EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 3, Article Number 6538 ISSN 1307-5543 – ejpam.com Published by New York Business Global An MPP Monounary Algebra Induced by an Endomorphism of the Direct Product of Two Chains Aveya Charoenpol1, Udom Chotwattakawanit2,∗ 1 Division of Mathematics, Faculty of Engineering, Rajamangala University of Technology Isan Khonkaen Campus, Khon Kaen 40000, Thailand. 2 Department of Mathematics, Faculty of Science, Khon Kaen University, Khon Kaen 40002, Thailand. Abstract. A finite modular lattice A is said to be MPP if there is an endomorphism, called an MPP endomorphism, whose the pre-period is equal to the length ofA. A monounary algebra (A, f) is said to be MPP if f is an MPP endomorphism of a lattice A, called an MPP corresponding lattice to (A, f). In this work, we show all MPP monounary algebras induced by endomorphisms of the direct products of two chains. 2020 Mathematics Subject Classifications: 06C05, 08A60, 08A35 Key Words and Phrases: Pre-period, Monounary algebra, Chain, Endomorphism 1. Introduction One of universal algebras which play important roles to simplify many problems in computer science is a monounary algebra. It is often considered as a special type of automata (see e.g. in [1, 2]). The advantage of monounary algebras is their easy visualization; especially, they can be represented as planar directed graphs. The important theories of unary and monounary algebras are shown in many monographs; for instance, [3]. Moreover, monounary algebras have closed relationships with all algebras via their endomorphisms. A monounary algebra is a set A equipped with a unary operation f : A → A and it is denoted by A = (A, f). Denote f0 is the identity map on A and fn = f ◦ fn−1 for all n ∈ N. A monounary algebra A is said to be connected if for each a, b ∈ A, there exist nonnegative integers n,m such that fn(a) = fm(b). An element a ∈ A is called a cyclic if fn(a) = a for some n ∈ N. The height of an element x ∈ A, denoted by ht(x), is the least non-negative integer i such that f i(x) is a cyclic element. The height of the finite monounary algebra A is defined by ht(A) := max {ht(x) | x ∈ A} . ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i3.6538 Email addresses: aveya.ch@rmuti.ac.th (A. Charoenpol), udomch@kku.ac.th (U. Chotwattakawanit) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) A. Charoenpol, U. Chotwattakawanit / Eur. J. Pure Appl. Math, 18 (3) (2025), 6538 2 of 13 In other words, ht(A) is the least non-negative integer λ(f) satisfying fλ(f)(A) = fλ(f)+1(A) and it is known as the pre-period of f (see [4]). For any algebra A, we can study the monounary algebra (A, f) induced by an endo- morphism f of A. Besides, an endomorphism is studied in any category and it is relevant to solve many problems in algebraic structures, relational structures and graphs; reader may look in [5–9]. If A is finite, one can see that |A| − 1 is an upper bound of λ(f); so, it is interesting to study the least upper bound as follows. The pre-period of algebra A is λ(A) = sup {λ(f) | f is an endomorphism of A} . In [10, 11], the authors focused on a finite lattice and showed that for a finite modular lattice L, its pre-period is less than or equal to the length of L where the length ℓ(L) of L is defined by |C| − 1 for the longest chain C in L. A finite modular lattice A is said to be MPP if there is an endomorphism f (called an MPP endomorphism) whose λ(f) = ℓ(A). A monounary algebra (A, f) is said to be MPP if there is an MPP endomorphism g of a lattice B such that (A, f) is isomorphic to (B, g). Such the lattice B is called an MPP corresponding lattice to (A, f). In the present work, we will show that all monounary algebras induced by an MPP endomorphism of the direct products of two chains (studied in [10]) are isomorphic to the monounary algebras (An, fn) and (Bn, gn) shown in the figures 1, 2 and 3. ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? �� ���� � ����� ��� ��� � ����� r r r r r r r r r r r r r r r r r r r r r rp p p p p p p p pppp ppp ppp ⟲ a1,n a1,n−1 a1,5 a1,4 a1,3 a1,2 a1,1 a1,0 a2,n a2,n−1 a2,5 a2,4 a2,3 a2,2 a3,n a3,n−1 a3,5 a3,4 an 2 ,n an 2 ,n−1 an 2 ,n−2 an 2 +1,n Figure 1: The graph of (An, fn) where n is an even natural number. 2. Basic concepts We denote the top and bottom of a lattice A by 1A and 0A (shortly, 1 and 0), respec- tively. A unary operation f on a lattice A = (A;∨,∧) is said to be an endomorphism of A. Charoenpol, U. Chotwattakawanit / Eur. J. Pure Appl. Math, 18 (3) (2025), 6538 3 of 13 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ��� ��� �� ���� ��� ��� ��� ��� r r r r r r r r r r r r r r r r r r r r r r r r r r rp p p p p p p p p p p pppp ppp ppp ⟲ a1,n a1,n−1 a1,n−2 a1,5 a1,4 a1,3 a1,2 a1,1 a1,0 a2,n a2,n−1 a2,n−2 a2,5 a2,4 a2,3 a2,2 a3,n a3,n−1 a3,n−2 a3,5 a3,4 an−1 2 ,n an−1 2 ,n−1 an−1 2 ,n−2 an−1 2 ,n−3 an+1 2 ,n an+1 2 ,n−1 Figure 2: The graph of (An, fn) where n is an odd natural number. A if f(a ∨ b) = f(a) ∨ f(b) and f(a ∧ b) = f(a) ∧ f(b) for all a, b ∈ A. The results in [11, Corollary 6] imply the following theorem. Theorem 1. [11] Let A be a finite modular lattice and f be an endomorphism of A. Then f is MPP if and only if f satisfies either 0 = fλ(f)(1) ≺ fλ(f)−1(1) ≺ . . . ≺ f(1) ≺ 1 (1) or 0 ≺ f(0) ≺ . . . ≺ fλ(f)−1(0) ≺ fλ(f)(0) = 1. (2) Corollary 1. Let f be an MPP endomorphism of a finite modular lattice A. (i) If f satisfies the condition (1), then f(0) = 0, ht(x) = min {n ∈ N ∪ {0} | fn(x) = 0} for all x ∈ A and ht(1) = ℓ(A). (ii) If f satisfies the condition (2), then f(1) = 1, ht(x) = min {n ∈ N ∪ {0} | fn(x) = 1} for all x ∈ A and ht(0) = ℓ(A). We denote the m-element chain by Cm = {1̄ ≺ 2̄ ≺ . . . ≺ m} for m ∈ N. For conve- nient, let a = 1 and b = m in Cm for all a ≤ 1 and b ≥ m. A. Charoenpol, U. Chotwattakawanit / Eur. J. Pure Appl. Math, 18 (3) (2025), 6538 4 of 13 ? ? ? � � � � ? � � r r r r r r r r r ppp ⟲ b2,1 b1,1 b1,2 b1,3 b1,n−1 b1,n b2,2 b2,3 b2,n Figure 3: The graph of (Bn, gn) for n ≥ 2. Theorem 2. [10] For each m ∈ N, the unary operation φm×2 on Cm × C2 defined by φm×2(i, j) = { (i− 1, 2) if i > 1, (1, 1) if i = 1. is an MPP endomorphism of Cm ×C2 fixing the bottom. By Theorem 1, the operations (seen in [10]) in the following theorem are MPP endo- morphisms. Theorem 3. For each m ∈ N, the operations ζ(m,m−1) : C 2 m → C2 m and ζ(m−1,m) : C 2 m → C2 m defined by ζ(m,m−1)(i, j) = (j, i− 1) and ζ(m−1,m)(i, j) = (j − 1, i) are MPP endomorphisms of C2 m fixing the bottom. Lemma 1. [10] For each m,n ≥ 3, if f is an MPP endomorphism of Cm ×Cn fixing the bottom, then either (i) f2k(m,n) = (m− k, n− k) and f2k+1(m,n) = (m− k, n− (k + 1)) for all 0 ≤ k ≤ min {m− 1, n− 2}, or (ii) f2k(m,n) = (m− k, n− k) and f2k+1(m,n) = (m− (k + 1), n− k) for all 0 ≤ k ≤ min {m− 2, n− 1}. Theorem 4. [10] Let m,n ∈ N. Then Cm ×Cn is MPP if and only if either m ≤ 2, n ≤ 2 or |m− n| ≤ 1. A. Charoenpol, U. Chotwattakawanit / Eur. J. Pure Appl. Math, 18 (3) (2025), 6538 5 of 13 3. All MPP endomorphisms of product of two chains It is well-known that if an algebra A is isomorphic to an algebra B under ϕ, the monoid End(A) of all endomorphisms of A is isomorphic to End(B) under the isomorphism Φ defined by Φ(f) = ϕ◦ f ◦ϕ−1 for all f ∈ End(A). Observe that the pre-period is invariant under Φ. Proposition 1. Let ϕ : A → B be an isomorphism between finite algebras A and B and f ∈ End(A). Then (i) ϕ ◦ f ◦ ϕ−1 ∈ End(B), (ii) λ(f) = λ(ϕ ◦ f ◦ ϕ−1), and (iii) ϕ is an isomorphism from (A, f) to ( B,ϕ ◦ f ◦ ϕ−1 ) . Proof. Let g = ϕ ◦ f ◦ ϕ−1. Since ϕ, f and ϕ−1 are homomorphisms, so is g. Hence, ϕ ◦ f ◦ ϕ−1 ∈ End(B). Moreover, gλ(f)(B) = ϕ ◦ fλ(f) ◦ ϕ−1(B) = ϕ ◦ fλ(f)(A) = ϕ ◦ fλ(f)+1(A) = ϕ ◦ fλ(f)+1 ◦ ϕ−1(B) = gλ(f)+1(B). So, λ(g) ≤ λ(f). Similarly, λ(g) ≥ λ(f). Thus λ(g) = λ(f). Since ϕ ◦ f = ϕ ◦ f ◦ ϕ−1 ◦ ϕ = g ◦ ◦ϕ, ϕ is an isomorphism from (A, f) to ( B,ϕ ◦ f ◦ ϕ−1 ) . Remark 1. Let m,n ∈ N. Then (i) ϕ : Cm ×Cn → (Cm ×Cn) ∂ defined by ϕ(i, j) = (m− i+ 1, n− j + 1) is an isomorphism where (Cm ×Cn) ∂ is the dual of Cm ×Cn. (ii) ψ : Cm ×Cn → Cn ×Cm defined by ψ(x, y) = (y, x) is an isomorphism. A. Charoenpol, U. Chotwattakawanit / Eur. J. Pure Appl. Math, 18 (3) (2025), 6538 6 of 13 For each f ∈ End(Cm ×Cn), we denote f∂ := ϕ ◦ f ◦ ϕ−1 and f⌣ := ψ ◦ f ◦ ψ−1. One can see that for each f ∈ End(Cm × Cn), f ⌣ ∈ End(Cn × Cm) and f fixes the bottom if and only if f∂ fixes the top. By Theorem 4 and Proposition 1, we will focus on MPP endomorphisms of Cm ×C2, Cm ×Cm and Cm ×Cm−1 fixing the bottom for m ∈ N \ {1}. Lemma 2. Let m ≥ 3 and f be an MPP endomorphism of Cm ×C2 fixing the bottom. Then (i) f(m, 2̄) = (m− 1, 2̄); (ii) if f(m− 1, 2̄) = (m− 1, 1̄), then m = 3; (iii) if there is t < m − 1 such that f(i, 2̄) = (i− 1, 2̄) for all i > t and f(t, 2̄) = (t, 1̄), then t = 1 and f(j, 1̄) = (j − 1, 2̄) for all j > t. Proof. (i) Assume that f(m, 2̄) = (m, 1̄). Then f(m, 1̄) = (m− 1, 1̄). Since f(1̄, 2̄) ≤ f(m, 2̄) = (m, 1̄), we get f(1̄, 2̄) = (k, 1̄) for some 1 ≤ k ≤ m. Since (1̄, 1̄) = f(1̄, 1̄) = f(1̄, 2̄) ∧ f(m, 1̄) = (k, 1̄) ∧ (m− 1, 1̄), we get k = 1. So, (m, 1̄) = f(m, 2̄) = f(1̄, 2̄) ∨ f(m, 1̄) = (1̄, 1̄) ∨ (m− 1, 1̄) = (m− 1, 1̄), a contradiction. By Theorem 1, f(m, 2̄) = (m− 1, 2̄). (ii) Suppose that f(m− 1, 2̄) = (m− 1, 1̄). By Theorem 1, f(m− 1, 1̄) = (m− 2, 1̄). Since f(m− 2, 2̄) ≤ f(m− 1, 2̄) = (m− 1, 1̄), we get f(m− 2, 2̄) = (k, 1̄) for some 1 ≤ k ≤ m− 1. Since (m− 1, 1̄) = f(m− 1, 2̄) = f(m− 1, 1̄) ∨ f(m− 2, 2̄) = (m− 2, 1̄) ∨ (k, 1̄), we get k = m− 1. Since f(m− 2, 1̄) = f(m− 1, 1̄) ∧ f(m− 2, 2̄) = (m− 2, 1̄) ∧ (m− 1, 1̄) = (m− 2, 1̄) and (1, 1̄) is the unique fixed point, m− 2 = 1; that is, m = 3. (iii) Suppose that there is t < m − 1 such that f(i, 2̄) = (i− 1, 2̄) for all i > t and f(t, 2̄) = (t, 1̄) and let j > t. Then (j − 1, 2̄) = f(j, 2̄) = f(t, 2̄) ∨ f(j, 1̄) = (t, 1̄) ∨ f(j, 1̄). A. Charoenpol, U. Chotwattakawanit / Eur. J. Pure Appl. Math, 18 (3) (2025), 6538 7 of 13 For j > t + 1, we have j − 1 > t which implies by the property of chain that f(j, 1̄) = (j − 1, 2̄) and f(t+ 1, 1̄) = f(t+ 2, 1̄) ∧ f(t+ 1, 2̄) = (t+ 1, 2̄) ∧ (t, 2̄) = (t, 2̄). Hence, f(t, 1̄) = f(t, 2̄) ∧ f(t+ 1, 1̄) = (t, 1̄) ∧ (t, 2̄) = (t, 1̄). Since (1, 1̄) is the unique fixed point, t = 1. Theorem 5. Let m ∈ N with m ≥ 2. (i) For m ≤ 3, ζ(m,m−1) ⇂Cm×C2 and φm×2 are all MPP endomorphisms of Cm × C2 fixing the bottom. (ii) For m > 3, φm×2 is the unique MPP endomorphism of Cm ×C2 fixing the bottom. (iii) For m ≥ 3, ζ(m,m−1) and ζ(m−1,m) are all MPP endomorphisms of Cm ×Cm fixing the bottom. (iv) For m > 3, ζ(m,m−1) ⇂Cm×Cm−1 is the unique MPP endomorphism of Cm × Cm−1 fixing the bottom. Proof. (i) Let f be an MPP endomorphism of C2 × C2 fixing the bottom. Then f(2̄, 2̄) = (2̄, 1̄) or f(2̄, 2̄) = (1̄, 2̄). Case f(2̄, 2̄) = (2̄, 1̄). Then f(2̄, 1̄) = (1̄, 1̄). Since (2̄, 1̄) = f(2̄, 2̄) = f(2̄, 1̄) ∨ f(1̄, 2̄) = (1̄, 1̄) ∨ f(1̄, 2̄), f(1̄, 2̄) = (2̄, 1̄) which implies that f = ζ(2,1). Case f(2̄, 2̄) = (1̄, 2̄). Similarly, f = ζ(1,2)(= φ2×2). In any cases, we are done for m = 2. Let f be an MPP endomorphism of C3 × C2 fixing the bottom. By Lemma 2 (i), f(3, 2̄) = (2, 2̄); and so, f(3, 1̄) ≤ (2, 2̄). Thus f ⇂C2×C2 is an MPP endomorphism of C2 ×C2 fixing the bottom. Case f ⇂C2×C2= ζ(2,1). Thus f(2̄, 2̄) = (2̄, 1̄) and f(2̄, 1̄) = (1̄, 1̄). Since (1̄, 1̄) = f(2̄, 1̄) = f(2̄, 2̄) ∧ f(3, 1̄) = (2̄, 1̄) ∧ f(3, 1̄) and (2̄, 2̄) = f(3, 2̄) = f(2̄, 2̄) ∨ f(3, 1̄) = (2̄, 1̄) ∨ f(3, 1̄), we get f(3, 1̄) = (1̄, 2̄). So, f = ζ(3,2) ⇂C3×C2 . Case f ⇂C2×C2= ζ(1,2). Thus f(2̄, 2̄) = (1̄, 2̄) and f(2̄, 1̄) = (1̄, 2̄). Since (1̄, 2̄) = f(2̄, 1̄) = f(2̄, 2̄) ∧ f(3, 1̄) = (1̄, 2̄) ∧ f(3, 1̄) and (2̄, 2̄) = f(3, 2̄) = f(2̄, 2̄) ∨ f(3, 1̄) = (1̄, 2̄) ∨ f(3, 1̄), A. Charoenpol, U. Chotwattakawanit / Eur. J. Pure Appl. Math, 18 (3) (2025), 6538 8 of 13 we get f(3, 1̄) = (2̄, 2̄). So, f = φ3×2. (ii) The proof follows directly from Lemma 2 (ii) and (iii). (iii) We will prove by the induction under the cardinality of the chain. By (i), this statement is true for m = 2. Let m ≥ 3 and f be an MPP endomorphism of Cm ×Cm fixing the bottom. By Lemma 1, we may assume that f2k(m,m) = (m− k,m− k) and f2k+1(m,m) = (m− k,m− (k + 1)).......(∗) for all 0 ≤ k ≤ m− 2. Then for each 0 ≤ k ≤ m− 2 and 1 ≤ i ≤ m− k, (m− k,m− (k + 1)) = f(m− k,m− k) = f(m− k,m− (k + 1)) ∨ f(i,m− k) = (m− (k + 1),m− (k + 1)) ∨ f(i,m− k) which implies that f(i,m− k) = (m− k, j) for some 1 ≤ j ≤ m− (k + 1) (3) and f(i,m− (k + 1)) = f(m− k,m− (k + 1)) ∧ f(i,m− k) = (m− (k + 1),m− (k + 1)) ∧ (m− k, j) = (m− (k + 1), j); and for i = m− (k + 1), we get by (∗) that j = m− (k + 2). Thus f(m− (k + 1),m− k) = (m− k,m− (k + 2)). (4) Besides, (m− k,m− (k + 1)) = f(m− k,m− k) = f(m− k, i) ∨ f(m− (k + 1),m− k) = f(m− k, i) ∨ (m− k,m− (k + 2)) implies f(m− k, i) = (j,m− (k + 1)) for some 1 ≤ j ≤ m− k. (5) By equations (3) and (5) (k = 1), Cm−1 × Cm−1 is closed under f . By the induction hypothesis, f ⇂Cm−1×Cm−1 is either ζ(m−1,m−2) or ζ(m−2,m−1). By the condition (∗), f ⇂Cm−1×Cm−1= ζ(m−1,m−2); that is, f(r, s) = (s, r − 1) for all 1 ≤ r, s ≤ m − 1. For each 1 ≤ i ≤ m− 1, (m− 1, i− 1) = f(i,m− 1) = f(i,m) ∧ f(m− 1,m− 1) = f(i,m) ∧ (m− 1,m− 2) implies by the equation (3) that f(i,m) = (m, i− 1). For each 1 ≤ i ≤ m− 1, A. Charoenpol, U. Chotwattakawanit / Eur. J. Pure Appl. Math, 18 (3) (2025), 6538 9 of 13 (i,m− 2) = f(m− 1, i) = f(m, i) ∧ f(m− 1,m− 1) = f(m, i) ∧ (m− 1,m− 2) implies by the equation (5) that f(m, i) = (i,m− 1). Hence, f = ζ(m,m−1). (iv) Let m > 3 and f be an MPP endomorphism of Cm × Cm−1 fixing the bottom. Suppose that f(m,m− 1) = (m,m− 2). By Lemma 1, f2(m−3)(m,m− 1) = (3, 2) and f2(m−3)+1(m,m− 1) = (3, 1). Again by Theorem 1, f2(m−2)(m,m− 1) = (2, 1) and f2(m−3)+1(m,m− 1) = (1, 1). Since (3, 1) = f(3, 2) = f(3, 1) ∨ f(2, 2) = (2, 1) ∨ f(2, 2), we get f(2, 2) = (3, 1). Since (1, 1) = f(2, 1) = f(3, 1) ∧ f(2, 2) = (2, 1) ∧ (3, 1) = (2, 1), we get 2 = 1, a contradiction. By Theorem 1, f(m,m− 1) = (m− 1,m− 1). By Lemma 1, f2k(m,m− 1) = (m− k,m− 1− k) and f2k+1(m,m− 1) = (m− (k + 1),m− 1− k) for all 0 ≤ k ≤ m−2. By the same arguments of proving (iii), we get f = ζ(m,m−1) ⇂Cm×Cm−1 . Example 1. All MPP endomorphisms (fixing the bottom) of C2×C2, C3×C2, C3×C3 and C4 ×C3 are shown in the figure 4, 5, 6 and 7, respectively. � �� � �� �� ?- r rr r � �� � �� � �� ?r rr r ⟲⟲ Figure 4: The MPP endomorphisms of C2 ×C2. � � � �� � � � �� �� �� ����� ?- r rr r r r � � � �� � � � �� � �� ? �� � r rr r r r ⟲⟲ Figure 5: The MPP endomorphisms of C3 ×C2. Corollary 2. Let m ∈ N with m ≥ 2. (i) For m ≤ 3, ζ(m,m−1) ⇂Cm×C2, φm×2, ζ(m,m−1) ⇂ ∂ Cm×C2 and φ∂ m×2 are all MPP endo- morphisms of Cm ×C2. (ii) For m > 3, φm×2 and φ∂ m×2 are all MPP endomorphisms of Cm ×C2. (iii) For m ≥ 3, ζ(m,m−1), ζ(m−1,m), ζ ∂ (m,m−1) and ζ ∂ (m−1,m) are all MPP endomorphisms of Cm ×Cm. A. Charoenpol, U. Chotwattakawanit / Eur. J. Pure Appl. Math, 18 (3) (2025), 6538 10 of 13 � � � �� � � � �� � � � �� �� �� @ @R ����� ? ? - - r rr r r r r r r ⟲ � � � �� � � � �� � � � �� � �� ? ������ ? �� r rr r r r r r r ⟲ ? Figure 6: The MPP endomorphisms of C3 ×C3. � � � � � �� � � � � � �� � � � � � �� �� �� �� ����� ? ? - @ @R �� �� ��� ������) -r r r r r r r r r r r r ⟲ Figure 7: The MPP endomorphism of C4 ×C3. (iv) For m > 3, ζ(m,m−1) ⇂Cm×Cm−1 and ζ(m,m−1) ⇂ ∂ Cm×Cm−1 are all MPP endomorphisms of Cm ×Cm−1. For each n ∈ N, we define monounary algebras (An, fn) and (Bn, gn) by An = {ac,h | 2c− 2 ≤ h ≤ n for some c ∈ N and h ∈ N0} , Bn = {bc,h | c ∈ {1, 2} and h ∈ {1, . . . , n}} , fn(ac,h) =  ac,h−1 if 2c− 2 < h, ac−1,h−1 if 2c− 2 = h, a1,0 if c = 1 and h = 0, and gn(bc,h) =  bc,h−1 if c = 1 and h ̸= 1, bc−1,h−1 if c = 2 and h ̸= 1, b2,1 if h = 1. One can observe that A. Charoenpol, U. Chotwattakawanit / Eur. J. Pure Appl. Math, 18 (3) (2025), 6538 11 of 13 f−1 n ({ac,h}) =  ∅ if h = n, {ac,h+1} if 2c− 1 < h < n, {ac,h+1, ac+1,h+1} if 2c− 1 = h, {a1,0, a1,1} if h = 0 for all ac,h ∈ An and g−1 n ({bc,h}) =  ∅ if either h = n or c = 2 and h ̸= 1, {b1,1, b2,1} if c = 2 and h = 1, {bc,h+1, ac+1,h+1} if c = 1 and h ̸= n for all bc,h ∈ Bn. For each n ∈ N, let ζn = ζ(m+1,m) if n = 2m and let ζn = ζ(m+1,m) ⇂Cm+1×Cm if n = 2m− 1. Theorem 6. All monounary algebras induced by an MPP endomorphism of the direct of two chains are isomorphic to either (An, fn) or (Bn, gn) for some n ∈ N. Proof. By Theorem 4, 5, and Proposition 1, it suffices to show that (An, fn) is iso- morphic to ( C⌈n+2 2 ⌉ × C⌈n+1 2 ⌉, ζn ) and (Bn, gn) is isomorphic to (Cn × C2, φn×2) for all n ∈ N. Let n ∈ N. Firstly, we will show that ϕn : An → C⌈n+2 2 ⌉ × C⌈n+1 2 ⌉ defined by ϕn(ac,h) =  (h2 + 2− c, h2 + 1) if h ∈ E, (h+3 2 , h+3 2 − c) if h ∈ O is an isomorphism. Let ac,h ∈ An. Case 1: h ∈ E. If ac,h = a1,0, then ϕn(fn(a1,0)) = ϕn(a1,0) = (1, 1) = ζn(1, 1) = ζn(ϕn(a1,0)). If 2c− 2 < h, then h 2 + 2− c > 1 and ϕn(fn(ac,h)) = ϕn(ac,h−1) = ( h+ 2 2 , h+ 2 2 − c) = ζn( h 2 + 2− c, h 2 + 1) = ζn(ϕn(ac,h)). If 2c− 2 = h, then ϕn(fn(ac,h)) = ϕn(ac−1,h−1) = ( h+ 2 2 , h+ 2 2 − c+ 1) = (c, 1) = ζn(1, c) = ζn( h 2 + 2− c, h 2 + 1) = ζn(ϕn(ac,h)). A. Charoenpol, U. Chotwattakawanit / Eur. J. Pure Appl. Math, 18 (3) (2025), 6538 12 of 13 Case 2: h ∈ O. Then 2c− 2 < h and h ̸= 0. Hence, h+3 2 > c+ 1 2 > 1 and ϕn(fn(ac,h)) = ϕn(ac,h−1) = ( h− 1 2 + 2− c, h− 1 2 + 1) = ζn( h+ 3 2 , h+ 3 2 − c) = ζn(ϕn(ac,h)). Finally, we will show that ψn : Bn → Cn × C2 defined by ψn(bc,h) = { (h, 2) if c = 1, (h, 1) if c = 2 is an isomorphism. Let bc,h ∈ Bn. Then ψn(bc,1) ∈ { (1, 1), (1, 2) } which implies that ψn(gn(bc,1)) = ψn(b2,1) = (1, 1) = φn×2(ψn(bc,1)) and for h ≥ 2 ψn(gn(b1,h)) = ψn(b1,h−1) = (h− 1, 2) = φn×2(h, 2) = φn×2(ψn(b1,h)) and ψn(gn(b2,h)) = ψn(b1,h−1) = (h− 1, 2) = φn×2(h, 1) = φn×2(ψn(b2,h)). Acknowledgements This work was financially supported by Academic Affairs Promotion Fund, Faculty of Science, Khon Kaen University, Fiscal year 2023(RAAPF). References [1] Miroslav Ciric and Stojan Bogdanovic. Lattices of subautomata and direct sum decompositions of automata. In Algebra Colloquium, volume 6, pages 71–88, 1999. [2] Klaus Denecke and Shelly L Wismath. Universal algebra and applications in theoret- ical computer science. Chapman and Hall/CRC, 2018. [3] Bjarni Jónsson. Topics in universal algebra, volume 250. Springer, 2006. [4] David Zupnik. Cayley functions. In Semigroup Forum, volume 3, pages 349–358. Springer, 1971. [5] Jie Fang and Zhong-Ju Sun. Semilattices with the strong endomorphism kernel prop- erty. Algebra universalis, 70(4):393–401, 2013. [6] Jaroslav Guričan and Miroslav Ploščica. The strong endomorphism kernel property for modular p-algebras and for distributive lattices. Algebra universalis, 75(2):243– 255, 2016. [7] Emı́lia Halušková. Some monounary algebras with ekp. Mathematica Bohemica, 145(4):401–414, 2020. A. Charoenpol, U. Chotwattakawanit / Eur. J. Pure Appl. Math, 18 (3) (2025), 6538 13 of 13 [8] BV Popov and OV Kovaleva. On a characterization of monounary algebras by their endomorphism semigroups. In Semigroup Forum, volume 73, pages 444–456. Springer, 2006. [9] Yeni Susanti and Joerg Koppitz. On endomorphisms of power-semigroups. Asian- European Journal of Mathematics, 10(03):1750058, 2017. [10] Aveya Charoenpol and Udom Chotwattakawanit. The maximum pre-period prop- erty of the direct product of chains. Asian-European Journal of Mathematics, 16(09):2350155, 2023. [11] Aveya Charoenpol and Udom Chotwattakawanit. The pre-period of the glued sum of finite modular lattices. Discussiones Mathematicae: General Algebra & Applications, 43(2):223–231, 2023.