This paper introduces the concept of homomorphism in intuitionistic fuzzy multigroups context. It also investigates Some homomorphic properties of intuitionistic fuzzy multigroups. It is shown that the homomorphic image and homomorphic preimage of intuitionistic fuzzy multigroups are also intuitionistic fuzzy multigroups. Finally, it presents some homomorphic properties of normalizer of intuitionistic fuzzy multigroups.
In modern mathematics, a set is a well-defined collection of distinct objects. Set theory was introduced by German mathematician George Ferdinand Ludwig Cantor(1845-1918). In classical sense, all mathematical notions including sets must be exact. However, if repeated occurrences of any object are allowed in a set, then the mathematical structure is called multiset [1]. Thus, a multiset differs from a set in the sense that each element has a multiplicity. An account of the development of multiset theory can be seen in [2,3,4,5]. Most of the real life situations are complex and modelling them we need a simplification of the complex system. The simplification must be in such a way that the information lost should be minimum. One way to do this is to allow some degree of uncertainty into it. To handle situations like this, Zadeh [6] proposed fuzzy sets. A fuzzy set has a membership function assigns to each element of the universe of discourse, a number from the unit interval [0,1] to indicate the degree of belongingness to the set under consideration. Fuzzy sets were introduced with a view to reconcile mathematical modelling and human knowledge in the engineering sciences. The theory of fuzzy sets has been applied to group theoretic notions [7,8,9,10].
Atanassov [11,12] introduced a generalized fuzzy sets called intuitionistic fuzzy set. In the same time, a theory called intuitionistic fuzzy set theory was independently introduced by Takeuti and Titani [13] as a theory developed in (a kind of) intuitionistic logic. Intuitionistic Fuzzy sets provide a flexible framework to explain uncertainty and vagueness. The theory of intuitionistic fuzzy sets has been applied to group theoretic notions [14,15,16]. As a generalization of multiset, Yager [17] introduced fuzzy multisets and suggested possible applications to relational databases. Shinoj et al., [18] has studied the structure of groups in fuzzy multisets. Several researches on fuzzy multigroup theory have been conducted as seen in [19,20,21,22,23,24,25]. The concept of intuitionistic fuzzy multiset was proposed in [26] as a study of intuiionistic fuzzy sets in multiset framework. Some works have been done on both the theory and applications of intuitionistic fuzzy multisets [27,28,29,30,31,32,33,34]. In a way to apply intuitionistic fuzzy multisets to group theory, Shinoj and John [35] proposed intuitionistic fuzzy multigroups. Adamu et al., [36] developed the concept of normal sub-intuitionistic fuzzy multigroups and investigate some of its related algebraic structures.
The motivation of this work is to establish the idea of homomorphism in intuitionistic fuzzy multigroups. This paper introduces the concept of homomorphism in intuitionistic fuzzy multigroups context and investigated some of its properties. The outline are presented as follows: Section 2 presents some foundational concepts relevant to the study whereas the main results are reported in Section 3. Section 4 summarises and concludes the paper.
Definition 1. [26] Let \(X\) be a nonempty set. An intuitionistic fuzzy multiset \(A\) of \(X\) is characterized by two count membership function \(CM_A\) and count non membership function \(CN_A\) defined by \[CM_A\colon X\to Q \, \textrm{and}\, CN_A\colon X\to Q,\] where \(Q\) is the set of all crisp multisets drawn from the unit interval \([0,1]\) such that for each \(x\in X\), the membership sequence is defined as a decreasingly ordered sequence of elements in \(A\) which is denoted by \(\mu^1_A (x),\mu^2_A (x),…,\mu^p_A (x)\), where \(\mu^1_A (x)\geq\mu^2_A (x)\geq,…,\geq\mu^p_A (x)\) and the corresponding non membership sequence of elements in \(A\) is denoted by \((\nu^1_A(x),\nu^2_A(x),…,\nu^p_A(x))\) such that \(0\leq\mu^i_A (x)+nu^i_A(x)\leq 1 \) for every \(\in X\) and \(i=1,2,…,p\).
An IFMS \(A\) is denoted by
\[A=\lbrace (\mu^1_A (x),\mu^2_A (x),…,\mu^p_A (x)),(\nu^1_A(x),\nu^2_A(x),…,\nu^p_A(x))>: x\in X\rbrace.\]Definition 2. [26] Length of an element \(x\) in an IFMS \(A\) is defined as the cardinality of \(CM_A(x)\) or \(CN_A(x)\) for which \(0\leq\mu^i_A (x)+nu^i_A(x)\leq1\) and is denoted by \(L(x:A)\). That is \(L(x:A)=|CM_A(x)|=|CN_A(x)|\). If \(A\) and \(B\) are IFMSs drawn from \(X\), then \(L(x:A,B)=\max[L(x:A),L(x:B)]\). Alternatively we use \(L(x)\) for \(L(x:A,B)\).
Definition 3. [26] For any two IFMSs \(A\) and \(B\) of a set \(X\), the following operations and relations hold.
Definition 4. [28] Let \(X\) and \(Y\) be two non-empty sets and \(f\colon X\to Y\) be a mapping. Then
Definition 5. [35] Let \(X\) be a group. An intuitionistic fuzzy multiset \(G\) of \(X\) is an intuitionistic fuzzy multigroup (IFMG) of \(X\) if the counts (count membership and non-membership) of \(G\) satisfies the following two conditions:
Definition 6. [35] For any intuitionistic fuzzy multigroup \(A \in IFMG(X), \exists\) its inverse, \(A^{-1},\) defined by \[CM_{A^{-1}}(x) = CM_A(x^{-1}) \forall x \in X \, \textrm{and}\, CN_{A^{-1}}(x) = CN_A(x^{-1}) \forall x \in X.\] Certainly, \(A \in IFMG(X)\) if and only if \(A^{-1} \in IFMG(X)\).
Definition 7. Let \(X\), \(Y\) be two groups and let \(f\colon X\to Y\) be an isomorphism of groups. Suppose \(A\) and \(B\) are intuitionistic fuzzy multigroups of \(X\) and \(Y\), respectively. Then, \(f\) induces a homomorphism from \(A\) to \(B\) which satisfies
Definition 8. Let \(X\) and \(Y\) be groups and let \(A\in IFMG(X)\) and \(B \in IFMG(Y )\), respectively.
Remark 1. Let \(X\) and \(Y\) be groups and let \(A\in IFMG(X)\) and \(B \in IFMG(Y )\), respectively. Then
Definition 9. Let \(A\) be a intuitionistic fuzzy submultigroup of \(B\in IFMG(X)\). Then, the normalizer of \(A\) in \(B\) is given by \[N(A) =\{g \in X \mid CM_A(gy) = CM_A(yg), CN_A(gy) = CN_A(yg)\ \forall y \in X \}.\]
Proposition 1. Let \(f\colon X\to Y\) be a homomorphism. For \(A,B\in IFMG(X)\), if \(A\subseteq B\), then \(f(A)\subseteq f(B)\).
Proof. Let \(A,B \in IFMG(X)\) and \(f: X\rightarrow Y\). Suppose \(CM_{A}(x) \leq CM_B(x)\) and \(CN_{A}(x) \leq CN_B(x)\ \forall \;x\in X.\) Then it follows that \[CM_{f(A)}(y) = CM_A(f^{-1}(y)) \leq CM_B(f^{-1}(y))=CM_{f(B)}(y),\] and \[CN_{f(A)}(y) = CN_A(f^{-1}(y)) \leq CN_B(f^{-1}(y))=CN_{f(B)}(y)\ \forall\; y\in Y.\] Hence \(f(A)\subseteq f(B)\).
Proposition 2. Let \(X\), \(Y\) be two groups and \(f\) be a homomorphism of \(X\) into \(Y\) for \(A, B\ IFMG(Y),\) if \(A\subseteq B,\) then \(f^{-1}(A)\subseteq f^{-1}(B)\).
Proof. Given that \(A,B \in IFMG(X)\) and \(f: X\rightarrow Y.\) Suppose \(CM_{A}(y) \leq CM_B(y)\) and \(CN_{A}(y) \leq CN_B(y)\ \forall \ y\in Y\). Then we have \[CM_{f^{-1}(A)}(x)=CM_A(f(x))\leq CM_B(f(x))=CM_{f^{-1}(B)}(x),\] Similarly, \[CN_{f^{-1}(A)}(x)=CN_A(f(x))\leq CN_B(f(x))=CN_{f^{-1}(B)}(x) \forall x\in X.\]
Definition 10. Let \(f\) be a homomorphism of a group \(X\) into a group \(Y\) , and \(A \in IFMG(X)\). If for all \(x,y \in X\), \(f(x) = f(y)\) implies \(CM_A(x) = CM_A(y)\) and \(CN_A(x) = CN_A(y)\) then, \(A\) is \(f-\)invariant.
Lemma 1. Let \(f:X\rightarrow Y\) be groups homomorphism and \(A \in IFMG(X)\). If \( \forall x \in X, f(x) = f(y)\), then, \(A\) is \(f-\)invariant.
Proof. Suppose \(f(x) = f(y) \forall x \in X\). Then, \[CM_{f(A)}(f(x)) = CM_{f(A)}(f(y)) \ \text{and}\ CN_{f(A)}(f(x)) = CN_{f(A)}(f(y)).\] This implies \(CM_A(x) = CM_A(y)\) and \(CN_A(x) = CN_A(y)\). Hence, \(A\) is \(f-\)invariant.
Lemma 2. If \(f:X\rightarrow Y\) is a homomorphism and \(A \in IFMG(X)\). Then
Proof.
Proposition 3. Let \(X\) and \(Y\) be groups such that \(f:X\rightarrow Y\) is an isomorphic mapping. If \(A\in IFMG(X)\) and \( B\in IFMG(Y)\). Then
Proof. Recall that if \(f\) is an isomorphism, then \(f(x)=y\) \(\forall y \in Y\), consequently, \(f(A)=B\).
Proposition 4. Let \(f:X\rightarrow Y\) be a homomorphism of groups. If \(\{A_i\}_{i\in I} \in IFMG(X) \text{and} \{B_i\}_{i\in I} \in IFMG(Y) \) respectively. Then
Proof.
Proposition 5. Let \(X\) be group and \(f\colon X\to Y\) be an automorphism. If \(A \in IFMG(X)\), then \(f(A) = A\) if and only if \(f^{-1}(A) = A\). Consequently, \( f(A) = f^{-1}(A)\).
Proof. Let \(f(x) = x\; \forall x \in X\) since \(f\) is an automorphism. Suppose \(f(A) = A,\) we have \begin{eqnarray*} CM_{f(A)}(x) & = & CM_A(f^{-1}(x)) = CM_A(x)\\ & = & CM_A(f^{-1}(x)) = CM_{f(A)}(x). \end{eqnarray*} Similarly,\begin{eqnarray*} CN_{f(A)}(x) & = & CN_A(f^{-1}(x)) = CN_A(x)\\ & = & CN_A(f^{-1}(x)) = CN_{f(A)}(x). \end{eqnarray*} Thus \(f^{-1}(A) = A\).
Conversely, let \(f^{-1}(A) = A,\) we have
\begin{eqnarray*} CM_{f^{-1}(A)}(x) & = & CM_A(f(x)) = CM_A(x)\\ & = & CM_A(f^{-1}(x)) = CM_{f(A)}(x). \end{eqnarray*} Similarly, \begin{eqnarray*} CN_{f^{-1}(A)}(x) & = & CN_A(f(x)) = CN_A(x)\\ & = & CN_A(f^{-1}(x)) = CN_{f(A)}(x). \end{eqnarray*} Thus \(f(A) = A\). Hence \( f(A) = A \Leftrightarrow f^{-1}(A) = A\).Proposition 6. Let \(f\colon X\to Y\) be a homomorphism. If \(A \in IFMG(X)\), then \(f^{-1}(f(A)) = A\) whenever \(f\) is injective.
Proof. Suppose \(f\) is injective, then \(f(x) = y\; \forall x\in X\) and \(\forall y \in Y\). Now \begin{eqnarray*} CM_{f^{-1}(f(A))}(x) & = & CM_{f(A)}(f(x)) = CM_{f(A)}(y)\\ & = & CM_A(f^{-1}(y)) = CM_A(x). \end{eqnarray*} Also, \begin{eqnarray*} CN_{f^{-1}(f(A))}(x) & = & CN_{f(A)}(f(x)) = CN_{f(A)}(y)\\ & = & CN_A(f^{-1}(y)) = CN_A(x). \end{eqnarray*} Hence, \(f^{-1}(f(A)) = A.\)
Corollary 1. Let \(f\colon X\to Y\) be a homomorphism. If \(B \in IFMG(Y )\), then \(f(f^{-1}(B)) = B\) whenever \(f\) is surjective.
Proof. Similar to Proposition 6.
Proposition 7. Let \(X\), \(Y\) and \(Z\) be groups and \(f\colon X\to Y\) and \(f\colon Y\to Z\) be homomorphisms. If \(\lbrace A_i\rbrace_{i\in I} \in IFMG(X)\) and \(\lbrace B_i\rbrace_{i\in I} \in IFMG(Y)\) and \(i \in I\). Then
Proof.
Proposition 8. Let \(X\) and \(Y\) be groups and \(f\colon X\to Y\) be an isomorphism. Then, \(A \in IFMG(X)\) if and only if \(f(A)\in IFMG(Y)\).
Proof. Suppose \(A \in IFMG(X).\) Let \(x, y \in Y,\) then, \( \exists f(a) = x\) and \(f(b) = y\) since \(f\) is an isomorphism \(\forall a, b \in X\). We know that \[CM_B(x) = CM_A(f^{-1}(x)) = \bigvee_{a \in f^{-1}(x)}CM_A(a)\] and \[CN_B(x) = CN_A(f^{-1}(x)) = \bigwedge_{a \in f^{-1}(x)}CN_A(a).\] Also, \[CM_B(y) = CM_A(f^{-1}(y))= \bigvee_{a \in f^{-1}(y)}CM_A(b),\] and \[CN_B(y) = CN_A(f^{-1}(y))= \bigwedge_{a \in f^{-1}(y)}CN_A(b).\] Clearly, \(a \in f^{-1}(x)\neq \emptyset\) and \(b \in f^{-1}(y)\neq \emptyset.\) For \(a \in f^{-1}(x)\) and \(b \in f^{-1}(y)\) \(\implies x = f(a)\) and \(y = f(b)\). Thus, \[f(ab^{-1}) = f(a)f(b^{-1}) = f(a)(f(b))^{-1} = xy^{-1}.\] Let \(c = ab^{-1}\, \implies c \in f^{-1}(xy^{-1})\).
Now,
\begin{eqnarray*} CM_B(xy^{-1}) & = & \bigvee_{c \in f^{-1}(xy^{-1)}} CM_A(c)= CM_A(ab^{-1})\\ & \geq & CM_A(a) \wedge CM_A(b)= CM_{f^{-1}(B)}(a) \wedge CM_{f^{-1}(B)}(b)\\ & = & CM_B(f(a)) \wedge CM_B(f(b))\\ & = & CM_B(x) \wedge CM_B(y)\; \forall x, y \in Y. \end{eqnarray*} Similarly, \begin{eqnarray*} CN_B(xy^{-1}) & = & \bigwedge_{c \in f^{-1}(xy^{-1)}} CN_A(c)= CN_A(ab^{-1})\\ & \leq & CN_A(a) \vee CN_A(b)= CN_{f^{-1}(B)}(a) \vee CN_{f^{-1}(B)}(b)\\ & = & CN_B(f(a)) \vee CN_B(f(b))\\ & = & CN_B(x) \vee CN_B(y)\; \forall x, y \in Y. \end{eqnarray*} Hence, \(f(A) \in IFMG(Y)\).Conversely, let \(a, b \in X\) and suppose \(f(A) \in IFMG(Y)\). Then,
\begin{eqnarray*} CM_A(ab^{-1}) &= & CM_{f^{-1}(B)}(ab^{-1})= CM_B(f(ab^{-1}))\\ & = & CM_B(f(a)f(b^{-1}))= CM_B(f(a)(f(b))^{-1})\\ & \geq & CM_B(f(a)) \wedge CM_B(f(b))\\ & = & CM_{f^{-1}(B)}(a) \wedge CM_{f^{-1}(B)}(b)\\ & = & CM_A(a) \wedge CM_A(b). \end{eqnarray*} Similarly, \begin{eqnarray*} CN_A(ab^{-1}) &= & CN_{f^{-1}(B)}(ab^{-1})= CN_B(f(ab^{-1}))\\ & = & CN_B(f(a)f(b^{-1}))= CN_B(f(a)(f(b))^{-1})\\ & \leq & CN_B(f(a)) \vee CN_B(f(b))\\ & = & CN_{f^{-1}(B)}(a) \vee CN_{f^{-1}(B)}(b)\\ & = & CN_A(a) \vee CN_A(b). \end{eqnarray*} Hence, \(A \in IFMG(X)\).Proposition 9. Let \(X\) and \(Y\) be groups and \(f\colon X\to Y\) be an isomorphism. Then, \(B \in IFMG(X)\) if and only if \(f^{-1}(B) \in IFMG(Y)\).
Proof. Suppose \(B \in IFMG(Y)\). Since \(f^{-1}(B)\) is an inverse image of \(B\), then we get \begin{eqnarray*} CM_{f^{-1}(B)}(ab^{-1}) & = & CM_B(f(ab^{-1}))= CM_B(f(a)f(b^{-1}))\\ & = & CM_B(f(a)(f(b))^{-1})\geq CM_B(f(a)) \wedge CM_B(f(b))\\ & = & CM_{f^{-1}(B)}(a) \wedge CM_{f^{_1}(B)}(b), \end{eqnarray*} and \begin{eqnarray*} CN_{f^{-1}(B)}(ab^{-1}) & = & CN_B(f(ab^{-1}))= CN_B(f(a)f(b^{-1}))\\ & = & CN_B(f(a)(f(b))^{-1})\leq CM_B(f(a)) \vee CN_B(f(b))\\ & = & CN_{f^{-1}(B)}(a) \vee CN_{f^{_1}(B)}(b)\; a,b\in X. \end{eqnarray*} Hence \(f^{-1}(B) \in IFMG(X)\).
Conversely, suppose \(f^{-1}(B) \in IFMG(X).\) We get
\begin{eqnarray*} CM_B(xy^{-1}) & = & CM_{f(A)}(xy^{-1})= CM_A(f^{-1}(xy^{-1}))\\ & = & CM_A(f^{-1}(x)f^{-1}(y^{-1}))= CM_A(f^{-1}(x)(f^{-1}(y))^{-1})\\ & \geq & CM_A(f^{-1}(x)) \wedge CM_A(f^{-1}(y))= CM_{f(A)}(x) \wedge CM_{f(A)}(y)\\ & = & CM_B(x) \wedge CM_B(y), \end{eqnarray*} and \begin{eqnarray*} CN_B(xy^{-1}) & = & CN_{f(A)}(xy^{-1})= CN_A(f^{-1}(xy^{-1}))\\ & = & CN_A(f^{-1}(x)f^{-1}(y^{-1}))= CN_A(f^{-1}(x)(f^{-1}(y))^{-1})\\ & \leq & CN_A(f^{-1}(x)) \vee CN_A(f^{-1}(y))= CN_{f(A)}(x) \vee CN_{f(A)}(y)\\ & = & CN_B(x) \vee CN_B(y) \; \forall x,y\in Y. \end{eqnarray*} Hence, \(B \in IFMG(Y)\).Corollary 2. Let \(X\) and \(Y\) be groups and \(f:X\rightarrow Y\) be an isomorphism. Then, the following statements hold;
Proof. Straightforward from Propositions 8 and 9.
Corollary 3. Let \(X\) and \(Y\) be groups and \(f:X\rightarrow Y\) be an isomorphism. If \(\bigcap_{i\in I} A_i\in IFMG(X)\) and \(\bigcap_{i\in I} B_i\in IFMG(Y)\). Then,
Proof. Straightforward from Propositions 8 and 9.
Corollary 4. Let \(X\) and \(Y\) be groups and \(f:X\rightarrow Y\) be an isomorphism. If \(\bigcup_{i\in I} A_i\in IFMG(X)\) and \(\bigcup_{i\in I} B_i\in IFMG(Y)\). Then,
Proof. Straightforward from Propositions 3.
Proposition 10. Let \(f\) be a homomorphism of an abelian group \(X\) onto an abelian group \(Y\). Let \(A\) and \(B\) be intuitionistic fuzzy multigroups of \(X\) such that \(A \subseteq B\). Then, \(f(N(A))\subseteq N(f(A))\).
Proof. Let \(\in f(N(A))\). Then, \(\exists u \in N(A)\) such that \(f(u) = x\). For all \(y, z \in Y\), we have \begin{eqnarray*} CM_{f(A)}(xyx^{-1}) & = & CM_{A(f^{-1}}(xyx^{-1}))\\ & = & CM_A(f^{-1}(x)f^{-1}(y)f^{-1}(x^{-1}))\\ & = & CM_A(f^{-1}(x)f^{-1}(y)f^{-1}(x)^{-1})\\ & = & CM_A(f^{-1}(x)f^{-1}(y)(f^{-1}(x))^{-1})\\ & = & CM_A(f^{-1}(f(u))f^{-1}(f(v))(f^{-1}(f(u)))^{-1})\\ & = & CM_A(uvu^{-1})= CM_A(vuu^{-1})= CM_A(v)\\ & = & CM_A(f^{-1}(y)) = CM_{f(A)}(y), \end{eqnarray*} and similarly,\begin{eqnarray*} CN_{f(A)}(xyx^{-1}) & = & CN_{A(f^{-1}}(xyx^{-1}))\\ & = & CN_A(f^{-1}(x)f^{-1}(y)f^{-1}(x^{-1}))\\ & = & CN_A(f^{-1}(x)f^{-1}(y)f^{-1}(x)^{-1})\\ & = & CN_A(f^{-1}(x)f^{-1}(y)(f^{-1}(x))^{-1})\\ & = & CN_A(f^{-1}(f(u))f^{-1}(f(v))(f^{-1}(f(u)))^{-1})\\ & = & CN_A(uvu^{-1})=CN_A(vuu^{-1})= CN_A(v) \\ & = & CN_A(f^{-1}(y)) = CN_{f(A)}(y), \end{eqnarray*} where \(v \in X\) such that \(f(v) = y\). Thus, \(x \in N(f(A))\). Hence \(f(N(A))\subseteq N(f(A))\).
Proposition 11. Let \(f\colon X\to Y\) be a homomorphism of abelian groups \(X\) and \(Y\) . Let \(A\) and \(B\) be intuitionistic fuzzy multigroups of \(Y\) such that \(B \subseteq A.\) Then, \(f^{-1}(N(B)) = N(f^{-1}(B))\).
Proof. Let \(x \in f^{-1}(N(B))\). Then for all \(y \in X\), \begin{eqnarray*} CM_{f^{-1}(B)}(xyx^{-1}) & = & CM_B(f(xyx^{-1}))= CM_B(f(x)f(y)f(x^{-1}))\\ & = & CM_B(f(x)f(y)(f(x))^{-1})= CM_B(f(y)f(x)(f(x))^{-1})\\ & = & CM_B(f(y)) = CM_{f^{-1}(B)}(y). \end{eqnarray*} Similarly,\begin{eqnarray*} CN_{f^{-1}(B)}(xyx^{-1}) & = & CN_B(f(xyx^{-1}))= CN_B(f(x)f(y)f(x^{-1}))\\ & = & CN_B(f(x)f(y)(f(x))^{-1})= CN_B(f(y)f(x)(f(x))^{-1})\\ & = & CN_B(f(y)) = CN_{f^{-1}(B)}(y). \end{eqnarray*} Thus \(x \in N(f^{-1}(B))\). So \(f^{-1}(N(B)) \subseteq N(f^{-1}(B))\). Again, let \(x \in N(f^{-1}(B))\) and \(f(x) = u\). Then for all \(v \in Y\), \begin{eqnarray*} CM_B(uvu^{-1}) & = & CM_B(f(x)f(y)(f(x))^{-1})= CM_B(f(y)f(x)(f(x))^{-1})\\ & = & CM_B(f(y)) = C_B(v), \end{eqnarray*} and \begin{eqnarray*} CN_B(uvu^{-1}) & = & CN_B(f(x)f(y)(f(x))^{-1})= CN_B(f(y)f(x)(f(x))^{-1})\\ & = & CN_B(f(y)) = CN_B(v), \end{eqnarray*} where \(y \in X\) such that \(f(y) = v\). Clearly, \(u\in N(B)\), that is, \(x \in f^{-1}(N(B))\). Thus \(N(f^{-1}(B)) \subseteq f^{-1}(N(B))\). Hence \(f^{-1}(N(B)) = N(f^{-1}(B))\).
Proposition 12. Let \(f:X\rightarrow Y\) be an isomorphism and let \(A\) be a normal sub-intuitionistic fuzzy multigroup of \(B\in IFMG(X)\). Then, \(f(A)\) is a normal sub-intuitionistic fuzzy multigroup of \(f(B) \in IFMG(Y)\).
Proof. By Proposition 8, \(f(A), f(B)\in IFMG(Y)\) and so, \(f(A) \subseteq f(B)\). We show that \(f(A)\) is a normal sub-intuitionistic fuzzy multigroup of \(f(B)\). Let \(x, y \in Y\). Since \(f\) is an isomorphism, then for some \(a \in X\) we have \(f(a) = x\). Thus, \begin{eqnarray*} CM_{f(A)}(xyx^{-1}) & = & CM_A(b)\ \text{for}\ f(b) = xyx^{-1}, \forall b \in X\\ & = & CM_A(a^{-1}ba)\ \text{for} \ f(a^{-1}ba) = y\\ & \geq & CM_A(b)\ \text{for}\ f(b) = y, \forall a^{-1}ba \in X\\ & = & CM_A(f^{-1}(y))\ \text{for}\ f(b) = y\\ & = & CM_{f(A)}(y). \end{eqnarray*} Similarly, \begin{eqnarray*} CN_{f(A)}(xyx^{-1}) & = & CN_A(b)\ \text{for}\ f(b) = xyx^{-1}, \forall b \in X\\ & = & CN_A(a^{-1}ba)\ \text{for} \ f(a^{-1}ba) = y\\ & \leq & CN_A(b)\ \text{for}\ f(b) = y, \forall a^{-1}ba \in X\\ & = & CN_A(f^{-1}(y))\ \text{for}\ f(b) = y\\ & = & CN_{f(A)}(y). \end{eqnarray*} Hence, \(f(A)\) is a normal sub-intuitionistic fuzzy multigroup of \(f(B)\).
Proposition 13. Let \(Y\) be a group and \(A \in IFMG(Y)\). If \(f\) is an isomorphism of \(X\) onto \(Y\) and \(B\) is a normal sub-intuitionistic fuzzy multigroup of \(A\), then \(f^{-1}(B)\) is a normal sub-intuitionistic fuzzy multigroup of \(f^{-1}(A)\).
Proof. By Proposition 9, \(f^{-1}(A), f^{-1}(B) \in IFMG(X)\). Since \(B\) is an intuitionistic fuzzy submultigroup of \(A\), so \(f^{-1}(B)\subseteq f^{-1}(A)\). Let \(a, b \in X\), then we have \begin{eqnarray*} CM_{f^{-1}(B)}(aba^{-1})& = & CM_B(f(aba^{-1})) = CM_B(f(a)f(b)(f(a))^{-1})\\ & = & CM_B(f(a)(f(a))^{-1}f(b))\geq CM_B(e) \wedge CM_B(f(b))\\ & = & CM_{f^{-1}(B)}(b), \end{eqnarray*} \(\implies CM_{f^{-1}(B)}(aba^{-1}) \geq CM_{f^{-1}(B)}(b)\).
Similarly,
\begin{eqnarray*} CN_{f^{-1}(B)}(aba^{-1})& = & CN_B(f(aba^{-1})) = CN_B(f(a)f(b)(f(a))^{-1})\\ & = & CN_B(f(a)(f(a))^{-1}f(b))\leq CM_B(e) \vee CN_B(f(b))\\ & = & CN_{f^{-1}(B)}(b), \end{eqnarray*} \(\implies CN_{f^{-1}(B)}(aba^{-1}) \leq CN_{f^{-1}(B)}(b)\). This completes the proof.