Normal Bipolar Soft Subgroups

Abstract

In this study, we introduce the concept of normal bipolar soft subgroup and investigate some properties of normal bipolar soft subgroups. We also define notions of bipolar soft left and right cosets of a bipolar soft group and the concept of conjugate bipolar soft subgroup, and obtain some of their properties.

Keywords:

• Bipolar soft group
• bipolar soft level set
• normal bipolar soft subgroup
• coset
• conjugate bipolar soft subgroup

1. Introduction

The concept of soft sets was defined by Molodtsov [1] in 1999 as an important mathematical tool for dealing with problems involving uncertainty which researchers encounter in several areas such as operation research, game theory, analysis and many more. Set theoretical operations between soft sets were introduced by Maji et al. [2]. Ali et al. [3] defined some new operations on the soft sets, and Sezgin and Atagün [4] studied on soft set operations as well. The concept of soft group, which is the first algebraic structure of soft sets, was defined by Aktaş and Çağman [5] in 2007. The theory of soft sets has also some applications in algebra as in following studies: [6–18].

The concept of bipolar soft sets and operations of bipolar soft sets were introduced by Shabir and Naz [19]. Abdullah et al. [20] defined notion of bipolar fuzzy soft sets by combining soft sets and bipolar fuzzy sets defined by Zhang [21, 22], and presented set-theoretical operations of bipolar fuzzy soft sets. Naz and Shabir [23] proposed the concept of fuzzy bipolar soft sets, and investigated algebraic structures on fuzzy bipolar soft sets. Karaaslan and Karataş [24] redefined notions of bipolar soft sets to construct topological and algebraical structures of bipolar soft sets. Hayat and Mahmood [25] introduced BS-h-sum and BS-h-product of BS-sets, and discussed bipolar soft intersectional h-ideals in the union of two isomorphic hemi-rings. Akram et al. [26] defined the concept of bipolar fuzzy soft Γ-subsemigroup and bipolar fuzzy soft Γ-ideals in a Γ-semigroup. The concept of bipolar soft group was defined by Karaaslan et al. [27] and some properties of bipolar soft groups were obtained.

In this study, we obtain some results related to bipolar soft groups, and define some new concepts on bipolar soft groups such as normal bipolar soft subgroup, conjugate bipolar soft subgroups based on the definition of bipolar soft group given in [27]. We also define concepts of bipolar soft left(right) coset of bipolar soft groups. To be more understandable, we support these new concepts by relevant examples. Furthermore, we investigate some properties of normal bipolar soft subgroups and obtain some results.

2. Preliminaries

In this section, we recall some basic definitions related to bipolar soft sets (BS-sets) and bipolar soft groups (BS-groups) given in [27].

Definition 2.1

[27]

Let E be a parameter set, $S\subset E$ and $f:S\to E$ be an injective function. Then, $S\cup f(S)$ is called extended parameter set of S and denoted by ${\mathcal{E}}_S$.

If S = E, then extended parameter set of S will be denoted by $\mathcal{E}$.

Definition 2.2

[27]

Let E be a parameter set, $S\subseteq E$ and $S\cup f(S)={\mathcal{E}}_S$ such that $f:S\to {\mathcal{E}}_S$ be an injective function. If $F:S\to \mathcal{P}(U)$ and $G:f(S)\to \mathcal{P}(U)$ are two mappings such that $F(e)\cap G(f(e))=\mathrm{\varnothing }$, then triple $(F,G,E)$ is called bipolar soft set. We can represent a bipolar soft set $(F,G,E)$ defined by a mapping as follows: $f_S:E\to P(U)\times P(U)$ such that $F(e)=\mathrm{\varnothing }$ and $G(f(e))=U$ if $e\in E\setminus S$ and $e\in E\setminus {\mathcal{E}}_S$.

Also we can write a bipolar soft set $f_S$ as a set of triples following form: $f_S=(F,G,E)=\{(e,F(e),G(f(e))t):e\in E\mathrm{and}F(e)\cap G(f(e))=\mathrm{\varnothing }\}.$

If $F(e)=\mathrm{\varnothing }$ and $G(f(e))=U$ for any $e\in E$, then $(e,\mathrm{\varnothing },U)$ will not be appeared in the bipolar soft set $f_S$.

From now onward, we will denote the sets $F(e)$ and $G(f(e))$ with $f_S^{+}(e)$ and $f_S^{-}(e)$, respectively, and these sets will be called positive and negative soft sets of bipolar soft set $f_S$, respectively. Set of all bipolar soft sets over U will be denoted by $BS(U)$.

Note 2.1

Let $f_S=(f_S^{+},f_S^{-},E)$ be a bipolar soft set over U. We will say that $f_S(e)=(f_S^{+}(e),f_S^{-}(e))$ is image of parameter $e\in E$.

Set theoretical operations of bipolar soft sets can be found in [24] given as follow: Let $f_S$ and $f_T\in BS(U)$. Then,

1. If $f_S^{+}(e)=\mathrm{\varnothing }$ and $f_S^{-}(e)=U$ for all $e\in E$, $f_S$ is said to be a null bipolar soft set, denoted by ${\mathrm{\Phi }}_b=(\mathrm{\Phi },\hat{U},E)$.

2. If $f_S^{+}(e)=U$ and $f_S^{-}(e)=\mathrm{\Phi }$ for all $e\in E$, $f_S$ is said to be absolute bipolar soft set, denoted by ${\hat{U}}_b=(\hat{U},\mathrm{\Phi },E)$.

3. $f_S$ is bipolar soft subset of $f_T$, denoted by $f_S⊑f_T$, if $f_S^{+}(e)\subseteq f_T^{+}(e)$ and $f_S^{-}(e)\supseteq f_T^{-}(e)$ for all $e\in E$.

4. $f_S=f_T$, if $f_S⊑f_T$ and $f_T⊑f_S$.

5. Bipolar soft union of $f_S$ and $f_T$, denoted by $f_S\bigsqcup f_T$, is a soft set over U and defined by $(f_S^{+}\tilde{\cup }{f}_T^{+}):S\cup T\to P(U)$ such that $(f_S^{+}\tilde{\cup }{f}_T^{+})(e)=f_S^{+}(e)\cup f_T^{+}(e)$ and $(f_S^{-}\tilde{\cap }{f}_T^{-}):S\cup T\to P(U)$ such that $(f_S^{-}\tilde{\cap }{f}_T^{-})(e)=f_S^{-}(e)\cap f_T^{-}(e)$ for all $e\in E$.

6. Bipolar soft intersection of $f_S$ and $f_T$, denoted by $f_S\sqcap f_T$, is a soft set over U and defined by $(f_S^{+}\tilde{\cap }{f}_T^{+}):S\cap T\to P(U)$ such that $(f_S^{+}\tilde{\cap }{f}_T^{+})(e)=f_S^{+}(e)\cap f_T^{+}(e)$ and $(f_S^{-}\tilde{\cup }{f}_T^{-}):S\cap T\to P(U)$ such that $(f_S^{-}\tilde{\cup }{f}_T^{-})(e)=f_S^{-}(e)\cup f_T^{-}(e)$ for all $e\in E$.

7. Bipolar soft complement of $f_S$ is denoted by $f_S^{\tilde{c}}$ and defined by $f_S^{\hat{c}}:E\to P(U)\times P(U)$ such that $f_S^{\hat{c}}(e)=\{(e,f_S^{-}(e),f_S^{+}(e)):e\in E\}$.

Bipolar soft group structure over a group G is defined in [27] as follows:

Definition 2.3

[27]

Let G be a group, $f:G\to \mathcal{G}$ be injective function. Then, $f_G=(f_G^{+},f_G^{-},\mathcal{G})\in BS(U)$ is called a bipolar soft groupoid (BS-groupoid) over U if $f_G(ab)⊒f_G(a)\sqcap f_G(b)$ for all $a,b\in \mathcal{G}$.

Here, $f_G(ab)⊒f_G(a)\sqcap f_G(b)$ means that $f_G^{+}(ab)\supseteq f_G^{+}(a)\cap f_G^{+}(b)$ and $f_G^{-}(ab)\subseteq f_G^{-}(a)\cup f_G^{-}(b)$. Also $\mathcal{G}$ means that extended parameter set of group G.

Definition 2.4

[27]

Let $f_G$ be a BS-groupoid over U. If $f_G(a^{-1})=f_G(a)$ for all $a\in G$, then $f_G$ is called BS-group and denoted by $f_G$.

From now on, set of all bipolar soft groups in G over U will be denoted by ${\mathcal{B}\mathcal{S}}_G(U)$.

Definition 2.5

[27]

Let G be a group and $f_G$, $g_G$ be two BS-sets over U. Then, product of $f_G$ and $g_G$ is defined as follows: $(f_G\ast g_G)(a)=\bigsqcup \{f_G(b)\sqcap g_G(c):b,c\in G\mathrm{and}bc=a\}$ and inverse of $f_G$ is $f_G^{-1}(a)=f_G(a^{-1})$ for all $a\in G$.

Definition 2.6

[27]

Let $f_A$ be a BS-set over U. Then, $(\alpha ,\beta )$-level of BS-set $f_A$, denoted by $f_A^{(\alpha ,\beta )}$, is defined as follows: $f_A^{(\alpha ,\beta )}=\{a\in A:f_A^{+}(a)\supseteq \alpha \mathrm{and}{f}_A^{-}(a)\subseteq \beta \}.$ Here, $\alpha \cap \beta =\mathrm{\varnothing }$.

Note that if $\alpha =\mathrm{\varnothing }$ or $\beta =U$, then $f_A^{(\mathrm{\varnothing },U)}=\{a\in A:f_A^{+}(a)\ne \mathrm{\varnothing }\mathrm{and}{f}_A^{-}(a)\ne U\}$ is called support of $f_A$, and denoted by $Supp(f_A)$.

Now, we define the concept of bipolar soft point as a new concept in bipolar soft set theory.

Definition 2.7

Let $f_S$ be a BS-set over U. If for any $e\in S$, $f_S^{+}(e)\ne \mathrm{\varnothing }$ and $f_S^{-}(e)\subseteq U$ such that $f_S^{+}(e)\cap f_S^{-}(e)=\mathrm{\varnothing }$, and $f_S^{+}(e^{\mathrm{\prime }})=\mathrm{\varnothing }$ and $f_S^{-}(e^{\mathrm{\prime }})=U$ for all $e^{\mathrm{\prime }}\in S\setminus \{e\}$, then BS-set $f_S$ is called bipolar soft point (BS-point).

Example 2.8

Let $E=\{e_1,e_2,e_3\}$ and $U=\{u_1,u_2,u_3\}$. Then, BS-sets ${e_1}_A=\{(e_1,\{u_1\},\{u_3\})\}$, ${e_2}_B=\{(e_2,\{u_2\},\{u_1\})\}$, ${e_2}_C=\{(e_1,\{u_2\},\{u_3\})\}$ and ${e_3}_D=\{(e_1,\{u_1\},\{u_3\})\}$ are some BS-points.

Note that a BS-point is BS-set. For convenience, a BS-point $\{(e,f_S^{+}(e),f_S^{-}(e))\}$ will be denoted by $(e,f_S^{+}(e),f_S^{-}(e))$. In particular, BS-point $(e,U,\mathrm{\varnothing })$ will be denoted by $(U_e,{\mathrm{\varnothing }}_e)$.

Definition 2.9

Let $f_S\in BS(U)$ and $e_G=(e,g_S^{+}(e),g_S^{-}(e))$ be a BS-point. If $g_S^{+}(e)\subseteq f_S^{+}(e)$ and $g_S^{-}(e)\supseteq f_S^{-}(e)$, then it is said that $e_G$ is belong to BS-set $f_S$, and denoted by $e_G\hat{\in }{f}_S$.

Let us consider BS-set $f_S=\{(e_1,\{u_1,u_2\},\{u_3\}),(e_2,\{u_2\},\{u_1,u_3\}),(e_3,\{u_1\},\{u_3\})\}$. Then, ${e_1}_{A_1}=(e_1,\{u_2\},\{u_3\})\hat{\in }{f}_S$, ${e_1}_{A_2}=(e_1,\{u_1\},\{u_3\})\hat{\in }{f}_S$, ${e_1}_{A_1}=(e_1,\{u_3\},\{u_1\})\text{⧸}\hat{\in }{f}_S$.

3. Normal Bipolar Soft Subgroups

Definition 3.1

Let $f_G$ be a BS-set over U. If $f_G^{+}(ab)=f_G^{+}(ba)$ and $f_G^{-}(ab)=f_G^{-}(ba)$ for all $a,b\in G$, then $f_G$ is called an Abelian BS-set over U.

Theorem 3.2

Let $f_G$ be a BS-set over U. Then, the following assertions are equivalent:

(1)	$f_G$ is an Abelian BS-set over U.
(2)	$f_G^{+}(ab{a}^{-1})=f_G^{+}(b)$ and $f_G^{-}(ab{a}^{-1})=f_G^{-}(b)$ for all $a,b\in G$.
(3)	$f_G^{+}(ab{a}^{-1})\supseteq f_G^{+}(b)$ and $f_G^{-}(ab{a}^{-1})\subseteq f_G^{-}(b)$ for all $a,b\in G$.
(4)	$f_G^{+}(ab{a}^{-1})\subseteq f_G^{+}(b)$ and $f_G^{-}(ab{a}^{-1})\supseteq f_G^{-}(b)$ for all $a,b\in G$.
(5)	$f_S\ast f_T=f_T\ast f_S$ for all $f_S,f_T\in B{S}_G(U)$.

Proof.

$(1)\Rightarrow (2)$: Let $a,b\in G$. Since $f_G$ is an Abelian BS-set over U, $f_G^{+}(ab{a}^{-1})=f_G^{+}(a^{-1}\cdot ab)=f_G^{+}(b)$ and $f_G^{-}(ab{a}^{-1})=f_G^{-}(a^{-1}\cdot ab)=f_G^{-}(b)$.

$(2)\Rightarrow (3)$: The proof is clear.

$(3)\Rightarrow (4)$: $f_G^{+}(ab{a}^{-1})\subseteq f_G^{+}(a^{-1}\cdot ab{a}^{-1}\cdot (a^{-1}{)}^{-1})=f_G^{+}(b)$ and  $f_G^{-}(ab{a}^{-1})\supseteq f_G^{-}(a^{-1}\cdot ab{a}^{-1}\cdot (a^{-1}{)}^{-1})=f_G^{-}(b)$ for all $a,b\in G.$

$(4)\Rightarrow (1)$: Let $a,b\in G$. Then, $\begin{array}{rl}{f}_G^{+}(ab)& =f_G^{+}(ab\cdot a{a}^{-1})\\ & =f_G^{+}(a\cdot ba\cdot a^{-1})\\ & \subseteq f_G^{+}(ba)\\ & =f_G^{+}(ba\cdot b{b}^{-1})\\ & =f_G^{+}(b\cdot ab\cdot b^{-1})\\ & \subseteq f_G^{+}(ab)\end{array}$ and $\begin{array}{rl}{f}_G^{-}(ab)& =f_G^{-}(ab\cdot a{a}^{-1})\\ & =f_G^{-}(a\cdot ba\cdot a^{-1})\\ & \supseteq f_G^{-}(ba)\\ & =f_G^{-}(ba\cdot b{b}^{-1})\\ & =f_G^{-}(b\cdot ab\cdot b^{-1})\\ & \supseteq f_G^{-}(ab).\end{array}$ Therefore, $f_G^{+}(ab)=f_G^{+}(ba)$ and $f_G^{-}(ab)=f_G^{-}(ba)$.

$(1)\Rightarrow (5)$: Let $a\in G$. Then, $\begin{array}{rl}(f_S^{+}\ast f_T^{+})(a)& =\bigcup _{b\in G}\{f_S^{+}(a{b}^{-1})\cap f_S^{+}(b)\}\\ & =\bigcup _{b\in G}\{f_S^{+}(b^{-1}a)\cap f_T^{+}(b)\}\\ & =\bigcup _{b\in G}\{f_T^{+}(b)\cap f_S^{+}(b^{-1}a)\}\\ & =(f_T^{+}\ast f_S^{+})(a)\end{array}$ and $\begin{array}{rl}(f_S^{-}\ast f_T^{-})(a)& =\bigcap _{b\in G}\{f_S^{-}(a{b}^{-1})\cup f_S^{-}(b)\}\\ & =\bigcap _{b\in G}\{f_S^{-}(b^{-1}a)\cup f_S^{-}(b)\}\\ & =\bigcap _{b\in G}\{f_T^{-}(b)\cup f_S^{-}(b^{-1}a)\}\\ & =(f_T^{-}\ast f_S^{-})(a).\end{array}$ Thus, we have that $f_S\ast f_T=f_T\ast f_S$.

$(5)\Rightarrow (1)$: Now $(U,\mathrm{\varnothing }{)}_{a^{-1}}\ast f_G=f_G\ast (U,\mathrm{\varnothing }{)}_{a^{-1}}$ for all $a\in G$. Here, $(U,\mathrm{\varnothing }{)}_{a^{-1}}\ast f_G(b)$ means that $(U_{a^{-1}}\ast f_G^{+})(b)$ and $({\mathrm{\varnothing }}_{a^{-1}}\ast f_G^{-})(b)$ for all $a,b\in G$. Then, $\begin{array}{rl}(U_{a^{-1}}\ast f_G^{+})(b)& =\bigcup \{f_G^{+}(a^{-1})\cap f_G^{+}(ab)\}\\ & =\bigcup (U\cap f_G^{+}(ab))\\ & =f_G^{+}(ab),\\ ({\mathrm{\varnothing }}_{a^{-1}}\ast f_G^{-})(b)& =\bigcap \{f_G^{-}(a^{-1})\cup f_G^{-}(ab)\}\\ & =\bigcap (\mathrm{\varnothing }\cup f_G^{-}(ab))\\ & =f_G^{-}(ab),\end{array}$ and $\begin{array}{rl}(f_G^{+}\ast U_{a^{-1}})(b)& =\bigcup \{f_G^{+}(ba)\cap f_G^{+}(a^{-1})\}\\ & =\bigcup (f_G^{+}(ba)\cap U)\\ & =f_G^{+}(ba),\\ (f_G^{-}\ast {\mathrm{\varnothing }}_{a^{-1}})(b)& =\bigcap \{f_G^{-}(ba)\cup f_G^{-}(a^{-1})\}\\ & =\bigcap (f_G^{-}(ba)\cup \mathrm{\varnothing })\\ & =f_G^{-}(ba).\end{array}$ Since $(U,\mathrm{\varnothing }{)}_{a^{-1}}\ast f_G=f_G\ast (U,\mathrm{\varnothing }{)}_{a^{-1}}$ for all $a\in G$. Hence, $f_G(ab)=f_G(ba)$ for all $b,a\in G$.

Theorem 3.3

Let $f_G\in BS(U)$. Then, $f_G$ is a BS-group if and only if $f_G$ satisfies the following conditions:

(1)	$(f_G\ast f_G)⊑f_G$.
(2)	$f_G^{-1}⊑f_G$ (or $f_G⊑f_G^{-1}$ or $f_G^{-1}=f_G$ ).

Proof.

(1) For all $g\in G$, from Definition 2.5, we know that $\begin{array}{rl}(f_G^{+}\ast f_G^{+})(g)& =\bigcup \{f_G^{+}(g_1)\cap f_G^{+}(g_2):g_1{g}_2=g,g_1,g_2\in G\},\\ (f_G^{-}\ast f_G^{-})(g)& =\bigcap \{f_G^{-}(g_1)\cup f_G^{-}(g_2):g_1{g}_2=g,g_1,g_2\in G\}.\end{array}$ Since $f_G^{+}(g_1)\cap f_G^{+}(g_2)\subseteq f_G^{+}(g_1{g}_2)=f_G^{+}(g)$ and $f_G^{-}(g_1)\cup f_G^{+}(g_2)\supseteq f_G^{-}(g_1{g}_2)=f_G^{-}(g),$ $(f_G^{+}\ast f_G^{+})(g)=\bigcup \{f_G^{+}(g_1)\cap f_G^{+}(g_2):g_1{g}_2=g,g_1,g_2\in G\}\subseteq f_G^{+}(g)$ and $(f_G^{-}\ast f_G^{-})(g)=\bigcap \{f_G^{-}(g_1)\cup f_G^{-}(g_2):g_1{g}_2=g,g_1,g_2\in G\}\supseteq f_G^{-}(g).$ Thus, $(f_G\ast f_G)⊑f_G$.

(2) For all $g\in G$, we have $\begin{array}{rl}& {(f_G^{+})}^{-1}(g)=f_G^{+}(g^{-1}),\\ & {(f_G^{-})}^{-1}(g)=f_G^{-}(g^{-1}).\end{array}$ Since $f_G$ is a BS-group, ${(f_G^{+})}^{-1}(g)\subseteq f_G^{+}(g)$ and ${(f_G^{-})}^{-1}(g)\supseteq f_G^{+}(g)$. So $f_G^{-1}⊑f_G$ (or $f_G⊑f_G^{-1}$ or $f_G^{-1}=f_G$).

Theorem 3.4

Let $H\subseteq G$ and $f_H$ be a BS-group. Then, $f_H$ is Abelian BS-group in G if and only if $f_H([g_1,g_2])=f_H(e)$ for all $g_1,g_2\in G,$ where $[g_1,g_2]=g_1^{-1}{g}_2^{-1}{g}_1{g}_2$ is commutator of $g_1$ and $g_2$.

Proof.

Let $f_H^{+}([g_1,g_2])=f_H^{+}(e)$ and $f_H^{-}([g_1,g_2])=f_H^{-}(e)$ for all $g_1,g_2\in G$. Then, $\begin{array}{rl}{f}_H^{+}(g_1{g}_2)& =f_H^{+}((g_2{g}_1)(g_2{g}_1{)}^{-1}(g_1{g}_2))\\ & \supseteq f_H^{+}(g_2{g}_1)\cap f_H^{+}(g_1^{-1}{g}_2^{-1}{g}_1{g}_2)\\ & =f_H^{+}(g_2{g}_1)\cap f_H^{+}(e)\\ & =f_H^{+}(g_2{g}_1),\\ f_H^{-}(g_1{g}_2)& =f_H^{-}((g_2{g}_1)(g_2{g}_1{)}^{-1}(g_1{g}_2))\\ & \subseteq f_H^{-}(g_2{g}_1)\cup f_H^{-}(g_1^{-1}{g}_2^{-1}{g}_1{g}_2)\\ & =f_H^{-}(g_2{g}_1)\cup f_H^{-}(e)\\ & =f_H^{-}(g_2{g}_1),\end{array}$ and $\begin{array}{rl}{f}_H^{+}(g_2{g}_1)& =f_H^{+}((g_1{g}_2)(g_1{g}_2{)}^{-1}(g_2{g}_1))\\ & \supseteq f_H^{+}(g_1{g}_2)\cap f_H^{+}(g_2^{-1}{g}_1^{-1}{g}_2{g}_1)\\ & =f_H^{+}(g_1{g}_2)\cap f_H^{-}(e)\\ & =f_H^{+}(g_1{g}_2),\\ f_H^{-}(g_2{g}_1)& =f_H^{-}((g_1{g}_2)(g_1{g}_2{)}^{-1}(g_2{g}_1))\\ & \subseteq f_H^{-}(g_1{g}_2)\cup f_H^{-}(g_2^{-1}{g}_1^{-1}{g}_2{g}_1)\\ & =f_H^{-}(g_1{g}_2)\cup f_H^{-}(e)\\ & =f_H^{-}(g_1{g}_2).\end{array}$ Since $f_H^{+}(g_1{g}_2)=f_H^{+}(g_2{g}_1)$ and $f_H^{-}(g_1{g}_2)=f_H^{-}(g_2{g}_1)$, $f_H(g_1{g}_2)=f_H(g_2{g}_1)$.

Conversely, assume that $f_H$, is an abelian BS-group in G. Then, for all $g_1,g_2\in G$, using Theorem 3.9 in [27], we have $\begin{array}{rl}{f}_H(g_1^{-1}{g}_2^{-1}(g_1^{-1}{)}^{-1})& ⊒f_H(g_2^{-1})\\ f_H(g_1^{-1}{g}_2^{-1}{g}_1)& ⊒f_H(g_2^{-1})\\ f_H(g_1^{-1}{g}_2^{-1}{g}_1{g}_2{g}_2^{-1})& ⊒f_H(g_2^{-1})\\ f_H((g_1^{-1}{g}_2^{-1}{g}_1{g}_2)g_2^{-1})& ⊒f_H(g_2^{-1})\\ f_H([g_1,g_2]\cdot g_2^{-1})& ⊒f_H(g_2^{-1})\\ f_H([g_1,g_2])& ⊒f_H(e).\end{array}$ Hence, the proof is followed.

Theorem 3.5

Let $H\subseteq G$. Then, a BS-group $f_H$ is Abelian BS-set in G if and only if $f_H([g_1,g_2])⊒f_H(g_1),forall{g}_1,g_2\in G.$

Proof.

Suppose that a BS-group $f_H$ is Abelian BS-set in G. Let $g_1,g_2\in G$. Then, $\begin{array}{rl}{f}_H^{+}([g_1,g_2])& =f_H^{+}(g_1^{-1}{g}_2^{-1}{g}_1{g}_2)\\ & \supseteq f_H^{+}(g_1^{-1})\cap f_H^{+}(g_2^{-1}{g}_1{g}_2)\\ & =f_H^{+}(g_1)\cap f_H^{+}(g_1)\\ & =f_H^{+}(g_1)\end{array}$ and $\begin{array}{rl}{f}_H^{-}([g_1,g_2])& =f_H^{-}(g_1^{-1}{g}_2^{-1}{g}_1{g}_2)\\ & \subseteq f_H^{-}(g_1^{-1})\cup f_H^{-}(g_2^{-1}{g}_1{g}_2)\\ & =f_H^{-}(g_1)\cup f_H^{-}(g_1)\\ & =f_H^{-}(g_1).\end{array}$ Thus, we have $f_H([g_1,g_2])⊒f_H(g_1)$.

Conversely, assume that $f_H([g_1,g_2])⊒f_H(g_1)$, for all $g_1,g_2\in G$. Then, $\begin{array}{rl}{f}_H^{+}(g_1^{-1}{g}_2{g}_1)& =f_H^{+}(g_2{g}_2^{-1}{g}_1^{-1}{g}_2{g}_1)\\ & \supseteq f_H^{+}(g_2)\cap f_H^{+}([g_2,g_1])\\ & =f_H^{+}(g_2),\\ f_H^{-}(g_1^{-1}{g}_2{g}_1)& =f_H^{-}(g_2{g}_2^{-1}{g}_1^{-1}{g}_2{g}_1)\\ & \subseteq f_H^{-}(g_2)\cup f_H^{-}([g_2,g_1])\\ & =f_H^{-}(g_2),\end{array}$ by assumption and Theorem 3.2, $f_H$ is Abelian BS-set in G.

Definition 3.6

Let $f_G$ be a BS-group over U and $f_K$ be a BS-subgroup over U. If $f_K$ is Abelian bipolar soft subset of $f_G$, then $f_K$ is called a normal bipolar soft subgroup (NBS-subgroup), and denoted by $f_K\widehat{⊲}{f}_G$.

From now on set of all NBS-subgroups in G over U will be denoted by ${\mathcal{N}\mathcal{B}\mathcal{S}}_G(U)$.

Example 3.7

Assume that $U=\{u_1,u_2,u_3,u_4,u_5,u_6,u_7,u_8,u_9,u_{10}\}$ is a universal set and $G=\{\pm 1,\pm i,\pm j,\pm k\}$, quaternion group and $K=\{j,-1,-j,1\}$, subgroup of G, be the subset of parameter set G. We define the BS-group $f_G$ by $\begin{array}{l}{f}_G^{+}(1)=U,\\ f_G^{+}(-1)=U,\\ f_G^{+}(i)=\{u_1,u_2,u_3,u_4,u_8\},\\ f_G^{+}(-i)=\{u_1,u_2,u_3,u_4,u_8\},\\ f_G^{+}(j)=\{u_4,u_5,u_6,u_7,u_8\},\\ f_G^{+}(-j)=\{u_4,u_5,u_6,u_7,u_8\},\\ f_G^{+}(k)=\{u_4,u_8,u_9,u_{10}\},\\ f_G^{+}(-k)=\{u_4,u_8,u_9,u_{10}\},\end{array}\mathrm{and}\begin{array}{l}{f}_G^{-}(1)=\mathrm{\varnothing },\\ f_G^{-}(-1)=\mathrm{\varnothing },\\ f_G^{-}(i)=\{u_5,u_6,u_7,u_9\},\\ f_G^{-}(-i)=\{u_5,u_6,u_7,u_9\},\\ f_G^{-}(j)=\{u_1,u_2,u_3,u_9\},\\ f_G^{-}(-j)=\{u_1,u_2,u_3,u_9\},\\ f_G^{-}(k)=\{u_1,u_2,u_3,u_5,u_6,u_7\},\\ f_G^{-}(-k)=\{u_1,u_2,u_3,u_5,u_6,u_7\},\end{array}$ and we define a BS-set $f_K$ by $\begin{array}{l}{f}_K^{+}(1)=U,\\ f_K^{+}(-1)=U,\\ f_K^{+}(j)=\{u_4,u_5,u_6\},\\ f_K^{+}(-j)=\{u_4,u_5,u_6\},\end{array}\begin{array}{l}{f}_K^{-}(1)=\mathrm{\varnothing },\\ f_K^{-}(-1)=\mathrm{\varnothing },\\ f_K^{-}(j)=\{u_1,u_2,u_9\},\\ f_K^{-}(-j)=\{u_1,u_2,u_9\}.\end{array}$ It is clear that $f_K$ is a NBS-subgroup of BS-group $f_G$ over U.

Corollary 3.8

If G is an Abelian group, then any BS-group over G is normal.

Corollary 3.9

If $f_G$ is BS-group and $N◃G,$ then $f_G{|}_N$ is an NBS-subgroup.

Definition 3.10

Let $f_G$ be a BS-group over U. Then, for all $k\in G$, the set $N(f_G)=\{g\in G:f_G^{+}(gk)=f_G^{+}(kg)\mathrm{and}{f}_G^{-}(gk)=f_G^{-}(kg)\}$ is called normaliser of $f_G$ in G.

Example 3.11

Let us consider BS-group $f_G$ given in Example 3.7. Then, $N(f_G)=\{1,-1\}$.

Theorem 3.12

$N(f_G)$ is a subgroup of G and the restriction of $f_G$ to $N(f_G),$ $f_G{|}_{N(f_G)},$ is an NBS-subgroup of $N(f_G)$.

Proof.

We know that $e\in N(f_G)$. Let $g_1,g_2\in N(f_G)$. For any $g\in G$, we see that $\begin{array}{rl}{f}_G^{+}(g_1{g}_2^{-1}\cdot g)& =f_G^{+}(g_1\cdot g_2^{-1}g)\\ & =f_G^{+}(g_2^{-1}g\cdot g_1)\\ & =f_G^{+}(g_1^{-1}{g}^{-1}\cdot g_2)\\ & =f_G^{+}(g_2\cdot g_1^{-1}{g}^{-1})\\ & =f_G^{+}(g\cdot g_1{g}_2^{-1})\end{array}$ and $\begin{array}{rl}{f}_G^{-}(g_1{g}_2^{-1}\cdot g)& =f_G^{-}(g_1\cdot g_2^{-1}g)\\ & =f_G^{-}(g_2^{-1}g\cdot g_1)\\ & =f_G^{-}(g_1^{-1}{g}^{-1}\cdot g_2)\\ & =f_G^{-}(g_2\cdot g_1^{-1}{g}^{-1})\\ & =f_G^{-}(g\cdot g_1{g}_2^{-1}).\end{array}$ Hence, $g_1{g}_2^{-1}\in N(f_G)$. Thus, $N(f_G)$ is a subgroup of G. By Corollary  3.9, it is clear that $f_G{|}_{N(f_G)}$ is NBS-subgroup and $f_G{|}_{N(f_G)}(g_1{g}_2)=f_G{|}_{N(f_G)}(g_2{g}_1),$ $\mathrm{\forall }{g}_1,g_2\in N(f_G)$. Therefore, $f_G{|}_{N(f_G)}$ is a NBS-subgroup.

Theorem 3.13

Let $N_i\subseteq G$ and $\{f_{N_i}:i\in I\}$ be a family of NBS-subgroups in G. Then, ${\sqcap }_{i\in I}{f}_{G_i}$ is a NBS-subgroup.

Proof.

The proof is obvious from Definitions of intersection of BS-sets and NBS-subgroup.

Lemma 3.14

Let $S,T\subseteq G,$ $f_S\in {\mathcal{N}\mathcal{B}\mathcal{S}}_G(U)$ and $f_T$ be any BS-set over G. Then, $f_S\ast f_T=f_S\ast f_T.$

Proof.

For all $g\in G$, we know that $\begin{array}{rl}& (f_S^{+}\ast f_T^{+})(g)=\bigcup \{f_S^{+}(g_1)\cap f_T^{+}(g_2):g_1{g}_2=g,g_1,g_2\in G\},\\ & (f_S^{-}\ast f_T^{-})(g)=\bigcap \{f_S^{-}(g_1)\cup f_T^{-}(g_2):g_1{g}_2=g,g_1,g_2\in G\}.\end{array}$ Since $f_S$ be a normal bipolar soft subgroup and $g_1{g}_2=g$ implies $g_1=g{g}_2^{-1}$, then $\begin{array}{rl}(f_S^{+}\ast f_T^{+})(g)& =\bigcup \{f_S^{+}(g{g}_2^{-1})\cap f_T^{+}(g_2):(g{g}_2^{-1}{g}_2)=g\}\\ & =\bigcup \{f_T^{+}(g_2)\cap f_S^{+}(g_2^{-1}g):g_2(g_2^{-1}g)=g\}\\ & =(f_T^{+}\ast f_S^{+})(g),\\ (f_S^{-}\ast f_T^{-})(g)& =\bigcap \{f_S^{-}(g{g}_2^{-1})\cup f_T^{-}(g_2):(g{g}_2^{-1}{g}_2)=g\}\\ & =\bigcap \{f_T^{-}(g_2)\cup f_S^{-}(g_2^{-1}g):g_2(g_2^{-1}g)=g\}\\ & =(f_T^{-}\ast f_S^{-})(g).\end{array}$ Thus, $f_S\ast f_T=f_S\ast f_T$.

Theorem 3.15

Let $f_S\in {\mathcal{N}\mathcal{B}\mathcal{S}}_G(U)$ and $f_T\in {\mathcal{B}\mathcal{S}}_G(U)$. Then, $f_S\ast f_T$ is a BS-group.

Proof.

Let show that conditions in the Theorem 3.3 is satisfied. Firstly, $\begin{array}{rl}(f_S^{+}\ast f_T^{+})\ast (f_S^{+}\ast f_T^{+})& =f_S^{+}\ast (f_T^{+}\ast f_S^{+})\ast f_T^{+}\\ & =f_S^{+}\ast (f_S^{+}\ast f_T^{+})\ast f_T^{+}\\ & =(f_S^{+}\ast f_S^{+})\ast (f_T^{+}\ast f_T^{+})\\ & \subseteq (f_S^{+}\ast f_T^{+}),(\mathrm{by}\mathrm{Theorem}3.3)\end{array}$ and $\begin{array}{rl}(f_S^{-}\ast f_T^{-})\ast (f_S^{-}\ast f_T^{-})& =f_S^{-}\ast (f_T^{-}\ast f_S^{-})\ast f_T^{-}\\ & =f_S^{-}\ast (f_S^{-}\ast f_T^{-})\ast f_T^{+}\\ & =(f_S^{-}\ast f_S^{-})\ast (f_T^{-}\ast f_T^{-})\\ & \supseteq (f_S^{-}\ast f_T^{-}).(\mathrm{by}\mathrm{Theorem}3.3).\end{array}$ Thus, $(f_S\ast f_T)\ast (f_S\ast f_T)⊑(f_S\ast f_T)$.

Now, we will show that $(f_S\ast f_T)(g^{-1})⊑(f_S\ast f_T)(g)$ (or $(f_S\ast f_T)(g^{-1})⊒(f_S\ast f_T)(g)$ or $(f_S\ast f_T)(g^{-1})=(f_S\ast f_T)(g))$ for all $g\in G$. $\begin{array}{rl}(f_S^{+}\ast f_T^{+})(g^{-1})& =\bigcup \{f_S^{+}(g_1)\cap f_T^{+}(g_2):g_1{g}_2=g^{-1},g_1,g_2\in G\}\\ & =\bigcup \{f_S^{+}((g_1^{-1}{)}^{-1})\cap f_T^{+}((g_2^{-1}{)}^{-1}):g_2^{-1}{g}_1^{-1}=g,g_1,g_2\in G\}\\ & =\bigcup \{f_T^{+}(g_2^{-1})\cap f_S^{+}(g_1^{-1}):g_2^{-1}{g}_1^{-1}=g,g_1,g_2\in G\}\\ & \subseteq (f_T^{+}\ast f_S^{+})(g)\\ & =(f_S^{+}\ast f_T^{+})(g),(\mathrm{by}\mathrm{Lemma}3.14)\end{array}$ and $\begin{array}{rl}(f_S^{-}\ast f_T^{-})(g^{-1})& =\bigcap \{f_S^{-}(g_1)\cup f_T^{-}(g_2):g_1{g}_2=g^{-1},g_1,g_2\in G\}\\ & =\bigcap \{f_S^{-}((g_1^{-1}{)}^{-1})\cup f_T^{-}((g_2^{-1}{)}^{-1}):g_2^{-1}{g}_1^{-1}=g,g_1,g_2\in G\}\\ & =\bigcap \{f_T^{-}(g_2^{-1})\cup f_S^{-}(g_1^{-1}):g_2^{-1}{g}_1^{-1}=g,g_1,g_2\in G\}\\ & \supseteq (f_T^{-}\ast f_S^{-})(g)\\ & =(f_S^{-}\ast f_T^{-})(g).(\mathrm{by}\mathrm{Lemma}3.14).\end{array}$ Thus, $(f_S\ast f_T)(g^{-1})⊑(f_S\ast f_T)(g)$. By Theorem 3.3, $f_S\ast f_T$ is a BS-group.

Corollary 3.16

If $f_S,f_T\in {\mathcal{N}\mathcal{B}\mathcal{S}}_G(U),$ then $(f_S\ast f_T)\in {\mathcal{N}\mathcal{B}\mathcal{S}}_G(U)$.

Lemma 3.17

Let G be a group with unit element e and $f_G$ BS-set over U. Then, $f_G$ is BS-group if and only if $f_G^{(\alpha ,\beta )}$ is a subgroup of G $\mathrm{\forall }(\alpha ,\beta )\in \mathrm{Im}(f_G)\cup \{({\alpha }^{\prime },{\beta }^{\prime })\in P(U)\times P(U):{\alpha }^{\prime }\subseteq f_G^{+}(e)\mathrm{and}{\beta }^{\prime }\supseteq f_G^{-}(e)\}$.

Proof.

Let $f_G\in {\mathcal{B}\mathcal{S}}_G(U)$ and $\gamma =(\alpha ,\beta )\in \mathrm{Im}(f_G)$. Since $f_G^{+}(e)\supseteq f_G^{+}(g_1)$ and $f_G^{-}(e)\subseteq f_G^{-}(g_1)$ for all $g_1\in G,e\in f_G^{(\alpha ,\beta )}$. Therefore, $f_G^{(\alpha ,\beta )}\ne \mathrm{\varnothing }$. For $g_1,g_2\in f_G^{(\alpha ,\beta )}$, $f_G^{+}(g_1)\supseteq \alpha$, $f_G^{-}(g_1)\subseteq \beta$, $f_G^{+}(g_2)\supseteq \alpha$ and $f_G^{-}(g_2)\subseteq \beta$. Since $f_G$ is a BS-group, $f_G^{+}(g_1{g}_2^{-1})\supseteq f_G^{+}(g_1)\cap f_G^{+}(g_2)\supseteq \alpha \cap \alpha =\alpha$ and $f_G^{-}(g_1{g}_2^{-1})\subseteq f_G^{-}(g_1)\cup f_G^{-}(g_2)\subseteq \beta \cup \beta =\beta$. Hence, $g_1{g}_2^{-1}\in {f_G}^{(\alpha ,\beta )}$. Similarly, if ${\alpha }^{\prime }\subseteq f_G^{+}(e)$ and ${\beta }^{\prime }\supseteq f_G^{-}(e)$, then it can be shown that ${f_G}^{(\alpha ,\beta )}$ is an subgroup of G.

Conversely, let ${f_G}^{(\alpha ,\beta )}$ be an subgroup of G for all $(\alpha ,\beta )\in \mathrm{Im}(f_G)\cup \{({\alpha }^{\prime },{\beta }^{\prime })\in P(U)\times P(U):{\alpha }^{\prime }\subseteq f_G^{+}(g_1)\mathrm{and}{\beta }^{\prime }\supseteq f_G^{-}(g_1)$. Then, for all $(\alpha ,\beta )\in \mathrm{Im}(f_G)$, we must have $e\in {f_G}^{(\alpha ,\beta )}$ and so $f_G^{+}(e)\supseteq \alpha$ and $f_G^{-}(e)\subseteq \beta$. Suppose that $g_1,g_2\in G$ and $f_G^{+}(g_1)=\alpha ,$ $f_G^{-}(g_1)=\beta$, $f_G^{+}(g_2)={\alpha }^{\prime }$ and $f_G^{-}(g_2)={\beta }^{\prime }$. Let $({\alpha }^{″},{\beta }^{″})=(\alpha \cap {\alpha }^{\prime },\beta \cup {\beta }^{\prime })$. Then, $g_1,g_2\in f_G^{({\alpha }^{″},{\beta }^{″})}$ and ${\alpha }_G^{″}\subseteq f_G^{+}(e)$, ${\beta }^{″}\supseteq f_G^{-}(e)$. By hypothesis, $f_G^{({\alpha }^{″},{\beta }^{″})}$ is subgroup of G and so $g_1{g}_2^{-1}\in f_G^{({\alpha }^{″},{\beta }^{″})}$. Thus, $f_G^{+}(g_1{g}_2^{-1})\supseteq {\alpha }^{″}=(\alpha \cap {\alpha }^{\prime })=f_G^{+}(g_1)\cap f_G^{+}(g_2)$ and $f_G^{-}(g_1{g}_2^{-1})\subseteq {\beta }^{″}=(\beta \cup {\beta }^{\prime })=f_G^{-}(g_1)\cup f_G^{-}(g_2)$. Hence, $f_G$ is BS-group over G.

Theorem 3.18

Let G be a group with unit element e and let $f_G$ be a BS-set over U. Then, $f_G$ is an NBS-subgroup if and only if $f_G^{(\alpha ,\beta )}$ is a normal subgroup of G $\mathrm{\forall }(\alpha ,\beta )\in \mathrm{Im}(f_G)\cup \{({\alpha }^{\prime },{\beta }^{\prime })\in P(U)\times P(U):{\alpha }^{\prime }\subseteq f_G^{+}(e)\mathrm{and}{\beta }^{\prime }\supseteq f_G^{-}(e)\}$.

Proof.

$:\Rightarrow$ Let $f_G$ be an NBS-subgroup and let $(\alpha ,\beta )\in \mathrm{Im}(f_G)\cup \{({\alpha }^{\prime },{\beta }^{\prime })\in P(U)\times P(U):{\alpha }^{\prime }\subseteq f_G^{+}(e)\mathrm{and}{\beta }^{\prime }\supseteq f_G^{-}(e)\}$. Since $f_G$ is BS-group, $f_G^{(\alpha ,\beta )}$ is a subgroup of G. If $g\in G$ and $k\in f_G^{(\alpha ,\beta )}$, from condition $(2)$ of Theorem 3.2 $f_G^{+}(gk{g}^{-1})=f_G^{+}(k)\supseteq \alpha$ and $f_G^{-}(gk{g}^{-1})=f_G^{-}(k)\subseteq \beta$. Thus, $f_G^{(\alpha ,\beta )}$ is a normal subgroup of G.

$\Leftarrow$: Suppose that $f_G^{(\alpha ,\beta )}$ is a normal subgroup of G $\mathrm{\forall }(\alpha ,\beta )\in \mathrm{Im}(f_G)\cup \{({\alpha }^{\prime },{\beta }^{\prime })\in P(U)\times P(U):{\alpha }^{\prime }\subseteq f_G^{+}(e)\mathrm{and}{\beta }^{\prime }\supseteq f_G^{-}(e)\}$. It follows from Lemma 3.17 that $f_G$ is a BS-group. Let $g_1,g_2\in G$ and $\alpha =f_G^{+}(g_2)$ and $\beta =f_G^{-}(g_2)$. Then, $g_1\in f_G^{(\alpha ,\beta )}$ and so $g_1{g}_2{g}_1^{-1}\in f_G^{(\alpha ,\beta )}$. Thus, $f_G^{+}(g_1{g}_2{g}_1^{-1})\supseteq \alpha =f_G^{+}(g_2)$ and $f_G^{-}(g_1{g}_2{g}_1^{-1})\subseteq \beta =f_G^{-}(g_2)$. Then, $f_G$ satisfies condition $(3)$ of Theorem 3.2. Therefore, $f_G$ is a normal BS-group.

Definition 3.19

[27]

Let $f_G$ BS-set over U. Then, $e-$set of $f_G$, denoted by $e_{f_G}$, is defined as $e_{f_G}=\{g\in G:f_G^{+}(g)=f_G^{+}(e)\mathrm{and}{f}_G^{-}(g)=f_G^{-}(e)\}.$

Theorem 3.20

Let $f_G$ be a BS-group. Then, e−set of $f_G$ and $Supp(f_G)$ are normal subgroup of G.

Proof.

The proof is obvious from Lemma 3.17 and Theorem 3.2.

Definition 3.21

Let $f_G,h_G\in {\mathcal{B}\mathcal{S}}_G(U)$. If there exists $g\in G$ such that $f_G(a)=h_G(ga{g}^{-1})$ for all $a\in G$, then $f_G$ and $h_G$ are called conjugate bipolar soft subgroup (w.r.t g) and we write, $f_G=h_G^g$, where $h_G^g(a)=f_G(ga{g}^{-1})$ for all $a\in G$.

Example 3.22

Let $G=S_3$ and $U=\mathbb{Z}$. We define bipolar soft group $f_G$ by $\begin{array}{r}\begin{array}{l}{f}_G^{+}((1))=\mathbb{Z}\\ f_G^{+}((12))=\{3,7,8,15\}\\ f_G^{+}((13))=\{3,7,8,15\}\\ f_G^{+}((23))=\{3,7,8,15\}\\ f_G^{+}((123))=\{3,7,8,15,20\}\\ f_G^{+}((132))=\{3,7,8,15,20\}\end{array}\mathrm{and}\begin{array}{l}{f}_G^{-}((1))=\mathrm{\varnothing }\\ f_G^{-}((12))=\{0,2,5,11,16,19,21\}\\ f_G^{-}((13))=\{0,2,5,11,16,19,21\}\\ f_G^{-}((23))=\{0,2,5,11,16,19,21\}\\ f_G^{-}((123))=\{0,2,5,11,19\}\\ f_G^{-}((132))=\{0,2,5,11,19\}.\end{array}\end{array}$ The conjugate bipolar soft subgroup is given by $\begin{array}{rl}& \begin{array}{l}{h_G^{+}}^{(1)}((1))=f_G^{+}((1))=\mathbb{Z}\\ {h_G^{+}}^{(1)}((12))=f_G^{+}((12))=\{3,7,8,15\}\\ {h_G^{+}}^{(1)}((13))=f_G^{+}((13))=\{3,7,8,15\}\\ {h_G^{+}}^{(1)}((23))=f_G^{+}((23))=\{3,7,8,15\}\\ {h_G^{+}}^{(1)}((123))=f_G^{+}((123))=\{3,7,8,15,20\}\\ {h_G^{+}}^{(1)}((132))=f_G^{+}((132))=\{3,7,8,15,20\}\end{array}\mathrm{and}\begin{array}{l}{h_G^{-}}^{(1)}((1))=f_G^{-}((1))=\mathrm{\varnothing }\\ {h_G^{-}}^{(1)}((12))=f_G^{-}((12))=\{0,2,5,11,16,19,21\}\\ {h_G^{-}}^{(1)}((13))=f_G^{-}((13))=\{0,2,5,11,16,19,21\}\\ {h_G^{-}}^{(1)}((23))=f_G^{-}((23))=\{0,2,5,11,16,19,21\}\\ {h_G^{-}}^{(1)}((123))=f_G^{-}((123))=\{0,2,5,11,19\}\\ {h_G^{-}}^{(1)}((132))=f_G^{-}((132))=\{0,2,5,11,19\}\end{array}\\ & \begin{array}{l}{h_G^{+}}^{(12)}((1))=f_G^{+}((1))=\mathbb{Z}\\ {h_G^{+}}^{(12)}((12))=f_G^{+}((12))=\{3,7,8,15\}\\ {h_G^{+}}^{(12)}((13))=f_G^{+}((23))=\{3,7,8,15\}\\ {h_G^{+}}^{(12)}((23))=f_G^{+}((13))=\{3,7,8,15\}\\ {h_G^{+}}^{(12)}((123))=f_G^{+}((132))=\{3,7,8,15,20\}\\ {h_G^{+}}^{(12)}((132))=f_G^{+}((123))=\{3,7,8,15,20\}\end{array}\mathrm{and}\begin{array}{l}{h_G^{-}}^{(12)}((1))=f_G^{-}((1))=\mathrm{\varnothing }\\ {h_G^{-}}^{(12)}((12)=f_G^{-}((12))=\{0,2,5,11,16,19,21\}\\ {h_G^{-}}^{(12)}((13))=f_G^{-}((23))=\{0,2,5,11,16,19,21\}\\ {h_G^{-}}^{(12)}((23)=f_G^{-}((13))=\{0,2,5,11,16,19,21\}\\ {h_G^{-}}^{(12)}((123))=f_G^{-}((132))=\{0,2,5,11,19\}\\ {h_G^{-}}^{(12)}((132))=f_G^{-}((123))=\{0,2,5,11,19\}\end{array}\\ & \begin{array}{l}{h_G^{+}}^{(23)}((1))=f_G^{+}((1))=\mathbb{Z}\\ {h_G^{+}}^{(23)}((12))=f_G^{+}((13))=\{3,7,8,15\}\\ {h_G^{+}}^{(23)}((13))=f_G^{+}((12))=\{3,7,8,15\}\\ {h_G^{+}}^{(23)}((23))=y{f}_G^{+}((23))=\{3,7,8,15\}\\ {h_G^{+}}^{(23)}((123))=f_G^{+}((132))=\{3,7,8,15,20\}\\ {h_G^{+}}^{(23)}((132))=f_G^{+}((123))=\{3,7,8,15,20\}\end{array}\mathrm{and}\begin{array}{l}{h_G^{-}}^{(23)}((1))=f_G^{-}((1))=\mathrm{\varnothing }\\ {h_G^{-}}^{(23)}((12)=f_G^{-}((13))=\{0,2,5,11,16,19,21\}\\ {h_G^{-}}^{(23)}((13))=f_G^{-}((12))=\{0,2,5,11,16,19,21\}\\ {h_G^{-}}^{(23)}((23)=f{-}_G((23))=\{0,2,5,11,16,19,21\}\\ {h_G^{-}}^{(23)}((123))=f_G^{-}((132))=\{0,2,5,11,19\}\\ {h_G^{-}}^{(23)}((132))=f_G^{-}((123))=\{0,2,5,11,19\}\end{array}\\ & \begin{array}{l}{h_G^{+}}^{(13)}((1))=f_G^{+}((1))=\mathbb{Z}\\ {h_G^{+}}^{(13)}((12))=f_G^{+}((23))=\{3,7,8,15\}\\ {h_G^{+}}^{(13)}((13))=f_G^{+}((13))=\{3,7,8,15\}\\ {h_G^{+}}^{(13)}((23))=f_G^{+}((12))=\{3,7,8,15\}\\ {h_G^{+}}^{(13)}((123))=f_G^{+}((132))=\{3,7,8,15,20\}\\ {h_G^{+}}^{(13)}((132))=f_G^{+}((123))=\{3,7,8,15,20\}\end{array}\mathrm{and}\begin{array}{l}{h_G^{-}}^{(13)}((1))=f_G^{-}((1))=\mathrm{\varnothing }\\ {h_G^{-}}^{(13)}((12)=f_G^{-}((23))=\{0,2,5,11,16,19,21\}\\ {h_G^{-}}^{(13)}((13))=f_G^{-}((13))=\{0,2,5,11,16,19,21\}\\ {h_G^{-}}^{(13)}((23)=f_G^{-}((12))=\{0,2,5,11,16,19,21\}\\ {h_G^{-}}^{(13)}((123))=f_G^{-}((132))=\{0,2,5,11,19\}\\ {h_G^{-}}^{(13)}((132))=f_G^{-}((123))=\{0,2,5,11,19\}\end{array}\\ & \begin{array}{l}{h_G^{+}}^{(123)}((1))=f_G^{+}((1))=\mathbb{Z}\\ {h_G^{+}}^{(123)}((12))=f_G^{+}((23))=\{3,7,8,15\}\\ {h_G^{+}}^{(123)}((13))=f_G^{+}((12))=\{3,7,8,15\}\\ {h_G^{+}}^{(123)}((23))=f_G^{+}((13))=\{3,7,8,15\}\\ {h_G^{+}}^{(123)}((123))=f_G^{+}((123))=\{3,7,8,15,20\}\\ {h_G^{+}}^{(123)}((132))=f_G^{+}((132))=\{3,7,8,15,20\}\end{array}\mathrm{and}\begin{array}{l}{h_G^{-}}^{(123)}((1))=f_G^{-}((1))=\mathrm{\varnothing }\\ {h_G^{-}}^{(123)}((12)=f_G^{-}((23))=\{0,2,5,11,16,19,21\}\\ {h_G^{-}}^{(123)}((13))=f_G^{-}((12))=\{0,2,5,11,16,19,21\}\\ {h_G^{-}}^{(123)}((23)=f_G^{-}((13))=\{0,2,5,11,16,19,21\}\\ {h_G^{-}}^{(123)}((123))=f_G^{-}((123))=\{0,2,5,11,19\}\\ {h_G^{-}}^{(123)}((132))=f_G^{-}((132))=\{0,2,5,11,19\}\end{array}\\ & \begin{array}{l}{h_G^{+}}^{(132)}((1))=f_G^{+}((1))=\mathbb{Z}\\ {h_G^{+}}^{(132)}((12))=f_G^{+}((13))=\{3,7,8,15\}\\ {h_G^{+}}^{(132)}((13))=f_G^{+}((23))=\{3,7,8,15\}\\ {h_G^{+}}^{(132)}((23))=f_G^{+}((12))=\{3,7,8,15\}\\ {h_G^{+}}^{(132)}((123))=f_G^{+}((123))=\{3,7,8,15,19\}\\ {h_G^{+}}^{(132)}((132))=f_G^{+}((132))=\{3,7,8,15,19\}\end{array}\mathrm{and}\begin{array}{l}{h_G^{-}}^{(132)}((1))=f_G^{-}((1))=\mathrm{\varnothing }\\ {h_G^{-}}^{(132)}((12)=f_G^{-}((13))=\{0,2,5,11,16,19,21\}\\ {h_G^{-}}^{(132)}((13))=f_G^{-}((23))=\{0,2,5,11,16,19,21\}\\ {h_G^{-}}^{(132)}((23)=f_G^{-}((12))=\{0,2,5,11,16,19,21\}\\ {h_G^{-}}^{(132)}((123))=f_G^{-}((123))=\{0,2,5,11,19\}\\ {h_G^{-}}^{(132)}((132))=f_G^{-}((132))=\{0,2,5,11,19\}.\end{array}\end{array}$

Theorem 3.23

Let $f_G$ be a BS-group in G. Then, $f_G$ is a NBS-subgroup if and only if $f_G$ is constant on conjugacy classes of G.

Proof.

Suppose that $f_G$ is NBS-subgroup over G. Then, $f_G^{+}(g_1^{-1}{g}_2{g}_1)=f_G^{+}(g_2{g}_1{g}_1^{-1})=f_G^{+}(g_2)$ and $f_G^{-}(g_1^{-1}{g}_2{g}_1)=f_G^{-}(g_2{g}_1{g}_1^{-1})=f_G^{-}(g_2)$ $\mathrm{\forall }{g}_1,g_2\in G$. Conversely, suppose that $f_G$ is constant on each conjugacy class of G. Then, $f_G^{+}(g_1{g}_2)=f_G^{+}(g_1{g}_2{g}_1{g}_1^{-1})=f_G^{+}(g_1(g_2{g}_1)g_1^{-1})=f_G^{+}(g_2{g}_1)$ and $f_G^{-}(g_1{g}_2)=f_G^{-}(g_1{g}_2{g}_1{g}_1^{-1})=f_G^{-}(g_1(g_2{g}_1)g_1^{-1})=f_G^{-}(g_2{g}_1)$ $\mathrm{\forall }{g}_1,g_2\in G$. Thus, $f_G$ is normal.

4. Bipolar Soft Left(right) Coset of Bipolar Soft Groups

In this section, we define the concept of bipolar soft left(right) coset for BS-groups, and obtain some properties of them.

Definition 4.1

Let $f_G\in {\mathcal{B}\mathcal{S}}_G(U)$ and $g\in G$. Bipolar soft set $(U_g,{\mathrm{\varnothing }}_g)\ast f_G$ is referred to as bipolar soft left coset of $f_G$ and denoted by $g{f}_G$. Also $g{f}_G$ means that $\begin{array}{r}g{f}_G(k)=f_G(g^{-1}k)\mathrm{for}\mathrm{all}k\in G.\end{array}$ Here, $g{f}_G$ means that $(U_g\ast f_G^{+},{\mathrm{\varnothing }}_g\ast f_G^{-})$.

Definition 4.2

Let $f_G\in {\mathcal{B}\mathcal{S}}_G(U)$ and $g\in G$. BS-set $f_G\ast (U_g,{\mathrm{\varnothing }}_g)$ is referred to as bipolar soft right coset of $f_G$ and denoted by $f_{G}g$. Also $f_{G}g$ means that $\begin{array}{r}{f}_{G}g(k)=f_G(k{g}^{-1})\mathrm{for}\mathrm{all}k\in G.\end{array}$ Here, $f_{G}g$ means that $(f_G^{+}\ast U_g,f_G^{-}\ast {\mathrm{\varnothing }}_g)$.

Example 4.3

Let us consider BS-group given in Example 3.22. Then, the bipolar soft left cosets of $f_G$ is as follows: $\begin{array}{rl}(1)f_G& =\{(1,\mathbb{Z},\mathrm{\varnothing }),((12),\{3,7,8,15\},\{0,2,5,11,16,19,21\}),((123),\{3,7,8,15,20\},\\ & \{0,2,5,11,19\}),((23),\{3,7,8,15\},\{0,2,5,11,16,19,21\}),((132),\{3,7,8,\\ & 15,20\},\{0,2,5,11,19\}),((13),\{3,7,8,15\},\{0,2,5,11,16,19,21\})\},\\ (12)f_G& =\{(1,\{3,7,8,15\},\{0,2,5,11,16,19,21\}),((12),\mathbb{Z},\mathrm{\varnothing }),((123),\{3,7,8,15\},\\ & \{0,2,5,11,16,19,21\}),((23),\{3,7,8,15,20\},\{0,2,5,11,19\}),((132),\{3,7,\\ & 8,15\},\{0,2,5,11,16,19,21\}),((13),\{3,7,8,15,20\},\{0,2,5,11,19\})\},\\ (13)f_G& =\{(1,\{3,7,8,15\},\{0,2,5,11,16,19,21\}),((12),\{3,7,8,15,20\},\{0,2,5,11,\\ & 19\}),((123),\{3,7,8,15\},\{0,2,5,11,16,19,21\}),((23),\{3,7,8,15,20\},\{0,2,\\ & 5,11,19\}),((132),\{3,7,8,15\},\{0,2,5,11,16,19,21\}),((13),\mathbb{Z},\mathrm{\varnothing })\},\\ (23)f_G& =\{(1,\{3,7,8,15\},\{0,2,5,11,16,19,21\}),((12),\{3,7,8,15,20\},\{0,2,5,11,19\}),\\ & ((123),\{3,7,8,15\},\{0,2,5,11,16,19,21\}),((23),\mathbb{Z},\mathrm{\varnothing }),((132),\{3,7,8,15\},\\ & \{0,2,5,11,19\}),((13),\{3,7,8,15,20\},\{0,2,5,11,19\})\},\\ (123)f_G& =\{(1,\{3,7,8,15,20\},\{0,2,5,11,19\}),((12),\{3,7,8,15\},\{0,2,5,11,16,19,21\}),\\ & ((123),\mathbb{Z},\mathrm{\varnothing }),((23),\{3,7,8,15\},\{0,2,5,11,16,19,21\}),((132),\{3,7,8,15,20\},\\ & \{0,2,5,11,19\}),((13),\{3,7,8,15\},\{0,2,5,11,16,19,21\})\},\\ (132)f_G& =\{(1,\{3,7,8,15,20\},\{0,2,5,11,19\}),((12),\{3,7,8,15\},\{0,2,5,11,16,19,21\}),\\ & ((123),\{3,7,8,15,20\},\{0,2,5,11,19\}),((23),\{3,7,8,15\},\\ & \{0,2,5,11,16,19,21\}),((132),\mathbb{Z},\mathrm{\varnothing }),((13),\{3,7,8,15\},\{0,2,5,11,16,19,21\})\}.\end{array}$ Bipolar soft right cosets of $f_G$ is as follows: $\begin{array}{rl}{f}_G(1)& =\{(1,\mathbb{Z},\mathrm{\varnothing }),((12),\{3,7,8,15\},\{0,2,5,11,16,19,21\}),((123),\{3,7,8,15,20\},\\ & \{0,2,5,11,19\}),((23),\{3,7,8,15\},\{0,2,5,11,16,19,21\}),((132),\{3,7,8,15,20\},\\ & \{0,2,5,11,19\}),((13),\{3,7,8,15\},\{0,2,5,11,16,19,21\})\},\\ f_G(12)& =\{(1,\{3,7,8,15\},\{0,2,5,11,16,19,21\}),((12),\mathbb{Z},\mathrm{\varnothing }),((123),\{3,7,8,15\},\\ & \{0,2,5,11,16,19,21\}),((23),\{3,7,8,15,20\},\{0,2,5,11,19\}),((132),\{3,7,8,15\},\\ & \{0,2,5,11,16,19,21\}),((13),\{3,7,8,15,20\},\{0,2,5,11,19\})\},\\ f_G(13)& =\{(1,\{3,7,8,15\},\{0,2,5,11,16,19,21\}),((12),\{3,7,8,15,20\},\{0,2,5,11,19\}),\\ & ((123),\{3,7,8,15\},\{0,2,5,11,16,19,21\}),((23),\{3,7,8,15,20\},\{0,2,5,11,19\}),\\ & ((132),\{3,7,8,15\},\{0,2,5,11,16,19,21\}),((13),\mathbb{Z},\mathrm{\varnothing })\},\\ f_G(23)& =\{(1,\{3,7,8,15\},\{0,2,5,11,16,19,21\}),((12),\{3,7,8,15,20\},\{0,2,5,11,19\}),\\ & ((123),\{3,7,8,15\},\{0,2,5,11,16,19,21\}),((23),\mathbb{Z},\mathrm{\varnothing }),((132),\{3,7,8,15\},\\ & \{0,2,5,11,16,19,21\}),((13),\{3,7,8,15,20\},\{0,2,5,11,19\})\},\\ f_G(123)& =\{(1,\{3,7,8,15,20\},\{0,2,5,11,19\}),((12),\{3,7,8,15\},\{0,2,5,11,16,19,21\}),\\ & ((123),\mathbb{Z},\mathrm{\varnothing }),((23),\{3,7,8,15\},\{0,2,5,11,16,19,21\}),((132),\{3,7,8,15,20\},\\ & \{0,2,5,11,19\}),((13),\{3,7,8,15\},\{0,2,5,11,16,19,21\})\},\\ f_G(132)& =\{(1,\{3,7,8,15,20\},\{0,2,5,11,19\}),((12),\{3,7,8,15\},\{0,2,5,11,16,19,21\}),\\ & ((123),\{3,7,8,15,20\},\{0,2,5,11,19\}),((23),\{3,7,8,15\},\{0,2,5,11,16,19,21\}),\\ & ((132),\mathbb{Z},\mathrm{\varnothing }),((13),\{3,7,8,15\},\{0,2,5,11,16,19,21\})\}.\end{array}$ Note that a left coset may not be a BS-group. Since in $f_G(132)$, $f_G^{+}((132)(13))=f_G^{+}((12))=\{3,7,8,17,18,25\}⊉\{3,7,8,15\}=f_G^{+}((132))\cap f^{+}((13))$, $f_G(132)$ is not a BS-group.

Theorem 4.4

Let $f_G$ be BS-group. Then, for all $g_1,g_2\in G,$

(1)	$g_1{f}_G=g_2{f}_G\iff g_1{e}_{f_G}=g_2{e}_{f_G};$
(2)	$f_G{g}_1=f_G{g}_2\iff e_{f_G}{g}_1=e_{f_G}{g}_2$.

Proof.

(1) Suppose that $g_1{f}_G=g_2{f}_G$. Then, $g_1{f}_G(k)=g_2{f}_G(k)$ which means that $f_G^{+}(g_1^{-1}k)=f_G^{+}(g_2^{-1}k)$ and $f_G^{-}(g_1^{-1}k)=f_G^{-}(g_2^{-1}k)$ for all $k\in G$. If it is chosen as $k=g_2$, then $f_G^{+}(g_1^{-1}{g}_2)=f_G^{+}(g_2^{-1}{g}_2)=f_G^{+}(e)$ and $f_G^{-}(g_1^{-1}{g}_2)=f_G^{-}(g_2^{-1}{g}_2)=f_G^{-}(e)$. Hence, $g_1^{-1}{g}_2\in e_{f_G}$ and so $g_1{e}_{f_G}=g_2{e}_{f_G}$.

Conversely, let $g_1{e}_{f_G}=g_2{e}_{f_G}$. Then, $g_1^{-1}{g}_2\in e_{f_G}$ and $g_2^{-1}{g}_1\in e_{f_G}$. It follows that $\begin{array}{rl}{f}_G^{+}(g_1^{-1}k)& =f_G^{+}(g_1^{-1}{g}_2{g}_2^{-1}k)\\ & \supseteq f_G^{+}(g_1^{-1}{g}_2)\cap f_G^{+}(g_2^{-1}k)\\ & =f_G^{+}(e)\cap f_G^{+}(g_2^{-1}k)\\ & =f_G^{+}(g_2^{-1}k)\end{array}$ and $\begin{array}{rl}{f}_G^{-}(g_1^{-1}k)& =f_G^{-}(g_1^{-1}{g}_2{g}_2^{-1}k)\\ & \subseteq f_G^{-}(g_1^{-1}{g}_2)\cup f_G^{-}(g_2^{-1}k)\\ & =f_G^{-}(e)\cup f_G^{-}(g_2^{-1}k)\\ & =f_G^{-}(g_2^{-1}k).\end{array}$ Then, (1) $\begin{array}{rl}{f}_G(g_1^{-1}k)& ⊑f_G(g_2^{-1}k)\\ f_G^{+}(g_2^{-1}k)& =f_G^{+}(g_2^{-1}{g}_1{g}_1^{-1}k)\\ & \supseteq f_G^{+}(g_2^{-1}{g}_1)\cap f_G^{+}(g_1^{-1}k)\\ & =f_G^{+}(e)\cap f_G^{+}(g_1^{-1}k)\\ & =f_G^{+}(g_1^{-1}k)\end{array}$(1) and $\begin{array}{rl}{f}_G^{-}(g_2^{-1}k)& =f_G^{-}(g_2^{-1}{g}_1{g}_1^{-1}k)\\ & \subseteq f_G^{-}(g_2^{-1}{g}_1)\cup f_G^{-}(g_1^{-1}k)\\ & =f_G^{-}(e)\cup f_G^{-}(g_1^{-1}k)\\ & =f_G^{-}(g_1^{-1}k).\end{array}$ Then, (2) $f_G(g_2^{-1}k)⊑f_G(g_1^{-1}k).$(2) From Equations (1(1) $\begin{array}{rl}{f}_G(g_1^{-1}k)& ⊑f_G(g_2^{-1}k)\\ f_G^{+}(g_2^{-1}k)& =f_G^{+}(g_2^{-1}{g}_1{g}_1^{-1}k)\\ & \supseteq f_G^{+}(g_2^{-1}{g}_1)\cap f_G^{+}(g_1^{-1}k)\\ & =f_G^{+}(e)\cap f_G^{+}(g_1^{-1}k)\\ & =f_G^{+}(g_1^{-1}k)\end{array}$(1) ) and (2(2) $f_G(g_2^{-1}k)⊑f_G(g_1^{-1}k).$(2) ), for all $k\in G$, $f_G(g_1^{-1}k)=f_G(g_2^{-1}k)$ which shows that $g_1{f}_G=g_2{f}_G$.

(2) The proof can be made by similar way to proof of item 1.

Theorem 4.5

Let $f_G\in {\mathcal{N}\mathcal{B}\mathcal{S}}_G(U)$ and $g_1,g_2\in G$. If $g_1{f}_G=g_2{f}_G,$ then $f_G(g_1)=f_G(g_2)$.

Proof.

Suppose that $g_1{f}_G=g_2{f}_G$. By Theorem 4.4, $g_1^{-1}{g}_2\in e_{f_G}$ and $g_2^{-1}{g}_1\in e_{f_G}$. Since $f_G$ is NBS-subgroup it follows that $f_G^{+}(g_1)=f_G^{+}(g_2^{-1}{g}_1{g}_2)\supseteq f_G^{+}(g_2^{-1}{g}_1)\cap f_G(g_2)=f_G(e)\cap f_G(g_2)=f_G(g_2)$. Also $f_G^{-}(g_1)=f_G^{-}(g_2^{-1}{g}_1{g}_2)\subseteq f_G^{-}(g_2^{-1}{g}_1)\cup f_G^{-}(g_2)=f_G(e)\cap f_G^{-}(g_2)=f_G^{-}(g_2)$. Thus, $f_G(g_1)⊑f_G(g_2)$. Similarly $f_G(g_2)⊑f_G(g_1)$. Therefore, $f_G(g_1)=f_G(g_2)$.

Theorem 4.6

Let $f_G$ be a NBS-subgroup and $g_1,g_2\in G$. Then, $\begin{array}{r}(g_1{f}_G)\ast (g_2{f}_G)=(g_1{g}_2)f_G,forall{g}_1,g_2\in G.\end{array}$

Proof.

For all $g_1,g_2\in G$, we have $\begin{array}{rl}(g_1{f}_G^{+})\ast (g_2{f}_G^{+})& =(U_{g_1}\ast f_G^{+})\ast (U_{g_2}\ast f_G^{+})\\ & =U_{g_1}\ast (f_G^{+}\ast U_{g_2})\ast f_G^{+}\\ & =U_{g_1}\ast (f_G^{+}\ast f_G^{+})\ast U_{g_2}\\ & =U_{g_1}\ast f_G^{+}\ast U_{g_2}\\ & =U_{g_1}\ast (f_G^{+}\ast U_{g_2})\\ & =U_{g_1}\ast (U_{g_2}\ast f_G^{+})\\ & =(U_{g_1}\ast U_{g_2})\ast f_G^{+}\\ & =(g_1{g}_2)f_G^{+},\end{array}$ and it can be shown that $(g_1{f}_G^{-})\ast (g_2{f}_G^{-})=(g_1{g}_2{f}_G^{-})$. Thus, $(g_1{f}_G)\ast (g_2{f}_G)=(g_1{g}_2)f_G$.

Theorem 4.7

Let $f_G$ be a NBS-subgroup and $G/f_G=\{g{f}_G:g\in G\}$. Then, $(G/f_G,\ast )$ is a group.

Proof.

By Theorem 4.6, $G/f_G$ is closed under operation $\ast$. Also $\ast$ satisfies the associative law. Since $\begin{array}{rl}& f_G^{+}\ast (g{f}_G^{+})=(e{f}_G^{+})\ast (g{f}_G^{+})=(eg)f_G^{+}=g{f}_G^{+},g\in G,\\ & f_G^{-}\ast (g{f}_G^{-})=(e{f}_G^{-})\ast (g{f}_G^{-})=(eg)f_G^{-}=g{f}_G^{-},g\in G.\end{array}$ $e{f}_G^{+}=f_G$ is unit element of $(G/f_G)$.

Since $\begin{array}{rl}& (g^{-1}{f}_G^{+})\ast (g{f}_G^{+})=(g{g}^{-1})f_G^{+}=e{f}_G^{+}=f_G^{+},g\in G,\\ & (g^{-1}{f}_G^{-})\ast (g{f}_G^{-})=(g{g}^{-1})f_G^{-}=e{f}_G^{-}=f_G^{-},g\in G,\end{array}$ inverse of $g{f}_G$ is $g^{-1}{f}_G\in f_G$. Thus, $(G/f_G,\ast )$ is a group.

Theorem 4.8

Let $f_G^{\ast }$ be a BS-set over $G/f_G$ defined by $f_G^{\ast }(g{f}_G)=f_G(g)$ $\mathrm{\forall }g\in G$, where $f_G\in NBP(G)$. Then, $f_G^{\ast }$ is NBS-subgroup over $G/f_G$.

Proof.

From Theorem 4.5, $g_1{f}_G=g_2{f}_G$ implies that $f_G(g_1)=f_G(g_2)$. For all $g{f}_G\in G/f_G$ $\begin{array}{rl}{f_G^{\ast }}^{+}((g{f}_G^{+}{)}^{-1})& ={f_G^{\ast }}^{+}(g^{-1}{f}_G^{+})=f_G^{+}(g^{-1})=f_G^{+}(g)={f_G^{\ast }}^{+}(g{f}_G^{+}),\\ {f_G^{\ast }}^{-}((g{f}_G^{-}{)}^{-1})& ={f_G^{\ast }}^{-}(g^{-1}{f}_G^{-})=f_G^{-}(g^{-1})=f_G^{-}(g)={f_G^{\ast }}^{-}(g{f}_G^{-}).\end{array}$ For all $g_1{f}_G,g_2{f}_G\in (G/f_G)$ $\begin{array}{rl}{f_G^{\ast }}^{+}((g_1{f}_G^{+})\ast (g_2{f}_G^{+}))& ={f_G^{\ast }}^{+}(g_1{g}_2{f}_G^{+})\\ & =f_G^{+}(g_1{g}_2)\\ & \supseteq f_G^{+}(g_1)\cap f_G^{+}(g_2)\\ & ={f_G^{\ast }}^{+}(g_1{f}_G^{+})\cap {f_G^{\ast }}^{+}(g_1{f}_G^{+}),\\ {f_G^{\ast }}^{-}((g_1{f}_G^{-})\ast (g_2{f}_G^{-}))& ={f_G^{\ast }}^{-}(g_1{g}_2{f}_G^{-})\\ & =f_G^{-}(g_1{g}_2)\\ & \subseteq f_G^{-}(g_1)\cup f_G^{-}(g_2)\\ & ={f_G^{\ast }}^{-}(g_1{f}_G^{-})\cup {f_G^{\ast }}^{-}(g_1{f}_G^{-}).\end{array}$ Then, $f_G^{\ast }$ is BS-group in $G/f_G$. Moreover, for all $g_1,g_2\in G$, since $\begin{array}{rl}{f_G^{\ast }}^{+}((g_1{f}_G^{+})\ast (g_2{f}_G^{+}))& ={f_G^{\ast }}^{+}(g_1{g}_2{f}_G^{+})\\ & =f_G^{+}(g_1{g}_2)\\ & =f_G^{+}(g_2{g}_1)\\ & ={f_G^{\ast }}^{+}(g_2{g}_1{f}_G^{+})\\ & ={f_G^{\ast }}^{+}((g_2{f}_G^{+})\ast (g_1{f}_G^{+})),\\ {f_G^{\ast }}^{-}((g_1{f}_G^{-})\ast (g_2{f}_G^{-}))& ={f_G^{\ast }}^{-}(g_1{g}_2{f}_G^{-})\\ & =f_G^{-}(g_1{g}_2)\\ & =f_G^{-}(g_2{g}_1)\\ & ={f_G^{\ast }}^{-}(g_2{g}_1{f}_G^{-})\\ & ={f_G^{\ast }}^{-}((g_2{f}_G^{-})\ast (g_1{f}_G^{-})).\end{array}$ We have that $f_G^{\ast }\in {\mathcal{N}\mathcal{B}\mathcal{S}}_G(U)$.

Definition 4.9

The group $G/f_G$ defined in Theorem 4.7 is called quotient group (or factor group) of G relative to the NBS-subgroup $f_G$.

Theorem 4.10

Let $f_G$ be a BS-group and let N be a normal subgroup of G. Then, BS-set $f_S$ is BS-group over G/N and is defined as follows: $\begin{array}{r}{f}_S(gN)=⨆\{f_G(k):k\in gN\}\mathrm{\forall }g\in G.\end{array}$

Proof.

Now, $\mathrm{\forall }gN\in G/N;$ $\begin{array}{rl}{f}_S^{+}((gN{)}^{-1})& =f_S^{+}(g^{-1}N)\\ & =\bigcup \{f_G^{+}(k):k\in g^{-1}N\}\\ & =\bigcup \{f_G^{+}(t^{-1}):t^{-1}\in g^{-1}N\}\\ & =\bigcup \{f_G^{+}(t):t\in gN\}\\ & =f_S^{+}(gN),\\ f_S^{-}((gN{)}^{-1})& =f_S^{-}(g^{-1}N)\\ & =\bigcap \{f_G^{-}(k):k\in g^{-1}N\}\\ & =\bigcap \{f_G^{-}(t^{-1}):t^{-1}\in g^{-1}N\}\\ & =\bigcap \{f_G^{-}(t):t\in gN\}\\ & =f_S^{-}(gN)\end{array}$ and $\mathrm{\forall }{g}_{1}N,g_{2}N\in G/N;$ $\begin{array}{rl}{f}_S^{+}(g_{1}N{g}_{2}N)& =\bigcup \{f_G^{+}(k):k\in g_1{g}_{2}N\}\\ & =\bigcup \{f_G^{+}(qr):q\in g_{1}N,r\in g_{2}N\}\\ & \supseteq \bigcup \{f_G^{+}(q)\cap f_G^{+}(r):q\in g_{1}N,r\in g_{2}N\}\\ & =(\bigcup \{f_G^{+}(q):q\in g_{1}N\})\cap (\bigcup \{f_G^{+}(r):r\in g_{2}N\})\\ & =f_S^{+}(g_{1}N)\cap f_S^{+}(g_{2}N),\\ f_S^{-}(g_{1}N{g}_{2}N)& =\bigcap \{f_G^{-}(k):k\in g_1{g}_{2}N\}\\ & =\bigcap \{f_G^{-}(qr):q\in g_{1}N,r\in g_{2}N\}\\ & \subseteq \bigcap \{f_G^{-}(q)\cup f_G^{-}(r):q\in g_{1}N,r\in g_{2}N\}\\ & =(\bigcap \{f_G^{-}(q):q\in g_{1}N\})\cup (\bigcap \{f_G^{-}(r):r\in g_{2}N\})\\ & =f_S^{-}(g_{1}N)\cup f_S^{-}(g_{2}N).\end{array}$

Since $f_S(g_{1}N{g}_{2}N)⊒f_S(g_{1}N)\sqcap f_S(g_{2}N)$ and $f_S((gN{)}^{-1})=f_S(gN)$, BS-set $f_S$ is a NBS-subgroup over G/N.

Definition 4.11

BS-subgroup $f_S$ defined in Theorem 4.10 is called quotient BS-subgroup of BS-group $f_G$ of G relative to the normal subgroup N of G.

Example 4.12

Let us consider BS-group $f_G$ given in Example 3.22, and consider normal subgroup $K=\{(1),(123),(132)\}$ of G. We define BS-subgroup $f_S$ over G/K by $\begin{array}{rl}{f}_S^{+}((1)K)=f_S^{+}((123)K)=f_S^{+}((132)K)& =f_G^{+}((1))\cup f_G^{+}((123))\cup f_G^{+}((132))\\ & =\mathbb{Z},\\ f_S^{-}((1)K)=f_S^{-}((123)K)=f_S^{-}((132)K)& =f_G^{-}((1))\cap f_G^{-}((123))\cap f_G^{-}((132))\\ & =\mathrm{\varnothing },\end{array}$ and $\begin{array}{rl}{f}_S^{+}((12)K)=f_S^{+}((13)K)=f_S^{+}((23)K)& =f_G^{+}((12))\cup f_G^{+}((13))\cup f_G^{+}((23))\\ & =\{3,7,8,15,15\},\\ f_S^{-}((12)K)=f_S^{-}((13)K)=f_S^{-}((23)K)& =f_G^{-}((12))\cap f_G^{-}((13))\cap f_G^{-}((23))\\ & =\{0,2,5,11,16,19,21\}.\end{array}$ BS-subgroup $f_S$ can be written as follows: $\begin{array}{rl}{f}_S& =\{((1),\mathbb{Z},\mathrm{\varnothing }),((123),\mathbb{Z},\mathrm{\varnothing }),((132),\mathbb{Z},\mathrm{\varnothing }),((12),\{3,7,8,15\},\{0,2,5,11,16,19,21\}),\\ & ((13),\{3,7,8,15\},\{0,2,5,11,16,19,21\}),((23),\{3,7,8,15\},\{0,2,5,11,16,19,21\})\}.\end{array}$

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Additional information

Notes on contributors

Faruk Karaaslan

Dr. Faruk Karaaslan graduated from the Department of Mathematics, Erciyes University, Kayseri, Turkey, in 2003 (BSc). He earned his MSc and PhD degrees from Gaziosmanpaşa University, Tokat, Turkey, in 2007 and 2013. Dr Karaaslan is currently an Associate Professor at Department of Mathematics, Faculty of Science, Çankırı Karatekin University. He has contributed more than 55 research papers in reputable journals and 4 book section in reputable publishers. His research interests include mathematical theories for modelling uncertainty, fuzzy sets, soft sets, rough sets, intuitionistic fuzzy sets, neutrosophic sets and bipolar fuzzy sets. He also studies on decision making, algebraic structures, and graphs structures of the mentioned set theories.

Aman Ullah

Dr. Aman Ullah graduated from the Department of Mathematics, University of Peshawar, Pakistan in 2001. He earned his MPhil and PhD degrees from University of Malakand, Pakistan in 2008 and 2016. Currently, Dr. Aman Ullah is serving as an Assistant Professor at Department of Mathematics, University of Malakand, Pakistan. He has contributed more than 40 research papers in reputable journals and 1 book section with a reputable publisher. His research interests include fuzzy algebra, noncommutative algebra, soft set theory, fractional differential equations, mathematical modelling and fuzzy fractional and integers order differential equation.

Imtiaz Ahmad

Dr. Imtiaz Ahmad graduated from Pakistani institutions. He earned PhD degrees from the Department of Mathematical Sciences, University of Essex, United Kingdom in 2010. Currently, Dr. Ahmad is serving as an Associate Professor at Department of Mathematics, University of Malakand, Pakistan, previously he served this department as Lecturer and Assistant Professor since February 2003. He published more than 50 research articles in various national and international research journals, writing general articles for newspapers, supervised several MPhil and PhDs, have completed different research projects, and authored several textbooks.