On (λ, μ)-anti-fuzzy subgroups

Abstract

We introduced the notions of (λ, μ)-anti-fuzzy subgroups, studied some properties of them and discussed the product of them.

1 Introduction and preliminaries

Fuzzy sets was first introduced by Zadeh [1] and then the fuzzy sets have been used in the reconsideration of classical mathematics. Yuan et al. [2] introduced the concept of fuzzy subgroup with thresholds. A fuzzy subgroup with thresholds λ and μ is also called a (λ, μ)-fuzzy subgroup. Yao continued to research (λ, μ)-fuzzy normal subgroups, (λ, μ)-fuzzy quotient subgroups and (λ, μ)-fuzzy subrings in [3–5].

Shen researched anti-fuzzy subgroups in [6] and Dong [7] studied the product of anti-fuzzy subgroups.

By a fuzzy subset of a nonempty set X we mean a mapping from X to the unit interval 0[1]. If A is a fuzzy subset of X, then we denote A_(α)= {x ∈ X|A(x) < α} for all α ∈ 0[1].

Throughout this article, we will always assume that 0 ≤ λ < μ ≤ 1.

Let G, G₁, and G₂ always denote groups in the following. 1, 1₁, and 1₂ are identities of G, G₁, and G₂, respectively.

2 (λ, μ)-anti-fuzzy subgroups

Definition 1. A fuzzy set A of a group G is called a (λ, μ)-anti-fuzzy subgroup of G if ∀a, b, c ∈ G.

$$A\left(ab\right)\wedge \mu \le \left(A\left(a\right)\vee A\left(b\right)\right)\vee \lambda$$

and

$$A\left(c^{-\mathsf{\text{1}}}\right)\wedge \mu \le A\left(c\right)\vee \lambda$$

where c^-1 is the inverse element of c.

Proposition 1. If A is a (λ, μ)-anti-fuzzy subgroup of a group G, then A(1) ∧ μ ≤ A(x) ∨ λ for all × ∈ G, where 1 is the identity of G.

Proof. ∀x ∈G and let x⁻¹ be the inverse element of x. Then A(1) ∧ μ = A(xx⁻¹) ∧ μ = (A(xx⁻¹) ∧ μ) ∧ μ ≤ ((A(x) ∨ A(x⁻¹)) ∨ λ) ∧ μ = (A(x) ∧ μ) ∨ (A(x⁻¹) ∧ μ) ∨ (λ ∧ μ) ≤ A(x) ∨ (A(x) ∨ λ) ∨ λ = A(x) ∨ λ.

Theorem 1. Let A be a fuzzy subset of a group G. Then A is a (λ, μ)-anti-fuzzy subgroup of G if and only if A(x^-1y) ∧ μ ≤ (A(x) ∨ A(y)) ∨ λ, ∀x, y ∈ G.

Proof. Let A is a (λ, μ)-anti-fuzzy subgroup of G, then

$$\begin{array}{c}A\left(x^{-1}y\right)\wedge \mu =A\left(x^{-1}y\right)\wedge \mu \wedge \mu \\ \le \left(\left(A\left(x^{-1}\right)\vee A\left(y\right)\right)\vee \lambda \right)\wedge \mu \\ =\left(A\left(x^{-1}\right)\wedge \mu \vee A\left(y\right)\right)\vee \left(\lambda \wedge \mu \right)\\ \le \left(\left(A\left(x\right)\vee \lambda \right)\vee A\left(y\right)\right)\vee \lambda \\ =\left(A\left(x\right)\vee A\left(y\right)\right)\vee \lambda .\end{array}$$

Conversely, suppose A(x^-1y) ∧ μ ≤ (A(x) ∨ A(y)) ∨ λ, ∀x, y ∈ G, then

A(1) ∧ μ = A(x^-1x) ∧ μ ≤ A(x) ∨ A(x) ∨ λ = A(x) ∨ λ.

So

A(x⁻¹) ∧ μ = A(x⁻¹1) ∧ μ = A(x⁻¹1) ∧ μ ∧ μ ≤ (A(x) ∨ A(1) ∨ λ) ∧ μ = (A(1) ∧ μ) ∨ ((A(x) ∨ λ) ∧ μ) ≤ (A(x) ∨ λ) ∨ ((A(x) ∨ λ) ∧ μ) = A(x) ∨ λ.

A(xy) ∧ μ = A((x^-1)^-1 y) ∧ μ = A((x^-1)^-1y) ∧ μ ∧ μ ≤ (A(x⁻¹) ∨ A(y) ∨ λ) ∧ μ = (A(x⁻¹) ∧ μ) ∨ ((A(y) ∨ λ) ∧ μ) ≤ (A(x) ∨ λ) ∨ (A(y) ∨ λ) = (A(x) ∨ A(y)) ∨ λ.

So A is a (λ, μ)-anti-fuzzy subgroup of G.

Theorem 2. Let A be a fuzzy subset of a group G. Then the following are equivalent:

(1) A is a (λ, μ)-anti-fuzzy subgroup of G;

(2) A_(α)is a subgroup of G, for any α ∈ (λ, μ], where A_(α)≠ ∅.

Proof. "(1) ⇒ (2)"

Let A be a (λ, μ)-anti-fuzzy subgroup of G. For any α ∈ (λ, μ], such that A _α ≠ ∅, we need to show that x⁻¹ y ∈ A_(α), for all x,y ∈ A_(α).

Since A(x) < α and A(y) < α, Then A(x⁻¹ y) ∧ μ ≤ A(x) ∨ A(y) ∨ λ < α ∨ α ∨ λ = α ∨ λ = α. Note that α ≤ μ, we obtain A(x⁻¹ y) < α. So x⁻¹ y ∈ A_(α).

"(2) ⇒ (1)"

Conversely, let A_(α)is a subgroup of G. We need to prove that A(x⁻¹ y) ∧ μ ≤ A(x) ∨ A(y) ∨ λ, ∀x ∈ G. If there exist x₀, y₀ ∈ G such that $A\left(x_0^{-1}{y}_0\right)\wedge \mu =\alpha >A\left(x_0\right)\vee A\left(y_0\right)\vee \lambda$, then A(x₀) < α, A(y₀) < α and α ∈ (λ, μ]. Thus x₀ ∈ A _α and y₀ ∈ A _α . But $A\left(x_0^{-1}{y}_0\right)\ge \alpha ,\mathsf{\text{that}}\mathsf{\text{is}}{x}_0^{-1}{y}_0\notin A_{\left(\alpha \right)}.$ This is a contradiction with that A_(α)is a subgroup of G. Hence A(x^-1 y) ∧ μ ≤ A(x) ∨ A(y) ∨ λ holds for any x, y ∈ G.

Therefore, A is a (λ, μ)-anti-fuzzy subgroup of G.

We set inf ∅ = 1, where ∅ is the empty set.

Theorem 3. Let f: G₁ → G₂ be a homomorphism and let A be a (λ, μ)-anti-fuzzy subgroup of G₁. Then f(A) is a (λ, μ)-anti-fuzzy subgroup of G₂, where

$$f\left(A\right)\left(y\right)=\underset{x\in G_1}{\text{inf}}\left\{A\left(x\right)|f\left(x\right)=y\right\},{\forall }_y\in G_2.$$

Proof. If f ⁻¹(y₁) = ∅ or f⁻¹(y₂) = ∅ for any y₁, y₂ ∈ G₂, then $\left(f\left(A\right)\left(y_1^{-1}{y}_2\right)\right)\wedge \mu \le 1=\left(f\left(A\right)\left(y_1\right)\vee f\left(A\right)\left(y_2\right)\right)\vee \lambda .$

Suppose that f⁻¹(y₁) ≠ ∅, f⁻¹(y₂) = ∅ for any y₁, y₂ ∈ G₂. Then

For any y₁, y₂ ∈ G₂, we have

$$\begin{array}{c}f\left(A\right)\left(y_1^{-1}{y}_2\right)\wedge \mu =\underset{t\in G_1}{\text{inf}}\left\{A\left(t\right)|f\left(t\right)=y_1^{-1}{y}_2\right\}\wedge \mu \\ =\underset{t\in G_1}{\text{inf}}\left\{\left(A\left(t\right)\right)\wedge \mu |f\left(t\right)=y_1^{-1}{y}_2\right\}\\ \le \underset{x_1,x_2\in G_1}{\text{inf}}\left\{\left(A\left(x_1^{-1}{x}_2\right)\right)\wedge \mu |f\left(x_1\right)=y_2,f\left(x_2\right)=y_2\right\}\\ \le \underset{x_1,x_2\in G_1}{\text{inf}}\left\{\left(A\left(x_1\right)\vee A\left(x_2\right)\right)\vee \lambda |f\left(x_1\right)=y_1,f\left(x_2\right)=y_2\right\}\\ =\left(\underset{x_1\in s_1}{\text{inf}}\left\{A\left(x_1\right)|f\left(x_1\right)=y_1\right\}\vee \underset{x_2\in S_1}{\text{inf}}\left\{A\left(x_2\right)|f\left(x_2\right)=y_2\right\}\right)\vee \lambda \\ =\left(f\left(A\right)\left(y_1\right)\vee f\left(A\right)\left(y_2\right)\right)\vee \lambda .\end{array}$$

So, f(A) is a (λ, μ)-anti-fuzzy subgroup of G₂.

Theorem 4. Let f : G₁ → G₂ be a homomorphism and let B be a (λ, μ)-anti-fuzzy subgroup of G₂. Then f⁻¹(B) is a (λ, μ)-anti-fuzzy subgroup of G₁, where

$$f^{-1}\left(B\right)\left(x\right)=B\left(f\left(x\right)\right),{\forall }_x\in G_1.$$

Proof. For any x₁, x₂ ∈ G₁,

$$\begin{array}{c}{f}^{-1}\left(B\right)\left(x_1^{-1}{x}_2\right)\wedge \mu =B\left(f\left(x_1^{-1}{x}_2\right)\right)\wedge \mu \\ =B\left({\left(f\left(x_1\right)\right)}^{-1}f\left(x_2\right)\right)\wedge \mu \\ \le \left(B\left(f\left(x_1\right)\right)\vee B\left(f\left(x_2\right)\right)\right)\vee \lambda \\ =\left(f^{-1}\left(B\right)\left(x_1\right)\vee f^{-1}\left(B\right)\left(x_2\right)\right)\vee \lambda .\end{array}$$

So, f⁻¹(B) is a (λ, μ)-anti-fuzzy subgroup of G₁.

Let G₁ be a group with the identity 1₁ and G₂ be a group with the identity 1₂, then G₁ × G₂ is a group with the identity (1₁, 1₂) if we define (x₁, y₁) (x₂, y₂) = (x₁x₂, y₁y₂) for all (x₁, y₁), (x₂, y₂) ∈ G₁ × G₂. Moreover, the inverse element of any (x, a) ∈ G₁ × G₂ is (y, b) ∈ G₁ × G₂ if and only if y is the inverse element of x in G₁ and b is the inverse element of a in G₂.

Theorem 5. Let A, B be two (λ, μ)-anti-fuzzy subgroups of groups G₁ and G₂, respectively. The product of A and B, denoted by A × B, is a (λ, μ)-anti-fuzzy subgroup of the group G₁ × G₂, where

$$\left(A\times B\right)\left(x,y\right)=A\left(x\right)\vee B\left(y\right),\forall \left(x,y\right)\in G_1\times G_2.$$

Proof. Let (x⁻¹, a⁻¹) be the inverse element of (x, a) in G₁ × G₂. Then x⁻¹ is the inverse element of x in G₁ and a⁻¹ is the inverse element of a in G₂. Hence A(x⁻¹) ∧ μ ≤ A(x) ∨ λ and B(a⁻¹) ∧ μ ≤ B(a) ∨ λ. For all (y, b) ∈ G₁ × G₂. We have

$$\begin{array}{c}\left(\left(A\times B\right){\left(x,a\right)}^{-1}\left(y,b\right)\right)\wedge \mu =\left(\left(A\times B\right)\left(x^{-1},a^{-1}\right)\left(y,b\right)\right)\wedge \mu \\ =\left(A\left(x^{-1}y\right)\vee B\left(a^{-1}b\right)\right)\wedge \mu \\ =\left(A\left(x^{-1}y\right)\wedge \mu \right)\vee \left(B\left(a^{-1}b\right)\wedge \mu \right)\\ \le \left(A\left(x\right)\vee A\left(y\right)\vee \lambda \right)\vee \left(B\left(a\right)\vee B\left(b\right)\vee \lambda \right)\\ =\left(A\left(x\right)\vee B\left(a\right)\right)\vee \left(A\left(y\right)\vee B\left(b\right)\right)\vee \lambda \\ =\left(\left(A\times B\right)\left(x,a\right)\right)\vee \left(\left(A\times B\right)\left(y,b\right)\right)\vee \lambda .\end{array}$$

Hence A × B is a (λ, μ)-anti-fuzzy subgroup of G₁ × G₂.

Theorem 6. Let A and B be two fuzzy subsets of groups G₁ and G₂, respectively. If A × B is a (λ, μ)-anti-fuzzy subgroup of G₁ × G₂, then at least one of the following statements must hold.

$$A\left(1_1\right)\wedge \mu \le B\left(a\right)\vee \lambda ,{\forall }_a\in G_2$$

and

$$B\left(1_2\right)\wedge \mu \le A\left(x\right)\vee \lambda ,\forall x\in G_1.$$

Proof. Let A × B be a (λ, μ)-anti-fuzzy subgroup of the group G₁ × G₂.

By contraposition, suppose that none of the statements hold. Then we can find x ∈ G₁ and a ∈ G₂ such that A(x) ∨ λ < B(1₂) ∧ μ and B(a) ∨ λ < A(1₁) ∧ μ. Now

(A×B) (x, a) ∨ λ = (A(x)∨B(a))∨λ = (A(x)∨λ)∨(B(a)∨λ) < (A(1₁)∧μ) ∨ (B(1₂)∧μ) = (A×B) (1₁,1₂) ∧ μ.

Thus A × B is a (λ, μ)-anti-fuzzy subgroup of the group G₁ × G₂ satisfying (A × B)(x, a) ∨ λ < (A × B) (1₁, 1₂) ∧ μ. This is a contradict with that (1₁, 1₂) iss the identity of G₁ × G₂ .

Theorem 7. Let A and B be fuzzy subsets of groups G₁ and G₂, respectively, such that B(1₂) ∧μ ≤ A(x) ∨λ for all × ∈ G₁. If A × B is a (λ, μ)-anti-fuzzy subgroup of G₁ × G₂, then A is a (λ, μ)-anti-fuzzy subgroup of G₁ .

Proof. From B(1₂) ∧ μ ≤ A(x) ∨ λ we obtain that μ ≤ A(x) ∨ λ or B(1₂) ≤ A(x) ∨ λ, for all x ∈ G₁.

Let x, y ∈ G₁, then (x, 1₂), (y, 1₂) ∈ G₁ × G₂.

Two cases are possible:

• (1) If μ ≤ A(x) ∨ λ for all x ∈ G₁. Then

A(xy) ∧ μ ≤ μ ≤ A(x) ∨ λ ≤ (A(x) ∨ A(y)) ∨ λ

and A(1₁) ∧ μ ≤ μ ≤ A(x) ∨ λ.

• (2) If B(1₂) ≤ A(x) ∨ λ for all x ∈ G₁. Then

  $$\begin{array}{c}A\left(xy\right)\wedge \mu \le \left(A\left(xy\right)\vee B\left(1_2{1}_2\right)\right)\wedge \mu \\ =\left(\left(A\times B\right)\left(xy,1_2{1}_2\right)\right)\wedge \mu \\ =\left(\left(A\times B\right)\left(\left(x,1_2\right)\left(y,1_2\right)\right)\right)\wedge \mu \\ \le \left(\left(A\times B\right)\left(x,1_2\right)\vee \left(A\times B\right)\left(y,1_2\right)\right)\vee \lambda \\ =A\left(x\right)\vee B\left(1_2\right)\vee A\left(y\right)\vee B\left(1_2\right)\vee \lambda \\ =\left(A\left(x\right)\vee A\left(y\right)\right)\vee \lambda .\end{array}$$

and

$$\begin{array}{c}A\left(1_1\right)\wedge \mu \le \left(A\left(1_1\right)\vee B\left(1_2\right)\right)\wedge \mu \\ =\left(\left(A\times B\right)\left(1_1,1_2\right)\right)\wedge \mu \\ \le \left(A\times B\right)\left(x,1_2\right)\vee \lambda \\ =A\left(x\right)\vee B\left(1_2\right)\vee \lambda \\ =A\left(x\right)\vee \lambda .\end{array}$$

Hence A is a (λ, μ)-anti-fuzzy subgroup of G₁.

Analogously, we have

Theorem 8. Let A and B be fuzzy subsets of groups G₁ and G₂, respectively, such that A(1₁) ∧ μ ≤ B(a) ∨ λ for all a ∈ G₂. If A × B is a (λ, μ)-anti-fuzzy subgroup of G₁ × G₂, then B is a (λ, μ)-anti-fuzzy subgroup of G₂ .

From the previous theorems, we have the following corollary

Corollary 1. Let A and B be fuzzy subsets of groups G₁ and G₂, respectively. If A × B is a (λ, μ)-anti-fuzzy subgroup of G₁ × G₂, then either A is a (λ, μ)-anti-fuzzy subgroup of G₁ or B is a (λ, μ)-anti-fuzzy subgroup of G₂ .