EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 10, No. 4, 2017, 739-748 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global On Gamma Acts over Gamma Semigroups Hamid Rasouli1,∗, Ali Reza Shabani2 1 Department of Mathematics, Science and Research Branch, Islamic Azad University Tehran, Iran 2 Department of Mathematics, Imam Khomeini Maritime University of Nowshahr Nowshahr, Iran Abstract. For a semigroup S, actions of S on non-empty sets, namely S-acts, are of interest to consider for their applications in many branches of science. The well known generalization of a semigroup is the Γ-semigroup. The notion of a Γ-act over a Γ-semigroup is a generalization of actions over semigroups. In this paper, certain intrinsic and basic properties of Γ-acts including cyclic, indecomposable and free are studied as well. Among other results, it is shown that a Γ-act is free only if |Γ| = 1. 2010 Mathematics Subject Classifications: 20N99, 20M30, 20M05, 08A30 Key Words and Phrases: Γ-semigroup, Γ-act, Γ-congruence 1. Introduction Nobusawa [10] introduced the notion of a Γ-ring, which is more general than a ring. Then Barnes [2] studied Γ-rings in a different way than that of Nobusawa. Motivated by these generalizations of rings, Sen [12] defined the concept of a Γ-semigroup, as a generalization of a semigroup. The investigation on Γ-semigroups was done by certain mathematicians which are parallel to the results in semigroup theory, for example, one may see [11, 13, 14]. Recently on this area some new papers appeared, such as [3–5]. The algebraic structure of a module over a ring has also been generalized to the Γ-module over a Γ-ring in [1]. A useful algebraic structure in a variety of applications like algebraic automata theory, theoretical computer science and information theory is the notion of S- act over a semigroup S which is more general than a module over a ring (see, for example, [8]). A generalization of an S-act to the Γ-act over a Γ-semigroup can be found in [14] in connection with the consideration of radicals of Γ-semigroups. Here we study some properties of Γ-acts originating by the basic properties of S-acts. First, we describe a Γ-act in terms of a Γ-representation of a Γ-semigroup by Γ-transformations of a set. Then ∗Corresponding author. Email addresses: hrasouli@srbiau.ac.ir; hrasouli5@yahoo.com (H. Rasouli), ashabani@srbiau.ac.ir (A.R. Shabani) http://www.ejpam.com 739 c© 2017 EJPAM All rights reserved. H. Rasouli, A.R. Shabani / Eur. J. Pure Appl. Math, 10 (4) (2017), 739-748 740 some particular morphisms in the category of Γ-acts are characterized. Finally, some results concerning cyclic, indecomposable and free Γ-acts are presented. In the sequel we recall the definitions of S-act and Γ-semigroup. For a semigroup S, a non-empty set A together with a mapping µ : S ×A→ A where (s, a) 7→ sa := µ(s, a) is called a (left) S-act if for all s, t ∈ S and a ∈ A, (st)a = s(ta) holds. This is written as SA. For a monoid S with an identity 1, we add the condition 1a = a, for all a ∈ A. The definition of an S-act, in this form, first proposed by Hoehnke in [6, 7], with a different name in connection with the consideration of radicals of semigroups. For more information on this basic concept, see [9]. There are some different definitions for a Γ-semigroup in the literature (see for example [11–14]). Here we consider the one which is introduced in [11] as follows. Let S and Γ be non-empty sets. Then S is said to be a Γ-semigroup if there exists a mapping λ : S × Γ × S → S, writing λ(a, γ, b) as aγb satisfying the identity (aγb)βc = aγ(bβc) for all a, b, c ∈ S and γ, β ∈ Γ. Let S be a Γ-semigroup. An element e of S is said to be a left (right) identity of S if, eγs = s (sγe = s) for all s ∈ S and γ ∈ Γ. A left as well as right identify is an identity of S. A Γ-semigroup with an identity is called a Γ-monoid. A non-empty subset I of S satisfying SΓI ⊆ I is called a left Γ-ideal of S. By a left Γ-congruence on S we mean an equivalence relation ρ on S for which sρs′ implies (tγs)ρ(tγs′) for s, s′, t ∈ S and γ ∈ Γ. Let S and T be two Γ-semigroups with left identities e and e′, respectively. A map f : S → T satisfying f(e) = e′ and f(sγs′) = f(s)γf(s′) for all s, s′ ∈ S, γ ∈ Γ, is called a Γ-semigroup homomorphism. 2. The structure of Γ-S-acts The purpose of this section is to study some basic properties of Γ-S-acts. Let us first give some definitions. Let S be a Γ-semigroup and A be a non-empty set. Recall from [14] that if there exists a mapping λ : S×Γ×A→ A where (s, γ, a) 7→ sγa := λ(s, γ, a) such that (sγt)βa = sγ(tβa) for all a ∈ A, s, t ∈ S and γ, β ∈ Γ, and if S has a left identity e, eγa = a for every a ∈ A and γ ∈ Γ, then A is called a (left) Γ-S-act. If no confusion arises, a Γ-S-act A is simply called a Γ-act and is denoted by ΓA. A non-empty subset A′ of A is said to be a Γ-subact of A if SΓA′ ⊆ A′, that is, sγa′ ∈ A′ for all s ∈ S, a′ ∈ A′ and γ ∈ Γ. Clearly, S itself is a Γ-S-act with its Γ-operation as the Γ-action. Also any left Γ-ideal of S is a Γ-subact of S. Let A be a Γ-S-act. An element θ ∈ A is called a zero element of A if sγθ = θ for every s ∈ S and γ ∈ Γ. Let ΓA, ΓB be two Γ-S-acts. A mapping f : ΓA → ΓB is called a Γ- S-homomorphism, or simply Γ-homomorphism, if f(sγa) = sγf(a) for every s ∈ S, a ∈ A and γ ∈ Γ. If S is a Γ-monoid with identity 1 and A is a Γ-S-act, then for every s, t ∈ S and γ, β ∈ Γ, we have sγt = sβt and sγa = sβa. Indeed, sγt = (sβ1)γt = sβ(1γt) = sβt; and sγa = (sβ1)γa = sβ(1γa) = sβa. Then it is more interesting to consider Γ-S-acts for a Γ-semigroup S with a left identity (not necessarily an identity). Therefore, from now on, S stands for a Γ-semigroup with a left identity e unless otherwise stated. The idea of representing something by some other objects which are better known at H. Rasouli, A.R. Shabani / Eur. J. Pure Appl. Math, 10 (4) (2017), 739-748 741 least in some respects is quite familiar in mathematics. It is well known that every repre- sentation of a ring by endomorphisms of an abelian group gives a module over that ring and vice versa. Analogously, representations of semigroups (monoids) by transformations of sets give rise to the notion of acts over semigroups (monoids) (see [9, Proposition I.4.4]). In the same way, we here describe Γ-acts in terms of Γ-representations of Γ-semigroups. Let A be a non-empty set and A denote the set of all maps ϕ : Γ→ AA, the so called Γ-transformations of A, where AA is the monoid of all transformations of A with the usual composition of mappings as its operation. Now we have Lemma 1. The set A is a Γ-semigroup with a left (not necessarily right) identity under the Γ-operation ϕγϕ′ : Γ → AA defined by ϕγϕ′(γ′) := ϕ(γ)ϕ′(γ′) for all ϕ,ϕ′ ∈ A and γ, γ′ ∈ Γ. Proof. Let ϕ,ϕ′, ϕ′′ ∈ A and γ, γ′. Then (ϕγϕ′)γ′ϕ′′ = ϕγ(ϕ′γ′ϕ′′). Indeed, for any γ′′ ∈ Γ we have [(ϕγϕ′)γ′ϕ′′](γ′′) = (ϕγϕ′)(γ′)ϕ′′(γ′′) = ϕ(γ)ϕ′(γ′)ϕ′′(γ′′) = ϕ(γ)(ϕ′γ′ϕ′′)(γ′′) = [ϕγ(ϕ′γ′ϕ′′)](γ′′), as desired. Also the constant mapping ε : Γ→ AA which maps every element of Γ to idA is a left identity element of A which is not necessarily a right identity. In the following, the notion of representations of a semigroup by transformations of a set is generalized. Definition 1. A Γ-representation of a Γ-semigroup S by Γ-transformations of a non- empty set A is a Γ-semigroup homomorphism Φ : S → A, where A is the Γ-semigroup described in Lemma 1. Proposition 1. Every Γ-representation of a Γ-semigroup S by Γ-transformations of a non-empty set A in A turns A into a Γ-S-act. Conversely, for every Γ-S-act ΓA, there is an associated Γ-representation of S by Γ-transformations of A in A. Proof. Let A be a non-empty set and S be a Γ-semigroup. If Φ : S → A is a Γ- representation, define λ : S × Γ × A → A by sγa = λ(s, γ, a) := Φ(s)(γ)(a), for all s ∈ S, γ ∈ Γ, a ∈ A. Then A is a Γ-S-act. For this, let s, s′ ∈ S, γ, γ′ ∈ Γ, a ∈ A. We have (sγs′)γ′a = Φ(sγs′)(γ′)(a) = (Φ(s)γΦ(s′))(γ′)(a) = (Φ(s)(γ))(Φ(s′)(γ′)(a)) = (Φ(s)(γ))(s′γ′a) = sγ(s′γ′a). Also if e and ε are left identities of S and A, respectively, then eγa = Φ(e)(γ)(a) = ε(γ)(a) = idA(a) = a. For the converse, consider any Γ-S-act ΓA. Define Φ : S → A H. Rasouli, A.R. Shabani / Eur. J. Pure Appl. Math, 10 (4) (2017), 739-748 742 by Φ(s) = ϕs : Γ → AA, where ϕs(γ)(a) := sγa for all s ∈ S, γ ∈ Γ, a ∈ A. It must be shown that Φ is a Γ-semigroup homomorphism. Let s, s′ ∈ S and γ ∈ Γ. Then Φ(sγs′) = Φ(s)γΦ(s′), or equivalently, ϕsγs′ = ϕsγϕs′ . To see this, let β ∈ Γ and a ∈ A. We get ϕsγs′(β)(a) = (sγs′)βa = sγ(s′βa) = sγ(ϕs′(β)(a)) = ϕs(γ)ϕs′(β)(a) = (ϕsγϕs′)(β)(a). Therefore, Φ is a Γ-representation. Remark 1. (i) Every S-act A over a semigroup S can be generalized to a Γ-S-act over the induced Γ-semigroup S. Indeed, first note that a semigroup S can be made into a Γ-semigroup by setting sγt := st for every s, t ∈ S and γ ∈ Γ. Now define a mapping from S × Γ × A to A by sγa := sa for every s ∈ S, γ ∈ Γ and a ∈ A. Then A is a Γ-S-act. Moreover, if A is a Γ-S-act over a Γ-semigroup S and γ is a fixed element of Γ, then S is a semigroup under the operation st := sγt for all s, t ∈ S, and A with the action sa := sγa, for every s ∈ S and a ∈ A, is an S-act. (ii) A Γ-S-act can have more than one zero, for instance, any non-empty set A becomes a Γ-S-act by definition sγa = a for every a ∈ A, s ∈ S and γ ∈ Γ, i.e. all elements of A are zero. If S has a right zero z, i.e. sαz = z for any s ∈ S, α ∈ Γ, then every element zγa for a ∈ A and γ ∈ Γ is a zero element of A. Indeed, for every s ∈ S and α ∈ Γ, sα(zγa) = (sαz)γa = zγa. Note that every Γ-S-act A can be extended to a Γ-S-act with a zero θ by taking the disjoint union A ∪ {θ}. In the following, we give some examples of Γ-S-acts. Example 1. (i) Let S,Γ and M be the sets of all 3 × 2, 2 × 3 and 3 × 3 matrices over Z, respectively. Under the usual matrix products, S is a Γ-semigroup and M is a Γ-S-act but not an S-act. (ii) Let S = {5n + 4 : n ∈ N}, Γ = {5n + 1 : n ∈ N} and A = {5n : n ∈ N}. Under the usual addition of natural numbers, S is a Γ-semigroup and A is a Γ-S-act but not an S-act. (iii) If A is a Γ-S-act, then the power set of A, P (A), is a Γ-S-act under the Γ-action sγX =: {sγx | x ∈ X} for s ∈ S,X ∈ P (A) and γ ∈ Γ. (iv) Let S be a Γ-semigroup. Then the set of all 2 × 2 matrices over S is a Γ-S-act under the Γ-action: s1γ ( s s′ t t′ ) := ( s1γs s1γs ′ s1γt s1γt ′ ) for s1, s, s ′, t, t′ ∈ S and γ ∈ Γ. (v) Let S and T be Γ-semigroups. Clearly, the cartesian product S×T is a Γ-semigroup with the Γ-operation (s1, t1)γ(s2, t2) := (s1γs2, t1γt2) for every s1, s2 ∈ S, t1, t2 ∈ T and H. Rasouli, A.R. Shabani / Eur. J. Pure Appl. Math, 10 (4) (2017), 739-748 743 γ ∈ Γ. Now suppose S and T contain right zero elements z, z′, respectively. Then the set A = { ( s z z′ t ) : s ∈ S, t ∈ T} is a Γ-S × T -act under the Γ-action: (s1, t1)γ ( s2 z z′ t2 ) := ( s1γs2 z z′ t1γt2 ) for s1, s2 ∈ S, t1, t2 ∈ T and γ ∈ Γ. To see this, let s1, s2, s ∈ S, t1, t2, t ∈ T and α, γ ∈ Γ. We have: ((s1, t1)α(s2, t2))γ ( s z z′ t ) = (s1αs2, t1αt2)γ ( s z z′ t ) = ( (s1αs2)γs z z′ (t1αt2)γt ) = ( s1α(s2γs) z z′ t1α(t2γt) ) = (s1, t1)α((s2, t2)γ ( s z z′ t ) ). Hence, A is a Γ-S × U -act. We here generalize some basic properties of S-acts to Γ-S-acts. Since the composition of two Γ-homomorphisms is a Γ-homomorphism, and the iden- tity map on a Γ-act is a Γ-homomorphism, we conclude that all Γ-acts together with all Γ-homomorphisms between them forms a category which is denoted by Γ-S-Act, or sim- ply Γ-Act if no confusion arise. The notions of Γ-monomorphisms, Γ-epimorphisms and Γ-isomorphisms in their categorical forms are defined as monomorphisms, epimorphisms and isomorphisms, respectively, in the category Γ-Act. Here we characterize these no- tions in terms of injective, surjective and bijective Γ-homomorphisms. Let us list some preliminaries. If ΓA is a Γ-act, a ∈ ΓA and γ ∈ Γ, then the map λa,γ : ΓS → ΓA defined by λa,γ(s) = sγa for every s ∈ S is a Γ-homomorphism. To see this, for every t ∈ S and β ∈ Γ we have λa,γ(tβs) = (tβs)γa = tβ(sγa) = tβλa,γ(s). Let ΓA be a Γ-S-act. An equivalence relation ρ on A is called a Γ-S-congruence, or simply a Γ-congruence, on ΓA if aρa′ implies (sγa)ρ(sγa′) for every a, a′ ∈ ΓA, s ∈ S and γ ∈ Γ. The set ΓA ρ = {[a]ρ : a ∈ ΓA} with the Γ-action sγ[a]ρ = [sγa]ρ for every s ∈ S and γ ∈ Γ is called the factor Γ-act of ΓA by ρ, and the canonical surjection πρ : ΓA → ΓA ρ where a 7→ [a]ρ is called the canonical Γ-epimorphism. Also for a Γ- homomorphism f : ΓA→ ΓB, the Γ-congruence ρ = kerf on ΓA where aρa′ if and only if f(a) = f(a′), for all a, a′ ∈ A, is called the kernel Γ-congruence of f . For each Γ-subact ΓB of ΓA, the Rees Γ-congruence ρB on A is given as aρBa ′ if and only if a = a′ or a, a′ ∈ B, for any a, a′ ∈ A. The resulting factor Γ-act ΓA ρB is simply denoted by ΓA ΓB . H. Rasouli, A.R. Shabani / Eur. J. Pure Appl. Math, 10 (4) (2017), 739-748 744 Proposition 2. In the category Γ-Act, Γ-monomorphisms, Γ-epimorphisms and Γ-iso- morphisms are exactly injective, surjective and bijective Γ-homomorphisms, respectively. Proof. It is easy to see that every injective Γ-homomorphism is a Γ-monomorphism, every surjective Γ-homomorphism is a Γ-epimorphism, and every Γ-isomorphism is bijec- tive. Take any Γ-homomorphism f : ΓA → ΓB. Suppose f is a Γ-monomorphism and f(a) = f(a′) for any a, a′ ∈ A. We show that a = a′. Let γ ∈ Γ. Consider the Γ- homomorphisms λa,γ , λa′,γ : ΓS → ΓA. We claim that fλa,γ = fλa′,γ . For every s ∈ S, fλa,γ(s) = f(λa,γ(s)) = f(sγa) = sγf(a) = sγf(a′) = f(sγa′) = f(λa′,γ(s)) = fλa′,γ(s). Since f is a Γ-monomorphism, λa,γ = λa′,γ . Hence, a = eγa = λa,γ(e) = λa′,γ(e) = eγa′ = a′, where e is a left identity of S. Then f is injective. Now let f be a Γ-epimorphism. Clearly, Imf is a Γ-subact of B. Consider Γ-homomorphisms g, h : ΓB → ΓB Imf defined by g(b) = Imf and h(b) = [b]Imf for every b ∈ B, respectively. Clearly, gf = hf and then g = h because f is a Γ-epimorphism. It follows that for every b ∈ B, Imf = g(b) = h(b) = [b]Imf whence Imf = B, i.e. f is surjective. Finally, assume that f is bijective. It suffices to show that f−1 is a Γ-homomorphism. Let s ∈ S, γ ∈ Γ, b ∈ B. Then there exists a ∈ A such that f(a) = b and hence f−1(sγb) = f−1(sγf(a)) = f−1(f(sγa)) = sγa = sγf−1(b). This implies that f is a Γ-isomorphism. Remark 2. Let S be a Γ-semigroup, ΓA a Γ-S-act and f : ΓA→ ΓS a Γ-homomorphism. Then A is a Γ-semigroup under the Γ-operation aγa′ := f(a)γa′ for every a, a′ ∈ A and γ ∈ Γ. For this, let a, a′, a′′ ∈ A and α, γ ∈ Γ. Then (aαa′)γa′′ = (f(a)αa′)γa′′ = f(f(a)αa′)γa′′ = (f(a)αf(a′))γa′′ = f(a)α(f(a′)γa′′) = aα(f(a′)γa′′) = aα(a′γa′′). Theorem 2 (Homomorphism Theorem for Γ-Acts). Let f : ΓA→ ΓB be a Γ-homomorphism and ρ be a Γ-congruence on ΓA such that aρa′ implies f(a) = f(a′), i.e. ρ ≤ kerf . Then f ′ : ΓA ρ → ΓB with f ′([a]ρ) := f(a), a ∈ ΓA, is the unique Γ-homomorphism such that f ′πρ = f . If ρ=kerf , then f ′ is injective. Also if f is surjective, then so is f ′. Proof. The mapping f ′ is well-defined, because for every [a]ρ, [a ′]ρ ∈ ΓA ρ , [a]ρ = [a′]ρ ⇔ aρa′ ⇒ f(a) = f(a′)⇒ f ′([a]ρ) = f ′([a′]ρ). For every s ∈ S,γ ∈ Γ and a ∈ A, f ′(sγ[a]ρ) = f ′([sγa]ρ) = f(sγa) = sγf(a) = sγf ′([a]ρ). Hence, f ′ is a Γ-homomorphism. Also for every a ∈ ΓA, (f ′πρ)(a) = f ′(πρ(a)) = f ′([a]ρ) = f(a). Now we show that f ′ is unique. Suppose there exists f ′′ : ΓA ρ → ΓB such that f ′′πρ = f . This implies that f ′′πρ = f ′πρ. Since πρ is an epimorphism, f ′′ = f ′. The remainder is an easy verification. Corollary 1. Let f : ΓA→ ΓB be a Γ-epimorphism. Then ΓA kerf ∼= ΓB. 3. Cyclic, Indecomposable and Free Γ-S-Acts In this section we study the notions of cyclic, free and indecomposable Γ-S-acts and investigate their properties. For each Γ-act, a unique decomposition into indecomposable Γ-subacts is obtained. It is also proved that if a Γ-act is free, then Γ is a singleton. H. Rasouli, A.R. Shabani / Eur. J. Pure Appl. Math, 10 (4) (2017), 739-748 745 Definition 2. A subset U 6= ∅ of a Γ-S-act ΓA is said to be a generating set of ΓA if every element a ∈ A can be presented as a = sγu for some s ∈ S, u ∈ U and γ ∈ Γ. In this case, we write ΓA = 〈U〉 (or SΓU), where SΓU = {sγu : s ∈ S, γ ∈ Γ, u ∈ U}. For simplicity, we use the notations SγU and SΓu for S{γ}U and SΓ{u}, respectively. Also A is finitely generated if it has a finite generating set of elements. We call ΓA a cyclic Γ-S-act if ΓA = 〈a〉 (= SΓa) for some a ∈ ΓA. Not that ΓA = 〈A〉, i.e. ΓA is always a generating set of itself. Lemma 3. Let U be a non-empty subset of a Γ-act ΓA and a ∈ ΓA. Then the following assertions hold: (i) SΓa = Sγa for every γ ∈ Γ. (ii) Sγa = Sβa for every γ, β ∈ Γ. (iii) SΓU = SγU for every γ ∈ Γ. Proof. (i) Let γ ∈ Γ and a ∈ ΓA. Clearly, Sγa ⊆ SΓa. For the reverse inclusion, take any β ∈ Γ and s ∈ S. Then sβa = sβ(eγa) = (sβe)γa ∈ Sγa which implies that SΓa = Sγa. (ii) Let γ, β ∈ Γ. Using (i), we get Sγa = SΓa and Sβa = SΓa. Then Sγa = Sβa. (iii) Let γ ∈ Γ. It follows from (i) that SγU = ⋃ u∈U Sγu = ⋃ u∈U SΓu = SΓU . The above lemma presents a simple characterization for generating subsets of a Γ-act. In particular, one can consider a cyclic Γ-act ΓA = 〈a〉 as Sγa for any γ ∈ Γ. In the following, we characterize cyclic Γ-acts in terms of the factor Γ-acts of ΓS. Theorem 4. If a Γ-act ΓA is cyclic, then there exists a Γ-congruence ρ on ΓS such that ΓA ∼= ΓS ρ . The converse also holds provided S is a Γ-monoid. Proof. Let ΓA = Sγa for some a ∈ ΓA and γ ∈ Γ. Then the Γ-homomorphism λa,γ : ΓS → ΓA is obviously a Γ-epimorphism. Using Corollary 1, we get ΓA ∼= ΓS kerλa,γ . Then setting ρ = kerλa,γ we get the result. Conversely, if ρ is a Γ-congruence on a Γ- monoid ΓS with identity 1, then for every [s]ρ ∈ ΓS ρ and γ ∈ Γ, [s]ρ = [sγ1]ρ = sγ[1]ρ which shows that ΓS ρ = 〈[1]ρ〉. A Γ-act is called simple if it contains no proper Γ-subacts. It is clear that a simple act must be cyclic. Now we give conditions under which cyclic Γ-acts, principal left Γ-ideals and Rees factor Γ-acts of a Γ-monoid by left Γ-ideals are simple. Proposition 3. Let ρ be a left Γ-congruence on a Γ-monoid ΓS. The cyclic Γ-act ΓS ρ is simple if and only if [1]ρ ∩ Sγt 6= ∅ for any t ∈ S and γ ∈ Γ. Proof. For a left Γ-congruence ρ on ΓS, consider the canonical Γ-epimorphism π : ΓS → ΓS ρ . Let ΓS ρ be simple and t ∈ S, γ ∈ Γ. Since π(Sγt) is a Γ-subact of ΓS ρ and ΓS ρ is simple, π(Sγt) = ΓS ρ . Hence, there exists u ∈ Sγt such that π(u) = [1]ρ. Thus u ∈ [1]ρ and then [1]ρ ∩ Sγt 6= ∅. Conversely, let ΓA be a Γ-subact of ΓS ρ . Take any t ∈ π−1(A) and γ ∈ Γ. Using the assumption, there exists s ∈ S such that sγt ∈ [1]ρ. Now [1]ρ = π(sγt) = sγπ(t) ∈ ΓA. This implies that ΓA = ΓS ρ and hence ΓS ρ is simple. H. Rasouli, A.R. Shabani / Eur. J. Pure Appl. Math, 10 (4) (2017), 739-748 746 The following two statements are corollaries of the previous proposition. They can also be obtained straightforward from the definition of a simple Γ-act. Corollary 2. A principal left Γ-ideal Sγz, z ∈ S, γ ∈ Γ is a simple Γ-act if and only if z ∈ Sβtγz for all t ∈ S, β ∈ Γ. Corollary 3. Let I be a left Γ-ideal of S. The Rees factor Γ-act ΓS I is simple if and only if I = S. Definition 3. A Γ-act ΓA is called decomposable if there exist two Γ-subacts ΓB and ΓC of ΓA such that ΓA = ΓB∪ ΓC and ΓB∩ ΓC = ∅. In this case, the disjoint union ΓB ∪̇ ΓC is called a decomposition of ΓA. Otherwise, ΓA is called indecomposable. If we consider Γ-S-acts with unique zero θ, then we have to replace ∅ by {θ} to define decomposable and indecomposable Γ-acts with unique zero. Recall that every S-act has a unique decomposition into indecomposable subacts (see [9, I.5.10]). In the following, an analogous result is obtained for the decomposition of Γ-acts. To this end, first note the following: Proposition 4. Every cyclic Γ-act is indecomposable. Proof. Suppose ΓA = Sγa, γ ∈ Γ, a ∈ A, is cyclic and A = ΓB ∪̇ ΓC for some Γ- subacts ΓB and ΓC of ΓA. Then a = eγa ∈ ΓB, say, and then ΓA = Sγa ⊆ ΓB which is a contradiction. Lemma 5. Let Ai ⊆ ΓA, i ∈ I, be indecomposable Γ-subacts of an Γ-act ΓA such that ∩i∈IAi 6= ∅. Then ∪i∈IAi is an indecomposable Γ-subact of ΓA. Proof. First note that ∪i∈IAi is a Γ-subact of ΓA. Indeed, SΓAi ⊆ Ai for every i ∈ I whence SΓ(∪i∈IAi) = ∪i∈I(SΓAi) ⊆ ∪i∈IAi. Suppose there exists a decomposition ∪i∈IAi = ΓB ∪̇ ΓC. Take x ∈ ∩i∈IAi with x ∈ ΓB, say. Then x ∈ Ai ∩ ΓB for all i ∈ I. Since Ai = Ai ∩ (ΓB ∪̇ ΓC) = (Ai ∩ ΓB) ∪̇ (Ai ∩ ΓC) and Ai is indecomposable, Ai ∩ ΓC = ∅ for every i ∈ I. Thus ∪i∈IAi = ΓB which is a contradiction. Theorem 6. Every Γ-S-act ΓA has a unique decomposition into indecomposable Γ-subacts. Proof. Take x ∈ ΓA. Then Sγx, γ ∈ Γ, is indecomposable by Proposition 4. Using Lemma 5, we get Ux = ⋃ {ΓU ⊆ ΓA : ΓU is indecomposable and x ∈ ΓU} is an indecom- posable Γ-subact of ΓA. For x, y ∈ ΓA, Ux = Uy or Ux ∩ Uy = ∅. Indeed, z ∈ Ux ∩ Uy implies Ux, Uy ⊆ Uz. Thus x ∈ Ux ⊆ Uz, y ∈ Uy ⊆ Uz, i.e. Uz ⊆ Ux ∩ Uy. Therefore, Ux = Uy = Uz. Denote by A′ a representative subset of elements x ∈ ΓA with respect to the equivalence relation ∼ defined by x ∼ y if and only if Ux = Uy. Then ΓA = ⋃ x∈A′ Ux is the unique decomposition of ΓA into indecomposable subacts. Definition 4. A set U of generating elements of a Γ-S-act ΓA is said to be a basis of ΓA if every element a ∈ ΓA can be uniquely presented in the from a = sγu for some s ∈ S, u ∈ U and γ ∈ Γ, i.e. if a = sγu = s′γ′u′ for s, s′ ∈ S, u, u′ ∈ U and γ, γ′ ∈ Γ, then s = s′, u = u′ and γ = γ′. If a Γ-act ΓA has a basis U , then it is called a free Γ-act. REFERENCES 747 Proposition 5. Let f : ΓA→ ΓB be a Γ-homomorphism. (i) If ΓA if finitely generated then so is f(ΓA). (ii) If ΓA = 〈U〉 and g : ΓA → ΓB is a Γ-homomorphism, then f(u) = g(u) for every u ∈ U implies f = g. (iii) If f is a Γ-epimorphism and ΓA = 〈U〉, then ΓB = 〈f(U)〉. (iv) If is a Γ-isomorphism and ΓA is a free Γ-act, then so is ΓB. Proof. It is straightforward. The following result shows that there is no free Γ-act whenever |Γ| > 1. Theorem 7. If ΓA is a free Γ-act, then |Γ| = 1. Proof. Suppose ΓA is a free Γ-act with a basis U . Consider γ, γ′ ∈ Γ, s ∈ S and u ∈ U . Using Lemma 3(ii), sγu ∈ Sγu = Sγ′u and then sγu = s′γ′u′ for some s, s′ ∈ S and u, u′ ∈ U . Since U is a basis, γ = γ′. Remark 3. Let |Γ| = 1. Using Remark 1(i), the category Γ-S-Act is equivalent to the category of all acts over the induced semigroup S (containing a left identity). Therefore, any categorical property of such Γ-acts coincides with the analogous property of their corresponding acts. For a Γ = {γ} and a Γ-semigroup S with a left identity, in view of the constructing free acts over monoids as in [9, Construction I.5.14], a free Γ-S-act with a basis X 6= ∅ is isomorphic to S × Γ ×X with the action sγ(t, γ, x) := (sγt, γ, x) for all s, t ∈ S and x ∈ X. Furthermore, any free Γ-act is invariant under the cardinality of its bases and the universal property of freeness holds for free Γ-acts. Hence, every Γ-act is a factor Γ-act of a free Γ-act (see [9, Theorem I.5.15, Proposition I.5.16]). Acknowledgements The authors thank the referee for carefully reading the paper. References [1] R Ameri and R Sadeghi. Gamma modules. Ratio Mathematica, 20:127–147, 2010. [2] W E Barnes. On the Γ-rings of Nobusawa. Pacific J. Math., 18(3):411–422, 1966. [3] S Chattopadhyay. Right orthodox Γ-semigroup. Southeast Asian Bull. Math., 29:23– 30, 2005. [4] R Chinram and P Siammai. On Green’s relations for Γ-semigroups and redutive Γ-semigroups. Int. J. Algebra, 2:187–195, 2008. 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