Direct product of fuzzy multigroups under \(t\)-norms

1. Introduction

Theory of multisets is an important generalization of classical set theory which has emerged by violating a basic property of classical sets that an element can belong to a set only once. The term multiset (mset in short) as Knuth [1] notes, was first suggested by De Bruijn [2] in a private communication to him. The concept of fuzzy sets proposed by Zadeh [3] is a mathematical tool for representing vague concepts. The idea of fuzzy multisets was conceived by Yager [4] as the generalization of fuzzy sets in multisets framework.

The concept of fuzzy multigroups was introduced by Shinoj \emph{et al.,} [5] as an application of fuzzy multisets to group theory, and some properties of fuzzy multigroups were presented. Ejegwa introduced the concept of fuzzy multigroupoids and presented the idea of fuzzy submultigroups with a number of results and more properties of abelian fuzzy multigroups were explicated [6, 7, 8]. Also Ejegwa introduced direct product in fuzzy multigroup setting as an extension of direct product of fuzzy subgroups [9]. In mathematics, a t-norm (also \( T \)-norm or, unabbreviated, triangular norm) is a kind of binary operation used in the framework of probabilistic metric spaces and in multi-valued logic, specifically in fuzzy logic. A \( t\)-norm generalizes intersection in a lattice and conjunction in logic. The name triangular norm refers to the fact that in the framework of probabilistic metric spaces \( t\)-norms are used to generalize triangle inequality of ordinary metric spaces. The author by using norms, investigated some properties of fuzzy algebraic structures [10, 11, 12, 13, 14, 15]. The author [15] defined fuzzy multigroups under t-norms and some properties of them are explored and some related results are obtained.

In this paper, we introduce the concept of direct product of fuzzy multigroups under \(t\)-norms \((TFSM(G))\) and investigate some properties and results about them. We prove that direct products of \(TFSM(G)\) are also \(TFSM(G).\) Next we investigate and obtain some new results of strong upper alpha-cut, weak upper alpha-cut, strong lower alpha-cut and weak lower alpha-cut of direct product of fuzzy Multigroups under \(t\)-norms. Later we prove that if \( A,C \in TFMS(G)\) and \( B,D \in TFMS(H)\) such that \(A\) is conjugate to \(B\) and \(C\) is conjugate to \( D,\) then \(A\times C \) is conjugate to \( B \times D.\) Also \( A\) and \( B\) are commutative if and only if \( A \times B \) is a commutative. Finally, we define group homomorphisms on direct propduct of fuzzy multigroups under t-norms and we prove that image and pre image of direct propduct of fuzzy multigroups under t-norms is also fuzzy multigroups under t-norms.

2. Preliminaries

This section contains some basic definitions and preliminary results which will be needed in the sequel. For details we refer to [15, 16, 17, 18, 19, 20, 21, 22, 23, 24].

Definition 1. Let \(G\) be an arbitrary group with a multiplicative binary operation and identity \(e\). A fuzzy subset of \(G\), we mean a function from \(G\) into \( [0,1]\). The set of all fuzzy subsets of \(G\) is called the \( [0,1]\)-power set of \(G\) and is denoted \( [0,1]^G.\)

Definition 2. Let \( X\) be a set. A fuzzy multiset \( A\) of \( X\) is characterized by a count membership function \[CM_{A} : X \to [0, 1]\] of which the value is a multiset of the unit interval \(I = [0, 1].\) That is, \[CM_{A}(x) = \lbrace \mu^{1}, \mu^{2}, … , \mu^{n}, … \rbrace \forall x \in X,\] where \( \mu^{1}, \mu^{2}, … , \mu^{n}, … \in [0,1]\) such that \[(\mu^{1} \geq \mu^{2} \geq … \geq \mu^{n} \geq … ).\] Whenever the fuzzy multiset is finite, we write \[CM_{A}(x) = \lbrace \mu^{1}, \mu^{2}, … , \mu^{n} \rbrace,\] where \( \mu^{1}, \mu^{2}, … , \mu^{n} \in [0,1]\) such that \[(\mu^{1} \geq \mu^{2} \geq … \geq \mu^{n} ),\] or simply \[CM_{A}(x) = \lbrace \mu^{i} \rbrace,\] for \( \mu^{i} \in [0,1]\) and \(i = 1, 2, … ,n.\) Now, a fuzzy multiset \( A\) is given as \[ A= \left\lbrace \dfrac{CM_{A}(x)}{x} : x \in X \right\rbrace \;\;\text{or}\;\; A= \left\lbrace (CM_{A}(x),x) : x \in X \right\rbrace. \] The set of all fuzzy multisets is depicted by \(FMS(X).\)

Example 1. Consider the set \(X = \lbrace a, b, c \rbrace \). Then for \( CM_{A}(a) = \lbrace 1, 0.5, 0.4 \rbrace \), \( CM_{A}(b) = \lbrace 0.9, 0.6 \rbrace \) and \( CM_{A}(c) = \lbrace 0 \rbrace \) we get that \( A\) is a fuzzy multiset of \( X\) written as \(A= \left\lbrace \dfrac{1, 0.5, 0.4}{a}, \dfrac{0.9, 0.6}{b} \right\rbrace.\)

Definition 3. Let \(A, B \in FMS(X).\) Then \( A\) is called a fuzzy submultiset of \( B\) written as \(A \subseteq B\) if \(CM_{A}(x) \leq CM_{B}(x)\) for all \(x \in X.\) Also, if \(A \subseteq B\) and \(A \neq B,\) then \( A\) is called a proper fuzzy submultiset of \( B\) and denoted as \(A \subset B.\)

Definition 4. Let \(A \in FMS(X)\) and \( \alpha \in [0,1]. \) Then we define the following notions:

• \( A_{\star}=\lbrace x \in X \hspace{0.1cm}|\hspace{0.1cm} CM_{A}(x)>0 \rbrace.\)
• \( A^{\star}=\lbrace x \in X \hspace{0.1cm}|\hspace{0.1cm} CM_{A}(x)=CM_{A}(e_{X}) \rbrace \) where \( e_{X}\) is the identity element of \(X. \)
• \( A_{[\alpha]}=\lbrace x \in X \hspace{0.1cm}|\hspace{0.1cm} CM_{A}(x) \geq \alpha \rbrace \) is called strong upper alpha-cut of \(A.\)
• \( A_{(\alpha)}=\lbrace x \in X \hspace{0.1cm}|\hspace{0.1cm} CM_{A}(x) > \alpha \rbrace \) is called weak upper alpha-cut of \(A.\)
• \( A^{[\alpha]}=\lbrace x \in X \hspace{0.1cm}|\hspace{0.1cm} CM_{A}(x) \leq \alpha \rbrace \) is called strong lower alpha-cut of \(A.\)
• \( A^{(\alpha)}=\lbrace x \in X \hspace{0.1cm}|\hspace{0.1cm} CM_{A}(x) < \alpha \rbrace \) is called weak lower alpha-cut of \(A.\)

Definition 5. Let \(A, B \in FMG(X).\) We say that \( A\) is conjugate to \( B\) if for all \( x,y \in X \) we have \( CM_{A}(x)=CM_{B}\left(yxy^{-1}\right).\)

Definition 6. Let \(A\in FMG(X).\) We say that \( A\) is commutative if \( CM_{A}(xy)=CM_{A}(yx) \) for all \( x,y \in X.\)

Definition 7. A \(t\)-norm \(T\) is a function \(T : [0,1]\times [0,1] \to [0,1]\) having the following four properties:
(T1) \(T(x,1)=x\) (neutralelement),
(T2) \(T(x,y)\leq T(x,z)\) if \(y\leq z\) (monotonicity),
(T3) \(T(x,y)= T(y,x)\) (commutativity),
(T4) \(T(x,T(y,z))=T(T(x,y),z)\) (associativity),\\ for all \(x,y,z \in[0,1].\)
We say that \(T\) be idempotent if \(T(x,x)=x\) for all \(x\in [0,1].\)

It is clear that if \(x_{1}\geq x_{2}\) and \(y_{1}\geq y_{2}\), then \(T(x_{1},y_{1}) \geq T(x_{2},y_{2}).\)

Example 2. (1) Standard intersection \(t\)-norm \(T_m(x,y) = \min \{ x,y \}.\)
(2) Bounded sum \(t\)-norm \(T_b(x,y) =\max\{0, x+y- 1 \}.\)
(3) Algebraic product \(t\)-norm \(T_p(x, y) = xy. \)
(4) Drastic \(T\)-norm \(T_{D}(x,y) = \left\{ \begin{array}{rl} y &\text{if } x=1\\ x &\text{if } y=1\\ 0 &\text{otherwise. } \\ \end{array} \right. \)
(5) Nilpotent minimum \(t\)-norm \(T_{nM}(x,y) = \left\{ \begin{array}{rl} \min \lbrace x , y \rbrace &\text{if } x+y >1\\ 0 &\text{otherwise. } \\ \end{array} \right. \)
(6) Hamacher product \(t\)-norm \(T_{H_{0}}(x,y) = \left\{ \begin{array}{rl} 0 &\text{if } x=y =0\\ \frac{xy}{x+y-xy} &\text{otherwise. } \\ \end{array} \right. \)

The drastic \(t\)-norm is the pointwise smallest \(t\)-norm and the minimum is the pointwise largest \(t\)-norm: \(T_{D}(x,y) \leq T(x,y) \leq T_{\min} (x ,y)\) for all \( x,y \in [0,1].\)

Lemma 1. Let \(T\) be a \(t\)-norm. Then \[T(T(x,y),T(w,z))= T(T(x,w),T(y,z)),\] for all \(x,y,w,z\in [0,1].\)

Definition 8. Let \(A \in FMS(G).\) Then \( A \) is said to be a fuzzy multigroup of \( G\) under \(t\)-norm \(T\) if it satisfies the following two conditions:
(1) \(CM_{A}(xy) \geq T(CM_{A}(x),CM_{A}(y)),\)
(2) \(CM_{A}(x^{-1}) \geq CM_{A}(x),\) for all \(x,y\in G.\)
The set of all fuzzy multisets of \( G\) under \(t\)-norm \(T\) is depicted by \(TFMS(G).\)

Theorem 1. Let \( A \in TFMS(G).\) If \(T\) be idempotent, then for all \(x \in G \), we have and \(n\geq 1,\)
(1) \(CM_{A}(e)\geq CM_{A}(x)\);
(2) \(CM_{A}(x^n)\geq CM_{A}(x);\)
(3) \(CM_{A}(x)=CM_{A}(x^{-1}).\)

3. Direct product of fuzzy multigroups under \(t\)-norms

Definition 9. Let \( A \in TFMS(G)\) and \( B \in TFMS(H).\) The direct product of \(A\) and \(B\), denoted by \(A \times B\), is characterized by a count membership function \[CM_{A \times B} : G \times H \to [0, 1]\] such that \[CM_{A \times B}(x,y)=T(CM_{A}(x) ,CM_{B}(y))\] for all \( x \in G \) and \( y \in H. \)

Example 3. Let \(G = \left\lbrace 1, x \right\rbrace\) be a group, where \(x^{2} = 1\) and \(H =\lbrace e, a, b, c \rbrace\) be a Klein 4-group, where \(a^{2} = b^{2} = c^{2}= e.\) Suppose \[A= \left\lbrace \dfrac{0.9,0.8}{1}, \dfrac{0.7, 0.6}{x} \right\rbrace \] and \[B= \left\lbrace \dfrac{1, 0.85}{e}, \dfrac{0.35, 0.25}{a}, \dfrac{0.10, 0.50}{b}, \dfrac{0.8, 0.6}{c} \right\rbrace\] be fuzzy multigroups of \( G\) and \( H.\) Let \[G \times H=\left\lbrace (1,e),(1,a),(1,b),(1,c),(x,e),(x,a),(x,b),(x,c) \right\rbrace\] be a a group from the classical sense. Define \[A \times B= \left\lbrace \dfrac{0.9, 0.8}{(1,e)}, \dfrac{0.35, 0.25}{(1,a)}, \dfrac{0.10, 0.50}{(1,b)}, \dfrac{0.8, 0.6}{(1,c)}, \dfrac{0.7, 0.6}{(x,e)}, \dfrac{0.35, 0.25}{(x,a)}, \dfrac{0.10, 0.50}{(x,b)}, \dfrac{0.7, 0.6}{(x,c)} \right\rbrace \] and let \(T_m(x,y) = \min \{ x,y \}\) be a standard intersection \(t\)-norm for all \( x,y \in [0,1]. \) Then \[ A \times B \in TFMS(G \times H). \]

Proposition 1. Let \(A_{i}\in TFMS(G_{i})\) for \(i=1,2.\) Then \(A_{1}\times A_{2}\in TFMS(G_{1}\times G_{2}).\)

Proof. Let \((a_{1},b_{1}),(a_{2},b_{2})\in G_{1}\times G_{2}.\) Then \begin{align*} (CM_{A \times B})((a_{1},b_{1})(a_{2},b_{2}))&=(CM_{A \times B})(a_{1}a_{2},b_{1}b_{2}) =T(CM_{A}(a_{1}a_{2}),CM_{B}(b_{1}b_{2}))\\ &\geq T(T(CM_{A}(a_{1}),CM_{A}(a_{2})),T(CM_{B}(b_{1}),CM_{B}(b_{2})))\\ &=T(T(CM_{A}(a_{1}),CM_{B}(b_{1}),T(CM_{A}(a_{2}),CM_{B}(b_{2}))\\ &=T((CM_{A \times B})(a_{1},b_{1}),(CM_{A \times B})(a_{2},b_{2})). \end{align*} Also \begin{align*} (CM_{A \times B})(a_{1},b_{1})^{-1}&= (CM_{A \times B})(a_{1}^{-1},b_{1}^{-1})\\&= T(CM_{A}(a_{1}^{-1}),CM_{B}(b_{1}^{-1}))\geq T(CM_{A}(a_{1}),CM_{B}(b_{1})). \end{align*} Thus \(A_{1}\times A_{2}\in TFMS(G_{1}\times G_{2}).\)

Corollary 1. Let \( A \in TFMS(G)\) and \( B \in TFMS(H).\) Then \[A \times 1_{H},1_{G} \times B \in TFMS(G\times H).\]

Corollary 2. Let \(A_{i}\in TFMS(G_{i})\) for \(i=1,2,…,n\). Then \[A_{1}\times A_{2}\times… \times A_{n} \in TFMS(G_{1}\times G_{2}\times… \times G_{n}).\]

Proposition 2. Let \( A \in TFMS(G)\) and \( B \in TFMS(H)\) such that \(T\) be idempotent \(t\)-norm. Then for all \( \alpha \in [0,1]\) the following assertions hold:
(1) \( (A \times B)_{\star} =A_{\star} \times B_{\star}.\)
(2) \( (A \times B)^{\star} =A^{\star} \times B^{\star}.\)
(3) \( (A \times B)_{[\alpha]} =A_{[\alpha]} \times B_{[\alpha]}.\)
(4) \( (A \times B)_{(\alpha)} =A_{(\alpha)} \times B_{(\alpha)}.\)
(5) \( (A \times B)^{[\alpha]} =A^{[\alpha]} \times B^{[\alpha]}.\)
(6) \( (A \times B)^{(\alpha)} =A^{(\alpha)} \times B^{(\alpha)}.\)

Proof.

• (1) We know that \( (A \times B)_{\star} = \lbrace (x,y) \in G \times H \hspace{0.1cm}|\hspace{0.1cm} CM_{A \times B}(x,y)>0 \rbrace. \) Then \( (x,y) \in (A \times B)_{\star} \Longleftrightarrow CM_{A \times B}(x,y) >0 \Longleftrightarrow T(CM_{A}(x),CM_{B}(y)) > 0\Longleftrightarrow T(CM_{A}(x),CM_{B}(y)) > 0=T(0,0) \Longleftrightarrow CM_{A}(x)>0 \hspace{0.1cm} \textit{and} \hspace{0.1cm} CM_{B}(y)>0\Longleftrightarrow x\in A_{\star} \hspace{0.1cm} \textit{and} \hspace{0.1cm} y\in B_{\star} \Longleftrightarrow (x,y) \in A_{\star} \times B_{\star}\,. \) Hence \( (A \times B)_{\star} =A_{\star} \times B_{\star}.\)
• (2) As \( (A \times B)^{\star} = \lbrace (x,y) \in G \times H \hspace{0.1cm}|\hspace{0.1cm} CM_{A \times B}(x,y)= CM_{A \times B}(e_{G},e_{H}) \rbrace \) so \( (x,y) \in (A \times B)^{\star} \Longleftrightarrow CM_{A \times B}(x,y) =CM_{A \times B}(e_{G},e_{H})\Longleftrightarrow T(CM_{A}(x),CM_{B}(y))=T(CM_{A}(e_{G}),CM_{B}(e_{H}))\Longleftrightarrow CM_{A}(x)=CM_{A}(e_{G}) \hspace{0.1cm} \textit{and} \hspace{0.1cm} CM_{B}(y)=CM_{B}(e_{H})\Longleftrightarrow x\in A^{\star} \hspace{0.1cm} \textit{and} \hspace{0.1cm} y\in B^{\star} \Longleftrightarrow (x,y) \in A^{\star} \times B^{\star}\,. \) Thus \( (A \times B)^{\star} =A^{\star} \times B^{\star}.\)
• (3) Let \( (A \times B)_{[\alpha]} = \lbrace (x,y) \in G \times H \hspace{0.1cm}|\hspace{0.1cm} CM_{A \times B}(x,y) \geq \alpha \rbrace. \) Now \( (x,y) \in (A \times B)_{[\alpha]} \Longleftrightarrow CM_{A \times B}(x,y) \geq \alpha \Longleftrightarrow T(CM_{A}(x),CM_{B}(y)) \geq \alpha\Longleftrightarrow T(CM_{A}(x),CM_{B}(y)) \geq \alpha=T(\alpha,\alpha) \Longleftrightarrow CM_{A}(x)\geq \alpha \hspace{0.1cm} \textit{and} \hspace{0.1cm} CM_{B}(y)\geq \alpha\Longleftrightarrow x\in A_{[\alpha]} \hspace{0.1cm} \textit{and} \hspace{0.1cm} y\in B_{[\alpha]} \Longleftrightarrow (x,y) \in A_{[\alpha]} \times B_{[\alpha]}\,. \) Thus \( (A \times B)_{[\alpha]} =A_{[\alpha]} \times B_{[\alpha]}.\)
• (4) Since \( (A \times B)_{(\alpha)} = \lbrace (x,y) \in G \times H \hspace{0.1cm}|\hspace{0.1cm} CM_{A \times B}(x,y) > \alpha \rbrace\), so \( (x,y) \in (A \times B)_{(\alpha)} \Longleftrightarrow CM_{A \times B}(x,y) > \alpha \Longleftrightarrow T(CM_{A}(x),CM_{B}(y)) > \alpha\Longleftrightarrow T(CM_{A}(x),CM_{B}(y)) > \alpha=T(\alpha,\alpha) \Longleftrightarrow CM_{A}(x) > \alpha \hspace{0.1cm} \textit{and} \hspace{0.1cm} CM_{B}(y) > \alpha\Longleftrightarrow x\in A_{(\alpha)} \hspace{0.1cm} \textit{and} \hspace{0.1cm} y\in B_{(\alpha)} \Longleftrightarrow (x,y) \in A_{(\alpha)} \times B_{(\alpha)}\,. \) So \( (A \times B)_{(\alpha)} =A_{(\alpha)} \times B_{(\alpha)}.\)
• (5) Because \( (A \times B)^{[\alpha]} = \lbrace (x,y) \in G \times H \hspace{0.1cm}|\hspace{0.1cm} CM_{A \times B}(x,y) \leq \alpha \rbrace\), then \( (x,y) \in (A \times B)^{[\alpha]} \Longleftrightarrow CM_{A \times B}(x,y) \leq \alpha \Longleftrightarrow T(CM_{A}(x),CM_{B}(y)) \leq \alpha(x,y) \in (A \times B)^{[\alpha]} \Longleftrightarrow CM_{A \times B}(x,y) \leq \alpha \Longleftrightarrow T(CM_{A}(x),CM_{B}(y)) \leq \alpha\Longleftrightarrow T(CM_{A}(x),CM_{B}(y)) \leq \alpha=T(\alpha,\alpha) \Longleftrightarrow CM_{A}(x) \leq \alpha \hspace{0.1cm} \textit{and} \hspace{0.1cm} CM_{B}(y) \leq \alpha\Longleftrightarrow T(CM_{A}(x),CM_{B}(y)) \leq \alpha=T(\alpha,\alpha) \Longleftrightarrow CM_{A}(x) \leq \alpha \hspace{0.1cm} \textit{and} \hspace{0.1cm} CM_{B}(y) \leq \alpha\Longleftrightarrow x\in A^{[\alpha]} \hspace{0.1cm} \textit{and} \hspace{0.1cm} y\in B^{[\alpha]} \Longleftrightarrow (x,y) \in A^{[\alpha]} \times B^{[\alpha]}\,. \) Therefore \( (A \times B)^{[\alpha]} =A^{[\alpha]} \times B^{[\alpha]}. \)
• (6) Because of \( (A \times B)^{(\alpha)} = \lbrace (x,y) \in G \times H \hspace{0.1cm}|\hspace{0.1cm} CM_{A \times B}(x,y) < \alpha \rbrace\), then \( (x,y) \in (A \times B)^{(\alpha)} \Longleftrightarrow CM_{A \times B}(x,y) < \alpha \Longleftrightarrow T(CM_{A}(x),CM_{B}(y)) < \alpha \Longleftrightarrow T(CM_{A}(x),CM_{B}(y)) < \alpha=T(\alpha,\alpha) \Longleftrightarrow CM_{A}(x) < \alpha \hspace{0.1cm} \textit{and} \hspace{0.1cm} CM_{B}(y) < \alpha\Longleftrightarrow x\in A^{(\alpha)} \hspace{0.1cm} \textit{and} \hspace{0.1cm} y\in B^{(\alpha)} \Longleftrightarrow (x,y) \in A^{(\alpha)} \times B^{(\alpha)}\,. \) Hence \( (A \times B)^{(\alpha)} =A^{(\alpha)} \times B^{(\alpha)}.\)

Proposition 3. Let \( A \in TFMS(G)\) and \( B \in TFMS(H)\) such that \(T\) be idempotent \(t\)-norm. Then for all \((x,y) \in G \times H\) the following assertions hold:
(1) \( CM_{A \times B}(e_{G},e_{H}) \geq CM_{A \times B}(x,y),\)
(2) \( CM_{A \times B}((x,y)^{n}) \geq CM_{A \times B}(x,y), \)
(3) \(CM_{A \times B}(x,y)=CM_{A \times B}(x^{-1},y^{-1}).\)

Proof. Using Proposition \ref{prop1} we get that \( A \times B \in TFMS(G \times H).\) Now Theorem \ref{thm1} gives us that assertions are hold.

Proposition 4. Let \( A \in TFMS(G)\) and \( B \in TFMS(H)\) such that \(T\) be idempotent \(t\)-norm. Then for all \( \alpha \in [0,1]\) the following assertions hold:
(1) \( (A \times B)_{\star}\) is a subgroup of \(G \times H,\)
(2) \( (A \times B)^{\star} \) is a subgroup of \(G \times H,\)
(3) \( (A \times B)_{[\alpha]}\) is a subgroup of \(G \times H,\)
(4) \( (A \times B)_{(\alpha)}\) is a subgroup of \(G \times H.\)

Proof. (1) Let \( (x_{1},y_{1}),(x_{2},y_{2}) \in (A \times B)_{\star}\). We need to prove that \( (x_{1},y_{1})(x_{2},y_{2})^{-1} \in (A \times B)_{\star}.\) As \( (x_{1},y_{1}),(x_{2},y_{2}) \in (A \times B)_{\star}\), so \(CM_{A \times B}(x_{1},y_{1}) >0\) and \(CM_{A \times B}(x_{2},y_{2}) >0.\) Now \begin{align*} CM_{A \times B} ((x_{1},y_{1})(x_{2},y_{2})^{-1}) &=CM_{A \times B} ((x_{1},y_{1})(x^{-1}_{2},y^{-1}_{2}))\\&= CM_{A \times B} (x_{1}x^{-1}_{2},y_{1}y^{-1}_{2})=T(CM_{A}(x_{1}x^{-1}_{2}),CM_{B}(y_{1}y^{-1}_{2}))\\&\geq T(T(CM_{A}(x_{1}),CM_{A}(x^{-1}_{2})),T(CM_{B}(y_{1}),CM_{B}(y^{-1}_{2})))\\&\geq T(T(CM_{A}(x_{1}),CM_{A}(x_{2})),T(CM_{B}(y_{1}),CM_{B}(y_{2})))\\&=T(T(CM_{A}(x_{1}),CM_{B}(y_{1})),T(CM_{A}(x_{2}),CM_{B}(y_{2})))\\&=T(CM_{A \times B}(x_{1},y_{1}),CM_{A \times B}(x_{2},y_{2}))>T(0,0)=0\,. \end{align*} Thus \( CM_{A \times B} ((x_{1},y_{1})(x_{2},y_{2})^{-1}) >0 \), which means that \((x_{1},y_{1})(x_{2},y_{2})^{-1} \in (A \times B)_{\star}.\) Hence \( (A \times B)_{\star}\) is a subgroup of \(G \times H.\)
(2) Let \( (x_{1},y_{1}),(x_{2},y_{2}) \in (A \times B)_{\star}\). We need to prove that \( (x_{1},y_{1})(x_{2},y_{2})^{-1} \in (A \times B)^{\star}.\) Because \( (x_{1},y_{1}),(x_{2},y_{2}) \in (A \times B)^{\star}\) then \( CM_{A \times B}(x_{1},y_{1})=CM_{A \times B}(x_{2},y_{2}) =CM_{A \times B}(e_{G},e_{H})\), which means that \( T(CM_{A}(x_{1}),CM_{B}(y_{1}))=T(CM_{A}(x_{2}),CM_{B}(y_{2}))=T(CM_{A}(e_{G}),CM_{B}(e_{H})) \), so \(CM_{A}(x_{1})=CM_{A}(x_{2})=CM_{A}(e_{G}) \) and \(CM_{A}(y_{1})=CM_{A}(y_{2})=CM_{A}(e_{H}).\) Thus \begin{align*} CM_{A \times B} ((x_{1},y_{1})(x_{2},y_{2})^{-1}) &=CM_{A \times B} ((x_{1},y_{1})(x^{-1}_{2},y^{-1}_{2}))\\ &= CM_{A \times B} (x_{1}x^{-1}_{2},y_{1}y^{-1}_{2})=T(CM_{A}(x_{1}x^{-1}_{2}),CM_{B}(y_{1}y^{-1}_{2})) \\ &\geq T(T(CM_{A}(x_{1}),CM_{A}(x^{-1}_{2})),T(CM_{B}(y_{1}),CM_{B}(y^{-1}_{2})))\\ &\geq T(T(CM_{A}(x_{1}),CM_{A}(x_{2})),T(CM_{B}(y_{1}),CM_{B}(y_{2})))\\ &=T(T(CM_{A}(e_{G}),CM_{A}(e_{G})),T(CM_{B}(e_{H}),CM_{B}(e_{H})))\\ &=T(CM_{A}(e_{G}),CM_{B}(e_{H}))=CM_{A \times B}(e_{G},e_{H})\\ &\geq CM_{A \times B} ((x_{1},y_{1})(x_{2},y_{2})^{-1})\,. \hspace{0.1cm} \textit{(Proposition \ref{prop2}(1))} \end{align*}
Thus \(CM_{A \times B} ((x_{1},y_{1})(x_{2},y_{2})^{-1})=CM_{A \times B}(e_{G},e_{H})\), so \( (x_{1},y_{1})(x_{2},y_{2})^{-1} \in (A \times B)^{\star}.\) Hence we obtain that \( (A \times B)^{\star} \) is a subgroup of \(G \times H.\)
(3) Let \( (x_{1},y_{1}),(x_{2},y_{2}) \in (A \times B)_{[\alpha]}\). We need to show that \( (x_{1},y_{1})(x_{2},y_{2})^{-1} \in (A \times B)_{[\alpha]}.\) As \( (x_{1},y_{1}),(x_{2},y_{2}) \in (A \times B)_{[\alpha]}\) so \(CM_{A \times B}(x_{1},y_{1}) \geq \alpha\) and \(CM_{A \times B}(x_{2},y_{2}) \geq \alpha.\) Now \begin{align*} CM_{A \times B} ((x_{1},y_{1})(x_{2},y_{2})^{-1}) &=CM_{A \times B} ((x_{1},y_{1})(x^{-1}_{2},y^{-1}_{2}))\\ &= CM_{A \times B} (x_{1}x^{-1}_{2},y_{1}y^{-1}_{2})=T(CM_{A}(x_{1}x^{-1}_{2}),CM_{B}(y_{1}y^{-1}_{2}))\\ &\geq T(T(CM_{A}(x_{1}),CM_{A}(x^{-1}_{2})),T(CM_{B}(y_{1}),CM_{B}(y^{-1}_{2})))\\ &\geq T(T(CM_{A}(x_{1}),CM_{A}(x_{2})),T(CM_{B}(y_{1}),CM_{B}(y_{2})))\\ &=T(T(CM_{A}(x_{1}),CM_{B}(y_{1})),T(CM_{A}(x_{2}),CM_{B}(y_{2})))\\ &=T(CM_{A \times B}(x_{1},y_{1}),CM_{A \times B}(x_{2},y_{2})) \geq T(\alpha,\alpha)=\alpha\,. \end{align*} Thus \( CM_{A \times B} ((x_{1},y_{1})(x_{2},y_{2})^{-1}) \geq \alpha\) which means that \((x_{1},y_{1})(x_{2},y_{2})^{-1} \in (A \times B)_{[\alpha]}.\) Hence \( (A \times B)_{[\alpha]}\) is a subgroup of \(G \times H.\)
(4) If \( (x_{1},y_{1}),(x_{2},y_{2}) \in (A \times B)_{(\alpha)},\) then \(CM_{A \times B}(x_{1},y_{1}) > \alpha\) and \(CM_{A \times B}(x_{2},y_{2}) > \alpha.\) Now \begin{align*} CM_{A \times B} ((x_{1},y_{1})(x_{2},y_{2})^{-1}) &=CM_{A \times B} ((x_{1},y_{1})(x^{-1}_{2},y^{-1}_{2}))\\ &= CM_{A \times B} (x_{1}x^{-1}_{2},y_{1}y^{-1}_{2})=T(CM_{A}(x_{1}x^{-1}_{2}),CM_{B}(y_{1}y^{-1}_{2}))\\ &\geq T(T(CM_{A}(x_{1}),CM_{A}(x^{-1}_{2})),T(CM_{B}(y_{1}),CM_{B}(y^{-1}_{2}))) \\ &\geq T(T(CM_{A}(x_{1}),CM_{A}(x_{2})),T(CM_{B}(y_{1}),CM_{B}(y_{2})))\\ &=T(T(CM_{A}(x_{1}),CM_{B}(y_{1})),T(CM_{A}(x_{2}),CM_{B}(y_{2})))\\ &=T(CM_{A \times B}(x_{1},y_{1}),CM_{A \times B}(x_{2},y_{2})) > T(\alpha,\alpha)=\alpha\,. \end{align*} Thus \( CM_{A \times B} ((x_{1},y_{1})(x_{2},y_{2})^{-1}) > \alpha\) which means that \((x_{1},y_{1})(x_{2},y_{2})^{-1} \in (A \times B)_{(\alpha)}.\) Hence \( (A \times B)_{(\alpha)}\) is a subgroup of \(G \times H.\)

Proposition 5. Let \( A \in TFMS(G)\) and \( B \in TFMS(H).\) If \( A\times B \in TFMS(G \times H),\) then at least one of the following statements hold:
(1) \(CM_{B}(e_{H})) \geq CM_{A}(x)\) for all \( x \in G, \)
2) \(CM_{A}(e_{G})) \geq CM_{B}(y)\) for all \( y \in G. \)

Proof. Suppose that none of the statements holds, then we can find \( a \in G \) and \(b \in H \) such that \(CM_{A}(a) >CM_{B}(e_{H})\) and \(CM_{B}(b) >CM_{A}(e_{G}).\) Now \begin{align*}CM_{A \times B}(a,b)&=T(CM_{A}(a),CM_{B}(b))\\& > T(CM_{B}(e_{H}),CM_{A}(e_{G}))\\&=T(CM_{A}(e_{G}),CM_{B}(e_{H}))=CM_{A \times B}(e_{G},e_{H})\,.\end{align*} Thus \(CM_{A \times B}(a,b) > CM_{A \times B}(e_{G},e_{H})\), which is contradiction with Proposition \ref{prop2}(1), hence at least one of the statements hold.

Proposition 6. Let \( A \in FMS(G)\) and \( B \in FMS(H)\) such that \( A\times B \in TFMS(G \times H)\) and \(CM_{A}(x) \leq CM_{B}(e_{H})\) for all \( x \in G. \) Then \( A \in TFMS(G).\)

Proof. As \(CM_{A}(x) \leq CM_{B}(e_{H})\) for all \( x \in G\), so \(CM_{A}(y) \leq CM_{B}(e_{H})\) and \(CM_{A}(xy) \leq CM_{B}(e_{H})=CM_{B}(e_{H}e_{H})\) for all \( y \in G.\) Then \begin{align*} CM_{A}(xy) &=T(CM_{A}(xy),CM_{B}(e_{H}e_{H}))\\&=CM_{A \times B}(xy,e_{H}e_{H})\\ &=CM_{A \times B}((x,e_{H})(y,e_{H}))\\ &\geq T(CM_{A \times B}(x,e_{H}),CM_{A \times B}(y,e_{H}))\\ &=T(T(CM_{A}(x),CM_{B}(e_{H})),T(CM_{A}(y),CM_{B}(e_{H})))\\&=T(CM_{A}(x),CM_{A}(y))\,. \end{align*} Thus \[CM_{A}(xy) \geq T(CM_{A}(x),CM_{A}(y)).\] Also since \(CM_{A}(x) \leq CM_{B}(e_{H})\) for all \( x \in G\) so \(CM_{A}(x^{-1}) \leq CM_{B}(e_{H}).\) Thus \begin{align*}CM_{A}(x^{-1})&=T(CM_{A}(x^{-1}),CM_{A}(e_{H}))\\&=T(CM_{A}(x^{-1}),CM_{A}(e_{H}^{-1}))\\&=CM_{A \times B}((x,e_{H})^{-1}) \\&\geq CM_{A \times B}(x,e_{H}) \\&=T(CM_{A}(x),CM_{A}(e_{H}))=CM_{A}(x)\end{align*} and then \( CM_{A}(x^{-1}) \geq CM_{A}(x) .\) Therefore \( A \in TFMS(G).\)

Proposition 7. Let \( A \in FMS(G)\) and \( B \in FMS(H)\) such that \( A\times B \in TFMS(G \times H)\) and \(CM_{B}(x) \leq CM_{A}(e_{G})\) for all \( x \in H. \) Then \( B \in TFMS(H).\)

Proof. The proof is similar to Proposition \ref{prop19}.

Corollary 3. Let \( A \in FMS(G)\) and \( B \in FMS(H)\) such that \( A\times B \in TFMS(G \times H).\) Then either \( A \in TFMS(G)\) or \( B \in TFMS(H).\)

Proof. Using Proposition \ref{prop18} we get that \(CM_{B}(e_{H})) \geq CM_{A}(x)\) for all \( x \in G \) or \(CM_{A}(e_{G})) \geq CM_{B}(y)\) for all \( y \in G. \) Then from Proposition \ref{prop19} and Proposition \ref{prop20} we have that either \( A \in TFMS(G)\) or \( B \in TFMS(H).\)

Proposition 8. Let \( A,C \in TFMS(G)\) and \( B,D \in TFMS(H).\) If \( A \) is conjugate to \( B\) and \( C \) is conjugate to \( D,\) then \( A \times C \) is conjugate to \( B \times D.\)

Proof. As \( A \) is conjugate to \( B\) so \( CM_{A}(x)=CM_{C}(gxg^{-1}) \) and as \( B \) is conjugate to \( D\) so \( CM_{B}(y)=CM_{D}(hyh^{-1}) \) for all \( x,g \in G \) and \( y,h \in H. \) Now \begin{align*} CM_{A \times B}(x,y)&=T( CM_{A}(x),CM_{B}(y))\\&=T(CM_{C}(gxg^{-1}),CM_{D}(hyh^{-1}))\\ &=CM_{C \times D}(gxg^{-1},hyh^{-1})\\&=CM_{C \times D}((g,h)(x,y)(g^{-1},h^{-1}))\\ &=CM_{C \times D}((g,h)(x,y)(g,h)^{-1})\,. \end{align*} Thus \( CM_{A \times B}(x,y)=CM_{C \times D}((g,h)(x,y)(g,h)^{-1}) \) which means that \( A \times C \) is conjugate to \( B \times D.\)

Proposition 9. Let \( A \in TFMS(G)\) and \( B \in TFMS(H).\) Then \( A\) and \( B\) are commutative if and only if \( A \times B \) is a commutative.

Proof. Let \(x_{1},y_{1} \in G\) and \(x_{2},y_{2} \in H\) such that \( x=(x_{1},x_{2}) \in G \times H\) and \( y=(y_{1},y_{2}) \in G \times H.\) Let \( A\) and \( B\) are commutative then \( CM_{A}(x_{1}y_{1}) =CM_{A}(y_{1}x_{1}) \) and \( CM_{B}(x_{2}y_{2}) =CM_{B}(y_{2}x_{2}).\) Which implies \begin{align*} CM_{A \times B}(xy)&=CM_{A \times B}((x_{1},x_{2})(y_{1},y_{2}))\\&=CM_{A \times B}(x_{1}y_{1},x_{2}y_{2})\\ &=T(CM_{A}(x_{1}y_{1}),CM_{ B}(x_{2}y_{2}))\\&=T(CM_{A}(y_{1}x_{1}),CM_{ B}(y_{2}x_{2}))\\ &=CM_{A \times B}(y_{1}x_{1},y_{2}x_{2})\\&=CM_{A \times B}((y_{1},y_{2})(x_{1},x_{2}))\\&=CM_{A \times B}(yx)\,. \end{align*} Thus \( CM_{A \times B}(xy)=CM_{A \times B}(yx) \) and then \( A \times B \) is a commutative. Conversely, suppose that \( A \times B \) is a commutative. Then \( CM_{A \times B}(xy)=CM_{A \times B}(yx) \Longleftrightarrow CM_{A \times B}((x_{1},x_{2})(y_{1},y_{2}))=CM_{A \times B}((y_{1},y_{2})(x_{1},x_{2})) \Longleftrightarrow CM_{A \times B}(x_{1}y_{1},x_{2}y_{2})=CM_{A \times B}(y_{1}x_{1},y_{2}x_{2}) \Longleftrightarrow T(CM_{A}(x_{1}y_{1}),CM_{B}(x_{2}y_{2}))=T(CM_{A}(y_{1}x_{1}) ,CM_{B}(y_{2}x_{2})) \Longleftrightarrow CM_{A}(x_{1}y_{1})=CM_{A}(y_{1}x_{1})\) and \(CM_{B}(x_{2}y_{2})=CM_{B}(y_{2}x_{2}) \) which gives us that \( A\) and \( B\) are commutative.

Definition 10. Let \( G \times H\) and \( I \times J\) be groups and \(f : G \times H \to I \times J\) be a homomorphism. Let \( A \times B \in FMS(G \times H)\) and \( C \times D \in FMS(I \times J).\) Define \( f(A \times B) \in FMS(I \times J)\) and \( f^{-1}(C \times D) \in FMS(G \times H)\) as: \begin{align*} f(CM_{A \times B})(i,j)&=(CM_{f(A \times B)})(i,j)= \left\{ \begin{array}{rl} \sup \lbrace CM_{A \times B}(g,h) \hspace{0.1cm}|\hspace{0.1cm} g \in G, h \in H , f(g,h)=(i,j) \rbrace &\text{if }f^{-1}(i,j) \neq \emptyset \\ 0 &\text{otherwise } \\ \end{array} \right. \end{align*} and \[f^{-1}(CM_{C \times D}(g,h))=CM_{f^{-1}(C \times D)}(g,h)=CM_{C \times D}(f(g,h))\] for all \( (g,h) \in G \times H. \)

Proposition 10. Let \( G \times H\) and \( I \times J\) be groups and \(f : G \times H \to I \times J\) be an epimorphism. If \( A \in TFMS(G), B \in TFMS( H)\) and \(A \times B \in TFMS(G \times H),\) then \( f(A \times B) \in TFMS( I \times J).\)

Proof. (1) Let \( X=(i_{1},j_{1}) \in I \times J \) and \( Y=(i_{2},j_{2}) \in I \times J\) such that \[ f^{-1}(XY)=f^{-1}((i_{1},j_{1})(i_{2},j_{2}))=f^{-1}(i_{1}i_{2},j_{1}j_{2}) \neq \emptyset. \] Then \begin{align*} f(A \times B)(XY)&=f(A \times B)((i_{1},j_{1})(i_{2},j_{2}))\\ &=f(A \times B)(i_{1}i_{2},j_{1}j_{2})\\ &=\sup \lbrace CM_{A \times B}(g_{1}g_{2},h_{1}h_{2}) \hspace{0.1cm}|\hspace{0.1cm} g_{1},g_{2} \in G, h_{1},h_{2} \in H , f(g_{1}g_{2},h_{1}h_{2})=(i_{1}i_{2},j_{1}j_{2}) \rbrace\\ &= \sup \lbrace CM_{A \times B}(g_{1}g_{2},h_{1}h_{2}) \hspace{0.1cm}|\hspace{0.1cm} g_{1},g_{2} \in G, h_{1},h_{2} \in H , (f(g_{1}g_{2}),f(h_{1}h_{2}))=(i_{1}i_{2},j_{1}j_{2}) \rbrace\\ &=\sup \lbrace CM_{A \times B}(g_{1}g_{2},h_{1}h_{2}) \hspace{0.1cm}|\hspace{0.1cm} g_{1},g_{2} \in G, h_{1},h_{2} \in H , f(g_{1}g_{2})=i_{1}i_{2},f(h_{1}h_{2})=j_{1}j_{2} \rbrace\\ &=\sup \lbrace T(CM_{A}(g_{1}g_{2}) ,CM_{B}(h_{1}h_{2}) ) \hspace{0.1cm}|\hspace{0.1cm} g_{1},g_{2} \in G, h_{1},h_{2} \in H , f(g_{1}g_{2})=i_{1}i_{2},f(h_{1}h_{2})=j_{1}j_{2} \rbrace\\ &=\sup \lbrace T(CM_{A}(g_{1}g_{2}) ,CM_{B}(h_{1}h_{2}) ) \hspace{0.1cm}|\hspace{0.1cm} g_{1},g_{2} \in G, h_{1},h_{2} \in H , f(g_{1}g_{2})=i_{1}i_{2},f(h_{1}h_{2})=j_{1}j_{2} \rbrace\\ &\geq \sup \lbrace T(T(CM_{A}(g_{1}),CM_{A}(g_{2})) ,T(CM_{B}(h_{1}),CM_{B}(h_{2})) ) \hspace{0.1cm}|\hspace{0.1cm} f(g_{1}g_{2})=i_{1}i_{2},f(h_{1}h_{2})=j_{1}j_{2} \rbrace\\ &=\sup \lbrace T(T(CM_{A}(g_{1}),CM_{B}(h_{1})) ,T(CM_{A}(g_{2}),CM_{B}(h_{2})) ) \hspace{0.1cm}|\hspace{0.1cm} f(g_{1}g_{2})=i_{1}i_{2},f(h_{1}h_{2})=j_{1}j_{2} \rbrace \end{align*} \begin{align*} &=\sup \lbrace T(T(CM_{A}(g_{1}),CM_{B}(h_{1})) ,T(CM_{A}(g_{2}),CM_{B}(h_{2})) )\hspace{0.1cm}|\hspace{0.1cm} f(g_{1})=i_{1},f(g_{2})=i_{2},f(h_{1})=j_{1},f(h_{2})=j_{2} \rbrace\\ &=\sup \lbrace T(CM_{A \times B}(g_{1},h_{1}), CM_{A \times B}(g_{2},h_{2})) \hspace{0.1cm}|\hspace{0.1cm} f(g_{1})=i_{1},f(g_{2})=i_{2},f(h_{1})=j_{1},f(h_{2})=j_{2} \rbrace\\ &=T(\sup \lbrace CM_{A \times B}(g_{1},h_{1}) \hspace{0.1cm}|\hspace{0.1cm} f(g_{1},h_{1})=(i_{1},j_{1}) \rbrace, \sup \lbrace CM_{A \times B}(g_{2},h_{2}) \hspace{0.1cm}|\hspace{0.1cm} f(g_{2},h_{2})=(i_{2},j_{2})\rbrace)\\ &=T(f(A \times B)(i_{1},j_{1}),f(A \times B)(i_{2},j_{2}))=T(f(A \times B)(X),f(A \times B)(Y))\,.\end{align*} Thus \[ f(A \times B)(XY) \geq T(f(A \times B)(X),f(A \times B)(Y)).\]
(2) Let \( X=(i,j) \in I \times J \) then \begin{align*} f(A \times B)(X^{-1})&=f(A \times B)((i,j)^{-1})\\&=f(A \times B)(i^{-1},j^{-1})\\ &= \sup \lbrace CM_{A \times B}(g^{-1},h^{-1}) \hspace{0.1cm}|\hspace{0.1cm} g\in G, h\in H , f(g^{-1},h^{-1})=(i^{-1},j^{-1}) \rbrace\\ &=\sup \lbrace CM_{A \times B}(g^{-1},h^{-1}) \hspace{0.1cm}|\hspace{0.1cm} g\in G, h\in H , (f(g^{-1}),f(h^{-1}))=(i^{-1},j^{-1}) \rbrace \\ &=\sup \lbrace CM_{A \times B}(g^{-1},h^{-1}) \hspace{0.1cm}|\hspace{0.1cm} g\in G, h\in H , f(g^{-1})=i^{-1},f(h^{-1}))=j^{-1} \rbrace\\ &=\sup \lbrace T(CM_{A}(g^{-1}) , CM_{ B}(h^{-1}) ) \hspace{0.1cm}|\hspace{0.1cm} g\in G, h\in H , f(g^{-1})=i^{-1},f(h^{-1}))=j^{-1} \rbrace\\ &\geq \sup \lbrace T(CM_{A}(g) , CM_{ B}(h) ) \hspace{0.1cm}|\hspace{0.1cm} g\in G, h\in H , f^{-1}(g)=i^{-1},f^{-1}(h)=j^{-1} \rbrace\\ &=\sup \lbrace T(CM_{A}(g) , CM_{ B}(h) ) \hspace{0.1cm}|\hspace{0.1cm} g\in G, h\in H , f(g)=i,f(h)=j \rbrace\\ &=\sup \lbrace CM_{A \times B}(g,h) \hspace{0.1cm}|\hspace{0.1cm} (g,h) \in G\times H , f(g,h)=(i,j)\rbrace\\ &=f(A \times B)(i,j)=f(A \times B)(X) \end{align*} and then \(f(A \times B)(X^{-1}) \geq f(A \times B)(X).\) Therefore \( f(A \times B) \in TFMS( I \times J).\)

Proposition 11. Let \( G \times H\) and \( I \times J\) be groups and \(f : G \times H \to I \times J\) be a homomorphism. If \( C \in TFMS(I) \), \( D \in TFMS(J) \) and \( C \times D \in TFMS(I \times J),\) then \( f^{-1}(C \times D) \in TFMS(G \times H).\)

Proof. (1) Let \( X=(g_{1},h_{1}) \in G \times H\) and \( Y=(g_{2},h_{2}) \in G \times H.\) Then \begin{align*} f^{-1}(CM_{C \times D})(XY)&=f^{-1}(CM_{C \times D})((g_{1},h_{1}) (g_{2},h_{2}))\\ &=f^{-1}((CM_{C \times D})(g_{1}g_{2},h_{1}h_{2}))\\ &=CM_{C \times D}(f(g_{1}g_{2},h_{1}h_{2}))\\ &=CM_{C \times D}(f(g_{1}g_{2}),f(h_{1}h_{2}))\\ &=T(CM_{C}(f(g_{1}g_{2})),CM_{D}(f(h_{1}h_{2})))\\ &= T(CM_{C}(f(g_{1})f(g_{2})),CM_{D}(f(h_{1})f(h_{2}))) \\ & \geq T(T(CM_{C}(f(g_{1})),CM_{C}(f(g_{2}))),T(CM_{D}(f(h_{1})),CM_{D}(f(h_{2})))\\ &= T(T(CM_{C}(f(g_{1})),CM_{D}(f(h_{1}))),T(CM_{C}(f(g_{2}),CM_{D}(f(h_{2})))\\ &=T(CM_{C \times D}(f(g_{1}),f(h_{1})),CM_{C \times D}(f(g_{2}),f(h_{2})))\\&=T(CM_{C \times D}(f(g_{1},h_{1})),CM_{C \times D}(f(g_{2},h_{2})))\\&=T(f^{-1}(CM_{C \times D})(g_{1},h_{1}),f^{-1}(CM_{C \times D})(g_{2},h_{2}))\\ &=T(f^{-1}(CM_{C \times D})(X),f^{-1}(CM_{C \times D})(Y))\,. \end{align*} Thus \[f^{-1}(CM_{C \times D})(XY) \geq T(f^{-1}(CM_{C \times D})(X),f^{-1}(CM_{C \times D})(Y)).\]
(2) Let \( X=(g,h) \in G \times H,\) then \begin{align*}f^{-1}(CM_{C \times D})(X^{-1})&=f^{-1}(CM_{C \times D})((g_{1},h_{1}) ^{-1})\end{align*} \begin{align*}&=CM_{C \times D}(f(g,h)^{-1})\\ &=CM_{C \times D}(f(g^{-1},h^{-1}))\\ &=CM_{C \times D}(f^{-1}(g),f^{-1}(h))\\ &=T(CM_{C}(f^{-1}(g)),CM_{D}(f^{-1}(h)))\\ &\geq T(CM_{C}(f(g)),CM_{D}(f(h)))\\ &=CM_{C \times D}(f(g),f(h))\\ &=CM_{C \times D}(f(g,h))\\ &=f^{-1}(CM_{C \times D})(g,h)\\ &=f^{-1}(CM_{C \times D})(X)\end{align*} and then \(f^{-1}(CM_{C \times D})(X^{-1}) \geq f^{-1}(CM_{C \times D})(X).\) Thus \( f^{-1}(C \times D) \in TFMS(G \times H).\)

Acknowledgments

We would like to thank the referees for carefully reading the manuscript and making several helpful comments to increase the quality of the paper.

Author Contributions

All author contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflicts of Interest:

The author declares no conflict of interest.