Symmetries of Power Digraphs 274.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.2 The Symmetric Power Digraphs of order p . . . . . . . . . . . . . . 284.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

5 On the Heights of the Power Digraphs 615.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 615.2 Heights in a Power Digraph . . . . . . . . . . . . . . . . . . . . . . 625.3 Classification of G(n, k) with respect to the Heights . . . . . . . . . 695.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

6 Isolated Fixed Points and Primality of Fermat Numbers 796.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 796.2 Isolated Fixed Points . . . . . . . . . . . . . . . . . . . . . . . . . 79

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6.3 Classification of G(n, k) withrespect to the Components . . . . . . . . . . . . . . . . . . . . . . 84

6.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

7 Conclusions and Future Directions 937.1 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 937.2 Further Research Problems . . . . . . . . . . . . . . . . . . . . . . 93

Bibliography 95

Appendix 99

List of Figures

1.1 The iteration digraph on 12 vertices and f (x) = x2 . . . . . . . . . . 31.2 The iteration digraph on 12 vertices and f (x) = x2 + 1 . . . . . . . . 31.3 The power digraph G(2 · 11 + 1, 2), where 11 is a Sophie Germain

prime . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 The power digraph = G(2 ·23+1, 2), where 23 is a Sophie Germain

prime . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.1 The non trivial component of G(F2, 2) having binary structure . . . 102.2 The isomorphic forests attached with the cycle vertices 1 and 18 . . 102.3 The symmetric digraph G(24F2, 2) = G(272, 2) . . . . . . . . . . . 132.4 The non-symmetric digraph G(23F2, 2) = G(136, 2) . . . . . . . . . 14

3.1 The power digraph G(14, 2) . . . . . . . . . . . . . . . . . . . . . . 163.2 The cycle of an iteration . . . . . . . . . . . . . . . . . . . . . . . 183.3 The regular power digraph G(14,5) . . . . . . . . . . . . . . . . . . 233.4 The semi-regular power digraph G(25,5) of degree 5 . . . . . . . . 24

4.1 The symmetric power digraph G(25, 25) of order 5 . . . . . . . . . 284.2 The symmetric power digraph G(27, 45) of order 3 . . . . . . . . . 294.3 The non-symmetric power digraph G(35, 8) . . . . . . . . . . . . . 304.4 The power digraph G(45, 5) not symmetric of order 3 . . . . . . . . 384.5 The power digraph G(75, 3) not symmetric of order 5 . . . . . . . . 544.6 the power digraph G(99, 5) not symmetric of order 5 . . . . . . . . 554.7 The power digraph G(49, 35) . . . . . . . . . . . . . . . . . . . . . 564.8 The power digraph G(103, 35) . . . . . . . . . . . . . . . . . . . . 574.9 The power digraph G(29, 15) . . . . . . . . . . . . . . . . . . . . . 584.10 The power digraph G(25, 15) . . . . . . . . . . . . . . . . . . . . . 59

5.1 The power digraph G(27,30) . . . . . . . . . . . . . . . . . . . . . 695.2 The power digraph G(36, 9) with height(G1(n, k)) =1 . . . . . . . . 725.3 The power digraph G(81, 9) having height 2 . . . . . . . . . . . . . 735.4 The power digraph G(40, 16) having height 1 . . . . . . . . . . . . 745.5 The power digraph G1(57, 3) having height 2 . . . . . . . . . . . . 77

6.1 The power digraph G(3,7) . . . . . . . . . . . . . . . . . . . . . . 806.2 The power digraph G(5,7) . . . . . . . . . . . . . . . . . . . . . . 80

xi

6.3 The power digraph G(15,7) . . . . . . . . . . . . . . . . . . . . . . 806.4 The power digraph G(18, 5) with 4 isolated fixed points . . . . . . . 826.5 The power digraph G(18, 3) with no isolated fixed point . . . . . . . 826.6 The power digraph G(28,15) having two isolated fixed point . . . . 846.7 The power digraph G(33, 6) having exactly two components . . . . . 886.8 The power digraph G(M5, 30) having exactly two components . . . 896.9 The power digraph G(33, 5) with more than two components . . . . 906.10 The power digraph G(41, 5) having more than two components . . . 91

xii

Nomenclature

Symbol DescriptionN The set of natural numbers.Zn The set of integers modulo n.Fm A Fermat number of the form 22m

+ 1.Mp A Mersenne number of the form 2p − 1.λ(n) The Carmichael lambda function.ϕ(n) The Euler function.νp(x) The highest power of p in the prime factorization of x.⌈x⌉ The smallest integer greater than or equal to x.ω(n) The number of distinct prime divisors of an integer n.ordna The multiplicative order of an element a in Zn.lcm(n,m) The least common multiple of n and m.gcd(n,m) The greatest common divisor of n and m.∑

1≤i≤k Ai Sum of the terms from A1 to Ak.∏1≤i≤k Ai Product of the terms from A1 to Ak.∪1≤i≤k S i Union of the sets from S 1 to S k.

a | b a divides b.a - b a does not divide b.ai ∥ b ai | b but ai+1 - b.a , b a is not equal to b.A � B A is isomorphic to B.G(n, k) A power digraph with n vertices associated with

the congruence xk ≡ y (mod n) .G(n, k) ×G(m, k) The product of digraphs G(n, k) and G(m, k).G1(n, k) The subdigraph of G(n, k) containing all vertices

relatively prime to n.G2(n, k) The subdigraph of G(n, k) containing all vertices

not relatively prime to n.Comp(x) The component of G(n, k) containing the vertex x.At(G(n, k)) The number of cycles of length t in G(n, k).

xiii

Symbol DescriptionS (C) The number of components in a specific

equivalence class, C.K(C) The length of the unique cycle of each component

in an equivalence class, C.M(C) The indegree of cycle vertices of each component

in an equivalence class, C.N(n, k, b) The indegree of a vertex b in G(n, k).height(a) The length of the shortest directed path from a to the nearest

cycle vertex of Comp(a).height(C) The maximum height of the vertices in a component C.

xiv

List of Publications

1. Ahmad, Uzma, and Syed M. Husnine. 2011. “Characterization of PowerDigraphs Modulo n.” Comment. Math. univ. Carolin 52(3): 359 − 367.

2. Ahmad, Uzma, and Syed M. Husnine. 2012. “On the Heights of PowerDigraphs Modulo n.” Czechoslovak Math. J. 62(137): 541 − 556.

3. Husnine, Syed M., Uzma Ahmad, and Lawrence Somer. 2011. “On Symmetriesof Power Digraphs.” Util. Mathematica 85: 257 − 271.

xv

Chapter 1

Introduction

1.1 History

Modular exponentiation is one of the significant areas of number theory. It becamethe focus of research for many number theorists since the time of Fermat. Muchwork has been done on the solutions of the modular exponentiation xk ≡ a ( mod n),where x, k, a and n are positive integers. In mathematics, it is extensively used inthe divisibility problems, the integer factorizations and the primality testing etc. Itis extremely useful in computer science, particularly in the field of cryptography asmany cryptographic schemes are based on modular exponentiation. It strengthensthe public key systems such as RSA and Diffie-Hellman. A large class of symmetrickey algorithms such as AES, IDEA, and RC4 make use of these congruences.

Many primality testing algorithms are based on the iterative modular expone-ntiation. To ascertain the primality of Fermat numbers Fm = 22m

+ 1, the Pepin’salgorithm is based on the iteration g : y → y2(mod Fm) (Pepin 1877). Accordingto this test, Fm is a prime if and only if a2m−1 ≡ −1(mod Fm), where a ∈ {3, 5}.An algorithm called Lucas-Lehmer test for the primality of Mersenne numbers Mp,where p is a prime was proposed in (Lehmer 1930). The iteratively defined modularsequences y1 = 4, yk+1 = y2

k − 2 (mod Mp), where k ≥ 1 were used in this paper.It was shown that Mp is a prime if and only if yp−1 ≡ 0 (mod Mp). An algorithmbased on discrete iteration yi = yi + b (mod n) for integer factorization, where b israndomly selected, is developed in (Pollard 1975).

One of the major applications of iteratively defined modular exponentiationis the generation of pseudo-random numbers which is considered to be of fundamen-tal importance in cryptography. Now a days most of security systems are formulatedby using powerful cryptographic algorithms. However, the implementation of pseu-do-random processes in the generation of secret measures for passwords, cryptogra-phic keys and similar quantities give rise to pseudo-security. A Pseudo-randomnumber generator named as Blum Blum Shub (BBS) was proposed in (Blum et al.

1

1986). It is of the form xn+1 ≡ x2n mod M, where M = pq is the product of two

distinct big primes each congruent to 3 mod 4 and x0 is a quadratic residue mod M.The BBS generator has several worth mentioning characteristics. It was shown thatunder certain hypothesis, BBS has the advantageous cryptographic characteristic ofunpredictability. In this work, the cycle length of the sequences produced by thisgenerator was investigated. They found a technique to ensure that the sequencesproduce by this generator are of maximum possible cycle lengths. The properties ofBBS generator were further studied in (Cusick 1995; Vazirani and Vazirani 1984;Chor et al. 1986) which supports the implementation of BBS in different cryptograp-hic applications. In view of the fact that the security of the BBS generator dependson the confidentiality of the factorization of N, which is not secured, the moregeneral generator of the form xn+1 ≡ xk

n mod M (Power generator), where k andM are any positive integers greater than 1 was proposed in (Brennan and Geist1998). They established how to compute all cycle lengths and the number of cyclesof each possible length in the case when some information about prime factorizationof M is known. The results proved by Brennan and Giest can help in selecting amodulus for BBS generator. The statistical properties of cycle structure of a powergenerator in terms of the number of cycles, upper bounds for the length of the cycle,distribution and linear complexity were further studied in (Friedlander et al. 1999;Griffin and Shparlinski 2000; Friedlander et al. 2001; Shparlinski 2001; Friedlanderand Shparlinski 2001; Chou and Shparlinski 2004; Martin and Pomerance 2005;El-Mahassni 2008; Sha 2011).

1.2 Problem Statement and MotivationsIn all the papers, mentioned in previous section, ZM and the power generator werestudied from the perspective of arithmetic dynamical systems. However, if we lookat ZM as the set of vertices and sketch a directed edge from x to y if xk ≡ y ( mod M),then we obtain a digraph. There are several important results and consequencesabout the congruences xn+1 ≡ xk

n (mod M) from the stance of graph theory.

Definition 1.2.1. Let f : H → H be any map. An iteration digraph of f is a directedgraph whose vertices are the elements in H such that there exists precisely one edgefrom y to f (y) for all y ∈ H.

An iteration digraph on 12 vertices corresponding to f (x) = x2 is shownin Figure 1.1, whereas an iteration digraph with the same number of vertices andf (x) = x2 + 1 is given in Figure 1.2.

Definition 1.2.2. An iteration digraph with Zn = {0, 1, ..., n−1} as the set of verticesand E = {(a, b) : ak ≡ b (modn)} as the edge set, where n and k are positiveintegers is said to be a power digraph modulo n and is denoted by G(n, k).

The power digraphs provide a remarkable class of digraphs which are notvertex-transitive. These digraphs not only give rise to the attractive class of graphs

2

Figure 1.1: The iteration digraph on 12 vertices and f (x) = x2

Figure 1.2: The iteration digraph on 12 vertices and f (x) = x2 + 1

but they also link three major branches of discrete mathematics namely, numbertheory, cryptography and graph theory.

The objective of this thesis is to generalize many characteristics of a powerdigraph G(n, k) for some integers n and k. The following six major problems areinvestigated:

Problem 1: To obtain the necessary and sufficient conditions for G(n, k) tobe symmetric of order p, where p is a prime divisor of n such that p2|n.

Problem 2: To derive the explicit formulae for the determination of theheights of vertices, components and power digraph itself in terms of the parametersn and k.

Problem 3: To find the criteria for the existence of isolated fixed points ofG(n, k) in terms of n and k.

Problem 4: To classify those power digraphs G(n, k) which consist of exactly

3

two components containing fixed points 0 and 1.

Problem 5: The classification of those power digraphs in which each vertexof indegree 0 is at height q.

Problem 6: To find the criteria for testing the primality of Fermat numbersin terms of the cycles or components of the power digraphs.

The necessary and sufficient conditions for the existence of symmetry ofpower digraphs G(n, k) for the general values of n and k are not found in theliterature except for the case when n is square-free. Therefore, the case when ncontains a square factor is considered i.e. n = pαq1q2 · · · qm such that p, qi aredistinct odd primes and α > 1. The conditions on p, qi, α and k are determinedunder which the digraphs G(n, k) are symmetric of order p. The existing results onthe symmetry of G(n, k) are extended in this thesis.

Heights of vertices in a digraph have an essential role in characterizing thepower digraphs. For example, the digraphs in which every cycle vertex has height1 and indegree 2 give rise to the interesting class of digraphs that look like littlesuns or ship wheels. It is shown that in a power digraph G(2p + 1, 2), where p isa Sophie Germain prime, its non trivial components seem like little suns (Krızekand Somer 2004). Figure 1.3 exhibits a digraph G(2 · 11 + 1, 2) in which its uniquenon trivial component looks like a little suns. However, a digraph G(47, 2) has twonon trivial sun like components (Figure 1.4). Moreover, if a component of a powerdigraph has height one and indegree of each of its cycle vertex ci is greater than 2then T (ci) ∪ ci is a directed star graph (K1,N(n,k,ci)−1). In Figure 4.2, for all the cyclevertices ci, T (ci) ∪ ci = K1,8.

There are some properties of heights which were established for G(n, 2) in(Carlip and Mincheva 2008; Somer and Krızek 2004, 2006) but not for the generalvalues of k. These results on the heights are generalized in this research work.The existence of isolated fixed points for k = 2 was studied in (Somer and Krızek2004) and for k = 3 in (Skowronek-Kaziw 2008). In this thesis, this study has beencompleted for all natural numbers.

1.3 Organization of ThesisIn Chapter 1, a brief introduction with some motivations for the research work andits impact on the field is detailed. Chapter 2 elaborates the contribution of manyMathematicians on the power digraphs which is available in the literature. Chapter3 comprises of some relevant preliminaries to make this dissertation self contained.The proofs of some important known results are also given in this chapter for betterunderstanding of the readers. Chapter 4 deals with the new results on the symmetryof power digraphs G(n, k) with n = pαq1 · · · qm, where p, qi are odd prime divisorsof n and α > 1. In section 4.2, the symmetry of those digraphs are considered in

4

Figure 1.3: The power digraph G(2 ·11+1, 2), where 11 is a Sophie Germain prime

which some direct factors of a digraph are regular. Then the classification of non-symmetric digraphs is presented. Further the necessary and sufficient conditionson n and k for the symmetry of power digraphs of order p are presented. Thework presented in this chapter is published in the paper (Husnine et al. 2011).Chapter 5 contains some new results on the heights of power digraph G(n, k). Afterintroducing the chapter in section 5.1, the height of any vertex in G(n, k) in termsof the heights of corresponding vertices in direct factors of G(n, k) is given. Theformulae for the heights of the vertices and components in G1(n, k) and G2(n, k)in terms of n and k are also derived in this section. The classification of powerdigraphs with respect to heights is given in section 5.3. The power digraphs G(n, pα)in which each component has height 1 are classified in this section. Moreover,the power digraphs G(n, p) in which each vertex of indegree 0 is at height q arecharacterized. A formula for the number of vertices at a specific height is alsogiven in this section. The contents of this chapter are published in (Ahmad andHusnine 2012). Chapter 6 elaborates some new results on the isolated fixed pointsand the classification of G(n, k) with exactly two components. In section 6.2, theexistence of an isolated fixed point in G(n, k) in terms of existence of the isolatedfixed points in direct factors of G(n, k) are given. The necessary and sufficientconditions for the existence of isolated fixed points in G(n, k) in terms of n and k

5

are established. Section 6.3 contains the classification of G(n, k) with respect to thecomponents. The semi-regular digraphs of degree 2t for some integer t consisting oftwo components are characterized. The primality of Fermat numbers is then relatedwith the number of components and semi-regularity of digraph. The necessary andsufficient conditions on n and k so that a power digraph G(n, k) consists of exactlytwo components are obtained. This complete the classification of power digraphswith exactly two components. The work presented in this chapter is published in(Ahmad and Husnine 2011). Chapter 7 contains concluding remarks and someopen problems.

For the basic concepts and definitions, we follow (Burton 2007) for numbertheory and (Deo 1990; Chartrand and Oellermann 1993) for graph theory. Allthe figures in the thesis are created with the help of computational mathematicalpackage MATLAB and displayed by using the Graphviz visualization tool. Thecopies of published papers are provided in the appendix.

6

Figure 1.4: The power digraph = G(2 · 23 + 1, 2), where 23 is a Sophie Germainprime

7

Chapter 2

Literature Review

Bryant made the first use of power graphs to characterize the abelian groups of orderat most 10932, up to isomorphism (Bryant 1967). This idea of associating graphswas further applied to other algebraic structures like finite rings and fields. Chasseinvestigated the structure of the iterative maps f : y → y2 + d on commutativefields (Chasse 1984, 1986). The iteration digraphs have a fascinating connectionwith a very strong dynamical structure in number theory namely, the Fermat primesof the form 22m

+ 1. Fermat primes, due to its highly interesting properties andapplication in geometry, pseudo-random number generation, signal processing andmost importantly in number theory influenced the researchers to work on themin detail (Krızek et al. 2001). The iteration digraphs of Fermat primes underthe squaring map was examined in (Szalay 1992). In particular, Szalay provedthat G(Fm, 2) has a special feature of having a binary structure. This is shown inFigure 2.1. It was proved that G(n, 2) is symmetric of order 2 if n ≡ 2 (mod 4) orn ≡ 4 (mod 8). Blanton et al. made some further progress and studied the iterationdigraphs G(p, 2), where p is any prime (Blanton et al. 1992a,b). With the helpof power digraphs G(n, 2), many new and old results about primitive roots werestudied. They proved that G(p, 2) has exactly two components if and only if p = Fm.In this work the cycle structure of G(p, 2) was also studied.

The iteration digraphs of squaring maps on prime field GF(p), where p isany prime were investigated independently in (Rogers 1996; Flores 1994) . Rogersestablished a formula for the decomposition of an iteration digraph into its cyclesand the trees attached with them by assuming various hypotheses (like Artin’shypothesis). The complete structure of G(p, k) was characterized in (Lucheta et al.1996). They proved many results concerning the existence of cycles, their lengths,the number of cycles, the number of non cycle vertices and the indegrees of vertices.They also determined the forest structure completely and proved that the forestsattached with the cycle vertices of G1(p, k) are isomorphic. This is illustrated inFigure 2.2. Soon after, Wilson gave a generalization of some of the results of(Lucheta et al. 1996) on the configuration of cycles, the indegrees of vertices and

9

the forest structure of cycle vertices of G1(n, k) (Wilson 1998).

Figure 2.1: The non trivial component of G(F2, 2) having binary structure

Figure 2.2: The isomorphic forests attached with the cycle vertices 1 and 18

The notion of Carmichael lambda function λ(n) was used for the first timein studying the power digraphs G(n, 2) in (Somer and Krızek 2004). With the helpof λ(n) and ϕ(n) they proved many old as well as new results about the existenceof cycles, the isolated fixed points, the number of cycles, the levels of components,the number of components and the primality of Fermat numbers. In particular, theyproved that there are no isolated cycles of G(n, 2) of length greater than one. Further,it was established that G(n, 2) has an isolated fixed point a , 0, where a = n/2 ifand only if 2 | n and n is square-free. They showed that the digraphs G(n, 2) aresemi-regular if and only if n = 2k for k = 0, 1, 2, 4. They also proved that for t > 1,At(G1(n, 2)) > 1 whenever At(G2(n, 2)) > 1. It was shown that the Fermat number

10

Fm is composite if and only if there exists a cycle in G(Fm, 2) of length greater thanone. They also proved that the digraph G(n, 2) has exactly 2 components if and onlyif n is a Fermat prime or n is a power of 2. Some necessary and sufficient conditionsfor a digraphs to have 3 and 4 components were also presented in this work.

The iteration digraphs G(n, 2), where n = 2p + 1 such that p is a SophieGermain prime were studied in (Krızek and Somer 2004). The structure of G(p, 2)and the digraphs obtained from another iteration x→ x2−2 ( mod p) were investiga-ted in (Vasiga and Shallit 2004). They obtained the asymptotic estimates for thenumber of cycle elements and the number of tail elements. They also studiedthe iteration digraphs of Fermat primes and Mersenne primes under two differentiterative maps. Somer and Krızek explored the structure of G2(n, 2) by makinguse of the properties of Carmichael lambda function and generalized some of theirearlier results in (Somer and Krızek 2006, 2007). In particular, they establisheda formula for the indegrees of vertices of G2(n, k). They also gave a completecharacterization of the semi-regular power digraphs.

The heights and levels of the vertices and components as well as some ofrelated properties for G(n, 2) were studied in (Carlip and Mincheva 2008; Lchetaet al. 1996; Somer and Krızek 2004, 2006). Lucheta et al. proved that if h0 is theleast value of h such that p − 1 | kht, then h0 is the height of the forests in G(p, k).Somer and Krızek showed that the height and level of each component of G1(n, 2)is ν2(λ(n)) and ν2(λ(n)) + 1 respectively. Then they proved that if n has form

n = pe11 pe2

2 .....perr ,

where p1 < p2 < .... < pr and ei > 0 , n′ varies over all positive integers suchthat n′ < n, n′ | n and gcd(n/n′, n′) = 1, then the height of any vertex in G2(n, 2)is max(max1≤i≤r ⌈log2 ei⌉, e), where e is the maximum value of ν2(λ(n′)). In thisdissertation, an attempt is made to generalize these results from G(n, 2) and G(p, k)to G(n, k) where n ≥ 1 and k ≥ 2.

The structure of power digraphs G(n, 3) were studied in (Skowronek-Kaziow2008, 2009). Some conditions for regularity and semi-regularity for G(n, 3) wereestablished in these papers. Moreover the structure of G(2k, 3) and G(3k, 3) werealso presented. It was shown that every component of a digraph G(n, 3) is a cycle ifand only if 3 does not divide the Euler totient function ϕ(n) and n is a square-free.It was proved that G1(2k, 3) contains only cycles and G2(2k, 3) is a tree with theroot in 0. Besides G1(3k, 3) contains two ternary trees with roots in 1 and 3k − 1 andG2(3k, 3) is a tree with the root in 0. All digraphs with 3 components were describedin these papers. Skowronek-Kaziow established a formula for the number of fixedpoints of G(n, 3). A criterion for 0 to be an isolated fixed point of G(n, 3) wasdetermined. In this thesis, the general criteria for the existence of an isolated fixedpoint are determined. The results of (Somer and Krızek 2004; Skowronek-Kaziow2009) are generalized.

Skowronek-Kaziow also discussed the symmetry of G(n, 3) and proved thatG(n, 3) is symmetric of order 2 if and only if n ≡ 2 (mod 4). The symmetry of

11

G(2l p, 2), where p is a Fermat prime was studied in (Carlip and Mincheva 2008).They showed that G(2l p, 2) is symmetric when l = 4 and is not symmetric whenl = 3 or l ≥ 5. They established that the converse of the result on symmetry of adigraph G(n, 2) proved in (Szalay 1992) does not hold. Figure 2.3 shows a digraphG(24F2, 2) = G(272, 2) which is symmetric of order 2. Whereas, Figure 2.4, exhibita digraph G(23F2, 2) = G(136, 2) which is not symmetric of order 2.

Somer and Krızek provided some sufficient conditions on n and k for thesymmetry of G(n, k) (Somer and Krızek 2009). In this paper, they used the conceptof product of digraphs for the first time for investigating the power digraphs G(n, k).Some necessary and sufficient conditions for G(n, k) to be symmetric of order pfor a square-free positive integer n by using the same techniques were presented in(Kramer-Miller 2009). A natural question arises whether it is possible to find thesymmetry of G(n, k) for any positive integers n and k. In this dissertation, effortshave been made to extend these results for the general values of n and k, where n isno more square-free.

12

Figure 2.3: The symmetric digraph G(24F2, 2) = G(272, 2)

13

Figure 2.4: The non-symmetric digraph G(23F2, 2) = G(136, 2)

14

Chapter 3

Preliminaries

This chapter contains some basic concepts about the power digraphs G(n, k). Inparticular, Carmichael lambda-function λ(n), partition of power digraphs, the comp-onents, the direct product of digraphs, the cycles and the indegrees of vertices arediscussed. Various related theorems have been inserted for the better understandingof the readers. Moreover, proofs of some important results are detailed. All theconcepts and terminologies are elaborated with the examples. From now onward,H = Zn = {0, 1, · · · , n − 1} and f (x) = xk, where n and k are positive integers.

3.1 Carmichael Lambda-Function λ(n)

The Carmichael lambda-function λ(n) was first introduced in (Carmichael 1910). Itappears to be a generalization of the Euler totient function ϕ(n). The values of theCarmichael lambda-function λ(n) are

λ(1) = 1,λ(2) = 1,λ(4) = 2,λ(2k) = 2k−2 for k ≥ 3,λ(pk) = (p − 1)pk−1, for any odd prime p and k ≥ 1,

λ(pe11 pe2

2 · · · perr ) = lcm(λ(pe1

1 ), λ(pe22 ), · · · , λ(per

r )),

where p1, p2, · · · , pr are distinct primes and ei ≥ 1 for all i. It is clear from the

definition that λ(n) ≤ ϕ(n) and that λ(n) | ϕ(n) for all n. It is observed that λ(n) =ϕ(n) if and only if n ∈ {1, 2, 4, ps, 2ps}, where p is any odd prime and s ≥ 1. It isalso easy to show that ϕ(n) = 2i if and only if λ(n) = 2 j, where j ≤ i.

The vital property of Carmichael lambda-function is that it takes a broaderview of the Euler’s Theorem. This is described in the following theorem:

15

Theorem 3.1.1. (Carmichael 1910) Let a, n ∈ N. Then

aλ(n) ≡ 1(mod n)

if and only if gcd (a, n) = 1. Moreover, there exists an integer g such that

ordna = λ(n),

where ordng denotes the multiplicative order of g modulo n.

We can write λ(n) asλ(n) = uv, (3.1.1)

where u is the highest factor of λ(n) relatively prime to k. Henceforth, this factorizationof λ(n) will be used.

3.2 Partition of a Power DigraphThe subdigraph of G(n, k) containing all the vertices relatively prime to n is denotedby G1(n, k) whereas, the subdigraph of G(n, k) containing all the vertices not relativelyprime to n is denoted by G2(n, k). It is clear that G1(n, k) and G2(n, k) are disconnected.Moreover, G(n, k) = G1(n, k) ∪ G2(n, k). For example in Figure 3.1, the union offirst four components make G2(n, k) and last two components constitute G1(n, k).

Figure 3.1: The power digraph G(14, 2)

3.3 Product of DigraphsLet n = n1n2, where gcd(n1, n2) = 1. We know by the Chinese Remainder Theoremthat a vertex a in G(n, k) corresponds to an ordered pair (a1, a2), where 0 ≤ a1 < n1

and 0 ≤ a2 < n2. By the same token we obtain that ak corresponds to (ak1, a

k2).

16

The product of digraphs, G(n1, k) and G(n2, k) is defined as follows: a vertex inG(n1, k)×G(n2, k) is an ordered pair (a1, a2) such that a1 ∈ G(n1, k) and a2 ∈ G(n2, k).Also there is an edge from (a1, a2) to (b1, b2) if and only if there is an edge from a1

to b1 in G(n1, k) and there is an edge from a2 to b2 in G(n2, k). This implies that(a1, a2) has an edge leading to (ak

1, ak2). It follows that G(n, k) � G(n1, k) ×G(n2, k).

Further, it can be asserted that if ω(n) denotes the number of distinct prime divisorsof n and

n = pe11 pe2

2 · · · perr , where p1 < p2 < · · · < pr and ei > 0 i.e. r = ω(n), then

(3.3.1)

G(n, k) � G(pe11 , k) ×G(pe2

2 , k) × · · · ×G(perr , k). (3.3.2)

Also it is easy to see that

G1(n, k) � G1(pe11 , k) ×G1(pe2

2 , k) × · · · ×G1(perr , k). (3.3.3)

3.4 Cycles and ComponentsIn this section, the properties of cycles and components of power digraphs G(n, k)are discussed in detail with examples and related results.Initiating with an arbitrary element x0 of H, we create the sequence of successiveelements of H by

xi+1 = f (xi) i = 0, 1, · · ·

As H contains finite number of elements, so for each x0 there must exist a leastpositive integer l such that f l(x0) ∈ {x0, f (x0), · · · f l−1(x0)}. Let k be the least non-negative integer such that f l(x0) = f k(x0). If c = l − k, then f k(x0) = f k+c (x0). Theset of elements {x0, f (x0), · · · , f k(x0)} form the tail whereas, { f k(x0), · · · , f k+c−1(x0)}constitute the cycle of length c. This is shown in Figure 3.2.

Definition 3.4.1. A cycle is a path from one vertex to itself and a cycle is a t-cycleif it contains exactly t vertices.

Definition 3.4.2. A cycle of length one is termed as a fixed point.

Remark 3.4.1. Corresponding to every cycle vertex c of a power digraph thereis a tree T (c) whose root is c and it contains the non cycle vertices d satisfyingdk j ≡ c (mod n) for some i ∈ N but dk j−1

is not congruent to any cycle vertex. It isimportant to note that T (c) does not contain c.

Definition 3.4.3. A component of G(n, k) is a subdigraph which is the largest conne-cted subgraph of the associated nondirected graph.

17

Figure 3.2: The cycle of an iteration

For example G(14, 2) consists of 6 components, two of them are isolatedfixed points, two components contain fixed points whereas, the remaining two comp-onents contain cycles of length 2 as shown in Figure 3.1.

Moreover, it is clear that 0 and 1 are always fixed points of G(n, k). Sinceeach vertex has outdegree one and the component has only a finite number ofvertices, it follows that each component contains a unique cycle. Thus the numberof components in a power digraph is equal to the number of its cycle.

It was exhibited that G(Fm, 2) consists of only two components each contai-ning a fixed point, where Fm is a Fermat prime and m ≥ 1 is any integer (Szalay1992). The cycle lengths and the number of cycles of G(p, 2), where p is any primewere characterized in (Blanton et al. 1992a). The cycle structure of G(p, k) wasdiscussed in (Lucheta et al. 1996). The generalization of these results about cyclesfrom the power digraphs G(p, k) to G(n, k) were presented in (Wilson 1998). In thiswork the following results about the existence of cycles, their lengths and the treesattached to the cycle vertices in G1(n, k) were proved by using the notations:If n = 2a∏m

i=1 paii with ai ≥ 1 and a ≥ 0 then,

δ1 =

{0, if a = 0, 1,1, if a ≥ 2 (3.4.1)

δ2 =

{0, if a < 31, if a ≥ 3 (3.4.2)

and

L = lcm(2δ1 , 2δ2(a−2), pa1−11 (p1 − 1), · · · , pam−1

m (pm − 1))

18

Lemma 3.4.1. (Wilson 1998) The vertex b is a cycle vertex if and only if ordnb | u,where u is the largest factor of L relatively prime to k.

Corollary 3.4.2. (Wilson 1998) There are gcd(2, u)δ1 gcd(2a−2, u)δ2∏m

i=1 gcd(u, pai−1i

(pi − 1)) cycle vertices in G1(n, k), where u is the largest factor of L relatively primeto k.

Theorem 3.4.3. (Wilson 1998) The length l(d) of a cycle of order d in G1(n, k) isthe smallest natural number l such that d | (kl − 1), i.e, l(d) = orddk.

Corollary 3.4.4. (Wilson 1998) The longest cycle in G1(n, k) has length equal toorduk, where u is the largest factor of L relatively prime to k.

Theorem 3.4.5. (Wilson 1998) Let c1 and c2 be any two cycle vertices in G1(n, k)and T (c1) and T (c2) be the trees attached to c1 and c2 respectively. Then T (c1) �T (c2).

Corollary 3.4.6. (Wilson 1998) Let t ≥ 1 be a fixed integer. Then any two componentsin G1(n, k) containing t-cycles are isomorphic.

The iteration digraphs G(n, 2) were again investigated in (Somer and Krızek2004). In this particular paper, they proved the results on the cycles by makinguse of Carmichael lambda function λ(n) instead of L. The same authors proved thefollowing results on the configuration of cycles in G(n, k) by using the propertiesof Carmichael lambda function λ(n) and the digraph products (Somer and Krızek2009).

Theorem 3.4.7. (Somer and Krızek 2009) Let n = n1n2, where gcd(n1, n2) = 1 anda = (a1, a2) be a vertex in G(n, k) � G(n1, k) × G(n2, k). Then a is a cycle vertex ifand only if a1 is a cycle vertex in G(n1, k) and a2 is a cycle vertex in G(n2, k).

Proof. First suppose that a = (a1, a2) is a cycle vertex in G(n, k). Then there existsome l ∈ N such that

akl ≡ a (mod n).

This impliesakl= (a1, a2)kl

= (akl

1 , akl

2 ) = (a1, a2).

Henceakl

1 ≡ a1(mod n1) and akl

2 ≡ a2(mod n1)

Therefore, a1 and a2 are the cycle vertices of G(n1, k) and G(n2, k) respectively.Conversely, suppose that a1 and a2 are the cycle vertices of G(n1, k) and

G(n2, k) respectively. Then there exist positive integers l1 and l2 such that

akl1

1 ≡ a1(mod n1) and akl2

2 ≡ a2(mod n1)

Let l3 = lcm(l1, l2). Then

akl3

1 ≡ a1(mod n1) and akl3

2 ≡ a2(mod n1)

19

This shows(c1, c2)kh3

= (ckl3

1 , ckl3

2 ) = (c1, c2),

which shows (c1, c2) is a cycle vertex in G(n, k). �

Theorem 3.4.8. (Somer and Krızek 2009) Let n be an integer having factorizationas given in (3.3.1). Then

At(G1(n, k)) =1t[

r∏i=1

(δi gcd(λ(peii ), kt − 1)) −

∑d|t,d,t

dAd(G(n, k))], (3.4.3)

where δi = 2 if 2 | kt − 1 and 8 | peii , and δi = 1 otherwise.

Proof. Let a be a vertex in a t-cycle of G1(n, k). Then t is the least positive integerfor which a satisfies the congruence

xkt − x ≡ x (xkt−1 − 1) ≡ 0 (mod n). (3.4.4)

Since gcd(a, n) = 1, it follows from (3.4.4) that a is in a t-cycle of G1(n, k) if andonly if t is the least positive integer such that a satisfies the congruence

xkt−1 ≡ 1 (mod n). (3.4.5)

We note that the vertex b in G1(n, k) satisfies the congruence (3.4.4) if and only ifb is in a d-cycle of G1(n, k) for some integer d dividing k. Since gcd(b, n) = 1,congruence (3.4.4) is satisfied by b if and only if x = b satisfies the congruence(3.4.5).

Clearly, the total number of vertices in the d-cycles of G1(n, k) is equal todAd(G1 (n, k)). It now follows from (3.4.5) and Theorem 3.5.1 that

N(n, kt − 1, 1) =∑d|t

dAd(G1(n, k)) =r∏

i=1

δi gcd(λ(pαii ), kt − 1).

Solving for At(G1(n, k)), we obtain (3.4.3). �

Theorem 3.4.9. (Somer and Krızek 2009) Let n be an integer having factorizationas given in (3.3.1). Then

At(G(n, k)) =1t[

r∏i=1

(δi gcd(λ(peii ), kt − 1) + 1) −

∑d|t,d,t

dAd(G(n, k))], (3.4.6)

where δi = 2 if 2 | kt − 1 and 8 | peii , and δi = 1 otherwise.

Proof. Leth(x) = xkt − x = x (xkt−1 − 1).

Then

M(h(x), n) =r∏

i=1

M(h(x), pαii ) (3.4.7)

20

Since gcd(a, akt−1 − 1) = 1 for all nonnegative integers a, we see by the proof ofTheorem 3.4.8 that

M(h(x), pαii ) = N(pαi

i , kt − 1, 1) + 1 = δi gcd(λ(pαi

i ), kt − 1) + 1 (3.4.8)

for i = 1, 2, · · · , r. Using a similar argument as that in the proof of Theorem 3.4.8,Eq. (3.4.6) is seen to follow from 3.4.7 and 3.4.8 �

About the same time Kramer obtained the following results on the cycles inG(n, k).

Lemma 3.4.10. (Kramer-Miller 2009) Let n = n1n2, where gcd(n1, n2) = 1 andJ(n1, k) be a component of G(n1, k) and L(n2, k) be a component of G(n2, k). Supposes is the length of L(n2, k)’s cycle and let t be the length of J(n1, k)’s cycle. ThenC(n, k) � J(n1, k)×L(n2, k) is a subdigraph of G(n, k) consisting of gcd(s, t) compon-ents, each having cycles of length lcm(s, t).

Lemma 3.4.11. (Kramer-Miller 2009) Let n = n1n2, where gcd(n1, n2) = 1 andJ(n1, k) be a component of G(n1, k) and L(n2, k) be a component of G(n2, k). Supposes is the length of L(n2, k)’s cycle and let t be the length of J(n1, k)’s cycle. ThenC(n, k) � J(n1, k) × L(n2, k) is a subdigraph of G(n, k) consisting of components,whose cycle length is greater than or equal to each max(s, t).

The following theorem about the components in the product of power digraphis proved in (Somer and Krızek 2009).

Theorem 3.4.12. (Somer and Krızek 2009) Let n = n1n2, where gcd(n1, n2) = 1and J(n1, k) be a union of components of G(n1, k) and let L(n2, k) be a union ofcomponents of G(n2, k). Then J(n1, k)×L(n2, k) is a union of components of G(n, k) �G(n1, k) × G(n2, k). Moreover, if L(n2, k) =

∪i Li(n2, k), where Li(n2, k) are distinct

components of G(n2, k) for all i, then

J(n1, k) × L(n2, k) =∪

i

J(n1, k) × Li(n2, k), (3.4.9)

where the union in equation (3.4.9) is a disjoint union.

Proof. We first prove that J(n1, k) × L(n2, k) is a union of components of G(n1, k) ×G(n2, k). Let a = (a1, a2) be a vertex of J(n1, k) × L(n2, k). Let c = (c1, c2) be acycle vertex in G(n, k) = G(n1, k) × G(n2, k) for which akh ≡ c(mod n) for somenonnegative integer h. It suffices to show that c is a vertex of J(n1, k) × L(n2, k) andthat each vertex b = (b1, b2) in G(n1, k) × G(n2, k) satisfying bkh ≡ c (mod n) forsome i ≥ 0 is also in J(n1, k) × L(n2, k). We note that

(c1, c2) = (a1, a2)kh= (akh

1 , akh

2 ).

Since the component of G(n1, k) containing the vertex a1 is also a component inJ(n1, k), it follows that c1 ≡ akh

1 (mod n1) is a vertex in J(n1, k). Similarly, c2 ≡

21

akh

2 ( mod n2) is a vertex in L(n2, k). Thus c = (c1, c2) is a vertex in J(n1, k)×L(n2, k).Now let b = (b1, b2) be a vertex in G(n1, k) ×G(n2, k) such that

bkh= (bkh

1 , bkh

2 ) = (c1, c2).

for some i ≥ 0. Since c1 is a cycle vertex in J(n1, k) by Theorem 3.4.7 and sincethe component of G(n1, k) containing the vertex c1 is also a component of J(n1, k),it follows that b1 is a vertex in J(n1, k). Similarly, b2 is a vertex in L(n2, k). Henceb = (b1, b2) is a vertex in J(n1, k) × L(n2, k).

We finally prove that (3.4.9) is also satisfied. Since the components

L1(n2, k), L2(n2, k), · · · , Lm(n2, k)

are disjoint, we see that the union in (3.4.9) is a disjoint union. Clearly, if ( j, l)is a vertex in J(n1, k) × L(n2, k), then j ∈ J(n1, k) and l ∈ Li(n2, k) for some i ∈{1, 2, · · · ,m}. Thus the set on the left-hand side of (3.4.9) is the union of the sets onthe right-hand side. �

3.5 Indegree of the VerticesThis section contains all previous results and literature about indegree of the verticesin G(n, k).

Definition 3.5.1. The indegree of x, denoted by indegn(x) is the number of directededges coming into a vertex x, and the number of edges coming out of x is referredto as the outdegree of x denoted by outdegn(x).

Since power digraph is based on a function, outdegree of each vertex inG(n, k) is one. It is easy to see that G(n, k) with n vertices also has exactly ndirected edges. It is also clear that cycle vertices have positive indegree. If N(n, k, b)represent the number of incongruent solutions of the congruence yk ≡ b(mod n).Then clearly,

N(n, k, b) = indegn(b) (3.5.1)

By using (3.3.1), (3.5.1) and Chinese Remainder Theorem, we can write

N(n, k, b) = indegnb =r∏

i=1

N(peii , k, b). (3.5.2)

Definition 3.5.2. A digraph G(n, k) is said to be regular if every vertex of G(n, k)has same indegree.

Definition 3.5.3. A digraph G(n, k) is said to be semi-regular of degree j if everyvertex of G(n, k) has indegree j or 0.

22

It is noted that a regular digraph does not contain any vertex of indegree0. Also it is easy to see that for each vertex x of regular digraph, indegn(x) =outdegn(x) = 1. In view of the fact that every component of power digraph G(n, k)contains a unique cycle, it is observed that G(n, k) is regular if and only if everycomponent of G(n, k) is itself a cycle. Figure 3.3 demonstrates the regular digraphwhereas, Figure 3.4 is an example of a power digraph which is semi-regular ofdegree 5 but not regular.

Figure 3.3: The regular power digraph G(14,5)

The indegree of any vertex in G(Fm, 2), where Fm is Fermat prime and m ≥ 1is any integer was proved to be 0 or 2 (Szalay 1992). It was also shown that theindegree of any vertex in G(p, 2), where p is any odd prime is also 0 or 2 (Blantonet al. 1992b). Later on, it was proved that the indegree of any vertex of G(p, k) is 0or gcd(k, p − 1) (Lucheta et al. 1996). The following Theorem is proved in (Wilson1998).

Theorem 3.5.1. (Wilson 1998) Let n be an integer having factorization as given in(3.3.1) and a be a vertex of positive indegree in G1(n, k) . Then

indeg(a) = N(n, k, a) =r∏

i=1

εi gcd(λ(peii ), k),

where εi = 2 if 2 | k and 8 | peii , and εi = 1 otherwise.

The indegree of the vertices of G2(n, 2) and semi-regularity of G(n, 2) werefurther studied in (Somer and Krızek 2004, 2006, 2007). They presented a generaliz-ation of their earlier results on the indegrees of the vertices of G2(n, k). They also

23

Figure 3.4: The semi-regular power digraph G(25,5) of degree 5

obtained the necessary and sufficient conditions for G(n, k) to be regular or semi-regular. Some of these results are given below:

Corollary 3.5.2. (Somer and Krızek 2007) Let p be a prime and α ≥ 1, k ≥ 2 beintegers. Then N(pα, k, 0) = pα−⌈

αk ⌉.

Theorem 3.5.3. (Somer and Krızek 2007) Let n ≥ 1 and k ≥ 2 be integers. Then

1. G1(n, k) is regular if and only if gcd(λ(n), k) = 1;

2. G2(n, k) is regular if and only if either n is square-free and gcd(λ(n), k) = 1or n = p, where p is prime;

3. G(n, k) is regular if and only if n is square-free and gcd(λ(n), k) = 1.

The following results on indegree of vertices were proved in (Kramer-Miller2009)

24

Lemma 3.5.4. (Kramer-Miller 2009) Let n = n1n2, where (n1, n2) = 1 and a =(a1, a2) be a vertex in G(n, k) = G(n1, k) ×G(n2, k). Then

N(n, k, a) = N(n1, k, a1) · N(n2, k, a2).

Lemma 3.5.5. (Kramer-Miller 2009) Let n = n1n2, where (n1, n2) = 1 and a =(a1, a2) be a vertex in G(n, k) = G(n1, k) ×G(n2, k). Then

N(n, k, a) ≥ max{N(n1, k, a1) · N(n2, k, a2)}.

25

Chapter 4

Symmetries of Power Digraphs

4.1 IntroductionThe symmetric graphs are considered to be one of the important families in graphtheory. Power digraphs provide a large class of digraphs that are symmetric ofcertain orders. This chapter presents the extensions of some of the results givenin (Kramer-Miller 2009; Somer and Krızek 2009) by finding some necessary andsufficient conditions for G(n, k) to be symmetric of order p, where n = pαq1 · · · qm

such that p , qi are odd primes and α > 1.

Definition 4.1.1. A power Digraph G(n, k) is said to be symmetric of order r if wecan partition G(n, k) into subdigraphs, each containing r isomorphic components.

Figure 4.1 illustrates a symmetric power digraph G(25, 25) of order 5, Figure4.2 demonstrates a symmetric power digraph G(27, 45) of order 3 whereas, Figure4.3 shows a power digraph G(35, 8) not symmetric of any order.

The following notations and conventions will be followed in the proceedings ofthis chapter.Let n be an odd integer defined as

n = pαq1q2 · · · qm, where p, qi are distinct primes andα > 1. (4.1.1)

Then it is easy to see that G2(pα, k) consists of only one component containing thevertex 0 which is a fixed point. From now onward this component of G2(pα, k)containing the fixed point 0 will be referred to as Comp(0).We note that digraph isomorphism on the components of G(n, k) defines an equivale-nce relation. The number of components in a specific equivalence class, C, isdenoted by S (C). Each component in an equivalence class C has a cycle of thesame length and this length is denoted by K(C). Also if J is a component in theequivalence class C then every cycle vertex in J has the same indegree. This sameindegree is represented by M(C). In terms of equivalence class, the digraph G(n, k)is symmetric of order m if m | S (C) for each equivalence class C.

27

Figure 4.1: The symmetric power digraph G(25, 25) of order 5

4.2 The Symmetric Power Digraphs of order p

The symmetries of G(n, k) for positive integers n and k are investigated in thissection. To obtain the main results, a power digraph is decomposed as the directproduct of smaller digraphs. There arises two cases; either there exists a subdigraphof certain direct factor which is symmetric of order m or no direct factor of a powerdigraph has a subdigraph that is symmetric of order m. The following lemma willconsider the first case and rest of the lemmas and theorems deal with the other case.Lemma 4.2.1 plays an important role in the analysis of symmetries of the powerdigraphs.

Lemma 4.2.1. Let n = n1n2, where gcd(n1, n2) = 1. Suppose J is a subdigraph ofG(n1, k) that is symmetric of order m and I is a subdigraph of G(n2, k). Then J × Iis symmetric of order m.

28

Figure 4.2: The symmetric power digraph G(27, 45) of order 3

Proof. Let A1, A2, · · · , At be the equivalence classes of J. Then we can write

J =t∪

i=1

Ai.

From Theorem 3.4.12,

J × I = (t∪

i=1

Ai) × I (4.2.1)

=

t∪i=1

(Ai × I).

To prove that J × I is symmetric of order p, we will show that for anycomponent C ⊆ Ai × I, the number of components isomorphic to C is a multiple ofp. Since J is symmetric of order p,

S (Ai) = bp,

for some positive integer b and 1 ≤ i ≤ t. Let C1,C2, · · ·Cbp be the components of

29

Figure 4.3: The non-symmetric power digraph G(35, 8)

30

Ai. Thus again

Ai =

bp∪i=1

Ci.

From Theorem 3.4.12, we have

Ai × I = (bp∪j=1

C j) × I

=

bp∪j=1

(C j × I).

This implies that C is isomorphic to a component C∗ in C jk×I for some 1 ≤ jk ≤ bp.Now we will determine the number of components isomorphic to C∗ in Ai × I. Forthis let R be the equivalence class of C jk × I containing C∗. Then C jk × I containsS (R) components isomorphic to C∗. As

C jk � C j for all 1 ≤ j ≤ bp

therefore,C jk × I � C j × I for all 1 ≤ j ≤ bp.

Hence C j × I contains S (R) components isomorphic to C∗ for every 1 ≤ j ≤ bp.Since there are bp components in Ai,

Ai × I =bp∪j=1

(C j × I)

contains bp.S (R) components isomorphic to C∗. Therefore, from (4.2.1) and thefact that Ai is arbitrary, the number of components isomorphic to C in J × I is amultiple of p. This completes the proof. �

Lemma 4.2.2. Let n=p1 p2 · · · ps be an odd number. Then G(n, k) is a permutationon H if and only if gcd(pi − 1, k) = 1 for all 1 ≤ i ≤ s.

Proof. Suppose gcd(pi−1, k) = 1 for all 1 ≤ i ≤ s and a ∈ G(n, k). Then by (3.5.2),

N(n, k, a) = N(p1, k, a) · N(p2, k, a) · · ·N(ps, k, a).

Note that if gcd(pi, a) = 1, by Theorem 3.5.1,

N(pi, k, a) = gcd(λ(pi), k)

Hence

N(n, k, a) =s∏

i=1

gcd(λ(pi), k)

=

s∏i=1

gcd(pi − 1, k)

= 1.

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While if pi|a, thenN(pi, k, a) = N(pi, k, 0) = 1.

It now follows thatN(pi, k, a) = 1 for all a ∈ G(n, k).

This implies that indeg na = 1. Also we know that outdeg na = 1. Thus themapping f (x) ≡ xk (mod n) is bijective. Therefore, G(n, k) is a permutation on H.The converse is obvious. �

Lemma 4.2.3. If G(pα, k) (α > 1) is not symmetric of order p, then S (Comp(0)) = 1or A1(G(pα, k)) + 1.

Proof. Let S (Comp(0)) denotes the number of components isomorphic to Comp(0)and J ⊆ G(pα, k) such that J � Comp(0). Now there arises two cases:Case 1:Suppose there does not exist any k ⊆ G1(pα, k) such that K � J � Comp(0). Then Jis isomorphic to some component in G2(pα, k). Since G2(pα, k) consists of only onecomponent namely, Comp(0), J = Comp(0) which shows that S (Comp(0)) = 1.Case 2:Now suppose J ⊆ G1(n, k). Since K(Comp(0)) = 1 and J � Comp(0), K(J) =1. From Corollary 3.4.6 any two components in G1(pα, k) having fixed points areisomorphic. Thus Comp(0) is isomorphic to all those components in G1(n, k) havingfixed points. Since the number of components with fixed points are equal to thenumber of fixed points therefore,

S (Comp(0)) = 1 + A1(G(pα, k)).

This completes the proof. �

We recall that p and qi always denote distinct odd primes.

Theorem 4.2.4. Suppose that G(pα, k) is not symmetric of order p. Let N(pα, k, 0) ,N(pα, k, 1) (α > 1). If W is a permutation, then G(pα, k) ×W is symmetric of orderp if and only if W is symmetric of order p .

Proof. Suppose W is symmetric of order p. Then G(pα, k)×W is symmetric of orderp by Lemma 4.2.1.

Conversely, suppose G(pα, k)×W is symmetric of order p, whereas W is notsymmetric of order p. Now we can write

W =∪

i

Ci,

where Ci is a cycle for all i, since W, being a permutation, consists entirely of cycles.Consider a subdigraph Y of W such that

Y = {Ci : K(Ci) = t and p | At(W)}.

32

Then obviously

Z = W − Y = {C j : K(C j) = s, p - As(W)}.Since W is not symmetric of order p, Z is nonempty. It is clear that Y and Z aredisjoint and

W = Y ∪ Z.

So we can write,

G(pα, k) ×W = G(pα, k) × (Y ∪ Z).

Thus, by Theorem 3.4.12, we have

G(pα, k) ×W = (G(pα, k) × Y) ∪ (G(pα, k) × Z).

Since Y , by definition, is symmetric of order p, by Lemma 4.2.1 G(pα, k) × Y issymmetric of order p. To show that G(pα, k) ×W is not symmetric of order p, it issufficient to prove that G(pα, k) × Z is not symmetric of order p.

Let u denotes the length of the smallest cycle in Z and Z1 be the subdigraphof Z containing all the cycles of length u and Z2 = Z − Z1. Take A ⊆ Z1 suchthatK(A) = u, then

Comp(0) × A ⊆ G(pα, k) × Z1.

From Lemma 3.4.10, Comp(0) × A consists of

gcd(K(Comp(0)),K(A)) = gcd(1, u) = 1

component containing a cycle of length

lcm(K(Comp(0)),K(A)) = lcm(1, u) = u.

Moreover, from Lemma 3.5.4,

M(Comp(0) × A) = M(Comp(0)) · M(A)= N(pα, k, 0) · M(A).

Now from Lemma 3.5.2 and the fact that A, being a component of permutation, hasindegree 1, we have

M(Comp(0) × A) = pα−⌈αk ⌉ · 1

= pα−⌈αk ⌉.

Now we will show that the number of components isomorphic to Comp(0) × A inG(pα, k) × Z is not a multiple of p. Again we have

G(pα, k) × Z = G(pα, k) × (Z1 ∪ Z2)= (G(pα, k) × Z1) ∪ (G(pα, k) × Z2).

33

First we will show that G(pα, k) × Z2 does not contain any component isomorphicto Comp(0) × A. For this suppose C∗ is any component of G(pα, k) × Z2. Then

C∗ ⊆ C × ζ for some C ⊆ G(pα, k) and ζ ⊆ Z2.

Now by the definition of Z2 , K(ζ) , u. Since u is the length of the smallest cycle inZ = Z1 ∪ Z2,

K(ζ) > u. (4.2.2)

We see by Corollary 3.4.11 that the cycle length of C∗ is

K(C∗) = lcm(K(C),K(ζ)) ≥ K(ζ). (4.2.3)

Equations (4.2.2) and (4.2.3) implies

K(C∗) ≥ K(ζ) > u.

Which shows that the cycle length of C∗ is greater than u. As C∗ is arbitrarycomponent so the cycle length of every component of G(pα, k) × Z2 is greater thanu. Since K(Comp(0) × A) = u, G(pα, k) × Z2 does not contain any componentisomorphic to Comp(0) × A.

Now consider G(pα, k)×Z1. Let E denotes the equivalence class of Comp(0)in G(pα, k) and T = G(pα, k) − E. Then we can write,

G(pα, k) × Z1 = (E ∪ T ) × Z1

= (E × Z1) ∪ (T × Z1).

Now we will show that T × Z1 does not contain any component isomorphic toComp(0) × A. As

N(pα, k, 0) , N(pα, k, 1),

and 1 is fixed point of G1(pα, k), therefore, Comp(0) is not isomorphic to anycomponent in G1(pα, k). Thus it follows by Lemma 4.2.3 that E only containsComp(0). By Lemma 3.5.4, the indegree of any cycle vertex in T × Z1 is

M(T × Z1) = M(T ) · M(Z1).

As the indegree of Z1, being a component of permutation, is 1,

M(T × Z1) = M(T ) · 1 = M(T ). (4.2.4)

Now as E only contains Comp(0), T = G(pα, k)−E = G1(pα, k). Also we know thatG1(pα, k) is semi-regular of degree N(pα, k, 1) by Theorem 3.5.1 and M(Comp(0)) =N(pα, k, 0), therefore,

M(T ) = N(pα, k, 1) , N(pα, k, 0). (4.2.5)

Hence from (4.2.4) and (4.2.5),

M(T × Z1) , N(pα, k, 0).

34

Whereas,M(Comp(0) × A) = N(pα, k, 0).

Thus T × Z1 contains no components isomorphic to Comp(0) × A.Now we will consider E ×Z1. Let B1, B2, · · · , Bh be the cycles of length u in

Z1 such that h = Au(Z) and p - h, hence we can write

E × Z1 = E ×h∪

i=1

Bi

=

h∪i=1

(E × Bi).

Since A ⊆ Z1, A � Bi, for all 1 ≤ i ≤ h. Thus

Comp(0) × A = E × A � E × Bi for all 1 ≤ i ≤ h.

This implies Comp(0)×A is isomorphic to each component in E×Z1. Since E×Bi, byLemma 3.4.10, consists of only one component, there are h components in E × Z1.Thus the number of components isomorphic to Comp(0) × A in E × Z1 is not amultiple of p.Hence G(pα, k)×Z1 is not symmetric of order p, which further impliesthat G(pα, k) ×W is not symmetric of order p. This leads to a contradiction. �

The following corollaries follow immediately from Theorem 4.2.4, Theorem3.5.1 and Corollary 3.5.2.

Corollary 4.2.5. Let n be defined as in (4.1.1) and G(pα, k) be not symmetric oforder p. Suppose further N(pα, k, 0) , N(pα, k, 1). If gcd(qi − 1, k) = 1 for all1 ≤ i ≤ m, then G(n, k) is not symmetric of order p.

Proof. Sincegcd(qi − 1, k) = 1 for all 1 ≤ i ≤ m,

from Lemma 4.2.2, G(q1q2 · · · qm, k) is a permutation. Since

p - q1q2 · · · qm,

G(q1q2 · · · qm, k) is not symmetric of order p. Thus the result follows from Theorem4.2.4 �

Corollary 4.2.6. Let n be defined as in (4.1.1), gcd(p − 1, k) = 1 = gcd(p, k) andgcd(qi − 1, k) = 1 for all 1 ≤ i ≤ m. Then G(n, k) is not symmetric of order p.

Proof. From Theorem 3.5.1,

N(pα, k, 1) = gcd(λ(pα), k)= gcd(pα−1(p − 1), k).

35

Since gcd(p − 1, k) = 1 = gcd(p, k),

N(pα, k, 1) = 1.

Whereas from Lemma 3.5.2,

N(pα, k, 0) = pα−⌈αk ⌉.

Since α > 1, α − ⌈αk ⌉ ≥ 1. Thus

N(pα, k, 0) , 1.

Which shows N(pα, k, 1) , N(pα, k, 0). Now as

gcd(qi − 1, k) = 1 for all 1 ≤ i ≤ m,

therefore, from Lemma 4.2.2, G(q1q2 · · · qm, k) is a permutation not symmetric oforder p. Thus the result follows from Theorem 4.2.4 �

Corollary 4.2.7. Let n be defined as in (4.1.1), where gcd(p−1, k) , 1 and gcd(qi−1, k) = 1 for all 1 ≤ i ≤ m. Then G(n, k) is not symmetric of order p .

Proof. From Theorem 3.5.1,

N(pα, k, 1) = gcd(pα−1(p − 1), k). (4.2.6)

Since gcd(p−1, k) , 1 and p - gcd(p−1, k), N(pα, k, 1) must contain a prime factorother than p. Hence N(pα, k, 1) can not be written as a power of p. On the otherhand, from Lemma 3.5.2,

N(pα, k, 0) = pα−⌈αk ⌉.

Since α > 1, α − ⌈αk ⌉ ≥ 1. Thus N(n, k, 0) is a power of p. This implies

N(pα, k, 1) , N(pα, k, 0).

As gcd(qi − 1, k) = 1 for all 1 ≤ i ≤ m so by Lemma 4.2.2, G(q1q2 · · · qm, k) is apermutation not symmetric of order p. Hence from Theorem 4.2.4, G(n, k) is notsymmetric of order p. �

Corollary 4.2.8. Let n be defined as in (4.1.1), where pα−1|k, gcd(p − 1, k) , 1 andgcd(qi − 1, k) = 1 for all 1 ≤ i ≤ m. Then G(n, k) is not symmetric of order p .

Proof. Since pα−1|k, gcd(p − 1, k) , 1,

N(pα, k, 1) = gcd(λ(pα), k)= pα−1 gcd(p − 1, k).

As p is odd and gcd(p, p−1) = 1, therefore N(pα, k, 1) is not a power of p.Whereas,N(n, k, 0) is a power of p. Since gcd(qi − 1, k) = 1 for all 1 ≤ i ≤ m, by Lemma4.2.2 and Theorem 4.2.4, G(n, k) is not symmetric of order p. �

36

Corollary 4.2.9. Let n be defined as in (4.1.1), where p - k and gcd(qi − 1, k) = 1for all 1 ≤ i ≤ r. Then G(n, k) is not symmetric of order p .

Proof. As p - k so gcd(pα−1, k) = 1. Therefore,

N(pα, k, 1) = gcd(pα−1(p − 1), k)= gcd(p − 1, k).

This implies either N(pα, k, 1) = 1 or N(pα, k, 1) = gcd(p−1, k) , 1. But in the lattercase, N(pα, k, 1) can not be written as a power of p. As α > 1 so N(pα, k, 0) , 1.Also N(pα, k, 0) is a power of p. Hence in either case N(n, k, pα) , N(pα, k, 0). Thuscorollary follows from Lemma 4.2.2 and Theorem 4.2.4. �

Example 4.2.10. let n = 45 = 32 · 5 = p2 · q1 and k = 5. we note that gcd(p, k) =gcd(3, 5) = 1, gcd(p−1, k) = gcd(2, 5) = 1 and gcd(q1−1, k) = gcd(4, 5) = 1. FromTheorem 4.2.4 and Cor 4.2.6, G(45, 5) is not symmetric of order 3. In fact G(45, 5)is symmetric of order 5. This is shown in Figure 4.4.

Example 4.2.11. let n = 75 = 52 · 3 = p2 · q1 and k = 3. we note that p = 5 - 3 = kand gcd(q1 − 1, k) = gcd(2, 3) = 1. From Theorem 4.2.4 and Cor 4.2.9, G(75, 3) isnot symmetric of order 5. In fact G(75, 3) is symmetric of order 3. This is shown inFigure 4.5.

Lemma 4.2.12. Let n be defined as in (4.1.1). Suppose N(pα, k, 0) , N(pα, k, 1),and p - gcd(qi − 1, k) for 1 ≤ i ≤ m. Suppose further there exists 0 ≤ r ≤ m suchthat gcd(qi − 1, k) = 1 when i ≤ r and gcd(qi − 1, k) , 1 when i > r. If Ji is anycomponent of G(n, k) having a cycle of indegree N(pα, k, 0) i.e M(Ji) = N(pα, k, 0)then,∪

i

Ji � Comp(0) ×G(q1, k) × · · · ×G(qr, k) ×G2(qr+1, k) × · · · ×G2(qm, k).

Proof. Let J be any component in Comp(0)×G(q1, k)×· · ·×G(qr, k)×G2(qr+1, k)×· · · ×G2(qm, k). Then

J ⊆ Comp(0) × F1 × F2 × · · · × Fr × Ir+1 × · · · × Im,

where Fi ⊆ G(qi, k) for 1 ≤ i ≤ r and I j ⊆ G2(q j, k) where r + 1 ≤ j ≤ m. Thus, byLemma 3.5.4,

M(J) = M(Comp(0)) · M(F1) · · ·M(Fr) · M(Ir+1) · · ·M(Im). (4.2.7)

When i ≤ r,Fi ⊆ G(qi, k) and gcd(qi − 1, k) = 1.

By Theorem 3.5.3, G(qi, k) is regular. Hence

M(Fi) = 1.

37

Figure 4.4: The power digraph G(45, 5) not symmetric of order 3

When r + 1 ≤ j ≤ m,I j ⊆ G2(q j, k).

We know that 0 is the only cycle vertex in G2(q j, k) and 0 is also an isolated fixedpoint. Hence M(I j) = 1. By using (4.2.7),

M(J) = M(Comp(0)) = N(pα, k, 0).

Thus J is a component having cycle indegree N(pα, k, 0). This implies J is isomorphicto some component of

∪i Ji.

Conversely, suppose J ⊆ ∪i Ji but J is not isomorphic to any component of

Comp(0) ×G(q1, k) × · · · ×G(qr, k) ×G2(qr+1, k) × · · · ×G2(qm, k).

Since

n = pαq1q2 · · · qm, where p, qi are distinct primes andα > 1,

38

by direct product of power digraphs, we can write

G(n, k) � G(pα, k) ×G(q1, k) × · · · ×G(qr, k) ×G(qr+1, k) × · · · ×G(qm, k).

Now J ⊆ G(n, k) implies

J ⊆ J′ × F1 · · · × Fr × Ir+1 · · · × Im,

where J′ ⊆ G(pα, k) , Fi ⊆ G(qi, k) for 1 ≤ i ≤ r and I j ⊆ G(q j, k) for r + 1 ≤ j ≤ m.Therefore there arises three possibilities,

Case 1Suppose J′ ⊆ G1(pα, k), Fi ⊆ G(qi, k) for 1 ≤ i ≤ r and I j ⊆ G2(q j, k) wherer + 1 ≤ j ≤ m. Thus, by Lemma 3.5.4

M(J) = M(J′) · M(F1) · · ·M(Fr) · M(Ir+1) · · ·M(Im).

Now by using Theorem 3.5.1 and the fact that M(Fi) = 1 when 1 ≤ i ≤ r, M(I j) = 1when r + 1 ≤ j ≤ m, we get

M(J) = N(pα, k, 1) · 1 · 1 · · · 1= N(pα, k, 1), N(pα, k, 0).

Case 2Suppose J′ = Comp(0) and there exist I j ⊆ G1(q j, k) for some j > r. Let x be thesmallest integer such that x > r and Ix ⊆ G1(qx, k), then M(Ix) = 0 or gcd(qx − 1, k).Since we are considering the cycle vertices, M(Ix) = gcd(qx − 1, k) , 1. It followsthat

M(J) = N(pα, k, 0) · 1 · 1 · · ·∏i≥x

gcd(qi − 1, k)

= N(pα, k, 0) ·∏i≥x

gcd(qi − 1, k)

, N(pα, k, 0).

Case 3Suppose C ⊆ G1(pα, k) and there exist I j ⊆ G1(q j, k) for some j > r. Let x be thesmallest integer such that x > r and Ix ⊆ G1(qx, k), then M(Ix) = 0 or gcd(qx − 1, k).Since we are considering the cycle vertices, M(Ix) = gcd(qx − 1, k) , 1. Thus wesee that

M(J) = N(pα, k, 1) · 1 · · · 1 ·∏i≥x

gcd(qi − 1, k)

= N(pα, k, 1) ·∏i≥x

gcd(qi − 1, k).

39

Since p - gcd(qi − 1, k) and gcd(qi − 1, k) , 1 for i ≥ x, M(J) cannot be written as apower of p. But by Lemma 3.5.2,

N(pα, k, 0) = pα−⌈αk ⌉ , 1.

This shows thatM(J) , N(pα, k, 0).

All cases lead to a contradiction. �

Theorem 4.2.13. Let n be defined as in (4.1.1) and G(pα, k) be not symmetric oforder p. Suppose further N(pα, k, 0) , N(pα, k, 1) and p - gcd(qi − 1, k) for 1 ≤ i ≤m. Then G(n, k) is not symmetric of order p .

Proof. Since n = pαq1q2 · · · qm,

G(n, k) � G(pα, k) ×G(q1, k) × · · · ×G(qm, k).

There are two possibilities, either gcd(qi − 1, k) = 1 for all 1 ≤ i ≤ m or there existssome x such that gcd(qx − 1, k) , 1.If gcd(qi − 1, k) = 1 for all 1 ≤ i ≤ m, then by lemma 4.2.2, G(q1q2 · · · qr, k) is apermutation (regular). Now since p and qi are distinct primes,

p - q1q2 · · · qm.

This implies G(q1q2 · · · qr, k) is not symmetric of order p. Also G(pα, k) is notsymmetric of order p. Hence by Theorem 4.2.4, G(n, k) is not symmetric of orderp.Thus we may suppose there exists 0 < r < m such that gcd(qi − 1, k) = 1 when i ≤ rand gcd(qi − 1, k) , 1 when i > r. If Ji is any component of G(n, k) having a cycleof indegree N(pα, k, 0), i.e, M(Ji) = N(pα, k, 0), then by Lemma 4.2.12,∪

i

Ji � Comp(0) ×G(q1, k) × · · · ×G(qr, k) ×G2(qr+1, k) × · · · ×G2(qm, k).

Since G2(qi, k) for r + 1 ≤ i ≤ m contains only the 0 vertex, we can write as,

Comp(0) ×G(q1, k) × · · · ×G(qr, k) ×G2(qr+1, k) × · · · ×G2(qm, k)� Comp(0) ×G(q1, k) × · · · ×G(qr, k).

Hence ∪i

Ji � Comp(0) ×G(q1, k) × · · · ×G(qr, k),

where gcd(qi−1, k) = 1 for 1 ≤ i ≤ r.We note that G(q1, k)×· · ·×G(qr, k) is regularby Lemma 4.2.2. Thus G(q1, k) × · · · ×G(qr, k) entirely consists of isolated cycles.Consider a subdigraph Y of W = G(q1, k) × · · · ×G(qr, k) such that

Y = {Ci : K(Ci) = t and p | At(W)}.

40

Then obviously

Z = W − Y = {C j : K(C j) = s, p - As(W)}.As p - q1 · · · qr so W is not symmetric of order p. Hence Z is nonempty. It is clearthat Y and Z are disjoint and

W = Y ∪ Z.

So we can write,

Comp(0) ×W = Comp(0) × (Y ∪ Z).

Thus, by Theorem 3.4.12, we have

Comp(0) ×W = (Comp(0) × Y) ∪ (Comp(0) × Z).

Since Y by definition is symmetric of order p, by Lemma 4.2.1, we obtain Comp(0)×Y is symmetric of order p. To show that Comp(0) ×W is not symmetric of order p,it is sufficient to prove that Comp(0) × Z is not symmetric of order p.

Let u denotes the length of the smallest cycle in Z and Z1 be the subdigraphof Z containing all the cycles of length u and Z2 = Z − Z1. Take A ⊆ Z1 suchthatK(A) = u, then

Comp(0) × A ⊆ Comp(0) × Z1.

From Lemma 3.4.10, Comp(0) × A consists of

gcd(K(Comp(0)),K(A)) = gcd(1, u) = 1

component containing a cycle of length

lcm(K(Comp(0)),K(A)) = lcm(1, u) = u.

Moreover, from Lemma 3.5.4, Lemma 3.5.2 and the fact that A, being a componentof permutation, has indegree 1, we have

M(Comp(0) × A) = M(Comp(0)) · M(A)= N(pα, k, 0) · M(A).= pα−⌈

αk ⌉ · 1

= pα−⌈αk ⌉.

Now we will show that the number of components isomorphic to Comp(0) × A inComp(0) × Z is not a multiple of p. Again we have

Comp(0) × Z = Comp(0) × (Z1 ∪ Z2)= (Comp(0) × Z1) ∪ (Comp(0) × Z2).

First we will show that Comp(0) × Z2 does not contain any component isomorphicto Comp(0) × A. For this suppose C∗ is any component of Comp(0) × Z2. Then

C∗ ⊆ Comp(0) × ζ for some ζ ⊆ Z2.

41

Now by the definition of Z2, K(ζ) , u. Since u is the length of the smallest cycle inZ = Z1 ∪ Z2,

K(ζ) > u. (4.2.8)

We see by Corollary 3.4.11 that the cycle length of C∗ is

K(C∗) = lcm(K(Com(0)),K(ζ)) ≥ K(ζ). (4.2.9)

It is clear from (4.2.8) and (4.2.9)

K(C∗) ≥ K(ζ) > u. (4.2.10)

Which shows that the cycle length of C∗ is greater than u.As C∗ is arbitrary componentso cycle length of every component of Comp(0) × Z2 is greater than u. SinceK(Comp(0)×A) = u, Comp(0)×Z2 does not contain any component isomorphic toComp(0) × A.

Now consider Comp(0) × Z1. Let B1, B2, · · · , Bh be the cycles of length u inZ1 such that h = Au(Z) and p - h, Hence we can write

Comp(0) × Z1 = Comp(0) ×h∪

i=1

Bi

=

h∪i=1

(Comp(0) × Bi).

Since A ⊆ Z1, A � Bi, for all 1 ≤ i ≤ h. Thus

Comp(0) × A = Comp(0) × A � Comp(0) × Bi for all 1 ≤ i ≤ h.

Hence Comp(0) × A is isomorphic to each component in Comp(0) × Z1. SinceComp(0) × Bi, by Lemma 3.4.10, consists of only one component, there are hcomponents in Comp(0) ×Z1. Thus the number of components isomorphic to Comp(0)×A in Comp(0)×Z1 is not a multiple of p. Hence Comp(0)×Z1 is not symmetricof order p, which further implies that Comp(0) × G(q1, k) × · · · × G(qr, k) is notsymmetric of order p. Hence

∪i Ji is not symmetric of order p. Since

∪i Ji contains

all components of G(n, k) having cycle indegree N(pα, k, 0), G(n, k) is not symmetricof order p. �

The following corollaries are immediate consequences of Theorem 4.2.13,Theore-m 3.5.1 and Corollary 3.5.2.

Corollary 4.2.14. Let n be defined as in (4.1.1) and gcd(p − 1, k) = 1 = gcd(p, k)and p - (qi − 1) for all 1 ≤ i ≤ m. Then G(n, k) is not symmetric of order p.

Proof. Since gcd(p − 1, k) = 1 = gcd(p, k), from Theorem 3.5.1,

N(pα, k, 1) = gcd(λ(pα), k)= gcd(pα−1(p − 1), k) = 1.

42

Whereas, from Lemma 3.5.2, N(pα, k, 0) = pα−⌈αk ⌉. Since α > 1, N(n, k, 0) is a

power of p. This implies

N(pα, k, 1) , N(pα, k, 0).

Since gcd(p, k) = 1 and p - (qi − 1), p - gcd(qi − 1, k). Hence, from Theorem 4.2.13,G(n, k) is not symmetric of order p. �

Corollary 4.2.15. Let n be defined as in (4.1.1), gcd(p − 1, k) , 1 and p - (qi − 1)for all 1 ≤ i ≤ m. Then G(n, k) is not symmetric of order p.

Proof. From Theorem 3.5.1,

N(pα, k, 1) = gcd(λ(pα), k)= gcd(pα−1(p − 1), k).

Since gcd(p−1, k) , 1 and gcd(p, p−1) = 1, N(pα, k, 1) must contain a prime factorother than p. Thus N(pα, k, 1) can not be written as a power of p. On the other hand,from Lemma 3.5.2,

N(pα, k, 0) = pα−⌈αk ⌉.

Since α > 1, α − ⌈αk ⌉ ≥ 1. Thus N(pα, k, 0) is a power of p. This implies

N(pα, k, 1) , N(pα, k, 0).

Also p - gcd((qi − 1), k) for all 1 ≤ i ≤ m. By using Theorem 4.2.13, proof iscompleted. �

Corollary 4.2.16. Let n be defined as in (4.1.1). Further suppose p - k and p -(qi − 1) for all 1 ≤ i ≤ m. Then G(n, k) is not symmetric of order p.

Proof. Since p - k, from Theorem 3.5.1,

N(pα, k, 1) = gcd(λ(pα), k)= gcd(p − 1, k).

Since gcd(p, p − 1) = 1, N(pα, k, 1) can not be written as a power of p. Whereas,from Lemma 3.5.2, N(pα, k, 0) is a power of p. Thus

N(pα, k, 1) , N(pα, k, 0)

The Corollary follows from Theorem 4.2.13. �

Corollary 4.2.17. Let n be defined as in (4.1.1), pα−1|k, (p− 1)|k and p - (qi − 1) forall 1 ≤ i ≤ m. Then G(n, k) is not symmetric of order p.

Proof. From Theorem 3.5.1 and the fact that pα−1|k, (p − 1)|k, it follows

N(pα, k, 1) = pα−1(p − 1). (4.2.11)

Now from Corollary 3.5.2 and (4.2.11), N(pα, k, 1) , N(pα, k, 0). The Corollaryfollows from Theorem 4.2.13. �

43

Example 4.2.18. let n = 99 = 32 ·11 = p2 ·q1 and k = 5.We note that p = 3 - 5 = kand gcd(q1−1, k) = gcd(10, 5) = 5 so that p = 3 - 5 = gcd(q1−1, k). From Theorem4.2.4 and Corollary 4.2.9, G(99, 5) is not symmetric of order 5. This is shown inFigure 4.6.

Lemma 4.2.19. Let pα−1|k (α > 1) and gcd(p − 1, k) = 1. If c1 and c2 are cyclevertices of G(pα, k), then T (c1) � T (c2).

Proof. If c1 and c2 are cycle vertices of G1(pα, k), then by Theorem 3.4.5, T (c1) �T (c2). Therefore, we may assume that

c1 ∈ G1(pα, k) and 0 = c2 ∈ Comp(0) = G2(pα, k).

Since pα−1|k, α ≤ k and gcd(p − 1, k) = 1, by Theorem 3.5.1,

N(pα, k, c1) = gcd(λ(pα), k)= gcd(pα−1(p − 1), k)= pα−1.

By Corollary 3.5.2, N(pα, k, 0) = pα−1. Hence both the cycle vertices c1 and c2 havethe same indegree.

Now we know that the number of vertices not relatively prime to pα is

pα − ϕ(pα) = pα − pα−1(p − 1) = pα−1.

Hence G2(pα, k) contains pα−1 vertices. This implies that all the non cycle verticesof G2(pα, k) are at a distance 1 from c2 (i.e. height(Comp(0)) = 1). Since

N(pα, k, c1) = pα−1,

the number of vertices in T (c1) must be greater or equal to pα−1. Now if T (c1) �T (c2), then there must exists a non cycle vertex in component containing c1 whichis at a distance greater than 1 from c1 (i.e. height of that non cycle vertex is greaterthan 1). Thus T (c1) must contains more than pα−1 vertices.

Now from Corollary 3.4.2, there are p − 1 cycle vertices of G1(pα, k) soG(pα, k) contains p cycle vertices. Hence the total number of vertices in G(pα, k)will be greater than

pα−1 + (p − 1)pα−1 = pα

which leads to a contradiction. �

Lemma 4.2.20. If pα−1|k (α > 1) and gcd(p−1, k) = 1, then the length of the longestcycle in G(pα, k) is ordp−1k and all the cycle lengths in G(pα, k), divide ordp−1k.

44

Proof. From Corollary 3.4.3, the length of any cycle in G1(pα, k) is given as:

l(d) = orddk, (4.2.12)

where d is the divisor of the highest factor of λ(pα) relatively prime to k. Now fromCorollary 3.4.4, the length of the longest cycle of G1(pα, k) is orduk, where u isthe highest factor of λ(pα) = pα−1(p − 1) relatively prime to k. Since pα−1| k andgcd(p − 1, k) = 1, u = p − 1. Hence the length of the longest cycle in G(pα, k) is

ordp−1k. (4.2.13)

Since d is the divisor of the highest factor of λ(pα) relatively prime to k, d | p − 1.Now we know, from (Chartrand and Oellermann 1993), that for x, a, b ∈ N, if a | bthen ordax | ordbx. Thus from (4.2.12) and (4.2.13),

orddk | ordp−1k.

�

Lemma 4.2.21. Let W be a directed graph such that W is a permutation (regulardigraph). Suppose C is any cycle of W of length t such that, ordp−1k | t and p -At(W). If pα−1| k (α > 1) and gcd(p − 1, k) = 1, then G(pα, k) × C is symmetric oforder p .

Proof. Let C be any cycle of W of length t such that p - At(W). Suppose D1,D2, · · · ,Da

are the components of G(pα, k), where D1 = Comp(0) and Di ⊆ G1(n, k) for2 ≤ i ≤ a. Now we know that M(D1) = N(pα, k, 0). Therefore, from Corollary3.5.2,

M(D1) = pα−⌈αk ⌉.

Since pα−1|k, α < k. This implies M(D1) = pα−1.

On the other hand, for all Di, where 2 ≤ i ≤ a, from Theorem 3.5.1 and thefact that the cycle vertices have positive indegree, it is clear that

M(Di) = N(pα, k, 1)= gcd(λ(pα), k)= gcd(pα−1(p − 1), k).

Since pα−1|k and gcd(p − 1, k) = 1, M(Di) = pα−1 for 2 ≤ i ≤ a. Thus

M(Di) = pα−1 for 1 ≤ i ≤ a. (4.2.14)

If H denotes the number of cycle vertices of G(pα, k), that is, H =∑a

i=1 K(Di), thenH = p by the same argument as in the proof of Lemma 4.2.19. Now by Lemma4.2.20, K(Di)|ordp−1k for all 1 ≤ i ≤ a, but by our assumption ord p−1k | t. Thisimplies that

K(Di)|t for all 1 ≤ i ≤ a. (4.2.15)

45

Now G(pα, k) ×C, by Theorem 3.4.12, can be written as

G(pα, k) ×C = (a∪

i=1

Di) ×C (4.2.16)

=

a∪i=1

(Di ×C).

By Lemma 3.4.10 and (4.2.15), Di × C consists entirely of gcd(K(Di), t) = K(Di)components having cycle length lcm(K(Di), t) = t. Hence from (4.2.16), G(pα, k) ×C consists entirely of ΣiK(Di) = p components having cycle length t.

Now we will find the indegree of cycle vertices of components of G(pα, k)×C. Let C′ be any component of G(pα, k) × C. Then from Lemma 3.5.4, M(C′) =M(Di) · M(C). From (4.2.14) and the fact that C, being a permutation, has indegree1, we have

M(C′) = M(Di) = pα−1.

Thus each tree attached to a cycle vertex in G(pα, k) ×C has at least pα−1 vertices.

To complete the proof, we must show that all the components of G(pα, k)×Care isomorphic. As we have already proved that every componentC′ of G(pα, k)×Chas the same cycle indegree M(C′) and the cycle length K(C′). To prove isomorphismwe only have to show that all the cycle vertices in G(pα, k) × C have isomorphicattached tree.

We claim that each tree attached to a cycle vertex in G(pα, k)×C has exactlypα−1 vertices. In other words, distance of all the non cycle vertices from a cyclevertex is 1. Suppose to the contrary that there exists a cycle vertex in G(pα, k) × Cwhose associated tree has more than pα−1 vertices. As we have shown that G(pα, k)×C consists entirely of p components having cycle length t therefore, it containsexactly pt cycle vertices. Thus the number of vertices in G(pα, k) × C is greaterthan pα−1 · pt = pαt. This contradicts the fact that G(pα, k) × C has exactly pαtvertices. Thus each of the trees attached to cycle vertices in G(pα, k)×C has height1 and contains exactly pα−1 vertices. Thus all cycle vertices in G(pα, k) × C haveisomorphic attached tree. Consequently, G(pα, k) ×C is symmetric of order p. �

Lemma 4.2.22. Let W be a directed graph such that W is a permutation. Supposepα−1| k and gcd(p − 1, k) = 1 and for all t ∈ N ordp−1k | t whenever p - At(W). ThenG(pα, k) ×W is symmetric of order p, where α > 1 .

Proof. Suppose that for every t ∈ N, p - At(W) whenever ordp−1k | t. Since W is apermutation so all of its components Ci are cycles. Consider a subgraph Y of Wsuch that

Y = {Ci : L(Ci) = t and p|At(W)} .

Then obviously

Z = W − Y = {C j : L(C j) = s, p - As(W),ordp−1k | s}.

46

We can write by Theorem 3.4.12,

G(pα, k) ×W = G(pα, k) × (Y ∪ Z)= (G(pα, k) × Y) ∪ (G(pα, k) × Z).

Since Y , by definition, is symmetric of order p, it follows from Lemma 4.2.1 thatG(pα, k) × Y is symmetric of order p. To show that G(pα, k) × W is symmetric oforder p it is sufficient to prove that G(pα, k) × Z is symmetric of order p.Let E1, E2, · · · , Em be the cycles of Z. Then Z =

∪mi=1 Ei. Thus we can write,

G(pα, k) × Z = G(pα, k) ×m∪

i=1

Ei

=

m∪i=1

(G((pα, k) × Ei).

By Lemma 4.2.21, G(pα, k) × Ei is symmetric of order p for all 1 ≤ i ≤ m. HenceG(pα, k) × Z is symmetric of order p which further implies that G(pα, k) × W issymmetric of order p. �

Lemma 4.2.23. Let W be a directed graph such that W is a permutation. Supposepα−1|k, gcd(p − 1, k) = 1 and G(pα, k) is not symmetric of order p. If there existst ∈ N such that ordp−1k - t and p - At(W), then G(pα, k) × W is not symmetric oforder p where α > 1 .

Proof. Let Y be a subgraph of W consisting of the cycles of length t such thatp|At(W) or ordp−1k|t whenever p - At(W) and Z be the subdigraph of W consistingof the cycles of length t such that p - At(W) and ord p−1k - t. Now again we canwrite by Theorem 3.4.12,

G(pα, k) ×W = G(pα, k) × (Y ∪ Z)= (G(pα, k) × Y) ∪ (G(pα, k) × Z).

From Lemmas 4.2.1 and 4.2.22, G(pα, k)× Y is symmetric of order p. Now to showthat G(pα, k)×W is not symmetric of order p, it is sufficient to show that G(pα, k)×Zis not symmetric of order p.

Let u denotes the length of the smallest cycle in Z. Suppose Z1 is the subdigr-aph of Z containing all cycles of length u and Z2 = Z − Z1. If A ⊆ Z1, then byLemma 3.4.10, Comp(0) × A ⊆ G(pα, k) × Z1 consists of only one componentcontaining a cycle of length u. It follows from Lemma 3.5.4 and the fact thatA, being a component of permutation, has indegree 1 that M(Comp(0) × A) =M(Comp(0)) · M(A) = N(pα, k, 0). Since pα−1| k, α < k. Thus from Corollary3.5.2, M(Comp(0) × A) = pα−1.

Now we will show that the number of components isomorphic to Comp(0)×A in G(pα, k) × Z is not a multiple of p. Again we can write,

G(pα, k) × Z = G(pα, k) × (Z1 ∪ Z2)= (G(pα, k) × Z1) ∪ (G(pα, k) × Z2).

47

We can easily prove that G(pα, k) × Z2 does not contain any component isomorphicto Comp(0) × A by the same argument as used in Theorem 4.2.4. Now to show thatG(pα, k) × Z is not symmetric of order p, we need only to prove that G(pα, k) × Z1

is not symmetric of order p. Consider the subdigraph I of G(pα, k) containing thosecomponents whose cycle length divides u and let J = G(pα, k) − I. Since G(pα, k)contains a fixed point, we see that I , ϕ. Now we know that the length of longestcycle of G(pα, k), by Corollary 3.4.4, is ordp−1k. But by the hypothesis of Theorem,ordp−1k - u. Hence we find that J , ϕ. Thus we can write,

G(pα, k) × Z1 = (I ∪ J) × Z1

= (I × Z1) ∪ (J × Z1).

It is easy to verify that J × Z1 does not contain any component isomorphic toComp(0) × A, as all components of J × Z1 have cycle length K(J) which is greaterthan u. Let m be the number of cycle vertices in I. Since I and J are non empty,we see that 0 < m < p, hence p - m. Suppose Z1 contains h cycles of length usuch that p - h. Then I × Z1 contains mh components having cycle length u, cycleindegree pα−1 and the tree associated to each cycle vertex in I × Z1 is the same as inComp(0)× A by Lemma 4.2.19. Hence all mh components in I × Z1 are isomorphicto Comp(0) × A but p - mh. Thus G(pα, k) ×W is not symmetric of order p . �

Theorem 4.2.24. Let W be a directed graph such that W is a permutation. Supposepα−1| k (α > 1), gcd(p − 1, k) = 1 and G(pα, k) is not symmetric of order p. ThenG(pα, k) × W is symmetric of order p if and only if for every t ∈ N, ordp−1k | twhenever p - At(W).

Proof. This follows from Lemmas 4.2.22 and 4.2.23. �

Lemma 4.2.25. Let n = pαq1q2 · · · qs (α > 1), p - gcd(qi − 1, k) = 1 for all1 ≤ i ≤ s , pα−1|k and gcd(p − 1, k) = 1. Suppose there exists 0 < r ≤ s such thatgcd(qi − 1, k) = 1 when i ≤ r and gcd(qi − 1, k) , 1 when i > r. Further assume thatJi is any component of G(n, k) having cycle indegree pα−1, i.e, M(Ji) = pα−1. Thenwe have∪

i

Ji � G(pα, k) ×G(q1, k) × · · · ×G(qr, k) ×G2(qr+1, k) × · · · ×G2(qm, k).

Proof. Let J be any component in G(pα, k)×G(q1, k)× · · · ×G(qr, k)×G2(qr+1, k)×· · · ×G2(qm, k). Then

J ⊆ C × F1 × F2 × · · · × Fr × Ir+1 × · · · × Im,

where C ∈ G(pα, k), Fi ⊆ G(qi, k) for 1 ≤ i ≤ r and I j ⊆ G2(q j, k), where r + 1 ≤j ≤ m. Thus, by Lemma 3.5.4,

M(J) = M(C) · M(F1) · · ·M(Fr) · M(Ir+1) · · ·M(Im). (4.2.17)

When i ≤ r,Fi ⊆ G(qi, k) and gcd(qi − 1, k) = 1.

48

Since by Theorem 3.5.3, G(qi, k) is regular,

M(Fi) = 1

When r + 1 ≤ j ≤ m,I j ⊆ G2(q j, k)

We know that 0 is the only cycle vertex in G2(q j, k) and 0 is also an isolated fixedpoint. Hence M(I j) = 1. By using (4.2.17),

M(J) = M(C).

Now C ⊆ G(pα, k) so C ⊆ G1(pα, k) or C ⊆ G2(pα, k). If C ⊆ G1(pα, k), then for anycycle vertex c ∈ C

M(C) = N(pα, k, c).

Thus from Theorem 3.5.1,

M(C) = N(pα, k, c)= gcd(λ(pei

i ), k)= gcd(pα−1(p − 1), k).

Since pα−1|k and gcd(p − 1, k) = 1,

M(C) = pα−1.

Now if C ⊆ G2(pα, k), then Comp(0), being the only component in G2(pα, k), mustequal to C. Hence, from Lemma 3.5.2

M(C) = M(Comp(0)) = N(pα, k, 0) = pα−⌈αk ⌉.

Since pα − 1|k, α < k. This implies ⌈αk ⌉ = 1 which shows

M(C) = pα−⌈αk ⌉ = pα−1.

In both cases J is a component having cycle indegree pα−1 i.e.

M(C) = pα−1. (4.2.18)

This implies J is isomorphic to some component of∪

i Ji.Conversely, suppose J ⊆ ∪i Ji but J is not isomorphic to any component of

G(pα, k) ×G(q1, k) × · · · ×G(qr, k) ×G2(qr+1, k) × · · · ×G2(qm, k).

Since

n = pαq1q2 · · · qm, where p, qi are distinct primes andα > 1,

by direct product of power digraphs, we can write

G(n, k) � G(pα, k) ×G(q1, k) × · · · ×G(qr, k) ×G(qr+1, k) × · · · ×G(qm, k).

49

Now J ⊆ G(n, k) implies

J ⊆ C × F1 · · · × Fr × Ir+1 · · · × Im,

where C ⊆ G(pα, k) , Fi ⊆ G(qi, k) for 1 ≤ i ≤ r and I j ⊆ G(q j, k) for r + 1 ≤ j ≤ m. Since J is not isomorphic to any component of

G(pα, k) ×G(q1, k) × · · · ×G(qr, k) ×G2(qr+1, k) × · · · ×G2(qm, k),

there exist I j ⊆ G1(q j, k) for some j > r. Let x be the smallest integer such that x > rand Ix ⊆ G1(qx, k), then M(Ix) = 0 or gcd(qx − 1, k). Since we are considering thecycle vertices, M(Ix) = gcd(qx − 1, k) , 1. It follows that

M(J) = pα−1 · 1 · 1 · · ·∏i≥x

gcd(qi − 1, k)

= pα−1 ·∏i≥x

gcd(qi − 1, k)

Since p - gcd(qi − 1, k) and gcd(qi − 1, k) , 1 for i ≥ x, M(J) cannot be written as apower of p. Which shows that

M(J) , pα−1.

This lead to a contradiction. �

Note that if L = {p1, p2, · · · , pr} where each pi is a distinct odd prime, thenG(L, k) denotes the digraph G(p1 p2 · · · pr, k).

Theorem 4.2.26. Let n = pαq1q2 · · · qs ,1 < α ≤ k, p - gcd(qi−1, k) for all 1 ≤ i ≤ sand G(pα, k) not be symmetric of order p. Then G(n, k) is symmetric of order p ifand only if the following conditions are satisfied.

1. pα−1|k and gcd(p − 1, k) = 1.

2. The set L = {qi : gcd(qi − 1, k) = 1} is nonempty and for every t ∈ N such thatp - At(G(L, k)), ordp−1k | t.

Proof. Suppose both conditions are true. We need to show that G(n, k) is symmetricof order p. Let

Q = {q1, q2, · · · , qs},L = {qi1 , · · · , qi f }, and

Q − L = {q j1 , · · · , q je}.

Then we can write

G(n, k) � G(pα, k) ×G(qi1 · · · qi f , k) ×G(q j1 , · · · , q je , k). (4.2.19)

50

Since gcd(qi j − 1, k) = 1 for all qi j ∈ L, by Lemma 4.2.2, G(qi1 · · · qi f , k) is apermutation. Now it follows from Lemma 4.2.22 and condition 2 that

G(pα, k) ×G(qi1 · · · qi f , k)

is symmetric of order p and consequently, by Lemma 4.2.1, G(n, k) is symmetric oforder p.

Conversely, suppose G(n, k) is symmetric of order p but condition (1) or (2)is not satisfied. Firstly, suppose condition (1) is not true, that is, either pα−1 - k orgcd(p − 1, k) , 1.Suppose pα−1 - k. From Lemma 3.5.2,

N(pα, k, 0) = pα−⌈αk ⌉.

Since α ≤ k,N(pα, k, 0) = pα−1.

On the other hand, by Theorem 3.5.1 and the fact that pα−1 - k

N(pα, k, 1) = gcd(pα−1(p−1)) , pα−1.

Now suppose gcd(p − 1, k) , 1. Again from Lemma 3.5.2,

N(pα, k, 0) = pα−1.

But by Theorem 3.5.1, N(pα, k, 1) contains a prime factor other than p. In eithercase

N(pα, k, 0) , N(pα, k, 1).

Thus by Theorem 4.2.13, G(n, k) is not symmetric of order p, which is a contradiction.

Now suppose condition (1) is true but condition (2) is not true. Then eitherL is empty or there exists t ∈ N such that ordp−1k - t and p - At(W).Suppose L is empty. Then gcd(qi − 1, k) , 1 for all 1 ≤ i ≤ s. If Ji is a componentof G(n, k) having cycle indegree pα−1, then by Lemma 4.2.25,∪

i

Ji � G(pα, k) ×G2(q1, k) × · · · ×G2(qs, k)

Since G2(qi, k) contains only the vertex 0 for all 1 ≤ i ≤ s, we see that∪

i Ji �G(pα, k). It follows that the subdigraph of G(n, k) containing all components havingcycle indegree pα−1 is not symmetric of order p. Hence G(n, k) is not symmetric oforder p.

Now Suppose L is non empty but there exists t ∈ N such that ordp−1k - t andp - At(W). Again as G2(qi, k) contains only the vertex 0 for all 1 ≤ i ≤ s, therefore,

51

from equation (4.2.19) and Lemma 4.2.25, the subdigraph of G(n, k) containing allcomponents having cycle indegree pα−1 is isomorphic to

G(n, k) � G(pα, k) ×G(qi1 · · · qi f , k) · · · ×G2(q1, k) × · · · ×G2(qs, k)� G(pα, k) ×G(qi1 · · · qi f , k).

Which is not symmetric of order p by Lemma 4.2.23. Thus G(n, k) is not symmetricof order p which leads to a contradiction. �

It is not immediately clear that the hypothesis of Theorem 4.2.26 can in factbe satisfied. Example 38 given below shows that this indeed can occur.

Example 4.2.27. Let n = pαq1, where p = 7, α = 2 , q1 = 103 and k = 35. Considerthe digraph

G(n, k) = G(72 · 103, 35) = G(5047, 35)

Then conditions (1) and (2) of Theorem 4.2.26 are both satisfied for G(72 · 103, 35)and thus G(72 · 103, 35) is symmetric of order 7. In particular, G(72, 35) is notsymmetric of order 7, 7|k = 35 and gcd(q1, k) = gcd(103, 35) = 1. Moreovereach cycle in G(103, 35) has length equal to 1 or 2 , 7 | A1(G(103, 35)) = 35,7 - A2(G(103, 35)) = 34 and

ord635 = 2|A2(G(103, 35)) = 34

We note that, by Theorem 3.4.9, A1(G(72, 35)) = 3 and thus G(72, 35) is not symmetricof order 7. We also note that by Theorem 3.4.9, A1(G(103, 35)) = 35 and A2(G(103, 35)) =34, while by Corollary 3.4.4, the longest cycle in G(103, 35) has length 2, since

gcd(λ(103), 35) = gcd(102, 35) = 1 and ord10235 = 2.

The digraphs G(72, 35) and G(103, 35) are shown in Figures 4.7 and 4.8 respectively.

Example 4.2.28. Let n = pαq1 = 52 · 29 and k = 15. Consider the digraph

G(n, k) = G(52 · 29, 35) = G(725, 15)

From Theorem 3.4.9 A1(G(25, 15) = gcd(p(p− 1), k − 1)+ 1 = gcd(20, 14)+ 1 = 3.Sice 5 - A1(G(25, 15), G(52, 15) is not symmetric of order 5. We note that 5|k = 15,pα−1 = 5 | k and gcd(p − 1, k) = gcd(4, 15) = 1. Thus condition (1) is satisfied.Now gcd(q1 − 1, k) = gcd(28, 15) = 1 so the set L given in condition (2) of Theorem4.2.26 is non empty. From Corollary 3.4.4, the length of longest cycle in G(29, 15)is ord2815 = 2. Hence each cycle in G(29, 15) has length equal to 1 or 2. We notethat

A1(G(29, 15)) = gcd(28, 14) + 1 = 15,

A2(G(29, 15)) =12

(gcd(28, 224) + 1 − A1(G(29, 15))) = (29 − 15)/2 = 7.

Which shows 5 | A1(G(29, 15)) but 5 - A2(G(29, 15)). Now we have

ordp−1k = ord415 = 2|2 = t

Thus condition (2) is satisfied. Hence G(725, 15) is symmetric of order 5.

52

The power digraphs G(52, 35) and G(29, 35) are shown in Figures 4.10 and4.9, respectively.

4.3 SummaryThe existence of the symmetry of G(n, k) for n = pαq1 · · · qm, where p and qi

are distinct odd primes is explored in this chapter. First it is proved that if oneof the direct factors of a power digraph is symmetric of order p, then the wholeG(n, k) is also symmetric of order p. Secondly, the complete structure of G(pα, k) isdetermined. The symmetries of a digraph which is a direct product of G(pα, k) anda regular digraph are studied. With the help of this result it has been shown that ifG(q1 · · · qm, k) is regular and indegrees of 0 and 1 in G(pα, k) are not the same, thenG(n, k) is not symmetric of order p. Further, it is established that the condition “ Ifindegrees of 0 and 1 in G(pα, k) are not the same” is sufficient for the power digraphsnot to be symmetric of order p. Finally, the necessary and sufficient conditions forG(n, k) to be symmetric of order p are derived.

53

Figure 4.5: The power digraph G(75, 3) not symmetric of order 5

54

Figure 4.6: the power digraph G(99, 5) not symmetric of order 5

55

Figure 4.7: The power digraph G(49, 35)

56

Figure 4.8: The power digraph G(103, 35)

57

Figure 4.9: The power digraph G(29, 15)

58

Figure 4.10: The power digraph G(25, 15)

59

Chapter 5

On the Heights of the PowerDigraphs

5.1 IntroductionThe concept of the heights of vertices and components in a power digraph plays animportant role in describing its structure. With the help of the heights of verticesit can be possible to determine the configuration of the trees associated with thecycle vertices. It is also used in studying the symmetry of digraphs. Since thecomplete structure of a digraph depends heavily upon the heights of its vertices,it is worthwhile to determine the nature of the heights of all vertices in a powerdigraph.

Definition 5.1.1. Let Comp(a) be a component containing a vertex a ∈ G(n, k) andd(a, b) denote the length of the shortest directed path from a to b. Then height of avertex a ∈ C, denoted by, height(a) is defined as

height(a) = min{d(a, ci) | ci are cycle vertices of Comp(a)}.

Definition 5.1.2. The height of any component C ⊆ G(n, k) is defined to be

height(C) = maxa∈C

height(a),

and the height of G(n, k) as

height(G(n, k)) = maxC⊆G(n,k) height(C).

Definition 5.1.3. A vertex x ∈ G(n, k) is said to be at level j ≥ 1 if there exists adirected path of least length j which ends at x and does not contain any directededge from a cycle. A vertex is at level 0 if there does not exist such a path. Acomponent C ⊆ G(n, k) is said to have s levels if the highest level of a vertex in Cis s − 1.

61

It is obvious that the vertex at height 0 and level s − 1 is a cycle vertex andthe vertex at level 0 is a vertex having indegree 0. We can also see that for anycomponent C of G(n, k) having s levels,

s = height(C) + 1. (5.1.1)

It is observed that the digraphs G(n, 1), G(1, k) and G(2, k), where n ≥ 1 and k ≥ 1contain components which are isolated fixed points. Hence each vertex in thesedigraphs is at height 0. For the rest of the chapter it is assumed that n > 2, k ≥ 2 andthe following conventions are followed:Suppose

n = pe11 pe2

2 · · · perr , (5.1.2)

k = qδ11 qδ22 · · · qδss , (5.1.3)

with p1 < p2 < · · · < pr, ei > 0 and q1 < q2 < · · · < qs, δi > 0. If λ(n) = uv, whereu is the the largest divisor of λ(n) relatively prime to k, let us write

λ(n) = uqγ11 qγ2

2 · · · qγss , (5.1.4)

and for x ∈ G(n, k),x = ypc1

1 pc22 · · · p

crr , (5.1.5)

where γi ≥ 0, 1 ≤ i ≤ s, gcd(y, n) = 1 and ci ≥ 0 for all 1 ≤ i ≤ r. Further, we definedi and n1 by

di =

0, if ci = 0ei, if 1 ≤ ci < ei

ci, if ci ≥ ei.(5.1.6)

n1 =∏1≤i≤r

pei−min{ei,di}i , (5.1.7)

so that gcd (x, n1) = 1. Since ordn1 x|λ(n), ordn1 x can be written as

ordn1 x = u1qβ11 qβ2

2 · · · qβss , (5.1.8)

where 1 ≤ βi ≤ γi for all 1 ≤ i ≤ s and u1|u.

5.2 Heights in a Power DigraphTo study the heights of vertices and components in a power digraph, G(n, k) isdecomposed into direct product of the its direct factors. The height of any vertexin G(n, k) is obtained in terms of the heights of corresponding vertices in the directfactors of G(n, k). This is proved in the following theorem.

Theorem 5.2.1. Let n = ml, where gcd(m, l) = 1 and x = (x1, x2) be a vertex inG(n, k) � G(m, k) ×G(l, k). Then height(x) = max{height(x1), height(x2)}.

62

Proof. Let x = (x1, x2) be a vertex in G(n, k) � G(m, k) ×G(l, k) such that

height(x1) = j and height(x2) = w.

Then, by definition of height, there exist cycle vertices c1 and c2 in Comp(x1) ⊆G(m, k) and Comp(x2) ⊆ G(l, k), respectively such that

xk j

1 ≡ c1(mod m), and xkw

2 ≡ c2(mod l). (5.2.1)

Now suppose s = max{ j,w}. It follows that

(xks

1 , xks

2 ) ≡ (cks

1 , cks

2 ) (mod n).

As c1 and c2 are the cycle vertices in Comp(x1) and Comp(x2), respectively therefore,c′1 = cks

1 and c′2 = cks

2 are also the cycle vertices in Comp(x1) and Comp(x2),respectively. Thus

(xks

1 , xks

2 ) ≡ (c′1, c′2) (mod n),

xks ≡ c′(mod n),

where c′ is a cycle vertex in Comp(x) due to Theorem 3.4.7. Hence

height(x) ≤ s = max{height(x1), height (x2)}. (5.2.2)

On the other hand, let r = height(x). Then there exists a cycle vertex b = (b1, b2) inComp(a) such that

xkr ≡ b(mod n),

where xkr= (xkr

1 , xkr

2 ). This implies

xkr

1 ≡ b1(mod m) and xkr

2 ≡ b2(mod l). (5.2.3)

Now b1, b2 are cycle vertices due to Theorem 3.4.7. This along with the fact that j,w are least positive integers satisfying (5.2.1) lead to

j ≤ r and w ≤ r. (5.2.4)

This impliess = max{ j,w} ≤ r = height(x), (5.2.5)

(5.2.2) and (5.2.2) complete the proof. �

Lemma 5.2.2. Let n and k be positive integers defined as in (5.1.2) and (5.1.3),respectively. Suppose x is a vertex of a component C of G1(n, k). Then

height(x) = max1≤i≤s⌈βi

δi⌉ = max

1≤i≤s⌈νqi(ordnx)νqi(k)

⌉,

where βi and δi are defined in (5.1.7) and (5.1.3), respectively and νqi(x) denotes thehighest power of qi in x.

63

Proof. Suppose w = height(x). Then there exists a cycle vertex b ∈ C such that

xkw ≡ b(mod n).

Since gcd(x, n) = 1, gcd(b, n) = gcd(xkw, n) = 1. Thus we have,

ordnb = ordnxkw=

ordnxgcd(kw, ordnx)

.

Now from (5.1.8), we obtain

ordnb = ordnxkw=

u1qβ11 qβ2

2 · · · qβss

qmin(β1,δ1w)1 qmin(β2,δ2w)

2 · · · qmin(βs,δsw)s

. (5.2.6)

Now we will show that a vertex a ∈ G1(n, k) is a cycle vertex if and only ifgcd(k, ordna) = 1. For this let a be a cycle vertex then there exists a positive integert such that

akt ≡ a(mod n)a(akt−1 − 1) ≡ 0(mod n).

Since gcd(a, akt−1 − 1) = 1 and gcd(a, n) = 1, we have

akt−1 − 1 ≡ 0(mod n)akt−1 ≡ 1(mod n).

This implies ordna | kt − 1. As gcd(kt − 1, k) = 1, therefore gcd(k, ordna) = 1.Conversely, suppose gcd(k, ordna) = 1. From Theorem 3.1.1, we know that ordna |λ(n). Thus by (3.1.1), ordna | u. Thus a is a cycle vertex by Lemma 3.4.1. Henceto make sure that b is a cycle vertex, we must show that gcd(ordnb, k) = 1. From(5.2.6), it is possible only when

δiw ≥ βi, for 1 ≤ i ≤ s

w ≥ βi

δi.

As w, being the height, must be an integer, therefore

w ≥ ⌈βi

δi⌉, for 1 ≤ i ≤ s

w ≥ max1≤i≤s⌈βi

δi⌉ = max

1≤i≤s⌈νqi(ordna)νqi(k)

⌉.

Since we are considering the least distance,

w = max1≤i≤s⌈βi

δi⌉ = max

1≤i≤s⌈νqi(ordna)νqi(k)

⌉.

�

64

Lemma 5.2.3. If x is a cycle vertex of G2(n, k), where n is defined as in (5.1.2), thenpei

i |x whenever pi|x.

Proof. Suppose x ∈ G2(n, k) is a cycle vertex. Then there exists m ≥ 1, such that

xkm ≡ x(mod n),x(xkm−1 − 1) ≡ 0(mod n).

The Lemma holds due to the fact that gcd(xkm−1 − 1, x) = 1. �

Theorem 5.2.4. Let n and k be positive integers defined as in (5.1.2) and (5.1.3),respectively. If x is a vertex of a component C of G(n, k), then the height of x is

height(x) = max{max1≤i≤r⌈logk

di

ci⌉,max

1≤i≤s⌈βi

δi⌉} = max{max

1≤i≤r⌈logk

di

ci⌉,max

1≤i≤s⌈νqi(ordn1a)νqi(k)

⌉},

where ci, di, n1 and βi are defined as in (5.1.5), (5.1.6), (5.1.7) and (5.1.8), respectivelyand di

ci= 1 if di = ci = 0.

Proof. Let x and n1 be defined as in (5.1.5) and (5.1.7), n2 = n/n1 and height(x) = l.Suppose

x ≡ x1(mod n1) and x ≡ x2(mod n2). (5.2.7)

Then from Theorem 5.2.4,

l = max{height(x1), height(x2)} = max{g, h}. (5.2.8)

Thus to find l, we have to find height(x1) in G(n1, k) and height(x2) in G(n2, k). Sincegcd(x, n1) = gcd(x1, n1) = 1 and ord n1 x = ord n1 x1 by (5.2.7), Lemma 5.2.2 yields,

height(x1) = g = max1≤i≤s⌈βi

δi⌉ = max

1≤i≤s⌈νqi(ordn1 x)νqi(k)

⌉. (5.2.9)

Now to determine the height of x2, we have to find the least positive integer h forwhich there exists a cycle vertex b ∈ G2(n2, k) satisfying

xkh

≡ xkh

2 ≡ b(mod n2).

By using (5.1.5) we can write,

b ≡ xkh ≡ ykhpc1kh

1 pc2kh

2 · · · pcrkh

r (mod n2).

For b ∈ G2(n2, k) to be a cycle vertex, we must have due to Lemma 5.2.3, peii |b

whenever pi|x2. Thus for all such i

cikh ≥ ei. (5.2.10)

If ci = 0, peii - b and if ci ≥ ei, (5.2.10) is satisfied for h = 0. However if ci ≤ ei, then

from (5.2.10), h ≥ ⌈logkeici⌉. Thus we can write,

h ≥ max1≤i≤r⌈logk

di

ci⌉, (5.2.11)

where di are defined as in (5.1.6). The result follows from (5.2.8), (5.2.9), (5.2.11)and the fact that we need the least h. �

65

Theorem 5.2.5. Let n and k be positive integers defined as in (5.1.2) and (5.1.3),respectively. Then the height of any component C of G1(n, k) is

height(C) = max1≤i≤s⌈γi

δi⌉ = max

1≤i≤s⌈νqi(λ(n))νqi(k)

⌉,

where γi are defined as in (5.1.4).

Proof. Let x ∈ C be any vertex in G1(n, k), by Lemma 5.2.2, height(x) is given as

height(x) = max1≤i≤s⌈νqi(ordnx)νqi(k)

⌉.

Then by definition of height(C),

height(C) = maxx∈C

height(x).

= maxx

max1≤i≤s⌈νqi(ordnx)νqi(k)

⌉.

Since ordnx | λ(n), for all x ∈ C we have

height(C) ≤ max1≤i≤s⌈νqi(λ(n))νqi(k)

⌉. (5.2.12)

Since C is arbitrary, (5.2.12) holds for all components C of G1(n, k). Now we willshow that the equality in (5.2.12) holds.

By Theorem 3.1.1, we can find a vertex g ∈ G1(n, k) having order λ(n).Thus from (5.2.12), the height of component C′ containing g is max1≤i≤s⌈

νqi (λ(n))νqi (k) ⌉.

By Theorem 3.4.5, any two cycle vertices of G1(n, k) have same associated trees.Thus if there is a vertex at height h in C′, then there must exists a vertex at height hin C. This implies height(C) = height(C′) which completes the proof. �

The following corollaries are immediate consequences of Theorem 5.2.5.

Corollary 5.2.6. Each component of G1(n, k) has exactly max1≤i≤s⌈νqi (λ(n))νqi (k) ⌉+1 levels,

where n and k are defined as in (3.3.1) and (3.3.2), respectively.

Proof. Let C be any component of G1(n, k) at level s. Then from 5.1.1, s = height(C)+1. By Theorem 5.2.5, proof is completed. �

Corollary 5.2.7. The height and level of any component C of G1(n, p) from its cycleis νp(λ(n)) and νp(λ(n)) + 1, respectively, where n ≥ 1 and p is any prime.

Proof. Let C be any component of G1(n, k). Since k = p, from Theorem 5.2.5,

height(C) = max1≤i≤s⌈νqi(λ(n))νqi(k)

⌉

= ⌈νp(λ(n))

1⌉

= νp(λ(n)).

Now from 5.2.5, the component C has level νp(λ(n)) + 1. �

66

The Corollary 5.2.7 was proved in (Somer and Krızek 1994) for p = 2.

Lemma 5.2.8. The height of Comp(0) in G(n, k) is max1≤i≤r ⌈logk ei⌉.

Proof. For any vertex a ∈ Comp(0), we have pi | a if and only if pi | n. Now theexpressions for a and n1 given in (5.1.5) and (5.1.7), respectively imply that

ci ≥ 1 where 1 ≤ i ≤ r

andn1 = 1 for all vertices in Comp(0).

By Theorem 5.2.4,

height(a) = max{max1≤i≤r⌈logk

ei

ci⌉, 1}

= max1≤i≤r⌈logk

ei

ci⌉.

Thus

height(Comp(0)) = maxa∈Comp(0)

max1≤i≤r⌈logk

ei

ci⌉.

Since ci ≥ 1,⌈logk

ei

ci⌉ ≤ ⌈logk ei⌉.

Henceheight(Comp(0)) ≤ max

1≤i≤r⌈logk ei⌉.

The existence of the vertex p1 p2 · · · pr and all those vertices in which at least oneci = 1 completes the proof. �

Theorem 5.2.9. Let n and k be positive integers defined as in (5.1.2) and (5.1.3),respectively. Then the height of G2(n, k) is

max{max1≤i≤r⌈logk ei⌉, max

n1 |n,gcd(n/n1,n1)=1max1≤i≤s⌈νqi(λ(n1))νqi(k)

⌉}.

Proof. Consider x ∈ C of G2(n, k). Then Theorem 5.2.4 yields

height(x) = max{max1≤i≤r⌈logk

di

ci⌉,max

1≤i≤s⌈νqi(ordn1a)νqi(k)

⌉}.

Since dici≤ ei for all 1 ≤ i ≤ r, it follows that

max1≤i≤r⌈logk

di

ci⌉ ≤ max

1≤i≤r⌈logk ei⌉. (5.2.13)

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Consider n1 defined as in (5.1.7). Then Theorem 5.2.5 gives

max1≤i≤s⌈νqi(ordn1a)δi

⌉ ≤ max1≤i≤s⌈νqi(λ(n1))νqi(k)

⌉

≤ maxn1 |n,gcd(n/n1,n1)=1

max1≤i≤s⌈νqi(λ(n1))νqi(k)

⌉. (5.2.14)

Thus from (5.2.13) and (5.2.14), we obtain

height(x) ≤ max{max1≤i≤r⌈logk ei⌉, max

n1 |n,gcd(n/n1,n1)=1max1≤i≤s⌈νqi(λ(n1))νqi(k)

⌉}. (5.2.15)

The proof is completed if there exist vertices in G2(n, k) with heights max1≤i≤r⌈logk ei⌉or max1≤i≤s⌈

νqi (λ(n1))δi⌉ for all n1|n such that gcd(n/n1, n1) = 1.

The existence of vertex having height, max1≤i≤r⌈logk ei⌉ is shown by Lemma5.2.8. By Theorem 5.2.5, for every n1 | n,

height(G1(n1, k)) = max1≤i≤s⌈νqi(λ(n1))δi

⌉.

Thus we can find a vertex a in G1(n1, k) such that

height(a) = height(G1(n1, k)).

As we know that we can always find a fixed point in G2(n, k) (e.g. 0 is alwaysa fixed point of G2(n, k)) therefore, consider a fixed point b in G2(n/n1, k). Sincegcd(n/n1, n1) = 1, by using (3.3.2) and Theorem 5.2.4, we can find a vertex c =(a, b) ∈ G2(n, k) such that

height(c) = max{height(a), height(b)}.

Since b is a cycle vertex, height (b) = 0 and

height(c) = height(a) = max1≤i≤s⌈νqi(λ(n1))δi

⌉}.

The result follows immediately. �

Lemma 5.2.2, Theorem 5.2.5, Lemma 5.2.8 and Theorem 5.2.9 are illustratedby the following Example.

Example 5.2.10. Let n = 27 = 33 and k = 30. Consider the power digraph

G(n, k) = G(27, 30) = G(33, 2 · 3 · 5).

For 23, 16, 19 ∈ G1(27, 30),ord2723 = 18 = 32 · 2, ord2716 = 32 and ord2719 = 3.Now from Lemma 5.2.4,

height(23) = max{⌈21⌉, ⌈1

1⌉} = 2,

68

Figure 5.1: The power digraph G(27,30)

height(16) = ⌈21⌉ = 2,

height(19) = ⌈11⌉ = 1

and similarly, heights of other vertices can be determined. From Theorem 5.2.5,

height(G1(27, 30)) = max{⌈ν2(λ(27))1

⌉, ⌈ν3(λ(27))1

⌉, ⌈ν5(λ(27))1

⌉} = 2.

and by Lemma 5.2.8,

height(Comp(0)) = max{⌈log30 3⌉} = 1.

Thus height (G(27, 30)) = max(1, 2) = 2. The digraph G(27, 30) is given in Figure5.1.

5.3 Classification of G(n, k) with respect to the HeightsThis section contains the results on the classification of those power digraphs inwhich each vertex of indegree 0 is at a specific height q. The first two theoremsprovide some necessary and sufficient conditions so that the digraphs G1(n, pα) andG1(n, 2α) have height 1. The third theorem establishes the necessary and sufficientconditions on n and α so that each vertex of indegree 0 of a digraph G(n, p) is atheight q.

69

Theorem 5.3.1. Suppose G1(n, pα) is not regular, where p is an odd prime. Thenthe height of G1(n, pα) is 1 if and only if n is of one of the following forms

1. n = 2w0 pw1∏

1≤i≤t phii where w0 ≥ 0, 1 < w1 ≤ α + 1 and pα+1 - pi − 1 for any

i.

2. n = 2w0 pw1∏

1≤i≤t phii where w0 ≥ 0, w1 ∈ {0, 1}, p | pi − 1 for some i but

pα+1 - pi − 1 for any i.

Proof. Suppose every vertex of indegree 0 is at height 1 and

n = 2w0 pw1∏

1≤i≤t phii .

Suppose on the contrary that n is not of the given form. Then there arise followingcases:Case 1:Suppose w1 > 1 but w1 > α + 1 or pα+1 | pi − 1 for some i. Then pα+1 | λ(pw1) orpα+1 | λ(phi

i ). Since λ(pw1) and λ(phii ) divide λ(n), pα+1 | λ(n). Thus Theorem 5.2.5

yields,

height(G1(n, k)) = ⌈νp(λ(n))νp(k)

⌉

= ⌈νp(λ(n))α

⌉

≥ ⌈α + 1α⌉ ≥ 2.

Hence there must exists some vertex of indegree 0 at height greater than or equal to2 which contradicts our assumption. This establishes (1).Case 2:Now suppose w1 ∈ {0, 1} but p - pi − 1 for any i. Then p - λ(phi

i ) for any i. Also

λ(2w0) = 2w0 and λ(pw1) = 1 or p − 1.

In either case p - λ(n). This implies

gcd(λ(n), k) = gcd(λ(n), pα) = 1.

From Theorem 3.5.3, it follows that G1(n, k) is regular i.e. G1(n, k) does not containany vertex of indegree 0. This implies all vertices of G1(n, k) are cycle vertices.Hence height of G1(n, k) is 0 which contradicts our assumption. Thus we maysuppose p | pi − 1 but pα+1 | pi − 1 for some i. Again by the same argument aswe have used in case 1, we can show the existence of a vertex having indegree 0at height greater than or equal to 2 which provides a contradiction. Thus (2) isestablished.

The converse is straight forward due to Theorem 5.2.5. �

Theorem 5.3.2. Suppose G1(n, 2α) is not regular. Then the height of G1(n, 2α) is 1if and only if n = 2w0

∏1≤i≤t phi

i , where 0 ≤ w0 ≤ α + 2 and 2α+1 - pi − 1 for any i.

70

Proof. Suppose the height of G1(n, 2α) is 1 and

n = 2w0∏

1≤i≤t phii .

But suppose to the contrary that either w0 > α + 2 or 2α+1 | pi − 1 for some i. Then2α+1 | λ(2w0) or 2α+1 | λ(phi

i ). In either case 2α+1 | λ(n). This along with Theorem5.2.5 yields

height(G1(n, k)) = ⌈ν2(λ(n))ν2(k)

⌉

= ⌈ν2(λ(n))α⌉

≥ ⌈α + 1α⌉ ≥ 2.

Therefore, there must exists a vertex having indegree 0 at height greater than orequal to 2 which provides a contradiction.Converse can be proved by making use of Theorem 5.2.5. �

Theorem 5.3.1 is illustrated by the Examples 5.3.3, 5.3.4 and Theorem 5.3.2by an example 5.3.7.

Example 5.3.3. Let n = 36 = 22 ·32 = 2w0 · pw1 and k = 9 = 32 = pα, so that w0 = 2,w1 = 2 and α = 2. The power digraph G1(36, 9) satisfies the conditions given inTheorem 5.3.1. Thus

height(G1(36, 9)) = 1.

We also note that λ(36) = 6 and

gcd(λ(n), pα) = gcd(λ(36), 9) , 1,

so that G1(n, pα), by Theorem 3.5.3, is not regular. See Figure 5.2 for detailedstructure of G(36, 9).

Example 5.3.4. Let n = 81 = 34 = pw1 and k = 9 = 32 = pα, where w1 = 4and α = 2. We note that w1 > 3 = α + 1, the digraph G1(81, 9) does not satisfythe conditions given in Theorem 5.3.1, height (G1(81, 9)) , 1. In fact from Theorem5.2.5, height (G1(81, 9)) = 2. This is shown in figure 5.4.

Example 5.3.5. Let n = 40 = 23 · 5 = 2w0 ·1 and k = 16 = 24 = 2α, where w0 = 3and α = 4. We note that w0 < 5 = α + 1 and 2α+1 = 25 - 4 = p1 − 1. Thus thedigraph G1(40, 16) satisfy the conditions given in Theorem 5.3.1,

height (G1(40, 16)) = 1.

We also note that λ(40) = 16 and

gcd(λ(n), 2α) = gcd(λ(40), 16) , 1

so that G1(n, 2α), by Theorem 3.5.3, is not regular. This is shown in Figure 5.4.

71

Figure 5.2: The power digraph G(36, 9) with height(G1(n, k)) =1

72

Figure 5.3: The power digraph G(81, 9) having height 2

73

Figure 5.4: The power digraph G(40, 16) having height 1

74

Theorem 5.3.6. Suppose G1(n, p) is not regular. Then every vertex of indegree 0 inG1(n, p) is at height q ≥ 1 if and only if n has one of the following forms

1. n = 2w0 pw1∏

1≤i≤t phii where w0 ≥ 0, 0 ≤ w1 ≤ q, p|pi − 1 for some i and

pq ∥ pi − 1 for all such pi’s.

2. n = 2w0 pq+1∏1≤i≤t phi

i where w0 ≥ 0 and either p - pi−1 for any i or pq ∥ pi−1for all pi such that p|pi − 1.

Proof. Let every vertex of indegree 0 in G1(n, p) be at height q ≥ 1 and

n = 2w0 pw1∏

1≤i≤t phii .

Let m = 2w0∏

1≤i≤t phii and l = pw1 . Since gcd(m, l) = 1, by (3.3.2),

G(n, p) � G(m, p) ×G(l, p).

But suppose to the contrary, n is not of the given forms. Then there arises followingcases:Case1:Let 0 ≤ w1 ≤ q and there is no i for which p |pi − 1. Then p - λ(Phi

i ) for all i. Alsop - λ(2w0). This implies p - λ(m). Hence by Theorem 3.5.3 it follows that G1(m, p)is regular and has height 0. Now from Theorem 5.2.5,

height(G1(l, p))) = ⌈νp(λ(l))νp(k)

⌉

= ⌈νp(λ(l))

1⌉

= w1 − 1< q.

Thus 0 ≤ height (G1(l, p)) ≤ q − 1.We see by (3.3.3) and Theorem 5.2.4 that,

height(G1(n, p)) = max(height(G1(m, p)), height(G1(l, p)))= max(height(G1(m, p)), 0)= height(G1(m, p))≤ q − 1.

Hence 0 ≤ height (G1(n, p)) ≤ q − 1. This implies that the height of every vertexof indegree 0 is less than or equal to q − 1 which contradicts the assumption thatevery vertex of indegree 0 in G1(n, p) is at height q ≥ 1. Thus there must exist somei for which p | pi − 1. Now suppose for one such p j, pd||p j − 1 where d , q. ThenTheorem 5.2.5 yields that

height(G(ph j

j , p) = ⌈d1⌉) = d.

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Which shows that there is a vertex a ∈ G1(ph j

j , p) of indegree 0 such that

height(a) = height(G1(ph j

j , p)) = d.

This impliesakd ≡ a j(mod ph j

j ),

for some cycle vertex a j in G1(ph j

j , p). Now consider the fixed points bi in G1(phii , p)

for all i , j, c0 in G1(2w0 , p) and c1 in G1(pw1 , p). Then from (3.3.2) and Theorem5.2.4, c = (c0, c1, b1, · · · , b j−1, a, · · · bt) ∈ G1(n, p) and

height(c) = max{height(c0), height(c1), height(b1), · · ·· · · , height(b j−1), height(a), · · · , height(bt)}.

Note that being fixed points, height(c0) = height(c1) = height(bi) = 0, where i , j.Thus height(c) = height(a) = d , q. Now we will determine the indegree of c.From Lemma 3.5.4,

N(n, p, c) = N(2w0 , p, c0) · N(pw1 , p, c1) · N(ph11 , p, c0) · · ·N(ph j−1

j−1 , p, b1) · · ·· · ·N(ph j

j , p, a) · · ·N(phtt , p, bt).

Since N(ph j

j , p, a) = 0, N(n, p, c) = 0. This implies c is a vertex having indegree 0and height d , q which contradicts our assumption.Case 2:Now suppose w1 = q + 1 and p|p j − 1 for some 1 ≤ j ≤ t but pd||p j − 1, , whered , q. Again by the same argument as above, we can prove the existence of a vertexof indegree 0 which is not at height q which leads to a contradiction.Case 3:It remains to show that w1 can not exceed q+1. Indeed if w1 > q+1, Theorem 5.2.5forces,

height(G1(pw1 , p))) = ⌈νp(λ(pw1))νp(k)

⌉

= ⌈νp(λ(pw1))

1⌉

= w1 − 1> q.

Thus there exists a vertex a ∈ G1(ph j

j , p) having indegree 0 and height(a) = height(G1(pw1 , p)) > q. Now by Theorem 5.2.4 and Lemma 3.5.4, we can show that thereexists a vertex in G1(n, p) at height greater than q and indegree 0 which contradictsthe assumption.All three cases leads to a contradiction. Hence n is one of the given forms.The converse is easy to prove using Theorem 5.2.5. �

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The following examples illustrate Theorem 5.3.6.

Example 5.3.7. Let n = 57 = 3 · 19 = p · p1 and k = p = 3, where p = 3, p1 = 19and w1 = 1. We note that p2 = 9∥p1 − 1 = 18, the digraph G1(57, 3) satisfies theconditions given in Theorem 5.3.6. Hence every vertex of indegree 0 is at height 2.We also note that gcd(λ(n), p) = gcd(λ(57), 3) = (18, 3) , 1. Thus from Theorem3.5.3, G1(57, 3) is not regular.

Figure 5.5: The power digraph G1(57, 3) having height 2

Example 5.3.8. Let n = 133 = 7 · 19 = p1 · p2 and k = p = 3, where p1 = 7 andp2 = 19. We can see p2∥p2 − 1 and p|p1 − 1 = 6 but p2 - p1 − 1. Thus the digraphG1(133, 3) does not satisfy the conditions given in Theorem 5.3.1. This implies everyvertex of indegree 0 is not at height 2. In fact there are some vertices of indegree0 which are at height 1, for example by Theorem 3.5.1, N(133, 3, 26) = 0 but byTheorem 5.2.2, height(26) = 1.

Theorem 5.3.9. Let n and k be positive integers defined as in (5.1.2) and (5.1.3),respectively such that p1 = 2. Then the number of vertices in G1(n, k) at height q isgiven by

gcd(2, u)δ1 gcd(2e1−2, u)δ2r∏

i=2

gcd(λ(peii ), u)[N(n, kq, b) − N(n, kq−1, b)],

77

where, δ1 = 0 if e1 = 0, 1 otherwise δ1 = 1, δ2 = 0 if e1 < 3 otherwise δ2 = 1and u is the largest divisor of λ(n) relatively prime to k and b is any cycle vertex ofG1(n, k).

Proof. From Theorem 3.4.5 we know that T (c1) � T (c2) for all the cycle verticesc1 and c2 of G1(n, k). Thus in order to find the number of vertices at height q inG1(n, k), we first find the number of vertices at height q from one cycle vertex, sayb. In other words we have to find the number of solutions of following congruence,where q is the least non negative integer.

xkq ≡ b(mod n). (5.3.1)

or equivalently, we have to find those vertices which satisfy (5.3.1) but do not satisfythe following congruence

xkq−1 ≡ b(mod n).

Hence the number of vertices at height q from b in the component containing b is

N(n, kq, b) − N(n, kq−1, b). (5.3.2)

Now from Theorem 3.4.2, the number of cycle vertices in G1(n, k) are

gcd(2, u)δ1 gcd(2e1−2, u)δ2r∏

i=2

gcd(λ(peii ), u) (5.3.3)

The result follows from (5.3.2) and (5.3.3) . �

5.4 SummaryThis chapter presents a generalization of the the work done in (Somer and Krızek2004, 2006) on the heights of vertices. To achieve the desired goal, the height ofany vertex in G(n, k) is obtained in terms of the heights of corresponding verticesin the direct factors of G(n, k). Several formulae for the heights of any vertex andcomponent in G1(n, k) in terms of n and k are established. With the help of theseresults, the formulae for the determination of the heights of any vertex and compone-nt in G2(n, k) in terms of n and k are also obtained. Further, the necessary andsufficient conditions on n such that each vertex of indegree 0 of a certain subdigraphof G(n, k) is at height q ≥ 1 are determined. Finally, an expression for the numberof vertices at a specific height is derived.

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Chapter 6

Isolated Fixed Points and Primalityof Fermat Numbers

6.1 Introduction

The cycle structures of power digraphs received much attention in the recent yearsdue to its fundamental role in many applications, for example, in pseudo-randomnumber generation, primality testing of Fermat numbers and many other types ofnumbers. Szalay exhibited that G(Fm, 2) consists of only two components eachcontaining a fixed point, where Fm is a Fermat prime for some natural number m(Szalay 1992). It was also shown that G(n, 2) has two components if and only if nis a Fermat prime or a power of 2 (Somer and Krızek 2004).

Definition 6.1.1. A vertex a is said to be an isolated fixed point if it is a fixed pointand there is no non-cycle vertex b such that bk ≡ a(mod n). In other words a hasindegree 1.

6.2 Isolated Fixed Points

This section deals with the study of isolated fixed points in G(n, k). It is easy to seethat if n is a square-free integer, then 0 is an isolated fixed point of G(n, k). Nowif G1(n, k) is regular then, 1 is an isolated fixed point of G(n, k). It is known thatfor k = 1, a digraph G(n, k) consists of isolated fixed points only. In this section,attempt has been made to sort out this problem for the general values of n and k. Thefollowing lemma is very important in the proofs of the main results of this chapter

Lemma 6.2.1. Let n = ml, where gcd(m, l) = 1 and x = (x1, x2) be a vertex inG(n, k) � G(m, k) ×G(l, k). Then x is an isolated fixed point of G(n, k) if and only ifx1 and x2 are isolated fixed points of G(m, k) and G(l, k), respectively.

79

Proof. Let x be an isolated fixed point. Then x is a cycle of length one and N(n, k, x) =1. From Theorems 3.4.7 and 3.4.10, x1 and x2 are fixed points of G(m, k) and G(l, k),respectively. Also by Theorem 3.5.4,

N(m, k, x1) = 1 = N(l, k, x2).

Hence x1 and x2 are isolated fixed points in G(m, k) and G(l, k), respectively. Converseis similar. �

Example 6.2.2. Let n = 15 = 3 · 5 and k = 7. Take 6 ∈ G(15, 7), we can see that67 ≡ 6(mod 15) and from Theorem 3.5.1, N(15, 7, 6) = 1. Thus 6 = (0, 1) is anisolated fixed point of G(15, 7) � G(3, 7) × G(5, 7). Since from Theorem 3.5.1 andCorollary 3.5.2, N(3, 7, 0) = 1 and N(5, 7, 1) = 1 respectively , 0 and 1 are isolatedfixed points of G(3, 7) and G(5, 7). This is shown in figure 6.10, 6.2 and 6.3.

Figure 6.1: The power digraph G(3,7)

Figure 6.2: The power digraph G(5,7)

Figure 6.3: The power digraph G(15,7)

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Theorem 6.2.3. The power digraph G(n, k), where n is defined as in (3.3.1) and k ≥2, has at least one isolated fixed point if and only if either ei = 1 or gcd(λ(pei

i ), k) = 1for all 1 ≤ i ≤ r in prime factorization of n.

Proof. Suppose G(n, k) has an isolated fixed point a. For all peii ∥ n,where 1 ≤ i ≤ r,

either ei = 1 or ei > 1.

Suppose to the contrary that there exists 1 ≤ j ≤ r such that

gcd(λ(pe j

j ), k) , 1 and e j > 1.

Since a is a fixed point, by Theorems 3.4.7, Theorem 3.4.10 and equation (3.3.1)there exist fixed points ai ∈ G(pei

i , k) for all 1 ≤ i ≤ r such that

a = (a1, · · · , a j, · · · , ar).

Now from Theorem 3.5.4, we can write

N(n, k, a) =r∏

i=1

N(n, k, ai). (6.2.1)

If a j ∈ G1(pe j

j , k), then

N(pe j

j , k, a j) = gcd(λ(pe j

j ), k) , 1. (6.2.2)

Thus in this case from equation (6.2.1) and (6.2.2) , N(n, k, a) , 1. This contradictsthe fact that a is an isolated fixed point. Hence we may suppose a j ∈ G2(pe j

j , k).Now we know that G2(pe j

j , k) consists of one component containing fixed point 0.Thus

a j ≡ 0(mod pe j

j ).

From Lemma 3.5.2,

N(pe j

j , k, a j) = N(pe j

j , k, 0) = pe j−⌈e jk ⌉

j

Since e j > 1 and k ≥ 2, e j − ⌈ e j

k ⌉ > 0. Thus N(pe j

j , k, a j) , 1. Now from equation(6.2.1), it follows that N(n, k, a) , 1 which again is a contradiction.

Conversely, suppose for all peii ∥ n, where 1 ≤ i ≤ r, either ei = 1 or

gcd(λ(peii ), k) = 1. If ei = 1, 0 is an isolated fixed point in G(pi, k). If ei > 1

and gcd(λ(peii ), k) = 1, 1 is an isolated fixed point in G(pei

i , k). Now consider a =(a1, a2, · · · , ar), where

ai =

{0, if ei = 11 if ei > 1 (6.2.3)

From Lemma 6.2.1, a is an isolated fixed point of G(n, k). �

81

Example 6.2.4. Suppose n = 18 = 2 · 32 and k = 5. Since G(18, 5) satisfies allthe conditions of Theorem 6.2.3, G(18, 5) contains at least one isolated fixed point.Figure 6.4 shows that 1, 8, 10 and 17 are isolated fixed points of G(18, 5). Whereas,if we take n = 18 but k = 3, then

gcd(λ(32), k) = gcd(λ(32), 3) = 3 , 1.

In this case G(18, 3) does not satisfies the conditions of Theorem 6.2.3. HenceG(18, 3) does not contain any isolated fixed point. This is illustrated in Figure 6.5.

Figure 6.4: The power digraph G(18, 5) with 4 isolated fixed points

Figure 6.5: The power digraph G(18, 3) with no isolated fixed point

Corollary 6.2.5. Suppose k is even and n > 2 is defined as in (3.3.1). The powerdigraph G(n, k) has at least one isolated fixed point if and only if n is square-free.

Proof. We know that 2 | λ(peii ) for all 1 ≤ i ≤ r. Since k is even,

gcd(λ(peii ), k) , 1,

for any 1 ≤ i ≤ r. Hence from Theorem 6.2.3, ei = 1 for all 1 ≤ i ≤ r which impliesn is square-free.

Conversely, if n is square-free, 0 is an isolated fixed point of G(n, k). �

82

Corollary 6.2.6. Suppose G1(n, k) is not regular and n is not square-free. The powerdigraph G(n, k), where n is defined as in (3.3.1) and k ≥ 2, has isolated fixed pointif and only if the following statements are satisfied.

1. k must be odd.

2. The sets l = {peii | ei > 1 and gcd(λ(pei

i , k) = 1)} and m = {pe j

j | e j = 1} arenon empty. Also G(n, k) � G(l, k) ×G(m, k).

3. The digraph G1(m, k) is not regular.

Proof. Suppose G(n, k) has an isolated fixed point a. If k is even, then from Corollary6.2.5, n is square-free which is a contradiction. This established (1). Now fromTheorem 6.2.3 either

ei = 1 or gcd(λ(peii ), k) = 1

for all 1 ≤ i ≤ r in prime factorization of n. If

gcd(λ(peii ), k) = 1 for all

1≤ i ≤ r,then from Theorem 3.5.3, G1(n, k) is regular which is not possible. Also ifei = 1 for all 1 ≤ i ≤ r, then n become square-free which again is a contradiction.Therefore, there must exists 1 ≤ s < r such that

ei = 1 for all 1 ≤ i ≤ s and gcd(λ(peii ), k) = 1 for all i > s.

Hence the sets l and m are non empty. Since l and m are disjoint,

G(n, k) � G(l, k) ×G(m, k).

Now if G1(m, k) is regular, then

gcd(λ(pi), k) = gcd(pi − 1, k) = 1 for all 1 ≤ i ≤ s.

Thus from Theorem 3.5.3, G1(n, k) = G1(l, k) × G1(m, k) is also regular which is acontradiction.

Conversely, suppose all three conditions are true. Since l is non empty andG1(l, k) is regular, 1 is an isolated fixed point in G(l, k). Again since m is nonempty,0 is an isolated fixed point of G2(m, k). Thus from Lemma 6.2.1, a = (1, 0) is anisolated fixed point of G(n, k) � G(l, k) ×G(m, k). �

Example 6.2.7. Let n = 28 = 22 · 7 and k = 15. Here we can see the sets l = {22}and m = {7} are non empty. Since gcd(λ(4), 15) = 1 and gcd(λ(7), 15) = 3 , 1, fromTheorem 3.5.3, G1(l, k) is regular and G1(m, k) is not regular. Thus G(28, 15) satisfyconditions 1, 2 and 3 of Theorem 6.2.3. Hence G(28, 15) contains an isolated fixedpoint. It is shown in figure 6.6.

83

Figure 6.6: The power digraph G(28,15) having two isolated fixed point

6.3 Classification of G(n, k) withrespect to the Components

Theorem 6.3.1. The power digraph G(n, k), where n > 2 and k ≥ 2 are positiveintegers exhibits the following properties

1. G(n, k) consists of exactly two components containing the fixed points 0 and1.

2. G1(n, k) is semi-regular of degree 2d for some d ≥ 1

if and only if k is even and n = 2l or n = Fm, where l ≥ 2, m ≥ 1 are integers andFm = 22m

+ 1 is a Fermat prime.

Proof. Suppose the power digraph G(n, k) exhibits the above properties (1) and (2).Since 0 and 1 are the fixed points of G(n, k), G2(n, k) and G1(n, k) both consist ofone component containing the fixed points 0 and 1, respectively.

84

First suppose k is odd then 2|k − 1. Since n > 2, 2|pi − 1 for all 1 ≤ i ≤ r.Thus

2|λ(peii ) for all1 ≤ i ≤ r.

Hence from Theorem 3.4.9,

A1(G(n, k)) =r∏

i=1

(gcd(λ(peii ), k − 1) + 1),

≥ 2 + 1 = 3.

This along with the fact that each component of G(n, k) contains a unique cycleimplies that the number of components of G(n, k) are greater than or equal to 3which contradicts (1). Thus k must be even.

We know that the Euler function ϕ(n) is a power of 2 if and only n =2lFm1 Fm2 · · · Fms . Also it is easy to show that ϕ(n) = 2i if and only if λ(n) = 2 j,where j ≤ i.

Now we claim that n must be of the form

2lFm1 Fm2 · · · Fms ,

where l ≥ 0 and Fmi are Fermat primes for all i. For if

n , 2lFm1 Fm2 · · · Fms ,

then λ(n) is not a power of 2. Therefore, there exists an odd prime divisor p of λ(n).Then by definition of λ(n) there exists i, where 1 ≤ i ≤ r such that p is a primedivisor of λ(pei

i ). If p |k, by Theorem 3.5.1, either

N(n, k, a) = 0 or p |N(n, k, a) for all a ∈ G1(n, k)

which contradicts (2). Thus we may suppose p - k. Now p is a factor of λ(n) whichis relatively prime to k. Thus from Theorem 3.4.1 there exists a cycle of length t inG1(n, k) such that

kt ≡ 1(mod p).

If t = 1, then p | k − 1. Now from Theorem 3.4.9,

A1(G(n, k)) =r∏

i=1

(gcd(λ(peii ), k − 1) + 1).

Since p is a prime divisor of λ(peii ) for some i,

A1(G(n, k)) ≥ p + 1.

This contradicts (1). Hence we may suppose t > 1. But then there exists acomponent containing a cycle of length t > 1. which again contradict (1). Thusin any case, we get a contradiction. Hence

n = 2lFm1 Fm2 · · · Fms ,

85

where l ≥ 0 and Fmi are Fermat primes for all i.

Now since G2(n, k) consists of only one component containing a fixed point0, n must be of the form pα, where p is any prime and α ≥ 1. Thus n = 2l or n = Fm,where l ≥ 2, m ≥ 1 are integers and Fm = 22m

+ 1 is a Fermat prime.

Conversely, suppose k is even and n = 2l or n = Fm, where l ≥ 2, m ≥ 1 areintegers and Fm = 22m

+ 1 is a Fermat prime. It is easy to see that λ(n) is a powerof 2. Property (2) can be proved from Theorem 3.5.1. To prove property (1), wefirst show that G1(n, k) does not contain any cycle of length greater than 1. FromLemma 3.4.1, Theorem 3.4.3 and the fact that the greatest divisor of λ(n) which isrelatively prime to k is 1, it follows that all cycles of G1(n, k) are fixed points. Nowfrom Theorem 3.4.8 and the fact that 2 - k − 1

A1(G1(n, k)) = gcd(λ(n), k − 1) = 1.

Since the number of components in G1(n, k) is equal to the number of cycles inG1(n, k) so G1(n, k) consists of only one component containing 1. This along withthe fact that G2(n, k) always consists of one component whenever n is power of aprime completes the proof. �

Remark 6.3.1. In Theorem 6.3.1, we have taken n > 2 as for n = 2, power digraphG(2, k) always consists of two components which are isolated fixed points. It doesnot depend on value of k.We also note that property (2) is not satisfied in this case.

Corollary 6.3.2. Let n be a positive integer and k = 2s, where s ≥ 1. A powerdigraph G(n, k) consists of exactly two components containing fixed points 0 and1 if and only if n = 2l or n = Fm, where Fm = 22m

+ 1 is a Fermat prime for all1 ≤ i ≤ s and l ≥ 1.

Proof. Since k = 2s, from Theorem 3.5.1

N(n, k, a) =r∏

i=1

gcd(λ(peii ), k) = 2d

for some d ≥ 1 or N(n, k, a) = 0. Hence G1(n, k) is semi-regular of degree 2d forsome d ≥ 1. Corollary follows from Theorem 6.3.1. �

Corollary 6.3.3. Let k be even integer (k ≥ 2). A Fermat number Fm = 22m+ 1 is

prime if and only if following are satisfied

1. G(Fm, k) consists of two components containing fixed points 0 and 1,

2. G1(Fm, k) is semi-regular of degree 2d for some 1 ≤ d ≤ 2m.

Proof. It is straight forward from Theorem 6.3.1. �

Corollary 6.3.4. Let n be a positive integer and k = 2s, where s ≥ 1. A Fermatnumber Fm = 22m

+ 1 is a prime if and only if G(Fm, k) consists of two componentscontaining the fixed points 0 and 1.

86

Proof. It can be proved from Theorem 3.5.1 and Corollary 6.3.3. �

Corollaries 6.3.2 and 6.3.4 for s = 1 have been proved in (Somer Krızek2004).

Theorem 6.3.5. Let n > 2 be a positive integer and k = qβ11 · · · q

βss be the prime

decomposition of k. The power digraph G(n, k) consists of two components if andonly if k is even and n is one of the following form

1. n = p, where p = 1 +∏

1≤i≤s qγii is prime, where γi ≥ 0 for all i.

2. n = qαj for some 1 ≤ j ≤ s and q j = 1 +∏

1≤i≤s,i, j qγii , where γi ≥ 0 for all i.

Proof. Suppose the power digraph G(n, k) consists of two components. If k is oddthen 2 | k − 1. Also since n > 2, 2 | λ(pei

i ) for all 1 ≤ i ≤ r. Hence from Theorem3.4.9,

A1(G(n, k)) =r∏

i=1

(gcd(λ(peii ), k − 1) + 1)

≥ 3

This along with the fact that the number of components is equal to the numberof cycles in a power digraph implies that the number of components of G(n, k) isgreater than or equal to 3 which is a contradiction. Hence k must be even.

As the vertices 0 and 1 belong to G(n, k) therefore, both of its componentscontain fixed points and there does not exist any other component containing thecycle of length greater than 1. Since G2(n, k) itself is a component containing 0therefore, n must be of the form n = pα, where p is any prime. Suppose on contraryn does not satisfy the conditions given in (1) and (2). There arises following casesCase 1:If n = pα, where p , qi for any 1 ≤ i ≤ s and α > 1 then

p | λ(n) = λ(pα) = pα−1(p − 1).

We can see that p - k which shows that p is a factor of λ(n) relatively prime to k.Thus from Lemma 3.4.1 and Theorem 3.4.3, there exists a cycle of length t suchthat

kt ≡ 1(mod p). (6.3.1)

The fact that there does not exist any other component containing the cycle of lengthgreater than 1 forces t = 1. But then p | k − 1 from (6.3.1). Consequently fromTheorem 3.4.9,

A1(G(n, k)) ≥ p + 1.

This further implies that the number of components of G(n, k) are greater than orequal to p + 1 which is a contradiction.Case 2:Now suppose n = p, where p is any prime or n = qαj for some 1 ≤ j ≤ s but there

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exists prime divisors p1 , qi and p2 , qi for any i such that p1 | p−1 and p2 | q j−1.Then p1 and p2 are prime divisors of λ(n) relatively prime to k. Now again by thesame argument as in case 1, we find the contradiction.

Conversely, suppose k is even and n is one of the form given in (1) and(2). We note that in either case λ(n) does not contain any prime factor which isrelatively prime to k. The only factor of λ(n) relatively prime to k is u = 1. We cansee k ≡ 1(mod u). Thus from Theorem 3.4.3, every cycle of G1(n, k) is of length 1that is fixed point. Also G2(n, k) consists of only one component containing a fixedpoint 0. Now from Theorem 3.4.9 there are two fixed points. This implies G(n, k)consists of two components which completes the proof. �

The following examples illustrate theorem 6.3.5.

Example 6.3.6. Let n = 33 = 27 and k = 6 = 2 · 3 = q1 · q2. Here n = q32 and

q2 − 1 = 2 = q1. All the conditions of Theorem 6.3.5are satisfied. Thus G(33, 6)consists of two components. This is shown in Figure 6.7.

Figure 6.7: The power digraph G(33, 6) having exactly two components

Example 6.3.7. Let p = M5 = 31, where M5 is Mersenne prime and k = 30 =2.3.5 = q1 · q2 · q3. Here p − 1 = 30 = 2.3.5 = q1 · q2 · q3. All the conditions ofTheorem 6.3.5are satisfied. Thus G(M5, 30) consists of two components. Figure 6.8exhibits the digraph G(M5, 30).

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Figure 6.8: The power digraph G(M5, 30) having exactly two components

Example 6.3.8. Let n = 33 = 27 and k = 5 = q1. Here n = 33 , qα1 and3 − 1 = 2 , qγ1

1 . The conditions of Theorem 6.3.5are not satisfied. Thus G(33, 5)does not consists of two components. In fact G(33, 5) contains 7 components. Thisis shown in Figure 6.7.

Example 6.3.9. Let p = 41 and k = 5 = q1. Here p − 1 = 40 = 23 · 5 , qγ11 . The

conditions of Theorem 6.3.5are not satisfied. Thus G(41, 5) does not consist of twocomponents. In fact G(41, 5) contains 7 components. This is shown in Figure 6.7.

6.4 SummaryIn this chapter, some necessary and sufficient conditions on n and k are determinedfor the digraph G(n, k) having at least one isolated fixed point. The first lemmaestablishes that an isolated fixed point exists in G(n, k) if and only if this happensin the corresponding direct factors of G(n, k). The classification of semi-regularpower digraphs with respect to its components is achieved. It is shown that n is aFermat prime or some power of 2 if and only if a power digraph G(n, k) is semi-regular (of degree 2α) with exactly two components. With the help of these results,

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Figure 6.9: The power digraph G(33, 5) with more than two components

the primality of a Fermat prime is tested. Finally, the power digraphs consisting ofexactly two components are completely classified.

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Figure 6.10: The power digraph G(41, 5) having more than two components

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Chapter 7

Conclusions and Future Directions

7.1 ConclusionsIn this dissertation, different properties of certain power digraphs are investigated.Much attention has been given to the symmetry of power digraphs, the heights ofvertices, the isolated fixed points and the classification of G(n, k) with respect tothe heights of vertices and the number of components. The symmetry of G(n, k)has been studied for some integers of the form n = pαq1 · · · qr, where p, qi areprimes and α > 1. Some conditions on p, qi, α and k have been determinedunder which the digraphs G(n, k) are not symmetric of order p. Further, certainnecessary and sufficient conditions for the existence of symmetric power digraphsof order p are established. These results extend the results of (Somer and Krızek2009; Kramer-Miller 2009). Investigation of the vertices and components of powerdigraphs culminated in the derivation of several formulae for their heights. Thepower digraphs G1(n, pα) with components of height 1 are classified for all primep. Further, some necessary and sufficient conditions on n such that each vertex ofindegree 0 of a certain subdigraph of G(n, p) is at height q ≥ 1 are obtained. Anexpression for the number of vertices at a specific height is derived. The generalcriteria such that a digraph G(n, k) has at least one isolated fixed point are alsoestablished. Finally, the necessary and sufficient conditions on n and k such that thedigraph G(n, k) contains exactly two components are also obtained. The last resultcompletes the classification of G(n, k) with exactly two components.

7.2 Further Research ProblemsThe following problems seem to be unsolved till the time of writing this thesis.They are interesting enough to be investigated.

1. The symmetry of power digraphs modulo n is still largely open to investigation.The symmetric power digraphs modulo n for even n have been completely

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characterized in (Deng and Yuan 2011). The case of an odd n with more thanone relatively prime square factors can be considered for future research. Thework done by (Kramer-Miller 2009), (Somer and Kek 2009), (Deng and Yuan2011) and (Husnine et al. 2011) can be helpful.

2. To find the number of power digraphs, consisting of n vertices, up to isomorp-hism.

3. To characterize the digraphs in which each of its non trivial component lookslike a directed star graph or a sun graph.

4. One can study the properties of power digraphs defined on other algebraicstructures, for example, a finite field or a finite group.

5. To study the different graphical properties like chromatic number, matchingproblems or labeling of power digraphs.

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