Open Access

On (λ, μ)-anti-fuzzy subgroups

Journal of Inequalities and Applications20122012:78

https://doi.org/10.1186/1029-242X-2012-78

Received: 19 December 2011

Accepted: 3 April 2012

Published: 3 April 2012

Abstract

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

Keywords

product(λ, μ)-fuzzysubgroupideal

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 [35].

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 a b μ ( A a A b ) λ
and
A c - 1 μ A c λ

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
A ( x - 1 y ) μ = A ( x - 1 y ) 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 .

Conversely, suppose A(x-1y) μ ≤ (A(x) A(y)) λ, x, y G, then

A(1) μ = A(x-1x) μA(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) μA(x) A(y) λ < α α λ = α λ = α. Note that αμ, 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) μA(x) A(y) λ, x G. If there exist x0, y0 G such that A x 0 - 1 y 0 μ = α > A x 0 A y 0 λ , then A(x0) < α, A(y0) < α and α (λ, μ]. Thus x0 A α and y0 A α . But A x 0 - 1 y 0 α , that is x 0 - 1 y 0 A α . This is a contradiction with that A(α)is a subgroup of G. Hence A(x-1 y) μA(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: G1G2 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 A y = inf x G 1 A x | f x = y , y G 2 .

Proof. If f −1(y1) = or f−1(y2) = for any y1, y2 G2, then f A y 1 - 1 y 2 μ 1 = f A y 1 f A y 2 λ .

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

For any y1, y2 G2, we have
f A y 1 - 1 y 2 μ = inf t G 1 A t | f t = y 1 - 1 y 2 μ = inf t G 1 A t μ | f t = y 1 - 1 y 2 inf x 1 , x 2 G 1 A x 1 - 1 x 2 μ | f x 1 = y 2 , f x 2 = y 2 inf x 1 , x 2 G 1 A x 1 A x 2 λ | f x 1 = y 1 , f x 2 = y 2 = ( inf x 1 s 1 A x 1 | f x 1 = y 1 inf x 2 S 1 A x 2 | f x 2 = y 2 ) λ = f A y 1 f A y 2 λ .

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

Theorem 4. Let f : G1G2 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 B x = B f x , x G 1 .
Proof. For any x1, x2 G1,
f - 1 B x 1 - 1 x 2 μ = B f x 1 - 1 x 2 μ = B f x 1 - 1 f x 2 μ B f x 1 B f x 2 λ = f - 1 B x 1 f - 1 B x 2 λ .

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
A × B x , y = A x B y , x , y G 1 × 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) μA(x) λ and B(a−1) μB(a) λ. For all (y, b) G1 × G2. We have
A × B x , a - 1 y , b μ = A × B x - 1 , a - 1 y , b μ = A x - 1 y B a - 1 b μ = A x - 1 y μ B a - 1 b μ A x A y λ B a B b λ = A x B a A y B b λ = A × B x , a A × B y , b λ .

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 1 1 μ B a λ , a G 2
and
B 1 2 μ A x λ , x 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) μA(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) μA(x) λ we obtain that μA(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. (1)

    If μA(x) λ for all x G1. Then

     

A(xy) μμA(x) λ ≤ (A(x) A(y)) λ

and A(11) μμA(x) λ.
  1. (2)
    If B(12) ≤ A(x) λ for all x G1. Then
    A x y μ A x y B 1 2 1 2 μ = A × B x y , 1 2 1 2 μ = A × B x , 1 2 y , 1 2 μ A × B x , 1 2 A × B y , 1 2 λ = A x B 1 2 A y B 1 2 λ = A x A y λ .
     
and
A 1 1 μ A 1 1 B 1 2 μ = A × B 1 1 , 1 2 μ A × B x , 1 2 λ = A x B 1 2 λ = A x λ .

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 .

Declarations

Acknowledgements

YF wished to thank Prof. Michela for her help with the language.

Authors’ Affiliations

(1)
School of Mathematics and Statistics, Chongqing Three Gorges University
(2)
School of Mathematics Science, Liaocheng University

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Copyright

© Feng and Yao; licensee Springer. 2012

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