Research ArticleProduct of the Generalized 𝐿-Subgroups

Dilek Bayrak and Sultan Yamak

Department of Mathematics, Namik Kemal University, 59000 Tekirdag, Turkey

Correspondence should be addressed to Dilek Bayrak; dbayrak@nku.edu.tr

Received 6 July 2015; Accepted 8 February 2016

Academic Editor: Ali Jaballah

Copyright © 2016 D. Bayrak and S. Yamak. This is an open access article distributed under the Creative Commons AttributionLicense, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properlycited.

We introduce the notion of (𝜆, 𝜇)-product of 𝐿-subsets. We give a necessary and sufficient condition for (𝜆, 𝜇)-𝐿-subgroup of aproduct of groups to be (𝜆, 𝜇)-product of (𝜆, 𝜇)-𝐿-subgroups.

1. Introduction

Starting from 1980, by the concept of quasi-coincidence ofa fuzzy point with a fuzzy set given by Pu and Liu [1], thegeneralized subalgebraic structures of algebraic structureshave been investigated again. Using the concept mentionedabove, Bhakat and Das [2, 3] gave the definition of (𝛼, 𝛽)-fuzzy subgroup, where 𝛼, 𝛽 are any two of {∈, 𝑞, ∈∨𝑞, ∈ ∧𝑞}with 𝛼 =∈∧𝑞. The (∈, ∈∨𝑞)-fuzzy subgroup is an importantand useful generalization of fuzzy subgroups that were laidby Rosenfeld in [4]. After this, many other researchers usedthe idea of the generalized fuzzy sets that give several charac-terization results in different branches of algebra (see [5–10]).In recent years,many researchersmake generalizationswhichare referred to as (𝜆, 𝜇)-fuzzy substructures and (∈𝜆, ∈𝜆∨𝑞𝜇)-fuzzy substructures on this topic (see [11–15]).

Identifying the subgroups of a Cartesian product ofgroups plays an essential role in studying group theory. Manyimportant results on characterization of Cartesian product ofsubgroups, fuzzy subgroups, and𝑇𝐿-fuzzy subgroups exist inliterature. Chon obtained a necessary and sufficient conditionfor a fuzzy subgroup of a Cartesian product of groups tobe product of fuzzy subgroups under minimum operation[16]. Later, some necessary and sufficient conditions for a𝑇𝐿-subgroup of a Cartesian product of groups to be a 𝑇-product of 𝑇𝐿-subgroups were given by Yamak et al. [17]. Asubgroup of a Cartesian product of groups is characterizedby subgroups in the same study. Consequently, it seems to beinteresting to extend this study to generalized 𝐿-subgroups.In this paper, we introduce the notion of the (𝜆, 𝜇)-product

of 𝐿-subsets and investigate some properties of the (𝜆, 𝜇)-product of 𝐿-subgroups. Also, we give a necessary andsufficient condition for (𝜆, 𝜇)-𝐿-subgroup of a Cartesianproduct of groups to be a product of (𝜆, 𝜇)-𝐿-subgroups.

2. Preliminaries

In this section, we start by giving some known definitions andnotations. Throughout this paper, unless otherwise stated, 𝐺always stands for any given groupwith amultiplicative binaryoperation, an identity 𝑒 and 𝐿 denote a complete lattice withtop and bottom elements 1, 0, respectively.

An 𝐿-subset of𝑋 is any function from𝑋 into 𝐿, which isintroduced by Goguen [18] as a generalization of the notionof Zadeh’s fuzzy subset [19]. The class of 𝐿-subsets of 𝑋 willbe denoted by 𝐹(𝑋, 𝐿). In particular, if 𝐿 = [0, 1], then it isappropriate to replace 𝐿-subset with fuzzy subset. In this casethe set of all fuzzy subsets of 𝑋 is denoted by 𝐹(𝑋). Let 𝐴and 𝐵 be 𝐿-subsets of 𝑋. We say that 𝐴 is contained in 𝐵 if𝐴(𝑥) ≤ 𝐵(𝑥) for every 𝑥 ∈ 𝑋 and is denoted by 𝐴 ≤ 𝐵. Then≤ is a partial ordering on the set 𝐹(𝑋, 𝐿).

Definition 1 (see [20]). An 𝐿-subset of 𝐺 is called an 𝐿-subgroup of 𝐺 if, for all 𝑥, 𝑦 ∈ 𝐺, the following conditionshold:(G1) 𝐴(𝑥) ∧ 𝐴(𝑦) ≤ 𝐴(𝑥𝑦).(G2) 𝐴(𝑥) ≤ 𝐴(𝑥−1).In particular, when 𝐿 = [0, 1], an 𝐿-subgroup of 𝐺 is

referred to as a fuzzy subgroup of 𝐺.

Hindawi Publishing CorporationJournal of MathematicsVolume 2016, Article ID 4918948, 5 pageshttp://dx.doi.org/10.1155/2016/4918948

2 Journal of Mathematics

Definition 2 (see [12]). Let 𝜆, 𝜇 ∈ 𝐿 and 𝜆 < 𝜇. Let 𝐴 be an𝐿-subset of 𝐺. 𝐴 is called (𝜆, 𝜇)-𝐿-subgroup of 𝐺 if, for all𝑥, 𝑦 ∈ 𝐺, the following conditions hold:

(i) 𝐴(𝑥𝑦) ∨ 𝜆 ≥ 𝐴(𝑥) ∧ 𝐴(𝑦) ∧ 𝜇.

(ii) 𝐴(𝑥−1) ∨ 𝜆 ≥ 𝐴(𝑥) ∧ 𝜇.

Denote by 𝐹𝑆(𝜆, 𝜇, 𝐺, 𝐿) the set of all (𝜆, 𝜇)-𝐿-subgroups of𝐺. When 𝐿 = [0, 1], its counterpart is written as 𝐹𝑆(𝜆, 𝜇, 𝐺).

Unless otherwise stated, 𝐿 always represents any givendistributive lattice.

3. Product of (𝜆,𝜇)-𝐿-Subgroups

Definition 3 (see [16]). Let 𝐴 𝑖 be an 𝐿-subset of 𝐺𝑖 for each𝑖 = 1, 2, . . . , 𝑛. Then product of 𝐴 𝑖 (𝑖 = 1, 2, . . . , 𝑛) denotedby 𝐴1 × 𝐴2 × ⋅ ⋅ ⋅ × 𝐴𝑛 is defined to be the 𝐿-subset of 𝐺1 ×𝐺2 × ⋅ ⋅ ⋅ × 𝐺𝑛 that satisfies

(𝐴1 × 𝐴2 × ⋅ ⋅ ⋅ × 𝐴𝑛) (𝑥1, 𝑥2, . . . , 𝑥𝑛)

= 𝐴1 (𝑥1) ∧ 𝐴2 (𝑥2) ∧ ⋅ ⋅ ⋅ ∧ 𝐴𝑛 (𝑥𝑛) .(1)

Example 4. We define the fuzzy subsets𝐴 and 𝐵 ofZ andZ2,respectively, as follows:

𝐴 (𝑥) ={

{

{

0.5, 𝑥 ∈ 2Z,

0.3, otherwise,

𝐵 (𝑥) ={

{

{

0.7, 𝑥 = 0,

0.2, 𝑥 = 1.

(2)

We obtain that

𝐴 × 𝐵 (𝑥) =

{{{{

{{{{

{

0.5, 𝑥 ∈ 2Z × {0} ,

0.3, 𝑥 ∈ Z − {2Z} × {0} ,

0.2, otherwise.

(3)

Theorem5. Let𝐺1, 𝐺2, . . . , 𝐺𝑛 be groups and⋁𝑛

𝑖=1𝜆𝑖 < ⋀

𝑛

𝑖=1𝜇𝑖

such that𝐴 𝑖 is (𝜆𝑖, 𝜇𝑖)-𝐿-subgroup of𝐺𝑖 for each 𝑖 = 1, 2, . . . , 𝑛.Then 𝐴1 × 𝐴2 × ⋅ ⋅ ⋅ × 𝐴𝑛 is (⋁

𝑛

𝑖=1𝜆𝑖, ⋀𝑛

𝑖=1𝜇𝑖)-𝐿-subgroup of

𝐺1 × 𝐺2 × ⋅ ⋅ ⋅ × 𝐺𝑛.

Proof. Let (𝑥1, 𝑥2, . . . , 𝑥𝑛), (𝑦1, 𝑦2, . . . , 𝑦𝑛) ∈ 𝐺1×𝐺2×⋅ ⋅ ⋅×𝐺𝑛.Then𝐴1 × 𝐴2 × ⋅ ⋅ ⋅ × 𝐴𝑛 ((𝑥1, 𝑥2, . . . , 𝑥𝑛) ⋅ (𝑦1, 𝑦2, . . . , 𝑦𝑛))

∨

𝑛

⋁𝑖=1

𝜆𝑖 = 𝐴1 × 𝐴2 × ⋅ ⋅ ⋅

× 𝐴𝑛 (𝑥1𝑦1, 𝑥2𝑦2, . . . , 𝑥𝑛𝑦𝑛) ∨ 𝜆1 ∨

𝑛

⋁𝑖=1

𝜆𝑖

= (𝐴1 (𝑥1𝑦1) ∧ 𝐴2 (𝑥2𝑦2) ∧ ⋅ ⋅ ⋅ ∧ 𝐴𝑛 (𝑥𝑛𝑦𝑛))

∨

𝑛

⋁𝑖=1

𝜆𝑖 ≥ (𝐴1 (𝑥1𝑦1) ∨ 𝜆1) ∧ (𝐴2 (𝑥2𝑦2) ∨ 𝜆2)

∧ ⋅ ⋅ ⋅ ∧ (𝐴𝑛 (𝑥𝑛𝑦𝑛) ∨ 𝜆𝑛) ≥ 𝐴1 (𝑥1) ∧ 𝐴1 (𝑦1) ∧ 𝜇1

∧ 𝐴2 (𝑥2) ∧ 𝐴2 (𝑦2) ∧ 𝜇2 ∧ ⋅ ⋅ ⋅ ∧ 𝐴𝑛 (𝑥𝑛) ∧ 𝐴𝑛 (𝑦𝑛)

∧ 𝜇𝑛 = 𝐴1 × 𝐴2 × ⋅ ⋅ ⋅ × 𝐴𝑛 (𝑥1, 𝑥2, . . . , 𝑥𝑛) ∧ 𝐴1

× 𝐴2 × ⋅ ⋅ ⋅ × 𝐴𝑛 (𝑦1, 𝑦2, . . . , 𝑦𝑛) ∧

𝑛

⋀𝑖=1

𝜇𝑖.

(4)

Similarly, it can be shown that

𝐴1 × 𝐴2 × ⋅ ⋅ ⋅ × 𝐴𝑛 ((𝑥1, 𝑥2, . . . , 𝑥𝑛)−1) ∨

𝑛

⋁𝑖=1

𝜆𝑖

≥ 𝐴1 × 𝐴2 × ⋅ ⋅ ⋅ × 𝐴𝑛 (𝑥1, 𝑥2, . . . , 𝑥𝑛) ∧

𝑛

⋀𝑖=1

𝜇𝑖.

(5)

Corollary 6. If 𝐴1, 𝐴2, . . . , 𝐴𝑛 are (𝜆, 𝜇)-𝐿-subgroups of𝐺1, 𝐺2, . . . , 𝐺𝑛, respectively, then 𝐴1 × 𝐴2 × ⋅ ⋅ ⋅ × 𝐴𝑛 is (𝜆, 𝜇)-𝐿-subgroup of 𝐺1 × 𝐺2 × ⋅ ⋅ ⋅ × 𝐺𝑛.

Theorem 7 (Theorem 2.9, [16]). Let 𝐺1, 𝐺2, . . . , 𝐺𝑛 begroups, let 𝑒1, 𝑒2, . . . , 𝑒𝑛 be identities, respectively, and let𝐴 be a fuzzy subgroup in 𝐺1 × 𝐺2 × ⋅ ⋅ ⋅ × 𝐺𝑛. Then𝐴(𝑒1, 𝑒2, . . . , 𝑒𝑖−1, 𝑥𝑖, 𝑒𝑖+1, . . . , 𝑒𝑛) ≥ 𝐴(𝑥1, 𝑥2, . . . , 𝑥𝑛) for 𝑖 =1, 2, . . . , 𝑘−1, 𝑘+1, . . . , 𝑛 if and only if𝐴 = 𝐴1×𝐴2×⋅ ⋅ ⋅×𝐴𝑛,where 𝐴1, 𝐴2, . . . , 𝐴𝑛 are fuzzy subgroups of 𝐺1, 𝐺2, . . . , 𝐺𝑛,respectively.

The following example shows that Theorem 7 may not betrue for any (𝜆, 𝜇)-𝐿-subgroup.

Example 8. Consider

𝐴 (𝑥) =

{{{{{{{

{{{{{{{

{

0.8, 𝑥 = (0, 0) ,

0.7, 𝑥 = (1, 0) ,

0.6, 𝑥 = (0, 1) ,

0.5, 𝑥 = (1, 1) .

(6)

It is easy to see that 𝐴 is (0, 0.5)-fuzzy subgroup of Z2 × Z2.Since 𝐴(1, 0) ≥ 𝐴(1, 1) and 𝐴(0, 1) ≥ 𝐴(1, 1), 𝐴 satisfies the

Journal of Mathematics 3

condition of Theorem 7, but there do not exist 𝐴1, 𝐴2 fuzzysubgroups of Z2 × Z2 such that 𝐴 = 𝐴1 × 𝐴2.

In fact, suppose that there exist 𝐴1, 𝐴2 ∈ 𝐹𝑆(0, 0.5,Z2 ×Z2) such that 𝐴 = 𝐴1 ×𝐴2. Since 𝐴(1, 0) = 0.7 and 𝐴(0, 1) =0.6, we have 𝐴1(1) ≥ 0.7 and 𝐴2(1) ≥ 0.6. Hence

0.5 = 𝐴1 × 𝐴2 (1, 1) = 𝐴1 (1) ∧ 𝐴2 (1) ≥ 0.7 ∧ 0.6

= 0.6,

(7)

a contradiction.

Definition 9. Let 𝐴 𝑖 be an 𝐿-subset of 𝐺𝑖 for each 𝑖 =1, 2, . . . , 𝑛.Then (𝜆, 𝜇)-product of𝐴 𝑖 (𝑖 = 1, 2, . . . , 𝑛) denotedby 𝐴1×

𝜆

𝜇𝐴2×𝜆

𝜇⋅ ⋅ ⋅ ×𝜆𝜇𝐴𝑛 is defined to be an 𝐿-subset of 𝐺1 ×

𝐺2 × ⋅ ⋅ ⋅ × 𝐺𝑛 that satisfies

(𝐴1×𝜆

𝜇𝐴2×𝜆

𝜇⋅ ⋅ ⋅ ×𝜆

𝜇𝐴𝑛) (𝑥1, 𝑥2, . . . , 𝑥𝑛)

= (𝐴1 (𝑥1) ∧ 𝐴2 (𝑥2) ∧ ⋅ ⋅ ⋅ ∧ 𝐴𝑛 (𝑥𝑛) ∧ 𝜇) ∨ 𝜆

= (

𝑛

⋀𝑖=1

𝐴 𝑖 (𝑥𝑖) ∧ 𝜇) ∨ 𝜆.

(8)

Example 10. We define the fuzzy subsets 𝐴 and 𝐵 of Z andZ2, respectively, as in Example 4. Then (0.4, 0.6)-product of𝐴 and 𝐵 is as follows:

𝐴×0.4

0.6𝐵 (𝑥) =

{

{

{

0.5, 𝑥 ∈ 2Z × {0} ,

0.4, otherwise.(9)

Lemma 11. Let 𝐺1, 𝐺2, . . . , 𝐺𝑛 be groups. Then we have thefollowing:

(1) If 𝐴 is (𝜆, 𝜇)-𝐿-subgroup of 𝐺1 × 𝐺2×, . . . , 𝐺𝑛 and𝐴 𝑖(𝑥) = 𝐴(𝑒1, 𝑒2, . . . , 𝑒𝑖−1, 𝑥, 𝑒𝑖+1, . . . , 𝑒𝑛) for 𝑖 =

1, 2, . . . , 𝑛, then 𝐴 𝑖 is (𝜆, 𝜇)-𝐿-subgroup of 𝐺𝑖 for all𝑖 = 1, 2, . . . , 𝑛.

(2) If 𝐴 𝑖 is (𝜆, 𝜇)-𝐿-subgroup of 𝐺𝑖 for all 𝑖 = 1, 2, . . . , 𝑛,then 𝐴1×𝜆𝜇𝐴2×

𝜆

𝜇⋅ ⋅ ⋅ ×𝜆𝜇𝐴𝑛 is (𝜆, 𝜇)-𝐿-subgroup of 𝐺1 ×

𝐺2×, . . . , 𝐺𝑛.

Proof. (1) Let 𝑥, 𝑦 ∈ 𝐺𝑖. Since 𝐴 ∈ 𝐹𝑆(𝜆, 𝜇, 𝐺1 × 𝐺2×,

. . . , 𝐺𝑛, 𝐿), we have

𝐴 𝑖 (𝑥) ∧ 𝐴 𝑖 (𝑦) ∧ 𝜇

= 𝐴 (𝑒1, 𝑒2, . . . , 𝑒𝑖−1, 𝑥, 𝑒𝑖+1, . . . , 𝑒𝑛)

∧ 𝐴 (𝑒1, 𝑒2, . . . , 𝑒𝑖−1, 𝑦, 𝑒𝑖+1, . . . , 𝑒𝑛) ∧ 𝜇

≤ 𝐴 (𝑒1, 𝑒2, . . . , 𝑒𝑖−1, 𝑥 ⋅ 𝑦, 𝑒𝑖+1, . . . , 𝑒𝑛) ∨ 𝜆

= 𝐴 𝑖 (𝑥 ⋅ 𝑦) ∨ 𝜆.

(10)

Similarly, we can show that 𝐴 𝑖(𝑥) ∧ 𝜇 ≤ 𝐴 𝑖(𝑥−1) ∨ 𝜆. Hence,

𝐴 𝑖 ∈ 𝐹𝑆(𝜆, 𝜇, 𝐺𝑖, 𝐿) by Definition 2 for 𝑖 = 1, 2, . . . , 𝑛.

(2) Let (𝑥1, 𝑥2, . . . , 𝑥𝑛), (𝑦1, 𝑦2, . . . , 𝑦𝑛) ∈ 𝐺1×𝐺2×, . . . , 𝐺𝑛.Then,

𝐴1×𝜆

𝜇𝐴2×𝜆

𝜇⋅ ⋅ ⋅ ×𝜆

𝜇𝐴𝑛 ((𝑥1, 𝑥2, . . . , 𝑥𝑛) (𝑦1, 𝑦2, . . . , 𝑦𝑛))

∨ 𝜆 = 𝐴1×𝜆

𝜇𝐴2×𝜆

𝜇⋅ ⋅ ⋅ ×𝜆

𝜇𝐴𝑛 (𝑥1𝑦1, 𝑥2𝑦2, . . . , 𝑥𝑛𝑦𝑛)

∨ 𝜆 = (

𝑛

⋀𝑖=1

𝐴 𝑖 (𝑥𝑖𝑦𝑖) ∧ 𝜇) ∨ 𝜆 = ((𝐴1 (𝑥1𝑦1) ∨ 𝜆)

∧ (𝐴2 (𝑥2𝑦2) ∨ 𝜆) ∧ ⋅ ⋅ ⋅ ∧ (𝐴𝑛 (𝑥𝑛𝑦𝑛) ∨ 𝜆) ∧ 𝜇)

∨ 𝜆 ≥ (𝐴1 (𝑥1) ∧ 𝐴1 (𝑦1) ∧ 𝜇 ∧ 𝐴2 (𝑥2) ∧ 𝐴2 (𝑦2)

∧ 𝜇 ∧ ⋅ ⋅ ⋅ ∧ 𝐴𝑛 (𝑥𝑛) ∧ 𝐴𝑛 (𝑦𝑛) ∧ 𝜇) ∨ 𝜆

= ((

𝑛

⋀𝑖=1

𝐴 𝑖 (𝑥𝑖) ∧ 𝜇) ∨ 𝜆) ∧ ((

𝑛

⋀𝑖=1

𝐴 𝑖 (𝑦𝑖) ∧ 𝜇)

∨ 𝜆) ∧ 𝜇 = 𝐴1×𝜆

𝜇𝐴2×𝜆

𝜇⋅ ⋅ ⋅ ×𝜆

𝜇𝐴𝑛 (𝑥1, 𝑥2, . . . , 𝑥𝑛)

∧ 𝐴1×𝜆

𝜇𝐴2×𝜆

𝜇⋅ ⋅ ⋅ ×𝜆

𝜇𝐴𝑛 (𝑦1, 𝑦2, . . . , 𝑦𝑛) ∧ 𝜇.

(11)

Similarly, we can show that

𝐴1×𝜆

𝜇𝐴2×𝜆

𝜇⋅ ⋅ ⋅ ×𝜆

𝜇𝐴𝑛 ((𝑥1, 𝑥2, . . . , 𝑥𝑛)

−1) ∨ 𝜆

≥ 𝐴1×𝜆

𝜇𝐴2×𝜆

𝜇⋅ ⋅ ⋅ ×𝜆

𝜇𝐴𝑛 (𝑥1, 𝑥2, . . . , 𝑥𝑛) ∧ 𝜇.

(12)

Hence, 𝐴1×𝜆

𝜇𝐴2×𝜆

𝜇⋅ ⋅ ⋅ ×𝜆𝜇𝐴𝑛 is (𝜆, 𝜇)-𝐿-subgroup of 𝐺1 ×

𝐺2×, . . . , 𝐺𝑛.

Theorem 12. Let 𝐺1, 𝐺2, . . . , 𝐺𝑛 be groups and 𝐴 ∈ 𝐹𝑆(𝜆, 𝜇,𝐺1×𝐺2×⋅ ⋅ ⋅×𝐺𝑛, 𝐿).Then (𝐴(𝑒1, 𝑒2, . . . , 𝑒𝑖−1, 𝑥𝑖, 𝑒𝑖+1, . . . , 𝑒𝑛) ∧𝐴(𝑥1, 𝑥2, . . . , 𝑥𝑖−1, 𝑒𝑖, 𝑥𝑖+1, . . . , 𝑥𝑛)∧𝜇) ∨𝜆 = 𝐴(𝑥1, 𝑥2, . . . , 𝑥𝑛)

for 𝑖 = 1, 2, . . . , 𝑛 if and only if𝐴 = 𝐴1×𝜆𝜇𝐴2×𝜆

𝜇⋅ ⋅ ⋅ ×𝜆𝜇𝐴𝑛, where

𝐴 𝑖(𝑥) = 𝐴(𝑒1, 𝑒2, . . . , 𝑒𝑖−1, 𝑥, 𝑒𝑖+1, . . . , 𝑒𝑛).

Proof. Suppose that (𝐴(𝑒1, 𝑒2, . . . , 𝑒𝑖−1, 𝑥𝑖, 𝑒𝑖+1, . . . , 𝑒𝑛)∧ 𝐴(𝑥1,𝑥2, . . . , 𝑥𝑖−1, 𝑒𝑖, 𝑥𝑖+1, . . . , 𝑥𝑛) ∧ 𝜇) ∨ 𝜆 = 𝐴(𝑥1, 𝑥2, . . . , 𝑥𝑛) for𝑖 = 1, 2, . . . , 𝑛. Now, for any (𝑥1, 𝑥2, . . . , 𝑥𝑛) ∈ 𝐺1×𝐺2×⋅ ⋅ ⋅×𝐺𝑛,we have

𝐴 (𝑥1, 𝑥2, . . . , 𝑥𝑛) = (𝐴 (𝑥1, 𝑒2, . . . , 𝑒𝑛)

∧ 𝐴 (𝑒1, 𝑥2, . . . , 𝑥𝑛) ∧ 𝜇) ∨ 𝜆

= (𝐴1 (𝑥1)

∧ ((𝐴 (𝑒1, 𝑥2, . . . , 𝑒𝑛) ∧ 𝐴 (𝑒1, 𝑒2, 𝑥3, . . . , 𝑥𝑛) ∧ 𝜇)

∨ 𝜆) ∧ 𝜇) ∨ 𝜆

4 Journal of Mathematics

= (𝐴1 (𝑥1) ∧ 𝐴2 (𝑥2) ∧ 𝐴 (𝑒1, 𝑒2, 𝑥3, . . . , 𝑥𝑛) ∧ 𝜇)

∨ 𝜆

...

= (𝐴1 (𝑥1) ∧ 𝐴2 (𝑥2) ∧ ⋅ ⋅ ⋅ ∧ 𝐴 (𝑒1, 𝑒2, . . . , 𝑥𝑛) ∧ 𝜇)

∨ 𝜆

= 𝐴1×𝜆

𝜇𝐴2×𝜆

𝜇⋅ ⋅ ⋅ ×𝜆

𝜇𝐴𝑛 (𝑥1, 𝑥2, . . . , 𝑥𝑛) .

(13)

Hence we obtain that 𝐴 = 𝐴1×𝜆

𝜇𝐴2×𝜆

𝜇⋅ ⋅ ⋅ ×𝜆𝜇𝐴𝑛. Conversely,

assume that 𝐴 = 𝐴1×𝜆

𝜇𝐴2×𝜆

𝜇⋅ ⋅ ⋅ ×𝜆𝜇𝐴𝑛. Then, we obtain

𝐴 (𝑥1, 𝑥2, . . . , 𝑥𝑛) = 𝐴1×𝜆

𝜇𝐴2×𝜆

𝜇

⋅ ⋅ ⋅ ×𝜆

𝜇𝐴𝑛 (𝑥1, 𝑥2, . . . , 𝑥𝑛)

= (𝐴 (𝑥1, 𝑒2, . . . , 𝑒𝑛) ∧ 𝐴 (𝑒1, 𝑥2, . . . , 𝑒𝑛) ∧ ⋅ ⋅ ⋅

∧ 𝐴 (𝑒1, 𝑒2, . . . , 𝑥𝑛) ∧ 𝜇) ∨ 𝜆

≤ (𝐴 (𝑥1, 𝑒2, . . . , 𝑒𝑛) ∧ 𝐴 (𝑒1, 𝑥2, . . . , 𝑒𝑛) ∧ ⋅ ⋅ ⋅

∧ (𝐴 (𝑒1, 𝑒2, . . . , 𝑥𝑛−1, 𝑥𝑛) ∨ 𝜆) ∧ 𝜇) ∨ 𝜆

= (𝐴 (𝑥1, 𝑒2, . . . , 𝑒𝑛) ∧ 𝐴 (𝑒1, 𝑥2, . . . , 𝑒𝑛) ∧ ⋅ ⋅ ⋅

∧ 𝐴 (𝑒1, 𝑒2, . . . , 𝑥𝑛−1, 𝑥𝑛) ∧ 𝜇) ∨ 𝜆

...

≤ (𝐴 (𝑥1, 𝑒2, . . . , 𝑒𝑛) ∧ 𝐴 (𝑒1, 𝑥2, . . . , 𝑥𝑛) ∧ 𝜇) ∨ 𝜆.

(14)

On the other hand, (𝐴(𝑥1, 𝑒2, . . . , 𝑒𝑛)∧𝐴(𝑒1, 𝑥2, . . . , 𝑥𝑛)∧𝜇)∨𝜆 ≤ 𝐴(𝑥1, 𝑥2, . . . , 𝑥𝑛) ∨ 𝜆.

Since 𝐴(𝑥1, 𝑥2, . . . , 𝑥𝑛) ≥ 𝜆, (𝐴(𝑥1, 𝑒2, . . . , 𝑒𝑛) ∧ 𝐴(𝑒1, 𝑥2,. . . , 𝑥𝑛) ∧ 𝜇) ∨ 𝜆 ≤ 𝐴(𝑥1, 𝑥2, . . . , 𝑥𝑛).

Hence, 𝐴(𝑥1, 𝑥2, . . . , 𝑥𝑛) = (𝐴(𝑥1, 𝑒2, . . . , 𝑒𝑛) ∧ 𝐴(𝑒1, 𝑥2,. . . , 𝑥𝑛) ∧ 𝜇) ∨ 𝜆.

Similarly, we get (𝐴(𝑒1, 𝑒2, . . . , 𝑒𝑖−1, 𝑥𝑖, 𝑒𝑖+1, . . . , 𝑒𝑛) ∧𝐴(𝑥1, 𝑥2, . . . , 𝑥𝑖−1, 𝑒𝑖, 𝑥𝑖+1, . . . , 𝑥𝑛)∧𝜇)∨ 𝜆 = 𝐴(𝑥1, 𝑥2, . . . , 𝑥𝑛)

for 𝑖 = 2, 3, . . . , 𝑛.

Lemma 13. Let 𝐺1, 𝐺2, . . . , 𝐺𝑛 be groups, let 𝑒1, 𝑒2, . . . , 𝑒𝑛 beidentities of𝐺1, 𝐺2, . . . , 𝐺𝑛, respectively, and𝐴 ∈ 𝐹𝑆(𝜆, 𝜇, 𝐺1 ×𝐺2 × ⋅ ⋅ ⋅ × 𝐺𝑛). If (𝐴(𝑒1, 𝑒2, . . . , 𝑒𝑖−1, 𝑥𝑖, 𝑒𝑖+1, . . . , 𝑒𝑛) ∧ 𝜇) ∨ 𝜆 ≥𝐴(𝑥1, 𝑥2, . . . , 𝑥𝑛) for 𝑖 = 1, 2, . . . , 𝑘 − 1, 𝑘 + 1, . . . , 𝑛,then (𝐴(𝑒1, 𝑒2, . . . , 𝑒𝑘−1, 𝑥𝑘, 𝑒𝑘+1, . . . , 𝑒𝑛) ∧ 𝜇) ∨ 𝜆 ≥ 𝐴(𝑥1,

𝑥2, . . . , 𝑥𝑛).

Proof. By Definition 2, we observe that

(𝐴 (𝑒1, 𝑒2, . . . , 𝑒𝑘−1, 𝑥𝑘, 𝑒𝑘+1, . . . , 𝑒𝑛) ∧ 𝜇) ∨ 𝜆

= (𝐴 ((𝑥1, 𝑥2, . . . , 𝑥𝑛)

⋅ (𝑥−1

1, 𝑥−1

2, . . . , 𝑥

−1

𝑘−1, 𝑒𝑘, 𝑥−1

𝑘+1, . . . , 𝑥

−1

𝑛)) ∧ 𝜇) ∨ 𝜆

= (𝐴 ((𝑥1, 𝑥2, . . . , 𝑥𝑛)

⋅ (𝑥−1

1, 𝑥−1

2, . . . , 𝑥

−1

𝑘−1, 𝑒𝑘, 𝑥−1

𝑘+1, . . . , 𝑥

−1

𝑛)) ∨ 𝜆) ∧ 𝜇

≥ (𝐴 (𝑥1, 𝑥2, . . . , 𝑥𝑛)

∧ 𝐴 (𝑥−1

1, 𝑥−1

2, . . . , 𝑥

−1

𝑘−1, 𝑒𝑘, 𝑥−1

𝑘+1, . . . , 𝑥

−1

𝑛) ∧ 𝜇) ∨ 𝜆

= (𝐴 (𝑥1, 𝑥2, . . . , 𝑥𝑛) ∨ 𝜆)

∧ (𝐴 (𝑥−1

1, 𝑥−1

2, . . . , 𝑥

−1

𝑘−1, 𝑒𝑘, 𝑥−1

𝑘+1, . . . , 𝑥

−1

𝑛) ∨ 𝜆)

∧ 𝜇

≥ ((𝐴 (𝑥1, 𝑥2, . . . , 𝑥𝑛) ∨ 𝜆)

∧ 𝐴 (𝑥1, 𝑥2, . . . , 𝑥𝑘−1, 𝑒𝑘, 𝑥𝑘+1, . . . , 𝑥𝑛) ∧ 𝜇) ∨ 𝜆

= (𝐴 (𝑥1, 𝑥2, . . . , 𝑥𝑛) ∨ 𝜆)

∧ (𝐴 (𝑥1, 𝑥2, . . . , 𝑥𝑘−1, 𝑒𝑘, 𝑥𝑘+1, . . . , 𝑥𝑛) ∨ 𝜆) ∧ 𝜇

≥ ((𝐴 (𝑥1, 𝑥2, . . . , 𝑥𝑛) ∨ 𝜆)

∧ (𝐴 (𝑒1, . . . , 𝑒𝑘−2, 𝑥𝑘−1, 𝑒𝑘, . . . , 𝑒𝑛)

∧ 𝐴 (𝑥1, . . . , 𝑒𝑘−1, 𝑒𝑘, 𝑥𝑘+1, . . . , 𝑥𝑛) ∧ 𝜇)) ∨ 𝜆

= (𝐴 (𝑥1, 𝑥2, . . . , 𝑥𝑛) ∨ 𝜆)

∧ (𝐴 (𝑥1, . . . , 𝑒𝑘−1, 𝑒𝑘, 𝑥𝑘+1, . . . , 𝑥𝑛) ∧ 𝜇) ∨ 𝜆

...

≥ 𝐴 (𝑥1, 𝑥2, . . . , 𝑥𝑛) ∨ 𝜆

≥ 𝐴 (𝑥1, 𝑥2, . . . , 𝑥𝑛) .

(15)

Lemma 14. Let 𝐺1, 𝐺2, . . . , 𝐺𝑛 be groups and 𝐴 ∈ 𝐹𝑆(0,

𝜇, 𝐺1×𝐺2×⋅ ⋅ ⋅×𝐺𝑛, 𝐿).Then𝐴(𝑒1, 𝑒2, . . . , 𝑒𝑖−1, 𝑥𝑖, 𝑒𝑖+1, . . . , 𝑒𝑛)∧𝐴(𝑥1, 𝑥2, . . . , 𝑥𝑖−1, 𝑒𝑖, 𝑥𝑖+1, . . . , 𝑥𝑛) ∧ 𝜇 = 𝐴(𝑥1, 𝑥2, . . . , 𝑥𝑛) for𝑖 = 1, 2, . . . , 𝑛 if and only if 𝐴(𝑒1, 𝑒2, . . . , 𝑒𝑖−1, 𝑥𝑖, 𝑒𝑖+1, . . . , 𝑒𝑛) ∧𝜇 ≥ 𝐴(𝑥1, 𝑥2, . . . , 𝑥𝑛) for 𝑖 = 1, 2, . . . , 𝑘 − 1, 𝑘 + 1, . . . , 𝑛.

Proof. Now assume that 𝐴(𝑒1, 𝑒2, . . . , 𝑒𝑖−1, 𝑥𝑖, 𝑒𝑖+1, . . . , 𝑒𝑛) ∧𝐴(𝑥1, 𝑥2, . . . , 𝑥𝑖−1, 𝑒𝑖, 𝑥𝑖+1, . . . , 𝑥𝑛) ∧ 𝜇 = 𝐴(𝑥1, 𝑥2, . . . , 𝑥𝑛) for𝑖 = 1, 2, . . . , 𝑛. Next, for any (𝑥1, 𝑥2, . . . , 𝑥𝑛) ∈ 𝐺1×𝐺2×⋅ ⋅ ⋅×𝐺𝑛,we have

𝐴 (𝑥1, 𝑥2, . . . , 𝑥𝑛)

= 𝐴 (𝑒1, 𝑒2, . . . , 𝑒𝑖−1, 𝑥𝑖, 𝑒𝑖+1, . . . , 𝑒𝑛)

∧ 𝐴 (𝑥1, 𝑥2, . . . , 𝑥𝑖−1, 𝑒𝑖, 𝑥𝑖+1, . . . , 𝑥𝑛) ∧ 𝜇

≤ 𝐴 (𝑒1, 𝑒2, . . . , 𝑒𝑖−1, 𝑥𝑖, 𝑒𝑖+1, . . . , 𝑒𝑛) ∧ 𝜇.

(16)

Conversely, assume that 𝐴(𝑒1, 𝑒2, . . . , 𝑒𝑖−1, 𝑥𝑖, 𝑒𝑖+1, . . . , 𝑒𝑛) ∧𝜇 ≥ 𝐴(𝑥1, 𝑥2, . . . , 𝑥𝑛) for 𝑖 = 1, 2, . . . , 𝑘 − 1, 𝑘 + 1, . . . , 𝑛. By

Journal of Mathematics 5

Definition 2 and Lemma 11, we obtain that

𝐴 (𝑥1, 𝑥2, . . . , 𝑥𝑛) = 𝐴 (𝑥1, 𝑥2, . . . , 𝑥𝑛)

∧ 𝐴 (𝑥1, 𝑥2, . . . , 𝑥𝑛) ∧ ⋅ ⋅ ⋅ ∧ 𝐴 (𝑥1, 𝑥2, . . . , 𝑥𝑛)

≤ 𝐴 (𝑥1, 𝑒2, . . . , 𝑒𝑛) ∧ 𝐴 (𝑒1, 𝑥2, . . . , 𝑒𝑛) ∧ ⋅ ⋅ ⋅

∧ 𝐴 (𝑒1, 𝑒2, . . . , 𝑥𝑛) ∧ 𝜇

≤ 𝐴 (𝑥1, 𝑒2, . . . , 𝑒𝑛) ∧ 𝐴 (𝑒1, 𝑥2, . . . , 𝑒𝑛) ∧ ⋅ ⋅ ⋅

∧ 𝐴 (𝑒1, 𝑒2, . . . , 𝑥𝑛−1, 𝑥𝑛) ∧ 𝜇

...

≤ 𝐴 (𝑥1, 𝑒2, . . . , 𝑒𝑛) ∧ 𝐴 (𝑒1, 𝑥2, . . . , 𝑥𝑛) ∧ 𝜇

≤ 𝐴 (𝑥1, 𝑥2, . . . , 𝑥𝑛) .

(17)

Hence, 𝐴(𝑥1, 𝑥2, . . . , 𝑥𝑛) = 𝐴(𝑥1, 𝑒2, . . . , 𝑒𝑛) ∧ 𝐴(𝑒1, 𝑥2, . . . ,𝑥𝑛) ∧ 𝜇. Similarly, we get 𝐴(𝑒1, 𝑒2, . . . , 𝑒𝑖−1, 𝑥𝑖, 𝑒𝑖+1, . . . , 𝑒𝑛) ∧𝐴(𝑥1, 𝑥2, . . . , 𝑥𝑖−1, 𝑒𝑖, 𝑥𝑖+1, . . . , 𝑥𝑛) = 𝐴(𝑥1, 𝑥2, . . . , 𝑥𝑛) for 𝑖 =2, 3, . . . , 𝑛.

As a consequence of Theorem 12 and Lemma 14, we havethe following corollary.

Corollary 15. Let 𝐺1, 𝐺2, . . . , 𝐺𝑛 be groups and 𝐴 ∈ 𝐹𝑆(0, 𝜇,𝐺1 ×𝐺2 × ⋅ ⋅ ⋅ ×𝐺𝑛, 𝐿). Then𝐴(𝑒1, 𝑒2, . . . , 𝑒𝑖−1, 𝑥𝑖, 𝑒𝑖+1, . . . , 𝑒𝑛) ∧𝜇 ≥ 𝐴(𝑥1, 𝑥2, . . . , 𝑥𝑛) for 𝑖 = 1, 2, . . . , 𝑘 − 1, 𝑘 + 1, . . . , 𝑛 if andonly if 𝐴 = 𝐴1×𝜇𝐴2×𝜇 ⋅ ⋅ ⋅ ×𝜇𝐴𝑛, where 𝐴 𝑖(𝑥) = 𝐴(𝑒1,

𝑒2, . . . , 𝑒𝑖−1, 𝑥, 𝑒𝑖+1, . . . , 𝑒𝑛).

The following example shows that Corollary 15 may notbe true when 𝜆 = 0.

Example 16. Consider

𝐴 (𝑥) =

{{{{{{{

{{{{{{{

{

0.4, 𝑥 = (0, 0) ,

0.3, 𝑥 = (1, 0) ,

0.2, 𝑥 = (0, 1) ,

0.1, 𝑥 = (1, 1) .

(18)

𝐴 is (0.2, 0.5)-fuzzy subgroup of Z2 × Z2 and satisfies thenecessary condition of Corollary 15. But there is not any 𝐴1and 𝐴2, (0.2, 0.5)-fuzzy subgroup of Z2 × Z2, which hold𝐴 = 𝐴1×

0.2

0.5𝐴2.

4. Conclusion

In this study, we give a necessary and sufficient condition for(0, 𝜇)-𝐿-subgroup of a Cartesian product of groups to be aproduct of (0, 𝜇)-𝐿-subgroups. The results obtained are notvalid for 𝜆 = 0, and a counterexample is provided.

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper.

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