(Totally) L -ordered Groups

[1] Introduction

Zhang and Liu defined a kind of an L-frame by a pair (A ; i _A ), where A is a classical frame and i _A  : L→A is a frame morphism. For a stratified L-topological space (X ; δ), the pair (δ ; i _X ) is one of this kind of L-frames, where i _X  : L→δ, is a map which sends a ∈ L to the constant map with the value a. Conversely, a point of an L-frame (A ; iA) is a frame morphism p : (A ; i _A ) → (L ; id _L ) satisfying p ∘ i _A  = id _L and Lpt (A) denotes the set of all points of (A ; i _A ). Then {Φ _x  : Lpt (A) →L| ∀ p ∈ Lpt (A) ; Φ _x  (p) = p (x)} is a stratified L-topology on Lpt (A). By these two assignments, Zhang and Liu constructed an adjunction between SL - Top and L - Loc and consequently they established the Stone representation theorem for distributive lattices by means of this adjunction. They pointed out that, from the viewpoint of lattice theory, Rodabaugh’s fuzzy version of the Stone representation theory is just one and it has nothing different from the classical one. While in our opinion, Zhang-Liu’s L-frames preserve many features and also seem to have no strong difference from a crisp one. Recently, based on complete Heyting algebras, Fan and Zhang [4, 13], studied quantitative domains through fuzzy set theory. Their approach first defines a fuzzy partial order, specifically a degree function, on a non-empty set. Yao introduced the notion of L-frames. It is a complete L-ordered set with the meet operation having a right fuzzy adjoint. They established an adjunction between the category of stratified L-topological spaces and the category of L-locales, the opposite category of this kind of L-frames. In the follows, Borzooei et al., in [3] introduced the notions of L-ordered group, convex L-subgroup and quotient L-ordered group.

This paper are organized as follows. In Section 2, we recall some notions and results about L-ordered sets and L-ordered groups. In Section 3, we give some properties of L-ordered group and introduce two products in L-ordered groups. In Section 4, we introduce the notions of totally L- ordered group and strong totally L-ordered group and we state and prove some related results.

[2] Preliminary

A frame is a complete lattice L satisfying the following infinite distributive law: a ∧ ( ⋁ b ∈ B B ) = ⋁ b ∈ B ( a ∧ b )

Let L be a frame and for any a, b ∈ L, a→b is defined by a→b = ⋁ {c ∈ L|a ∧ c ≤ b}. Then for any x, y, z ∈ L and Y ⊂ L, the following conditions hold:

(x ∧ y) →z = x→ (y→z),

From now on, in this paper, we let L is a frame.

Let P be a set and e : P × P→L be a map. The pair (P, e) is called an L-ordered set if for all x, y, z ∈ P, (E1): e (x, x) =1,(E2): e (x, y) ∧ e (y, z) ≤ e (x, z), (E3): e (x, y) = e (y, x) =1 implies x = y. In an L-ordered set (P, e), the map e is called an L-order relation on P. If (P, ≤) is a poset, then (P, χ_≤) is an L-ordered set, where χ_≤ is the characteristic function of ≤. Moreover, for each L-ordered set (P, e), the set ≤ _e  = {(x, y) ∈ P × P|e (x, y) =1} is a partial order on P and so (P, ≤  _e ) is a poset. We denote L ^P the set of all L-subsets of P, that is L ^P  = {f|f : P→L}. A map f : (P, e _P ) → (Q, e _Q ) between two L-ordered sets is called monotone if for all x, y ∈ P, e _P  (x, y) ≤ e _Q  (f (x) , f (y)). Let (P, e) be an L-ordered set and S be an L-subset of P. Then support S is defined by the set supp (S) = {x ∈ P| 0 < S (x)}. Let (P, e) be an L-ordered set and S ∈ L ^P . An element x₀ ∈ P is called a join (meet) of S, in symbol x₀ = ⊔ S (x₀ = ⊓ S), if for all x ∈ P, (J1): S (x) ≤ e (x ; x₀) ((M1): S (x) ≤ e (x₀ ; x)), (J2):  ⋀ _y∈P (S (y) →e (y, x)) ≤ e (x₀, x)((M2):  ⋀ _y∈P (S (y) →e (x, y)) ≤ e (x, x₀))(See [12, 13]). If a join and a meet of S exist, then they are unique.

Theorem 2.1. [13] Let (P, e) be an L-ordered set and S ∈ L ^P . Then

x₀ = ⊔ S if and only if e (x₀, x) =  ⋀ _y∈P (S (y) →e (y, x)),

An L-ordered set (P, e) is called an L-lattice if for all x, y ∈ P, ⊔_χ{x,y} and ⊓_χ{x,y} exist.

Proposition 2.2. [3] Let (P, e) be an L-lattice. Then the following conditions hold for all x, y, a ∈ P:

(P, ≤) is a lattice,

Corollary 2.3. [3] An L-ordered set (P, e) is an L-lattice if and only if (P, ≤  _e ) is a lattice such that e (a, x ∧ y) = e (a, x) ∧ e (a, y) and e (x ∨ y, a) = e (x, a) ∧ e (y, a), for all x, y, a ∈ P.

An L-fuzzy ordered group (or L-ordered group, for short) is a group (G ; . , 1) together with an L-order relation e : G × G→L on G such that each translation maps () . a : G→G and a . () : G→G are monotone, or equivalently (FOG): e (x, y) ≤ e (b . x . a, b . y . a), for each x, y, a, b ∈ G. We use (G, e) to denote an L-ordered group G. An L-ordered group (G, e) is called an L-lattice ordered group, if (G, e) is an L-lattice (See [3]).

Corollary 2.4. If (G, e) is an L-ordered group, then (G, ≤  _e ) is an ordered group.

Converse of this corollary is not hold in general. We show this by an example.

Example 2.5. Let L = {0, a, 1} be a chain and consider ( ℤ , e ) that e is an L-order relation defined by: e ( x , y ) = { 1 if x ≤ y 0 if 0 ≤ x and x ≰ y a if x < 0 and x ≰ y

Obviously ( ℤ , e ) is an L-ordered set and ( ℤ , ≤ e ) is an ordered group, but ( ℤ , e ) is not an L-ordered group.

Let (G, e) be an L-ordered group, a ∈ G and S ∈ L ^G . We define a ∧ S by (a ∧ S) (y) = ∨ {S (x) |x ∈ G, a ∧ x = y}. a ∨ S and a . S is define similarly (See [3]).

Proposition 2.6. [3] Let (G, e) be an L-ordered group. Then the following conditions hold, for allx, y, a, b ∈ G:

For each x, y, a, b ∈ G, e (x, y) = e (b . x . a, b . y . a),

Let (G, e) be an L-ordered group. By an L-subgroup of (G, e) we mean an element S ∈ L ^G such that S (x) ∧ S (y) ≤ S (x . y), for all x, y ∈ G and S (x) = S (x^-1), for all x ∈ G. An L-subgroup S of (G, e) is callednormal if S (y) ≤ S (x . y . x^-1), for all x, y ∈ G. Clearly, if S is normal, then S (y) = S (x . y . x^-1), for all x, y ∈ G. Moreover, if S₁, ⋯ , Sn ∈ L ^G , then S₁ . ⋯ . S _n define by (S₁ . ⋯ . S _n ) (y) = ∨ {S (x₁) ∧ S (x₂) ∧ ⋯ ∧ S (x _n ) |x₁ . ⋯ . x _n  = y}, for all y ∈ G. A non-empty subset A of an ordered set X is said to be convex if for x, y ∈ A and z ∈ X, x ≤ z ≤ y implies z ∈ A. A subset A of lattice L is said to be filter if for any x, y ∈ L, x, y ∈ A implies x ∧ y ∈ A and x ∈ A and y ≥ x implies y ∈ A. A filter F of a Boolean algebra B is an ultrafilter if F is maximal with respect to the property that 0 ∉ F. For any set I, an ultrafilter of P (I) (powerset of I) is called an ultrafilter over I. Let {A _i } _i∈I be a family of algebras of a given type, and let U be an ultrafilter over I. Define θ _U on ∏ i ∈ I A i by (a, b) ∈ θ _U if and only if {i ∈ I|f (i) = g (i)} ∈ U. Let {A _i } _i∈I and U be an ultrafilter over I; we define the ultraproduct ∏ i ∈ I A i / U to be ∏ i ∈ I A i / θ U. The elements of ∏ i ∈ I A i / U are denoted by a/U; where a ∈ ∏ i ∈ I A i. Let (G, e) be an L-ordered group and S ∈ L ^G . By a positive cone of S, we mean a map S⁺ : G→L defined by S⁺ (x) = S (x) ∧ e (1 _G , x) for all x ∈ G. The positive cone of G is the positive cone of the map S _G  : G→L sending each element x ∈ G to e (1 _G , x). So, for each x ∈ G, (S _G ) ⁺ (x) = e (1 _G , x) ∧ e (1 _G , x) = e (1 _G , x) (See [3]).

Theorem 2.7. [3] Let S be the positive cone of (G, e). Then S has the following properties, for all x, y ∈ G:

S (x) ∧ S (y) ≤ S (x . y),

[3] Some properties of L-ordered groups

In this section, we give some properties of L-ordered groups, such as generated L-subgroup of an L-ordered group and distributivity in L-ordered groups. Then we introduce two products in L-ordered groups. Finally, we show that ultraproduct of L-ordered groups where L is finite chain is an L-ordered group.

Lemma 3.1. Let (G, e) be an L-ordered group and {S _i } _i∈I be a family of L-subgroups of (G, e). Then ⋀_i∈IS _i is an L-subgroup of (G, e) where ( ⋀ _i∈IS _i ) (x) =  ⋀ _i∈IS _i  (x), for x ∈ G.

Proof. For any x, y ∈ G, ( ⋀ _i∈IS _i ) (x^-1) =  ⋀ _i∈IS _i  (x^-1) allowbreak =  ⋀ _i∈IS _i  (x) = ( ⋀ _i∈IS _i ) (x). Moreover, ( ⋀ _i∈IS _i ) (x) ∧ ( ⋀ _i∈IS _i )(y) = ( ⋀ _i∈I (S _i  (x) ∧ S _i  (y)) ≤  ⋀ _i∈IS _i  (x . y) = ( ⋀ _i∈IS _i ) (x . y). Hence ⋀_i∈IS _i is an L-subgroup of G.□

Proposition 3.2. Let (G, e) be an L-ordered group, S be an L-subset of (G, e) and 〈S〉 be the L-subgroup generated by S, the smallest L-subgroup of (G, e) that contains S. Then 〈 S 〉 = ⋀ { A ∈ L G | A is an L - subgroup of G and S ⊂ A }

Proof. By Lemma 3.1, the proof is obvious.□

Example 3.3. Let L = {0, a, 1} be a chain and ( ℤ , e ) be an L-ordered group, where e : ℤ × ℤ → L is defined by: e ( x , y ) = { a if x > y 1 if x ≤ y

Let S ∈ L ^G where S (2) = a and S (x) =0 for all x ≠ 2. Then for any x ∈ ℤ 〈 S 〉 ( x ) = { a if x = 2 n 0 if x = 2 n + 1

In the next theorem, we determine L-subgroup generated by S ∈ L ^G .

Theorem 3.4. Let (G, e) be an L-ordered group and S ∈ L ^G . Then 〈 S 〉 ( x ) = ⋁ x = x 1 . ⋯ . x n ( ⋀ j = 1 n ( S ( x j ) ∨ S ( x j - 1 ) ) )

Proof. It is obvious that 〈S〉 (x) = 〈S〉 (x^-1), for any x ∈ G. Now for any x, y ∈ G, we have 〈 S 〉 ( x ) ∧ 〈 S 〉 ( y ) = ⋁ x = x 1 . ⋯ . x n ( ⋀ i = 1 n ( S ( x i ) ∨ S ( x i - 1 ) ) ) ∧ ⋁ y = y 1 . ⋯ . y m ( ⋀ i = 1 m ( S ( y i ) ∨ S ( y i - 1 ) ) ) = ⋁ x = x 1 . ⋯ . x n ⋁ y = y 1 . ⋯ . y m ( ⋀ i = 1 n ( S ( x i ) ∨ S ( x i - 1 ) ) ) ∧ ( ⋀ i = 1 m ( S ( y i ) ∨ S ( y i - 1 ) ) ) = ⋁ x = x 1 . ⋯ . x n ⋁ y = y 1 . ⋯ . y m ( ⋀ i = 1 n + m ( S ( z i ) ∨ S ( z i - 1 ) ) ( z i = x i for 1 ≤ i ≤ n , o . w . z i = y i - n ) ≤ ⋁ x . y = z 1 . ⋯ . x k ( ⋀ i = 1 k S ( z i ) ∨ S ( z i - 1 ) ) = 〈 S 〉 ( x . y ) Moreover, for any x ∈ G S ( x ) ≤ S ( x ) ∨ S ( x - 1 ) ≤ ⋁ x = x 1 . ⋯ . x n ( ∧ j ( S ( x j ) ∨ S ( x j - 1 ) ) ) = 〈 S 〉 ( x )

So 〈S〉 contain S. Let S′ ∈ L ^G be an L-subgroup of G where S (x) ≤ S′ (x) for any x ∈ G. Then 〈 S 〉 ( x ) = ⋁ x = x 1 . ⋯ . x n ( ⋀ j = 1 n ( S ( x j ) ∨ S ( x j - 1 ) ) ) ≤ ⋁ x = x 1 . ⋯ . x n ( ⋀ j = 1 n ( S ′ ( x j ) ∨ S ′ ( x j - 1 ) ) ) = ⋁ x = x 1 . ⋯ . x n ( ⋀ j = 1 n ( S ′ ( x j ) ) ) ≤ ⋁ x = x 1 . ⋯ . x n ( S ′ ( x 1 . ⋯ . x n ) ) = ⋁ x = x 1 . ⋯ . x n ( S ′ ( x ) ) = S ′ ( x )

Hence 〈S〉 (x) ≤ S′ (x) for any x ∈ G and so 〈S〉 is the smallest L-subgroup of (G, e) that containing S.□

Corollary 3.5. Let (G, e) be an L-ordered group, S ∈ L ^G and supp (S) = {a}. Then 〈S〉 (x) = S (a) if and only if x = na, for some n ∈ ℤ.

Theorem 3.6. Let (G, e) be an L-lattice ordered group, S ∈ L ^G such that ⊔S exists and ⋁_y∈GS (y) =1. Then a ∧ ⊔ S = ⊔ (a ∧ S), for all a ∈ G.

Proof. Let u = ⊔ S. Then for all x ∈ G, e (u, x) =  ⋀ _y∈G (S (y) →e (y, x)). Define T : G→L by T (y) = (a ∧ u) . (a ∧ S) ^-1 (y), for all y ∈ G. From Proposition 2.6, a ∧ ⊔ S = ⊔ ( a ∧ S ) ⇔ ( a ∧ u ) . ( ⊔ ( a ∧ S ) ) - 1 = 1 ⇔ ( a ∧ u ) . ⊓ ( a ∧ S ) - 1 = 1 ⇔ ⊓ ( ( a ∧ u ) . ( a ∧ S ) - 1 ) = 1

Hence, It is enough to show that ⋀_y∈G (T (y) →e (x, y)) ≤ e (x, 1) and T (x) ≤ e (1, x), for all x ∈ G. For each y ∈ G, we have T ( y ) = ( a ∧ u ) . ( a ∧ S ) - 1 ( y ) = ( a ∧ S ) - 1 ( ( a ∧ u ) - 1 y ) = ( a ∧ S ) ( y - 1 ( a ∧ u ) ) = ⋁ { z ∈ G | a ∧ z = y - 1 ( a ∧ u ) } S ( z ) = ⋁ { z ∈ G | y = ( a ∧ u ) ( a ∧ z ) - 1 } S ( z ) .

Let v ∈ G. Since ⊔S = u, for all y ∈ G, S (y) ≤ e (y, u) and so for each y ∈ {z ∈ G| v = (a ∧ u) (a ∧ z) ^-1}, S ( y ) ≤ e ( y , u ) ≤ e ( y ∧ a , u ∧ a ) = e ( 1 , ( u ∧ a ) ( y ∧ a ) - 1 ) = e ( 1 , v )

Now, we show that ⋀_y∈G (T (y) →e (x, y)) ≤ e (x, 1).

For all x ∈ G, ⋀ y ∈ G ( T ( y ) → e ( x , y ) ) = ⋀ y ∈ G ( ( ⋁ { z ∈ G | y = ( a ∧ u ) ( a ∧ z ) - 1 } S ( z ) ) → e ( x , y ) ) = ⋀ y ∈ G ⋀ { z ∈ G | y = ( a ∧ u ) ( a ∧ z ) - 1 } ( S ( z ) → e ( x , y ) ) = ⋀ y ∈ G ( S ( y ) → e ( x , ( a ∧ u ) ( a ∧ y ) - 1 ) ) . Moreover, for all x ∈ G, ⋀ y ∈ G ( S ( y ) → e ( x , uy - 1 ( a ∧ y ) ) ) = ⋀ y ∈ G ( S ( y ) → e ( x , uy - 1 a ∧ u ) ) ) = ⋀ y ∈ G ( S ( y ) → ( e ( x , uy - 1 a ) ∧ e ( x , u ) ) ) = ⋀ y ∈ G ( S ( y ) → e ( x , uy - 1 a ) ) ∧ ⋀ y ∈ G ( S ( y ) → e ( x , u ) ) = ⋀ y ∈ G ( S ( y ) → e ( y , ux - 1 a ) ) ∧ e ( x , u ) , since ⋁ y ∈ G S ( y ) = 1 = e ( u , ux - 1 a ) ∧ e ( x , u ) , by Theorem 2.1 = e ( x , a ) ∧ e ( x , u ) = e ( x , a ∧ u ) , by Proposition 2.2 . Put x ∈ G. For all z ∈ G, e ( x ( a ∧ z ) , a ∧ u ) = ⋀ y ∈ G ( S ( y ) → e ( x ( a ∧ z ) , uy - 1 ( a ∧ y ) ) ) ≤ S ( z ) → e ( x ( a ∧ z ) , uz - 1 ( a ∧ z ) ) = S ( z ) → e ( x , uz - 1 ) which implies that  ⋀ _z∈G (S (y) →e (x (a ∧ z) , a ∧ u)) ≤  ⋀ _z∈G (S (y) →e (x, uz^-1)) = e (x, 1). Therefore, ⊓T = 1 and so a ∧ ⊔ S = ⊔ (a ∧ S).□

Theorem 3.7. Let (G, e) be an L-lattice ordered group, S ∈ L ^G such that ⊓S exists and ⋁_y∈GS (y) =1. Then a ∨ ⊓ S = ⊓ (a ∨ S), for all a ∈ G.

Proof. The proof is similar to the proof ofTheorem 3.6.□

Definition 3.8. Let (P, e) be an L-ordered set. An element α ∈ L is called e-strong if for each x, y ∈ P, α ≤ e (x, y) implies that x ≤ y.

Clearly, if (P, e) is an L-ordered set, then the greatest element of L that is 1, is an e-strong element. Moreover, if t ∈ L is an e-strong element and t ≤ e (x, y) ∧ e (y, x), then x = y.

In the next theorem, we define two kind of products of L-ordered groups and then we show that they are L-ordered groups, too.

Theorem 3.9. Let (G₁, e₁) and (G₂, e₂) be two L-ordered groups. Then, for any x₁, x₂ ∈ G₁ and x₂, y₂ ∈ G₂;

(G₁ × G₂, e₁ ∧ e₂) is an L-ordered group, where (e₁ ∧ e₂) ((x₁, y₁) , (x₂, y₂)) = e₁ (x₁, x₂)∧e₂ (y₁, y₂),

Proof (i) The proof of (E1) and (E3) is clear.

(E2) For any (x₁, y₁) , (x₂, y₂) , (x₃, y₃) ∈ G₁ × G₂, we have ( e 1 ∧ e 2 ) ( ( x 1 , y 1 ) , ( x 2 , y 2 ) ) ∧ ( e 1 ∧ e 2 ) ( ( x 2 , y 2 ) , ( x 3 , y 3 ) ) = e 1 ( x 1 , x 2 ) ∧ e 2 ( y 1 , y 2 ) ∧ e 1 ( x 2 , x 3 ) ∧ e 2 ( y 2 , y 3 ) ≤ e 1 ( x 1 , x 3 ) ∧ e 2 ( y 1 , y 3 ) = ( e 1 ∧ e 2 ) ( ( x 1 , y 1 ) , ( x 3 , y 3 ) )

So (G₁ × G₂, e₁ ∧ e₂) is an L-ordered set.

Also, for any (x₁, y₁) , (x₂, y₂) , (a₁, a₂) , (b₁, b₂) ∈ G₁ × G₂ ( e 1 ∧ e 2 ) ( ( x 1 , y 1 ) , ( x 2 , y 2 ) ) = e 1 ( x 1 , x 2 ) ∧ e 2 ( y 1 , y 2 ) = e 1 ( a 1 . x 1 . b 1 , a 2 . y 1 . b 2 ) ∧ e 2 ( a 1 . x 2 . b 1 , a 2 . y 2 . b 2 ) = ( e 1 ∧ e 2 ) ( ( a 1 , a 2 ) . ( x 1 , y 1 ) . ( b 1 , b 2 ) , ( a 1 , a 2 ) . ( x 2 , y 2 ) . ( b 1 , b 2 ) )

Hence, (G₁ × G₂, e₁ ∧ e₂) is an L-ordered group.

(ii) For any (x₁, y₁) , (x₂, y₂) , (x₃, y₃) ∈ G₁ × G₂,

(E1): (e₁ × e₂) ((x₁, y₁) , (x₁, y₁)) = (e₁ (x₁, x₁) , e₂ (y₁, allowbreaky₁)) = (1, 1)

(E2): ( e 1 ∧ e 2 ) ( ( x 1 , y 1 ) , ( x 2 , y 2 ) ) ∧ ( e 1 ∧ e 2 ) ( ( x 2 , y 2 ) , ( x 3 , y 3 ) ) = ( e 1 ( x 1 , x 2 ) , e 2 ( y 1 , y 2 ) ) ∧ ( e 1 ( x 2 , x 3 ) , e 2 ( y 2 , y 3 ) ) = ( e 1 ( x 1 , x 2 ) ∧ e 1 ( x 2 , x 3 ) , e 2 ( y 1 , y 2 ) ) ∧ e 2 ( y 2 , y 3 ) ) ≤ ( e 1 ( x 1 , x 3 ) , e 2 ( y 1 , y 3 ) ) = ( e 1 ∧ e 2 ) ( ( x 1 , y 1 ) , ( x 3 , y 3 ) ) (E3): Let (e₁ ∧ e₂) ((x₁, y₁) , (x₂, y₂) = (e₁ ∧ e₂) ((x₂, allowbreaky₂) , (x₁, y₁)) = (1, 1). Then (e₁ (x₁, x₂) , e₂ (y₁, y₂)) allowbreak = (1, 1) and (e₁ (x₂, x₁) , e₂ (y₂, y₁)) = (1, 1). So e₍x₁, x₂) = e₁ (x₂, x₁) =1 and so we have x₁ = x₂. By the similar way we can show that y₁ = y₂. Therefore, (x₁, y₁) = (x₂, y₂). Hence G₁ × G₂, e₁ × e₂) is an L × L-ordered set.

Also, for any (x₁, y₁) , (x₂, y₂) , (a₁, a₂) , (b₁, b₂) ∈ G₁ × G₂, we have ( e 1 ∧ e 2 ) ( ( x 1 , y 1 ) , ( x 2 , y 2 ) ) = ( e 1 ( x 1 , x 2 ) , e 2 ( y 1 , y 2 ) ) = ( e 1 ( a 1 . x 1 . b 1 , a 2 . y 1 . b 2 ) , e 2 ( a 1 . x 2 . b 1 , a 2 . y 2 . b 2 ) ) = ( e 1 ∧ e 2 ) ( ( a 1 , a 2 ) . ( x 1 , y 1 ) . ( b 1 , b 2 ) , ( a 1 , a 2 ) . ( x 2 , y 2 ) . ( b 1 , b 2 ) )

Therefore, (G₁ × G₂, e₁ × e₂) is an L × L-ordered group.□

Example 3.10. Let ( ℤ , e ) be an L-ordered group where e is the characteristic function of ≤ on ℤ and L = {0, 1}. Then ( ℤ × ℤ , e ∧ e ) is an L-ordered group that (e ∧ e) ((x₁, y₁) , (x₂, y₂)) =1 if and only if e (x₁, x₂) =1 and e (y₁, y₂) =1. Also ( ℤ × ℤ , e × e ) is an L × L-ordered group that e ∧ e is define as below: ( e ∧ e ) ( ( x 1 , y 1 ) , ( x 2 , y 2 ) ) = { ( 0 , 0 ) if x 1 ≰ x 2 and y 1 ≰ y 2 ( 0 , 1 ) if x 1 ≰ x 2 and y 1 ≤ y 2 ( 1 , 0 ) if x 1 ≤ x 2 and y 1 ≰ y 2 ( 1 , 1 ) if x 1 ≤ x 2 and y 1 ≤ y 2

Theorem 3.11. Let {(G _i , e _i )} _i∈I be a family of L-ordered groups where L be a finite chain. Then the ultraproduct of this family is an L-ordered group.

Proof. Let U be an ultra filter of the power set of I. Define the relation θ on G = Π_i∈IG _i by (f, g) ∈ θ ⇔ {i ∈ I|f (i) = g (i)} ∈ U. By [5, Example 1.3.29], we know that θ is a group congruence on G. Set G/U = {[f] | f ∈ G}, where [f] = {g ∈ G|  (f, g) ∈ θ}. Define e : G × G→L, by e (f, g) = ⋀ _i∈Ie _i  (f (i) , g (i)). Similar to the proof of Theorem 3.9(i), it can be easily shown that (G, e) is an L-ordered group. For each f, g ∈ G, let e ¯ ( [ f ] , [ g ] ) be defined by ⋁_α∈[1]e (α, gf^-1). We claim that e ¯ is an L-ordered group on G/U. Clearly (E1) holds. Let f, g, h ∈ G. Then e ¯ ( [ f ] , [ g ] ) ∧ e ¯ ( [ g ] , [ h ] ) = ( ⋁ α ∈ [ 1 ] e ( α , gf - 1 ) ) ∧ ( ⋁ β ∈ [ 1 ] e ( β , hg - 1 ) ) = ⋁ β ∈ [ 1 ] ( ( ⋁ α ∈ [ 1 ] e ( α , gf - 1 ) ) ∧ e ( β , hg - 1 ) ) = ⋁ β ∈ [ 1 ] ⋁ α ∈ [ 1 ] ( e ( α , gf - 1 ) ∧ e ( β , hg - 1 ) ) = ⋁ β ∈ [ 1 ] ⋁ α ∈ [ 1 ] ( e ( α f , g ) ∧ e ( g , β - 1 h ) ) ≤ ⋁ β ∈ [ 1 ] ⋁ α ∈ [ 1 ] ( e ( α f , β - 1 h ) ) = ⋁ β ∈ [ 1 ] ⋁ α ∈ [ 1 ] ( e ( β α , hf - 1 ) ) .

Let α, β ∈ [1]. Then A = {i ∈ I| f (i) =1} ∈ U, B = {j ∈ I| g (j) =1} ∈ U, and A ∩ B ∈ U (since U is a filter). Moreover, A ∩ B ⊆ {i ∈ I| βα (i) =1} and so by the assumption {i ∈ I| βα (i) =1} ∈ U, which implies that αβ ∈ [1]. It follows that [1] = {αβ| α, β ∈ [1]}. Hence ⋁ β ∈ [ 1 ] ⋁ α ∈ [ 1 ] ( e ( β α , hf - 1 ) ) = ⋁ α ∈ [ 1 ] ( e ( α , hf - 1 ) ) = e ¯ ( [ f ] , [ h ] ) and so (E3) hold. Now, let f, g ∈ G such that e ¯ ( [ f ] , [ g ] ) = e ¯ ( [ g ] , [ f ] ) = 1. Then ⋁_α∈[1]e (α, gf^-1) allowbreak = ⋁ _α∈[1]e (α, fg^-1) =1. Since L is a finite chain, there are α, β ∈ [1] such that e (α, gf^-1) = e (β, fg^-1) =1. Since α, β ∈ [1], we get A = {i ∈ I| α (i) =1} ∈ U and B = {i ∈ I| β (i) =1} ∈ U and so A ∩ B ∈ U (note that A∩ B ≠ ∅). For each i ∈ A ∩ B, we have 1 = e _i  (α (i) , gf^-1 (i)) = e _i  (1, g (i) f (i) ^-1) = e _i  (f (i) , g (i)). In a similar way, we can see that 1 = e _i  (g (i) , f (i)), for all i ∈ I. Hence f (i) = g (i), for each i ∈ A ∩ B and then A ∩ B ⊆ {i ∈ I| f (i) = g (i)}. Since A ∩ B ∈ U and U is a filter, {i ∈ I| f (i) = g (i)} ∈ U. Thus [f] = [g]. From definition of e ¯, it is clear that, e ¯ ( [ α ] [ f ] [ β ] , [ α ] [ g ] [ β ] ) = e ¯ ( [ f ] , [ g ] ). Therefore, ( G / U ; e ¯ ) is an L-ordered group.□

Theorem 3.12. Let (G, e) be an L-ordered group, S ∈ L ^G such that supp (S) = {a₁, a₂, . . . , a _n } and ⊔S exists. Then ⋁ i = 1 n S ( a i ) is an e-strong element.

Proof. Suppose that x, y ∈ G such that ⋁ i = 1 n S ( a i ) ≤ e ( x , y ) and ⊔S = b, for some b ∈ G. Then for all x ∈ G, e ( b , x ) = ⋀ i = 1 n ( S ( a i ) → e ( a i , x ) ). For all i ∈ {1, 2, . . . , n}, we have S ( a i ) ≤ ⋁ i = 1 n S ( a i ) ≤ e ( x , y ) = e ( b , b . x - 1 . y ) = ⋀ i = 1 n ( S ( a i ) → e ( a i , b . x - 1 . y ) ) ≤ S ( a i ) → e ( a i , b . x - 1 . y )

Hence S (a _i ) →e (a _i , b . x^-1 . y) =1 and so e ( x , y ) = ⋀ i = 1 n ( S ( a i ) → e ( a i , x ) ) = 1. Therefore, ⋁ i = 1 n S ( a i ) is an e-strong element.□

Corollary 3.13. Let (G, e) be an L-ordered group, S ∈ L ^G such that supp (S) = {a} and ⊔S exists. Then S (a) is an e-strong element.

In the following proposition, we want to get the converse of Corollary 3.13.

Proposition 3.14. Let L be a chain, (G, e) be an L-ordered group, α ∈ L be e-strong and S : G→L define by S (a) = α and S (x) =0, for all x ∈ G - {a}. Then ⊔S exists and is equal to a.

Proof. It is enough to show that α→e (a, x) = e (a, x), for all x ∈ G. Let x ∈ G. If e (a, x) ≥ α, then by the assumption a ≤ x and so e (a, x) =1. Moreover, α→e (a, x) =1. Hence α→e (a, x) =1 = e (a, x). If e (a, x) < α, then α→e (a, x) = ∨ {u ∈ L| α ∧ u ≤ e (a, x)} = e (a, x). Therefore, α→e (a, x) = e (a, x), for all x ∈ G and so by Theorem 2.1, ⊔S = a.□

[4] Totally L-ordered group

In this section, we introduce the concept of totally L-ordered groups and study some of their properties.

Definition 4.1. Let (G, e) and (G′, e′) be two L-ordered groups. The map f : (G, e) → (G′, e′) is called an L-ordered group morphism (or L-morphism) if f is both a group homomorphism and a monotone map, that is for any x, y ∈ G, f (x . y) = f (x) . f (y), f (x^-1) = (f (x)) ^-1 and e (f (x) , f (y)) ≥ e (x, y). The map f is called an L-ordered group isomorphism (or L-isomorphism) if there exists an L-morphism g : (G′, e′) → (G, e) such that f ∘ g = Id_G′ and g ∘ f = Id _G .

Remark 4.2. Let (G, e) and (G′, e′) be two L-ordered groups. Suppose that f : (G, e) → (G′, e′) is an L-isomorphism. Then there exist an L-morphism g : (G′, e′) → (G, e) such that g ∘ f = Id _G and so for all x, y ∈ G, we have e ( x , y ) ≤ e ( f ( x ) , f ( y ) ) ≤ e ( g ∘ f ( x ) , g ∘ f ( y ) ) = e ( x , y )

Therefore e (x, y) = e (f (x) , f (y)), for all x, y ∈ G. Moreover, if f : (G, e) → (G′, e′) is an L-morphism such that f is one to one, onto and e (f (x) , f (y)) = e (x, y), for all x, y ∈ G, then f^-1 : G′ → G is an L-isomorphism, too. In fact, for each x, y ∈ G, e (f^-1 (x) , f^-1 (y)) = e (f ∘ f^-1 (x) , f ∘ f^-1 (y)) = e (x, y).

Definition 4.3. Let (T, e) be an L-ordered group. We called (T, e) is a totally L-ordered group if for any x, y ∈ T, e (x, y) ∨ e (y, x) =1.

Example 4.4. Let L′ = {0, 1} be a lattice, L = L′ × L′ and ( ℤ , e ) be an L-ordered group such that e : ℤ × ℤ → L is defined by: e ( x , y ) = { ( 0 , 1 ) if x < y ( 1 , 0 ) if x > y ( 1 , 1 ) if x = y

Then ( ℤ , e ) is a totally L-ordered group. But it is not an L-lattice ordered group, because (Z, ≤  _e ) is not a lattice. In fact, x ≤  _e y if and only if x = y and join and meet is not exist.

Proposition 4.5. Every totally L-ordered group (T, e) that range of e is a chain, is an L-lattice ordered group.

Proof. For any x, y ∈ G, e (x, y) ∨ e (y, x) =1. Range of e is a chain, so e (x, y) =1 or e (y, x) =1. So (T, ≤  _e ) is a totally ordered group and so (T, ≤  _e ) is a lattice ordered group. Therefore (T, e) is an L-lattice ordered group.□

Proposition 4.6. Let (T, e) be an L-ordered group. Then (T, e) is a totally L-ordered group if and only if for any x ∈ T, e (x, 1 _T ) ∨ e (1 _T , x) =1.

Proof. It is obvious that if (T, e) is a totally L-ordered group, then for any x ∈ T, e (x, 1 _T ) ∨ e (1 _T , x) =1. Conversely, let for any x ∈ T, e (x, 1 _T ) ∨ e (1 _T , x) =1 and a, b ∈ T. Then e (a . b^-1, 1 _T ) ∨ e (1 _T , a . b^-1) =1 and so by Proposition 2.6, e (a, b) ∨ e (b, a) =1.□

Corollary 4.7. Let (T, e) be an L-ordered group and T = 〈χ_{a}〉 for a ∈ T. Then T is a totally L-ordered group if and only if e (1 _T , a) ∨ e (a, 1 _T ) =1.

Proof. By Proposition 4.2, it is obvious that if (T, e) is a totally L-ordered group, then e (1 _T , a) ∨ e (a, 1 _T ) =1. Now, let e (1 _T , a) ∨ e (a, 1 _T ) =1. By Corollary 3.5, for any x, y ∈ T, there exists n ∈ ℤ such that y - x = na. Hence e ( x , y ) ∨ e ( y , x ) = e ( 1 T , y - x ) ∨ e ( y - x , 1 T ) = e ( 1 T , na ) ∨ e ( na , 1 T ) ≥ ( e ( 1 T , a ) ∧ ⋯ ∧ e ( ( n - 1 ) a , na ) ) ∨ ( e ( na , ( n - 1 ) a ) ∧ ⋯ ∧ e ( a , 1 T ) ) = e ( 1 T , a ) ∨ e ( a , 1 T ) = 1

Therefore, (T, e) is a totally L-ordered group.□

Theorem 4.8. Let (T, e₁) and (H, e₂) be two totally L-ordered groups and e : (T × H) × (T × H) →L is defined by e ((x, y) , (x′, y′)) = e₁ (x, x′) ∧ e₂ (y, y′). Then (T × H, e) is a totally L-ordered group if and only if for any a ∈ Im (e₁) and b ∈ Im (e₂), a ∨ b = 1.

Proof. Let (T × H, e) be a totally L-ordered group, a ∈ Im (e₁) and b ∈ Im (e₂). Then there exist x, x′ ∈ T and y, y′ ∈ H such that a = e₁ (x, x′) and b = e₂ (y′, y). Since (T × H, e) is a totally L-ordered group, we have 1 = e ( ( x , y ) , ( x ′ , y ′ ) ) ∨ e ( ( x ′ , y ′ ) , ( x , y ) ) = ( e 1 ( x , x ′ ) ∧ e 2 ( y , y ′ ) ) ∨ ( e 1 ( x ′ , x ) ∧ e 2 ( y ′ , y ) ) = ( e 1 ( x , x ′ ) ∨ e 1 ( x ′ , x ) ) ∧ ( e 1 ( x , x ′ ) ∨ e 2 ( y ′ , y ) ) ∧ ( e 2 ( y , y ′ ) ∨ e 1 ( x ′ , x ) ) ∧ ( e 2 ( y , y ′ ) ∨ e 2 ( y ′ , y ) ) = 1 ∧ ( e 1 ( x , x ′ ) ∨ e 2 ( y ′ , y ) ) ∧ ( e 2 ( y , y ′ ) ∨ e 1 ( x ′ , x ) ) ∧ 1 = ( e 1 ( x , x ′ ) ∨ e 2 ( y ′ , y ) ) ∧ ( e 2 ( y , y ′ ) ∨ e 1 ( x ′ , x ) ) ≤ e 1 ( x , x ′ ) ∨ e 2 ( y ′ , y ) = a ∨ b and so a ∨ b = 1.

Conversely, if for any a ∈ Im (e₁) and b ∈ Im (e₂), a ∨ b = 1, then for any x, x′ ∈ T and y, y′ ∈ H e ( ( x , y ) , ( x ′ , y ′ ) ) ∨ e ( ( x ′ , y ′ ) , ( x , y ) ) = ( e 1 ( x , x ′ ) ∨ e 2 ( y ′ , y ) ) ∧ ( e 2 ( y , y ′ ) ∨ e 1 ( x ′ , x ) ) = 1 ∧ 1 = 1

Hence (T × H, e) is a totally L-ordered group.□

Definition 4.9. Let (T, e) be an L-ordered group. We called (T, e) is a strong totally L-ordered group if for any x, y ∈ T, e (x, y) =1 or e (y, x) =1.

Example 4.10. (i) Let L = {0, a, 1} be a frame and ( ℝ × ℝ , e ) be an L-ordered group such that e : ( ℝ × ℝ ) × ( ℝ × ℝ ) → L is defined by: e ( ( x , y ) , ( x ′ , y ′ ) ) = { 1 if x < x ′ 1 if x = x ′ , y ≤ y ′ a if x = x ′ , y ′ < y 0 if x > x ′

Then ( ℝ × ℝ , e ) is a strong totally L-ordered group. Clearly (E1), (E3) and (FOG) hold. For any x , x ′ , y , y ′ , x ″ , y ″ ∈ ℝ, if x < x″ or x = x″ and y ≤ y″ then e ((x, y) , (x″, y″)) =1 and e ((x, y) , (x′, y′)) ∧ e ((x′, y′) , (x″, y″)) ≤ e ((x, y) , (x″, y″)), if x = x″ and y″ < y, then e ((x, y) , (x″, y″)) = a, if e ((x, y) , (x′, y′)) ∧ e ((x′, y′) , (x″, y″)) =1, then x ≤ x′ and x′ ≤ x″ and so x = x′ = x″. Now, e ((x, y) , (x′, y′)) ∧ e ((x′, y′) , (x″, y″)) =1 and x = x′ = x″ implies y ≤ y′ and y′ ≤ y″. Hence, y ≤ y″ and so e ((x, y) , (x″, y″)) =1, that is a contradiction. Now, let e ((x, y) , (x″, y″)) =0. Then x > x″. If x ≤ x′ and x′ ≤ x″, then x ≤ x″, that is a contradiction. So x > x′ or x′ > x″. By definition of e, e ((x, y) , (x′, y′)) =0 or e ((x, y) , (x′, y′)) =0. Therefore (E3) holds. Now, let e ((x, y) , (x′, y′)) ≤1 for x, x′, y, y′ ∈ R. If x > x′, then e ((x′, y′) , (x, y)) =1. Now, let x = x′. Then y′ < y and so e ((x′, y′) , (x, y)) =1. Therefore, (T, e) is a strong totally L-ordered group.

(ii) Let L = {0, a, b, 1} be a bounded lattice such that a ∨ b = 1. Then (Z, e₁) and (Z, e₂) are strong totallyL-ordered groups, where e₁ and e₂ are defined as follows, for any m , n ∈ ℤ e 1 ( m , n ) = { 1 if m ≤ n a if m > n e 2 ( m , n ) = { 1 if m ≤ n b if m > n

Obviously (E1) and (E3) hold. For any m , n , k ∈ ℤ, if m ≤ k, then e₁ (m, k) =1. So (E2) hold. Now, let m > k. Then e₁ (m, k) = a. We have e ((1, 2) , (2, 1)) = e₁ (1, 2) ∧ e₂ (2, 1) =1 ∧ b = b ≠ 1. Also, e ((2, 1) , (1, 2)) = a ≠ 1. So ( ℤ × ℤ , e = e 1 ∧ e 2 ) is not a strong totally L-ordered group.

Lemma 4.11. Every strong totally L-ordered group is an L-lattice ordered group.

Proof. Let (G, e) be a strong totally L-ordered group. For any x, y ∈ G, e (x, y) =1 or e (y, x) =1. So x ≤  _e y or x ≤  _e y = x. So (G, ≤  _e ) is a chain. In any chain, join and meet are exist, so (G, ≤  _e ) is a lattice. Therefore (G, e) is an L-lattice ordered group.□

Let (T, e) be a totally L-ordered group. Set B (T) = {f : (T, e) → (T, e) |f isatotally L - isomorphism}.

Theorem 4.12. Let (T, e) be a totally L-ordered group. Define e ¯ : B ( T ) × B ( T ) → L by e ¯ ( f , g ) = ⋀ t ∈ T e ( f ( t ) , g ( t ) )

Then ( B ( T ) , e ¯ ) with the composition of functions is an L-ordered group. If (T, e) is a strong totally L-ordered group, then ( B ( T ) , e ¯ ) is an L-lattice ordered group.

Proof. First we show that (B (T) , ∘) is a group, where ∘ is a composition of functions. Let f, g ∈ B (T). Then for all x, y ∈ T, e ( x , y ) = e ( g ( x ) , g ( y ) ) = e ( f ( g ( x ) ) , f ( g ( y ) ) )

Hence f ∘ g is an L-isomorphism and so f ∘ g ∈ B (T). Also, for each f ∈ B (T), by Remark 1, f^-1 is an L-isomorphism and so f^-1 ∈ B (T). Hence (B (T) , ∘) is a group. Now we show that e ¯ is an L-order relation.

(E1): e ¯ ( f , f ) = ⋀ t ∈ T e ( f ( t ) , f ( t ) ) = ⋀ 1 = 1, for any f ∈ B (T).

(E2): Let f, g, h ∈ B (T). Then e ¯ ( f , g ) ∧ e ¯ ( g , h ) = ( ⋀ t ∈ T e ( f ( t ) , g ( t ) ) ) ∧ ( ⋀ t ∈ T e ( g ( t ) , h ( t ) ) ) = ⋀ t ∈ T ( e ( f ( t ) , g ( t ) ) ∧ e ( g ( t ) , h ( t ) ) ) ≤ ⋀ t ∈ T e ( f ( t ) , h ( t ) ) By ( E 3 ) = e ¯ ( f , h )

(E3): If e ¯ ( f , g ) = e ¯ ( g , f ) = 1, for f, g ∈ B (T), then ⋀_t∈Te (f (t) , g (t)) =1. Hence for any t ∈ T we have e (f (t) , g (t)) =1 and e (g (t) , f (t)) =1 and so f (t) = g (t), for any t ∈ T. Therefore, f = g and so e ¯ is an L-ordered relation. Now, for any f, g, h ∈ B (T), since h is onto, then for any s ∈ T there is t ∈ T such that h (t) = s. Hence e ¯ ( f ∘ h , g ∘ h ) = ⋀ t ∈ T e ( f ( h ( t ) ) , g ( h ( t ) ) ) = ⋀ s ∈ T e ( f ( s ) , g ( s ) ) = e ¯ ( f , g )