Isomorphic Digraphs from Affine Maps of Finite Cyclic Groups

Abstract

Let 𝑛 be a positive integer. For any pair of integers 𝑘 and 𝑎, let 𝐺(𝑛,𝑘,𝑎) be the digraph whose set of vertices is {0,1,2,…,𝑛−1}, and there exists a directed edge from vertex 𝑥 to vertex 𝑦 if 𝑘𝑥+𝑎≡𝑦(mod𝑛). In this paper, we obtain a necessary and sufficient condition for which 𝐺(𝑛,𝑘₁,𝑎₁)≃𝐺(𝑛,𝑘₂,𝑎₂).

1. Introduction

Let 𝑋 be a nonempty set and let 𝑓 be a mapping from 𝑋 into itself. Then, we call 𝐺(𝑋,𝑓) the iteration digraph from 𝑓 of 𝑋, where the set of vertices of 𝐺(𝑋,𝑓) is 𝑋 and there is a directed edge from a vertex 𝑎∈𝐺(𝑋,𝑓) to a vertex 𝑏∈𝐺(𝑋,𝑓) if 𝑓(𝑎)=𝑏. In particular, there are many interesting structures on the digraphs when there is an algebraic structure on 𝑋.

Many people study the iteration graphs from endomorphisms of finite abelian groups. The authors of [1] studied the digraph from the endomorphism 𝑓(𝑥)=𝑥² of (ℤ/𝑝ℤ)^∗, where 𝑝 is a prime. These results were generalized to the digraph from any endomorphism of (ℤ/𝑝ℤ)^∗ in [2]. Sha and Hu [3] studied some elementary properties of the digraph from any endomorphism of 𝔽^∗_𝑞. They also obtained many elementary properties of the digraph from any endomorphism of finite cyclic groups in [3].

Let 𝑛 be a positive integer and lrt 𝑘,𝑎 be two integers. In this paper, we denote by 𝐺(𝑛,𝑘,𝑎) the digraph obtained from the affine mapping 𝑥↦𝑘𝑥+𝑎 of ℤ_𝑛, the ring of integers modulo 𝑛. In particular, we set 𝐺(𝑛,𝑘)=𝐺(𝑛,𝑘,𝑎) if 𝑛∣𝑎, which is from an endomorphism 𝑥↦𝑘𝑥.

We are mainly interested in the problem when 𝐺(𝑛,𝑘₁,𝑎₁)≃𝐺(𝑛,𝑘₂,𝑎₂). The authors in [2] gave an example 𝐺(10,2)≃𝐺(10,8). Let 𝑝 be a prime. Sha [4] showed that 𝐺(𝑝,𝑘₁)≃𝐺(𝑝,𝑘₂) if and only if ord_𝑝𝑘₁=ord_𝑝𝑘₂. In [5], Deng and Yuan gave a necessary and sufficient condition for 𝐺(𝑝−1,𝑘₁)≃𝐺(𝑝−1,𝑘₂), where 𝑝 is a prime. In this paper, we obtain a necessary and sufficient condition for when 𝐺(𝑛,𝑘₁,𝑎₁)≃𝐺(𝑛,𝑘₂,𝑎₂).

The paper is organized as follows. In Section 2, we study the 𝐺(𝑛,𝑘,𝑎), where 𝑛 is a power of a prime. The main results are proved in Section 3. We also determined the digraph from projective map on one-dimensional projective space over a finite field in the last section.

2. Digraph Product and the Local Case

Let 𝐺 be a digraph and let 𝑎 be a vertex in 𝐺. A component of a digraph is a subdigraph which is a maximal connected subgraph. Each component of a finite iteration graph 𝐺(𝑋,𝑓) contains a unique cycle. Let 𝑂_𝑖 denote this 𝑖-cycle. Let 𝐴₁(𝐺) denote the number of cycle of length 1 in 𝐺.

Let 𝐺₁,𝐺₂,…,𝐺_𝑟 be iteration digraphs. The product 𝐺₁×𝐺₂×⋯×𝐺_𝑟 is the digraph whose vertices are the ordered pairs (𝑎₁,𝑎₂,…,𝑎_𝑟), where 𝑎_𝑖∈𝐺_𝑖 and there is a directed edge from (𝑎₁,𝑎₂,…,𝑎_𝑟) to (𝑏₁,𝑏₂,…,𝑏_𝑟) if and only if there is a directed edge in 𝐺_𝑖 from 𝑎_𝑖 to 𝑏_𝑖 for each 1≤𝑖≤𝑟. We use the following notation: ^𝑟_𝑖=1𝐺_𝑖∶=𝐺₁×𝐺₂×⋯×𝐺_𝑟. (1)

Lemma 1 (see [6]). 𝑂_𝑠×𝑂_𝑡 is isomorphic to gcd(𝑠,𝑡)𝑂_{lcm[𝑠,𝑡]}, which is a union of gcd(𝑠,𝑡) cycles, each having length lcm[𝑠,𝑡].

Theorem 2. Let 𝑛=∏^𝑟_𝑖=1𝑝^𝑒𝑖_𝑖 be the prime factorization of 𝑛 and let 𝑘,𝑎 be two integers. Then 𝐺(𝑛,𝑘,𝑎)≅^𝑟_𝑖=1𝐺𝑝^𝑒𝑖_𝑖,𝑘,𝑎. (2)

Proof. Let 𝐺=𝐺(𝑛,𝑘,𝑎) and 𝐺_𝑖=𝐺(𝑝^𝑒𝑖_𝑖,𝑘,𝑎). Assume that 𝜑∶ℤ_𝑛→ℤ_𝑝^𝑒₁₁×ℤ_𝑝^𝑒₂₂×⋯×ℤ_𝑝^𝑒𝑟_𝑟 is the canonical isomorphism of rings. Then 𝜑 induces an isomorphism of digraphs.

2.1. The Case of Linear Maps

The structure of 𝐺(𝑛,𝑘) has been studied in [2, 4, 5]. We present some results here.

Lemma 3 (see [4]). Let 𝑘,𝑛 be positive integers. Then one has the following.(i)If 𝑝∣𝑘 for any prime 𝑝∣𝑛, then 𝐺(𝑛,𝑘) is a tree attached to the fixed point 0.(ii)If gcd(𝑛,𝑘)=1, then 𝐺(𝑛,𝑘)≃∪_𝑟𝑎_𝑟𝑂_𝑟, where 𝑎_𝑟=(1/𝑟)∑_𝑖∣𝑟𝜇(𝑖)gcd(𝑛,𝑘^{𝑟/𝑖−1}) and 𝜇 is the Möbius function.

Let 𝑣_𝑝(𝑛) be the 𝑝-adic valuation of 𝑛 for any prime 𝑝 and integer 𝑛. It is the largest power of 𝑝 dividing 𝑛 for any prime 𝑝 and integer 𝑛. We set 𝑣_𝑝(0)=∞ as usual. If 𝑋 is a set of primes and 𝑛 is a positive integer, we denote that 𝑛_𝑋=_𝑝∈𝑋𝑝^{𝑣𝑝(𝑛)}. (3)

Lemma 4. Let 𝑝 be a prime. Let 𝑘,𝑎 be positive integers such that 𝑘,𝑎≤𝑝^𝑒−1. The following statements are equivalent:(i)𝐺(𝑝^𝑒,𝑘,𝑎)≃𝐺(𝑝^𝑒,𝑘);(ii)gcd(𝑝^𝑒,𝑘−1,𝑎)=gcd(𝑝^𝑒,𝑘−1);(iii)𝑣_𝑝(𝑘−1)≤𝑣_𝑝(𝑎);(iv)𝐺(𝑝^𝑒,𝑘,𝑎) has a fixed point.

Proof. By hypothesis the statements (ii), (iii), and (iv) are equivalent to that the congruence equation (𝑘−1)𝑥≡𝑎(mod𝑝^𝑒) that has a solution. Suppose that 𝐺(𝑝^𝑒,𝑘,𝑎)≃𝐺(𝑝^𝑒,𝑘). Then 𝐺(𝑝^𝑒,𝑘,𝑎) has a fixed point 𝑧 since 𝐺(𝑝^𝑒,𝑘) has a fixed point 0. So 𝑘𝑧+𝑎≡𝑧(mod𝑝^𝑒). On the other hand, if 𝑘𝑧+𝑎≡𝑧(mod𝑝^𝑒), then 𝑥↦𝑥+𝑧 is an isomorphism of 𝐺(𝑝^𝑒,𝑘) and 𝐺(𝑝^𝑒,𝑘,𝑎).

Remark 5. Many people showed that the tree attached to any two cycle vertices in 𝐺(𝑛,𝑘) is isomorphic. Moreover, Deng and Yuan [5] showed that the tree attached to cycle vertices in 𝐺(𝑝−1,𝑘) is totally determined by gcd(𝑝−1,𝑘), where 𝑝 is a prime. In fact, if we replace 𝑝−1 by 𝑛 in the arguments in [5], then the result is also valid.

2.2. The Case of Affine Maps

Lemma 6. Let 𝑘=𝑙𝑝^𝑡+1 be a positive integer, where 𝑝 is a prime and 𝑝∤𝑙, 𝑡≥1.(i)If 𝑝>2 or 𝑡>1, then ord_𝑝𝑟𝑘=max{1,𝑝^𝑟−𝑡} for any integers 𝑡≥1.(ii)If 𝑝=2 and 𝑡=1, assuming that 2^𝑢‖𝑘²−1, then ord₂^𝑟2𝑘=1,𝑟=1,2,2≤𝑟≤𝑢,_{𝑟−𝑢+1},𝑟>𝑢. (4)

Proof. If 𝑟≤𝑡, then it is clear that ord_𝑝𝑟𝑘=1=max{1,𝑝^𝑟−𝑡}. Firstly, we assume that 𝑟>𝑡. By computation, 𝑝^{𝑚−𝑠𝑖}∣∣^𝑝𝑚_𝑖 for any 1≤𝑖≤𝑝^𝑚−1 and 𝑝^𝑠𝑖∣∣𝑖. But 𝑘^{𝑝𝑟−𝑡}−1=𝑙𝑝^𝑡+1^{𝑝𝑟−𝑡}−1=^{𝑝𝑟−𝑡}_𝑖=1𝑝^𝑟−𝑡𝑖𝑙𝑝^𝑡𝑖; (5) thus 𝑝^{𝑟−𝑡−𝑠𝑖+𝑡𝑖}∣∣^{𝑝𝑟−𝑡}_𝑖(𝑙𝑝^𝑡)^𝑖. It is clear that 𝑟−𝑡−𝑠_𝑖+𝑡_𝑖≥𝑟. So 𝑝^𝑟∣𝑘^{𝑝𝑟−𝑡}−1; considering that ord_𝑝𝑟𝑘∣𝑝^𝑟−𝑡. So ord_𝑝𝑟𝑘=𝑝 if 𝑟−𝑡=1. Suppose that 𝑟−𝑡>1. If 𝑝>2 or 𝑡>1, then 𝑘^{𝑝𝑟−𝑡−1}−1=𝑙𝑝^𝑡+1^{𝑝𝑟−𝑡−1}=−1_{𝑝𝑟−𝑡−1𝑖=1}𝑝^{𝑟−𝑡−1}𝑖𝑙𝑝^𝑡𝑖=𝑙𝑝^𝑟−1+^{𝑝𝑟−𝑡−1}_𝑖=2𝑝^{𝑟−𝑡−1}𝑖𝑙𝑝^𝑡𝑖. (6) Therefore, 𝑟−𝑡−1−𝑠_𝑖+𝑡_𝑖≥𝑟 for any 𝑖≥2. Hence, 𝑝^𝑟∣^{𝑝𝑟−𝑡−1}_𝑖(𝑙𝑝^𝑡)^𝑖 for 𝑖≥2 and 𝑝^𝑟∤𝑘^{𝑝𝑟−𝑡−1}−1. Thus, ord_𝑝𝑟𝑘=𝑝^𝑟−𝑡 in this case.
Suppose that 2‖𝑘−1and2^𝑢‖𝑘²−1, 𝑢≥3. It remains to show that ord₂^𝑟𝑘=2^{𝑟−𝑢+1} for 𝑟>𝑢. Let 𝑘²=𝑞2^𝑢+1, where 2∤𝑞. By the discussion above we get 2^𝑟‖𝑘^{2𝑟−𝑢+1}−1 and 2^𝑟−1‖𝑘^2𝑟−𝑢−1. Therefore, ord₂^𝑟𝑘=2^{𝑟−𝑢+1} for 𝑟>𝑢. This finishes the proof.

Lemma 7. Let 𝑝 be a prime. Let 𝑘,𝑎≤𝑝^𝑒−1 be two positive integers such that 𝑣_𝑝(𝑘−1)>𝑣_𝑝(𝑎). Then 𝐺(𝑝^𝑒,𝑘,𝑎)≃𝑝^𝑒−𝑠𝑂_𝑝𝑠, where𝑝^𝑠=ord_𝑝𝑝𝑝^{𝑒+𝑣(𝑘−1)−𝑣(𝑎)}𝑝𝑘=^{𝑒−𝑣𝑝(𝑎)},2∤𝑝𝑜𝑟𝑣_𝑝(𝑘−1)>1,2,𝑝=2,𝑣₂(𝑘−1)=1,𝑎𝑛𝑑𝑒+1≤𝑣₂(),2𝑘−1_{𝑒+2−𝑣2(𝑘2−1)},𝑝=2,𝑣₂(𝑘−1)=1,𝑎𝑛𝑑𝑒+1>𝑣₂(𝑘−1). (7)

Proof. Let 𝑓(𝑥)=𝑘𝑥+𝑎. Then 𝑓^𝑡(𝑥)=𝑘^𝑡𝑥+(∑^𝑡−1_𝑖=0𝑘^𝑖)𝑎. We consider the following congruence equation 𝑘^𝑡𝑥+^𝑡−1_𝑖=0𝑘^𝑖𝑎≡𝑥(mod𝑛). (8) Equation (8) has a solution if and only if gcd(𝑝^𝑒,𝑘^𝑡−1)∣gcd(𝑝^𝑒,(𝑘^𝑡−1)/(𝑘−1)⋅𝑎). In this case, 𝑝^𝑒∣𝑘^𝑡−1 and 𝑝^𝑒∣(𝑘^𝑡−1)/(𝑘−1)⋅𝑎, since 𝑣_𝑝(𝑘−1)>𝑣_𝑝(𝑎). Since 𝑡 is the minimal positive integer such that 𝑝^𝑒∣(𝑘^𝑡−1)/(𝑘−1)⋅𝑎, we get 𝑡=ord_𝑝𝑝𝑝^{𝑒+𝑣(𝑘−1)−𝑣(𝑎)}𝑘=𝑝^𝑠. Moreover, each integer is a solution of (8). Thus, 𝐺(𝑝^𝑒,𝑘,𝑎) is a union of cycles, each having length 𝑝^𝑠. The rest of the proof follows from Lemma 6.

The structure of digraph 𝐺(𝑝^𝑒,𝑘,𝑎) is clear now. We classify it as follows. Type (1): 𝐺(𝑝^𝑒,𝑘,𝑎)≃𝐺(𝑝^𝑒,𝑘) is a tree attached to a fixed point, which is determined by gcd(𝑝^𝑒,𝑘). Type (2): 𝐺(𝑝^𝑒,𝑘,𝑎)≃𝐺(𝑝^𝑒,𝑘) is a union of cycles, and the minimal length of these cycles is 1. Type (3): 𝐺(𝑝^𝑒,𝑘,𝑎)≃𝑝^𝑒−𝑠𝑂_𝑝𝑠, where 1≤𝑠≤𝑒.

Let 𝑋_𝑖={𝑝∣𝐺(𝑝^{𝑣𝑝(𝑛)},𝑘,𝑎)istype(𝑖)}. By the properties of digraph product we see that 𝐺(𝑛,𝑘,𝑎)≃𝐺𝑛_𝑋1,𝑘,𝑎×𝐺𝑛_𝑋2,𝑘,𝑎×𝐺𝑛_𝑋3,𝑘,𝑎, (9) where 𝐺(𝑛_𝑋1,𝑘,𝑎) is a tree attached to a fixed point, 𝐺(𝑛_𝑋2,𝑘,𝑎) is a union of cycles, and the minimal length of these cycles is 1, and 𝐺₃≃(𝑛_𝑋3/𝑙)𝑂_𝑙.

3. The Main Results

We begin with some lemmas before we prove our main results.

Lemma 8. Suppose that 𝑓 is a permutation of a finite set 𝑋 and 𝐺(𝑋,𝑓)≃∪^𝑛_𝑖=1𝑎_𝑖𝑂_𝑖. Then 𝐺(𝑋,𝑓^𝑟)≃∪^𝑛_𝑖=1gcd(𝑟,𝑖)𝑎_𝑖𝑂_{𝑖/gcd(𝑖,𝑟)} for each 𝑟≥1.

Proof. Assume that 𝐸={𝑥₁,𝑥₂,…,𝑥_𝑙} is a cycle of length 𝑙 in 𝐺(𝑋,𝑓) such that 𝑓(𝑥_𝑖)=𝑥_𝑖+1, for 𝑖=1,2,…,𝑙−1 and 𝑓(𝑥_𝑙)=𝑥₁. Suppose that 𝑥_𝑖 is contained in a cycle of length 𝑚 in 𝐺(𝑋,𝑓^𝑟). Then 𝑚=min{𝑗∶𝑓^𝑟𝑗(𝑥_𝑖)=𝑥_𝑖}=min{𝑗∶𝑙∣𝑟𝑗}=𝑙/gcd(𝑟,𝑙). Hence, 𝐸 splits into gcd(𝑟,𝑙) cycles with the same length 𝑙/gcd(𝑟,𝑙). This completes the proof.

Lemma 9. Let 𝑓_𝑖 be a permutation of a finite set 𝑋_𝑖, 𝑖=1,2. Then(i)𝐺(𝑋₁,𝑓₁)≃𝐺(𝑋₂,𝑓₂) if and only if the numbers of fixed point of 𝐺(𝑋₁,𝑓^𝑟₁) and 𝐺(𝑋₂,𝑓^𝑟₂) are equal for each 𝑟≥1;(ii)𝐺(𝑋₁,𝑓₁)×𝑂_𝑙≃𝐺(𝑋₂,𝑓₂)×𝑂_𝑙 if and only if 𝐺(𝑋₁,𝑓^𝑙₁)≃𝐺(𝑋₂,𝑓^𝑙₂) for each 𝑙≥1.

Proof. Suppose that 𝐺(𝑋₁,𝑓₁)≃∪𝑎_𝑖𝑂_𝑖 and 𝐺(𝑋₂,𝑓₂)≃∪𝑏_𝑗𝑂_𝑗.
Assume that 𝜑∶𝐺(𝑋₁,𝑓₁)→𝐺(𝑋₂,𝑓₂) is an isomorphism of digraphs. Then 𝑓₂(𝜑(𝑥))=𝜑(𝑓₁(𝑥)) for any 𝑥∈𝑋₁. By induction, 𝑓^𝑟₂(𝜑(𝑥))=𝜑(𝑓^𝑟₁(𝑥)). Hence, 𝜑 is also an isomorphism of 𝐺(𝑋₁,𝑓^𝑟₁) and 𝐺(𝑋₂,𝑓^𝑟₂). Thus, 𝐴₁(𝐺(𝑋,𝑓^𝑟₁))=𝐴₁(𝐺(𝑋,𝑓^𝑟₂)). Conversely, if 𝐺(𝑋₁,𝑓^𝑟₁)≃𝐺(𝑋₂,𝑓^𝑟₂), then 𝑎₁=𝑏₁. Assume that 𝑎_𝑖=𝑏_𝑖 for any 𝑖<𝑟. By Lemma 8, 𝐴₁(𝐺(𝑋₁,𝑓^𝑟₁))=∑_𝑖∣𝑟𝑖𝑎_𝑖 and 𝐴₁(𝐺(𝑋₂,𝑓^𝑟₂))=∑_𝑗∣𝑟𝑗𝑏_𝑗. Hence, ∑_𝑖∣𝑟𝑖𝑎_𝑖=∑_𝑗∣𝑟𝑗𝑏_𝑗 and 𝑎_𝑟=𝑏_𝑟. The proof of statement (i) is complete by induction.
It remains to prove (ii). Assume that 𝐺(𝑍,𝑔)≃𝑂_𝑙. Then 𝐺(𝑋_𝑖,𝑓_𝑖)×𝑂_𝑙≃𝐺(𝑋_𝑖×𝑍,𝑓_𝑖×𝑔), where 𝑓_𝑖×𝑔 maps (𝑥,𝑧) to (𝑓_𝑖(𝑥),𝑔(𝑧)) for any (𝑥,𝑧)∈𝑋_𝑖×𝑍. Let 𝐺^𝑟_𝑖=𝐺(𝑋_𝑖×𝑍,(𝑓_𝑖×𝑔)^𝑟). By Lemma 8, 𝐺^𝑟_𝑖≃𝐺𝑋_𝑖,𝑓^𝑟_𝑖×𝐺𝑍,𝑔^𝑟≃𝐺𝑋_𝑖,𝑓^𝑟_𝑖×gcd(𝑟,𝑙)𝑂_{𝑙/gcd(𝑟,𝑙)}. (10)
If 𝐺(𝑋₁,𝑓₁)×𝑂_𝑙≃𝐺(𝑋₂,𝑓₂)×𝑂_𝑙, then 𝐺^𝑟₁≃𝐺^𝑟₂ for each 𝑟≥1. Let 𝑟=𝑙 in (10). We get 𝐺(𝑋₁,𝑓^𝑙₁)≃𝐺(𝑋₂,𝑓^𝑙₂). Conversely, if 𝐺(𝑋₁,𝑓^𝑙₁)≃𝐺(𝑋₂,𝑓^𝑙₂), then by (10) 𝐴₁(𝐺^𝑟_𝑖)=0 if 𝑙∤𝑟. If 𝑙∣𝑟, then 𝐺^𝑟_𝑖≃𝐺𝑋_𝑖,𝑓^𝑟_𝑖×gcd(𝑟,𝑙)𝑂_{𝑙/gcd(𝑟,𝑙)}≃𝑙𝐺𝑋_𝑖,𝑓^𝑙_𝑖^𝑟/𝑙. (11) Hence, 𝐺^𝑟₁≃𝐺^𝑟₂ if 𝑙∣𝑟. Therefore, 𝐴₁(𝐺^𝑟₁)=𝐴₁(𝐺^𝑟₂) for any 𝑟≥1. By statement (i), 𝐺(𝑋₁,𝑓₁)×𝑂_𝑙≃𝐺(𝑋₂,𝑓₂)×𝑂_𝑙.

In [3], Sha and Hu showed that 𝐺(𝑝,𝑘₁)≃𝐺(𝑝,𝑘₂) if and only if ord_𝑝𝑘₁=ord_𝑝𝑘₂, where 𝑝 is a prime and 𝑝∤𝑘_𝑖. We generalized this result to the following.

Lemma 10. Let 𝑛, 𝑘₁ and 𝑘₂ be positive integers such that gcd(𝑛,𝑘_𝑖)=1, 𝑖=1,2. Then 𝐺(𝑛,𝑘₁)≃𝐺(𝑛,𝑘₂) if and only if ord_𝑡𝑘₁=ord_𝑡𝑘₂ for any divisor 𝑡 of 𝑛.

Proof. Let 𝐺^𝑟_𝑖=𝐺(𝑛,𝑘^𝑟_𝑖). By Lemma 3 and Möbius inverse transform 𝐴₁(𝐺^𝑟_𝑖)=gcd(𝑘^𝑟_𝑖−1,𝑛) for any 𝑟≥1. Suppose that ord_𝑡𝑘₁=ord_𝑡𝑘₂ for any divisor 𝑡 of 𝑛. If 𝑡₁∣gcd(𝑘^𝑟₁−1,𝑛), let 𝑠=ord_𝑡1𝑘₁=ord_𝑡1𝑘₂; then 𝑠∣𝑟. Thus 𝑡₁∣gcd(𝑘^𝑟₂−1,𝑛). By symmetry, we have 𝑡₂∣gcd(𝑘^𝑟₁−1,𝑛) if 𝑡₂∣gcd(𝑘^𝑟₂−1,𝑛). Thus, gcd(𝑘^𝑟₁−1,𝑛)=gcd(𝑘^𝑟₂−1,𝑛). By Lemma 9, 𝐺(𝑛,𝑘₁)≃𝐺(𝑛,𝑘₂).
Conversely, if 𝐺(𝑛,𝑘₁)≃𝐺(𝑛,𝑘₂), then 𝐺^𝑟₁≃𝐺^𝑟₂ and gcd(𝑘^𝑟₁−1,𝑛)=gcd(𝑘^𝑟₂−1,𝑛) for any 𝑟≥1. If 𝑡∣𝑛, let 𝑠_𝑖=ord_𝑡𝑘_𝑖; then 𝑡∣gcd(𝑘^𝑠1₁−1,𝑛)=gcd(𝑘^𝑠1₂−1,𝑛). Hence, 𝑠₂∣𝑠₁. By symmetry, 𝑠₁∣𝑠₂. Thus, 𝑠₁=𝑠₂. The proof is complete.

Theorem 11. Let 𝑛=∏^𝑟_𝑖=1𝑝^𝑒𝑖_𝑖 be the prime factorization of 𝑛 and let 𝑘_𝑖, 𝑎_𝑖 be integers, 𝑖=1,2. Suppose that 𝑌_𝑗={𝑝_𝑖∣𝐺(𝑝^𝑒𝑖_𝑖,𝑘_𝑗,𝑎_𝑗)𝑖𝑠𝑇𝑦𝑝𝑒(2)} and that the minimal length of cycles in 𝐺(𝑛,𝑘_𝑖,𝑎_𝑖) is 𝑙_𝑖. Then 𝐺(𝑛,𝑘₁,𝑎₁)≃𝐺(𝑛,𝑘₂,𝑎₂) if and only if the following two conditions are satisfied.(i)gcd(𝑛,𝑘₁)=gcd(𝑛,𝑘₂).(ii)𝑌₁=𝑌₂=𝑌, 𝑙₁=𝑙₂=𝑙, and ord_𝑡𝑘^𝑙₁=ord_𝑡𝑘^𝑙₂ for any divisor 𝑡 of 𝑛_𝑌.

Proof. Suppose that 𝑋_𝑗={𝑝_𝑖∣𝐺(𝑝^𝑒𝑖_𝑖,𝑘_𝑗,𝑎_𝑗)isType(1)} and that 𝑍_𝑗={𝑝_𝑖∣𝐺(𝑝^𝑒𝑖_𝑖,𝑘_𝑗,𝑎_𝑗)isType(3)}. We have the following decomposition: 𝐺𝑛,𝑘_𝑖,𝑎_𝑖≃𝐺𝑛_𝑋𝑖,𝑘_𝑖,𝑎_𝑖×𝐺𝑛_𝑌𝑖,𝑘_𝑖,𝑎_𝑖×𝐺𝑛_𝑍𝑖,𝑘_𝑖,𝑎_𝑖≃𝐺𝑛_𝑋𝑖,𝑘_𝑖×𝐺𝑛_𝑌𝑖,𝑘_𝑖×𝑛_𝑍𝑖𝑙_𝑖𝑂_𝑙𝑖, (12) where 𝐺(𝑛_𝑋𝑖,𝑘_𝑖,𝑎_𝑖) is a tree attached to a fixed point, 𝐺(𝑛_𝑌𝑖,𝑘_𝑖,𝑎_𝑖) is a union of cycles, and the minimal length of these cycles is 1.
Therefore, 𝐺(𝑛,𝑘₁,𝑎₁)≃𝐺(𝑛,𝑘₂,𝑎₂) if and only if the following two conditions hold.(iii)𝐺(𝑛_𝑋1,𝑘₁)≃𝐺(𝑛_𝑋2,𝑘₂).(iv)𝑛_𝑍1/𝑙₁=𝑛_𝑍2/𝑙₂ and 𝐺(𝑛_𝑌1,𝑘₁)×𝑂_𝑙1≃𝐺(𝑛_𝑌2,𝑘₂)×𝑂_𝑙2.
However, by Remark 5, (i) is equivalent to (iii). It remains to show that (ii) is equivalent to (iv). If (ii) is satisfied, then 𝑍₁=𝑍₂ and by Lemma 10  𝐺(𝑛_𝑌,𝑘^𝑙₁)≃𝐺(𝑛_𝑌,𝑘^𝑙₂). By Lemma 9, 𝐺(𝑛_𝑌,𝑘₁)×𝑂_𝑙≃𝐺(𝑛_𝑌,𝑘₂)×𝑂_𝑙. Conversely, if (iv) holds, we have 𝑙₁=𝑙₂=𝑙 since the minimal length of cycles in 𝐺(𝑛_𝑌𝑖,𝑘₁)×𝑂_𝑙𝑖 is 𝑙_𝑖. So 𝑛_𝑍1=𝑛_𝑍2, 𝑍₁=𝑍₂, 𝑌₁=𝑌₂=𝑌, and 𝐺(𝑛_𝑌,𝑘₁)×𝑂_𝑙≃𝐺(𝑛_𝑌,𝑘₂)×𝑂_𝑙. The rest of proof follows from Lemmas 9 and 10.

Remark 12. Suppose that 𝐺(𝑛,𝑘₁,𝑎₁)≃𝐺(𝑛,𝑘₂,𝑎₂). By Theorem 11, we see that both 𝐺(𝑝^{𝑣𝑝(𝑛)},𝑘₁,𝑎₁) and 𝐺(𝑝^{𝑣𝑝(𝑛)},𝑘₂,𝑎₂) are the same types for any prime 𝑝∣𝑛. Moreover, they are isomorphic if they are Type (1) or (3).

Example 13. Consider 𝐺(40,37,1), 𝐺(40,29,1), and 𝐺(40,29,3). One has 𝐺()()()()()≃𝑂40,37,1≃𝐺5,37,1×𝐺8,37,1≃𝐺5,2,1×𝐺8,5,1₁∪𝑂₄×𝑂₈≃5𝑂₈,𝐺()()()()()≃𝑂40,29,𝑖≃𝐺5,29,𝑖×𝐺8,29,𝑖≃𝐺5,4,𝑖×𝐺8,5,𝑖₁∪2𝑂₂×𝑂₈≃5𝑂₈, (13) where 𝑖=1,3. Hence, 𝐺(40,37,1)≃𝐺(40,29,1), but 𝐺(5,37,1)≃𝐺̸(5,29,1). This shows that even if 𝐺(𝑛,𝑘₁,𝑎₁)≃𝐺(𝑛,𝑘₂,𝑎₂), it may happen that 𝐺(𝑝^{𝑣𝑝(𝑛)},𝑘₁,𝑎₁)≃𝐺̸(𝑝^{𝑣𝑝(𝑛)},𝑘₂,𝑎₂). But we also see that 𝐺(40,29,1)≃𝐺(40,29,3), 𝐺(5,29,1)≃𝐺(5,29,3), and 𝐺(8,29,1)≃𝐺(8,29,3). In fact, we have the following result if 𝑘 is fixed.

Theorem 14. Let 𝑛=∏^𝑟_𝑖=1𝑝^𝑒i_𝑖 be the prime factorization of 𝑛. The following three statements are equivalent for any integers 𝑘, 𝑎₁, and 𝑎₂:(i)𝐺(𝑛,𝑘,𝑎₁)≃𝐺(𝑛,𝑘,𝑎₂);(ii)gcd(𝑛,𝑘−1,𝑎₁)=gcd(𝑛,𝑘−1,𝑎₂);(iii)𝐺(𝑝^𝑒𝑖_𝑖,𝑘,𝑎₁)≃𝐺(𝑝^𝑒𝑖_𝑖,𝑘,𝑎₂) for each 𝑖∈{1,2,…,𝑟}.

Proof. Let 𝑋_𝑗={𝑝_𝑖∣𝐺(𝑝^𝑒𝑖_𝑖,𝑘,𝑎_𝑗)isType(1)}, 𝑌_𝑗={𝑝_𝑖∣𝐺(𝑝^𝑒𝑖_𝑖,𝑘,𝑎_𝑗)isType(2)}, and 𝑍_𝑗={𝑝_𝑖∣𝐺(𝑝^𝑒𝑖_𝑖,𝑘,𝑎_𝑗)isType(3)}.
(i)⇒(ii). By Theorem 11  𝑋₁=𝑋₂=𝑋, 𝑌₁=𝑌₂=𝑌, and 𝑍₁=𝑍₂=𝑍. If 𝑝_𝑖∈𝑋∪𝑌, then gcd(𝑝^𝑒𝑖_𝑖,𝑘−1,𝑎_𝑗)=gcd(𝑝^𝑒𝑖_𝑖,𝑘−1), 𝑗=1,2. If 𝑝_𝑖∈𝑍, Suppose 𝐺(𝑝^𝑒𝑖_𝑖,𝑘,𝑎_𝑗)≃𝑝^{𝑒𝑖−𝑠𝑖}_𝑖𝑂_𝑝^𝑠_𝑖𝑖 for any 𝑝_𝑖∈𝑍, where 𝑝^𝑠𝑖_𝑖=ord_𝑝^𝑒_{𝑖𝑝𝑖𝑝𝑖𝑗}⁾_𝑖^{+𝑣(𝑘−1)−𝑣(𝑎}𝑘. By Lemma 7, 𝑣_𝑝𝑖(𝑎₁)=𝑣_𝑝𝑖(𝑎₂). Thus, gcd(𝑝^𝑒𝑖_𝑖,𝑘−1,𝑎₁)=gcd(𝑝^𝑒𝑖_𝑖,𝑘−1,𝑎₂) for any 𝑖. The result follows from that gcd(𝑛,𝑘−1,𝑎_𝑗)=∏^𝑟_𝑖=1gcd(𝑝^𝑒𝑖_𝑖,𝑘_𝑗,𝑎_𝑗), 𝑗=1,2.
(ii)⇒(iii). Assume that gcd(𝑝^𝑒𝑖_𝑖,𝑘−1,𝑎₁)=gcd(𝑝^𝑒𝑖_𝑖,𝑘−1,𝑎₂)=𝑝^𝑢𝑖_𝑖 for each 𝑖 with 𝑢_𝑖≤𝑒_𝑖. If 𝑣_𝑝𝑖(𝑘−1)=𝑢_𝑖, then by Lemma 4  𝐺(𝑝^𝑒𝑖_𝑖,𝑘,𝑎₁)≃𝐺(𝑝^𝑒𝑖_𝑖,𝑘,𝑎₂)≃𝐺(𝑝^𝑒𝑖_𝑖,𝑘). If 𝑣_𝑝𝑖(𝑘−1)>𝑢_𝑖, then 𝑣_𝑝𝑖(𝑎₁)=𝑣_𝑝𝑖(𝑎₂)=𝑢_𝑖. By Lemma 7 again 𝐺(𝑝^𝑒𝑖_𝑖,𝑘,𝑎_𝑗)≃𝑝^{𝑒𝑖−𝑠𝑖}_𝑖𝑂_𝑝^𝑠_𝑖𝑖, where 𝑝^𝑠𝑖_𝑖=ord_𝑝^𝑒_{𝑖𝑝𝑖𝑖𝑖}^{+𝑣(𝑘−1)−𝑢}𝑘.
(iii)⇒(i) follows immediately from (2).

4. The Digraph from Projective Map

Let ℙ¹(𝔽_𝑞)=𝔽²_𝑞⧵{(0,0)}/∼ be the one-dimensional projective space over 𝔽_𝑞, where (𝑎₁,𝑎₂)∼(𝑏₁,𝑏₂) if and only if there exists 𝜆∈𝔽^∗_𝑞 such that 𝜆𝑎_𝑖=𝑏_𝑖. With any invertible 2×2 matrix 𝐴 over 𝔽_𝑞, we associate the Möbius transformation of ℙ¹(𝔽_𝑞) as 𝑧↦𝑧𝐴. Let 𝑓(𝑥) be a polynomial over 𝔽_𝑞 with 𝑓(0)≠0. Recall that 𝑜(𝑓), the order of 𝑓(𝑥), is the minimal positive integer 𝑑 such that 𝑓(𝑥)∣𝑥^𝑑−1. We classify the digraph from Möbius transformation of ℙ¹(𝔽_𝑞) as follows.

Theorem 15. Let 𝐺=𝐺(ℙ¹(𝔽_𝑞),𝐴). Suppose that 𝑓(𝑥) and 𝑚(𝑥) are the characteristic polynomial and minimal polynomial of 𝐴, respectively. Then(i)if 𝑓(𝑥)=(𝑥−𝜆)² and 𝑚(𝑥)=𝑥−𝜆, then 𝐺≃(𝑞+1)𝑂₁;(ii)if 𝑓(𝑥)=𝑚(𝑥)=(𝑥−𝜆)², then 𝐺≃𝑂₁∪(𝑞/𝑝)𝑂_𝑝;(iii)if 𝑓(𝑥)=(𝑥−𝜆₁)(𝑥−𝜆₂) and 𝜆₁≠𝜆₂, then 𝐺≃2𝑂₁∪((𝑞−1)/𝑠)𝑂_𝑠, where 𝑠 is the multiplicative order of 𝜆₁𝜆₂⁻¹;(iv)if 𝑓(𝑥) is irreducible over 𝔽_𝑞[𝑥], then 𝐺≃((𝑞+1)/𝑠)𝑂_𝑠, where 𝑠=𝑜(𝑓)/gcd(𝑜(𝑓),𝑞−1).

Proof. (i)–(iii). We see that 𝐴 is similar to 𝜆𝐼₂, 𝜆10𝜆 and 𝜆₁00𝜆₂ in cases (i), (ii), and (iii), respectively. But the digraphs are isomorphic from similar matrices. The results follows from a simple calculation.
It remains to prove (iv). Let 𝕂=𝔽_𝑞[𝑥]/(𝑓(𝑥)). We can consider 𝐺 as a subdigraph of 𝐺₁=𝐺(𝕂,𝐴) under the natural inclusion 𝔽_𝑞⊆𝕂. Then 𝑓(𝑥)=(𝑥−𝜆)(𝑥−𝜆^𝑞) with some 𝜆 in 𝕂. By (iii), 𝐺₁≃2𝑂₁∪((𝑞²−1)/𝑡)𝑂_𝑡, where 𝑡 is the multiplicative order of 𝜆^𝑞−1 in 𝕂. By a direct computation, we see that 𝑧𝐴∈𝔽_𝑞 for 𝑧∈𝕂 if and only if 𝑧∈𝔽_𝑞. Hence, 𝐺 is a union of cycles of 𝐺₁. However, 𝐺 has no fixed point since 𝑓(𝑥) is irreducible. So the length of each cycle in 𝐺 is 𝑡, and 𝐺≃((𝑞+1)/𝑡)𝑂_𝑡. Note that 𝑡=min{𝑛>0∣𝜆^{𝑛(𝑞−1)}=1}, and that 𝑜(𝑓)=min{𝑛>0∣𝜆^𝑛=1}. Therefore, 𝑡=𝑜(𝑓)/gcd(𝑜(𝑓),𝑞−1). This finishes the proof.

Corollary 16. 𝐺(ℙ¹(𝔽_𝑞),𝐴₁)≃𝐺(ℙ¹(𝔽_𝑞),𝐴₂) if and only if they have the same number of fixed points.