Free Hom-groups, Hom-rings and Semisimple modules
Imed Basdouri, Sami Chouaibi, Abdenacer Makhlouf, Esmael Peyghan
aa r X i v : . [ m a t h . R A ] J a n Free Hom-groups, Hom-rings and Semisimple modules
Imed Basdouri ∗ , Sami Chouaibi † , Abdenacer Makhlouf ‡ , Esmael Peyghan § Faculty of Sciences, Department of Mathematics, Gafsa University, Tunisia. Faculty of Sciences, Department of Mathematics, Sfax University, Tunisia. Universit´e de Haute Alsace, IRIMAS-d´epartement de math´ematiques, Mulhouse, France. Department of Mathematics, Faculty of Science, Arak University, Arak, Iran.
January 12, 2021
Abstract
The purpose of this paper is to introduce and study a Hom-type generalization of rings. We providetheir basic properties and and some key constructions. Furthermore, we consider modules over Hom-ringsand characterize the category of simple modules and simple Hom-rings. In addition, we extend some classicalresults and concepts of groups to Hom-groups. We construct free regular Hom-group using Super-Leaf weightedtrees and discuss Normal Hom-subgroups, abelianization of regular Hom-group, universality of tensor productof Hom-groups and simple Hom-groups.
Keywords: Hom-group, free Hom-group, Hom-ring, Module over α -Hom-ring, tensor product, simple, semisimple. Introduction
Hom-type objects have been under intensive research in the last decade. The notion of Hom-Lie algebra wasintroduced by Hartwig, Larsson, and Silvestrov in [6] as part of a study of deformations of the Witt and theVirasoro algebras [1],[3],[4]. Hom-associative algebras were introduced in [11], where it is shown that the commu-tator bracket of a Hom-associative algebra gives rise to a Hom-Lie algebra and where a classification of Hom-Lieadmissible algebras is established. The first notion of Hom-group appeared first in [2]. Then it was introducedas a non-associative analogue of a group in [10], where the authors first gave a new construction of the universalenveloping algebra that is different from the one in [13]. This new construction leads to a Hom-Hopf algebrastructure on the universal enveloping algebra of a Hom-Lie algebra. One can associate a Hom-group to any Hom-Lie algebra by considering group-like elements in its universal enveloping algebra. Recently, M. Hassanzadehdeveloped representations and a (co)homology theory for Hom-groups in [7]. He also proved Lagrange’s theoremfor finite Hom-groups in [8]. In [9], Hom-Lie groups and relationship between Hom-Lie groups and Hom-Liealgebras were explored. The author studied integrable Hom-Lie algebra describe a Hom-Lie group action on asmooth manifold. They define a Hom-exponential (Hexp) map from the Hom-Lie algebra of a Hom-Lie group tothe Hom-Lie group and discuss the universality of this Hexp map. We also describe a Hom-Lie group action ona smooth manifold.The purpose of this paper is to introduce Hom-rings which are twisted versions of the ordinary rings. Wediscuss some of their properties and provide construction procedures using ordinary rings. Also, we introduce thenotion of modules on Hom-rings and characterize the category of simple modules and simple Hom-rings. Likewise,some developments of Hom-group theory are provide. In particular free regular Hom-groups are constructed andnormal Hom-subgroups explored. The paper is organized as follows: In Section 1, we recall some basic definitionsand results concerning Hom-groups. In Section 2, we provide a construction of free regular Hom-group based onsuper-leaf weighted trees. In Section 3, we introduce Normal Hom-subgroups and discuss their properties. In ∗ Email : [email protected] † Email : [email protected] ‡ Email : [email protected] § Email : [email protected]
We review some basics about Hom-groups from [10], [7] and [8], and provide some new properties about hom-groups. Also, we recall the notion of Hom-group homomorphism and show a Hom-group structure on the setof all Hom-group homomorphisms between two abelian regular Hom-groups. Finally, we discuss some classes ofHom-subgroups.
Definition 1.1.
A Hom-group is a quadruplet ( G, µ, e, α ) consisting of a set G with a distinguished element e of G , an operation µ : G × G → G and a set map α : G → G , such that the following axioms are satisfied:1. The product map µ : G × G → G and the set map α : G → G satisfy the Hom-associativity condition µ ( α ( g ) , µ ( h, k )) = µ ( µ ( g, h ) , α ( z )) . (1.1) For simplicity when there is no confusion we omit the multiplication sign µ.
2. The map α is multiplicative, i.e., α ( µ ( g, h )) = µ ( α ( g ) , α ( h )) .3. The element e is called unit and it satisfies the Hom-unitarity condition µ ( g, e ) = µ ( e, g ) = α ( g ) α ( e ) = e. (1.2)
4. the map g g − is an antimorphism; µ ( g, h ) − = µ ( h − , g − ) .
5. For any g ∈ G there exists a natural number k satisfying the Hom-invertibility condition α k ( µ ( g, g − )) = α k ( µ ( g − , g )) = e. (1.3) The smallest such k is called the invertibility index of g. If α is invertible, condition (1.3) can be simplified to the following condition:For every element g ∈ G , there exists an element g − which µ ( g, g − ) = µ ( g − , g ) = e. A Hom-group, such as α is invertible, is called regular Hom-group ([8],[9]). If moreover the product µ is commu-tative, we say that G is an abelian (commutative) Hom-group. In this case, the product µ will be denoted by + and the element e by . Therefore the multiplicativity axiom will change to α ( g + h ) = α ( g ) + α ( h ) , ∀ g, h ∈ G. Remarks 1.2.
1. From the Hom-unitarity and the Hom-associativity conditions, we can show that the multiplicativity of themap α is intuitive; α ( µ ( g, h )) = µ ( e, µ ( g, h )) = µ ( α ( e ) , µ ( g, h )) = µ ( µ ( e, g ) , α ( h )) = µ ( α ( g ) , α ( h ) .2. Since we have the antimorphism g g − , therefore by the definition, inverse of any element g ∈ G isunique although different elements may have different invertibility indexes.3. The inverse of the element e of a Hom-group is itself because α ( µ ( e, e )) = α ( e ) = e.
4. For any Hom-group ( G, µ, e, α ) , we have α ( g − ) = α ( g ) − and if the invertibility index of g is k , then theinvertibility index of α ( g ) is k − . Example 1.3.
Let ( G, µ, e ) be a group and α : G → G be a group homomorphism. We define a new product µ α : G × G → G given by µ α ( g, h ) = α ( µ ( g, h )) = µ ( α ( g ) , α ( h )) . Then ( G, µ α , e, α ) is a Hom-group denoted by G α . It is said the twist group of G . xample 1.4. Let ( G, µ G , e G , α G ) and ( H, µ H , e H , α H ) be two Hom-groups. Then ( G × H, µ, e, α ) is a Hom-groupsuch as: • the identity element e = ( e G , e H ) , • the map α is defined by α ( g, h ) = ( α G ( g ) , α H ( h )) , ∀ ( g, h ) ∈ G × H, • the multiplication is given by µ (( g , h ) , ( g , h )) = ( µ G ( g , g ) , µ H ( h , h )) , ∀ ( g i , h i ) ∈ G × H, i = 1 , . We have this new example.
Example 1.5.
Let ( Z , +) be the additive group of integers. We have End ( Z ) ≃ Z , i.e., any endomorphism α ∈ End ( Z ) is completely determined by the element a (1) ∈ Z and for every q ∈ Z there exists an endomorphism α ∈ End ( Z ) such that α (1) = q . In other words, the correspondence α α (1) is an isomorphism between End ( Z ) and Z . Then, for every q ∈ Z , we can define a Hom-group denoted by ( Z , + q , , α ) such as + q is definedby n + q m = q ( n + m ) , ∀ n, m ∈ Z and checks the following condition qn + q ( m + q p ) = ( n + q m ) + q qp = q ( n + m + p ) , ∀ n, m, p ∈ Z . (1.4) The Hom-group ( Z , + q , , α ) is called the q -additive group of integers. Lemma 1.6.
Let ( G, · , e, α ) be a Hom-group and g ∈ G . If g · g = α ( g ) and α ( g ) = g , then g = e. Proof.
Let h ∈ G such that h · g = e . Then α ( g ) = e · α ( g ) = ( h · g ) · α ( g ) = α ( h ) · ( g · g ) = α ( h ) · α ( g ) = α ( h · g ) = α ( e ) = e. From α ( g ) = g , we can find out that g = e. Proposition 1.7.
Let ( G, µ, e, α ) be a Hom-group and g be an invertible element of index n . If α i ( g ) h = α i ( g ) k, ∀ i ≥ n , then we have α ( h ) = α ( k ) . Proof.
Using Definition 1.1, we get α ( h ) = eα ( h ) = α i ( g − g ) α ( h ) = ( α i ( g − ) α i ( g ))) α ( h ) = α i +1 ( g − )( α i ( g ) h ) , ∀ i ≥ n. Applying α i ( g ) h = α i ( g ) k , the above equation reduces to the following: α ( h ) = α i +1 ( g − )( α i ( g ) k ) = α i ( g − g ) α ( k ) = eα ( k ) = α ( k ) . Corollary 1.8. If ( G, µ, e, α ) is a regular Hom-group, g is an invertible element and gh = gk , then h = k . Lemma 1.9.
Let ( G, µ, e, α ) be a Hom-group, g be an invertible element of index n and l be an invertible elementof index m . If ( α i ( g ) h )( kα j ( l )) = ( α i ( g ) k )( hα j ( l )) , ∀ i ≥ n, j ≥ m , then α ( hk ) = α ( kh ) . roof. Using Definition 1.1, we get α ( hk ) = α ( h ) α ( k )= (cid:16) α i +2 ( g − ) (cid:16) α i +1 ( g ) α ( h ) (cid:17)(cid:17)(cid:16)(cid:16) α ( k ) α j +1 ( l ) (cid:17) α j +2 ( l − ) (cid:17) = (cid:16) α i +2 ( g − ) (cid:16) α i +1 ( g ) α ( h ) (cid:17)(cid:17) α (cid:16) ( kα j ( l )) α j +1 ( l − ) (cid:17) = α i +3 ( g − ) (cid:16)(cid:16) α i +1 ( g ) α ( h ) (cid:17)(cid:16) ( kα j ( l )) α j +1 ( l − ) (cid:17)(cid:17) = α i +3 ( g − ) (cid:16)(cid:16) ( α i ( g ) h )( kα j ( l )) (cid:17) α j +2 ( l − ) (cid:17) = α i +3 ( g − ) (cid:16)(cid:16) ( α i ( g ) k )( hα j ( l )) (cid:17) α j +2 ( l − ) (cid:17) = α i +3 ( g − ) (cid:16)(cid:16) α i +1 ( g ) α ( k ) (cid:17)(cid:16) ( hα j ( l )) α j +1 ( l − ) (cid:17)(cid:17) = α i +3 ( g − ) (cid:16)(cid:16) α i +1 ( g ) α ( k ) (cid:17) α ( h ) (cid:17) = α i +3 ( g − ) (cid:16) α i +2 ( g ) α ( kh ) (cid:17)(cid:17) = α i +2 ( g − g ) α ( kh )= α ( kh ) . Corollary 1.10.
Let ( G, µ, e, α ) be a regular Hom-group and g , l be invertible elements. If ( gh )( kl ) = ( gk )( hl ) ,then hk = kh . Proposition 1.11.
Let ( G, µ, e, α ) be a Hom-group, g be an invertible element of index n and l be an invertible el-ement of index m . If ( α i ( g − ) α j ( h − ))( α i ( g ) α j ( h )) = e, ∀ i ≥ n, j ≥ m , then α i +5 ( g ) α j +5 ( h ) = α j +5 ( h ) α i +5 ( g ) . Proof.
We have e = α i ( g − g ) α j ( h − h ) . Lemma 1.9 gives us α ( α i ( g ) α j ( h − )) = α ( α j ( h − ) α i ( g )) . By multiplying this last equation on the left by α j ++4 ( h ) and using the hom-associativity of α , we get α i +5 ( g ) = α j +4 ( h ) (cid:16) α i +3 ( g ) α j +3 ( h − ) (cid:17) = (cid:16) α j +3 ( h ) α i +3 ( g ) (cid:17) α j +3 ( h − ) . Multiply this equation on the right by α j ++5 ( h ) , we find α i +5 ( g ) α j +5 ( h ) = α j +5 ( h ) α i +5 ( g ) . Corollary 1.12.
Let ( G, µ, e, α ) be a regular Hom-group and g , l be invertible elements of G . If ( g − h − )( gh ) = e ,then gh = hg . Proposition 1.13.
Let ( G, · , e, α ) be a regular abelian Hom-group. For all g, h, k and l in G , we have ( g · h ) · ( k · l ) = ( g · k ) · ( h · l ) . (1.5) Proof.
For all g, h, k and l in G , we have ( g · h ) · ( k · l ) = α ( g ) · ( h · α − ( k · l )) = α ( g ) · ( α − ( h · k ) · l ) = α ( g ) · ( α − ( k · h ) · l )= α ( g ) · ( k · α − ( h · l )) = ( g · k ) · ( h · l ) . Definition 1.14.
Let ( G, µ G , e G , α G ) and ( H, µ H , e H , α H ) be two Hom-groups. A homomorphism of Hom-groupsis a map f : G → H such that f ( µ G ( g, h )) = µ H ( f ( g ) , f ( h )) , ∀ g, h ∈ G and f ◦ α G = α H ◦ f. It is said to be a weak homomorphism if f ◦ α G = α H ◦ f . emark 1.15. Let f : G → H be a homomorphism of Hom-groups such that f ( e G ) = e H and g be an elementin G . Then, we have f ( α G ( g )) = f ( µ G ( e G , g )) = µ H ( f ( e G ) , f ( g )) = µ H ( e H , f ( g )) = α H ( f ( g )) . So any homomorphism of Hom-groups f : ( G, µ G , e G , α G ) → ( H, µ H , e H , α H ) such that f ( e G ) = e H is a weakhomomorphism. Definition 1.16.
Two Hom-groups ( G, µ G , e G , α G ) and ( H, µ H , e H , α H ) are called isomorphic if there exists abijective homomorphism of Hom-groups f : ( G, µ G , e G , α G ) → ( H, µ H , e H , α H ) . Proposition 1.17.
Let ( G, · , e, α ) be a regular Hom-group and define a map f : G → G by f ( g ) = g · g for each g ∈ G . Then G is an abelian Hom-group if and only if the map f is a Hom-group homomorphism. Proof.
Suppose that G is an abelian Hom-group. Let g, h be arbitrary elements in G . Then using Lemma 1.13,we have f ( g · h ) = ( g · h ) · ( g · h ) = ( g · g ) · ( h · h ) = f ( g ) · f ( h ) . Hence f is a Hom-group homomorphism from G to G . Conversely, suppose that f : G → G by f ( g ) = g · g isa Hom-group homomorphism. Let g, h be arbitrary elements in G . We prove that g · h = h · g . Since f is aHom-group homomorphism, we have f ( g · h ) = f ( g ) · f ( h ) , i.e., ( g · h ) · ( g · h ) = ( g · g ) · ( h · h ) . By Hom-associativityof the multiplication of G , the last equation is equivalent to α ( g ) · ( h · α − ( g · h )) = α ( g ) · ( g · α − ( h · h )) . Multiplying this by α ( g − ) on the left, we obtain α ( h ) · α ( g · h ) = α ( g ) · α ( h · h ) . By Hom-associativity we have α ( h · g ) · α ( h ) = α ( g · h ) · α ( h ) . Multiplying this by α ( h − ) on the right, we find α ( h · g ) = α ( g · h ) . Since α is bijective, we have g · h = h · g. So G is abelian. Proposition 1.18.
Let ( G, + G , e G , α G ) and ( H, + H , e H , α H ) be abelian regular Hom-groups.Then ( Hom ( G, H ) , + , , α ) is an abelian Hom-group, where Hom ( G, H ) is the set of all Hom-group homomor-phisms from G to H , f + g is defined by ( f + g )( x ) = f ( x ) + H g ( x ) , for all x ∈ G, (1.6) and α : Hom ( G, H ) → Hom ( G, H ) is defined by α ( f )( x ) = α H ( f ( x )) , for all x ∈ G. Proof.
According to Lemma 1.13 and from the commutativity of H , we can prove that f + g is again a Hom-grouphomomorphism. The other properties for Hom ( G, H ) to be an abelian Hom-group are consequences of the factthat ( H, + H , e H , α H ) is an abelian Hom-group. Definition 1.19.
Let ( G, µ, e, α ) be a Hom-group. A nonempty subset H of the Hom-group G is a Hom-subgroupof G if ( H, µ, e, α ) is a Hom-group. We use the notation H ≤ G to indicate that H is a Hom-subgroup of G. If H is a Hom-subgroup, we see that e the identity for G is also the identity for H . Consequently the followingtheorem is obvious: Theorem 1.20.
A subset H of the Hom-group ( G, · , e, α ) is a Hom-subgroup of G if and only if1. e ∈ H, h ∈ H ⇒ h − ∈ H, h , h ∈ H ⇒ h · h ∈ H . Remarks 1.21.
1. From the third item of the previous theorem, one notes that if H is a Hom-subgroup of G then α ( h ) = e · h ∈ H for all h ∈ H . Therefore α ( H ) ⊂ H.
2. Let ( G, µ, e, α ) be a Hom-group and H be a subset of G . A Hom-group structure on H is defined by theinclusion map. xample 1.22. Let G be a group and α : G → G be a group homomorphism. If H is a subgroup of G which α ( H ) ⊂ H , then H α is a Hom-subgroup of G α . Example 1.23.
Let ( G, µ G , e G , α G ) and ( H, µ H , e H , α H ) . If f : G → H is a Hom-group homomorphism suchthat f ( e G ) = e H , then the inverse image of each Hom-subgroup of H is a Hom-subgroup of G. Definition 1.24. [7] Let ( G, µ, e, α ) be a regular Hom-group. The center Z ( G ) of the Hom-group ( G, µ, e, α ) isthe set of all g ∈ G where gx = xg for all x ∈ G. Proposition 1.25. [7] ( Z ( G ) , µ, e, α ) is a Hom-subgroup of ( G, µ, e, α ) . Lemma 1.26.
Let ( G, µ, e, α ) be a regular Hom-group. Then, ∀ g ∈ Z ( G ) , we have α k ( g ) ∈ Z ( G ) , ∀ k ∈ N . Proof.
Let g ∈ Z ( G ) .1. For k = 1 , we have α ( g ) x = ( eg ) x = α ( e )( gα − ( x )) = e ( α − ( x ) g ) = ( eα − ( x )) α ( g ) = xα ( g ) . Then α ( g ) ∈ Z ( G ) .2. Suppose that for any k ∈ N we have α k ( g ) ∈ Z ( G ) and show that α k +1 ( g ) belongs to Z ( G ) ; α k +1 ( g ) x = ( eα k ( g )) x = α ( e )( α k ( g ) α − ( x )) = e ( α − ( x ) α k ( g )) = ( eα − ( x )) α k +1 ( g ) = xα k +1 ( g ) . Let (
G, µ, e ) be a group and (
G, µ α , e, α ) the associated twist group of ( G, µ, e ). Consider the set Z α ( G ) = { g ∈ G/α ( gx ) = α ( xg ) , ∀ x ∈ G } . Then, we have the following proposition
Proposition 1.27. If Z ( G ) is the center of the group ( G, µ, e ) , so we have1. Z ( G ) ⊂ α − ( α ( Z ( G ))) ⊂ Z α ( G ) . Z ( G ) = Z α ( G ) if α is injective.3. If α ( Z ( G )) ⊂ Z ( G ) and α is injective, then Z α ( G ) is a Hom-subgroup of ( G, µ α , e, α ) . Proof.
1. Let g ∈ α − ( α ( Z ( G ))) . Then, it exists h ∈ Z ( G ) such as α ( g ) = α ( h ) . Let us show that g is anelement in Z α ( G ) , i.e., α ( gx ) = α ( xg ) , ∀ x ∈ G.α ( gx ) = α ( g ) α ( x ) = α ( h ) α ( x ) = α ( hx ) = α ( xh ) = α ( x ) α ( h ) = α ( x ) α ( g ) = α ( xg ) . This shows that g ∈ Z α ( G ) .
2. Let g ∈ Z α ( G ) . So, α ( gx ) = α ( xg ) for all x ∈ G . Since α is injective, then gx = xg, ∀ x ∈ G . Therefore, Z α ( G ) ⊂ Z ( G ) .
3. According to Example 1.22 and the previous property, we can show that Z α ( G ) is a Hom-subgroup of ( G, µ α , e, α ) . − generator A planar tree is an oriented graph drawn on a plane with only one root. It is called binary when any vertex istrivalent, i.e., one root and two leaves. Usually we draw the root at the bottom of the tree and the leaves aredrawn at the top of it: RootLeaves 6or any natural number n ≥
1, let T n denote the set of planar binary trees with n leaves and one root. The first T n are depicted below. T = (cid:26) (cid:27) , T = , T = , , T = , , , , . An element in T n will be called an n -tree. When necessary we label the leaves of an n -tree by 1 , , , · ·· , n from left to right.Let ψ ∈ T n and ϕ ∈ T m be a pair of trees, the ( n + m )-tree ψ ∨ ϕ , called the grafting of ψ and ϕ , is obtained byjoining the roots of ψ and ϕ to create a new root. For instance, ψ ∨ ϕ = ψ ϕ . Remark 2.1.
Note that grafting is a nonassociative and non-commutative operation. For any tree ϕ ∈ T n , thereare unique integers p and q with p + q = n and trees ϕ ∈ T p and ϕ ∈ T q such that ϕ = ϕ ∨ ϕ . The tree ϕ is called the left subtree (L-subtree) of ϕ and ϕ is the right subtree (R-subtree) of ϕ . So any tree ϕ will berepresented as follows ϕ = L-subtree R-subtree . Definition 2.2.
Let ϕ = ϕ ∨ ϕ be a tree. Two successive leaves of ϕ are called1. left (resp. right) attached leaves (LAL resp. RAL) if they have the same node in the left (resp. right) subtreeof ϕ : LAL z}|{
RAL z}|{
2. left (resp. right) disjoint leaves (LDL resp. RDL) if they don’t have the same node in the left (resp. right)subtree of ϕ : LDL z}|{
RDL z}|{
3. disjoint leaves (DL) if the first leaf is the last leaf of the left subtree of ϕ and the other is the first leaf ofthe right subtree of ϕ : DL z}|{ , DL z}|{ . . .. . . , DL z}|{ . . .. . . , DL z}|{ Definition 2.3.
A bicolored Leaf n -tree is a n -tree such that each leaf is colored with black (b) or white (w). Theset of bicolored Leaf n -tree will be denoted by D n . Examples 2.4. efinition 2.5. A Super-Leaf weighted n -tree is a pair ( ϕ, a ) where: • ϕ is an element of D n , • a is an n − tuplet ( a , a , · · · , a n ) such that, for all ≤ i ≤ n, a i ∈ Z . We call the bicolored Leaf tree ϕ the underlying tree of the Super-Leaf weighted n -tree ( ϕ, a ) . For all ≤ i ≤ n ,the integer a i is said to be the weight of the leaf i . We will indeed barely use the notation ( ϕ, a ) at all, and find more convenient to picture a Super-leaf weighted n -tree ( ϕ, a , a , · · · , a n ) by drawing the tree ϕ and putting the weight a i next to each leaf. For example, here area Super-leaf weighted 2-tree and a Super-leaf weighted 3-tree:2 3 − n -trees. For example: − ∨ − n ≥
1, we let B n denote the set of all Super-Leaf weighted n -trees. Let B denote the union over n ∈ N ofthe sets B n together with an element that we call the unit and denote by . Note that the element is differentfrom the Super-Leaf weighted 1-tree 0 , or 0 , and which checks the following property = 0 0 = 0 0 . We then consider the free vector space T generated by the set B .We define on T two natural linear maps:1. α : T → T sending to and sending a Super-Leaf weighted n -tree ( ϕ, a , a , · · · , a n ) to the Super-Leafweighted n -trees ( ϕ, a + 1 , a + 2 , · · · , a n + 1) .
2. a product ∨ that, for any pair of Super-Leaf weighted trees, is just the grafting of these trees and such thatfor any Super-Leaf weighted = ϕ = ( ϕ, a ) ∨ ϕ = ϕ ∨ = α ( ϕ ) (2.1)For examples1. α (cid:16) (cid:17) = 2 32. α (cid:16) − (cid:17) = − . Remark 2.6.
Since α ( ) = , we can notice that, for all a ∈ Z ; = a a = a a Lemma 2.7.
The map α defined above is a morphism for the grafting of trees, that is α ( ϕ ∨ ψ ) = α ( ϕ ) ∨ α ( ψ ) , for any ϕ, ψ ∈ T . In addition, α is bijective and α − is defined, for all ( ϕ, a , a , · · · , a n ) ∈ T , by α − ( ϕ, a , a , · · · , a n ) = ( ϕ, a − , a − , · · · , a n − . Definition 2.8.
Two leaves of a Super-Leaf weighted n -tree are said inverse type leaves if one is colored by bthen the other is colored by w. We now define a reduction process which allows one to obtain a reduced Super-Leaf weighted Ω-type n -treesfrom an arbitrary Leaf weighted Ω-type n -tree. Definition 2.9.
Let ϕ be a Super-Leaf weighted n -tree. An elementary reduction of ϕ is defined as follows1. If the leaf i and the leaf i + 1 are two left attached and inverse type leaves of the same weight, then anelementary reduction of ϕ is obtained by replacing the leaf i and the leaf i + 1 by .
2. If the leaf i and the leaf i + 1 are two right attached and inverse type leaves of the same weight, then anelementary reduction of ϕ is obtained by replacing the leaf i and the leaf i + 1 by .
3. If the leaf i and the leaf i + 1 are two left disjoint and inverse type leaves of weights respectively b and b + 1 ,then an elementary reduction of ϕ is obtained by replacing the leaf i and the leaf i + 1 by .
4. If the leaf i and the leaf i + 1 are two right disjoint and inverse type leaves of weights respectively b + 1 and b , then an elementary reduction of ϕ is obtained by replacing the leaf i and the leaf i + 1 by .
5. If the leaf i and the leaf i + 1 are two DL1 and inverse type leaves of the same weight, then an elementaryreduction of ϕ is obtained by replacing the leaf i and the leaf i + 1 by .
6. If the leaf i and the leaf i + 1 are two DL2 and inverse type leaves of weights respectively b and b + 1 , thenan elementary reduction of ϕ is obtained by replacing the leaf i and the leaf i + 1 by .
7. If the leaf i and the leaf i + 1 are two DL3 and inverse type leaves of weights respectively b + 1 and b , thenan elementary reduction of ϕ is obtained by replacing the leaf i and the leaf i + 1 by .
8. If the leaf i and the leaf i + 1 are two DL4 and inverse type leaves of of the same weight, then an elementaryreduction of ϕ is obtained by replacing the leaf i and the leaf i + 1 by . Examples 2.10.
1. Suppose ϕ = ( ϕ ∨ a a ) ∨ ϕ for some Super-Leaf weighted trees ϕ , ϕ and a ∈ Z .Then the elementary reduction of ϕ results in the Super-Leaf weighted tree α ( ϕ ) ∨ ϕ .
2. If ϕ = ϕ ∨ ( ϕ ∨ a a ) for some Super-Leaf weighted trees ϕ , ϕ and a ∈ Z . Then the elementaryreduction of ϕ results in the Super-Leaf weighted tree ϕ ∨ α ( ϕ ) .
3. Suppose ϕ = (( ϕ ∨ a ) ∨ a + 1 ) ∨ ϕ for some Super-Leaf weighted trees ϕ , ϕ and a ∈ Z . Then theelementary reduction of ϕ results in the Super-Leaf weighted tree α ( ϕ ) ∨ ϕ .
4. If ϕ = ϕ ∨ ( a + 1 ∨ ( a ∨ ϕ )) for some Super-Leaf weighted trees ϕ , ϕ and a ∈ Z . Then the elementaryreduction of ϕ results in the Super-Leaf weighted tree ϕ ∨ α ( ϕ ) . . Suppose ϕ = ( ϕ ∨ a ) ∨ ( a ∨ ϕ ) for some Super-Leaf weighted trees ϕ , ϕ and a ∈ Z . Then the elementaryreduction of ϕ results in the Super-Leaf weighted tree α ( ϕ ) ∨ α ( ϕ ) .
6. Suppose ϕ = ( ϕ ∨ b a ) ∨ ( a + 1 ∨ ϕ ) for some Super-Leaf weighted trees ϕ , ϕ and a, b ∈ Z . Thenthe elementary reduction of ϕ results in the Super-Leaf weighted tree ( ϕ ∨ b + 1 ) ∨ α ( ϕ ) .
7. If ϕ = ( ϕ ∨ a + 1 ) ∨ ( a b ∨ ϕ ) for some Super-Leaf weighted trees ϕ , ϕ and a, b ∈ Z . Then theelementary reduction of ϕ results in the Super-Leaf weighted tree α ( ϕ ) ∨ ( b + 1 ∨ ϕ ) .8. Suppose ϕ = ( ϕ ∨ b a ) ∨ ( a c ∨ ϕ ) for some Super-Leaf weighted trees ϕ , ϕ and a, b, c ∈ Z . Thenthe elementary reduction of ϕ results in the Super-Leaf weighted tree ( ϕ ∨ b + 1 )( c + 1 ∨ ϕ ) . Definition 2.11.
Let ϕ be a Super-Leaf weighted tree. A reduction of ϕ (or a reduction process starting at ϕ )consists of consequent applications of elementary reductions starting at ϕ and ending at a reduced Super-Leafweighted tree: ϕ → ϕ → ϕ → · · · → ϕ n . The Super-Leaf weighted tree ϕ n is termed the reduced Super-Leaf weighted tree of ϕ and it is denoted ϕ. Example 2.12. c b a a b d by item of −−−−−−−−−−−→ Definition 2.9 c b+1 b+1 d by item of −−−−−−−−−−−→ Definition 2.9 c+1 d+1
In general, there are different possible reductions of a Super-Leaf weighted tree ϕ . Nevertheless, it turns outthat all possible reductions of ϕ end up with the same reduced Super-Leaf weighted tree. To see this we need thefollowing lemma. Lemma 2.13.
For any two elementary reductions ϕ → ϕ and ϕ → ϕ of a Super-Leaf weighted tree ϕ , there exist elementaryreductions ϕ → ϕ and ϕ → ϕ . Proof.
Let ϕ λ −→ ϕ and ϕ λ −→ ϕ be two elementary reductions of a Super-Leaf weighted tree ϕ . There are two possibleways to carry out the reductions λ and λ . Case1: Disjoint reductions
In this case ϕ = ψ ∨ ψ and λ ( ϕ ) = ψ ∨ ψ , λ ( ϕ ) = ψ ∨ ψ . Then ϕ λ −→ ψ ∨ ψ λ −→ ψ ∨ ψ ϕ λ −→ ψ ∨ ψ λ −→ ψ ∨ ψ . Hence the lemma holds. ase2: Overlapping reductions In this case ϕ can take the following form ϕ = ( ψ ∨ a a ) ∨ ( a + 1 ∨ ψ ) . Then, there are two different possible reductions of ϕ : • ϕ λ −→ α ( ψ ) ∨ ( a + 1 ∨ ψ ) , and • ϕ λ −→ ( ψ ∨ a + 1 ) ∨ α ( ψ ) . To have the lemma, we would like to show that λ ( ϕ ) = λ ( ϕ ) : λ ( ϕ ) = ( ψ ∨ a + 1 ) ∨ α ( ψ )= ( ψ ∨ a + 1 ) ∨ ( ∨ ψ )= ( ψ ∨ a + 1 ) ∨ ( a a ∨ ψ )= α ( ψ ) ∨ ( a + 1 ∨ ψ ) . Proposition 2.14.
Let ϕ be a Super-Leaf weighted tree. Then any two reductions of ϕ : ϕ → ϕ ′ → ϕ ′ → · · · → ϕ ′ n ϕ → ϕ ” → ϕ ” → · · · → ϕ ” m result in the same reduced Super-Leaf weighted tree of ϕ, i. e, ϕ ′ n = ϕ ” n . Proof.
By induction on ϕ. If ϕ = or ϕ is a Super-Leaf weighted -tree, then ϕ is reduced and there is nothing to prove.Let ϕ ∈ B r , r ≥ and ϕ → ϕ ′ → ϕ ′ → · · · → ϕ ′ n ϕ → ϕ ” → ϕ ” → · · · → ϕ ” m be two reductions of ϕ. Then by Lemma 2.13, there are elementary reductions ϕ ′ → ϕ and ϕ ” → ϕ . Considera reduction process for ϕ : ϕ → ϕ → ϕ → · · · → ϕ k . By induction all reduced Super-Leaf weighted trees of ϕ ′ are equal to each other, as well as all reduced Super-Leafweighted trees of ϕ ” . Since ϕ k is a reduced Super-Leaf weighted tree of both ϕ ′ and ϕ ” , then ϕ ′ n = ϕ k = ϕ ” m as desired. This proves the proposition. Let G be the set of all reduced Super-Leaf weighted trees.We now intend to define inverse of a Super-Leaf weighted n -trees to hold the structure of Hom-group on G . Wefirst define Mirror inverse type of a Super-Leaf weighted n -tree( n ≥ efinition 2.15. Mirror inverse type of a Super-Leaf weighted n -trees( n ≥ ) ( ϕ, V ) is another Super-Leafweighted n -trees M ( ϕ ) with left and right children of all nodes interchanged and inversing the type of every leaf.We note MRinv the application that sends each Super-Leaf weighted n -trees to its mirror inverse type. Examples 2.16. MRinv (cid:16) (cid:17) = 3 7 MRinv (cid:16) − (cid:17) = 3 1 − MRinv (cid:16) − (cid:17) = 4 2 6 − n -trees by the map ̥ : T → T sending to , a Super-Leafweighted Ω-type 1-tree a to a , a Super-Leaf weighted Ω-type 1-tree a to a and a super-Leaf weighted n -tree( n ≥
2) ( ϕ, a ) to it’s Mirror inverse.According to the definition 2.9, The proof of the following lemma is trivial:
Lemma 2.17.
For all ϕ and ψ in T , we have ̥ ( ϕ ∨ ψ ) = ̥ ( ψ ) ∨ ̥ ( ϕ ) . Theorem 2.18.
Let e α be the bijevtive map defined on G by, for all ϕ ∈ T , e α ( ϕ ) = α ( ϕ ) and e α − ( ϕ ) = α − ( ϕ ) .If we define a multiplication ” · ” on G by ϕ · ψ = ϕ ∨ ψ, ∀ ϕ, ψ ∈ G, we have that ( G, · , , e α ) is a regular Hom-group. G will be said the free regular Hom-group with -generator. Proof.
1. The map e α is multiplicative: e α ( ϕ · ψ ) = e α ( ϕ ∨ ψ ) = α ( ϕ ∨ ψ ) = α ( ϕ ) ∨ α ( ψ ) = α ( ϕ ) · α ( ψ ) = e α ( ϕ ) · e α ( ψ )
2. The multiplication ” · ” is Hom-associative: e α ( ϕ ) · ( ψ · φ ) = ( ϕ · ψ ) · e α ( φ ) , for any ϕ, ψ, φ ∈ G . To show this, it enough to prove that α ( ϕ ) ∨ ( ψ ∨ φ ) = ( ϕ ∨ ψ ) ∨ α ( φ ) , for given ϕ, ψ, φ ∈ T . We will observe, that each of the reduced Super-Leaf weighted trees α ( ϕ ) ∨ ( ψ ∨ φ ) , ( ϕ ∨ ψ ) ∨ α ( φ ) can be obtained from a same Super-Leaf weighted tree by a sequence of elementary reductions.We will show this by induction on ψ . f ψ = , there is nothing to prove.If ψ = a or a , we have α ( ϕ ) ∨ ( ψ ∨ φ ) = α ( ϕ ) ∨ ( a ∨ φ ) = ( ϕ ∨ ) ∨ ( a ∨ φ ) = ( ϕ ∨ a − a − a ∨ φ ) = ( ϕ ∨ a ) ∨ α ( φ ) . If ψ ∈ B n , then ψ can take the following form ψ = ψ ∨ ψ , where ψ ∈ B p and ψ ∈ B q , such that p + q = n. By Proposition 2.14, we have ψ = ψ ∨ ψ = ψ · ψ . Then, e α ( ϕ ) · ( ψ · φ ) = e α ( ϕ ) · (( ψ · ψ ) · φ )= e α ( ϕ ) · ( e α ( ψ ) · ( ψ · e α − ( φ ))= ( ϕ · e α ( ψ )) · ( e α ( ψ ) · φ ) , and ( ϕ · ψ ) · e α ( φ ) = ( ϕ · ( ψ · ( ψ )) · e α ( φ )= (( e α − ( ϕ ) · ψ ) · e α ( ψ )) · e α ( φ )= ( ϕ · e α ( ψ )) · ( e α ( ψ ) · φ ) . Hence by Proposition 2.14, we have e α ( ϕ ) · ( ψ · φ ) = ( ϕ · ψ ) · e α ( φ ) .
3. The inverse for every ϕ ∈ G , is defined by ̥ ( ϕ ) : ϕ · ̥ ( ϕ ) = ϕ ∨ ̥ ( ϕ ) = = . Let X be a nonempty set. We define T X by T X = ⊕ L n ≥ B n ⊗ X . Elements of B n ⊗ X shall be pictured forevery ϕ ∈ B n , x , · · · , x n ∈ X , by inserting, for all i = 1 , · · · , n the element x i at the top of the leaf with label x i : x x x x − x . Let now X − = { x − /x ∈ X } and consider the set X ∗ = X ∪ X − . An element y ∈ X ∗ can be defined as elementof B ⊗ X in the following way:1. If y = x ∈ X, we asoociate to y a Super-Leaf weighted 1-tree; y = x .
2. If y = x − ∈ X − , for x ∈ X , we asoociate to y a Super-Leaf weighted 1-tree; y = x . ∨ and α have natural extensions to T X , that we denote by same symbols.For α , it defines a map α : X ∗ → X ∗ as follows1. If y = x ∈ X, we can define α ( y ) by α ( y ) = x .
2. If y = x − ∈ X − , for x ∈ X , we can define α ( y ) by α ( y ) = x . Moreover, α a ( y ), for all y ∈ X ∗ and a ∈ Z is defined as follows.1. If y = x ∈ X, we can define α ( y ) by α a ( y ) = xa .
2. If y = x − ∈ X − , for x ∈ X , we can define α ( y ) by α a ( y ) = xa . So, the map α sends each element ϕ ⊗ ( x , · · · , x n ) of B n ⊗ X to α ( ϕ ) ⊗ ( x , · · · , x n ) . Also the reduction process and the map
MRinv have natural extensions to T X . They are defined as follows. Definition 2.19.
The reduction of an element of T X is defined by the same process as in Definition 2.9 but fortwo leaves labeled by the same x ∈ X. Example 2.20. x c x b x a x a x b x d by item of −−−−−−−−−−−→ Definition 2.9 x c x b + 1 x b + 1 x d by item of −−−−−−−−−−−→ Definition 2.9 x c + 1 x d + 1 Definition 2.21.
Let ϕ ⊗ ( x , · · · , x n ) be an element of B n ⊗ X . Then MRinv ( ϕ ⊗ ( x , · · · , x n )) = Minv ( ϕ ) ⊗ ( x n , · · · , x ) . Example 2.22. Minv (cid:16) x − x x (cid:17) = x x x − . Definition 2.23.
The set of all reduced elements of B n ⊗ X is the free regular Hom-group. Normal Hom-subgroups
We introduce in this section Normal Ho-subgroups and discuss their properties.
Definition 3.1.
Let ( G, µ G , e, α ) be a regular Hom-group. A Hom-subgroup H of G is called a normal Hom-subgroup of G if ∀ g ∈ G, gH = Hg. (3.1)
We indicate that H is a normal Hom-subgroup of G with the notation H ☎ G. Lemma 3.2.
Let H be a Hom-subgroup of a regular Hom-group ( G, µ, e, α ) . H is a normal Hom-subgroup if andonly if ∀ g ∈ G, ∀ h ∈ H, ( gh ) α ( g − ) ∈ H. (3.2) Proof.
Let H be a normal Hom-subgroup of a regular Hom-group ( G, µ G , e, α ) . Then, for all g ∈ G and h ∈ H ,we have gh = hg . Multiply this equation on the right by α ( g − ) , we obtain ( gh ) α ( g − ) = ( hg ) α ( g − ) = α ( h )( gg − ) = α ( h ) ∈ H. So, for all g ∈ G and h ∈ H , we have ( gh ) α ( g − ) = α ( g )( hg − ) ∈ H . Conversely, let g ∈ G and h ∈ H suchthat ( gh ) α ( g − ) ∈ H . There exists h ′ ∈ H such that ( gh ) α ( g − ) = α ( h ′ ) . By multiplying this last equality onthe right by α ( g ) , we find (cid:16) ( gh ) α ( g − ) (cid:17) α ( g ) = α ( h ′ ) α ( g ) , which gives us α ( gh ) α ( g − g ) = α ( h ′ g ) , or α ( gh ) = α ( h ′ g ) . So, the invrtibility of the map α gives that gh = h ′ g . Therefore, we have gH = Hg . The following lemmas give examples of normal subgroups.
Lemma 3.3.
Let ( G, µ, e, α ) be a regular Hom-group and Z ( G ) = { g ∈ G : gx = xg ∀ x ∈ G } be the center of theHom-group G . Then Z ( G ) is a normal Hom-subgroup. Proof.
Let g be an element in G and h ∈ Z ( G ) . Then ( gh ) α ( g − ) = ( hg ) α ( g − ) = α ( h )( gg − ) = α ( h ) e = α ( h ) . According to Lemma 1.26, one can show that ( gh ) α ( g − ) ∈ Z ( G ) . Definition 3.4.
Let ( G, · , e, α ) be a regular Hom-group and H ≤ G. The centralizer C G ( H ) and the normalizer N G ( H ) of the Hom-subgroup H are defined respectively as follows • C G ( H ) = { g ∈ G : ∀ h ∈ H, ( gh ) α ( g − ) = α ( h ) } , • N G ( H ) = { g ∈ G : ∀ h ∈ H, ( gh ) α ( g − ) ∈ H } . Lemma 3.5. C G ( H ) and N G ( H ) are two Hom-subgroups of G and C G ( H ) ☎ N G ( H ) . Proof.
Let us show first that C G ( H ) ≤ G ; • For all h ∈ H , we have ( e · h ) α ( e ) = α ( h ) e = α ( h ) . Then e ∈ C G ( H ) . • Let g, g ′ ∈ C G ( H ) . We want to prove that gg ′ ∈ C G ( H );(( gg ′ ) h ) α ( g ′− g − ) = ( α ( g )( g ′ α − ( h ))) α ( g ′− g − ) = α ( g )(( g ′ α − ( h ))( g ′− g − ))= α ( g )((( α − ( g ′ ) α − ( h )) g ′− ) α ( g − )) = α ( g )( α − (( g ′ α − ( h )) α ( g ′− )) α ( g − ))= α ( g )( hα ( g − )) = ( α ( g ) h ) α ( g − ) = α (( gα − ( h )) α ( g − ))= α ( h ) . Consequently, we have gg ′ ∈ C G ( H ) . Let g ∈ C G ( H ) . We obtain ( g − h ) α ( g ) = ( g − ( α − ( gh ) α − ( g − ))) α ( g ) = ( g − ( α − ( g ) α − ( hg − ))) α ( g )= ( α − ( g − g ) α − ( hg − )) α ( g ) = ( hg − ) α ( g ) = α ( h )( g − g )= α ( h ) . So, for all g ∈ C G ( H ) , we have g − ∈ C G ( H ) . Therefore, C G ( H ) ≤ G . Similarly, we can show that N G ( H ) ≤ G. Now, we prove that C G ( H ) ☎ N G ( H ) . Let n ∈ N G ( H ) and c ∈ C G ( H ) , for all h ∈ H , we have [(( nc ) α ( n − )) h ] α [ α ( n )( c − n − ] = [ α ( nc )( α ( n − ) α − ( h ))] α [ α ( n )( c − n − )]= α ( nc )[( α ( n − ) α − ( h ))( α ( n )( c − n − ))]= α ( nc )[(( n − α − ( h )) α ( n )) α ( c − n − )]= α ( nc )( h ′ α ( c − n − ) where h ′ = ( n − α − ( h )) α ( n )= α ( nc )(( α − ( h ′ ) α ( c − )) α ( n − ))= ( α ( nc )( α − ( h ′ ) α ( c − ))) α ( n − )= ( α ( n )( α ( c )( α − ( h ′ ) c − ))) α ( n − )= ( α ( n ) h ′ ) α ( n − ) . So ( nc ) α ( n − ) is in C G ( H ) . Indeed ( α ( n ) h ′ ) α ( n − ) = ( α ( n )(( n − α − ( h )) α ( n ))) α ( n − )= (( α ( n )( n − α − ( h ))) α ( n )) α ( n − )= ( hα ( n )) α ( n − )= α ( h ) . Lemma 3.6.
Every Hom-subgroup of an abelian Hom-group is normal.
Lemma 3.7.
Let ( G × H, · , e, α ) be the Hom-group defined in (1.4). Then ( G × { e H } , · , e, α ) is a normal Hom-subgroup of ( G × H, · , e, α ) Proof.
For all g , g ∈ G and h ∈ H , we have (( g , h )( g , e H )) α ( g − , h − ) = (( g g ) α G ( g − ) , α H ( hh − )) = (( g g ) α G ( g − ) , e H ) . So, we have the lemma.
Proposition 3.8.
Let f : G → H be a Hom-group homomorphism between the Hom-group ( G, · , e G , α G ) and theHom-group ( H, · , e H , α H ) . If f ( e G ) = e H , then Ker ( f ) = { g ∈ G/f ( g ) = e H } is a normal Hom-subgroup of G. Proof.
According to [7],
Ker ( f ) is a Hom-subgroup of G. Let now g ∈ G and h ∈ Ker ( f ) . We have f (( gh ) α G ( g − )) = ( f ( g ) f ( h )) f ( α G ( g − )) = ( f ( g ) e H ) α H ( f ( g − )) = α H ( f ( g ) f ( g − )) = α H ( e H ) = e H . Therefore
Ker ( f ) ☎ G. A consequence of this proposition is the following corollary
Corollary 3.9.
Let ( G, · , e, α ) be a Hom-group. Then Ker ( α ) ☎ G. Proposition 3.10.
Let f : G → H be a Hom-group homomorphism between the Hom-group ( G, · , e G , α G ) andthe Hom-group ( H, · , e H , α H ) such that f ( e G ) = e H .1. If N ☎ G , then f ( N ) ☎ f ( G ) .
2. If M ☎ f ( G ) , then f − ( M ) ☎ G. roof.
1. Let h ∈ f ( G ) and n ∈ N . If g ∈ G such that f ( g ) = h, we have ( hf ( n )) α H ( h − ) = ( f ( g ) f ( n )) α H ( f ( g − )) = f ( gn ) f ( α G ( g − )) = f (( gn ) α G ( g − )) ∈ f ( N ) , since N ☎ G . So f ( N ) ☎ f ( G ) .
2. Let g ∈ G and n ∈ f − ( M ) , we have f (( gn ) α G ( g − )) = ( f ( g ) f ( n )) f ( α G ( g − )) = ( f ( g ) f ( n )) α H ( f ( g − )) ∈ M, since M ☎ f ( G ) , proving the proposition. Definition 3.11. If H is a Hom-subgroup of a Hom-group ( G, · , e, α ) and g ∈ G then gH = { gh/h ∈ H } , is a left coset of H . Lemma 3.12.
Let H be a Hom-subgroup of a regular Hom-group ( G, · , e, α ) and g, g ′ ∈ G . Suppose α ( H ) = H. Then gH = g ′ H if and only if g − g ′ ∈ H. Proof.
Suppose gH = g ′ H . Then α ( g ′ ) ∈ g ′ H and so α ( g ′ ) ∈ gH . Thus α ( g ′ ) = gh for some h ∈ H and we seethat g − g ′ = α − ( h ) ∈ H , since α ( H ) = H. Conversely, suppose g − g ′ ∈ H. Then g − g ′ = h , for some h ∈ H .Thus α ( g ′ ) = α ( g ) h and consequently α ( g ′ ) H = ( α ( g ) h ) α ( H ) = α ( g )( hH ) . Observe that hH = H because h ∈ H . Therefore gH = g ′ H , since α ( g ) H = α ( gh ) . Theorem 3.13.
Let H be a Hom-subgroup of a regular Hom-group ( G, · , e, α ) and consider the set G/H = { gH/g ∈ G } . If H ☎ G , then ( G/H, · , eH, e α ) is a Hom-group, where • e α : G/H → G/H is defined by e α ( gH ) = α ( g ) H , • g H · g H = ( g g ) H. Proof.
1. For all g , g ∈ G , we have ( g H )( g H ) = α ( g )( Hα − ( g H )) = α ( g )( α − ( Hg ) H ) = α ( g )( α − ( g H ) H )= α ( g )( g α − ( HH )) = ( g g )( HH ) = ( g g ) H.
2. Since the multiplication of G is Hom-associative, then the multiplication of G/H is Hom-associative.3. eH is the identity in G/H .4. Then inverse of gH is g − H. Definition 3.14.
Let ( G, µ, e, α ) be a regular Hom-group. The Commutator Hom-subgroup N = [ G, G ] of G isthe set of the elements [ g, h ] = ( g − h − )( gh ) , ∀ g, h ∈ G. (3.3) Proposition 3.15.
The Commutator subgroup N = [ G, G ] of a regular Hom-group ( G, µ, e, α ) is a normal Hom-subgroup. roof. Let s, g, h ∈ G . Then we have [( sg ) α ( s − ) , ( sh ) α ( s − )] = (cid:16)(cid:16) α ( s )( g − s − ) (cid:17)(cid:16) α ( s )( h − s − ) (cid:17)(cid:17)(cid:16)(cid:16) ( sg ) α ( s − ) (cid:17)(cid:16) ( sh ) α ( s − ) (cid:17)(cid:17) = (cid:16) α ( s ) (cid:16) ( g − s − ) (cid:16) sα − ( h − s − ) (cid:17)(cid:17)(cid:17)(cid:16)(cid:16) ( sg ) α ( s − ) (cid:17)(cid:16) α ( s )( hs − ) (cid:17)(cid:17) = (cid:16) α ( s ) (cid:16) ( g − s − ) (cid:16) α − ( sh − ) s − (cid:17)(cid:17)(cid:16) α ( sg ) (cid:16) α ( s − )( sα − ( hs − )) (cid:17)(cid:17) = (cid:16) α ( s ) (cid:16) α − (( g − s − )( sh − )) α ( s − ) (cid:17)(cid:17)(cid:16) α ( sg ) (cid:16) ( s − s )( hs − ) (cid:17)(cid:17) = (cid:16) α ( s ) (cid:16) ( g − h − ) α ( s − ) (cid:17)(cid:17)(cid:16) α ( sh ) α ( hs − ) (cid:17) = α ( s ) (cid:16)(cid:16) ( g − h − ) α ( s − ) (cid:17)(cid:16) ( sg )( hs − ) (cid:17)(cid:17) = α ( s ) (cid:16) α ( g − h − ) (cid:16) α ( s − ) α − (cid:16) ( sg )( hs − ) (cid:17)(cid:17)(cid:17) = α ( s ) (cid:16) α ( g − h − ) (cid:16)(cid:16) s − α − ( sg ) (cid:17) ( hs − ) (cid:17)(cid:17) = α ( s ) (cid:16) α ( g − h − ) (cid:16)(cid:16) α − ( s − s ) g (cid:17) ( hs − ) (cid:17)(cid:17) = α ( s ) (cid:16) α ( g − h − ) (cid:16) α ( g )( hs − ) (cid:17)(cid:17) = α ( s ) (cid:16) α ( g − h − ) (cid:16) ( gh ) α ( s − ) (cid:17)(cid:17) = α ( s ) (cid:16)(cid:16) ( g − h − )( gh ) (cid:17) α ( s − ) (cid:17) = α ( s ) (cid:16) [ g, h ] α ( s − ) (cid:17) = (cid:16) α ( s )[ g, h ] (cid:17) α ( s − ) . So, we can notice that ( s [ g, h ]) α ( s − ) = [( α − ( s ) g ) α − ( s − ) , ( α − ( s ) h ) α − ( s − )] . This completes the proof. Definition 3.16.
Let ( G, µ, e, α ) be a regular Hom-group and N = [ G, G ] be the Commutator Hom-subgroup of G . We define the quotient Hom-group ( G ab = G/N, µ ′ , e α ) by1. µ ′ ( gN, hN ) = µ ( g, h ) N, e α ( gN ) = α ( g ) N. The Hom-group G ab is called the Abelianization of the Hom-group G. Lemma 3.17.
Let ( G, µ, e, α ) be a regular Hom-group and N = [ G, G ] be the Commutator subgroup of G . Thenthe Hom-group ( G ab , µ ′ , e α ) is abelian. Moreover, if H is another normal Hom-subgroup and the quotient G/H isabelian, then N ⊂ H. Proof.
From Lemma 1.11, one can show that ( G ab , µ ′ , e α ) is abelian. The proof of the second part is straightfor-ward. Lemma 3.18.
Let ( G, µ, e, α ) be a regular Hom-group and ( G ab , µ ′ , e α ) its Abelianization. Then the map π : G → G ab is an homomorphism of Hom-groups. It is called the canonical Hom-projection of G onto G ab . Proof.
The proof is straightforward.
Proposition 3.19.
Let ( G, µ G , e G , α G ) be a regular Hom-group, ( G ab , µ ′ , f α G ) its Abelization and ( H, µ H , e H , α H ) be a regular abelain Hom-group. If f : G → H is a Hom-group Homomorphism, then there exists a unique Hom-group homomorphism e f : G ab → H such that e f ◦ π = f. Proof.
We can deduce that
Ker ( f ) is a normal Hom-subgroup of G . Moreover, one can check that N ⊂ Ker ( f ) . This allows us to define a map e f : G ab → H by setting e f ( gN ) = f ( g ) . To see that this is well-defined, supposethat gN = g ′ N , so that gg ′− ∈ N ⊂ Ker ( f ) . Then we have f ( g ) f ( g ′ ) − = f ( gg ′− ) = e H , which implies that f ( g ) = f ( g ′ ) . Also note that e f : G ab → H is a homomorphism because • e f (( gg ′ ) N ) = f ( gg ′ ) = f ( g ) f ( g ′ ) = e f ( gN ) e f ( g ′ N ) , • e f ( f α G ( gN )) = e f ( α G ( g ) N ) = f ( α G ( g )) = α H ( f ( g )) = α H ( e f ( gN )) , and note that e f ◦ π = f , since for all g ∈ G we have e f ◦ π ( g ) = e f ( gN ) = f ( g ) . Finally, suppose that F : G ab → H is another morphism satisfying F ◦ π = f. Then for all g ∈ G we have e f ( gN ) = f ( g ) = F ( π ( g )) = F ( gN ) . So that F = e f as desired. Simple Hom-groups
Definition 4.1.
A Hom-group is simple if it is nontrivial and has no proper nontrivial normal Hom-subgroups.
The following theorem gives a characterization of simple Hom-groups.
Theorem 4.2.
Nontrivial simple Hom-groups are regular.
Proof.
Let ( G, µ, e, α ) be a simple Hom-group. According to Corollary 3.9, we have ker ( α ) ☎ G . Since ( G, µ, e, α ) is a simple Hom-group, then Ker ( α ) = { e } or Ker ( α ) = G . The Hom-group G is nontrivial, so Ker ( α ) = G .Thus α is bijective. Definition 4.3.
Given a Hom-group ( G, µ, e, α ) , a Hom-subgroup H is a maximal subgroup if:1. H ☎ G,
2. If K ≤ G is a normal Hom-subgroup and H ≤ K , then K = H or K = G , i.e., the only normal Hom-subgroup of G which contains H as a proper Hom-subgroup is G. Proposition 4.4.
A normal Hom-subgroup H is a maximal Hom-subgroup if and only if G/H is a simple Hom-group.
Proof.
We will start proving that if H is not maximal normal, then G/H is not simple. So, assume H is anormal Hom-subgroup which is not maximal, that is, there is K ☎ G such that H (cid:12) K (cid:12) G. Let π : G → G/H bethe quotient map. Since π is a Hom-group homomorphism, then π ( K ) is normal. Since H (cid:12) K (cid:12) G , this is anontrivial subgroup of G/H , showing that
G/H is not simple.Now, we prove the converse. Assume that
G/H is not simple and let e K be a nontrivial normal Hom-subgroup of G/H . Now consider the set K = π − ( e K ) . According to Proposition 3.10, we have K ☎ G. So it remains to claimthat K contains H . Let h ∈ H , then α ( h ) ∈ H . So hH = eH and therefore, π ( h ) = eH . Consequently, we have h ∈ π − ( eH ) and H ⊂ π − ( eH ) ⊂ K. Definition 5.1.
Let ( A, α A ) and ( B, α B ) be two hom-groups and ( C, α C ) is another Hom-group (say in additivenotation even if it is not assumed to be commutative). A map f : A × B → C from the cartesian product of anytwo Hom-groups A and B into a third C is called a Hom-bilinear map if;1. f ( a a , α B ( b )) = f ( a , b ) + f ( a , b ) ,2. f ( α A ( a ) , b b ) = f ( a, b ) + f ( a, b ) ,3. f ( α A ( a ) , α B ( b )) = α C ◦ f ( a, b ) ,for all a, a , a ∈ A and b, b , b ∈ B . The set of all such Hom-bilinear map from A × B to C is denoted by HL ( A, B ; C ) . Remark 5.2. If α A = α B = α C = Id, we have the definition of a bilinear map.
Lemma 5.3.
Let f be a Hom-bilinear map from the Cartesian product of two regular Hom-groups ( A, α A ) and ( B, α B ) into a third regular Hom-group ( C, α C ) (say in additive notation). Then f ( e A , b ) = f ( a, e b ) = 0 , ∀ a ∈ A and b ∈ B, where e A is the neutral element of A and e B is the neutral element of B. Proof.
For all b ∈ B , we have α C ( f ( e A , b )) = f ( α A ( e A ) , α B ( b )) = f ( e A e A , α B ( b )) = f ( e A , b ) + f ( e A , b ) . Let c be the additive inverse of f ( e A , b ) in C .Then α C ( c + f ( e A , b )) = α C ( c ) + α C ( f ( e A , b )) = α C ( c ) + ( f ( e A , b ) + f ( e A , b ))= ( c + f ( e A , b )) + α C ( f ( e A , b )) = 0 + α C ( f ( e A , b )) = α C ( f ( e A , b )) . Since α C is injective, we have f ( e A , b ) = 0 . The same holds for f ( a, e b ) = 0 . orollary 5.4. Let f : A × B → C be a Hom-bilinear map from the Cartesian product of two regular Hom-groups ( A, α A ) and ( B, α B ) into a third regular Hom-group ( C, α C ) (say in additive notation). Then for all a ∈ A and b ∈ B , we have f ( a − , b ) = − f ( a, b ) and f ( a, b − ) = − f ( a, b ) . Lemma 5.5.
Let f be a Hom-bilinear map from the Cartesian product of two regular Hom-groups ( A, α A ) and ( B, α B ) into a third regular Hom-group ( C, α C ) (the product of C is denoted additively even if it is notcommutative). Then, for all a , a ∈ A and b , b ∈ B , we have f ( a , b ) + f ( a , b ) = f ( a , b ) + f ( a , b ) , (5.1) i.e., any two elements of the image of f commute. Proof.
Let a , a ∈ A and b , b ∈ B . Then we have f ( α A ( a + a ) , α B ( b + b )) = f ( α A ( a + a ) , α B ( b ) + α B ( b ))= f ( α A ( a + a ) , α B ( b )) + f ( α A ( a + a ) , α B ( b ))= (cid:16) f ( α A ( a ) , α B ( b )) + f ( α A ( a ) , α B ( b )) (cid:17) + (cid:16) f ( α A ( a ) , α B ( b )) + f ( α A ( a ) , α B ( b )) (cid:17) = α C (cid:16)(cid:16) f ( a , b ) + f ( a , b ) (cid:17) + (cid:16) f ( a , b ) + f ( a , b ) (cid:17)(cid:17) . On other hand we have f ( α A ( a + a ) , α B ( b + b )) = f ( α A ( a ) + α A ( a )) , α B ( b + b ))= f ( α A ( a ) , α B ( b + b )) + f ( α A ( a ) , α B ( b + b ))= (cid:16) f ( α A ( a ) , α B ( b )) + f ( α A ( a ) , α B ( b )) (cid:17) + (cid:16) f ( α A ( a ) , α B ( b )) + f ( α A ( a ) , α B ( b )) (cid:17) = α C (cid:16)(cid:16) f ( a , b ) + f ( a , b ) (cid:17) + (cid:16) f ( a , b ) + f ( a , b ) (cid:17)(cid:17) . So, it follows from Lemma 1.9 that f ( a , b ) + f ( a , b ) = f ( a , b ) + f ( a , b ) . Lemma 5.6.
Let f be a Hom-bilinear map from the Cartesian product of two regular Hom-groups ( A, α A ) and ( B, α B ) into a third regular Hom-group ( C, α C ) (the product of C is denoted additively even if it is notcommutative). Then,1. for all b ∈ B , the kernel of the map f ( · , b ) : a ∈ A → f ( a, b ) ∈ C contains [ A, A ] ,2. for all a ∈ A , the kernel of the map f ( a, · ) : b ∈ B → f ( a, b ) ∈ C contains [ B, B ] . Proof.
By straightforward calculation, we have f ([ a , a ] , b ) = − ( f ( a , α − B ( b )) + f ( a , α − B ( b ))) + ( f ( a , α − B ( b )) + f ( a , α − B ( b ))) , ∀ [ a , a ] ∈ [ A, A ] , b ∈ B, and f ( a, [ b , b ]) = − ( f ( α − A ( a ) , b ) + f ( α − A ( a ) , b )) + ( f ( α − A ( a ) , b ) + f ( α − A ( a ) , b )) , ∀ [ b , b ] ∈ [ B, B ] , a ∈ A. Definition 5.7.
Let ( A, α A ) and ( B, α B ) be two regular Hom-groups. A tensor product of A and B is a regularHom-group ( A ⊗ B, α A ⊗ B ) which is equipped with a Hom-bilinear map θ : A × B → A ⊗ B, such that, for every regular Hom-group ( C, α C ) and every Hom-bilinear map f : A × B → C , there exists a uniqueHom-group homomorphism e f : A ⊗ B → C such that the diagram: × B θ / / f ❋❋❋❋❋❋❋❋❋ A ⊗ B e f { { ①①①①①①①①① C , commutes, that is e f ◦ θ = f. Proposition 5.8.
Let ( A, α A ) and ( B, α B ) be two regular Hom-groups. Tensor products of A and B are uniqueup to unique isomorphism. Lemma 5.9.
Let ( A, α A ) and ( B, α B ) be two Hom-groups. We have ( A ⊗ B, θ, α A ⊗ B ) ∼ = ( B ⊗ A, θ ′ , α B ⊗ A ) . Proof.
Defining the maps f : A × B → B × A ( a, b ) f ( a, b ) = ( b, a ) , g : B × A → A × B ( b, a ) f ( b, a ) = ( a, b ) , we have f ◦ ( α A × α B ) = ( α B × α A ) ◦ f and g ◦ ( α B × α A ) = ( α A × α B ) ◦ g. From the universel property of θ and θ ′ , we have the following diagram: A × B θ (cid:15) (cid:15) f / / B × A g / / θ ′ (cid:15) (cid:15) A × B θ (cid:15) (cid:15) A ⊗ B f ′ / / B ⊗ A g ′ / / A ⊗ B The Hom-bilinearity of θ ′ ◦ f and θ ◦ g ensure the existence of homomorphisms f ′ and g ′ . Now g ◦ f is theidentity map of A × B , hence g ′ ◦ f ′ is the identity map of A ⊗ B . Similarly f ′ ◦ g ′ is the identity of B ⊗ A , thus ( A ⊗ B, θ, α A ⊗ B ) ∼ = ( B ⊗ A, θ ′ , α B ⊗ A ) . Proposition 5.10.
Let ( A, α A ) and ( B, α B ) be two regular Hom-groups. Then A ⊗ B ∼ = A ab × B ab . In particular, A ⊗ B is a regular abelian Hom-group. Proof.
According to Lemma 5.6, one can show that any Hom-bilinear map f : A × B → C from the Cartesianproduct of two regular Hom-groups ( A, α A ) and ( B, α B ) into a third regular Hom-group ( C, α C ) is determined bya unique Hom-bilinear f ′ : A ab × B ab → C such that f = f ′ ◦ γ , where γ : A × B → A ab × B ab is the canonicalprojection map. In this section, we introduce Hom-rings and their properties.
The word Hom-ring was mentioned for the first time by A. Gohr and Y. Fregier in [5]. They defined Hom-ringsas follows. Let A be a set together with two binary operations + : A × A → A and · : A × A → A , one self-map α : A → A and a special element 0 ∈ A . Then ( A, + , · , , α ) is called a Hom-ring if: • ( A, + ,
0) is an abelian group. • The multiplication is distributive on both sides. • α is an abelian group homomorphism. • α and · satisfy the Hom-associativity condition; α ( x ) · ( y · z ) = ( x · y ) · α ( z ).In this section, we define different types of Hom-rings based on Hom-groups and provide various ways ofconstructing them. Definition 6.1.
A Hom-ring of type (1) is a tuple ( A, + , · , , α, β ) consisting of a set A together with two binaryoperations + : A × A → A (the addition) and · : A × A → A (the multiplication) and two set maps α, β : A → A ,such that: . ( A, + , , α ) is an abelian Hom-group.2. β is an endomorphism of the abelian Hom-group ( A, + , , α ) , i.e, β ( x + y ) = β ( x ) + β ( y ) , ∀ x, y ∈ A, and α ◦ β = β ◦ α. α and β are multiplicative maps , i.e, α ( xy ) = α ( x ) α ( y ) and β ( xy ) = β ( x ) β ( y ) , for all x, y ∈ A.
4. The map β and the product · satisfy the Hom-associativity condition; β ( x ) · ( y · z ) = ( x · y ) · β ( z ) . (6.1)
5. The multiplication is Hom-distributive over the addition on both sides;(a) α ( x ) · ( y + z ) = x · y + x · z, (6.2) (b) ( y + z ) · α ( x ) = y · x + z · x, (6.3) for all x, y, z ∈ A. If ( A, + , · , , α, β ) admits a unit element ∈ A satisfying the following properties1. x · · x = β ( x ) , ∀ x ∈ A, (6.4) α (1) = β (1) = 1 , (6.5) the Hom-ring A is said to be unitary. Remark 6.2.
Note that if α = id , we automatically have the definition of Hom-ring in [5].When α = β , a Hom-ring of type (1) should be reffered to as ” α -Hom-ring of type (1) ”. Proposition 6.3.
Let ( A, + , · , , α, β ) be a tuple, where ( A, + , · , is an abelian Hom-group and α is a multi-plicative map. If ( A, + , · , , α, β ) satisfies (6.1), (6.2) or (6.3), (6.4) and (6.5) , then β is an endomorphism ofthe abelian hom-group ( A, + , , α ) and a multiplicative map. Proof.
Suppose that ( A, + , · , , α, β ) satisfies (6.1), (6.2), (6.4) and (6.5), so for any x, y ∈ A , we have β ( x + y ) = 1 · ( x + y ) = α (1) · ( x + y ) = 1 · x + 1 · y = β ( x ) + β ( y ) , and α ◦ β ( x ) = α ( β ( x )) = α (1 · x ) = α (1) · α ( x ) = 1 · α ( x ) = β ( α ( x )) . The multiplicativity of β is obtained from (6.1) β ( x · y ) = ( x · y ) · x · y ) · β (1) = β ( x ) · ( y ·
1) = β ( x ) β ( y ) . Remark 6.4.
As a consequence of the previous proposition, a unitary Hom-ring of type (1) can be defined as atuple ( A, + , · , , α, β ) such that ( A, + , , α ) is an abelian Hom-group, α is a multiplicative map and ( A, + , · , , α, β ) satisfies the identities (6.1), (6.2), (6.3), (6.4) and (6.5). Now, we are going to introduce another type of Hom-ring as follows
Definition 6.5.
A Hom-ring of type (2) is a tuple ( A, + , · , , α, β ) consisting of a set A together with two binaryoperations + : A × A → A (the addition) and · : A × A → A (the multiplication) and two set maps α, β : A → A ,such that:1. ( A, + , , α ) is an abelian Hom-group. . β is an endomorphism of the abelian Hom-group ( A, + , , α ) , i.e., β ( x + y ) = β ( x ) + β ( y ) , ∀ x, y ∈ A, and α ◦ β = β ◦ α. α and β are multiplicative maps, i.e., α ( xy ) = α ( x ) α ( y ) and β ( xy ) = β ( x ) β ( y ) , for all x, y ∈ A.
4. The maps α, β and the product · satisfy β ( α ( x )) · ( β ( α ( y )) · β ( z )) = ( α ( x ) · β ( α ( y ))) · β ( α ( z )) .
5. The multiplication is Hom-distributive over the addition on both sides;(a) α ( x ) · β ( y + z ) = α ( x ) · β ( y ) + α ( x ) · β ( z ) ,(b) α ( y + z ) · α ( β ( x )) = α ( y ) · β ( x ) + α ( z ) · β ( x ) , for all x, y, z ∈ A. If ( A, + , · , , α, β ) admits a unit element ∈ A satisfying the following identities1. x · β ( x ) and · x = α ( x ) , ∀ x ∈ A, α (1) = β (1) = 1 , the Hom-ring A is said to be unitary. Definition 6.6.
Let ( A, + , · , , α, β ) be a Hom-ring of type (2) . It is said to be an ” α -Hom-ring of type (2) ” if α = β. Definition 6.7.
A Hom-ring ( A, + , · , , α, β ) is called a commutative Hom-ring if for all x, y ∈ A we have x · y = y · x . Also, a Hom-ring is said to be regular if the twist maps α and β are bijective. The inverse ofan element x in a Hom-ring ( A, + , · , , α, β ) with respect to the addition is named the additive inverse and it isdenoted by ( − x ) . Now we have introduced these two types of Hom-ring, it’s natural to seek for relations among them andprovide a way for constructing them.
Lemma 6.8.
Let ( A, + , · , , α, β ) be a Hom-ring of type (1) . One has for all x, y, z ∈ A :1. β ( α ( x )) · ( β ( α ( y )) · β ( z )) = α ( α ( x ) · β ( y )) · β ( z ) . ( α ( x ) · β ( α ( y ))) · β ( α ( z )) = β ◦ α ( α ( x ) · ( y · z )) . α ( x ) · β ( y + z ) = α ( x ) · β ( y ) + α ( x ) · β ( z ) . α ( y + z ) · α ( β ( x )) = α ( y ) · β ( x ) + α ( z ) · β ( x ) . Lemma 6.9.
Let ( A, + , · , , α, β ) be a regular Hom-ring of type (2) . Then, for all x, y, z ∈ A :1. β ( x ) · ( y · z ) = ( x · y ) · α ( z ) . ( x · y ) · β ( z ) = β ( x ) · ( yβ ( α − ( z ))) .3. α ( x ) · ( y + z ) = x · y + x · z .4. ( y + z ) · α ( x ) = y · x + z · x. The desired equivalence is then a simple corollary of this two lemmas:
Proposition 6.10.
Regular α -Hom-rings of type (1) and (2) are equivalent. The following provides a source of examples for Hom-ring;
Proposition 6.11.
Let ( A, + , · , be a ring and α, β : A → A two commuting homomorphism of of rings.Defining a new addition e + : A × A → A and a new multiplication e · : A × A → A given by • x e + y = α ( x + y ) = α ( x ) + α ( y ) , • x e · y = β ( x · y ) = β ( x ) · β ( y ) , for all x, y ∈ A , hen ( A, e + , e · , , α, β ) is a Hom-ring of type (1) . We denote this by A α,β and it is called the twist ring of A. Proof.
The proof is straightforward and left to the reader.
Proposition 6.12.
Let ( A, + , · , , α, β ) be a regular Hom-ring of type (1) . Defining a new addition + ′ : A × A → A and a new multiplication · ′ : A × A → A given by • x + ′ y = α − ( x + y ) = α − ( x ) + α − ( y ) , • x · ′ y = β − ( x · y ) = β − ( x ) · β − ( y ) , for all x, y ∈ A ,then ( A, + ′ , · ′ , is a ring. It is called the compatible ring of the Hom-ring ( A, + , · , , α, β ) . Proposition 6.13.
Let ( A, + , · , be a ring and α, β : A → A two commuting homomorphism of rings. Defininga new addition e + : A × A → A and a new multiplication e · : A × A → A given by • x e + y = α ( x + y ) = α ( x ) + α ( y ) , • x e · y = β ( x ) · α ( y ) , for all x, y ∈ A ,then ( A, e + , e · , , α, β ) is a Hom-ring of type (2) . We denote this by A α,β and it is called the twist ring of A. Proposition 6.14.
Let ( A, + , · , , α, β ) be a regular Hom-ring of type (2) . Defining a new addition + ′ : A × A → A and a new multiplication · ′ : A × A → A given by • x + ′ y = α − ( x + y ) = α − ( x ) + α − ( y ) , • x · ′ y = β − ( x ) · α − ( y ) , for all x, y ∈ A ,then ( A, + ′ , · ′ , is a ring, which is called the compatible ring of the Hom-ring ( A, + , · , , α, β ) . In what follows we provide some properties and define Hom-ring homomorphisms and Hom-subrings.
Proposition 6.15.
Let ( A, + , · , , α, β ) be a unitary regular hom-ring of type (1) or of type (2) .
1. For all a ∈ A , we have a · · a = 0 .
2. Let a, b ∈ A . Then , the additive inverse of ( a · b ) is ( − a ) · b = a · ( − b ) = − ( a · b ) . One can notes that α ( − a ) = − α ( a ) . Proof.
We have · α ( a ) = (0 + 0) · α ( a ) = 0 · a + 0 · a. Let c be the additive inverse of · a , i.e., c + 0 · a = 0 . Then we have α ( c + 0 · a ) = α ( c ) + 0 · α ( a ) = α ( c ) + (0 · a + 0 · a )) = ( c + 0 · a ) + α (0 · a ) = α (0 · a ) . Since α is injective, then · a = 0 . The second property is a consequence of the first property. Definition 6.16.
Let ( A, + A , · A , A , α A , β A ) and ( B, + B , · B , B , α B , β B ) be two unitary hom-rings. • A homomorphism of Hom-rings is a map f : A → B such that1. f (0 A ) = 0 B and f (1 A ) = 1 B f ( x + A y ) = f ( x ) + B f ( y ) f ( x · A y ) = f ( x ) · B f ( y ) . f ◦ α A = α B ◦ f , and f ◦ β A = β B ◦ f. • The map f is called a weak homomorphism if it satisfies the second and the third condition. • The Hom-rings ( A, + A , · A , A , α A , β A ) and ( B, + B , · B , B , α B , β B ) are called isomorphic if f is bijective. • If f is the inclusion map, we say that ( A, + , · , A , α, β ) is a Hom-subring of ( B, + , · , B , α, β ) . emark 6.17. Let ( A, + , · , A , α, β ) be a Hom-subring of ( B, + , · , B , α, β ) . Then for all x ∈ A , we have (ob-viously) α ( x ) ∈ A and β ( x ) ∈ A . Also, a Hom-subring of a Hom-ring ( A, + , · , A , α, β ) is a Hom-subgroup of ( A, + , A , α ) stable by the multiplication and the map β . Definition 6.18.
Let ( A, + , · , , α, β ) be a unitary regular Hom-ring of type (1) (resp. regular Hom-ring of type (2) ). The center Z ( A ) (resp. center Z ′ ( A ) ) of A is the set of all x ∈ A where x · a = a · x (resp. x · β ( a ) = α ( a ) · x ),for all a ∈ A . Proposition 6.19.
Let ( A, + , · , , α ) be a unitary regular Hom-ring.1. If ( A, + , · , , α ) is of type (1) , then Z ( A ) is a Hom-subring of A .2. If ( A, + , · , , α ) is of type (2) , then Z ′ ( A ) is a Hom-subring of A . Proof.
1. Let ( A, + , · , , α ) be a Hom-ring of type (1) We have, for all a ∈ A , · a = a · and · a = a · β ( a ) . Then and are in Z ( A ) . Let x, y ∈ Z ( A ) . Then for all a ∈ A , we have(a) ( x + y ) · a = ( x + y ) · α ( α − ( a )) = x · α − ( a ) + y · α − ( a ) = α − ( a ) · x + α − ( a ) · y = a · ( x + y ) . Thus Z ( A ) is closed under addition of A .(b) If x ∈ Z ( A ) and a ∈ A then − ( x · a ) = − ( a · x ) . Furthermore, we have − ( x · a ) = ( − x ) · a and − ( a · x ) = a · ( − x ) . So − x ∈ Z ( A ) .(c) ( x · y ) · a = ( x · y ) · α ( α − ( a )) = α ( x ) · ( y · α − ( a )) = α ( x ) · ( α − ( a ) · y ) = ( x · α − ( a )) · α ( y ) =( α − ( a ) · x ) · α ( y ) = a · ( x · y ) . Thus Z ( A ) is closed under multiplication of A .2. Let ( A, + , · , , α ) be a Hom-ring of type (2) . We have, for all a ∈ A , · β ( a ) = α ( a ) · and · β ( a ) = α ( a ) · β ◦ α ( a ) . Then and are in Z ′ ( A ) . Let x, y ∈ Z ′ ( A ) . Then for all a ∈ A , we have(a) ( x + y ) · β ( a ) = α − ( α − ( x ) + α − ( y )) · β ( α ( α − ( a ))) = x · β ( α − ( a )) + y · β ( α − ( a )) = a · x + a · y = α ( a ) · ( x + y ) . Thus Z ′ ( A ) is closed under addition of A .(b) If x ∈ Z ′ ( A ) and a ∈ A then − ( x · a ) = − ( a · x ) . Furthermore, we have − ( x · a ) = ( − x ) · a and − ( a · x ) = a · ( − x ) . Then − x ∈ Z ′ ( A ) . (c) ( x · y ) · β ( a ) = β ( x ) · ( y · β ( α − ( a ))) = β ( x ) · ( a · y ) = ( x · a ) · α ( y ) = α ( a ) · ( x · y ) . Thus Z ′ ( A ) is closedunder multiplication of A . Proposition 6.20.
Let ( A, + , α A ) be a regular abelian Hom-group and consider the group homomorphisms from A into A . Then ( End ( A ) , + , ◦ , , α End ( A ) , id ) is a Hom-ring of type (1) such that1. The addition of two such homomorphisms may be defined pointwise to produce another group homomorphism.Explicitly, given two such homomorphisms f and g , the sum of f and g is the homomorphism ( f + g )( x ) = f ( x ) + g ( x ) ,2. The composition ◦ of two such homomorphisms f and g is explicitly ( f ◦ g )( x ) = f ( g ( x )) ,3. α End ( A ) ( f ) = α A ◦ f . Proof.
The proof is straightforward and left to the reader.
Polynomial Hom-Rings:
Let A be a ring and R = A [ X , · · · , X n ] be the polynomial ring in n variables.Let us recall that any element of R is uniquely written as a linear combination, with coefficients in A , of monomes X n = X m ...X m n n , where m runs over the set of sequences of n nonnegative integers.Take a ring endomorphism of A , α : A → A ⊂ R. Then a ring endomorphism e α of R is uniquely determined bythe n polynomials e α ( X i ) = P r a i,r X r = P r a i,r X r ...X r n n for 1 ≤ i ≤ n. Let e α ( P ) = P m α ( a m ) e α ( X ) m ... e α ( X n ) m n , for P = P m a m X m ...X m n n ∈ R. For P = P m a m X m ...X m n n and Q = P m b m X m ...X m n n , define b + by P b + Q = X m α ( a m + b m ) e α ( X ) m ... e α ( X n ) m n , P = P m a m X m ...X m n n and Q = P l b l X l ...X l n n , define b · by P b · Q = X d X m + l = d α ( a m ) α ( b l ) e α ( X ) d ... e α ( X n ) d n . Then R e α = ( R, b + , b · , e α ) is a α − Hom-ring of type (1).
Twisted Group-Ring:
Let G be a group and K G the corresponding group-ring over K . the set K G is generated by { e g : g ∈ G } .If α : G → G is a group homomorphism, then it can be extended to a ringendomorphism of K G by setting α (cid:16) X g ∈ G a g e g (cid:17) = X g ∈ G a g α ( e g ) = X g ∈ G a g e α ( g ) . Consider the usual ring structure on K G . Then, we define a α − Hom-ring of type (1) , ( K G, b + , b · , e α ) over K G bysetting: X g ∈ G a g e g b + X g ∈ G b g e g = X g ∈ G ( a g + b g ) e α ( g ) , ( X g ∈ G a g e g ) b · ( X h ∈ G b h e h ) = X gh ∈ G a g b g e α ( gh ) . α -Hom-rings of type (1) We introduce in the following the structure of module over α -Hom-ring of type (1) . The properties of modulesfor other types Hom-ring will be discussed in a forthcoming paper.In the rest of this article, the word Hom-ring means an α -Hom-ring of type (1) . Definition 7.1.
Let ( A, + A , · , , α ) be a unitary Hom-ring with unit A . A left A -module M consists of an abelianHom-group ( M, + M , β ) and an operation λ l : A ⊗ M → M such that for all a, b in A and m , m in M , wehave: λ l ( α ( a ) ⊗ ( m + M m )) = λ l ( a ⊗ m ) + M λ l ( a ⊗ m ) , (7.1) λ l (( a + A b ) ⊗ β M ( m )) = λ l ( a ⊗ m ) + M λ l ( b ⊗ m ) , (7.2) λ l (( a · b ) ⊗ β ( m )) = λ l ( α ( a ) ⊗ λ l ( b ⊗ m ) , (7.3) λ l (1 A ⊗ m ) = β ( m ) . (7.4) The operation of the Hom-ring on M is called scalar multiplication, and it is usually written as am for a in A and m in M . The notation A M indicates a left A -module M . If β is invertible, such left module A M is saidregular. A right A -module M or M A is defined similarly, except that the ring acts on the right; i.e., the operationtakes the form λ r : M ⊗ A → M , and the above axioms are written with a scalar multiplication on the right. Example 7.2. If A is a unitary Hom-ring, then we may consider A as either a left or right A-module, where theaction is given by the multiplication. Proposition 7.3.
Let ( A, + A , · , , α ) be a unitary Hom-ring. If A is commutative, then these left modules arethe same as right modules. Proof.
Let M be a right A -module, i.e., there exists λ r : M ⊗ A −→ M by λ r ( m ⊗ a ) = ma . Now, we define λ l : A ⊗ M −→ M by λ l ( a ⊗ m ) = ma . Using the commutativity of the Hom-ring A , we can verify that λ l definesa left module structure for M over A. Definition 7.4. If R and S are two Hom-rings, then an R − S -bimodule is an abelian Hom- group M such that:1. M is a left R -module and a right S -module,2. ( rm ) α S ( s ) = α R ( r )( ms ) . efinition 7.5. A bimodule is both a left A -module with action λ l and a right A -module with action λ r satisfyingthe compatibility condition λ r ◦ ( λ l ⊗ α ) = λ l ◦ ( α ⊗ λ r ) . (7.5) It is simply called A -module. Proposition 7.6.
Let ( A, + A , · , A , α ) be a unitary Hom-ring and ( M, + M , M , β ) be a regular left A -module.1. For all m in M , we have A m = 0 M .
2. If moreover a ∈ A admits an additive inverse of index r ≤ n , then the additive inverse of am on M is − ( am ) = ( − a ) m = a ( − m ) . We can observe that β ( − m ) = − β ( m ) . Proof.
We can show that the first property in the same way as the first property of Proposition 6.1. For thesecond property, we have α n − ( a ) β n ( m ) + α n − ( − a ) β n ( m ) = α n ( a + ( − a )) β n +1 ( m ) = 0 β n − ( m ) = 0 . In the same way we check the other identity.
Remark 7.7. If M is a module over a Hom-ring A , then two equations (7.3) and (7.4) give that β ( am ) = α ( a ) β ( m ) . (7.6) One can get, for all m element in right (regular) A -module M , m A = 0 M . Proposition 7.8.
Let ( A, + A , · A , , α ) be a unitary regular Hom-ring, ( M, + M , β ) a regular A -module (left), ( A, + ′ A , · ′ ) the compatible ring. If λ l : A −→ End ( M ) a λ l ( a ) , is the action of the Hom-ring A on M , then ( M, + ′ M = β − ◦ + M ) is a (left) ( A, + ′ A , · ′ ) -module with respect the action λ ′ l = β − λ l . Proof.
For the action λ ′ l , the conditions for ( M, + ′ M ) to be an ( A, + ′ A , · ′ ) -module is written in the form β − ( a ( m + M m )) = β − ( am + M am ) , (7.7) β − (( a + A b ) m )) = β − ( am + M bm ) , (7.8) β − ( aβ − ( bm ) = β − ( α − ( a · A b ) m ) , (7.9) β − (1 A m ) = m . (7.10) Also, one can check that equation (7.6) is equivalent to λ l α ( a ) = βλ l ( a ) β − . (7.11) Then the result can easily follow.
From Proposition 7.8, we can get a method of building (left) regular modules of a unitary regular Hom-ring:
Theorem 7.9.
Let ( A, + A , · A , α ) be a unitary regular Hom-ring with the compatible ring ( A, + ′ A , · ′ A ) .1. If ( M, + M , β ) is a (left) regular A -module then β ( a ✄ m ) = α ( a ) ✄ β ( m ) , ∀ a ∈ A, m ∈ M, (7.12) where a ✄ m = β − ( am ) defines a ( A, + ′ A , · ′ A ) -module structure for ( M, β − ◦ + M ) , it is called the compatible A -module of ( M, + M , β ) .2. Suppose that ( M, + ′ M ) is a (left) module over ( A, + ′ A , · ′ A ) for the action a ✄ m . If it exists an automorphismof the group ( M, + ′ M ) such that (7.12) is satisfied, then ( M, + M = β ◦ + ′ M , β ) is a (left) module over ( A, + A , · A , α ) with respect to the action given by am = β ( a ✄ m ) , for all a ∈ A and m ∈ M . roof. ( M, + M , β ) is a (left) regular A -module, i.e., β is invertible. Then from (7.6), we can get easily (7.12).2. Let a, b ∈ A and m , m ∈ M , from (7.12) and the properties of ( M, + ′ M ) as a(left) ( A, + ′ A , · ′ A ) -module,we have α ( a )( m + M m ) = β ( α ( a ) ✄ β ( m + ′ M m ))= α ( a ) ✄ ( β ( m ) + ′ M β ( m )) = α ( a ) ✄ β ( m ) + ′ M α ( a ) ✄ β ( m )= β ( α ( a ) ✄ β ( m )) + ′ M β ( α ( a ) ✄ β ( m )) = α ( a ) ✄ β ( m )) + M β ( α ( a ) ✄ β ( m )= β ( a ✄ m ) + M β ( a ✄ m )= am + M am , (7.13)( a + A b ) β ( m ) = β (( a + A b ) ✄ β ( m )) = α (( a + ′ A b )) ✄ β ( m )= α ( a ) ✄ β ( m ) + ′ M α ( b ) ✄ β ( m ) = β ( α ( a ) ✄ β ( m )) + ′ M β ( α ( b ) ✄ β ( m ))= α ( a ) ✄ β ( m )) + M β ( α ( b ) ✄ β ( m ) = β ( a ✄ m ) + M β ( b ✄ m )= am + M bm , (7.14) α ( a )( bm ) = β ( α ( a ) ✄ ( α ( b ) ✄ β ( m ))) = α ( a ) ✄ β ( α ( b ) ✄ β ( m )) = α ( a ) ✄ ( α ( b ) ✄ β ( m ))= ( α ( a ) · ′ α ( b )) ✄ β ( m ) = α ( a · b ) ✄ β ( m )) = β (( a · b ) ✄ m )= ( a · b ) β ( m ) , (7.15)1 A m = β (1 A ✄ m ) = α (1 A ) ✄ β ( m ) = 1 A ✄ β ( m ) = β ( m ) . (7.16) Therefore ( M, + M = β ◦ + ′ M , β ) is a (left) module over ( A, + A , · A , α ) with respect the action defined by am = β ( a ✄ m ) , for all a ∈ A and m ∈ M. Definition 7.10.
Let ( R, α R ) be a Hom-ring. Let ( A, α A ) be a right R -module, ( B, α B ) a left R -module and ( C, α C ) be an abelian Hom-group (all abelian Hom-groups having the law of composition denoted by + ). A map f : A × B → C from the Cartesian product of A and B into C is called Hom-R-Bilinear if1. f ( a + a , α B ( b )) = f ( a , b ) + f ( a , b ) ,2. f ( α A ( a ) , b + b ) = f ( a, b ) + f ( a, b ) ,3. f ( α A ( a ) , α B ( b )) = α C ◦ f ( a, b ) ,4. f ( ar, α B ( b )) = f ( α A ( a ) , rb ) ,for all a, a , a ∈ A, b, b , b ∈ B and r ∈ R . The set of all such Hom-R-bilinear maps from A × B to C isdenoted by HL R ( A, B ; C ) . Definition 7.11.
Let ( R, α R ) be a Hom-ring, ( A, α A ) be a right regular R -module and ( B, α B ) be a left regular R -module. The tensor product of A and B , which is denoted by A ⊗ R B , is a regular abelian Hom-group togetherwith a Hom-R-Bilinear map θ : A × B → A ⊗ R B, such that the following universal property holds: for any Hom-R-bilinear map f : A × R B → C , there exists aunique Hom-module homomorphism e f : A ⊗ R B → C such that the diagram: A × B θ / / f " " ❋❋❋❋❋❋❋❋❋ A ⊗ R B e f { { ✇✇✇✇✇✇✇✇✇ C , commutes, that is e f ◦ θ = f. roposition 7.12. Let
R, S be Hom-rings. If A is a regular R − S -bimodule and B is a left regular S -module,then A ⊗ S B is a left R -module satisfying r ( a ⊗ b ) = ( ra ) ⊗ b. Proof.
For each s ∈ S , we have α R ( r )( as ) ⊗ α B ( b ) = ( ra ) α S ( s ) ⊗ α B ( b ) = α A ( ra ) ⊗ ( α S ( s ) n ) = ( α R ( r ) α A ( a )) ⊗ ( α S ( s ) n ) . Definition 8.1.
Let A be a Hom-ring and ( M, β ) and ( N, β ) be two (left) A -modules. A homomrophism ofHom-groups f : N → M is said to be a homomorphism of A -modules if f ( n ) = α A ( a ) f ( n ) , for all a ∈ A and n in M. If f is the inclusion map, we say that N is an A -submodule of M. Remark 8.2. An A -submodule of M is a Hom-subgroup N of M which is itself a module under the action ofHom-ring elements. Note that, for any A -module M , both { } and M are trivial A -submodules of M . Let us now recall that thedirect sum of two Hom-groups ( H, β ) and ( K, β ) ([8]) is the set H ⊕ K = { ( h, k ) | h ∈ H, k ∈ K } , together with the component-wise operations of the groups H and K (that is, if ( h , k ) , ( h , k ) ∈ H ⊕ K , then( h , k )( h , k ) = ( h h , k k )) and the map β ⊕ β such as β ⊕ β ( h, k ) = ( β ( h ) , β ( k )) , ∀ ( h, k ) ∈ H ⊕ K .There is a similar structure with modules. Definition 8.3.
Let ( M, β ) and ( N, β ) be modules over a Hom-ring A . We define the direct sum of M and N by M M N = { m ⊕ n | m ∈ M, n ∈ M } . (8.1) The structure of A -module of M L N is given by ( m ⊕ n ) ⊕ ( m ⊕ n ) = ( m + M m ) ⊕ ( n + N n ) , (8.2)( β ⊕ β )( m ⊕ n ) = β ( m ) ⊕ β ( n ) , (8.3) a ( m ⊕ n ) = am ⊕ an, where a ∈ A. (8.4) Definition 8.4.
Let A be a unitary (regular) Hom-ring. An A -module M is said to be simple if it is non-zero,and has no submodules except { } and itself. If M is a direct sum of certain simple submodules, M is calledsemisimple. Lemma 8.5.
Let ( M, + M , β ) be a (left) module over a unitary Hom-ring A . Then ( kerβ, + M , β ) is an A -submodule. Proof.
For all m ∈ Kerβ , by β ( am ) = α ( a ) β ( m ) = 0 , for all a ∈ A , it is easy to show the lemma. Theorem 8.6.
Let ( M, + M , β ) be a (left)simple module over a unitary Hom-ring A . Then M is regular, i.e., the map β isinvertible. Proof.
According to Lemma 8.5, ( kerβ, + M , β ) is an A -submodule. Since the module M is simple, either kerβ = { } or kerβ = M . The A -module M is simple, then kerβ = { } . So β is invertible. Proposition 8.7.
Let ( A, + A , · A , α ) be a unitary regular Hom-ring with the compatible ring ( A, + ′ A , · ′ A ) . Supposethat ( M, + ′ M ) is a simple (left) module over ( A, + ′ A , · ′ A ) for the action a ✄ m . If it exists an automorphismof the group ( M, + ′ M ) such that (7.12) is satisfied, then ( M, + M = β ◦ + ′ M , β ) is a simple (left) module over ( A, + A , · A , α ) with respect to the action given by am = β ( a ✄ m ) , for all a ∈ A and m ∈ M . roof. Indeed, let N be a non-zero A -submodule of ( M, + M = β ◦ + ′ M , β ) , that is N ⊂ M and ( N, + M = β ◦ + ′ M , β ) is an A -module with respect to the action defined by a ⊗ n an, ∀ a ∈ A, n ∈ N . According to Theorem 7.9, ( N, + ′ M ) is a module over the ring ( A, + ′ A , · ′ A ) with respect to the action a ⊗ n a ✄ n = β − ( an ) , where ✄ defines the action of the ring ( A, + ′ A , · ′ A ) on ( M, + ′ M ) . Since ( M, + ′ M ) is simple, then N = M as groups withrespect to the addition + ′ M . So from the property β ( N ) = N , we can show the result. The following theorem gives a characterization of (left) modules of a regular unitary Hom-ring.
Theorem 8.8.
Let ( A, + A , · A , α ) be a unitary regular Hom-ring with the compatible ring ( A, + ′ A , · ′ A ) and ( M, + M , β ) be a simple regular A -module. Then the compatible group ( M, + ′ M = β − ◦ + M ) is a semisimple module over thering ( A, + ′ A , · ′ A ) . Proof.
Suppose that ( M, + M , β ) is a regular A -module for the action a ⊗ n am, ∀ a ∈ A, m ∈ M , then accordingto Theorem 7.9, ( M, + ′ M = β − ◦ + M ) is a module over ( A, + ′ A , · ′ A ) for the action a ⊗ n a ✄ m = β − ( am ) . Let m be a non-zero element in M and let k be the smallest integer such that β k − ( m ) = m and β k ( m ) = m . Thegroup of ( M, + ′ M = β − ◦ + M ) , N = A ✄ m L A ✄ β ( m ) L · · · L A ✄ β k − ( m ) is a submodule of ( M, + ′ M ) overthe ring ( A, + ′ A , · ′ A ) . Furthermore, we have β ( N ) = N = Am L Aβ ( m ) L · · · L Aβ k − ( m ) . Consequently N isa submodule of ( M, + M , β ) over ( A, + A , · A , α ) . So the simplicity of ( M, + M , β ) gives M = N as a module over ( M, + ′ M = β − ◦ + M ) . Therefore, as a module over ( A, + ′ A , · ′ A ) , we have M = N = A ✄ m L A ✄ β ( m ) L · ·· L A ✄ β k − ( m ) , which shows the semisimplicity of the module ( M, + ′ M = β − ◦ + M ) over ( A, + ′ A , · ′ A ) for theaction a ⊗ n a ✄ m = β − ( am ) . Definition 8.9.
A Hom-ring A is called semisimple (resp. simple) if A is semisimple (resp. simple) as an A -module. Theorem 8.10.
Simples unitary Hom-rings are regular Hom-rings.
Proof.
The proof is straightforward.
Theorem 8.11.
Let ( A, + A , · A , α ) be a unitary Hom-ring with the compatible ring ( A, + ′ A , · ′ A ) . Then the com-patible ring ( A, + ′ A , · ′ A ) is semisimple. Proof.
The proof is straightforward.
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