The structure of power digraph connected with the congruence a ≡ b(mod n)

We assign to each positive integer n a digraph Γ(n, 11) whose set of vertices is Zn = {0, 1, 2, ..., n − 1} and there exists exactly one directed edge from a to b if and only if a ≡ b(mod n), where a, b ∈ Zn. Let Γ1(n, 11) be the subdigraph induced by the vertices which are coprime to n. We discuss when the subdigraph Γ1(n, 11) is regular or semi-regular. A formula for the number of fixed points of Γ(n, 11) is established. A necessary and sufficient condition for the symmetry of the digraph Γ(n, 11) is proved. Moreover, using Carmichaeĺs lambda function, the number of components and conditions for the existence of cycles in the digraph Γ(n, 11) is presented.


Introduction
In the year 1967 [2], S. Bryant introduced the concept of power digraph modulo n. In this paper, for each element x of a group, a point P (x) was assigned and joined with P (x 2 ) by a directed line arising a graph called the graph of the group. Subsequently, in 1992 [1], E.L. Blanton et al. described the structure of power digraph Γ(n, 2) using quadratic congruence in which there is a directed edge from a to b if and only if a 2 ≡ b(mod n) and they have proved that the number theoretic properties of n and (n−1) are closely associated with certain geometric properties of the digraph Γ(n, 2). In [16], L. Szalay has studied the number of fixed points, the number of binary trees and gave the condition for the digraph Γ(n, 2) to be symmetric. In 1996 [11], C. Lucheta et al. studied the structure of the power digraph using the congruence a k ≡ b(mod p), where k ≥ 2 is a positive integer. In 2004 ( [10,14]), L. Somer et al. established the necessary and sufficient for the existence of isolated fixed points and examined when a digraph is semiregular using quadratic congruence. In 2009 [13], J. Skowronek-Kaziw et al. defined a digraph Γ(n, 3) using the congruence a 3 ≡ b(mod n) in which the formula for the number of fixed points of Γ(n, 3) is established. Also, the necessary and sufficient condition for the symmetry of the digraph Γ(n, 3) is proved. In 2016 [12], M. Rahmati studied the digraph Γ(n, 5) which is constructed using the congruence a 5 ≡ b(mod n). In this paper, the author has determined the number of fixed points and discussed the structure of Γ(n, 5) for n = 2 k & 5 k , where k is a natural number. In 2019 [6], M. Haris Mateen et al. discussed the digraph Γ(n, 7) in association with the congruence a 7 ≡ b(mod n) and gave the explicit formula of fixed points. In the same paper, the author also discussed the condition for which the digraph Γ(n, 7) has exactly 7 components. In 2021 [8], M. Haris Mateen et al. classified cyclic vertices and enumerate components of power digraph for the classes of numbers 2 r and p r where r ∈ Z + . Inspired by these works, in the present paper, we try to propose some refreshing characterizations of the digraph Γ(n, 11) endowed with the congruence relation a 11 ≡ b(mod n) to study its structures. We organize our paper as follows: In section 2, we provide some definitions and results from number theory. In section 3, we have introduced the digraph Γ(n, 11) and discussed its structure and properties. Also, established a formula for number of fixed points of the digraph Γ(n, 11). Finally, in section 4, we have shown some applications of Carmichael lambda function. Throughout the paper all notations are usual. For example, the greatest common divisor of two integers m and n is denoted by gcd(m, n), the least common divisor of two integers m and n is denoted by lcm(m, n), the order of a modulo n is denoted by ord n (a) etc.

Preliminaries
Definition 2.1 [3] Let n be a fixed positive integer. Two integers a and b are said to be congruent modulo n if n divides a − b and this is denoted by a ≡ b(mod n).
Definition 2.2 [15] For each pair of integers n(≥ 1) and k(≥ 2), a power digraph modulo n denoted by Γ(n, k) is a digraph with vertex set Z n = {0, 1, 2, ..., n − 1} and the ordered pair (a, b) is a directed edge of Γ(n, k) from a to b if and only if a k ≡ b(mod n), where a, b ∈ Z n . Theorem 2.3 [3] If p is a prime and f (x )= a n x n +a n−1 x n−1 +. . .+a 1 x+a 0 ; a n 6 ≡ 0(mod p) is a polynomial of degree n ≥ 1 with integral coefficients, then the congruence f (x ) ≡ 0(mod p) has at most n incongruent solutions modulo p.
Corollary 2.4 [3] If p is a prime number and d|p − 1, then the congruence x d − 1 ≡ 0(mod p) has exactly d number of solutions.
Theorem 2.5 [3] Let n be an integer possessing a primitive root let gcd(a, n) = 1, then the congruence x k ≡ a(mod n) has a solution if and only if where d = gcd(k, ϕ(n)); if it has a solution, there are exactly d number of solutions modulo n.
Corollary 2.6 [3] Let p be a prime and gcd(a, p) = 1. Then the congruence x k ≡ a(mod p) has a solution if and only if a Theorem 2.7 [3] Let a be an odd integer, then x 2 ≡ a(mod 2 n ), for n ≥ 3, has a solution if and only if a ≡ 1(mod 8).

Structure and Properties of the Digraph Γ(n, 11)
For a positive integer n, Z n = {0, 1, 2, ..., n − 1} denotes the complete set of residues modulo n. We consider a directed graph Γ(n, 11) whose vertex set is Z n and any two vertices a, b ∈ Z n are connected by exactly one directed edge from a to b iff For example, consider the digraph Γ(17, 11) as shown below: If there exists a finite sequence {a 1 , a 2 , a 3 , ..., a t } of distinct numbers in Z n such that . . .
Then we say that the numbers a 1 , a 2 , a 3 , · · · , a t constitutes a cycle of length t. We call a cycle of length t as a t-cycle and a cycle of length 1 is named as a fixed point. For instance, the digraph Γ(13, 11) contains five 2-cyclecs and three fixed points. We denote the number of directed edges coming to the vertex a by indeg(a) and the number of directed edges leaving the vertex a by outdeg(a), where a ∈ Z n . Since the residue of a number modulo n is unique, the outdegree of each vertex is one and indegree of each vertex is a non-negative integer. For an isolated fixed point a, indeg(a)=1=outdeg(a).
Note that, a component of a digraph is a subdigraph which is a maximal connected subgraph of the associated nondirected graph (for details see [5]). As the outdegree of each vertex of the digraph Γ(n, 11) is equal to 1, so the number of components of Γ(n, 11) is equal to the number of all cycles. The cycles may or may not be isolated (see figure 1, figure 2 and figure 4). Moreover, by using the properties of the congruence relation (1) and the definition of cycle, it can be established that the digraph Γ(k 11 t − k, 11) has a t-cycle containing the vertex k, for an arbitrary natural number t ≥ 1. For example, Γ(n, 11) has a 3-cycle containing the vertex 2 iff n|2 11 3 − 2 = 2(2 1330 − 1) = 2(2 665 − 1)(2 665 + 1) Let, n = 2 665 − 1, then there is a 3-cycle containing the vertex 2 in the digraph Γ(2 665 − 1, 11) (see figure3). The diagraph Γ(n, 11) is called symmetric if its set of components can be split into two sets in such a way that there exists a bijection between these two sets such that the corresponding digraph are isomorphic. We call a digraph regular if the indegree of each vertex is equal to 1. Every component of such digraph is a cycle. A digraph is semi-regular if there exists a positive integer d such that each vertex either has indegree d or 0.
Moreover, if n is an arbitrary natural number and f is a polynomial with the integer coefficients, then the function ρ f (n)=|{0 ≤ m ≤ n − 1: f (m) = 0(mod n)}| is a multiplicative function (see [7]).

Lemma 3.1
The numbers 0, 1 & (n−1) are fixed points of Γ(n, 11). Moreover, 0 is an isolated fixed point of Γ(n, 11) if and only if n is square-free.
Proof: Clearly, Therefore, the numbers 0, 1 & (n − 1) are fixed points of Γ(n, 11). Let, 0 be an isolated fixed point of Γ(n, 11). If n is not square-free then p 2 |n, for some prime p and This shows that n p is mapped into 0 and so 0 is not an isolated fixed point, a contradiction. Hence, n is square-free. Conversely, if n is square free, then there exists no k, 1 ≤ k ≤ (n − 1) such that n|k 11 . So, there is no such k such that k 11 ≡ 0(mod n) and hence 0 is an isolated fixed point of Γ(n, 11). 3. The number k is an isolated fixed point if and only if n − k is an isolated fixed point.
Proof: We observe that, Also, for an even n we have Therefore, the statement (i) is true. Moreover, The last three observations establish the statements (ii), (iii), and (iv) of the lemma respectively. 1. The number of undecic roots (if they exist) of any undecic residue in Γ 1 (n, 11) is equal to the number of undecic roots of 1(mod n) i.e., each vertex of the digraph Γ 1 (n, 11) has same positive indegree d or 0.
It is clear that the congruence x 11 ≡ 1(mod n) is equivalent to the system of congruences . . .
Our aim is to find the number of solutions of the congruence x 11 ≡ 1(mod n). Clearly, 1 is the trivial solution of this congruence. Now, let us suppose that a(6 = 1) be a nontrivial solution of the congruence x 11 ≡ 1(mod n), then Therefore, we have gcd(a, p)=1 and ord p a = 11. Moreover, Z * p is a cyclic group of order p − 1, so 11|p − 1 which implies p ≡ 1(mod 11). Thus, there exists a non-trivial solution of x 11 ≡ 1(mod p) if and only if p ≡ 1(mod 11).
If p = 11 then by corollary 2.4, we can conclude that each of the congruence x 11 ≡ 1(mod p) has 11 solutions and by corollary 2.6, x 11 ≡ 1(mod p α ), α > 1 has also 11 solutions.
If p=11 then by corollary 2.6, the congruence x 11 ≡ 1(mod p) have exactly one solution x = 1.
Conversely, let 11|ϕ(n), then 11 2 |n or there exists a prime p ≡ 1(mod 11) such that p|n. Thus, by lemma 3.3, it follows that indeg(a) = 0 or 11 ω(n) , for each vertex a in Γ 1 (n, 11) and hence Γ 1 (n, 11) is semiregular. Proof: Let the digraph Γ(n, 11) be cyclic. By definition, a digraph is cyclic if all of its components are cyclic. Moreover, if all the components of the digraph Γ(n, 11) are cycles, then the digraph Γ(n, 11) is regular. Therefore, it is clear that Γ 1 (n, 11) is regular and hence 11 6 |ϕ(n) and n is square free (by lemma 3.4 and lemma 3.1).
Conversely, let us assume that n be square free and 11 6 |ϕ(n). Then by lemma 3.4, Γ 1 (n, 11) is regular. So, for the digraph Γ(n, 11) to be regular, it is sufficient to prove that the digraph Γ 2 (n, 11) is regular. Let, a be any vertex of Γ 2 (n, 11) such that a 6 = 0. Then gcd(a, n) = d(> 1) and n = d · n d .
Let, p be a prime such that p|n, then p|d or p | n d . If p|d, then the solution u of the congruence u 11 ≡ a(mod n) satisfies u ≡ 0(mod p). Hence, u 11 ≡ a ≡ 0(mod p), for each prime p such that p|d. Thus, the solution u is unique by lemma 3.1.
The following theorem gives a formula for the number of fixed points of the digraph Γ(n, 11).
Conversely, let n ≡ 2(mod 4). Clearly, n is even and n 2 is odd. Let C 1 and C 2 be two disjoint components of Γ(n, 11) containing the elements 0 and n 2 respectively. Then by lemma 3.2, Γ(n, 11) − {C 1 , C 2 } is symmetric. So, in order to prove that Γ(n, 11) is symmetric, it is enough to prove that Case-I: Let n be square free, then 0 is an isolated fixed point in C 1 . For C 1 ∼ = C 2 , it is enough to show n 2 is also an isolated fixed point in C 2 .
Suppose n 2 is not an isolated fixed point, then there exist k(6 = 0) ∈ Z n such that Then (n − k) 11 ≡ n 2 (mod n). As n | k 11 − n 2 & n | k 11 + n 2 this gives n | 2k 11 . Since, n is square free, so n = 2k. Thus n is even and k = n 2 . Therefore, we get ( n 2 ) 11 ≡ n 2 (mod n). This shows that n 2 is an isolated fixed point in C 2 and hence C 1 ∼ = C 2 .
Case-II: Let n be not square free, then 0 is not an isolated fixed point in C 1 . So there exist k ∈ Z n such that k 11 ≡ 0(mod n) which implies k 11 is even and hence k is even. Let k = 2 a · b, where a ≥ 1 and b is an odd number. As k 11 ≡ 0(mod n) and b is odd, so b 11 must be an odd multiple of n 2 . Therefore b 11 ≡ n 2 (mod n).
Again, if d 11 ≡ n 2 (mod n) then this imply l 11 ≡ 0(mod n), which is not true and hence d 11 6 ≡ n 2 (mod n) i.e., n 6 |d 11 − n 2 . Moreover, n | l 121 , d 121 is an odd multiple of n 2 and d 121 ≡ n 2 (mod n). Thus, there exists f ∈ Z n such that d 11 ≡ f (mod n) and f 11 ≡ n 2 (mod n).
The following theorem generalizes the well-known Eulerś Theorem which says that a ϕ(n) ≡ 1(mod n) if and only if gcd(a, n) = 1. It shows that λ(n) is the least possible order modulo n. Theorem 4.3 (Carmichaelś Theorem, see [4] and [9]). Let a, n ∈ N. Then a λ(n) ≡ 1(mod n) if and only if gcd(a, n) = 1.
Moreover, there exists an integer g such that ord n g = λ(n); where ord n g denotes the multiplicative order of g(mod n).
Assume now that λ(n) has the following prime power factorization: where q 1 < q 2 < q 3 < . . . < q r are primes and l j > 0. It is evident that from the definition of λ that q 1 = 2, if n > 2. Proof: Suppose there exists a cycle of length t in the digraph Γ(n, 11). Let a be a vertex of a t−cycle in the digraph Γ(n, 11), then t is the least +ve integer such that a 11 t ≡ a(mod n).
Case-I: If gcd(a, n) = 1 then a ∈ Γ 1 (n, 11). Since, the vertices of Γ 1 (n, 11) forms a group of order ϕ(n) with respect to the multiplication modulo n so a −1 exists and we have where t is such least positive integer. ⇒ ord d (11) = t Now, by Carmichaelś Theorem we have a λ(n) ≡ 1(mod n) (as gcd(a, n) = 1). Also, ord n (a) = d so d | λ(n). Again, d | 11 t − 1, so d must be even. Thus, if gcd(a, n) = 1 and a is a vertex of a t-cycle in the digraph Γ(n, 11), then t = ord d (11), where d is an even positive divisor of λ(n).
Conversely, let t = ord d (11), for some even positive divisor d of λ(n). By Carmichaelś Theorem, there exists an integer g such that ord n g = λ(n) that is g λ(n) ≡ 1(mod n). Let a = g λ(n) d (as d | λ(n) so λ(n) d is an integer) then a d ≡ 1(mod n) ⇒ ord n (a) = d. Again, since t = ord d (11) so t is the least positive integer such that 11 t ≡ 1(mod d) ⇒ d | 11 t − 1. Thus, t is the least positive integer such that a 11 t − 1 ≡ 1(mod n) which gives a 11 t ≡ a(mod n). Hence, a is a vertex of some t-cycle in the digraph Γ(n, 11).  Proof: Let us assume that the number of components of the digraph Γ(n, 11) is 3, then there exist at most 3 fixed points. By theorem 3.7, either n = 4 or n is the power of some odd prime number. Clearly, there is no t-cycle for t > 1. Otherwise, there are more than 3 components of Γ(n, 11). Hence, by theorem 4.4, d 6 |11 t − 1, for every t > 1 and every even divisor d(if it exists) > 2 of the Carmichael λ-function λ(n). Therefore, 11 | d and λ(n) = 2 · 11 s , for some natural number s. Hence, n must be 4 or a prime number of the form n = 2 · 11 k + 1, k ≥ 0.
Conversely, we assume that n = 4 or n is a prime number of the form n = 2 · 11 k + 1, for some integer k ≥ 0. If n = 4, then clearly there are 3 components. If n is a prime number of the form n = 2 · 11 k + 1, then by theorem 3.7, we have exactly 3 fixed points and λ(n) = 2 · 11 k . If there are more than 3 components, then clearly, there exist a cycle of length t > 1 and t = ord d (11), for some even positive divisor d of λ(n). Then t is the least positive integer such that 11 t ≡ 1(mod d) and d | 11 t − 1. Also, d | λ(n) i.e., d | 2 · 11 k , so we get d = 2. Thus, t = ord d (11) = ord 2 (11) = 1, which is a contradiction. Hence, the only cycles of Γ(n, 11) are the fixed points at 0,1 and at (n − 1).

Theorem 4.7
The number of components of the digraph Γ(n, 11) is 11 if n = 11 k , k ≥ 1.

Conclusions
In this paper, we tried to study the structure of the digraph Γ(n, 11) associated with the congruence a 11 ≡ b(mod n). We discussed some conditions under which the digraph Γ 1 (n, 11) is regular or semiregular. Cyclic and symmetric nature of the digraph Γ(n, 11) has been discussed. We also established a formula to find the number of fixed points of the digraph Γ(n, 11). Finally, we studied about the existence of t-cycle with the help of Carmichael lambda function.

Conflict of Interest
We certify that there is no actual or potential conflict of interest in relation to this paper.