Research Article Product of the Generalized...
Transcript of Research Article Product of the Generalized...
Research ArticleProduct of the Generalized πΏ-Subgroups
Dilek Bayrak and Sultan Yamak
Department of Mathematics, Namik Kemal University, 59000 Tekirdag, Turkey
Correspondence should be addressed to Dilek Bayrak; [email protected]
Received 6 July 2015; Accepted 8 February 2016
Academic Editor: Ali Jaballah
Copyright Β© 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 (π, π)-product of πΏ-subsets. We give a necessary and sufficient condition for (π, π)-πΏ-subgroup of aproduct of groups to be (π, π)-product of (π, π)-πΏ-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 (πΌ, π½)-fuzzy subgroup, where πΌ, π½ are any two of {β, π, ββ¨π, β β§π}with πΌ =ββ§π. The (β, ββ¨π)-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 [5β10]).In recent years,many researchersmake generalizationswhichare referred to as (π, π)-fuzzy substructures and (βπ, βπβ¨ππ)-fuzzy substructures on this topic (see [11β15]).
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, andππΏ-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 aππΏ-subgroup of a Cartesian product of groups to be a π-product of ππΏ-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 πΏ-subgroups.In this paper, we introduce the notion of the (π, π)-product
of πΏ-subsets and investigate some properties of the (π, π)-product of πΏ-subgroups. Also, we give a necessary andsufficient condition for (π, π)-πΏ-subgroup of a Cartesianproduct of groups to be a product of (π, π)-πΏ-subgroups.
2. Preliminaries
In this section, we start by giving some known definitions andnotations. Throughout this paper, unless otherwise stated, πΊalways stands for any given groupwith amultiplicative binaryoperation, an identity π and πΏ denote a complete lattice withtop and bottom elements 1, 0, respectively.
An πΏ-subset ofπ is any function fromπ into πΏ, which isintroduced by Goguen [18] as a generalization of the notionof Zadehβs fuzzy subset [19]. The class of πΏ-subsets of π willbe denoted by πΉ(π, πΏ). In particular, if πΏ = [0, 1], then it isappropriate to replace πΏ-subset with fuzzy subset. In this casethe set of all fuzzy subsets of π is denoted by πΉ(π). Let π΄and π΅ be πΏ-subsets of π. We say that π΄ is contained in π΅ ifπ΄(π₯) β€ π΅(π₯) for every π₯ β π and is denoted by π΄ β€ π΅. Thenβ€ is a partial ordering on the set πΉ(π, πΏ).
Definition 1 (see [20]). An πΏ-subset of πΊ is called an πΏ-subgroup of πΊ if, for all π₯, π¦ β πΊ, the following conditionshold:(G1) π΄(π₯) β§ π΄(π¦) β€ π΄(π₯π¦).(G2) π΄(π₯) β€ π΄(π₯β1).In particular, when πΏ = [0, 1], an πΏ-subgroup of πΊ is
referred to as a fuzzy subgroup of πΊ.
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 π, π β πΏ and π < π. Let π΄ be anπΏ-subset of πΊ. π΄ is called (π, π)-πΏ-subgroup of πΊ if, for allπ₯, π¦ β πΊ, the following conditions hold:
(i) π΄(π₯π¦) β¨ π β₯ π΄(π₯) β§ π΄(π¦) β§ π.
(ii) π΄(π₯β1) β¨ π β₯ π΄(π₯) β§ π.
Denote by πΉπ(π, π, πΊ, πΏ) the set of all (π, π)-πΏ-subgroups ofπΊ. When πΏ = [0, 1], its counterpart is written as πΉπ(π, π, πΊ).
Unless otherwise stated, πΏ always represents any givendistributive lattice.
3. Product of (π,π)-πΏ-Subgroups
Definition 3 (see [16]). Let π΄ π be an πΏ-subset of πΊπ for eachπ = 1, 2, . . . , π. Then product of π΄ π (π = 1, 2, . . . , π) denotedby π΄1 Γ π΄2 Γ β β β Γ π΄π is defined to be the πΏ-subset of πΊ1 ΓπΊ2 Γ β β β Γ πΊπ that satisfies
(π΄1 Γ π΄2 Γ β β β Γ π΄π) (π₯1, π₯2, . . . , π₯π)
= π΄1 (π₯1) β§ π΄2 (π₯2) β§ β β β β§ π΄π (π₯π) .(1)
Example 4. We define the fuzzy subsetsπ΄ and π΅ ofZ andZ2,respectively, as follows:
π΄ (π₯) ={
{
{
0.5, π₯ β 2Z,
0.3, otherwise,
π΅ (π₯) ={
{
{
0.7, π₯ = 0,
0.2, π₯ = 1.
(2)
We obtain that
π΄ Γ π΅ (π₯) =
{{{{
{{{{
{
0.5, π₯ β 2Z Γ {0} ,
0.3, π₯ β Z β {2Z} Γ {0} ,
0.2, otherwise.
(3)
Theorem5. LetπΊ1, πΊ2, . . . , πΊπ be groups andβπ
π=1ππ < β
π
π=1ππ
such thatπ΄ π is (ππ, ππ)-πΏ-subgroup ofπΊπ for each π = 1, 2, . . . , π.Then π΄1 Γ π΄2 Γ β β β Γ π΄π is (β
π
π=1ππ, βπ
π=1ππ)-πΏ-subgroup of
πΊ1 Γ πΊ2 Γ β β β Γ πΊπ.
Proof. Let (π₯1, π₯2, . . . , π₯π), (π¦1, π¦2, . . . , π¦π) β πΊ1ΓπΊ2Γβ β β ΓπΊπ.Thenπ΄1 Γ π΄2 Γ β β β Γ π΄π ((π₯1, π₯2, . . . , π₯π) β (π¦1, π¦2, . . . , π¦π))
β¨
π
βπ=1
ππ = π΄1 Γ π΄2 Γ β β β
Γ π΄π (π₯1π¦1, π₯2π¦2, . . . , π₯ππ¦π) β¨ π1 β¨
π
βπ=1
ππ
= (π΄1 (π₯1π¦1) β§ π΄2 (π₯2π¦2) β§ β β β β§ π΄π (π₯ππ¦π))
β¨
π
βπ=1
ππ β₯ (π΄1 (π₯1π¦1) β¨ π1) β§ (π΄2 (π₯2π¦2) β¨ π2)
β§ β β β β§ (π΄π (π₯ππ¦π) β¨ ππ) β₯ π΄1 (π₯1) β§ π΄1 (π¦1) β§ π1
β§ π΄2 (π₯2) β§ π΄2 (π¦2) β§ π2 β§ β β β β§ π΄π (π₯π) β§ π΄π (π¦π)
β§ ππ = π΄1 Γ π΄2 Γ β β β Γ π΄π (π₯1, π₯2, . . . , π₯π) β§ π΄1
Γ π΄2 Γ β β β Γ π΄π (π¦1, π¦2, . . . , π¦π) β§
π
βπ=1
ππ.
(4)
Similarly, it can be shown that
π΄1 Γ π΄2 Γ β β β Γ π΄π ((π₯1, π₯2, . . . , π₯π)β1) β¨
π
βπ=1
ππ
β₯ π΄1 Γ π΄2 Γ β β β Γ π΄π (π₯1, π₯2, . . . , π₯π) β§
π
βπ=1
ππ.
(5)
Corollary 6. If π΄1, π΄2, . . . , π΄π are (π, π)-πΏ-subgroups ofπΊ1, πΊ2, . . . , πΊπ, respectively, then π΄1 Γ π΄2 Γ β β β Γ π΄π is (π, π)-πΏ-subgroup of πΊ1 Γ πΊ2 Γ β β β Γ πΊπ.
Theorem 7 (Theorem 2.9, [16]). Let πΊ1, πΊ2, . . . , πΊπ begroups, let π1, π2, . . . , ππ be identities, respectively, and letπ΄ be a fuzzy subgroup in πΊ1 Γ πΊ2 Γ β β β Γ πΊπ. Thenπ΄(π1, π2, . . . , ππβ1, π₯π, ππ+1, . . . , ππ) β₯ π΄(π₯1, π₯2, . . . , π₯π) for π =1, 2, . . . , πβ1, π+1, . . . , π if and only ifπ΄ = π΄1Γπ΄2Γβ β β Γπ΄π,where π΄1, π΄2, . . . , π΄π are fuzzy subgroups of πΊ1, πΊ2, . . . , πΊπ,respectively.
The following example shows that Theorem 7 may not betrue for any (π, π)-πΏ-subgroup.
Example 8. Consider
π΄ (π₯) =
{{{{{{{
{{{{{{{
{
0.8, π₯ = (0, 0) ,
0.7, π₯ = (1, 0) ,
0.6, π₯ = (0, 1) ,
0.5, π₯ = (1, 1) .
(6)
It is easy to see that π΄ is (0, 0.5)-fuzzy subgroup of Z2 Γ Z2.Since π΄(1, 0) β₯ π΄(1, 1) and π΄(0, 1) β₯ π΄(1, 1), π΄ satisfies the
Journal of Mathematics 3
condition of Theorem 7, but there do not exist π΄1, π΄2 fuzzysubgroups of Z2 Γ Z2 such that π΄ = π΄1 Γ π΄2.
In fact, suppose that there exist π΄1, π΄2 β πΉπ(0, 0.5,Z2 ΓZ2) such that π΄ = π΄1 Γπ΄2. Since π΄(1, 0) = 0.7 and π΄(0, 1) =0.6, we have π΄1(1) β₯ 0.7 and π΄2(1) β₯ 0.6. Hence
0.5 = π΄1 Γ π΄2 (1, 1) = π΄1 (1) β§ π΄2 (1) β₯ 0.7 β§ 0.6
= 0.6,
(7)
a contradiction.
Definition 9. Let π΄ π be an πΏ-subset of πΊπ for each π =1, 2, . . . , π.Then (π, π)-product ofπ΄ π (π = 1, 2, . . . , π) denotedby π΄1Γ
π
ππ΄2Γπ
πβ β β Γπππ΄π is defined to be an πΏ-subset of πΊ1 Γ
πΊ2 Γ β β β Γ πΊπ that satisfies
(π΄1Γπ
ππ΄2Γπ
πβ β β Γπ
ππ΄π) (π₯1, π₯2, . . . , π₯π)
= (π΄1 (π₯1) β§ π΄2 (π₯2) β§ β β β β§ π΄π (π₯π) β§ π) β¨ π
= (
π
βπ=1
π΄ π (π₯π) β§ π) β¨ π.
(8)
Example 10. We define the fuzzy subsets π΄ and π΅ of Z andZ2, respectively, as in Example 4. Then (0.4, 0.6)-product ofπ΄ and π΅ is as follows:
π΄Γ0.4
0.6π΅ (π₯) =
{
{
{
0.5, π₯ β 2Z Γ {0} ,
0.4, otherwise.(9)
Lemma 11. Let πΊ1, πΊ2, . . . , πΊπ be groups. Then we have thefollowing:
(1) If π΄ is (π, π)-πΏ-subgroup of πΊ1 Γ πΊ2Γ, . . . , πΊπ andπ΄ π(π₯) = π΄(π1, π2, . . . , ππβ1, π₯, ππ+1, . . . , ππ) for π =
1, 2, . . . , π, then π΄ π is (π, π)-πΏ-subgroup of πΊπ for allπ = 1, 2, . . . , π.
(2) If π΄ π is (π, π)-πΏ-subgroup of πΊπ for all π = 1, 2, . . . , π,then π΄1Γπππ΄2Γ
π
πβ β β Γπππ΄π is (π, π)-πΏ-subgroup of πΊ1 Γ
πΊ2Γ, . . . , πΊπ.
Proof. (1) Let π₯, π¦ β πΊπ. Since π΄ β πΉπ(π, π, πΊ1 Γ πΊ2Γ,
. . . , πΊπ, πΏ), we have
π΄ π (π₯) β§ π΄ π (π¦) β§ π
= π΄ (π1, π2, . . . , ππβ1, π₯, ππ+1, . . . , ππ)
β§ π΄ (π1, π2, . . . , ππβ1, π¦, ππ+1, . . . , ππ) β§ π
β€ π΄ (π1, π2, . . . , ππβ1, π₯ β π¦, ππ+1, . . . , ππ) β¨ π
= π΄ π (π₯ β π¦) β¨ π.
(10)
Similarly, we can show that π΄ π(π₯) β§ π β€ π΄ π(π₯β1) β¨ π. Hence,
π΄ π β πΉπ(π, π, πΊπ, πΏ) by Definition 2 for π = 1, 2, . . . , π.
(2) Let (π₯1, π₯2, . . . , π₯π), (π¦1, π¦2, . . . , π¦π) β πΊ1ΓπΊ2Γ, . . . , πΊπ.Then,
π΄1Γπ
ππ΄2Γπ
πβ β β Γπ
ππ΄π ((π₯1, π₯2, . . . , π₯π) (π¦1, π¦2, . . . , π¦π))
β¨ π = π΄1Γπ
ππ΄2Γπ
πβ β β Γπ
ππ΄π (π₯1π¦1, π₯2π¦2, . . . , π₯ππ¦π)
β¨ π = (
π
βπ=1
π΄ π (π₯ππ¦π) β§ π) β¨ π = ((π΄1 (π₯1π¦1) β¨ π)
β§ (π΄2 (π₯2π¦2) β¨ π) β§ β β β β§ (π΄π (π₯ππ¦π) β¨ π) β§ π)
β¨ π β₯ (π΄1 (π₯1) β§ π΄1 (π¦1) β§ π β§ π΄2 (π₯2) β§ π΄2 (π¦2)
β§ π β§ β β β β§ π΄π (π₯π) β§ π΄π (π¦π) β§ π) β¨ π
= ((
π
βπ=1
π΄ π (π₯π) β§ π) β¨ π) β§ ((
π
βπ=1
π΄ π (π¦π) β§ π)
β¨ π) β§ π = π΄1Γπ
ππ΄2Γπ
πβ β β Γπ
ππ΄π (π₯1, π₯2, . . . , π₯π)
β§ π΄1Γπ
ππ΄2Γπ
πβ β β Γπ
ππ΄π (π¦1, π¦2, . . . , π¦π) β§ π.
(11)
Similarly, we can show that
π΄1Γπ
ππ΄2Γπ
πβ β β Γπ
ππ΄π ((π₯1, π₯2, . . . , π₯π)
β1) β¨ π
β₯ π΄1Γπ
ππ΄2Γπ
πβ β β Γπ
ππ΄π (π₯1, π₯2, . . . , π₯π) β§ π.
(12)
Hence, π΄1Γπ
ππ΄2Γπ
πβ β β Γπππ΄π is (π, π)-πΏ-subgroup of πΊ1 Γ
πΊ2Γ, . . . , πΊπ.
Theorem 12. Let πΊ1, πΊ2, . . . , πΊπ be groups and π΄ β πΉπ(π, π,πΊ1ΓπΊ2Γβ β β ΓπΊπ, πΏ).Then (π΄(π1, π2, . . . , ππβ1, π₯π, ππ+1, . . . , ππ) β§π΄(π₯1, π₯2, . . . , π₯πβ1, ππ, π₯π+1, . . . , π₯π)β§π) β¨π = π΄(π₯1, π₯2, . . . , π₯π)
for π = 1, 2, . . . , π if and only ifπ΄ = π΄1Γπππ΄2Γπ
πβ β β Γπππ΄π, where
π΄ π(π₯) = π΄(π1, π2, . . . , ππβ1, π₯, ππ+1, . . . , ππ).
Proof. Suppose that (π΄(π1, π2, . . . , ππβ1, π₯π, ππ+1, . . . , ππ)β§ π΄(π₯1,π₯2, . . . , π₯πβ1, ππ, π₯π+1, . . . , π₯π) β§ π) β¨ π = π΄(π₯1, π₯2, . . . , π₯π) forπ = 1, 2, . . . , π. Now, for any (π₯1, π₯2, . . . , π₯π) β πΊ1ΓπΊ2Γβ β β ΓπΊπ,we have
π΄ (π₯1, π₯2, . . . , π₯π) = (π΄ (π₯1, π2, . . . , ππ)
β§ π΄ (π1, π₯2, . . . , π₯π) β§ π) β¨ π
= (π΄1 (π₯1)
β§ ((π΄ (π1, π₯2, . . . , ππ) β§ π΄ (π1, π2, π₯3, . . . , π₯π) β§ π)
β¨ π) β§ π) β¨ π
4 Journal of Mathematics
= (π΄1 (π₯1) β§ π΄2 (π₯2) β§ π΄ (π1, π2, π₯3, . . . , π₯π) β§ π)
β¨ π
...
= (π΄1 (π₯1) β§ π΄2 (π₯2) β§ β β β β§ π΄ (π1, π2, . . . , π₯π) β§ π)
β¨ π
= π΄1Γπ
ππ΄2Γπ
πβ β β Γπ
ππ΄π (π₯1, π₯2, . . . , π₯π) .
(13)
Hence we obtain that π΄ = π΄1Γπ
ππ΄2Γπ
πβ β β Γπππ΄π. Conversely,
assume that π΄ = π΄1Γπ
ππ΄2Γπ
πβ β β Γπππ΄π. Then, we obtain
π΄ (π₯1, π₯2, . . . , π₯π) = π΄1Γπ
ππ΄2Γπ
π
β β β Γπ
ππ΄π (π₯1, π₯2, . . . , π₯π)
= (π΄ (π₯1, π2, . . . , ππ) β§ π΄ (π1, π₯2, . . . , ππ) β§ β β β
β§ π΄ (π1, π2, . . . , π₯π) β§ π) β¨ π
β€ (π΄ (π₯1, π2, . . . , ππ) β§ π΄ (π1, π₯2, . . . , ππ) β§ β β β
β§ (π΄ (π1, π2, . . . , π₯πβ1, π₯π) β¨ π) β§ π) β¨ π
= (π΄ (π₯1, π2, . . . , ππ) β§ π΄ (π1, π₯2, . . . , ππ) β§ β β β
β§ π΄ (π1, π2, . . . , π₯πβ1, π₯π) β§ π) β¨ π
...
β€ (π΄ (π₯1, π2, . . . , ππ) β§ π΄ (π1, π₯2, . . . , π₯π) β§ π) β¨ π.
(14)
On the other hand, (π΄(π₯1, π2, . . . , ππ)β§π΄(π1, π₯2, . . . , π₯π)β§π)β¨π β€ π΄(π₯1, π₯2, . . . , π₯π) β¨ π.
Since π΄(π₯1, π₯2, . . . , π₯π) β₯ π, (π΄(π₯1, π2, . . . , ππ) β§ π΄(π1, π₯2,. . . , π₯π) β§ π) β¨ π β€ π΄(π₯1, π₯2, . . . , π₯π).
Hence, π΄(π₯1, π₯2, . . . , π₯π) = (π΄(π₯1, π2, . . . , ππ) β§ π΄(π1, π₯2,. . . , π₯π) β§ π) β¨ π.
Similarly, we get (π΄(π1, π2, . . . , ππβ1, π₯π, ππ+1, . . . , ππ) β§π΄(π₯1, π₯2, . . . , π₯πβ1, ππ, π₯π+1, . . . , π₯π)β§π)β¨ π = π΄(π₯1, π₯2, . . . , π₯π)
for π = 2, 3, . . . , π.
Lemma 13. Let πΊ1, πΊ2, . . . , πΊπ be groups, let π1, π2, . . . , ππ beidentities ofπΊ1, πΊ2, . . . , πΊπ, respectively, andπ΄ β πΉπ(π, π, πΊ1 ΓπΊ2 Γ β β β Γ πΊπ). If (π΄(π1, π2, . . . , ππβ1, π₯π, ππ+1, . . . , ππ) β§ π) β¨ π β₯π΄(π₯1, π₯2, . . . , π₯π) for π = 1, 2, . . . , π β 1, π + 1, . . . , π,then (π΄(π1, π2, . . . , ππβ1, π₯π, ππ+1, . . . , ππ) β§ π) β¨ π β₯ π΄(π₯1,
π₯2, . . . , π₯π).
Proof. By Definition 2, we observe that
(π΄ (π1, π2, . . . , ππβ1, π₯π, ππ+1, . . . , ππ) β§ π) β¨ π
= (π΄ ((π₯1, π₯2, . . . , π₯π)
β (π₯β1
1, π₯β1
2, . . . , π₯
β1
πβ1, ππ, π₯β1
π+1, . . . , π₯
β1
π)) β§ π) β¨ π
= (π΄ ((π₯1, π₯2, . . . , π₯π)
β (π₯β1
1, π₯β1
2, . . . , π₯
β1
πβ1, ππ, π₯β1
π+1, . . . , π₯
β1
π)) β¨ π) β§ π
β₯ (π΄ (π₯1, π₯2, . . . , π₯π)
β§ π΄ (π₯β1
1, π₯β1
2, . . . , π₯
β1
πβ1, ππ, π₯β1
π+1, . . . , π₯
β1
π) β§ π) β¨ π
= (π΄ (π₯1, π₯2, . . . , π₯π) β¨ π)
β§ (π΄ (π₯β1
1, π₯β1
2, . . . , π₯
β1
πβ1, ππ, π₯β1
π+1, . . . , π₯
β1
π) β¨ π)
β§ π
β₯ ((π΄ (π₯1, π₯2, . . . , π₯π) β¨ π)
β§ π΄ (π₯1, π₯2, . . . , π₯πβ1, ππ, π₯π+1, . . . , π₯π) β§ π) β¨ π
= (π΄ (π₯1, π₯2, . . . , π₯π) β¨ π)
β§ (π΄ (π₯1, π₯2, . . . , π₯πβ1, ππ, π₯π+1, . . . , π₯π) β¨ π) β§ π
β₯ ((π΄ (π₯1, π₯2, . . . , π₯π) β¨ π)
β§ (π΄ (π1, . . . , ππβ2, π₯πβ1, ππ, . . . , ππ)
β§ π΄ (π₯1, . . . , ππβ1, ππ, π₯π+1, . . . , π₯π) β§ π)) β¨ π
= (π΄ (π₯1, π₯2, . . . , π₯π) β¨ π)
β§ (π΄ (π₯1, . . . , ππβ1, ππ, π₯π+1, . . . , π₯π) β§ π) β¨ π
...
β₯ π΄ (π₯1, π₯2, . . . , π₯π) β¨ π
β₯ π΄ (π₯1, π₯2, . . . , π₯π) .
(15)
Lemma 14. Let πΊ1, πΊ2, . . . , πΊπ be groups and π΄ β πΉπ(0,
π, πΊ1ΓπΊ2Γβ β β ΓπΊπ, πΏ).Thenπ΄(π1, π2, . . . , ππβ1, π₯π, ππ+1, . . . , ππ)β§π΄(π₯1, π₯2, . . . , π₯πβ1, ππ, π₯π+1, . . . , π₯π) β§ π = π΄(π₯1, π₯2, . . . , π₯π) forπ = 1, 2, . . . , π if and only if π΄(π1, π2, . . . , ππβ1, π₯π, ππ+1, . . . , ππ) β§π β₯ π΄(π₯1, π₯2, . . . , π₯π) for π = 1, 2, . . . , π β 1, π + 1, . . . , π.
Proof. Now assume that π΄(π1, π2, . . . , ππβ1, π₯π, ππ+1, . . . , ππ) β§π΄(π₯1, π₯2, . . . , π₯πβ1, ππ, π₯π+1, . . . , π₯π) β§ π = π΄(π₯1, π₯2, . . . , π₯π) forπ = 1, 2, . . . , π. Next, for any (π₯1, π₯2, . . . , π₯π) β πΊ1ΓπΊ2Γβ β β ΓπΊπ,we have
π΄ (π₯1, π₯2, . . . , π₯π)
= π΄ (π1, π2, . . . , ππβ1, π₯π, ππ+1, . . . , ππ)
β§ π΄ (π₯1, π₯2, . . . , π₯πβ1, ππ, π₯π+1, . . . , π₯π) β§ π
β€ π΄ (π1, π2, . . . , ππβ1, π₯π, ππ+1, . . . , ππ) β§ π.
(16)
Conversely, assume that π΄(π1, π2, . . . , ππβ1, π₯π, ππ+1, . . . , ππ) β§π β₯ π΄(π₯1, π₯2, . . . , π₯π) for π = 1, 2, . . . , π β 1, π + 1, . . . , π. By
Journal of Mathematics 5
Definition 2 and Lemma 11, we obtain that
π΄ (π₯1, π₯2, . . . , π₯π) = π΄ (π₯1, π₯2, . . . , π₯π)
β§ π΄ (π₯1, π₯2, . . . , π₯π) β§ β β β β§ π΄ (π₯1, π₯2, . . . , π₯π)
β€ π΄ (π₯1, π2, . . . , ππ) β§ π΄ (π1, π₯2, . . . , ππ) β§ β β β
β§ π΄ (π1, π2, . . . , π₯π) β§ π
β€ π΄ (π₯1, π2, . . . , ππ) β§ π΄ (π1, π₯2, . . . , ππ) β§ β β β
β§ π΄ (π1, π2, . . . , π₯πβ1, π₯π) β§ π
...
β€ π΄ (π₯1, π2, . . . , ππ) β§ π΄ (π1, π₯2, . . . , π₯π) β§ π
β€ π΄ (π₯1, π₯2, . . . , π₯π) .
(17)
Hence, π΄(π₯1, π₯2, . . . , π₯π) = π΄(π₯1, π2, . . . , ππ) β§ π΄(π1, π₯2, . . . ,π₯π) β§ π. Similarly, we get π΄(π1, π2, . . . , ππβ1, π₯π, ππ+1, . . . , ππ) β§π΄(π₯1, π₯2, . . . , π₯πβ1, ππ, π₯π+1, . . . , π₯π) = π΄(π₯1, π₯2, . . . , π₯π) for π =2, 3, . . . , π.
As a consequence of Theorem 12 and Lemma 14, we havethe following corollary.
Corollary 15. Let πΊ1, πΊ2, . . . , πΊπ be groups and π΄ β πΉπ(0, π,πΊ1 ΓπΊ2 Γ β β β ΓπΊπ, πΏ). Thenπ΄(π1, π2, . . . , ππβ1, π₯π, ππ+1, . . . , ππ) β§π β₯ π΄(π₯1, π₯2, . . . , π₯π) for π = 1, 2, . . . , π β 1, π + 1, . . . , π if andonly if π΄ = π΄1Γππ΄2Γπ β β β Γππ΄π, where π΄ π(π₯) = π΄(π1,
π2, . . . , ππβ1, π₯, ππ+1, . . . , ππ).
The following example shows that Corollary 15 may notbe true when π = 0.
Example 16. Consider
π΄ (π₯) =
{{{{{{{
{{{{{{{
{
0.4, π₯ = (0, 0) ,
0.3, π₯ = (1, 0) ,
0.2, π₯ = (0, 1) ,
0.1, π₯ = (1, 1) .
(18)
π΄ is (0.2, 0.5)-fuzzy subgroup of Z2 Γ Z2 and satisfies thenecessary condition of Corollary 15. But there is not any π΄1and π΄2, (0.2, 0.5)-fuzzy subgroup of Z2 Γ Z2, which holdπ΄ = π΄1Γ
0.2
0.5π΄2.
4. Conclusion
In this study, we give a necessary and sufficient condition for(0, π)-πΏ-subgroup of a Cartesian product of groups to be aproduct of (0, π)-πΏ-subgroups. The results obtained are notvalid for π = 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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