Research ArticleThe Homomorphisms and Operations of Rough Groups

Fei Li1 and Zhenliang Zhang2

1 Department of Mathematics Beijing Forestry University Beijing 100083 China2Department of Mathematics Kunming Institute of Technology Kunming Yunnan 650000 China

Correspondence should be addressed to Fei Li lifei 1004163com

Received 28 April 2014 Accepted 18 May 2014 Published 5 June 2014

Academic Editor Yunqiang Yin

Copyright copy 2014 F Li and Z Zhang This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Based on the light relation between a normal subgroup and a complete congruence relation of a group we consider thehomomorphism problem of rough groups and rough quotient groups and investigate their operational properties Some new resultsare obtained

1 Introduction

Rough set theory proposed by Pawlak [1] is an extension ofset theory for study of information systems characterized byinexact and uncertain information It has been demonstratedto be useful in fields such as knowledge discovery datamining decision analysis pattern recognition and algebra

Rough set theory includes three basic elements theuniverse set the binary relations and a subset describedby a pair of ordinary sets In the past few years moststudies have been focusing on the binary relations and thesubsets many interesting and constructive extensions tobinary relations and the subsets have been proposed [1ndash8]But the researchers have paid little attention to another basicelement the universe set In real world some universe hasbeen given operations such as the set of natural numbers andthe set of real numbers So it is very natural to askwhat wouldhappen if we substitute an algebraic structure for the universeset Biswas and Nanda [9] generalized the universe of roughsets to groups and introduced the notion of rough subgroupsand some new properties of rough approximations have beenobtained Jiang et al [10] investigated the product structure offuzzy rough sets on a group and provided some new algebraicstructures Yin et al [11] studied fuzzy roughness of 119899-aryhypergroups based on a complete residuated lattice Xiao andZhang [12] studied the rough sets on a semigroup and pro-posed two new algebraic structuresmdashrough prime ideals andrough fuzzy prime ideals In [6] a new algebraic definitionfor pseudo-Cayley graphs containing Cayley graphs has been

proposed a rough approximation was expanded to pseudo-Cayley graphs and some new properties have been obtainedFor more other papers on this line please refer to [12ndash24]which have greatly enriched the theoretical research of roughsets

The aim of this paper is to investigate the homomorphismproblem of rough group and rough quotient groups The restof the paper is organized as follows In Section 2 we recallsome basic notions and results which will be used throughoutthe paper In Section 3 the homomorphism problems ofrough groups and rough quotient groups are studied andsome related properties are discussed In Section 4 con-gruence relation and the operation of rough groups areinvestigated

2 Preliminaries

Let 119866 [25] be a group with unit element 119890 and let 120588 bean equivalence relation on 119866 If forall(119886119887) (119888119889) isin 120588 we have(119886119888 119887119889) isin 120588 then 120588 is called a congruence relation and[119886]120588 indicates congruence class of 119886 about 120588 Furthermore if[119886]120588[119887]120588 = [119886119887]120588 then 120588 is called complete

Lemma 1 (see [1]) Let 120588 be a congruence relation of 119866 Then119873120588 = 119886 isin 119866 | (119886 119890) isin 120588 is a normal subgroup of 119866 and(119886 119887) isin 120588 hArr 119886 isin 119887119873120588 contrarily if 119873 is a normal subgroupof 119866 one can define a congruence relation 120588119873 where (119886 119887) isin

120588119873 hArr 119886 isin 119887119873120588 then 120588119873 is a congruence relation of 119866 and119873120588 = 119886 isin 119866 | (119886 119890) isin 120588119873

Hindawi Publishing Corporatione Scientific World JournalVolume 2014 Article ID 507972 6 pageshttpdxdoiorg1011552014507972

2 The Scientific World Journal

Lemma 1 indicates the congruence relation of 119866 and thenormal subgroup of 119866 is one to one correspondence

Theorem 2 Let 119873 be a normal subgroup of 119866 then 120588119873 is acomplete congruence relation

Proof Consider forall119886 119887 isin 119866[119886]120588119873[119887]120588119873

= (119886119873)(119887119873) = 119886119887119873 =

[119886119887]120588119873 Therefore 120588119873 is a complete congruence relation

According to Theorem 2 we know the complete congru-ence relation of 119866 and the normal subgroup of 119866 is also oneto one correspondence

Definition 3 (see [1]) Let 120588 be an equivalence relation on 119866

and 119860 a nonempty subset of 119866 Then the sets

120588 (119860) = cup [119909]120588 | [119909]120588 sube 119860

120588 (119860) = cup [119909]120588 | [119909]120588 cap 119860 = 0

(1)

are called respectively the 120588-lower and 120588-upper approxima-tions of the set 119860 And 120588(119860) = (120588(119860) 120588(119860)) is called a roughset with respect to 120588

Definition 4 (see [1]) Let 120588 be an equivalence relation on 119866

and 119860 a nonempty subset of 119866 Then

120588 (119860)

120588

= [119909]120588 | [119909]120588 sube 119860

120588 (119860)

120588

= [119909]120588 | [119909]120588 cap 119860 = 0

(2)

are called respectively the lower and upper rough quotientof 119860

Lemma 5 (see [1]) Let 120588 be an equivalence relation on 119866 andfor all 119860 119861 sube 119866 one has the following

(1) 120588(119860) sube 119860 sube 120588(119860)(2) 120588(119860 cup 119861) = 120588(119860) cup 120588(119861) 120588(119860 cap 119861) sube 120588(119860) cap 120588(119861)(3) 120588(119860 cup 119861) supe 120588(119860) cup 120588(119861) 120588(119860 cap 119861) = 120588(119860) cap 120588(119861)(4) If 119860 sube 119861 then 120588(119860) sube 120588(119861) 120588(119860) sube 120588(119861)

Lemma 6 Let 120588 be an equivalence relation on 119866 and forall119860 119861 sube

119866 one has the following(1) 120588(119860)120588 sube 119860120588 sube 120588(119860)120588(2) 120588(119860cup119861)120588 = 120588(119860)120588cup120588(119861)120588 120588(119860cap119861)120588 sube 120588(119860)120588cap

120588(119861)120588(3) 120588(119860cup119861)120588 supe 120588(119860)120588cup120588(119861)120588 120588(119860cap119861)120588 = 120588(119860)120588cap

120588(119861)120588(4) If 119860 sube 119861 then 120588(119860)120588 sube 120588(119861)120588 120588(119860)120588 sube 120588(119861)120588

3 The Homomorphism Problem of RoughGroups and Rough Quotient Groups

Definition 7 Let 120588 be a complete congruence relation on 119866

and 119860 sube 119866 if 120588(119860) is a subgroup (a normal subgroup)

of 119866 then 119860 is called the lower rough group (lower roughnormal subgroup) of 119866 and 120588(119860)120588 is called the lower roughquotient group (lower rough normal quotient group) if 120588(119860)is a subgroup (a normal subgroup) of 119866 then 119860 is called theupper rough group (upper rough normal subgroup) of119866 and120588(119860)120588 is called the upper roughquotient group (upper roughnormal quotient group) if 120588(119860) 120588(119860) are all the subgroup(normal subgroup) of119866 then119860 is called rough group (roughnormal subgroup) of 119866

Theorem 8 Let 120588 be a complete congruence relation on 119866119860 sube 119866 and 119873120588 sube 119860 If 119860 is a subgroup (normal subgroup)of 119866 then 119860 is rough group (rough normal subgroup) of 119866 and120588(119860)120588 120588(119860)120588 are respectively lower rough quotient group(lower rough normal quotient group) and upper rough quotientgroup (upper rough normal quotient group)

Corollary 9 Let 120588 be a complete congruence relation on 119866119860 sube 119866 and119873120588 sube 119860 Then 120588(119860)(120588(119860)) is a subgroup (normalsubgroup) of 119866 hArr 120588(119860)120588(120588(119860)120588) is a subgroup (normalsubgroup) of 119866120588

Corollary 10 Let 120588 be a complete congruence relation on 119866119860 sube 119866 and 119873120588 sube 119860 If 119860 is a subgroup (normal subgroup) of119866 then

(1) 120588(119860) lt 119860 lt 120588(119860) lt 119866(120588(119860) ⊲ 119860 ⊲ 120588(119860) ⊲ 119866)

(2) 120588(119860)120588 lt 119860120588 lt 120588(119860)120588 lt 119866120588(120588(119860)120588 ⊲ 119860120588 ⊲ 120588(119860)120588 ⊲ 119866120588)

Corollary 11 Let 120588 be a complete congruence relation on 119866

and let 119860 be a subgroup of 119866 and 119873120588 sube 119860 Then 120588(119860)119860 cong

(120588120588(119860))(119860120588(119860))

Proof Based on the third isomorphism theorem of group itis easy to prove this corollary

Lemma 12 Let 119866 and 119879 be two groups 119891 119866 rarr 119879 is ahomomorphism If 120588 is a congruence relation on 119866 and 119873120588 supe

Ker119891 then 119891(120588) is a congruence relation on 119879 Further if 119891 isan injective then

(1) 119891(120588) = 120588119891(119873120588)

(2) 119891(119873120588) = 119873119891(120588)where 119891(120588) = (119891(119886) 119891(119887)) | (119886 119887) isin 120588

Proof It is easy to prove 119891(120588) is a congruence relation on 119879(1) Because 119873120588 is a normal subgroup on 119866 then 119891(119873120588)

is the normal subgroup on 119879 So forall(119891(119886) 119891(119887)) isin

119891(120588) hArr (119886 119887) isin 120588 hArr 119886 isin 119887119873120588 hArr 119891(119886) isin

119891(119887)119891(119873120588) hArr (119891(119886) 119891(119887)) isin 120588119891(119873120588) so119891(120588) = 120588119891(119873120588)

(2) It follows immediately from (1)

Theorem 13 Let 119891 119866 rarr 119879 be an injective homomorphismIf 120588 is a complete congruence relation on 119866 and 119873120588 supe Ker119891then 119891(120588) is a complete congruence relation on 119879

The Scientific World Journal 3

Proof According to Lemma 12 it is easy to prove 119891(120588)

is a congruence relation on 119879 Now we prove it is com-plete forall119886 119887 isin 119866 we have [119891(119886)]

119891(120588)[119891(119887)]

119891(120588)=

(119891(119886)119891(119873120588))(119891(119887)119891(119873120588)) = 119891(119886119873120588119887119873120588) = 119891(119886119887)119891(119873120588)

= [119891(119886119887)]119891(120588)

= [119891(119886)119891(119887)]119891(120588)

Therefore 119891(120588) is complete

Lemma 14 Let 119891 119866 rarr 119879 be a surjective homomorphismIf 120588 is a congruence relation on 119879 then 119891

minus1(120588) is a congruence

relation on 119866 Further if 119891 is an injective then

(1) 119891minus1(120588) = 120588119891minus1(119873120588)

(2) 119891minus1(119873120588) = 119873119891minus1(120588)

where 119891minus1(120588) = (119886 119887) | (119891(119886) 119891(119887)) isin 119891(120588)

Proof It is easy to prove 119891minus1(120588) is a congruence relation on119866

(1) Because119873120588 is a normal subgroup on 119879 then 119891minus1(119873120588)

is the normal subgroup on 119866 So forall(119886 119887) isin 119891minus1(120588) rArr

(119891(119886) 119891(119887)) isin 120588 rArr 119891(119886) isin 119891(119887)119873120588 and because 119891is injective then 119886 isin 119887119891

minus1(119873120588) rArr (119886 119887) isin 120588119891minus1(119873120588)

that is 119891minus1(120588) sube 120588119891minus1(119873120588)

On the contrary (119886 119887) isin

120588119891minus1(119873120588)rArr 119886 isin 119887119891

minus1(119873120588) rArr 119891(119886) isin 119891(119887)119873120588 rArr

(119891(119886) 119891(119887)) isin 120588 rArr (119886 119887) isin 119891minus1(120588) that is 120588119891minus1(119873120588) sube

119891minus1(120588) hence 119891minus1(120588) = 120588119891minus1(119873120588)

(2) It follows immediately from (1)

Theorem 15 Let 119891 119866 rarr 119879 be an injective homomorphismIf 120588 is a complete congruence relation on 119879 then 119891

minus1(120588) is a

complete congruence relation on 119866

Proof According to Lemma 14 it is easy to prove 119891minus1(120588)

is a congruence relation on 119866 Now we prove it is com-plete forall119886 119887 isin 119866 we have [119886]119891minus1(120588)[119887]119891minus1(120588) =

(119886119891minus1(119873120588))(119887119891

minus1(119873120588)) = 119886119887119891

minus1(119873120588) = [119886119887]119891minus1(119873120588)

Therefore 119891minus1(120588) is complete

Lemma 16 Let 119891 119866 rarr 119879 be a surjective homomorphism If120588 is a congruence relation on 119866 119860 sube 119866 and Ker119891 sube 119873120588 sube 119860then

(1) 119891(120588(119860)) = 119891(120588)(119891(119860))(2) if 119891 is injective then 119891(120588(119860)) = 119891(120588)(119891(119860))

Proof (1) Consider forall119887 isin 119891(120588(119860)) rArr exist119886 isin 120588(119860) 119891(119886) = 119887 rArr

[119886]120588 cap 119860 = 0 119891(119886) = 119887 rArr exist1198861015840isin [119886]120588 119886

1015840isin 119860 rArr (119886 119886

1015840) isin 120588

1198861015840isin 119860 rArr (119891(119886) 119891(119886

1015840)) isin 119891(120588) 119891(1198861015840) isin 119891(119860) rArr 119891(119886

1015840) isin

[119891(119886)]119891(120588)

119891(1198861015840) isin 119891(119860) rArr [119891(119886)]119891(120588)

cap 119891(119860) = 0 rArr 119891(119886) =119887 isin 119891(120588)(119891(119860)) that is 119891(120588(119860)) sube 119891(120588)(119891(119860)) on thecontrary forall119887 isin 119891(120588)(119891(119860)) rArr exist119886 isin 119866 119891(119886) = 119887 [119887]119891(120588) cap119891(119860) = 0 rArr exist119887

1015840isin 119891(119860) 1198871015840 isin [119887]119891(120588) rArr exist119886

1015840isin 119860 119891(1198861015840) = 119887

1015840

119891(1198861015840) isin [119887]119891(120588) rArr (119891(119886

1015840)119891(119886)) isin 119891(120588) 1198861015840 isin 119860 rArr (119886

1015840 119886) isin 120588

1198861015840isin 119860 rArr 119886

1015840isin [119886]120588 119886

1015840isin 119860 rArr [119886]120588 cap 119860 = 0 rArr 119886 isin 120588(119860) rArr

119891(119886) = 119887 isin 119891(120588(119860)) that is 119891(120588(119860)) supe 119891(120588)(119891(119860)) so119891(120588(119860)) = 119891(120588)(119891(119860))

(2) Consider forall119887 isin 119891(120588(119860)) hArr exist119886 isin 120588(119860) 119891(119886) = 119887 hArr

[119886]120588 = 119886119873120588 sube 119860 hArr 119891([119886]120588) = 119891(119886)119891(119873120588) sube 119891(119860) hArr 119887119873119891(120588) =[119887]119891(120588) sube 119891(119860) hArr 119887 isin 119891(120588)(119891(119860)) so 119891(120588(119860)) = 119891(120588)(119891(119860))

Theorem 17 Let 119891 119866 rarr 119879 be a surjective homomorphismlet 120588 be a congruence relation on 119866 and let 119860 be a subgroup of119866 and Ker119891 sube 119873120588 sube 119860 Then

(1) 120588(119860)120588 cong 119891(120588)(119891(119860))119891(120588)

(2) if 119891 is injective then 120588(119860)120588 cong 119891(120588)(119891(119860))119891(120588)

Proof (1) Suppose that 119860 is a subgroup of 119866 then 120588(119860) is asubgroup on119866 and119891(120588)(119891(119860)) is a subgroup on119879 accordingto Lemma 16 119891(120588)(119891(119860)) = 119891(120588(119860)) therefore 120588(119860)120588 cong

119891(120588)(119891(119860))119891(120588)(2) If 119891 is injective according to Lemma 16 we have

119891(120588)(119891(119860)) = 119891(120588(119860)) therefore 120588(119860)120588 cong 119891(120588)(119891(119860))119891(120588)

Corollary 18 Let 119891 119866 rarr 119879 be a surjective homomorphismlet 120588 be a complete congruence relation of 119866 based on Ker119891and let 119860 be a subgroup of 119866 and 119860 supe Ker119891 Then

(1) 120588(119860)120588 cong 119891(120588(119860))

(2) if 119891 is injective then 120588(119860)120588 cong 119891(120588(119860))

Lemma 19 Let 119891 119866 rarr 119879 be a surjective homomorphismand let 120588 be a congruence relation on 119879 and 119861 sube 119879 Then

(1) 119891minus1(120588(119861)) = 119891minus1(120588)(119891minus1(119861))

(2) if 119891 is an injective then 119891minus1(120588(119861)) = 119891

minus1(120588)(119891minus1(119861))

Proof (1) Consider forall119886 isin 119891minus1(120588(119861)) hArr 119891(119886) isin 120588(119861) hArr

[119891(119886)]120588cap 119861 = 0 hArr exist119886

1015840isin 119866 119891(1198861015840) isin 119861 119891(119886

1015840) isin [119891(119886)]

120588hArr

1198861015840isin 119891minus1(119861) (119891(1198861015840) 119891(119886)) isin 120588 hArr 119886

1015840isin 119891minus1(119861) (1198861015840 119886) isin

119891minus1(120588) hArr [119886]119891minus1(120588) cap 119891

minus1(119861) = 0 hArr 119886 isin 119891

minus1(120588)(119891minus1(119861)) and

hence 119891minus1(120588(119861)) = 119891minus1(120588)(119891minus1(119861))

(2) Consider forall119886 isin 119891minus1(120588(119861)) hArr 119891(119886) isin 120588(119861) hArr

[119891(119886)]120588= 119891([119886]119891minus1(120588)) sube 119861 hArr [119886]119891minus1(120588) sube 119891

minus1(119861) hArr 119886 isin

119891minus1(120588)(119891minus1(119861)) so 119891

minus1(120588(119861)) = 119891

minus1(120588)(119891minus1(119861))

Theorem 20 Let 119891 119866 rarr 119879 be a surjective homomorphismand let 120588 be a congruence relation on 119879 and 119861 is a subgroup of119879 Then

(1) 119891minus1(120588)(119891minus1(119861))119891minus1(120588) cong 120588(119861)

(2) if 119891 is an injective then 119891minus1(120588)(119891minus1(119861))119891

minus1(120588) cong

120588(119861)

4 The Scientific World Journal

Proof By Lemma 19 and the first isomorphism theorem ofgroup we have

119891minus1(120588)(119891minus1(119861))119891

minus1(120588) = 119891

minus1(120588(119861))119891

minus1(120588) cong 120588(119861)

119891minus1(120588)(119891minus1(119861))119891

minus1(120588) = 119891

minus1(120588(119861))119891

minus1(120588) cong 120588(119861)

4 Congruence Relation and the Operation ofRough Group

Lemma 21 Let 120588 984858 be the congruence relations on 119866 Then119873120588cap984858 = 119873120588 cap 119873984858

Proof Consider forall119886 isin 119873120588cap984858 hArr (119886 119890) isin 120588 cap 984858 hArr (119886 119890) isin 120588(119886 119890) isin 984858 hArr 119886 isin 119873120588 119886 isin 119873984858 hArr 119886 isin 119873120588 cap 119873984858 and therefore119873120588cap984858 = 119873120588 cap 119873984858

Lemma 22 Let 120588 984858 be two congruence relations on 119866 and 119886 isin

119866 Then 119886(119873120588 cap 119873984858) = 119886119873120588 cap 119886119873984858

Proof It is easy to prove that 119886(119873120588cap119873984858) sube 119886119873120588cap119886119873984858 On thecontrary forall119887 isin 119886119873120588 cap 119886119873984858 rArr 119887 isin 119886119873120588 119887 isin 119886119873984858 rArr (119886 119887) isin 120588(119886 119887) isin 984858 rArr (119886 119887) isin 120588 cap 984858 rArr 119887 isin 119886119873120588cap984858 = 119886(119873120588 cap 119873984858)that is 119886119873120588 cap 119886119873984858 sube 119886(119873120588 cap 119873984858) therefore 119886119873120588 cap 119886119873984858 =

119886(119873120588 cap 119873984858)

Lemma 23 Let 120588 984858 be two congruence relations on 119866 and 119886 isin

119866 Then [119886]120588cap984858 = [119886]120588 cap [119886]984858

Proof Consider forall119887 isin [119886]120588cap984858 hArr 119887 isin 119886119873120588cap984858 = 119886(119873120588 cap 119873984858) =

119886119873120588 cap 119886119873984858 hArr 119887 isin 119886119873120588 119887 isin 119886119873984858 hArr 119887 isin [119886]120588 cap [119886]984858 hence[119886]120588cap984858 = [119886]120588 cap [119886]984858

Theorem 24 Let 120588 984858 be two complete congruence relations on119866 Then 120588 cap 984858 is a complete congruence relation on 119866

Proof Because 119873120588 119873984858 are both the normal subgroups of 119866then119873120588cap984858 = 119873120588 cap119873984858 is a normal subgroup of 119866 so119873120588cap984858 is acomplete congruence relation

Theorem 25 Let 120588 984858 be two congruence relations on 119866 and119860 sube 119866 Then

(1) 120588 cap 984858(119860) sube 120588(119860) cap 984858(119860)(2) 120588 cap 984858(119860) supe 120588(119860) cup 984858(119860) supe 120588(119860) cap 984858(119860)

Proof (1) Consider forall119886 isin 120588 cap 984858(119860) rArr [119886]120588cap984858 cap119860 = 0 Because[119886]120588cap984858 = [119886]120588 cap [119886]984858 then [119886]120588cap984858 sube [119886]120588([119886]984858) rArr [119886]120588 cap119860 = 0[119886]984858 cap 119860 = 0 rArr 119886 isin 120588(119860) 119886 isin 984858(119860) rArr 119886 isin 120588(119860) cap 984858(119860)therefore 120588 cap 984858(119860) sube 120588(119860) cap 984858(119860)

(2) Consider forall119886 isin 120588(119860) cup 984858(119860) rArr 119886 isin 120588(119860) or 119886 isin

984858(119860) rArr [119886]120588 sube 119860 or [119886]984858 sube 119860 rArr [119886]120588cap984858 = [119886]120588 cap [119886]984858 sube

119860 rArr 119886 isin 120588 cap 984858(119860) and therefore 120588 cap 984858(119860) supe 120588(119860) cup 984858(119860) supe

120588(119860) cap 984858(119860)

Lemma 26 (see [25]) Let 120588 984858 be two congruence relations on119866 Then 120588 ∘ 984858 is a congruence relation on 119866 hArr 120588 ∘ 984858 = 984858 ∘ 120588

Theorem 27 Let 120588 984858 be two complete congruence relations on119866 and 120588∘984858 = 984858∘120588 Then 120588∘984858 is a complete congruence relationon 119866

Proof By Lemma 26 120588 ∘ 984858 is the congruence relation and[119886]120588∘984858[119887]120588∘984858 = (119886119873120588∘984858)(119887119873120588∘984858) = 119886119887119873120588∘984858 = [119886119887]120588∘984858 hence120588 ∘ 984858 is the complete congruence relation

Lemma 28 Let 120588 984858 be two congruence relations on 119866 Then119873120588∘984858 = 119873120588119873984858

Proof Consider forall119886 isin 119873120588cap984858 rArr (119886 119890) isin 120588 cap 984858 rArr exist119887 isin 119866(119886 119887) isin 120588 (119887 119890) isin 984858 rArr 119886 isin 119887119873120588 119887 isin 119873984858 rArr 119886119887 isin

119887119873120588119873984858 rArr 119886 isin 119873120588119873984858 that is119873120588∘984858 sube 119873120588119873984858 On the contraryforall119886 isin 119873120588119873984858 rArr exist119887 isin 119873120588 119888 isin 119873984858 119886 = 119887119888 rArr (119887 119890) isin 120588 (119888 119890) isin984858 rArr (119887119888 119888) isin 120588 (119888 119890) isin 984858 rArr (119887119888 119890) isin 120588 ∘ 984858 rArr 119887119888 = 119886 isin 119873120588∘984858that is119873120588∘984858 supe 119873120588119873984858 hence119873120588∘984858 = 119873120588119873984858

Lemma 29 Let 120588 984858 be two congruence relations on 119866 and119886 119887 isin 119866 Then [119886119887]120588∘984858 = [119886]120588[119887]984858

Proof Consider [119886119887]120588∘984858 = 119886119887119873120588∘984858 = 119886119887119873120588119873984858 =

(119886119873120588)(119887119873984858) = [119886]120588[119887]984858

Theorem 30 Let 120588 984858 be two congruence relations on 119866 and120588 ∘ 984858 = 984858 ∘ 120588 If 119860 is a subgroup of 119866 and119873120588 119873984858 sube 119860 Then

(1) 120588 ∘ 984858(119860) supe 120588(119860)984858(119860)(2) 120588 ∘ 984858(119860) = 120588(119860)984858(119860)

Proof (1) Consider forall119886 isin 120588(119860)984858(119860) rArr exist119887 isin 120588(119860) 119888 isin 984858(119860)119886 = 119887119888 rArr [119887]120588 cap 119860 = 0 [119888]984858 cap 119860 = 0 119886 = 119887119888 rArr exist119887

1015840isin [119887]120588

1198871015840isin 119860 1198881015840 isin [119888]984858 119888

1015840isin 119860 rArr 119887

10158401198881015840isin [119887]120588[119888]984858 = [119887119888]120588∘984858 =

[119886]120588∘984858 11988710158401198881015840isin 119860119860 = 119860 rArr [119886]120588∘984858 cap 119860 = 0 rArr 119886 isin 120588 ∘ 984858(119860) so

120588 ∘ 984858(119860) supe 120588(119860)984858(119860)(2) Consider forall119886 isin 120588 ∘ 984858(119860) rArr [119886]120588∘984858 = [119886]120588[119890]984858 sube 119860

because [119886]120588 = [119886]120588[119890] sube [119886]120588[119890]984858 sube 119860 and 119873984858 = [119890]119886 sube

119860 rArr 119886 isin 120588(119860) 119890 isin 984858(119860) rArr 119886 = 119886119890 isin 120588(119860)984858(119860) that is120588 ∘ 984858(119860) sube 120588(119860)984858(119860) On the contrary forall119886 isin 120588(119860)984858(119860) rArr

exist119887 isin 120588(119860) 119888 isin 984858(119860) 119886 = 119887119888 rArr [119887]120588 sube 119860 [119888]984858 sube 119860 119886 = 119887119888 rArr

[119886]120588 ∘ 984858 = [119887119888]120588 ∘ 984858 = [119887]120588[119888]984858 sube 119860119860 = 119860 rArr 119886 isin 120588 ∘ 984858(119860) thatis 120588(119860)984858(119860) sube 120588 ∘ 984858(119860) so 120588 ∘ 984858(119860) = 120588(119860)984858(119860)

Theorem 31 Let 120588 be complete congruence relation on 119866 and119860 119861 sube 119866 Then

(1) 120588(119860119861) = 120588(119860)120588(119861)(2) 120588(119860119861) = 120588(119860)120588(119861)

Proof (1) Consider forall119886 isin 120588(119860119861) rArr [119886]120588 cap 119860119861 = 0 rArr

exist119887 isin [119886]120588 119887 isin 119860119861 rArr 119886 isin 119887119873120588 exist119888 isin 119860 119889 isin 119861119887 = 119888119889 rArr 119886 isin 119888119889119873120588 = (119888119873120588)(119889119873120588) and [119888]120588 cap 119860 = 0[119889]120588 cap 119861 = 0 rArr exist119888

1015840isin 119873120588 119889

1015840isin 119889119873120588 119886 = 119888

10158401198891015840 because

[1198881015840]120588 = [119888]120588 [119889

1015840]120588 = [119889]120588 rArr [119888

1015840]120588 cap 119860 = 0 [1198891015840]120588 cap 119861 = 0

119886 = 11988810158401198891015840rArr 119886 = 119887

10158401198881015840isin 120588(119860)120588(119861) that is 120588(119860119861) sube 120588(119860)120588(119861)

On the contrary forall119886 isin 120588(119860)120588(119861) rArr exist119887 isin 120588(119860) 119888 isin 120588(119861)119886 = 119887119888 rArr [119887]120588 cap 119860 = 0 [119888]120588 cap 119861 = 0 rArr exist119887

1015840isin [119887]120588 119887

1015840isin 119860

The Scientific World Journal 5

1198881015840isin [119888]120588 119888

1015840isin 119861 rArr 119887

10158401198881015840isin [119887]120588[119888]120588 = [119887119888]120588 = [119886]120588

11988710158401198881015840isin 119860119861 rArr [119886]120588 cap119860119861 = 0 rArr 119886 isin 120588(119860119861) that is 120588(119860)120588(119861) sube

120588(119860119861) Therefore 120588(119860119861) = 120588(119860)120588(119861)(2) forall119886 isin 120588(119860119861) rArr [119886]120588 sube 119860119861 rArr exist119887 isin 119860 119888 isin 119861 119886 =

119887119888 rArr [119887]120588[119888]120588 = [119887119888]120588 = [119886]120588 sube 119860119861 if exist1198871015840 isin [119887]120588 1198871015840ni 119860 119888

1015840isin

[119888]120588 1198881015840ni 119861 then 119887

10158401198881015840isin [119887]120588[119888]120588 119887

10158401198881015840ni 119860119861 Contradiction

that is [119887]120588 sube 119860 [119888]120588 sube 119861 so 119886 = 119887119888 isin 120588(119860)120588(119861) thatis 120588(119860119861) sube 120588(119860)120588(119861) On the contrary forall119886 isin 120588(119860)120588(119861) rArr

exist119887 isin 120588(119860) 119888 isin 120588(119861) 119886 = 119887119888 rArr [119887]120588 sube 119860 [119888]120588 sube 119861 119886 =

119887119888 rArr [119886]120588 = [119887119888]120588 = [119887]120588[119888]120588 sube 119860119861 rArr 119886 isin 120588(119860119861) that is120588(119860)120588(119861) sube 120588(119860119861) so 120588(119860119861) = 120588(119860)120588(119861)

Corollary 32 Let 120588 be a complete congruence relation on 119866

and 119860 sube 119866 119899 isin 119873 Then(1) 120588(119860119899) = (120588(119860))

119899(2) 120588(119860119899) = (120588(119860))

119899

Corollary 33 Let 120588 be a complete congruence relation on 119866

and let 119860 be a subgroup of 119866 and 119899 isin 119873 Then(1) 120588119899(119860) = (120588(119860))

119899= 120588(119860

119899) = 120588(119860)

(2) 120588119899(119860) = (120588(119860))119899= 120588(119860

119899) = 120588(119860)

Proof Because 119860 is a subgroup then 120588(119860) 120588(119860) are both thesubgroups of 119866 so (120588(119860))

119899= 120588(119860) (120588(119860))119899 = 120588(119860) 119860119899 = 119860

and 120588 is an equivalence relation then 120588119899= 120588 hence we can

get (1) and (2)

Corollary 34 Let 120588 be a complete congruence relation on 119866

and let 119860 119861 be two subgroups of 119866 Then(1) 120588(119860 cup 119861) = 120588(119860) cup 120588(119861) sube 120588(119860119861) = 120588(119860)120588(119861)(2) 120588(119860 cup 119861) sube 120588(119860119861) = 120588(119860)120588(119861)

Proof (1) Because 119860 = 119860(119890) sube 119860119861 and 119861 sube 119860119861 so 119860 cup 119861 sube

119860119861 hence 120588(119860 cup 119861) sube 120588(119860119861) and 120588(119860 cup 119861) = 120588(119860) cup 120588(119861)therefore (1) holds (2) Because 119860 cup 119861 sube 119860119861 so 120588(119860 cup 119861) sube

120588(119860119861) = 120588(119860)120588(119861)

Conflict of Interests

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

Acknowledgments

This research was supported by the Fundamental ResearchFunds for the Central Universities (no YX2014-08) and theNSFC of China (Grant no 61370193)

References

[1] Z Pawlak ldquoRough setsrdquo International Journal of Computer ampInformation Sciences vol 11 no 5 pp 341ndash356 1982

[2] G Liu and W Zhu ldquoThe algebraic structures of generalizedrough set theoryrdquo Information Sciences vol 178 no 21 pp 4105ndash4113 2008

[3] J N Mordeson ldquoRough set theory applied to (fuzzy) idealtheoryrdquo Fuzzy Sets and Systems vol 121 no 2 pp 315ndash324 2001

[4] Z Pawlak and A Skowron ldquoRough sets some extensionsrdquoInformation Sciences vol 177 no 1 pp 28ndash40 2007

[5] Z Pawlak andA Skowron ldquoRough sets andBoolean reasoningrdquoInformation Sciences vol 177 no 1 pp 41ndash73 2007

[6] M H Shahzamanian M Shirmohammadi and B DavvazldquoRoughness inCayley graphsrdquo Information Sciences vol 180 no17 pp 3362ndash3372 2010

[7] Z Pawlak and A Skowron ldquoRudiments of rough setsrdquo Informa-tion Sciences vol 177 no 1 pp 3ndash27 2007

[8] V Leoreanu-Fotea and B Davvaz ldquoRoughness in n-ary hyper-groupsrdquo Information Sciences vol 178 no 21 pp 4114ndash41242008

[9] R Biswas and S Nanda ldquoRough groups and rough subgroupsrdquoBulletin of the Polish Academy of Sciences Mathematics vol 42pp 251ndash254 1994

[10] J S Jiang C X Wu and D G Chen ldquoThe product structureof fuzzy rough sets on a group and the rough T-fuzzy grouprdquoInformation Sciences vol 175 no 1-2 pp 97ndash107 2005

[11] Y Q Yin J M Zhan and P Corsini ldquoFuzzy roughness of n-ary hypergroups based on a complete residuated latticerdquoNeuralComputing and Applications vol 20 no 1 pp 41ndash57 2011

[12] Q M Xiao and Z L Zhang ldquoRough prime ideals and roughfuzzy prime ideals in semigroupsrdquo Information Sciences vol 176no 6 pp 725ndash733 2006

[13] N Kuroki ldquoRough ideals in semigroupsrdquo Information Sciencesvol 100 no 1ndash4 pp 139ndash163 1997

[14] N Kuroki and P PWang ldquoThe lower and upper approximationsin a fuzzy grouprdquo Information Sciences vol 90 no 1ndash4 pp 203ndash220 1996

[15] K C Gupta and M K Kantroo ldquoGeneralized product of fuzzysubsets of a ringrdquo Fuzzy Sets and Systems vol 117 no 3 pp 419ndash429 2001

[16] O Kazanci and B Davvaz ldquoOn the structure of rough prime(primary) ideals and rough fuzzy prime (primary) ideals incommutative ringsrdquo Information Sciences vol 178 no 5 pp1343ndash1354 2008

[17] N Kuroki and J N Mordeson ldquoStructure of rough sets andrough groupsrdquo Journal of Fuzzy Mathematics vol 5 no 1 pp183ndash191 1997

[18] F Li Y Yin and L Lu ldquo(120599 119879)-fuzzy rough approximationoperators and the TL-fuzzy rough ideals on a ringrdquo InformationSciences vol 177 no 21 pp 4711ndash4726 2007

[19] Y Q Yin and X K Huang ldquoFuzzy roughness in hyperringsbased on a complete residuated latticerdquo International Journal ofFuzzy Systems vol 13 no 3 pp 185ndash194 2011

[20] W J Liu Y D Cu and H X Li ldquoRough fuzzy ideals andrough fuzzy prime ideals in semigroupsrdquo Fuzzy Systems andMathematics vol 21 no 3 pp 127ndash132 2007

[21] J L Zhang and Z L Zhang ldquoRough subgroups and roughsubringsrdquo Pure and Applied Mathematics vol 20 pp 92ndash962004

[22] J L Zhang and Z L Zhang ldquoFuzzy rough subgroupsrdquo FuzzySystems and Mathematics vol 21 no 3 pp 127ndash132 2007

[23] Z L Zhang J L Zhang and Q M Xiao Fuzzy Algebra andRough Algebra Wuhan University Press Wuhan China 2007

6 The Scientific World Journal

[24] B Davvaz and V Leoreanu-Fotea ldquoApplications of intervalvalued fuzzy n-ary polygroups with respect to t-norms (t-conorms)rdquo Computers and Mathematics with Applications vol57 no 8 pp 1413ndash1424 2009

[25] DW ShengAbstractAlgebra Science PublishingPress BeijingChina 2001

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 The Scientific World Journal

Lemma 1 indicates the congruence relation of 119866 and thenormal subgroup of 119866 is one to one correspondence

Theorem 2 Let 119873 be a normal subgroup of 119866 then 120588119873 is acomplete congruence relation

Proof Consider forall119886 119887 isin 119866[119886]120588119873[119887]120588119873

= (119886119873)(119887119873) = 119886119887119873 =

[119886119887]120588119873 Therefore 120588119873 is a complete congruence relation

According to Theorem 2 we know the complete congru-ence relation of 119866 and the normal subgroup of 119866 is also oneto one correspondence

Definition 3 (see [1]) Let 120588 be an equivalence relation on 119866

and 119860 a nonempty subset of 119866 Then the sets

120588 (119860) = cup [119909]120588 | [119909]120588 sube 119860

120588 (119860) = cup [119909]120588 | [119909]120588 cap 119860 = 0

(1)

are called respectively the 120588-lower and 120588-upper approxima-tions of the set 119860 And 120588(119860) = (120588(119860) 120588(119860)) is called a roughset with respect to 120588

Definition 4 (see [1]) Let 120588 be an equivalence relation on 119866

and 119860 a nonempty subset of 119866 Then

120588 (119860)

120588

= [119909]120588 | [119909]120588 sube 119860

120588 (119860)

120588

= [119909]120588 | [119909]120588 cap 119860 = 0

(2)

are called respectively the lower and upper rough quotientof 119860

Lemma 5 (see [1]) Let 120588 be an equivalence relation on 119866 andfor all 119860 119861 sube 119866 one has the following

(1) 120588(119860) sube 119860 sube 120588(119860)(2) 120588(119860 cup 119861) = 120588(119860) cup 120588(119861) 120588(119860 cap 119861) sube 120588(119860) cap 120588(119861)(3) 120588(119860 cup 119861) supe 120588(119860) cup 120588(119861) 120588(119860 cap 119861) = 120588(119860) cap 120588(119861)(4) If 119860 sube 119861 then 120588(119860) sube 120588(119861) 120588(119860) sube 120588(119861)

Lemma 6 Let 120588 be an equivalence relation on 119866 and forall119860 119861 sube

119866 one has the following(1) 120588(119860)120588 sube 119860120588 sube 120588(119860)120588(2) 120588(119860cup119861)120588 = 120588(119860)120588cup120588(119861)120588 120588(119860cap119861)120588 sube 120588(119860)120588cap

120588(119861)120588(3) 120588(119860cup119861)120588 supe 120588(119860)120588cup120588(119861)120588 120588(119860cap119861)120588 = 120588(119860)120588cap

120588(119861)120588(4) If 119860 sube 119861 then 120588(119860)120588 sube 120588(119861)120588 120588(119860)120588 sube 120588(119861)120588

3 The Homomorphism Problem of RoughGroups and Rough Quotient Groups

Definition 7 Let 120588 be a complete congruence relation on 119866

and 119860 sube 119866 if 120588(119860) is a subgroup (a normal subgroup)

of 119866 then 119860 is called the lower rough group (lower roughnormal subgroup) of 119866 and 120588(119860)120588 is called the lower roughquotient group (lower rough normal quotient group) if 120588(119860)is a subgroup (a normal subgroup) of 119866 then 119860 is called theupper rough group (upper rough normal subgroup) of119866 and120588(119860)120588 is called the upper roughquotient group (upper roughnormal quotient group) if 120588(119860) 120588(119860) are all the subgroup(normal subgroup) of119866 then119860 is called rough group (roughnormal subgroup) of 119866

Theorem 8 Let 120588 be a complete congruence relation on 119866119860 sube 119866 and 119873120588 sube 119860 If 119860 is a subgroup (normal subgroup)of 119866 then 119860 is rough group (rough normal subgroup) of 119866 and120588(119860)120588 120588(119860)120588 are respectively lower rough quotient group(lower rough normal quotient group) and upper rough quotientgroup (upper rough normal quotient group)

Corollary 9 Let 120588 be a complete congruence relation on 119866119860 sube 119866 and119873120588 sube 119860 Then 120588(119860)(120588(119860)) is a subgroup (normalsubgroup) of 119866 hArr 120588(119860)120588(120588(119860)120588) is a subgroup (normalsubgroup) of 119866120588

Corollary 10 Let 120588 be a complete congruence relation on 119866119860 sube 119866 and 119873120588 sube 119860 If 119860 is a subgroup (normal subgroup) of119866 then

(1) 120588(119860) lt 119860 lt 120588(119860) lt 119866(120588(119860) ⊲ 119860 ⊲ 120588(119860) ⊲ 119866)

(2) 120588(119860)120588 lt 119860120588 lt 120588(119860)120588 lt 119866120588(120588(119860)120588 ⊲ 119860120588 ⊲ 120588(119860)120588 ⊲ 119866120588)

Corollary 11 Let 120588 be a complete congruence relation on 119866

and let 119860 be a subgroup of 119866 and 119873120588 sube 119860 Then 120588(119860)119860 cong

(120588120588(119860))(119860120588(119860))

Proof Based on the third isomorphism theorem of group itis easy to prove this corollary

Lemma 12 Let 119866 and 119879 be two groups 119891 119866 rarr 119879 is ahomomorphism If 120588 is a congruence relation on 119866 and 119873120588 supe

Ker119891 then 119891(120588) is a congruence relation on 119879 Further if 119891 isan injective then

(1) 119891(120588) = 120588119891(119873120588)

(2) 119891(119873120588) = 119873119891(120588)where 119891(120588) = (119891(119886) 119891(119887)) | (119886 119887) isin 120588

Proof It is easy to prove 119891(120588) is a congruence relation on 119879(1) Because 119873120588 is a normal subgroup on 119866 then 119891(119873120588)

is the normal subgroup on 119879 So forall(119891(119886) 119891(119887)) isin

119891(120588) hArr (119886 119887) isin 120588 hArr 119886 isin 119887119873120588 hArr 119891(119886) isin

119891(119887)119891(119873120588) hArr (119891(119886) 119891(119887)) isin 120588119891(119873120588) so119891(120588) = 120588119891(119873120588)

(2) It follows immediately from (1)

Theorem 13 Let 119891 119866 rarr 119879 be an injective homomorphismIf 120588 is a complete congruence relation on 119866 and 119873120588 supe Ker119891then 119891(120588) is a complete congruence relation on 119879

The Scientific World Journal 3

Proof According to Lemma 12 it is easy to prove 119891(120588)

is a congruence relation on 119879 Now we prove it is com-plete forall119886 119887 isin 119866 we have [119891(119886)]

119891(120588)[119891(119887)]

119891(120588)=

(119891(119886)119891(119873120588))(119891(119887)119891(119873120588)) = 119891(119886119873120588119887119873120588) = 119891(119886119887)119891(119873120588)

= [119891(119886119887)]119891(120588)

= [119891(119886)119891(119887)]119891(120588)

Therefore 119891(120588) is complete

Lemma 14 Let 119891 119866 rarr 119879 be a surjective homomorphismIf 120588 is a congruence relation on 119879 then 119891

minus1(120588) is a congruence

relation on 119866 Further if 119891 is an injective then

(1) 119891minus1(120588) = 120588119891minus1(119873120588)

(2) 119891minus1(119873120588) = 119873119891minus1(120588)

where 119891minus1(120588) = (119886 119887) | (119891(119886) 119891(119887)) isin 119891(120588)

Proof It is easy to prove 119891minus1(120588) is a congruence relation on119866

(1) Because119873120588 is a normal subgroup on 119879 then 119891minus1(119873120588)

is the normal subgroup on 119866 So forall(119886 119887) isin 119891minus1(120588) rArr

(119891(119886) 119891(119887)) isin 120588 rArr 119891(119886) isin 119891(119887)119873120588 and because 119891is injective then 119886 isin 119887119891

minus1(119873120588) rArr (119886 119887) isin 120588119891minus1(119873120588)

that is 119891minus1(120588) sube 120588119891minus1(119873120588)

On the contrary (119886 119887) isin

120588119891minus1(119873120588)rArr 119886 isin 119887119891

minus1(119873120588) rArr 119891(119886) isin 119891(119887)119873120588 rArr

(119891(119886) 119891(119887)) isin 120588 rArr (119886 119887) isin 119891minus1(120588) that is 120588119891minus1(119873120588) sube

119891minus1(120588) hence 119891minus1(120588) = 120588119891minus1(119873120588)

(2) It follows immediately from (1)

Theorem 15 Let 119891 119866 rarr 119879 be an injective homomorphismIf 120588 is a complete congruence relation on 119879 then 119891

minus1(120588) is a

complete congruence relation on 119866

Proof According to Lemma 14 it is easy to prove 119891minus1(120588)

is a congruence relation on 119866 Now we prove it is com-plete forall119886 119887 isin 119866 we have [119886]119891minus1(120588)[119887]119891minus1(120588) =

(119886119891minus1(119873120588))(119887119891

minus1(119873120588)) = 119886119887119891

minus1(119873120588) = [119886119887]119891minus1(119873120588)

Therefore 119891minus1(120588) is complete

Lemma 16 Let 119891 119866 rarr 119879 be a surjective homomorphism If120588 is a congruence relation on 119866 119860 sube 119866 and Ker119891 sube 119873120588 sube 119860then

(1) 119891(120588(119860)) = 119891(120588)(119891(119860))(2) if 119891 is injective then 119891(120588(119860)) = 119891(120588)(119891(119860))

Proof (1) Consider forall119887 isin 119891(120588(119860)) rArr exist119886 isin 120588(119860) 119891(119886) = 119887 rArr

[119886]120588 cap 119860 = 0 119891(119886) = 119887 rArr exist1198861015840isin [119886]120588 119886

1015840isin 119860 rArr (119886 119886

1015840) isin 120588

1198861015840isin 119860 rArr (119891(119886) 119891(119886

1015840)) isin 119891(120588) 119891(1198861015840) isin 119891(119860) rArr 119891(119886

1015840) isin

[119891(119886)]119891(120588)

119891(1198861015840) isin 119891(119860) rArr [119891(119886)]119891(120588)

cap 119891(119860) = 0 rArr 119891(119886) =119887 isin 119891(120588)(119891(119860)) that is 119891(120588(119860)) sube 119891(120588)(119891(119860)) on thecontrary forall119887 isin 119891(120588)(119891(119860)) rArr exist119886 isin 119866 119891(119886) = 119887 [119887]119891(120588) cap119891(119860) = 0 rArr exist119887

1015840isin 119891(119860) 1198871015840 isin [119887]119891(120588) rArr exist119886

1015840isin 119860 119891(1198861015840) = 119887

1015840

119891(1198861015840) isin [119887]119891(120588) rArr (119891(119886

1015840)119891(119886)) isin 119891(120588) 1198861015840 isin 119860 rArr (119886

1015840 119886) isin 120588

1198861015840isin 119860 rArr 119886

1015840isin [119886]120588 119886

1015840isin 119860 rArr [119886]120588 cap 119860 = 0 rArr 119886 isin 120588(119860) rArr

119891(119886) = 119887 isin 119891(120588(119860)) that is 119891(120588(119860)) supe 119891(120588)(119891(119860)) so119891(120588(119860)) = 119891(120588)(119891(119860))

(2) Consider forall119887 isin 119891(120588(119860)) hArr exist119886 isin 120588(119860) 119891(119886) = 119887 hArr

[119886]120588 = 119886119873120588 sube 119860 hArr 119891([119886]120588) = 119891(119886)119891(119873120588) sube 119891(119860) hArr 119887119873119891(120588) =[119887]119891(120588) sube 119891(119860) hArr 119887 isin 119891(120588)(119891(119860)) so 119891(120588(119860)) = 119891(120588)(119891(119860))

Theorem 17 Let 119891 119866 rarr 119879 be a surjective homomorphismlet 120588 be a congruence relation on 119866 and let 119860 be a subgroup of119866 and Ker119891 sube 119873120588 sube 119860 Then

(1) 120588(119860)120588 cong 119891(120588)(119891(119860))119891(120588)

(2) if 119891 is injective then 120588(119860)120588 cong 119891(120588)(119891(119860))119891(120588)

Proof (1) Suppose that 119860 is a subgroup of 119866 then 120588(119860) is asubgroup on119866 and119891(120588)(119891(119860)) is a subgroup on119879 accordingto Lemma 16 119891(120588)(119891(119860)) = 119891(120588(119860)) therefore 120588(119860)120588 cong

119891(120588)(119891(119860))119891(120588)(2) If 119891 is injective according to Lemma 16 we have

119891(120588)(119891(119860)) = 119891(120588(119860)) therefore 120588(119860)120588 cong 119891(120588)(119891(119860))119891(120588)

Corollary 18 Let 119891 119866 rarr 119879 be a surjective homomorphismlet 120588 be a complete congruence relation of 119866 based on Ker119891and let 119860 be a subgroup of 119866 and 119860 supe Ker119891 Then

(1) 120588(119860)120588 cong 119891(120588(119860))

(2) if 119891 is injective then 120588(119860)120588 cong 119891(120588(119860))

Lemma 19 Let 119891 119866 rarr 119879 be a surjective homomorphismand let 120588 be a congruence relation on 119879 and 119861 sube 119879 Then

(1) 119891minus1(120588(119861)) = 119891minus1(120588)(119891minus1(119861))

(2) if 119891 is an injective then 119891minus1(120588(119861)) = 119891

minus1(120588)(119891minus1(119861))

Proof (1) Consider forall119886 isin 119891minus1(120588(119861)) hArr 119891(119886) isin 120588(119861) hArr

[119891(119886)]120588cap 119861 = 0 hArr exist119886

1015840isin 119866 119891(1198861015840) isin 119861 119891(119886

1015840) isin [119891(119886)]

120588hArr

1198861015840isin 119891minus1(119861) (119891(1198861015840) 119891(119886)) isin 120588 hArr 119886

1015840isin 119891minus1(119861) (1198861015840 119886) isin

119891minus1(120588) hArr [119886]119891minus1(120588) cap 119891

minus1(119861) = 0 hArr 119886 isin 119891

minus1(120588)(119891minus1(119861)) and

hence 119891minus1(120588(119861)) = 119891minus1(120588)(119891minus1(119861))

(2) Consider forall119886 isin 119891minus1(120588(119861)) hArr 119891(119886) isin 120588(119861) hArr

[119891(119886)]120588= 119891([119886]119891minus1(120588)) sube 119861 hArr [119886]119891minus1(120588) sube 119891

minus1(119861) hArr 119886 isin

119891minus1(120588)(119891minus1(119861)) so 119891

minus1(120588(119861)) = 119891

minus1(120588)(119891minus1(119861))

Theorem 20 Let 119891 119866 rarr 119879 be a surjective homomorphismand let 120588 be a congruence relation on 119879 and 119861 is a subgroup of119879 Then

(1) 119891minus1(120588)(119891minus1(119861))119891minus1(120588) cong 120588(119861)

(2) if 119891 is an injective then 119891minus1(120588)(119891minus1(119861))119891

minus1(120588) cong

120588(119861)

4 The Scientific World Journal

Proof By Lemma 19 and the first isomorphism theorem ofgroup we have

119891minus1(120588)(119891minus1(119861))119891

minus1(120588) = 119891

minus1(120588(119861))119891

minus1(120588) cong 120588(119861)

119891minus1(120588)(119891minus1(119861))119891

minus1(120588) = 119891

minus1(120588(119861))119891

minus1(120588) cong 120588(119861)

4 Congruence Relation and the Operation ofRough Group

Lemma 21 Let 120588 984858 be the congruence relations on 119866 Then119873120588cap984858 = 119873120588 cap 119873984858

Proof Consider forall119886 isin 119873120588cap984858 hArr (119886 119890) isin 120588 cap 984858 hArr (119886 119890) isin 120588(119886 119890) isin 984858 hArr 119886 isin 119873120588 119886 isin 119873984858 hArr 119886 isin 119873120588 cap 119873984858 and therefore119873120588cap984858 = 119873120588 cap 119873984858

Lemma 22 Let 120588 984858 be two congruence relations on 119866 and 119886 isin

119866 Then 119886(119873120588 cap 119873984858) = 119886119873120588 cap 119886119873984858

Proof It is easy to prove that 119886(119873120588cap119873984858) sube 119886119873120588cap119886119873984858 On thecontrary forall119887 isin 119886119873120588 cap 119886119873984858 rArr 119887 isin 119886119873120588 119887 isin 119886119873984858 rArr (119886 119887) isin 120588(119886 119887) isin 984858 rArr (119886 119887) isin 120588 cap 984858 rArr 119887 isin 119886119873120588cap984858 = 119886(119873120588 cap 119873984858)that is 119886119873120588 cap 119886119873984858 sube 119886(119873120588 cap 119873984858) therefore 119886119873120588 cap 119886119873984858 =

119886(119873120588 cap 119873984858)

Lemma 23 Let 120588 984858 be two congruence relations on 119866 and 119886 isin

119866 Then [119886]120588cap984858 = [119886]120588 cap [119886]984858

Proof Consider forall119887 isin [119886]120588cap984858 hArr 119887 isin 119886119873120588cap984858 = 119886(119873120588 cap 119873984858) =

119886119873120588 cap 119886119873984858 hArr 119887 isin 119886119873120588 119887 isin 119886119873984858 hArr 119887 isin [119886]120588 cap [119886]984858 hence[119886]120588cap984858 = [119886]120588 cap [119886]984858

Theorem 24 Let 120588 984858 be two complete congruence relations on119866 Then 120588 cap 984858 is a complete congruence relation on 119866

Proof Because 119873120588 119873984858 are both the normal subgroups of 119866then119873120588cap984858 = 119873120588 cap119873984858 is a normal subgroup of 119866 so119873120588cap984858 is acomplete congruence relation

Theorem 25 Let 120588 984858 be two congruence relations on 119866 and119860 sube 119866 Then

(1) 120588 cap 984858(119860) sube 120588(119860) cap 984858(119860)(2) 120588 cap 984858(119860) supe 120588(119860) cup 984858(119860) supe 120588(119860) cap 984858(119860)

Proof (1) Consider forall119886 isin 120588 cap 984858(119860) rArr [119886]120588cap984858 cap119860 = 0 Because[119886]120588cap984858 = [119886]120588 cap [119886]984858 then [119886]120588cap984858 sube [119886]120588([119886]984858) rArr [119886]120588 cap119860 = 0[119886]984858 cap 119860 = 0 rArr 119886 isin 120588(119860) 119886 isin 984858(119860) rArr 119886 isin 120588(119860) cap 984858(119860)therefore 120588 cap 984858(119860) sube 120588(119860) cap 984858(119860)

(2) Consider forall119886 isin 120588(119860) cup 984858(119860) rArr 119886 isin 120588(119860) or 119886 isin

984858(119860) rArr [119886]120588 sube 119860 or [119886]984858 sube 119860 rArr [119886]120588cap984858 = [119886]120588 cap [119886]984858 sube

119860 rArr 119886 isin 120588 cap 984858(119860) and therefore 120588 cap 984858(119860) supe 120588(119860) cup 984858(119860) supe

120588(119860) cap 984858(119860)

Lemma 26 (see [25]) Let 120588 984858 be two congruence relations on119866 Then 120588 ∘ 984858 is a congruence relation on 119866 hArr 120588 ∘ 984858 = 984858 ∘ 120588

Theorem 27 Let 120588 984858 be two complete congruence relations on119866 and 120588∘984858 = 984858∘120588 Then 120588∘984858 is a complete congruence relationon 119866

Proof By Lemma 26 120588 ∘ 984858 is the congruence relation and[119886]120588∘984858[119887]120588∘984858 = (119886119873120588∘984858)(119887119873120588∘984858) = 119886119887119873120588∘984858 = [119886119887]120588∘984858 hence120588 ∘ 984858 is the complete congruence relation

Lemma 28 Let 120588 984858 be two congruence relations on 119866 Then119873120588∘984858 = 119873120588119873984858

Proof Consider forall119886 isin 119873120588cap984858 rArr (119886 119890) isin 120588 cap 984858 rArr exist119887 isin 119866(119886 119887) isin 120588 (119887 119890) isin 984858 rArr 119886 isin 119887119873120588 119887 isin 119873984858 rArr 119886119887 isin

119887119873120588119873984858 rArr 119886 isin 119873120588119873984858 that is119873120588∘984858 sube 119873120588119873984858 On the contraryforall119886 isin 119873120588119873984858 rArr exist119887 isin 119873120588 119888 isin 119873984858 119886 = 119887119888 rArr (119887 119890) isin 120588 (119888 119890) isin984858 rArr (119887119888 119888) isin 120588 (119888 119890) isin 984858 rArr (119887119888 119890) isin 120588 ∘ 984858 rArr 119887119888 = 119886 isin 119873120588∘984858that is119873120588∘984858 supe 119873120588119873984858 hence119873120588∘984858 = 119873120588119873984858

Lemma 29 Let 120588 984858 be two congruence relations on 119866 and119886 119887 isin 119866 Then [119886119887]120588∘984858 = [119886]120588[119887]984858

Proof Consider [119886119887]120588∘984858 = 119886119887119873120588∘984858 = 119886119887119873120588119873984858 =

(119886119873120588)(119887119873984858) = [119886]120588[119887]984858

Theorem 30 Let 120588 984858 be two congruence relations on 119866 and120588 ∘ 984858 = 984858 ∘ 120588 If 119860 is a subgroup of 119866 and119873120588 119873984858 sube 119860 Then

(1) 120588 ∘ 984858(119860) supe 120588(119860)984858(119860)(2) 120588 ∘ 984858(119860) = 120588(119860)984858(119860)

Proof (1) Consider forall119886 isin 120588(119860)984858(119860) rArr exist119887 isin 120588(119860) 119888 isin 984858(119860)119886 = 119887119888 rArr [119887]120588 cap 119860 = 0 [119888]984858 cap 119860 = 0 119886 = 119887119888 rArr exist119887

1015840isin [119887]120588

1198871015840isin 119860 1198881015840 isin [119888]984858 119888

1015840isin 119860 rArr 119887

10158401198881015840isin [119887]120588[119888]984858 = [119887119888]120588∘984858 =

[119886]120588∘984858 11988710158401198881015840isin 119860119860 = 119860 rArr [119886]120588∘984858 cap 119860 = 0 rArr 119886 isin 120588 ∘ 984858(119860) so

120588 ∘ 984858(119860) supe 120588(119860)984858(119860)(2) Consider forall119886 isin 120588 ∘ 984858(119860) rArr [119886]120588∘984858 = [119886]120588[119890]984858 sube 119860

because [119886]120588 = [119886]120588[119890] sube [119886]120588[119890]984858 sube 119860 and 119873984858 = [119890]119886 sube

119860 rArr 119886 isin 120588(119860) 119890 isin 984858(119860) rArr 119886 = 119886119890 isin 120588(119860)984858(119860) that is120588 ∘ 984858(119860) sube 120588(119860)984858(119860) On the contrary forall119886 isin 120588(119860)984858(119860) rArr

exist119887 isin 120588(119860) 119888 isin 984858(119860) 119886 = 119887119888 rArr [119887]120588 sube 119860 [119888]984858 sube 119860 119886 = 119887119888 rArr

[119886]120588 ∘ 984858 = [119887119888]120588 ∘ 984858 = [119887]120588[119888]984858 sube 119860119860 = 119860 rArr 119886 isin 120588 ∘ 984858(119860) thatis 120588(119860)984858(119860) sube 120588 ∘ 984858(119860) so 120588 ∘ 984858(119860) = 120588(119860)984858(119860)

Theorem 31 Let 120588 be complete congruence relation on 119866 and119860 119861 sube 119866 Then

(1) 120588(119860119861) = 120588(119860)120588(119861)(2) 120588(119860119861) = 120588(119860)120588(119861)

Proof (1) Consider forall119886 isin 120588(119860119861) rArr [119886]120588 cap 119860119861 = 0 rArr

exist119887 isin [119886]120588 119887 isin 119860119861 rArr 119886 isin 119887119873120588 exist119888 isin 119860 119889 isin 119861119887 = 119888119889 rArr 119886 isin 119888119889119873120588 = (119888119873120588)(119889119873120588) and [119888]120588 cap 119860 = 0[119889]120588 cap 119861 = 0 rArr exist119888

1015840isin 119873120588 119889

1015840isin 119889119873120588 119886 = 119888

10158401198891015840 because

[1198881015840]120588 = [119888]120588 [119889

1015840]120588 = [119889]120588 rArr [119888

1015840]120588 cap 119860 = 0 [1198891015840]120588 cap 119861 = 0

119886 = 11988810158401198891015840rArr 119886 = 119887

10158401198881015840isin 120588(119860)120588(119861) that is 120588(119860119861) sube 120588(119860)120588(119861)

On the contrary forall119886 isin 120588(119860)120588(119861) rArr exist119887 isin 120588(119860) 119888 isin 120588(119861)119886 = 119887119888 rArr [119887]120588 cap 119860 = 0 [119888]120588 cap 119861 = 0 rArr exist119887

1015840isin [119887]120588 119887

1015840isin 119860

The Scientific World Journal 5

1198881015840isin [119888]120588 119888

1015840isin 119861 rArr 119887

10158401198881015840isin [119887]120588[119888]120588 = [119887119888]120588 = [119886]120588

11988710158401198881015840isin 119860119861 rArr [119886]120588 cap119860119861 = 0 rArr 119886 isin 120588(119860119861) that is 120588(119860)120588(119861) sube

120588(119860119861) Therefore 120588(119860119861) = 120588(119860)120588(119861)(2) forall119886 isin 120588(119860119861) rArr [119886]120588 sube 119860119861 rArr exist119887 isin 119860 119888 isin 119861 119886 =

119887119888 rArr [119887]120588[119888]120588 = [119887119888]120588 = [119886]120588 sube 119860119861 if exist1198871015840 isin [119887]120588 1198871015840ni 119860 119888

1015840isin

[119888]120588 1198881015840ni 119861 then 119887

10158401198881015840isin [119887]120588[119888]120588 119887

10158401198881015840ni 119860119861 Contradiction

that is [119887]120588 sube 119860 [119888]120588 sube 119861 so 119886 = 119887119888 isin 120588(119860)120588(119861) thatis 120588(119860119861) sube 120588(119860)120588(119861) On the contrary forall119886 isin 120588(119860)120588(119861) rArr

exist119887 isin 120588(119860) 119888 isin 120588(119861) 119886 = 119887119888 rArr [119887]120588 sube 119860 [119888]120588 sube 119861 119886 =

119887119888 rArr [119886]120588 = [119887119888]120588 = [119887]120588[119888]120588 sube 119860119861 rArr 119886 isin 120588(119860119861) that is120588(119860)120588(119861) sube 120588(119860119861) so 120588(119860119861) = 120588(119860)120588(119861)

Corollary 32 Let 120588 be a complete congruence relation on 119866

and 119860 sube 119866 119899 isin 119873 Then(1) 120588(119860119899) = (120588(119860))

119899(2) 120588(119860119899) = (120588(119860))

119899

Corollary 33 Let 120588 be a complete congruence relation on 119866

and let 119860 be a subgroup of 119866 and 119899 isin 119873 Then(1) 120588119899(119860) = (120588(119860))

119899= 120588(119860

119899) = 120588(119860)

(2) 120588119899(119860) = (120588(119860))119899= 120588(119860

119899) = 120588(119860)

Proof Because 119860 is a subgroup then 120588(119860) 120588(119860) are both thesubgroups of 119866 so (120588(119860))

119899= 120588(119860) (120588(119860))119899 = 120588(119860) 119860119899 = 119860

and 120588 is an equivalence relation then 120588119899= 120588 hence we can

get (1) and (2)

Corollary 34 Let 120588 be a complete congruence relation on 119866

and let 119860 119861 be two subgroups of 119866 Then(1) 120588(119860 cup 119861) = 120588(119860) cup 120588(119861) sube 120588(119860119861) = 120588(119860)120588(119861)(2) 120588(119860 cup 119861) sube 120588(119860119861) = 120588(119860)120588(119861)

Proof (1) Because 119860 = 119860(119890) sube 119860119861 and 119861 sube 119860119861 so 119860 cup 119861 sube

119860119861 hence 120588(119860 cup 119861) sube 120588(119860119861) and 120588(119860 cup 119861) = 120588(119860) cup 120588(119861)therefore (1) holds (2) Because 119860 cup 119861 sube 119860119861 so 120588(119860 cup 119861) sube

120588(119860119861) = 120588(119860)120588(119861)

Conflict of Interests

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

Acknowledgments

This research was supported by the Fundamental ResearchFunds for the Central Universities (no YX2014-08) and theNSFC of China (Grant no 61370193)

References

[1] Z Pawlak ldquoRough setsrdquo International Journal of Computer ampInformation Sciences vol 11 no 5 pp 341ndash356 1982

[2] G Liu and W Zhu ldquoThe algebraic structures of generalizedrough set theoryrdquo Information Sciences vol 178 no 21 pp 4105ndash4113 2008

[3] J N Mordeson ldquoRough set theory applied to (fuzzy) idealtheoryrdquo Fuzzy Sets and Systems vol 121 no 2 pp 315ndash324 2001

[4] Z Pawlak and A Skowron ldquoRough sets some extensionsrdquoInformation Sciences vol 177 no 1 pp 28ndash40 2007

[5] Z Pawlak andA Skowron ldquoRough sets andBoolean reasoningrdquoInformation Sciences vol 177 no 1 pp 41ndash73 2007

[6] M H Shahzamanian M Shirmohammadi and B DavvazldquoRoughness inCayley graphsrdquo Information Sciences vol 180 no17 pp 3362ndash3372 2010

[7] Z Pawlak and A Skowron ldquoRudiments of rough setsrdquo Informa-tion Sciences vol 177 no 1 pp 3ndash27 2007

[8] V Leoreanu-Fotea and B Davvaz ldquoRoughness in n-ary hyper-groupsrdquo Information Sciences vol 178 no 21 pp 4114ndash41242008

[9] R Biswas and S Nanda ldquoRough groups and rough subgroupsrdquoBulletin of the Polish Academy of Sciences Mathematics vol 42pp 251ndash254 1994

[10] J S Jiang C X Wu and D G Chen ldquoThe product structureof fuzzy rough sets on a group and the rough T-fuzzy grouprdquoInformation Sciences vol 175 no 1-2 pp 97ndash107 2005

[11] Y Q Yin J M Zhan and P Corsini ldquoFuzzy roughness of n-ary hypergroups based on a complete residuated latticerdquoNeuralComputing and Applications vol 20 no 1 pp 41ndash57 2011

[12] Q M Xiao and Z L Zhang ldquoRough prime ideals and roughfuzzy prime ideals in semigroupsrdquo Information Sciences vol 176no 6 pp 725ndash733 2006

[13] N Kuroki ldquoRough ideals in semigroupsrdquo Information Sciencesvol 100 no 1ndash4 pp 139ndash163 1997

[14] N Kuroki and P PWang ldquoThe lower and upper approximationsin a fuzzy grouprdquo Information Sciences vol 90 no 1ndash4 pp 203ndash220 1996

[15] K C Gupta and M K Kantroo ldquoGeneralized product of fuzzysubsets of a ringrdquo Fuzzy Sets and Systems vol 117 no 3 pp 419ndash429 2001

[16] O Kazanci and B Davvaz ldquoOn the structure of rough prime(primary) ideals and rough fuzzy prime (primary) ideals incommutative ringsrdquo Information Sciences vol 178 no 5 pp1343ndash1354 2008

[17] N Kuroki and J N Mordeson ldquoStructure of rough sets andrough groupsrdquo Journal of Fuzzy Mathematics vol 5 no 1 pp183ndash191 1997

[18] F Li Y Yin and L Lu ldquo(120599 119879)-fuzzy rough approximationoperators and the TL-fuzzy rough ideals on a ringrdquo InformationSciences vol 177 no 21 pp 4711ndash4726 2007

[19] Y Q Yin and X K Huang ldquoFuzzy roughness in hyperringsbased on a complete residuated latticerdquo International Journal ofFuzzy Systems vol 13 no 3 pp 185ndash194 2011

[20] W J Liu Y D Cu and H X Li ldquoRough fuzzy ideals andrough fuzzy prime ideals in semigroupsrdquo Fuzzy Systems andMathematics vol 21 no 3 pp 127ndash132 2007

[21] J L Zhang and Z L Zhang ldquoRough subgroups and roughsubringsrdquo Pure and Applied Mathematics vol 20 pp 92ndash962004

[22] J L Zhang and Z L Zhang ldquoFuzzy rough subgroupsrdquo FuzzySystems and Mathematics vol 21 no 3 pp 127ndash132 2007

[23] Z L Zhang J L Zhang and Q M Xiao Fuzzy Algebra andRough Algebra Wuhan University Press Wuhan China 2007

6 The Scientific World Journal

[24] B Davvaz and V Leoreanu-Fotea ldquoApplications of intervalvalued fuzzy n-ary polygroups with respect to t-norms (t-conorms)rdquo Computers and Mathematics with Applications vol57 no 8 pp 1413ndash1424 2009

[25] DW ShengAbstractAlgebra Science PublishingPress BeijingChina 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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Mathematical PhysicsAdvances in

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OptimizationJournal of

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International Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

The Scientific World Journal 3

Proof According to Lemma 12 it is easy to prove 119891(120588)

is a congruence relation on 119879 Now we prove it is com-plete forall119886 119887 isin 119866 we have [119891(119886)]

119891(120588)[119891(119887)]

119891(120588)=

(119891(119886)119891(119873120588))(119891(119887)119891(119873120588)) = 119891(119886119873120588119887119873120588) = 119891(119886119887)119891(119873120588)

= [119891(119886119887)]119891(120588)

= [119891(119886)119891(119887)]119891(120588)

Therefore 119891(120588) is complete

Lemma 14 Let 119891 119866 rarr 119879 be a surjective homomorphismIf 120588 is a congruence relation on 119879 then 119891

minus1(120588) is a congruence

relation on 119866 Further if 119891 is an injective then

(1) 119891minus1(120588) = 120588119891minus1(119873120588)

(2) 119891minus1(119873120588) = 119873119891minus1(120588)

where 119891minus1(120588) = (119886 119887) | (119891(119886) 119891(119887)) isin 119891(120588)

Proof It is easy to prove 119891minus1(120588) is a congruence relation on119866

(1) Because119873120588 is a normal subgroup on 119879 then 119891minus1(119873120588)

is the normal subgroup on 119866 So forall(119886 119887) isin 119891minus1(120588) rArr

(119891(119886) 119891(119887)) isin 120588 rArr 119891(119886) isin 119891(119887)119873120588 and because 119891is injective then 119886 isin 119887119891

minus1(119873120588) rArr (119886 119887) isin 120588119891minus1(119873120588)

that is 119891minus1(120588) sube 120588119891minus1(119873120588)

On the contrary (119886 119887) isin

120588119891minus1(119873120588)rArr 119886 isin 119887119891

minus1(119873120588) rArr 119891(119886) isin 119891(119887)119873120588 rArr

(119891(119886) 119891(119887)) isin 120588 rArr (119886 119887) isin 119891minus1(120588) that is 120588119891minus1(119873120588) sube

119891minus1(120588) hence 119891minus1(120588) = 120588119891minus1(119873120588)

(2) It follows immediately from (1)

Theorem 15 Let 119891 119866 rarr 119879 be an injective homomorphismIf 120588 is a complete congruence relation on 119879 then 119891

minus1(120588) is a

complete congruence relation on 119866

Proof According to Lemma 14 it is easy to prove 119891minus1(120588)

is a congruence relation on 119866 Now we prove it is com-plete forall119886 119887 isin 119866 we have [119886]119891minus1(120588)[119887]119891minus1(120588) =

(119886119891minus1(119873120588))(119887119891

minus1(119873120588)) = 119886119887119891

minus1(119873120588) = [119886119887]119891minus1(119873120588)

Therefore 119891minus1(120588) is complete

Lemma 16 Let 119891 119866 rarr 119879 be a surjective homomorphism If120588 is a congruence relation on 119866 119860 sube 119866 and Ker119891 sube 119873120588 sube 119860then

(1) 119891(120588(119860)) = 119891(120588)(119891(119860))(2) if 119891 is injective then 119891(120588(119860)) = 119891(120588)(119891(119860))

Proof (1) Consider forall119887 isin 119891(120588(119860)) rArr exist119886 isin 120588(119860) 119891(119886) = 119887 rArr

[119886]120588 cap 119860 = 0 119891(119886) = 119887 rArr exist1198861015840isin [119886]120588 119886

1015840isin 119860 rArr (119886 119886

1015840) isin 120588

1198861015840isin 119860 rArr (119891(119886) 119891(119886

1015840)) isin 119891(120588) 119891(1198861015840) isin 119891(119860) rArr 119891(119886

1015840) isin

[119891(119886)]119891(120588)

119891(1198861015840) isin 119891(119860) rArr [119891(119886)]119891(120588)

cap 119891(119860) = 0 rArr 119891(119886) =119887 isin 119891(120588)(119891(119860)) that is 119891(120588(119860)) sube 119891(120588)(119891(119860)) on thecontrary forall119887 isin 119891(120588)(119891(119860)) rArr exist119886 isin 119866 119891(119886) = 119887 [119887]119891(120588) cap119891(119860) = 0 rArr exist119887

1015840isin 119891(119860) 1198871015840 isin [119887]119891(120588) rArr exist119886

1015840isin 119860 119891(1198861015840) = 119887

1015840

119891(1198861015840) isin [119887]119891(120588) rArr (119891(119886

1015840)119891(119886)) isin 119891(120588) 1198861015840 isin 119860 rArr (119886

1015840 119886) isin 120588

1198861015840isin 119860 rArr 119886

1015840isin [119886]120588 119886

1015840isin 119860 rArr [119886]120588 cap 119860 = 0 rArr 119886 isin 120588(119860) rArr

119891(119886) = 119887 isin 119891(120588(119860)) that is 119891(120588(119860)) supe 119891(120588)(119891(119860)) so119891(120588(119860)) = 119891(120588)(119891(119860))

(2) Consider forall119887 isin 119891(120588(119860)) hArr exist119886 isin 120588(119860) 119891(119886) = 119887 hArr

[119886]120588 = 119886119873120588 sube 119860 hArr 119891([119886]120588) = 119891(119886)119891(119873120588) sube 119891(119860) hArr 119887119873119891(120588) =[119887]119891(120588) sube 119891(119860) hArr 119887 isin 119891(120588)(119891(119860)) so 119891(120588(119860)) = 119891(120588)(119891(119860))

Theorem 17 Let 119891 119866 rarr 119879 be a surjective homomorphismlet 120588 be a congruence relation on 119866 and let 119860 be a subgroup of119866 and Ker119891 sube 119873120588 sube 119860 Then

(1) 120588(119860)120588 cong 119891(120588)(119891(119860))119891(120588)

(2) if 119891 is injective then 120588(119860)120588 cong 119891(120588)(119891(119860))119891(120588)

Proof (1) Suppose that 119860 is a subgroup of 119866 then 120588(119860) is asubgroup on119866 and119891(120588)(119891(119860)) is a subgroup on119879 accordingto Lemma 16 119891(120588)(119891(119860)) = 119891(120588(119860)) therefore 120588(119860)120588 cong

119891(120588)(119891(119860))119891(120588)(2) If 119891 is injective according to Lemma 16 we have

119891(120588)(119891(119860)) = 119891(120588(119860)) therefore 120588(119860)120588 cong 119891(120588)(119891(119860))119891(120588)

Corollary 18 Let 119891 119866 rarr 119879 be a surjective homomorphismlet 120588 be a complete congruence relation of 119866 based on Ker119891and let 119860 be a subgroup of 119866 and 119860 supe Ker119891 Then

(1) 120588(119860)120588 cong 119891(120588(119860))

(2) if 119891 is injective then 120588(119860)120588 cong 119891(120588(119860))

Lemma 19 Let 119891 119866 rarr 119879 be a surjective homomorphismand let 120588 be a congruence relation on 119879 and 119861 sube 119879 Then

(1) 119891minus1(120588(119861)) = 119891minus1(120588)(119891minus1(119861))

(2) if 119891 is an injective then 119891minus1(120588(119861)) = 119891

minus1(120588)(119891minus1(119861))

Proof (1) Consider forall119886 isin 119891minus1(120588(119861)) hArr 119891(119886) isin 120588(119861) hArr

[119891(119886)]120588cap 119861 = 0 hArr exist119886

1015840isin 119866 119891(1198861015840) isin 119861 119891(119886

1015840) isin [119891(119886)]

120588hArr

1198861015840isin 119891minus1(119861) (119891(1198861015840) 119891(119886)) isin 120588 hArr 119886

1015840isin 119891minus1(119861) (1198861015840 119886) isin

119891minus1(120588) hArr [119886]119891minus1(120588) cap 119891

minus1(119861) = 0 hArr 119886 isin 119891

minus1(120588)(119891minus1(119861)) and

hence 119891minus1(120588(119861)) = 119891minus1(120588)(119891minus1(119861))

(2) Consider forall119886 isin 119891minus1(120588(119861)) hArr 119891(119886) isin 120588(119861) hArr

[119891(119886)]120588= 119891([119886]119891minus1(120588)) sube 119861 hArr [119886]119891minus1(120588) sube 119891

minus1(119861) hArr 119886 isin

119891minus1(120588)(119891minus1(119861)) so 119891

minus1(120588(119861)) = 119891

minus1(120588)(119891minus1(119861))

Theorem 20 Let 119891 119866 rarr 119879 be a surjective homomorphismand let 120588 be a congruence relation on 119879 and 119861 is a subgroup of119879 Then

(1) 119891minus1(120588)(119891minus1(119861))119891minus1(120588) cong 120588(119861)

(2) if 119891 is an injective then 119891minus1(120588)(119891minus1(119861))119891

minus1(120588) cong

120588(119861)

4 The Scientific World Journal

Proof By Lemma 19 and the first isomorphism theorem ofgroup we have

119891minus1(120588)(119891minus1(119861))119891

minus1(120588) = 119891

minus1(120588(119861))119891

minus1(120588) cong 120588(119861)

119891minus1(120588)(119891minus1(119861))119891

minus1(120588) = 119891

minus1(120588(119861))119891

minus1(120588) cong 120588(119861)

4 Congruence Relation and the Operation ofRough Group

Lemma 21 Let 120588 984858 be the congruence relations on 119866 Then119873120588cap984858 = 119873120588 cap 119873984858

Proof Consider forall119886 isin 119873120588cap984858 hArr (119886 119890) isin 120588 cap 984858 hArr (119886 119890) isin 120588(119886 119890) isin 984858 hArr 119886 isin 119873120588 119886 isin 119873984858 hArr 119886 isin 119873120588 cap 119873984858 and therefore119873120588cap984858 = 119873120588 cap 119873984858

Lemma 22 Let 120588 984858 be two congruence relations on 119866 and 119886 isin

119866 Then 119886(119873120588 cap 119873984858) = 119886119873120588 cap 119886119873984858

Proof It is easy to prove that 119886(119873120588cap119873984858) sube 119886119873120588cap119886119873984858 On thecontrary forall119887 isin 119886119873120588 cap 119886119873984858 rArr 119887 isin 119886119873120588 119887 isin 119886119873984858 rArr (119886 119887) isin 120588(119886 119887) isin 984858 rArr (119886 119887) isin 120588 cap 984858 rArr 119887 isin 119886119873120588cap984858 = 119886(119873120588 cap 119873984858)that is 119886119873120588 cap 119886119873984858 sube 119886(119873120588 cap 119873984858) therefore 119886119873120588 cap 119886119873984858 =

119886(119873120588 cap 119873984858)

Lemma 23 Let 120588 984858 be two congruence relations on 119866 and 119886 isin

119866 Then [119886]120588cap984858 = [119886]120588 cap [119886]984858

Proof Consider forall119887 isin [119886]120588cap984858 hArr 119887 isin 119886119873120588cap984858 = 119886(119873120588 cap 119873984858) =

119886119873120588 cap 119886119873984858 hArr 119887 isin 119886119873120588 119887 isin 119886119873984858 hArr 119887 isin [119886]120588 cap [119886]984858 hence[119886]120588cap984858 = [119886]120588 cap [119886]984858

Theorem 24 Let 120588 984858 be two complete congruence relations on119866 Then 120588 cap 984858 is a complete congruence relation on 119866

Proof Because 119873120588 119873984858 are both the normal subgroups of 119866then119873120588cap984858 = 119873120588 cap119873984858 is a normal subgroup of 119866 so119873120588cap984858 is acomplete congruence relation

Theorem 25 Let 120588 984858 be two congruence relations on 119866 and119860 sube 119866 Then

(1) 120588 cap 984858(119860) sube 120588(119860) cap 984858(119860)(2) 120588 cap 984858(119860) supe 120588(119860) cup 984858(119860) supe 120588(119860) cap 984858(119860)

Proof (1) Consider forall119886 isin 120588 cap 984858(119860) rArr [119886]120588cap984858 cap119860 = 0 Because[119886]120588cap984858 = [119886]120588 cap [119886]984858 then [119886]120588cap984858 sube [119886]120588([119886]984858) rArr [119886]120588 cap119860 = 0[119886]984858 cap 119860 = 0 rArr 119886 isin 120588(119860) 119886 isin 984858(119860) rArr 119886 isin 120588(119860) cap 984858(119860)therefore 120588 cap 984858(119860) sube 120588(119860) cap 984858(119860)

(2) Consider forall119886 isin 120588(119860) cup 984858(119860) rArr 119886 isin 120588(119860) or 119886 isin

984858(119860) rArr [119886]120588 sube 119860 or [119886]984858 sube 119860 rArr [119886]120588cap984858 = [119886]120588 cap [119886]984858 sube

119860 rArr 119886 isin 120588 cap 984858(119860) and therefore 120588 cap 984858(119860) supe 120588(119860) cup 984858(119860) supe

120588(119860) cap 984858(119860)

Lemma 26 (see [25]) Let 120588 984858 be two congruence relations on119866 Then 120588 ∘ 984858 is a congruence relation on 119866 hArr 120588 ∘ 984858 = 984858 ∘ 120588

Theorem 27 Let 120588 984858 be two complete congruence relations on119866 and 120588∘984858 = 984858∘120588 Then 120588∘984858 is a complete congruence relationon 119866

Proof By Lemma 26 120588 ∘ 984858 is the congruence relation and[119886]120588∘984858[119887]120588∘984858 = (119886119873120588∘984858)(119887119873120588∘984858) = 119886119887119873120588∘984858 = [119886119887]120588∘984858 hence120588 ∘ 984858 is the complete congruence relation

Lemma 28 Let 120588 984858 be two congruence relations on 119866 Then119873120588∘984858 = 119873120588119873984858

Proof Consider forall119886 isin 119873120588cap984858 rArr (119886 119890) isin 120588 cap 984858 rArr exist119887 isin 119866(119886 119887) isin 120588 (119887 119890) isin 984858 rArr 119886 isin 119887119873120588 119887 isin 119873984858 rArr 119886119887 isin

119887119873120588119873984858 rArr 119886 isin 119873120588119873984858 that is119873120588∘984858 sube 119873120588119873984858 On the contraryforall119886 isin 119873120588119873984858 rArr exist119887 isin 119873120588 119888 isin 119873984858 119886 = 119887119888 rArr (119887 119890) isin 120588 (119888 119890) isin984858 rArr (119887119888 119888) isin 120588 (119888 119890) isin 984858 rArr (119887119888 119890) isin 120588 ∘ 984858 rArr 119887119888 = 119886 isin 119873120588∘984858that is119873120588∘984858 supe 119873120588119873984858 hence119873120588∘984858 = 119873120588119873984858

Lemma 29 Let 120588 984858 be two congruence relations on 119866 and119886 119887 isin 119866 Then [119886119887]120588∘984858 = [119886]120588[119887]984858

Proof Consider [119886119887]120588∘984858 = 119886119887119873120588∘984858 = 119886119887119873120588119873984858 =

(119886119873120588)(119887119873984858) = [119886]120588[119887]984858

Theorem 30 Let 120588 984858 be two congruence relations on 119866 and120588 ∘ 984858 = 984858 ∘ 120588 If 119860 is a subgroup of 119866 and119873120588 119873984858 sube 119860 Then

(1) 120588 ∘ 984858(119860) supe 120588(119860)984858(119860)(2) 120588 ∘ 984858(119860) = 120588(119860)984858(119860)

Proof (1) Consider forall119886 isin 120588(119860)984858(119860) rArr exist119887 isin 120588(119860) 119888 isin 984858(119860)119886 = 119887119888 rArr [119887]120588 cap 119860 = 0 [119888]984858 cap 119860 = 0 119886 = 119887119888 rArr exist119887

1015840isin [119887]120588

1198871015840isin 119860 1198881015840 isin [119888]984858 119888

1015840isin 119860 rArr 119887

10158401198881015840isin [119887]120588[119888]984858 = [119887119888]120588∘984858 =

[119886]120588∘984858 11988710158401198881015840isin 119860119860 = 119860 rArr [119886]120588∘984858 cap 119860 = 0 rArr 119886 isin 120588 ∘ 984858(119860) so

120588 ∘ 984858(119860) supe 120588(119860)984858(119860)(2) Consider forall119886 isin 120588 ∘ 984858(119860) rArr [119886]120588∘984858 = [119886]120588[119890]984858 sube 119860

because [119886]120588 = [119886]120588[119890] sube [119886]120588[119890]984858 sube 119860 and 119873984858 = [119890]119886 sube

119860 rArr 119886 isin 120588(119860) 119890 isin 984858(119860) rArr 119886 = 119886119890 isin 120588(119860)984858(119860) that is120588 ∘ 984858(119860) sube 120588(119860)984858(119860) On the contrary forall119886 isin 120588(119860)984858(119860) rArr

exist119887 isin 120588(119860) 119888 isin 984858(119860) 119886 = 119887119888 rArr [119887]120588 sube 119860 [119888]984858 sube 119860 119886 = 119887119888 rArr

[119886]120588 ∘ 984858 = [119887119888]120588 ∘ 984858 = [119887]120588[119888]984858 sube 119860119860 = 119860 rArr 119886 isin 120588 ∘ 984858(119860) thatis 120588(119860)984858(119860) sube 120588 ∘ 984858(119860) so 120588 ∘ 984858(119860) = 120588(119860)984858(119860)

Theorem 31 Let 120588 be complete congruence relation on 119866 and119860 119861 sube 119866 Then

(1) 120588(119860119861) = 120588(119860)120588(119861)(2) 120588(119860119861) = 120588(119860)120588(119861)

Proof (1) Consider forall119886 isin 120588(119860119861) rArr [119886]120588 cap 119860119861 = 0 rArr

exist119887 isin [119886]120588 119887 isin 119860119861 rArr 119886 isin 119887119873120588 exist119888 isin 119860 119889 isin 119861119887 = 119888119889 rArr 119886 isin 119888119889119873120588 = (119888119873120588)(119889119873120588) and [119888]120588 cap 119860 = 0[119889]120588 cap 119861 = 0 rArr exist119888

1015840isin 119873120588 119889

1015840isin 119889119873120588 119886 = 119888

10158401198891015840 because

[1198881015840]120588 = [119888]120588 [119889

1015840]120588 = [119889]120588 rArr [119888

1015840]120588 cap 119860 = 0 [1198891015840]120588 cap 119861 = 0

119886 = 11988810158401198891015840rArr 119886 = 119887

10158401198881015840isin 120588(119860)120588(119861) that is 120588(119860119861) sube 120588(119860)120588(119861)

On the contrary forall119886 isin 120588(119860)120588(119861) rArr exist119887 isin 120588(119860) 119888 isin 120588(119861)119886 = 119887119888 rArr [119887]120588 cap 119860 = 0 [119888]120588 cap 119861 = 0 rArr exist119887

1015840isin [119887]120588 119887

1015840isin 119860

The Scientific World Journal 5

1198881015840isin [119888]120588 119888

1015840isin 119861 rArr 119887

10158401198881015840isin [119887]120588[119888]120588 = [119887119888]120588 = [119886]120588

11988710158401198881015840isin 119860119861 rArr [119886]120588 cap119860119861 = 0 rArr 119886 isin 120588(119860119861) that is 120588(119860)120588(119861) sube

120588(119860119861) Therefore 120588(119860119861) = 120588(119860)120588(119861)(2) forall119886 isin 120588(119860119861) rArr [119886]120588 sube 119860119861 rArr exist119887 isin 119860 119888 isin 119861 119886 =

119887119888 rArr [119887]120588[119888]120588 = [119887119888]120588 = [119886]120588 sube 119860119861 if exist1198871015840 isin [119887]120588 1198871015840ni 119860 119888

1015840isin

[119888]120588 1198881015840ni 119861 then 119887

10158401198881015840isin [119887]120588[119888]120588 119887

10158401198881015840ni 119860119861 Contradiction

that is [119887]120588 sube 119860 [119888]120588 sube 119861 so 119886 = 119887119888 isin 120588(119860)120588(119861) thatis 120588(119860119861) sube 120588(119860)120588(119861) On the contrary forall119886 isin 120588(119860)120588(119861) rArr

exist119887 isin 120588(119860) 119888 isin 120588(119861) 119886 = 119887119888 rArr [119887]120588 sube 119860 [119888]120588 sube 119861 119886 =

119887119888 rArr [119886]120588 = [119887119888]120588 = [119887]120588[119888]120588 sube 119860119861 rArr 119886 isin 120588(119860119861) that is120588(119860)120588(119861) sube 120588(119860119861) so 120588(119860119861) = 120588(119860)120588(119861)

Corollary 32 Let 120588 be a complete congruence relation on 119866

and 119860 sube 119866 119899 isin 119873 Then(1) 120588(119860119899) = (120588(119860))

119899(2) 120588(119860119899) = (120588(119860))

119899

Corollary 33 Let 120588 be a complete congruence relation on 119866

and let 119860 be a subgroup of 119866 and 119899 isin 119873 Then(1) 120588119899(119860) = (120588(119860))

119899= 120588(119860

119899) = 120588(119860)

(2) 120588119899(119860) = (120588(119860))119899= 120588(119860

119899) = 120588(119860)

Proof Because 119860 is a subgroup then 120588(119860) 120588(119860) are both thesubgroups of 119866 so (120588(119860))

119899= 120588(119860) (120588(119860))119899 = 120588(119860) 119860119899 = 119860

and 120588 is an equivalence relation then 120588119899= 120588 hence we can

get (1) and (2)

Corollary 34 Let 120588 be a complete congruence relation on 119866

and let 119860 119861 be two subgroups of 119866 Then(1) 120588(119860 cup 119861) = 120588(119860) cup 120588(119861) sube 120588(119860119861) = 120588(119860)120588(119861)(2) 120588(119860 cup 119861) sube 120588(119860119861) = 120588(119860)120588(119861)

Proof (1) Because 119860 = 119860(119890) sube 119860119861 and 119861 sube 119860119861 so 119860 cup 119861 sube

119860119861 hence 120588(119860 cup 119861) sube 120588(119860119861) and 120588(119860 cup 119861) = 120588(119860) cup 120588(119861)therefore (1) holds (2) Because 119860 cup 119861 sube 119860119861 so 120588(119860 cup 119861) sube

120588(119860119861) = 120588(119860)120588(119861)

Conflict of Interests

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

Acknowledgments

This research was supported by the Fundamental ResearchFunds for the Central Universities (no YX2014-08) and theNSFC of China (Grant no 61370193)

References

[1] Z Pawlak ldquoRough setsrdquo International Journal of Computer ampInformation Sciences vol 11 no 5 pp 341ndash356 1982

[2] G Liu and W Zhu ldquoThe algebraic structures of generalizedrough set theoryrdquo Information Sciences vol 178 no 21 pp 4105ndash4113 2008

[3] J N Mordeson ldquoRough set theory applied to (fuzzy) idealtheoryrdquo Fuzzy Sets and Systems vol 121 no 2 pp 315ndash324 2001

[4] Z Pawlak and A Skowron ldquoRough sets some extensionsrdquoInformation Sciences vol 177 no 1 pp 28ndash40 2007

[5] Z Pawlak andA Skowron ldquoRough sets andBoolean reasoningrdquoInformation Sciences vol 177 no 1 pp 41ndash73 2007

[6] M H Shahzamanian M Shirmohammadi and B DavvazldquoRoughness inCayley graphsrdquo Information Sciences vol 180 no17 pp 3362ndash3372 2010

[7] Z Pawlak and A Skowron ldquoRudiments of rough setsrdquo Informa-tion Sciences vol 177 no 1 pp 3ndash27 2007

[8] V Leoreanu-Fotea and B Davvaz ldquoRoughness in n-ary hyper-groupsrdquo Information Sciences vol 178 no 21 pp 4114ndash41242008

[9] R Biswas and S Nanda ldquoRough groups and rough subgroupsrdquoBulletin of the Polish Academy of Sciences Mathematics vol 42pp 251ndash254 1994

[10] J S Jiang C X Wu and D G Chen ldquoThe product structureof fuzzy rough sets on a group and the rough T-fuzzy grouprdquoInformation Sciences vol 175 no 1-2 pp 97ndash107 2005

[11] Y Q Yin J M Zhan and P Corsini ldquoFuzzy roughness of n-ary hypergroups based on a complete residuated latticerdquoNeuralComputing and Applications vol 20 no 1 pp 41ndash57 2011

[12] Q M Xiao and Z L Zhang ldquoRough prime ideals and roughfuzzy prime ideals in semigroupsrdquo Information Sciences vol 176no 6 pp 725ndash733 2006

[13] N Kuroki ldquoRough ideals in semigroupsrdquo Information Sciencesvol 100 no 1ndash4 pp 139ndash163 1997

[14] N Kuroki and P PWang ldquoThe lower and upper approximationsin a fuzzy grouprdquo Information Sciences vol 90 no 1ndash4 pp 203ndash220 1996

[15] K C Gupta and M K Kantroo ldquoGeneralized product of fuzzysubsets of a ringrdquo Fuzzy Sets and Systems vol 117 no 3 pp 419ndash429 2001

[16] O Kazanci and B Davvaz ldquoOn the structure of rough prime(primary) ideals and rough fuzzy prime (primary) ideals incommutative ringsrdquo Information Sciences vol 178 no 5 pp1343ndash1354 2008

[17] N Kuroki and J N Mordeson ldquoStructure of rough sets andrough groupsrdquo Journal of Fuzzy Mathematics vol 5 no 1 pp183ndash191 1997

[18] F Li Y Yin and L Lu ldquo(120599 119879)-fuzzy rough approximationoperators and the TL-fuzzy rough ideals on a ringrdquo InformationSciences vol 177 no 21 pp 4711ndash4726 2007

[19] Y Q Yin and X K Huang ldquoFuzzy roughness in hyperringsbased on a complete residuated latticerdquo International Journal ofFuzzy Systems vol 13 no 3 pp 185ndash194 2011

[20] W J Liu Y D Cu and H X Li ldquoRough fuzzy ideals andrough fuzzy prime ideals in semigroupsrdquo Fuzzy Systems andMathematics vol 21 no 3 pp 127ndash132 2007

[21] J L Zhang and Z L Zhang ldquoRough subgroups and roughsubringsrdquo Pure and Applied Mathematics vol 20 pp 92ndash962004

[22] J L Zhang and Z L Zhang ldquoFuzzy rough subgroupsrdquo FuzzySystems and Mathematics vol 21 no 3 pp 127ndash132 2007

[23] Z L Zhang J L Zhang and Q M Xiao Fuzzy Algebra andRough Algebra Wuhan University Press Wuhan China 2007

6 The Scientific World Journal

[24] B Davvaz and V Leoreanu-Fotea ldquoApplications of intervalvalued fuzzy n-ary polygroups with respect to t-norms (t-conorms)rdquo Computers and Mathematics with Applications vol57 no 8 pp 1413ndash1424 2009

[25] DW ShengAbstractAlgebra Science PublishingPress BeijingChina 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 The Scientific World Journal

Proof By Lemma 19 and the first isomorphism theorem ofgroup we have

119891minus1(120588)(119891minus1(119861))119891

minus1(120588) = 119891

minus1(120588(119861))119891

minus1(120588) cong 120588(119861)

119891minus1(120588)(119891minus1(119861))119891

minus1(120588) = 119891

minus1(120588(119861))119891

minus1(120588) cong 120588(119861)

4 Congruence Relation and the Operation ofRough Group

Lemma 21 Let 120588 984858 be the congruence relations on 119866 Then119873120588cap984858 = 119873120588 cap 119873984858

Proof Consider forall119886 isin 119873120588cap984858 hArr (119886 119890) isin 120588 cap 984858 hArr (119886 119890) isin 120588(119886 119890) isin 984858 hArr 119886 isin 119873120588 119886 isin 119873984858 hArr 119886 isin 119873120588 cap 119873984858 and therefore119873120588cap984858 = 119873120588 cap 119873984858

Lemma 22 Let 120588 984858 be two congruence relations on 119866 and 119886 isin

119866 Then 119886(119873120588 cap 119873984858) = 119886119873120588 cap 119886119873984858

Proof It is easy to prove that 119886(119873120588cap119873984858) sube 119886119873120588cap119886119873984858 On thecontrary forall119887 isin 119886119873120588 cap 119886119873984858 rArr 119887 isin 119886119873120588 119887 isin 119886119873984858 rArr (119886 119887) isin 120588(119886 119887) isin 984858 rArr (119886 119887) isin 120588 cap 984858 rArr 119887 isin 119886119873120588cap984858 = 119886(119873120588 cap 119873984858)that is 119886119873120588 cap 119886119873984858 sube 119886(119873120588 cap 119873984858) therefore 119886119873120588 cap 119886119873984858 =

119886(119873120588 cap 119873984858)

Lemma 23 Let 120588 984858 be two congruence relations on 119866 and 119886 isin

119866 Then [119886]120588cap984858 = [119886]120588 cap [119886]984858

Proof Consider forall119887 isin [119886]120588cap984858 hArr 119887 isin 119886119873120588cap984858 = 119886(119873120588 cap 119873984858) =

119886119873120588 cap 119886119873984858 hArr 119887 isin 119886119873120588 119887 isin 119886119873984858 hArr 119887 isin [119886]120588 cap [119886]984858 hence[119886]120588cap984858 = [119886]120588 cap [119886]984858

Theorem 24 Let 120588 984858 be two complete congruence relations on119866 Then 120588 cap 984858 is a complete congruence relation on 119866

Proof Because 119873120588 119873984858 are both the normal subgroups of 119866then119873120588cap984858 = 119873120588 cap119873984858 is a normal subgroup of 119866 so119873120588cap984858 is acomplete congruence relation

Theorem 25 Let 120588 984858 be two congruence relations on 119866 and119860 sube 119866 Then

(1) 120588 cap 984858(119860) sube 120588(119860) cap 984858(119860)(2) 120588 cap 984858(119860) supe 120588(119860) cup 984858(119860) supe 120588(119860) cap 984858(119860)

Proof (1) Consider forall119886 isin 120588 cap 984858(119860) rArr [119886]120588cap984858 cap119860 = 0 Because[119886]120588cap984858 = [119886]120588 cap [119886]984858 then [119886]120588cap984858 sube [119886]120588([119886]984858) rArr [119886]120588 cap119860 = 0[119886]984858 cap 119860 = 0 rArr 119886 isin 120588(119860) 119886 isin 984858(119860) rArr 119886 isin 120588(119860) cap 984858(119860)therefore 120588 cap 984858(119860) sube 120588(119860) cap 984858(119860)

(2) Consider forall119886 isin 120588(119860) cup 984858(119860) rArr 119886 isin 120588(119860) or 119886 isin

984858(119860) rArr [119886]120588 sube 119860 or [119886]984858 sube 119860 rArr [119886]120588cap984858 = [119886]120588 cap [119886]984858 sube

119860 rArr 119886 isin 120588 cap 984858(119860) and therefore 120588 cap 984858(119860) supe 120588(119860) cup 984858(119860) supe

120588(119860) cap 984858(119860)

Lemma 26 (see [25]) Let 120588 984858 be two congruence relations on119866 Then 120588 ∘ 984858 is a congruence relation on 119866 hArr 120588 ∘ 984858 = 984858 ∘ 120588

Theorem 27 Let 120588 984858 be two complete congruence relations on119866 and 120588∘984858 = 984858∘120588 Then 120588∘984858 is a complete congruence relationon 119866

Proof By Lemma 26 120588 ∘ 984858 is the congruence relation and[119886]120588∘984858[119887]120588∘984858 = (119886119873120588∘984858)(119887119873120588∘984858) = 119886119887119873120588∘984858 = [119886119887]120588∘984858 hence120588 ∘ 984858 is the complete congruence relation

Lemma 28 Let 120588 984858 be two congruence relations on 119866 Then119873120588∘984858 = 119873120588119873984858

Proof Consider forall119886 isin 119873120588cap984858 rArr (119886 119890) isin 120588 cap 984858 rArr exist119887 isin 119866(119886 119887) isin 120588 (119887 119890) isin 984858 rArr 119886 isin 119887119873120588 119887 isin 119873984858 rArr 119886119887 isin

119887119873120588119873984858 rArr 119886 isin 119873120588119873984858 that is119873120588∘984858 sube 119873120588119873984858 On the contraryforall119886 isin 119873120588119873984858 rArr exist119887 isin 119873120588 119888 isin 119873984858 119886 = 119887119888 rArr (119887 119890) isin 120588 (119888 119890) isin984858 rArr (119887119888 119888) isin 120588 (119888 119890) isin 984858 rArr (119887119888 119890) isin 120588 ∘ 984858 rArr 119887119888 = 119886 isin 119873120588∘984858that is119873120588∘984858 supe 119873120588119873984858 hence119873120588∘984858 = 119873120588119873984858

Lemma 29 Let 120588 984858 be two congruence relations on 119866 and119886 119887 isin 119866 Then [119886119887]120588∘984858 = [119886]120588[119887]984858

Proof Consider [119886119887]120588∘984858 = 119886119887119873120588∘984858 = 119886119887119873120588119873984858 =

(119886119873120588)(119887119873984858) = [119886]120588[119887]984858

Theorem 30 Let 120588 984858 be two congruence relations on 119866 and120588 ∘ 984858 = 984858 ∘ 120588 If 119860 is a subgroup of 119866 and119873120588 119873984858 sube 119860 Then

(1) 120588 ∘ 984858(119860) supe 120588(119860)984858(119860)(2) 120588 ∘ 984858(119860) = 120588(119860)984858(119860)

Proof (1) Consider forall119886 isin 120588(119860)984858(119860) rArr exist119887 isin 120588(119860) 119888 isin 984858(119860)119886 = 119887119888 rArr [119887]120588 cap 119860 = 0 [119888]984858 cap 119860 = 0 119886 = 119887119888 rArr exist119887

1015840isin [119887]120588

1198871015840isin 119860 1198881015840 isin [119888]984858 119888

1015840isin 119860 rArr 119887

10158401198881015840isin [119887]120588[119888]984858 = [119887119888]120588∘984858 =

[119886]120588∘984858 11988710158401198881015840isin 119860119860 = 119860 rArr [119886]120588∘984858 cap 119860 = 0 rArr 119886 isin 120588 ∘ 984858(119860) so

120588 ∘ 984858(119860) supe 120588(119860)984858(119860)(2) Consider forall119886 isin 120588 ∘ 984858(119860) rArr [119886]120588∘984858 = [119886]120588[119890]984858 sube 119860

because [119886]120588 = [119886]120588[119890] sube [119886]120588[119890]984858 sube 119860 and 119873984858 = [119890]119886 sube

119860 rArr 119886 isin 120588(119860) 119890 isin 984858(119860) rArr 119886 = 119886119890 isin 120588(119860)984858(119860) that is120588 ∘ 984858(119860) sube 120588(119860)984858(119860) On the contrary forall119886 isin 120588(119860)984858(119860) rArr

exist119887 isin 120588(119860) 119888 isin 984858(119860) 119886 = 119887119888 rArr [119887]120588 sube 119860 [119888]984858 sube 119860 119886 = 119887119888 rArr

[119886]120588 ∘ 984858 = [119887119888]120588 ∘ 984858 = [119887]120588[119888]984858 sube 119860119860 = 119860 rArr 119886 isin 120588 ∘ 984858(119860) thatis 120588(119860)984858(119860) sube 120588 ∘ 984858(119860) so 120588 ∘ 984858(119860) = 120588(119860)984858(119860)

Theorem 31 Let 120588 be complete congruence relation on 119866 and119860 119861 sube 119866 Then

(1) 120588(119860119861) = 120588(119860)120588(119861)(2) 120588(119860119861) = 120588(119860)120588(119861)

Proof (1) Consider forall119886 isin 120588(119860119861) rArr [119886]120588 cap 119860119861 = 0 rArr

exist119887 isin [119886]120588 119887 isin 119860119861 rArr 119886 isin 119887119873120588 exist119888 isin 119860 119889 isin 119861119887 = 119888119889 rArr 119886 isin 119888119889119873120588 = (119888119873120588)(119889119873120588) and [119888]120588 cap 119860 = 0[119889]120588 cap 119861 = 0 rArr exist119888

1015840isin 119873120588 119889

1015840isin 119889119873120588 119886 = 119888

10158401198891015840 because

[1198881015840]120588 = [119888]120588 [119889

1015840]120588 = [119889]120588 rArr [119888

1015840]120588 cap 119860 = 0 [1198891015840]120588 cap 119861 = 0

119886 = 11988810158401198891015840rArr 119886 = 119887

10158401198881015840isin 120588(119860)120588(119861) that is 120588(119860119861) sube 120588(119860)120588(119861)

On the contrary forall119886 isin 120588(119860)120588(119861) rArr exist119887 isin 120588(119860) 119888 isin 120588(119861)119886 = 119887119888 rArr [119887]120588 cap 119860 = 0 [119888]120588 cap 119861 = 0 rArr exist119887

1015840isin [119887]120588 119887

1015840isin 119860

The Scientific World Journal 5

1198881015840isin [119888]120588 119888

1015840isin 119861 rArr 119887

10158401198881015840isin [119887]120588[119888]120588 = [119887119888]120588 = [119886]120588

11988710158401198881015840isin 119860119861 rArr [119886]120588 cap119860119861 = 0 rArr 119886 isin 120588(119860119861) that is 120588(119860)120588(119861) sube

120588(119860119861) Therefore 120588(119860119861) = 120588(119860)120588(119861)(2) forall119886 isin 120588(119860119861) rArr [119886]120588 sube 119860119861 rArr exist119887 isin 119860 119888 isin 119861 119886 =

119887119888 rArr [119887]120588[119888]120588 = [119887119888]120588 = [119886]120588 sube 119860119861 if exist1198871015840 isin [119887]120588 1198871015840ni 119860 119888

1015840isin

[119888]120588 1198881015840ni 119861 then 119887

10158401198881015840isin [119887]120588[119888]120588 119887

10158401198881015840ni 119860119861 Contradiction

that is [119887]120588 sube 119860 [119888]120588 sube 119861 so 119886 = 119887119888 isin 120588(119860)120588(119861) thatis 120588(119860119861) sube 120588(119860)120588(119861) On the contrary forall119886 isin 120588(119860)120588(119861) rArr

exist119887 isin 120588(119860) 119888 isin 120588(119861) 119886 = 119887119888 rArr [119887]120588 sube 119860 [119888]120588 sube 119861 119886 =

119887119888 rArr [119886]120588 = [119887119888]120588 = [119887]120588[119888]120588 sube 119860119861 rArr 119886 isin 120588(119860119861) that is120588(119860)120588(119861) sube 120588(119860119861) so 120588(119860119861) = 120588(119860)120588(119861)

Corollary 32 Let 120588 be a complete congruence relation on 119866

and 119860 sube 119866 119899 isin 119873 Then(1) 120588(119860119899) = (120588(119860))

119899(2) 120588(119860119899) = (120588(119860))

119899

Corollary 33 Let 120588 be a complete congruence relation on 119866

and let 119860 be a subgroup of 119866 and 119899 isin 119873 Then(1) 120588119899(119860) = (120588(119860))

119899= 120588(119860

119899) = 120588(119860)

(2) 120588119899(119860) = (120588(119860))119899= 120588(119860

119899) = 120588(119860)

Proof Because 119860 is a subgroup then 120588(119860) 120588(119860) are both thesubgroups of 119866 so (120588(119860))

119899= 120588(119860) (120588(119860))119899 = 120588(119860) 119860119899 = 119860

and 120588 is an equivalence relation then 120588119899= 120588 hence we can

get (1) and (2)

Corollary 34 Let 120588 be a complete congruence relation on 119866

and let 119860 119861 be two subgroups of 119866 Then(1) 120588(119860 cup 119861) = 120588(119860) cup 120588(119861) sube 120588(119860119861) = 120588(119860)120588(119861)(2) 120588(119860 cup 119861) sube 120588(119860119861) = 120588(119860)120588(119861)

Proof (1) Because 119860 = 119860(119890) sube 119860119861 and 119861 sube 119860119861 so 119860 cup 119861 sube

119860119861 hence 120588(119860 cup 119861) sube 120588(119860119861) and 120588(119860 cup 119861) = 120588(119860) cup 120588(119861)therefore (1) holds (2) Because 119860 cup 119861 sube 119860119861 so 120588(119860 cup 119861) sube

120588(119860119861) = 120588(119860)120588(119861)

Conflict of Interests

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

Acknowledgments

This research was supported by the Fundamental ResearchFunds for the Central Universities (no YX2014-08) and theNSFC of China (Grant no 61370193)

References

[1] Z Pawlak ldquoRough setsrdquo International Journal of Computer ampInformation Sciences vol 11 no 5 pp 341ndash356 1982

[2] G Liu and W Zhu ldquoThe algebraic structures of generalizedrough set theoryrdquo Information Sciences vol 178 no 21 pp 4105ndash4113 2008

[3] J N Mordeson ldquoRough set theory applied to (fuzzy) idealtheoryrdquo Fuzzy Sets and Systems vol 121 no 2 pp 315ndash324 2001

[4] Z Pawlak and A Skowron ldquoRough sets some extensionsrdquoInformation Sciences vol 177 no 1 pp 28ndash40 2007

[5] Z Pawlak andA Skowron ldquoRough sets andBoolean reasoningrdquoInformation Sciences vol 177 no 1 pp 41ndash73 2007

[6] M H Shahzamanian M Shirmohammadi and B DavvazldquoRoughness inCayley graphsrdquo Information Sciences vol 180 no17 pp 3362ndash3372 2010

[7] Z Pawlak and A Skowron ldquoRudiments of rough setsrdquo Informa-tion Sciences vol 177 no 1 pp 3ndash27 2007

[8] V Leoreanu-Fotea and B Davvaz ldquoRoughness in n-ary hyper-groupsrdquo Information Sciences vol 178 no 21 pp 4114ndash41242008

[9] R Biswas and S Nanda ldquoRough groups and rough subgroupsrdquoBulletin of the Polish Academy of Sciences Mathematics vol 42pp 251ndash254 1994

[10] J S Jiang C X Wu and D G Chen ldquoThe product structureof fuzzy rough sets on a group and the rough T-fuzzy grouprdquoInformation Sciences vol 175 no 1-2 pp 97ndash107 2005

[11] Y Q Yin J M Zhan and P Corsini ldquoFuzzy roughness of n-ary hypergroups based on a complete residuated latticerdquoNeuralComputing and Applications vol 20 no 1 pp 41ndash57 2011

[12] Q M Xiao and Z L Zhang ldquoRough prime ideals and roughfuzzy prime ideals in semigroupsrdquo Information Sciences vol 176no 6 pp 725ndash733 2006

[13] N Kuroki ldquoRough ideals in semigroupsrdquo Information Sciencesvol 100 no 1ndash4 pp 139ndash163 1997

[14] N Kuroki and P PWang ldquoThe lower and upper approximationsin a fuzzy grouprdquo Information Sciences vol 90 no 1ndash4 pp 203ndash220 1996

[15] K C Gupta and M K Kantroo ldquoGeneralized product of fuzzysubsets of a ringrdquo Fuzzy Sets and Systems vol 117 no 3 pp 419ndash429 2001

[16] O Kazanci and B Davvaz ldquoOn the structure of rough prime(primary) ideals and rough fuzzy prime (primary) ideals incommutative ringsrdquo Information Sciences vol 178 no 5 pp1343ndash1354 2008

[17] N Kuroki and J N Mordeson ldquoStructure of rough sets andrough groupsrdquo Journal of Fuzzy Mathematics vol 5 no 1 pp183ndash191 1997

[18] F Li Y Yin and L Lu ldquo(120599 119879)-fuzzy rough approximationoperators and the TL-fuzzy rough ideals on a ringrdquo InformationSciences vol 177 no 21 pp 4711ndash4726 2007

[19] Y Q Yin and X K Huang ldquoFuzzy roughness in hyperringsbased on a complete residuated latticerdquo International Journal ofFuzzy Systems vol 13 no 3 pp 185ndash194 2011

[20] W J Liu Y D Cu and H X Li ldquoRough fuzzy ideals andrough fuzzy prime ideals in semigroupsrdquo Fuzzy Systems andMathematics vol 21 no 3 pp 127ndash132 2007

[21] J L Zhang and Z L Zhang ldquoRough subgroups and roughsubringsrdquo Pure and Applied Mathematics vol 20 pp 92ndash962004

[22] J L Zhang and Z L Zhang ldquoFuzzy rough subgroupsrdquo FuzzySystems and Mathematics vol 21 no 3 pp 127ndash132 2007

[23] Z L Zhang J L Zhang and Q M Xiao Fuzzy Algebra andRough Algebra Wuhan University Press Wuhan China 2007

6 The Scientific World Journal

[24] B Davvaz and V Leoreanu-Fotea ldquoApplications of intervalvalued fuzzy n-ary polygroups with respect to t-norms (t-conorms)rdquo Computers and Mathematics with Applications vol57 no 8 pp 1413ndash1424 2009

[25] DW ShengAbstractAlgebra Science PublishingPress BeijingChina 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

The Scientific World Journal 5

1198881015840isin [119888]120588 119888

1015840isin 119861 rArr 119887

10158401198881015840isin [119887]120588[119888]120588 = [119887119888]120588 = [119886]120588

11988710158401198881015840isin 119860119861 rArr [119886]120588 cap119860119861 = 0 rArr 119886 isin 120588(119860119861) that is 120588(119860)120588(119861) sube

120588(119860119861) Therefore 120588(119860119861) = 120588(119860)120588(119861)(2) forall119886 isin 120588(119860119861) rArr [119886]120588 sube 119860119861 rArr exist119887 isin 119860 119888 isin 119861 119886 =

119887119888 rArr [119887]120588[119888]120588 = [119887119888]120588 = [119886]120588 sube 119860119861 if exist1198871015840 isin [119887]120588 1198871015840ni 119860 119888

1015840isin

[119888]120588 1198881015840ni 119861 then 119887

10158401198881015840isin [119887]120588[119888]120588 119887

10158401198881015840ni 119860119861 Contradiction

that is [119887]120588 sube 119860 [119888]120588 sube 119861 so 119886 = 119887119888 isin 120588(119860)120588(119861) thatis 120588(119860119861) sube 120588(119860)120588(119861) On the contrary forall119886 isin 120588(119860)120588(119861) rArr

exist119887 isin 120588(119860) 119888 isin 120588(119861) 119886 = 119887119888 rArr [119887]120588 sube 119860 [119888]120588 sube 119861 119886 =

119887119888 rArr [119886]120588 = [119887119888]120588 = [119887]120588[119888]120588 sube 119860119861 rArr 119886 isin 120588(119860119861) that is120588(119860)120588(119861) sube 120588(119860119861) so 120588(119860119861) = 120588(119860)120588(119861)

Corollary 32 Let 120588 be a complete congruence relation on 119866

and 119860 sube 119866 119899 isin 119873 Then(1) 120588(119860119899) = (120588(119860))

119899(2) 120588(119860119899) = (120588(119860))

119899

Corollary 33 Let 120588 be a complete congruence relation on 119866

and let 119860 be a subgroup of 119866 and 119899 isin 119873 Then(1) 120588119899(119860) = (120588(119860))

119899= 120588(119860

119899) = 120588(119860)

(2) 120588119899(119860) = (120588(119860))119899= 120588(119860

119899) = 120588(119860)

Proof Because 119860 is a subgroup then 120588(119860) 120588(119860) are both thesubgroups of 119866 so (120588(119860))

119899= 120588(119860) (120588(119860))119899 = 120588(119860) 119860119899 = 119860

and 120588 is an equivalence relation then 120588119899= 120588 hence we can

get (1) and (2)

Corollary 34 Let 120588 be a complete congruence relation on 119866

and let 119860 119861 be two subgroups of 119866 Then(1) 120588(119860 cup 119861) = 120588(119860) cup 120588(119861) sube 120588(119860119861) = 120588(119860)120588(119861)(2) 120588(119860 cup 119861) sube 120588(119860119861) = 120588(119860)120588(119861)

Proof (1) Because 119860 = 119860(119890) sube 119860119861 and 119861 sube 119860119861 so 119860 cup 119861 sube

119860119861 hence 120588(119860 cup 119861) sube 120588(119860119861) and 120588(119860 cup 119861) = 120588(119860) cup 120588(119861)therefore (1) holds (2) Because 119860 cup 119861 sube 119860119861 so 120588(119860 cup 119861) sube

120588(119860119861) = 120588(119860)120588(119861)

Conflict of Interests

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

Acknowledgments

This research was supported by the Fundamental ResearchFunds for the Central Universities (no YX2014-08) and theNSFC of China (Grant no 61370193)

References

[1] Z Pawlak ldquoRough setsrdquo International Journal of Computer ampInformation Sciences vol 11 no 5 pp 341ndash356 1982

[2] G Liu and W Zhu ldquoThe algebraic structures of generalizedrough set theoryrdquo Information Sciences vol 178 no 21 pp 4105ndash4113 2008

[3] J N Mordeson ldquoRough set theory applied to (fuzzy) idealtheoryrdquo Fuzzy Sets and Systems vol 121 no 2 pp 315ndash324 2001

[4] Z Pawlak and A Skowron ldquoRough sets some extensionsrdquoInformation Sciences vol 177 no 1 pp 28ndash40 2007

[5] Z Pawlak andA Skowron ldquoRough sets andBoolean reasoningrdquoInformation Sciences vol 177 no 1 pp 41ndash73 2007

[6] M H Shahzamanian M Shirmohammadi and B DavvazldquoRoughness inCayley graphsrdquo Information Sciences vol 180 no17 pp 3362ndash3372 2010

[7] Z Pawlak and A Skowron ldquoRudiments of rough setsrdquo Informa-tion Sciences vol 177 no 1 pp 3ndash27 2007

[8] V Leoreanu-Fotea and B Davvaz ldquoRoughness in n-ary hyper-groupsrdquo Information Sciences vol 178 no 21 pp 4114ndash41242008

[9] R Biswas and S Nanda ldquoRough groups and rough subgroupsrdquoBulletin of the Polish Academy of Sciences Mathematics vol 42pp 251ndash254 1994

[10] J S Jiang C X Wu and D G Chen ldquoThe product structureof fuzzy rough sets on a group and the rough T-fuzzy grouprdquoInformation Sciences vol 175 no 1-2 pp 97ndash107 2005

[11] Y Q Yin J M Zhan and P Corsini ldquoFuzzy roughness of n-ary hypergroups based on a complete residuated latticerdquoNeuralComputing and Applications vol 20 no 1 pp 41ndash57 2011

[12] Q M Xiao and Z L Zhang ldquoRough prime ideals and roughfuzzy prime ideals in semigroupsrdquo Information Sciences vol 176no 6 pp 725ndash733 2006

[13] N Kuroki ldquoRough ideals in semigroupsrdquo Information Sciencesvol 100 no 1ndash4 pp 139ndash163 1997

[14] N Kuroki and P PWang ldquoThe lower and upper approximationsin a fuzzy grouprdquo Information Sciences vol 90 no 1ndash4 pp 203ndash220 1996

[15] K C Gupta and M K Kantroo ldquoGeneralized product of fuzzysubsets of a ringrdquo Fuzzy Sets and Systems vol 117 no 3 pp 419ndash429 2001

[16] O Kazanci and B Davvaz ldquoOn the structure of rough prime(primary) ideals and rough fuzzy prime (primary) ideals incommutative ringsrdquo Information Sciences vol 178 no 5 pp1343ndash1354 2008

[17] N Kuroki and J N Mordeson ldquoStructure of rough sets andrough groupsrdquo Journal of Fuzzy Mathematics vol 5 no 1 pp183ndash191 1997

[18] F Li Y Yin and L Lu ldquo(120599 119879)-fuzzy rough approximationoperators and the TL-fuzzy rough ideals on a ringrdquo InformationSciences vol 177 no 21 pp 4711ndash4726 2007

[19] Y Q Yin and X K Huang ldquoFuzzy roughness in hyperringsbased on a complete residuated latticerdquo International Journal ofFuzzy Systems vol 13 no 3 pp 185ndash194 2011

[20] W J Liu Y D Cu and H X Li ldquoRough fuzzy ideals andrough fuzzy prime ideals in semigroupsrdquo Fuzzy Systems andMathematics vol 21 no 3 pp 127ndash132 2007

[21] J L Zhang and Z L Zhang ldquoRough subgroups and roughsubringsrdquo Pure and Applied Mathematics vol 20 pp 92ndash962004

[22] J L Zhang and Z L Zhang ldquoFuzzy rough subgroupsrdquo FuzzySystems and Mathematics vol 21 no 3 pp 127ndash132 2007

[23] Z L Zhang J L Zhang and Q M Xiao Fuzzy Algebra andRough Algebra Wuhan University Press Wuhan China 2007

6 The Scientific World Journal

[24] B Davvaz and V Leoreanu-Fotea ldquoApplications of intervalvalued fuzzy n-ary polygroups with respect to t-norms (t-conorms)rdquo Computers and Mathematics with Applications vol57 no 8 pp 1413ndash1424 2009

[25] DW ShengAbstractAlgebra Science PublishingPress BeijingChina 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 The Scientific World Journal

[24] B Davvaz and V Leoreanu-Fotea ldquoApplications of intervalvalued fuzzy n-ary polygroups with respect to t-norms (t-conorms)rdquo Computers and Mathematics with Applications vol57 no 8 pp 1413ndash1424 2009

[25] DW ShengAbstractAlgebra Science PublishingPress BeijingChina 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of