structure of a munn semigroup of finite rank every stable order of

Ukrainian Mathematical Journal, Vol. 61, No. 1, 2009

STRUCTURE OF A MUNN SEMIGROUP OF FINITE RANK EVERY STABLE ORDER OF WHICH IS FUNDAMENTAL OR ANTIFUNDAMENTAL V. D. Derech

UDC 512.534.5

We describe the structure of a Munn semigroup of finite rank every stable order of which is fundamental or antifundamental.

The Munn semigroup (see [1]), i.e., the inverse semigroup of all isomorphisms between principal ideals of a semilattice with respect to the ordinary operation of composition of binary relations, plays a fundamental role in the theory of representations of inverse semigroups (see [2, p. 170]). This explains the importance of comprehensive investigation of these semigroups and their classification. In the present paper, we investigate the structure of a Munn semigroup of finite rank every order of which is fundamental or antifundamental (for definitions, see Sec. 1). The main result of this paper is Theorem 1. 1. Main Definitions and Terminology A semilattice E is called a semilattice of finite length if there exists a natural number n such that the length of any chain from E does not exceed n. Let E be a semilattice of finite length. It is obvious that it contains the least element, which we denote by 0. Let TE denote the Munn semigroup, i.e., the inverse semigroup of all isomorphisms between the principal ideals of the semilattice E with respect to the ordinary operation of composition of binary relations. It is ⎛ 0⎞ clear that the transformation ⎜ ⎟ is the least element of the semigroup TE . Denote the domain of definition ⎝ 0⎠ and the range of values of the transformation f ∈ TE by dom ( f ) and im ( f ) , respectively. Let S be an arbitrary semigroup and let N0 be the set of all nonnegative integers. A function rank : S → N0 is called a rank function on the semigroup S if rank (ab) ≤ min (rank (a) , rank (b)) for any a, b ∈ S . The number rank ( x ) is called the rank of the element x. Let S be an inverse semigroup whose lattice of idempotents has finite length. The function rank (a) = –1 h aa , where h aa –1 is the height of the idempotent aa –1 in the semilattice of idempotents of the semi-

(

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(

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group S, is a rank function (see [3]). We say that an inverse semigroup is a semigroup of finite rank if the semilattice of its idempotents has finite length. Let f ∈ TE . If dom ( f ) = aE, then, by definition, rank ( f ) = rank (a) , where rank (a) is the height of the element a in the semilattice E. In [4], it was proved that rank : TE → N0 is a rank function. An order relation τ on an arbitrary semigroup S is called fundamental (see [5] or [6, p. 289]) if the ordered semigroup (S; τ) is O-isomorphic to a certain semigroup of partial transformations of a set ordered by inVinnytsya National Technical University, Vinnytsya, Ukraine. Translated from Ukrains’kyi Matematychnyi Zhurnal, Vol. 61, No. 1, pp. 52–60, January, 2009. Original article submitted January 29, 2008. 0041–5995/09/6101–0057

© 2009

Springer Science+Business Media, Inc.

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clusion. If τ is a fundamental order relation on a semigroup S, then the order relation τ –1 is called antifundamental. For a survey of results concerning fundamental orders on inverse semigroups, see [6]. Let P be an ordered set with least element 0. Let ≺ denote the covering relation. If 0 ≺ a , then the element a is called an atom of the ordered set P. If E is a nontrivial semilattice of finite length, then it obviously contains atoms. One says that an element b ∈ E is a union of atoms if there exists a subset C of the set of atoms such that sup C = b. An ideal I of a semigroup S is called dense if any homomorphism of the semigroup S injective on the ideal I is also injective on the entire semigroup S. For the other necessary definitions from the theory of semigroups, see [7]. 2. Main Result It is clear that the structure of the Munn semigroup TE is completely determined by the structure of the semilattice E. For this reason, we formulate the main result of the present paper in terms of the semilattice E. First, we consider the simplest case, namely, we assume that rank (b) ≤ 1 for any element b ∈ E . If rank (b) = 0 for any b ∈ E , then the semilattice E is trivial. Hence, the Munn semigroup TE is also trivial. If the semilattice E contains an element of rank 1, then the semigroup TE is obviously a Brandt semigroup, and, furthermore, its structural group is one-element. The stable orders of a Brandt semigroup with trivial structural group can easily be described (see the lemma below). This result can be regarded as a part of mathematical folklore. Lemma 1. If B is a Brandt semigroup whose structural group is trivial, then the stable orders of the semigroup B are exhausted by the following ones: Δ, {0} × B ∪ Δ, and B × {0} ∪ Δ , where 0 is the zero of the Brandt semigroup and Δ is the identity transformation on it. Proof. The statement of the lemma can easily be verified. The situation becomes much more complicated if the semilattice E contains at least one element x for which rank ( x ) ≥ 2. In what follows, we consider semilattices of exactly this type. We now formulate the main result of the present paper. Theorem 1. Let E be a semilattice of finite length and let TE be the corresponding Munn semigroup. The following conditions are equivalent: (i) any stable order on the semigroup TE is fundamental or antifundamental; (ii) every nonzero element of the semilattice E is a union of atoms; (iii) the ideal I1 = { f ∈ TE rank ( f ) ≤ 1} is dense in the semigroup TE . To prove the theorem, we need several lemmas. Lemma 2. Suppose that a semilattice P contains elements a, b, and c such that a ≺ c , b ≺ c , and a ≠ b . Then c = sup {a, b} .

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Proof. Let an element m ∈ P be such that a ≤ m and b ≤ m . Then am = a and bm = b ; moreover, by assumption, we have ac = a and bc = b. Hence, acm = am = a. In other words, we have a ≤ cm. Let a = cm. Since bc = b, we have bcm = bm = b or ba = b, i.e., b ≤ a . 1. If a = b, then this contradicts the assumption. 2. If b < a , then b < a < c, which contradicts the assumption. Thus, a < cm but cm ≤ c. Since a ≺ c , we have cm = c. The last equality means that c ≤ m . Thus, c = sup {a, b} . Lemma 2 is proved. Lemma 3. If a semilattice P of finite length contains a nonzero element that is not a union of atoms, then there exists an element c ∈ P that covers exactly one nonzero element. Proof. Let K denote the set of nonzero elements that cannot be represented as a union of atoms. In the set K, we choose the element (denote it by c) with the least rank. It is obvious that rank (c) ≥ 2. Let us show that the element c covers exactly one nonzero element. Assume the contrary, i.e., assume that the element c covers two different elements (say, x and y). Since rank ( x ) < rank (c) and rank ( y) < rank (c) , each of the elements x and y is a union of atoms. By virtue of Lemma 2, we have c = sup {x, y} = x ∨ y . Since x = sup A1 and y = sup A2 , where A1 ⊆ A and A2 ⊆ A (here, A denotes the set of all atoms of the semilattice P ), we have c = x ∨ y = sup A1 ∨ sup A2 = sup ( A1 ∪ A2 ) . Therefore, the element c is a union of atoms, a contradiction. Lemma 3 is proved. Lemma 4. Let E be a semilattice of finite length. If an element c covers exactly one nonzero element b, then A(c) = A(b), where A(c) and A(b) are the atoms of the elements c and b, respectively. Proof. Since b ≺ c , we have A(b) ⊆ A(c). Let a ∈ A(c). Then a < c. Consider an arbitrary maximum chain that connects a and c. This chain contains an element m such that m ≠ 0 and m ≺ c . By assumption, the element c covers exactly one nonzero element (namely, the element b). Therefore, m = b. This yields a ≤ b. Hence, A(c) ⊆ A(b). Thus, A(c) = A(b). Lemma 4 is proved. Further, let E be a semilattice of finite length that contains a nonzero element that is not a union of atoms. Then, by virtue of Lemma 3, there exists an element c ( rank (c) ≥ 2) that covers exactly one element b. Let Δ c and Δ b denote the identity transformations of the principal ideals cE and bE, respectively. It is obvious that Δ b ⊂ Δ c . On the Munn semigroup TE , we consider the binary relation ρ = { f ° Δ c ° ϕ, f  Δ b ° ϕ Let ρt denote the transitive closure of the binary relation ρ.

f, ϕ ∈ TE } .

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Lemma 5. If (α, β) ∈ ρt , then β ⊆ α . Proof. Since ρt is the transitive closure of the binary relation ρ, there exist τ1, τ2 , … , τ n ∈ TE such that (α, τ1 ) ∈ ρ, (τ1, τ 2 ) ∈ ρ, … , (τ n – 1, τ n ) ∈ ρ, and (τ n , β) ∈ ρ. Since Δ b ⊂ Δ c , we have f  Δ b  ϕ ⊆ f ° Δ c ° ϕ for any f, ϕ ∈TE . Thus, β ⊆ τ n ⊆ τ n –1 ⊆ … ⊆ τ 2 ⊆ τ1 ⊆ α. Therefore, β ⊆ α . Lemma 5 is proved. We can now prove the implication (i) → (ii). Assume that any stable order on the semigroup TE is fundamental or antifundamental. It is necessary to prove that any nonzero element of the semilattice E is a union of atoms. Assume the contrary, i.e., assume that the semilattice E contains a nonzero element that cannot be represented in the form of a union of atoms. Then, according to Lemma 3, there exists an element c that covers exactly one nonzero element (say, b). In other words, b ≺ c and, furthermore, rank (c) ≥ 2. On TE , we consider the binary relation ⎧⎛ 0⎞ ⎫ Σ = ⎨ ⎜ ⎟ ⎬ × I1 ∪ ρt ∪ Δ, ⎩⎝ 0⎠ ⎭ where ρ = { f ° Δ c ° ϕ, f  Δ b ° ϕ f, ϕ ∈ TE } , I1 = { f ∈ TE rank ( f ) ≤ 1} , and Δ = { ψ, ψ ψ ∈ TE } . It is obvious that the binary relation Σ is reflexive. Furthermore, it is obvious that the binary relation ρ is stable. It is easy to show that the transitive closure of a stable binary relation is also stable. Thus, ρt is a stable binary relation. This implies that Σ is also a stable binary relation. We now show that the binary relation Σ is transitive. Consider all possible cases. Case 1. (α, β) ∈ ρt and (β, γ ) ∈ ρt . Since ρt is a transitive binary relation, we have (α, γ ) ∈ ρt . ⎧⎛ 0⎞ ⎫ Case 2. (α, β) ∈ ρt and (β, γ ) ∈ ⎨⎜ ⎟ ⎬ × I1 . ⎩⎝ 0⎠ ⎭ In this case, we have ⎛ 0⎞ β = ⎜ ⎟. ⎝ 0⎠ Let us show that ⎛ 0⎞ α = ⎜ ⎟. ⎝ 0⎠ Since (α, β) ∈ ρt , there exist τ1, τ2 , … , τ n ∈ TE such that (α, τ1 ) ∈ ρ, (τ1, τ 2 ) ∈ ρ, … , (τ n – 1, τ n ) ∈ ρ, (τ n , β) ∈ ρ. Since (τ n , β) ∈ ρ, there exist f, ϕ ∈TE such that τ n = f ° Δ c ° ϕ and β = f ° Δ b ° ϕ. Let

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im ( f ) = mE and dom (ϕ) = kE. We show that mE ∩ bE ∩ kE = mbkE = {0}. Assume the contrary, i.e., assume that there exists an element z such that z ≠ 0 and z ∈ mE ∩ bE ∩ kE. This implies that there exist nonzero elements x, y ∈ E such that ⎛ x⎞ ⎜⎝ z ⎟⎠ ∈ f ,

⎛ z⎞ ⎜⎝ z ⎟⎠ ∈ Δ b ,

and

⎛ z⎞ ⎜⎝ y ⎟⎠ ∈ϕ .

Thus, ⎛ ⎜⎝

x⎞ ∈ f ° Δ b ° ϕ, y ⎟⎠

and, furthermore, ⎛ x⎞ ⎛ 0⎞ ⎜ ⎟ ≠ ⎜ ⎟, ⎝ y⎠ ⎝ 0⎠ i.e., we arrive at a contradiction. Thus, mE ∩ bE ∩ kE = mbkE = {0}, whence mbk = 0. Since b ≺ c and the element c covers a single element, this implies that mck = 0 and mE ∩ cE ∩ kE = {0}. Thus, ⎛ 0⎞ f ° Δc ° ϕ = ⎜ ⎟ = τn . ⎝ 0⎠ Since ( τ n – 1, τ n ) ∈ ρ

and

⎛ 0⎞ τn = ⎜ ⎟ , ⎝ 0⎠

we can prove as above that ⎛ 0⎞ τ n –1 = ⎜ ⎟ . ⎝ 0⎠ By analogy, we get ⎛ 0⎞ τ n –1 = … = τ2 = τ1 = α = ⎜ ⎟ . ⎝ 0⎠ Thus, in case 2, we have ⎛ 0⎞ α = β = ⎜ ⎟. ⎝ 0⎠ Therefore, (a, γ ) ∈Σ .

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⎧⎛ 0⎞ ⎫ Case 3. (α, β) ∈ ⎨⎜ ⎟ ⎬ × I1 and (β, γ ) ∈ ρt . ⎩⎝ 0⎠ ⎭ In this case, we have ⎛ 0⎞ α = ⎜ ⎟ ⎝ 0⎠

β ∈ I1 .

and

If ⎛ 0⎞ β = ⎜ ⎟, ⎝ 0⎠ then it is easy to show that ⎛ 0⎞ γ = ⎜ ⎟. ⎝ 0⎠ Now let ⎛ 0 a1 ⎞ β = ⎜ , ⎝ 0 a2 ⎟⎠ where a1 and a2 are atoms of the semilattice E. Since (β , γ ) ∈ ρt , there exist μ1 , μ 2 , … , μ n ∈ TE such that (β, μ1 ) ∈ ρ, (μ1, μ 2 ) ∈ ρ, … , (μ n – 1 , μ n ) ∈ ρ, (μ n , γ ) ∈ ρ. Thus, (β, μ1 ) ∈ ρ

and

⎛ 0 a1 ⎞ . β = ⎜ ⎝ 0 a2 ⎟⎠

We show that μ1 = β. Since (β, μ1 ) ∈ ρ, there exist f, ϕ ∈TE such that β = f ° Δ c ° ϕ and μ 1 = f ° Δ b ° ϕ. Since Δ b ⊂ Δ c , we have μ1 ⊆ β . Further, ⎛ 0 a1 ⎞ = f ° Δc ° ϕ . β = ⎜ ⎝ 0 a2 ⎟⎠ Therefore, there exists an atom a ∈ E such that ⎛ 0 a1 ⎞ ⎜⎝ 0 a ⎟⎠ ⊆ f,

⎛ 0 a⎞ ⎜⎝ 0 a ⎟⎠ ⊆ Δ c ,

and

⎛0 a ⎞ ⎜⎝ 0 a ⎟⎠ ⊆ ϕ . 2

⎛ 0 a⎞ is an atom of the semilattice E(TE ) and, furthermore, It is obvious that ⎜ ⎝ 0 a ⎟⎠

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⎛ 0 a⎞ ⎜⎝ 0 a ⎟⎠ ⊆ Δ c . Since Δ b ≺ Δ c , by virtue of Lemma 4 we have ⎛ 0 a⎞ ⎜⎝ 0 a ⎟⎠ ⊆ Δ b . Hence, ⎛ 0 a1 ⎞ ⎜⎝ 0 a ⎟⎠ ⊆ μ1 . 2 Moreover, as noted above, we have μ1 ⊆ β . Therefore, ⎛ 0 a1 ⎞ = β. μ1 = ⎜ ⎝ 0 a2 ⎟⎠ Since (μ1, μ 2 ) ∈ ρ, (μ 2 , μ 3 ) ∈ ρ, … , and (μ n , γ ) ∈ ρ, reasoning by analogy we obtain μ 2 = μ3 = … = μ n = γ = β. Thus, (α, γ ) ∈Σ . Lemma 5 is proved. We now show that the binary relation Σ is antisymmetric. Consider all possible cases. Case A. (α, β) ∈ ρt and (β, α) ∈ ρt . By virtue of Lemma 5, we have β ⊆ α and α ⊆ β , whence α = β. ⎧⎛ 0⎞ ⎫ Case B. (α, β) ∈ ρt and (β, α) ∈ ⎨ ⎜ ⎟ ⎬ × I1 . ⎩⎝ 0⎠ ⎭ Since ⎛ 0⎞ β = ⎜ ⎟, ⎝ 0⎠ it follows from the arguments presented above that ⎛ 0⎞ α = ⎜ ⎟. ⎝ 0⎠

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Therefore, α = β. Thus, the binary relation Σ is a stable order on TE . It is known (see [6, p. 303]) that the stable order Ω on an inverse semigroup is fundamental if and only if Ω ⊆ ω, where ω is a canonical order. However, it is obvious that Σ ⊄ ω and Σ ⊄ ω –1, a contradiction. Further, we prove the implication (ii) → (i). Namely, assume that any nonzero element of the semilattice E is a union of atoms. It is necessary to prove that any stable order on the inverse semigroup TE is fundamental or antifundamental. Prior to the proof, we formulate several necessary results. Result 1 [8, p. 564]. For an ideal I of an inverse semigroup, the following properties are equivalent: (a) I is a left-reductive ideal; (b) I is a right-reductive ideal; (c) I is a reductive ideal. Result 2 [8, p. 564]. For an ideal I of an inverse semigroup S, the following properties are equivalent: (a) I is a dense ideal; (b) I is a ∨-basis ideal; (c) I is a reductive ideal.

(Property (b) means that every element b ∈ S can be represented in the form b = sup A, where A ⊆ I ). Further, let S be an inverse semigroup of finite rank with zero 0 and let b ∈ S be an arbitrary element of the semigroup S. Let R1(b) denote the set {x ∈ S x ≤ b ∧ rank (x) ≤ 1} . Result 3 [9]. Let S be an inverse semigroup of finite rank with zero. A homomorphism F : b 哫 R1(b) is an isomorphism if and only if the ideal I1 = {x ∈ S rank (x) ≤ 1} is dense. Result 4 [9]. Let S be an inverse semigroup of finite rank with zero. If the ideal I1 = { f ∈ TE rank (f) ≤ 1} is dense, then the equivalence R1(b) ⊆ R1(c) ¤ b ≤ c holds for any b, c ∈ S . First, we show that the ideal I1 = { f ∈ TE rank (f) ≤ 1} is dense. To establish this, we prove several lemmas.

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Lemma 6. Let E be a semilattice of finite length every nonzero element of which is a union of atoms. If ϕ ∈TE is such that R1(ϕ) ⊆ E(TE ), then ϕ ∈ E(TE ). Proof. We prove the lemma by induction on rank. If rank (ϕ) = 1, then it is obvious that ϕ ∈ E(TE ). Assume that any element α of rank ≤ k from the semigroup TE such that R1(α) ⊆ E(TE ) is an idempotent. We prove that any element ψ of rank k + 1 such that R1(ψ ) ⊆ E(TE ) is also an idempotent. Let dom ( ψ ) = bE. Since any nonzero element of the semilattice E is a union of atoms and, furthermore, rank (b) ≥ 2, the element b covers at least two different elements (say, x and y ). We choose an element z so that z < b . Then rank (Δ z ) ≤ k and, hence, rank (Δ z  ψ ) ≤ k. Since Δ z  ψ ⊂ ψ, we get R1(Δ z ° ψ) ⊆ R1(ψ ) ⊆ E(TE ). By induction hypothesis, this implies that Δ z  ψ is an idempotent. Thus, zψ = z. We now prove that bψ = b. Assume the contrary, i.e., bψ ≠ b. Let bψ = c. It is obvious that the covering relation remains true in the case of isomorphism. Therefore, since x ≺ b and y ≺ b , we have x ≺ c and y ≺ c . This implies that bc = x and, simultaneously, bc = y, a contradiction. Thus, bψ = b. Consequently, ψ is an idempotent. Lemma 6 is proved. Further, let P be a semilattice of finite length. For an element b ≠ 0, let A(b) denote the set {x ∈ P x ≤ b ∧ rank (x) = 1} . Lemma 7. Let P be a semilattice of finite length such that any nonzero element of it is a union of atoms. Then sup A(b) = b for any element b ≠ 0. Proof. By assumption, any element of the semilattice P is a union of atoms. Therefore, b = sup B, where B is a certain subset of the set of atoms. It is obvious that B ⊆ A(b). Let an element c be such that x ≤ c for any x ∈ A(b). Then c is the upper bound of the set B. Since b = sup B, we have b ≤ c . Thus, b = sup A(b) . Lemma 7 is proved. Lemma 8. Let E be a semilattice of finite length every nonzero element of which is a union of atoms, let TE be the corresponding Munn semigroup, and let f, ϕ ∈TE be such that R1( f ) = R1(ϕ) . Then dom( f ) = dom (ϕ) and im( f ) = im (ϕ). Proof. Let dom( f ) = bS and dom (ϕ) = cS. We show that A(b) = A(c). Assume that a ∈ A(b). Then a ≤ b , whence a ∈ dom( f ) . Let a f = a1 . Since a is an atom, we conclude that a1 is also an atom. Therefore, ⎛0 a⎞ ⎜⎝ 0 a ⎟⎠ ∈ TE 1

and

⎛0 a⎞ ⎜⎝ 0 a ⎟⎠ ∈ R1 ( f ) . 1

Since R1( f ) = R1(ϕ) , we have ⎛0 a⎞ ⎜⎝ 0 a ⎟⎠ ∈ R1 (ϕ ) . 1

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Hence, ⎛0 a⎞ ⎜⎝ 0 a ⎟⎠ ⊆ ϕ. 1 Then a ∈ dom (ϕ), i.e., a ∈ cE and, hence, a ≤ c . Furthermore, rank (a) = 1. Thus, a ∈ A(c). Therefore, A(b) ⊆ A(c). By analogy, we prove that A(c) ⊆ A(b). This yields A(c) = A(b). Then, by virtue of Lemma 7, we have b = c. Thus, bS = cS, i.e., dom( f ) = dom (ϕ). By analogy, we prove that im( f ) = im (ϕ). Lemma 8 is proved. Lemma 9. Let E be a semilattice of finite length every nonzero element of which is a union of atoms and let TE be the corresponding Munn semigroup. If f, ϕ ∈TE are such that is an idempotent.

R1( f ) = R1(ϕ) , then

Proof. Let ⎛ 0 a1 ⎞ –1 ⎜⎝ 0 a ⎟⎠ ⊆ f  ϕ , 2 where a1 and a2 are atoms of the semilattice E. Then there exists an atom a ∈ E such that ⎛ a1 ⎞ ⎜⎝ a ⎟⎠ ∈ f

and

⎛ a⎞ –1 ⎜⎝ a ⎟⎠ ∈ ϕ , 2

whence ⎛ a2 ⎞ ⎜⎝ a ⎟⎠ ∈ ϕ. Thus, ⎛ 0 a1 ⎞ ⎜⎝ 0 a ⎟⎠ ∈ R1 ( f )

and

⎛ 0 a2 ⎞ ⎜⎝ 0 a ⎟⎠ ∈ R1 (ϕ ) .

Since R1( f ) = R1(ϕ) , we have ⎛ 0 a1 ⎞ ⎜⎝ 0 a ⎟⎠ ∈ R 1(ϕ) . Therefore, ⎛ 0 a1 ⎞ ⎜⎝ 0 a ⎟⎠ ⊆ ϕ .

f  ϕ –1

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Since ⎛ 0 a2 ⎞ ⎜⎝ 0 a ⎟⎠ ⊆ ϕ, it follows from two last inclusions that a1 = a2 . Thus, R1( f  ϕ –1 ) ⊆ E(TE ). Therefore, by virtue of Lemma 6, f  ϕ –1 is an idempotent. Lemma 9 is proved. Lemma 10. Let S be an arbitrary inverse semigroup. If elements b and c are such that bb –1 = cc –1 , b –1b = c –1c , and bc –1 ∈ E(S), then b = c. Proof. By assumption, we have bc –1 bc –1 = bc –1 and, therefore, bc –1 bc –1c = bc –1c, which implies that bc –1 bb –1 b = bb –1 b, or bc –1b = b. It follows from the last equality that bc –1 bb –1 = bb –1 . Using the equality bb –1 = cc –1 , we get bc –1 cc –1 = cc –1 and, hence, bc –1 = cc –1 . Therefore, bc –1c = cc –1 c = c. Furthermore, bc –1c = bb –1 b = b, whence b = c. Lemma 10 is proved. Lemma 11. Let E be a semilattice of finite length any nonzero element of which is a union of atoms and let TE be the corresponding Munn semigroup. If f, ϕ ∈TE are such that R1( f ) = R1(ϕ) , then f = ϕ. According to Lemma 8, we have dom( f ) = dom (ϕ) and im( f ) = im (ϕ). By virtue of Lemma 9, f ° ϕ –1 is an idempotent. Then, using Lemma 10, we get f = ϕ. Lemma 11 is proved. We can now pass to the proof of the implication (ii) → (i). Let E be a semilattice of finite length every nonzero element of which is a union of atoms. Let Ω be a stable order on the semigroup TE different from equality. In view of the result presented in [6, p. 303], we must prove that Ω ⊆ ω or Ω ⊆ ω –1 , where ω is the canonical order on TE . Since the binary relation Ω is different from equality, there exist α, λ ∈TE such that (α, λ) ∈ Ω and α ≠ λ. In [9], it was proved in the general form that a function F : σ 哫 R1(σ) , where σ ∈TE , is a homomorphism from the semigroup TE into the supersemigroup P( I1 ) , i.e., the semigroup of all nonempty subsets of the ideal I1 = { f ∈ TE rank( f ) ≤ 1} with respect to ordinary global multiplication. According to Lemma 11, this homomorphism is injective, and, hence (see Result 3), the ideal I1 is dense. It follows from Result 2 that the ideal I1 is reductive, and, therefore, according to Result 1, it is right- and left-reductive. Since α ≠ λ, there exists β ∈ I1 such that β  α ≠ β  λ . It is obvious that the ideal I1 is a Brandt semigroup with trivial structural group. Therefore, by virtue of Lemma 1, only the following two cases are possible: ⎛ 0⎞ (a) β  α = ⎜ ⎟ ⎝ 0⎠

and rank (β  λ) = 1;

⎛ 0⎞ (b) rank (β  α) = 1 and β  λ = ⎜ ⎟ . ⎝ 0⎠

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For definiteness, assume that case (a) is realized. Then ⎛ ⎛ 0⎞ ⎞ ⎜⎝ ⎜⎝ 0 ⎟⎠ , β  λ ⎟⎠ ∈ Ω .

(1)

Let us prove that, in this case, one has Ω ⊆ ω, where ω is the canonical order on the inverse semigroup TE . Assume that (ϕ, η) ∈ Ω. It is necessary to prove that ϕ ⊆ η, or, in view of Result 4, that R1(ϕ) ⊆ R1(η) . Assume the contrary, i.e., assume that there exists an element ψ ∈ R1(ϕ) , ψ ∉ R1(η) . Then ⎛ 0⎞ ψ ≠ ⎜ ⎟ ⎝ 0⎠ and, hence, rank (ψ ) = 1. Since ψ ∈ R1(ϕ) , we have ψ ⊆ ϕ. Therefore, ψ  ψ –1  ϕ = ψ. Since (ϕ, η) ∈ Ω, we conclude that (ψ ° ψ –1 ° ϕ, ψ ° ψ –1 ° η) ∈ Ω. Consequently, (ψ , ψ ° ψ –1 ° η) ∈ Ω, whence

(ψ, ψ  ψ –1  η  ψ –1  ψ) ∈Ω.

(2)

Consider the element ψ ° ψ –1 ° η ° ψ –1 ° ψ. The following two cases are possible: ⎛ 0⎞ (a) ψ ° ψ –1 ° η ° ψ –1 ° ψ = ⎜ ⎟ ; ⎝ 0⎠ (b) rank (ψ ° ψ –1 ° η ° ψ –1 ° ψ ) = 1. If ⎛ 0⎞ ψ ° ψ –1 ° η ° ψ –1 ° ψ = ⎜ ⎟ , ⎝ 0⎠ then ⎛ ⎜⎝ ψ,

⎛ 0⎞ ⎞ ⎜⎝ 0 ⎟⎠ ⎟⎠ ∈ Ω,

and, furthermore, as noted above, ⎛ 0⎞ ψ ≠ ⎜ ⎟. ⎝ 0⎠ Since the ideal I1 is a 0-simple semigroup, we get I1 ° ψ ° I1 = I1 . Hence, there exist elements τ, ξ ∈TE such that β  λ = τ ° ψ ° ξ. Thus, we have [see relation (1)]

S TRUCTURE OF A MUNN SEMIGROUP OF FINITE RANK

⎛ ⎜⎝ β  λ,

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⎛ 0⎞ ⎞ ⎜⎝ 0 ⎟⎠ ⎟⎠ ∈ Ω

and

⎛ ⎛ 0⎞ ⎞ ⎜⎝ ⎜⎝ 0 ⎟⎠ , β  λ ⎟⎠ ∈ Ω.

This yields ⎛ 0⎞ βλ = ⎜ ⎟, ⎝ 0⎠ i.e., we arrive at a contradiction. Now assume that rank (ψ ° ψ –1 ° η ° ψ –1  ψ ) = 1. In view of relation (2), it follows from Lemma 1 that ψ ° ψ –1 ° η ° ψ –1 ° ψ = ψ. However, ψ ° ψ –1 ° η ° ψ –1 ° ψ ⊆ η, i.e., ψ ⊆ η . Therefore, ψ ∈ R1(η) , a contradiction. Thus, R1(ϕ) ⊆ R1(η) . This implies (see Result 4) that ϕ ⊆ η, i.e., (ϕ, η) ∈ ω. Thus, we have proved the equivalence of conditions (i) and (ii) in Theorem 1. The implication (ii) → (iii) follows directly from Lemma 11 and Result 3. Let us substantiate the implication (iii) → (ii). To this end, we prove one more statement. Lemma 12. Let S be an inverse semigroup of finite length and let idempotents b and c be such that R1(b) = R1(c). Then, for any x that belongs to the ideal I 1 = {z ∈ S rank (z) ≤ 1} , the equality x b = x c is true. Proof. Let x ∈ I1 . Then it is obvious that x –1x b ∈ R1(b). Since R1(b) = R1(c) by assumption, we conclude that x –1x b ∈ R1(c). Then x –1x bc = x –1x b. Hence, x b = x x –1 xb = x x –1 xbc = xbc . By analogy, since x –1x c ∈ R1(c) and R1(b) = R1(c), we have Hence,

x –1x c ∈ R1(b). Then

x c = x x –1 xc = x x –1 xcb = xbc .

(3) x –1x cb = x –1x c.

(4)

It follows from (3) and (4) that x b = x c. Lemma 12 is proved. We now prove the implication (iii) → (ii) by contradiction. Assume that the semilattice has a nonzero element that can be represented as a union of atoms. Then, according to Lemma 3, there exists an element (denoted by c) that covers exactly one nonzero element (say, b) and is such that rank (c) ≥ 2. Consider Δ b and Δ c that belong to E(TE ). Since the semilattice E is isomorphic to the semilattice E(TE ), we conclude that Δ c covers exactly one idempotent, namely Δ b . By virtue of Lemma 4, R1(Δ b ) = R1(Δ c ) . It follows from Lemma 12 that α  Δ b = α  Δ c for any α ∈ I1 = { f ∈ S rank ( f ) ≤ 1} . However, Δ b ≠ Δ c . Therefore, the ideal I1 is not left-reductive, and, hence (see Results 1 and 2), the ideal I1 is not dense, a contradiction. The implication (iii) → (ii) is proved. Further, the semigroup S is called permutable if any two congruences of it commute with respect to the ordinary operation of superposition of binary relations. As above, let E denote a semilattice of finite length (at

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least 2). It was proved in [10, p. 746] that every nonzero ideal of the permutable Munn semigroup TE is dense. Using this result, we obtain the following corollaries of Theorem 1: Corollary 1. Every stable order on a permutable Munn semigroup is fundamental or antifundamental. Corollary 2. Every nonzero idempotent of a permutable Munn semigroup is a union of atoms. REFERENCES 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

W. D. Munn, “Fundamental inverse semigroups,” Quart. J. Math. Oxford, 21, 157–170 (1970). M. Petrich, Inverse Semigroups, Wiley, New York (1984). V. D. Derech, “Congruences of a permutable inverse semigroup of finite rank,” Ukr. Mat. Zh., 57, No. 4, 469–473 (2005). V. D. Derech, “On permutable congruences on antigroups of finite rank,” Ukr. Mat. Zh., 56, No. 3, 346–351 (2004). B. M. Schein, “Representation of ordered semigroups,” Mat. Sb., 65, No. 2, 188–197 (1964). S. M. Goberstein, “Fundamental order relations on inverse semigroups and on their generalizations,” Semigroup Forum, 21, 285–328 (1980). A. H. Clifford and G. B. Preston, The Algebraic Theory of Semigroups [Russian translation], Vols. 1, 2, Mir, Moscow (1972). B. M. Schein, “Completions, translational hulls and ideal extensions of inverse semigroups,” Czech. Math. J., 23, 575–610 (1973). V. D. Derech, “On maximal stable orders on an inverse semigroup of finite rank with zero,” Ukr. Mat. Zh., 60, No. 8, 1035–1041 (2008). V. D. Derech, “Structure of a permutable Munn semigroup of finite rank,” Ukr. Mat. Zh., 58, No. 6, 742–746 (2006).