Anti complex fuzzy subgroups under \(s\)-norms

1. Introduction

Group theory has applications in physics, chemistry, and computer science, and even puzzles like Rubik's Cube can be represented using group theory. Fuzzy sets, proposed by Zadeh [1], are sets whose elements have degrees of membership. Rosenfeld [2] introduced fuzzy sets in the realm of group theory and formulated the concepts of fuzzy subgroups of a group. Many authors have worked on fuzzy group theory [2,3,4], especially, some authors considered the fuzzy subgroups with respect to norms [5,6,7]. Alsarahead and Ahmad [8] defined the complex fuzzy subgroup and investigate some of its characteristics. The author by using norms, investigated some properties of fuzzy algebraic structures [9,10,11].

In this paper, by using \(s\)-norms, we define and investigate some properties of anti complex fuzzy subgroups of group \( G \) under \(s\)-norm \(S\) as \(ACFS(G).\) Also we define the composition and intersection of two \( \mu_{1},\mu_{2} \in ACFS(G)\) and obtain some of their characteristics. Later, we introduce and investigate the normality of \( \mu \in ACFS(G)\) denoted by \(NACFS(G).\) Finally, we introduce the normality between two \( \mu_{1},\mu_{2} \in ACFS(G)\) as \( \mu_{1} \bowtie \mu_{2} \) and investigate some important properties of them. By using a group homomorphism \( f: G \to H,\) we prove that if \( \mu \in ACFS(G)\) and \( \nu \in ACFS(H),\) then \( f(\mu) \in ACFS(H)\) and \( f^{-1}(\nu) \in ACFS(G).\) Also if \( \mu \in NACFS(G)\) and \( \nu \in NACFS(H),\) then we prove that \( f(\mu) \in NACFS(H)\) and \( f^{-1}(\nu) \in NACFS(G).\) Also we show that if \( \mu_{1},\mu_{2} \in ACFS(G)\) such that \(\mu_{1} \bowtie \mu_{2},\) then we show that \(f(\mu_{1}) \bowtie f(\mu_{2})\) and if \( \mu_{1},\mu_{2} \in ACFS(H)\) such that \(\mu_{1} \bowtie \mu_{2},\) then we obtain \(f^{-1}(\mu_{1}) \bowtie f^{-1}(\mu_{2}).\)

2. Preliminaries

The following definitions and preliminaries are required in the sequel of our work and hence presented in brief. For details we refer readers to [6,12,13,14,15].

Definition 1. A group is a non-empty set \( G\), on which there is a binary operation \((a,b) \to ab\) such that

• if \( a \) and \( b\) belong to \( G \) then \( ab\) is also in \( G\) (closure),
• \(a(bc) = (ab)c\) for all \(a,b,c \in G\) (associativity),
• there is an element \( e_{G} \in G\) such that \(ae_{G} = ee_{G}a = a\) for all \(a \in G\) (identity),
• if \(a \in G,\) then there is an element \(a^{-1} \in G\) such that \(aa^{-1} = a^{-1}a =e_{G}\) (inverse).

One can easily check that this implies the unicity of the identity and of the inverse. A group \( G \) is called abelian if the binary operation is commutative, i.e., \(ab = ba\) for all \(a,b \in G.\)

Remark 1. There are two standard notations for the binary group operation: either the additive notation, that is \((a,b) \to a + b\) in which case the identity is denoted by \( 0\), or the multiplicative notation, that is \((a,b) \to ab\) for which the identity is denoted by \( e. \)

Definition 2. Let \(G\) be an arbitrary group with a multiplicative binary operation and identity \(e\). By 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 3. Let \( X\) be a nonempty set. A complex fuzzy set \( A\) on \( X\) is an object having the form \(A =\lbrace (x, \mu_{A}(x)) | x \in X \rbrace,\) where \( \mu_{A}\) denotes the degree of membership function that assigns each element \( x \in X \), a complex number \( \mu_{A}(x)) \) lies within the unit circle in the complex plane. We shall assume that \( \mu_{A}(x) \) will be represented by \(r_{A}(x)e^{iw_{A}(x)}\) , where \(i = \sqrt{-1},\) and \( r:X \to [0,1] \) and \( w:X \to [0,2\pi] \). Note that by setting \( w(x)=0 \) in the definition above, we return back to the traditional fuzzy subset. Let \( \mu_{1}=r_{1}e^{w_{1}} \) and \( \mu_{2}=r_{2}e^{w_{2}} \) be two complex numbers lie within the unit circle in the complex plane. By \( \mu_{1} \leq \mu_{2} \), we mean \(r_{1} \leq r_{2} \) and \( w_{1} \leq w_{2}.\)

Definition 4. An \(s\)-norm \(S\) is a function \(S : [0,1]\times [0,1] \to [0,1]\) having the following four properties:

• \(S(x,0)=x\),
• \(S(x,y)\leq S(x,z)\) if \(y\leq z\),
• \(S(x,y)= S(y,x)\),
• \(S(x,S(y,z))=S(S(x,y),z)\)

for all \(x,y,z \in[0,1].\)

We say that \(S\) is idempotent if for all \(x \in [0,1]\),\( S(x, x) =x.\)

Example 1. The basic \(s\)-norms are \(S_m(x,y) = \max \{ x,y \}\),\(S_b(x,y) = \min\{1, x+y \}\) and \(S_p(x, y) = x+y-xy \) for all \(x,y \in [0,1].\) \(S_m\) is standard union, \(S_b\) is bounded sum, \(S_p\) is algebraic sum.

Lemma 1. Let \(S\) be an \(s\)-norm. Then \[S(S(x,y),S(w,z))= S(S(x,w),S(y,z)), \] for all \(x,y,w,z\in [0,1].\)

3. Anti complex fuzzy subgroups under \(s\)-norms

Definition 5. Let \( G\) be a group and \( \mu: G \to [0,1] \) be a complex fuzzy set on \(G.\) Then \( \mu=re^{iw}\) is said to be an anti complex fuzzy subgroup of \( G \) under \(s\)-norm \(S\) if the following conditions hold:

• \(r(xy) \leq S(r(x),r(y)),\)
• \(r(x^{-1}) \leq r(x),\)
• \(w(xy) \leq \max \lbrace w(x),w(y) \rbrace,\)
• \(w(x^{-1}) \leq w(x),\)

for all \( x,y \in G. \) The set of all anti complex fuzzy subgroups of \( G \) under \(s\)-norm \(S\) is denoted by \(ACFS(G)\).

Example 2. Let \(G = \{ 0,a,b,c\} \) be the Klein's group. Every element is its own inverse, and the product of any two distinct non-identity elements is the remaining non-identity element. Thus the Klein 4-group admits the elegant presentation \(a^2=b^2=c^2=abc=0.\) Define \( r: G \to [0,1] \) by \begin{equation*} r(x) = \left\{ \begin{array}{rl} 0.35 &\text{if }x=a, \\ 0.45 &\text{if }x=b, \\ 0.65 &\text{if }x=c, \\ 0.85 &\text{if }x=0,\\ \end{array} \right. \end{equation*} and \( w: G \to [0,2\pi] \) by \begin{equation*} w(x) = \left\{ \begin{array}{rl} 0.4 \pi &\text{if }x=a, \\ 0.4 \pi &\text{if }x=b, \\ 0.5 \pi &\text{if }x=c, \\ 0.6 \pi &\text{if }x=0. \\ \end{array} \right. \end{equation*} Let \(S(a, b) = S_p(a, b) =a+b-ab \) for all \(a,b\in [0,1]\), then \(\mu(x) =r(x)e^{iw(x)}\in ACFS(G)\) for all \( x \in G. \)

Proposition 1. Let \(\mu=re^{iw}\in ACFS(G)\) such that \(C\) be idempotent \(s\)-norm. Then

• \(\mu(e) \leq \mu(x)\) for all \(x \in G \),
• \(\mu(x^n) \leq \mu(x)\) for all \(x \in G \) and \(n\geq 1,\)
• \(\mu(x)=\mu(x^{-1})\) for all \(x \in G.\)

Proof. Let \(\mu=re^{iw}\in ACFS(G)\) and \(x \in G \) and \(n\geq 1\). Then \begin{eqnarray*} (1) r(e)&=&r(xx^{-1}) \leq S(r(x),r(x^{-1})) \leq S(r(x),r(x))=r(x),\\ w(e)&=&w(xx^{-1}) \leq \max \lbrace w (x),w(x^{-1}) \rbrace \leq \max \lbrace w (x),w(x) \rbrace=w(x),\\ \mu(e)&=&r(e)e^{iw(x)} \leq r(x)e^{iw(x)} = \mu(x),\\ r(x^n)&=&r(\underbrace{xx...x}_{n}) \leq S(\underbrace{r(x),r(x),...,r(x)}_{n})= r(x),\\ w(x^n)&=&w(\underbrace{xx...x}_{n}) \leq \max \lbrace \underbrace{r(x),r(x),...,r(x)}_{n} \rbrace= w(x),\\ \mu(x^n)&=&r(x^n)e^{iw(x^n)} \leq r(x)e^{iw(x)}=\mu(x),\\ r(x)&=&r((x^{-1}))^{-1} \leq r(x^{-1}) \leq r(x),\\ r(x)&=& r(x^{-1}),\\ w(x)&=&w((x^{-1}))^{-1} \leq w(x^{-1}) \leq w(x),\\ w(x)&=&w(x^{-1}),\\ \mu(x)&=&r(x)e^{iw(x)}=r(x^{-1})e^{iw(x^{-1})}=\mu(x^{-1}). \end{eqnarray*}

Proposition 2. Let \(\mu=re^{iw}\in ACFS(G)\) and \(x\in G\) such that \(S\) be idempotent \(s\)-norm. Then \(\mu (xy)=\mu(y)\) \(\forall y \in G\) if and only if \(\mu (x)=\mu(e). \)

Proof. Let \(\mu(xy)= \mu(y)\) \(\forall y \in G.\) As \(y=e\), so \(\mu (x)=\mu(e). \) Conversely, let \(\mu (x)=\mu(e)\), then \(r(x)=r(e) \) and \(w(x)=w(e)\). From Proposition 1, we get \(r(x) \leq r(y) \) and \( r(x) \leq r(xy) \). Also \(w(x) \leq w(y) \) and \( w(x) \leq w(xy).\) Now \(r(xy) \leq S(r(x),r(y)) \leq S(r(y),r(y))=r(y)=r(x^{-1}xy) \leq S(r(x),r(xy)) \leq S(r(xy),r(xy))=r(xy).\)

Also \(w(xy) \leq \max \lbrace w(x),w(y) \rbrace \leq \max \lbrace w(y),w(y) \rbrace=w(y)=w(x^{-1}xy) \leq \max \lbrace w(x),w(xy) \rbrace \leq \max \lbrace w(xy),w(xy) \rbrace=w(xy).\) Therefore \(\mu(xy)=r(xy)e^{iw(xy)}=r(y)e^{iw(y)}=\mu(y).\)

Definition 6. Let \(G\) be a set and \(\mu_{1}=r_{1}e^{iw_{1}}\), \(\mu_{2}=r_{2}e^{iw_{2}}\) be two complex fuzzy sets on \(G.\) Denote the composition of \( \mu_{1} \) and \( \mu_{2} \) as \( \mu_{1} \circ \mu_{2}=(r_{1} \circ r_{2})e^{i(w_{1} \circ w_{2})}\) such that \( r_{1} \circ r_{2} : G \to [0,1] \) and \( w_{1} \circ w_{2}: G \to [0,2\pi] \) and define by \((\mu_{1} \circ \mu_{2})(x)=(r_{1} \circ r_{2})(x)e^{i(w_{1} \circ w_{2})(x)}(x)\) such that \begin{equation*} (r_{1} \circ r_{2})(x) = \left\{ \begin{array}{rl} \inf\limits_{x=ab}S(r_{1}(a),r_{2}(b)) &\text{if }x=ab, \\ 0 &\text{if }x \neq ab, \end{array} \right. \end{equation*} and \begin{equation*} (w_{1} \circ w_{2})(x) = \left\{ \begin{array}{rl} \max\limits_{x=ab} \lbrace w_{1}(a),w_{2}(b) \rbrace &\text{if }x=ab, \\ 0 &\text{if }x \neq ab, \end{array} \right. \end{equation*} thus \((\mu_{1} \circ \mu_{2})(x)=\inf_{x=ab}S(r_{1}(a),r_{2}(b))e^.{i \max_{x=ab}} \{ w_{1}(a),w_{2}(b) \} .\)

Proposition 3. Let \(\mu^{-1}\) be the inverse of \(\mu\) such that \(\mu^{-1}(x)=\mu (x^{-1}).\) Then \(\mu\in ACFS(G)\) if and only if \(\mu\) satisfies the following conditions:

• (1) \( \mu \leq \mu \circ \mu ;\)
• (2) \(\mu^{-1}=\mu.\)

Proof. Let \(x,y,z \in G\) with \(x=yz\) and \(\mu\in ACFS(G).\) Then \begin{eqnarray*} r(x)&=&r (yz) \leq S(r(y),r(z))=(r \circ r)(x),\\ w(x)&=&w (yz) \leq \max \lbrace r(y),r(z) \rbrace=(w \circ w)(x),\\ \mu(x)&=&r(x)e^{iw(x)} \leq (r \circ r)(x)e^{i(w \circ w)(x)}=(\mu \circ \mu)(x), \end{eqnarray*} so \(\mu\leq \mu \circ \mu.\)

Also from Proposition 1, for all \( x \in G \), we have that \( \mu^{-1}(x)=\mu(x^{-1})=\mu(x) \) and so \(\mu^{-1}=\mu.\) Conversely let \( \mu \leq \mu \circ \mu \) and \(\mu^{-1}=\mu. \) We prove that \(\mu\in ACFS(G).\) As \( \mu \leq \mu \circ \mu \) so \(r(x)\leq (r \circ r)(x)\) and \(w(x)\leq (w \circ w)(x)\). Thus

\[r(yz)=r(x)\leq (r \circ r)(x)=\inf_{x=yz}S(r(y),r(z))\leq S(r(y),r(z)) \] and \[w(yz)=w(x)\geq (w \circ w)(x)=\max_{x=yz} \lbrace w(y),w(z) \rbrace \geq \lbrace w(y),w(z) \rbrace. \] Since \(\mu^{-1}=\mu\) so \( r^{-1}(x) =r(x)\) and \( w^{-1}(x) =w(x).\) Therefore \( r(x^{-1})=r^{-1}(x) =r(x) \) and \( w(x^{-1})=w^{-1}(x) =w(x).\) Then \(\mu\in ACFS(G).\)

Corollary 1. Let \(\mu_{1},\mu_{2} \in ACFS(G)\) and \(G\) be commutative group. Then \(\mu_{1} \circ \mu_{2} \in ACFS(G)\) if and only if \(\mu_{1}\circ \mu_{2} = \mu_{2}\circ \mu_{1}.\)

Proof. As \(\mu_{1},\mu_{2} \in CFST(G)\) and \(\mu_{1} \circ \mu_{2} \in ACFS(G)\), so from Proposition 3, we get \(\mu^{-1}_{1} =\mu_{1} \) and \(\mu^{-1}_{2} =\mu_{2} \) and \( (\mu_{2}\circ \mu_{1})^{-1}=\mu_{2}\circ \mu_{1}. \) Then \( \mu_{1}\circ \mu_{2}=\mu^{-1}_{1}\circ \mu^{-1}_{2}=(\mu_{2}\circ \mu_{1})^{-1}=\mu_{2}\circ \mu_{1}.\) Conversely, let \(\mu_{1}\circ \mu_{2} = \mu_{2}\circ \mu_{1}\), then \((\mu_{1} \circ \mu_{2})\circ(\mu_{1} \circ \mu_{2})=\mu_{1} \circ (\mu_{2} \circ \mu_{1}) \circ \mu_{2}= \mu_{1} \circ (\mu_{1} \circ \mu_{2}) \circ \mu_{2}= (\mu_{1} \circ \mu_{1}) \circ (\mu_{2} \circ \mu_{2}) \geq \mu_{1} \circ \mu_{2}.\) Also \((\mu_{1} \circ \mu_{2})^{-1} =( \mu_{2} \circ \mu_{1})^{-1}=\mu_{1}^{-1} \circ \mu_{2}^{-1}=\mu_{1} \circ \mu_{2}.\) Then the Proposition 3 gives us that \(\mu_{1} \circ \mu_{2} \in ACFS(G).\)

Definition 7. Let \(\mu_{1}=r_{1}e^{iw_{1}} \in ACFS(G)\) and \(\mu_{2}=r_{2}e^{iw_{2}} \in ACFS(G).\) Define the intersection \(\mu_{1} \cap \mu_{2}\) as \( \mu_{1} \cap \mu_{2} = r_{1}e^{iw_{1}} \cap r_{2}e^{iw_{2}}=(r_{1} \cap r_{2})e^{i(w_{1} \cap w_{2})}\) such that \(r_{1} \cap r_{2}: G \to [0,1] \) and \( w_{1} \cap w_{2}: G \to [0,2\pi] \) and for all \( x \in G \), define \((r_{1} \cap r_{2})(x)=S(r_{1}(x),r_{2}(x))\) and \((w_{1} \cap r_{2})(x)=\max \lbrace w_{1}(x),w_{2}(x) \rbrace.\)

Proposition 4. Let \(\mu_{1}=r_{1}e^{iw_{1}} \in ACFS(G)\) and \(\mu_{2}=r_{2}e^{iw_{2}} \in ACFS(G).\) Then \(\mu_{1} \cap \mu_{2} \in ACFS(G).\)

Proof.

• (1) Let \( g_{1},g_{2} \in G.\) Then \begin{eqnarray*} (r_{1} \cap r_{2})(g_{1}g_{2})&=&S(r_{1}(g_{1}g_{2}) ,r_{2}(g_{1}g_{2})) \leq S(S(r_{1}(g_{1}) ,r_{1}(g_{2})),S(r_{2}(g_{1}) ,r_{2}(g_{2})))\\ &=&S(S(r_{1}(g_{1}) ,r_{2}(g_{1})),T(r_{1}(g_{2}) ,r_{2}(g_{2}))) \\ &=&S((r_{1} \cap r_{2})(g_{1}),(r_{1} \cap r_{2})(g_{2})), \end{eqnarray*} and thus \((r_{1} \cap r_{2})(g_{1}g_{2}) \leq S((r_{1} \cap r_{2})(g_{1}),(r_{1} \cap r_{2})(g_{2})).\)
• (2) If \( g \in G,\) then \begin{eqnarray*} (r_{1} \cap r_{2})(g^{-1})=S(r_{1}(g^{-1}),r_{2}(g^{-1})) &\leq& S(r_{1}(g),r_{2}(g))=(r_{1} \cap r_{2})(g)\\ (r_{1} \cap r_{2})(g^{-1}) &\geq& (r_{1} \cap r_{2})(g). \end{eqnarray*}
• (3) Let \( g_{1},g_{2} \in G.\) Then \begin{eqnarray*} (w_{1} \cap w_{2})(g_{1}g_{2})&=&\max \lbrace w_{1}(g_{1}g_{2}) ,w_{2}(g_{1}g_{2}) \rbrace\\ &\leq& \max \lbrace \max \lbrace w_{1}(g_{1}) ,w_{1}(g_{2}) \rbrace,\max \lbrace w_{2}(g_{1}) ,w_{2}(g_{2}) \rbrace \rbrace\\ &=&\max \lbrace \max \lbrace w_{1}(g_{1}) ,w_{2}(g_{1}) \rbrace,\max \lbrace w_{1}(g_{2}) ,w_{2}(g_{2})) \rbrace \\ &=&\max \lbrace (w_{1} \cap w_{2})(g_{1}),(w_{1} \cap w_{2})(g_{2}))\\ (w_{1} \cap w_{2})(g_{1}g_{2}) &\leq& \max \lbrace (w_{1} \cap w_{2})(g_{1}),(w_{1} \cap w_{2})(g_{2})). \end{eqnarray*}
• (4) Let \( g \in G \) so \begin{eqnarray*} (w_{1} \cap w_{2})(g^{-1})&=&\max \lbrace w_{1}(g^{-1}), w_{21}(g^{-1}) \rbrace\\ &\leq& \max \lbrace w_{1}(g), w_{2}(g) \rbrace\\ &=& (w_{1} \cap w_{2})(g)\\ (w_{1} \cap w_{2})(g^{-1}) &\leq& (w_{1} \cap w_{2})(g). \end{eqnarray*}

Thus from (1)-(4) we give that \(\mu_{1} \cap \mu_{2} \in ACFS(G).\)

Corollary 2. Let \(I_{n}=\{1,2,...,n\}.\) If \(\{\mu_{i}\hspace{0.1cm} | \hspace{0.1cm}i\in I_{n}\} \subseteq ACFS(G)\) then \(\mu=\cap_{i\in I_{n}}\mu_{i}\in ACFS(G).\)

Definition 8. \(\mu\in ACFS(G)\) is called normal if for all \(x,y\in G\) we have that \(\mu(xyx^{-1}) = \mu (y).\) The set of all normal anti complex fuzzy subgroups of \( G \) under \(s\)-norm \(S\) ia denoted by \(NACFS(G)\).

Proposition 5. Let \(\mu_{1}=r_{1}e^{iw_{1}} \in NACFS(G)\) and \(\mu_{2}=r_{2}e^{iw_{2}} \in NACFS(G).\) Then \(\mu_{1} \cap \mu_{2} \in NACFS(G).\)

Proof. From Proposition 4 we will have that \(\mu_{1} \cap \mu_{2} \in ACFS(G).\) Let \( g_{1} , g_{2} \in G \) then \begin{eqnarray*} (r_{1} \cap r_{2})(g_{1}g_{2}g^{-1}_{1})&=&S(r_{1} (g_{1}g_{2}g^{-1}_{1}),r_{2} (g_{1}g_{2}g^{-1}_{1}))=S(r_{1} (g_{2}),r_{2} (g_{2}))=(r_{1} \cap r_{2})(g_{2})\\ (w_{1} \cap w_{2})(g_{1}g_{2}g^{-1}_{1})&=&\max \lbrace w_{1} (g_{1}g_{2}g^{-1}_{1}),w_{2} (g_{1}g_{2}g^{-1}_{1}) \rbrace=\max \lbrace w_{1} (g_{2}),w_{2} (g_{2}) \rbrace =(w_{1} \cap w_{2})(g_{2})\\ (\mu_{1} \cap \mu_{2})(g_{1}g_{2}g^{-1}_{1})&=&(r_{1} \cap r_{2})(g_{1}g_{2}g^{-1}_{1})e^{i(w_{1} \cap w_{2})(g_{1}g_{2}g^{-1}_{1})}=(r_{1} \cap r_{2})(g_{2})e^{i(w_{1} \cap w_{2})(g_{2})}=(\mu_{1} \cap \mu_{2})(g_{2}) \end{eqnarray*} and therefore \(\mu_{1} \cap \mu_{2} \in NACFS(G).\)

Corollary 3. Let \(I_{n}=\{1,2,...,n\}.\) If \(\{\mu_{i}\hspace{0.1cm} | \hspace{0.1cm}i\in I_{n}\} \subseteq NACFS(G)\), then \(\mu=\cap_{i\in I_{n}}\mu_{i}\in NACFS(G).\)

Definition 9. Let \(\mu_{1}=r_{1}e^{iw_{1}} \in ACFS(G)\) and \(\mu_{2}=r_{2}e^{iw_{2}} \in ACFS(G)\) such that \( \mu_{1} \subseteq \mu_{2}.\) We say that \(\mu_{1}\) is normal of the \(\mu_{2}\), written \(\mu_{1} \bowtie \mu_{2}\), if \(r_{1}(g_{1}g_{2}g_{1}^{-1}) \leq S(r_{1}(g_{2}), r_{2}(g_{1})) \hspace{0.1cm} and \hspace{0.1cm} w_{1}(g_{1}g_{2}g_{1}^{-1}) \leq \max \lbrace w_{1}(g_{2}), w_{2}(g_{1}) \rbrace \) for all \( g_{1} , g_{2} \in G. \)

Proposition 6. If \(S\) be idempotent \(s\)-norm, then every \(\mu=re^{iw} \in ACFS(G)\) will be normal of itself.

Proof. Let \( g_{1},g_{2} \in G \) and \(\mu=re^{iw} \in ACFS(G),\) then \begin{eqnarray*} r(g_{1}g_{2}g^{-1}_{1}) &\leq& S(r(g_{1}) ,r(g_{2}g^{-1}_{1}))\\ &\leq& S(r(g_{1}),S(r(g_{2}),r(g^{-1}_{1})))\\ &\leq& S(r(g_{1}),S(r(g_{2}),r(g_{1})))\\ &=&S(r(g_{2}),S(r(g_{1}),r(g_{1})))\\ &=&S(r(g_{2}),r(g_{1}))\\ (g_{1}g_{2}g^{-1}_{1}) &\leq& S(r(g_{2}),r(g_{1})).\end{eqnarray*} Also \begin{eqnarray*} w(g_{1}g_{2}g^{-1}_{1}) &\leq& \max \lbrace w(g_{1}) ,w(g_{2}g^{-1}_{1})) \rbrace\\ &\leq& \max \lbrace w(g_{1}),\max \lbrace w(g_{2}),w(g^{-1}_{1}) \rbrace \rbrace \\ &\leq& \max \lbrace w(g_{1}),\max \lbrace w(g_{2}),w(g_{1})))\\ &=&\max \lbrace w(g_{2}),\max \lbrace w(g_{1}),w(g_{1}) \rbrace \rbrace\\ & =&\max \lbrace w(g_{2}),w(g_{1}) \rbrace \\ w(g_{1}g_{2}g^{-1}_{1}) &\leq& \max \lbrace w(g_{2}),w(g_{1}) \rbrace.\end{eqnarray*} Therefore \( \mu=re^{iw} \bowtie \mu=re^{iw}.\)

Proposition 7. Let \(\mu_{1} =r_{1}e^{iw_{1}} \in NACFS(G)\) and \(\mu_{2} =r_{2}e^{iw_{2}} \in ACFS(G)\) such that \(S\) be idempotent \(s\)-norm, then \(\mu_{1} \cap\mu_{2} \bowtie \mu_{2}.\)

Proof. As Proposition 4 \((\mu_{1} \cap\mu_{2}) \leq \mu_{2}\) and \((\mu_{1} \cap\mu_{2})\in ACFS(G).\) Let \( g_{1},g_{2} \in G \) and \( \mu_{1} \cap\mu_{2}=(r_{1} \cap r_{2})e^{i(w_{1} \cap w_{2})},\) then \begin{eqnarray*} (r_{1} \cap r_{2})(g_{1}g_{2}g^{-1}_{1})&=&S(r_{1}(g_{1}g_{2}g^{-1}_{1}), r_{2}(g_{1}g_{2}g^{-1}_{1}))\\ &=&S(r_{1}(g_{2}), r_{2}(g_{1}g_{2}g^{-1}_{1}))\\ &\leq& S(r_{1}(g_{2}),S( r_{2}(g_{1}g_{2}), r_{2}(g^{-1}_{1})))\\ &\leq& S(r_{1}(g_{2}),S( r_{2}(g_{1}g_{2}), r_{2}(g_{1})))\\ &\leq& S(r_{1}(g_{2}),S(S(r_{2}(g_{1}),r_{2}(g_{2})), r_{2}(g_{1})))\\ &=&S(r_{1}(g_{2}),S(S(r_{2}(g_{1}),r_{2}(g_{1})), r_{2}(g_{2})))\\ &=&S(r_{1}(g_{2}),S(r_{2}(g_{1}), r_{2}(g_{2})))\\ &=&S(S(r_{1}(g_{2}),r_{2}(g_{2})),r_{2}(g_{1}))\\ &=&S((r_{1} \cap r_{2})(g_{2}),r_{2}(g_{1})),\end{eqnarray*} and thus \( (r_{1} \cap r_{2})(g_{1}g_{2}g^{-1}_{1}) \leq S((r_{1} \cap r_{2})(g_{2}),r_{2}(g_{1})).\)

Also

\begin{eqnarray*} (w_{1} \cap w_{2})(g_{1}g_{2}g^{-1}_{1})&=&\max \lbrace w_{1}(g_{1}g_{2}g^{-1}_{1}), w_{2}(g_{1}g_{2}g^{-1}_{1}) \rbrace\\ &=&\max \lbrace w_{1}(g_{2}), w_{2}(g_{1}g_{2}g^{-1}_{1}) \rbrace \\ &\leq& \max \lbrace w_{1}(g_{2}),\max \lbrace w_{2}(g_{1}g_{2}), w_{2}(g^{-1}_{1}) \rbrace \rbrace\\ &\leq& \max \lbrace w_{1}(g_{2}),\max \lbrace w_{2}(g_{1}g_{2}), w_{2}(g_{1}) \rbrace \rbrace \\ &\leq& \max \lbrace w_{1}(g_{2}),\max \lbrace \max \lbrace w_{2}(g_{1}),w_{2}(g_{2}) \rbrace, w_{2}(g_{1}) \rbrace \rbrace\\ &=&\max \lbrace w_{1}(g_{2}),\max \lbrace \max \lbrace w_{2}(g_{1}),w_{2}(g_{1})) \rbrace, w_{2}(g_{2}) \rbrace \rbrace \\ &=&\max \lbrace w_{1}(g_{2}),\max \lbrace w_{2}(g_{1}), w_{2}(g_{2}) \rbrace \rbrace \\ &=&\max \lbrace \max \lbrace w_{1}(g_{2}),w_{2}(g_{2}) \rbrace,w_{2}(g_{1}) \rbrace \\ &=&\max \lbrace (w_{1} \cap w_{2})(g_{2}),w_{2}(g_{1}) \rbrace,\end{eqnarray*} and then \( (w_{1} \cap w_{2})(g_{1}g_{2}g^{-1}_{1}) \leq \max \lbrace (w_{1} \cap w_{2})(g_{2}),w_{2}(g_{1}) \rbrace.\) Therefore \( \mu_{1} \cap\mu_{2}=(r_{1} \cap r_{2})e^{i(w_{1} \cap w_{2})} \bowtie \mu_{2}. \)

Proposition 8. Let \(\mu_{1} =r_{1}e^{iw_{1}} \in ACFS(G)\) and \(\mu_{2} =r_{2}e^{iw_{2}} \in ACFS(G)\) and \(\mu_{3} =r_{3}e^{iw_{3}} \in ACFS(G)\) and \(S\) be idempotent \(s\)-norm. If \( \mu_{1} \bowtie \mu_{3}\) and \( \mu_{2} \bowtie \mu_{3},\) then \( \mu_{1} \cap \mu_{2} \bowtie \mu_{3}.\)

Proof. By Proposition 4, we have \( \mu_{1} \cap \mu_{2} \in ACFS(G)\) and \( \mu_{1} \cap \mu_{2} \leq \mu_{3}. \) Let \( g_{1},g_{2} \in G.\) As \( \mu_{1}\bowtie \mu_{3}\), so \( r_{1}(g_{1}g_{2}g^{-1}_{1}) \leq S(r_{1}(g_{2}),r_{3}(g_{1}))\) and \( w_{1}(g_{1}g_{2}g^{-1}_{1}) \leq \max \lbrace r_{1}(g_{2}),r_{3}(g_{1}) \rbrace\) and as \( \mu_{2} \bowtie \mu_{3}\) so \( r_{2}(g_{1}g_{2}g^{-1}_{1}) \leq S(r_{2}(g_{2}),r_{3}(g_{1}))\) and \( w_{2}(g_{1}g_{2}g^{-1}_{1}) \leq \max \lbrace w_{2}(g_{2}),w_{3}(g_{1}) \rbrace.\) Now \begin{eqnarray*} (r_{1} \cap r_{2})(g_{1}g_{2}g^{-1}_{1})&=&S(r_{1} (g_{1}g_{2}g^{-1}_{1}),r_{2} (g_{1}g_{2}g^{-1}_{1}))\\ &\leq& S(S(r_{1}(g_{2}),r_{3}(g_{1})),S(r_{2}(g_{2}),r_{3}(g_{1})))\\ &=&S(S(r_{1}(g_{2}),r_{2}(g_{2})),S(r_{3}(g_{1}),r_{3}(g_{1})))\\ &=&S(S(r_{1}(g_{2}),r_{2}(g_{2})),r_{3}(g_{1}))\\ &=&S((r_{1} \cap r_{2})(g_{2}),r_{3}(g_{1})),\end{eqnarray*} and then \( (r_{1} \cap r_{2})(g_{1}g_{2}g^{-1}_{1}) \leq S((r_{1} \cap r_{2})(g_{2}),r_{3}(g_{1})).\) Also\begin{eqnarray*} (w_{1} \cap w_{2})(g_{1}g_{2}g^{-1}_{1})&=&\max \lbrace w_{1} (g_{1}g_{2}g^{-1}_{1}),w_{2} (g_{1}g_{2}g^{-1}_{1}) \rbrace\\ &\geq& \max \lbrace \max \lbrace w_{1}(g_{2}),w_{3}(g_{1}) \rbrace ,\max \lbrace w_{2}(g_{2}),w_{3}(g_{1}) \rbrace \rbrace\\ &=&\max \lbrace \max \lbrace w_{1}(g_{2}),w_{2}(g_{2}) \rbrace ,\max \lbrace w_{3}(g_{1}),w_{3}(g_{1})\rbrace \rbrace \\ &=&\max \lbrace \max \lbrace w_{1}(g_{2}),w_{2}(g_{2}) \rbrace,w_{3}(g_{1}) \rbrace\\ &=&\max \lbrace (w_{1} \cap w_{2})(g_{2}),w_{3}(g_{1}) \rbrace,\end{eqnarray*} and so \( (w_{1} \cap w_{2})(g_{1}g_{2}g^{-1}_{1}) \leq \max \lbrace (w_{1} \cap w_{2})(g_{2}),w_{3}(g_{1}) \rbrace.\) Thus \( \mu_{1} \cap\mu_{2}=(r_{1} \cap r_{2})e^{i(w_{1} \cap w_{2})} \bowtie \mu_{3}. \)

Corollary 4. Let \(I_{n}=\{1,2,...,n\}\) and \(\{\mu_{i}\hspace{0.1cm} | \hspace{0.1cm}i\in I_{n}\} \subseteq ACFS(G)\) such that \(\{\mu_{i}\hspace{0.1cm} | \hspace{0.1cm}i\in I_{n}\} \bowtie \xi.\) Then \(\mu=\cap_{i\in I_{n}}\mu_{i} \bowtie \xi.\)

4. Group homomorphisms and anti complex fuzzy subgroups under \(s\)-norms

Definition 10. Let \(f:G \to H\) be a mapping such that \( \mu_{G}=r_{G}e^{iw_{G}}\) and \( \mu_{H}=r_{H}e^{iw_{H}}\) be two complex fuzzy sets on \(G\) and \(H,\) respectively. Define \( f(\mu_{G}): H \to [0,1]\) as \(f(\mu_{G})=f(r_{G}e^{iw_{G}})=f(r_{G})e^{if(w_{G})}\) such that for all \( h \in H \) we define \(f(r_{G})(h)=\inf \lbrace r_{G}(g) \hspace{0.1cm}|\hspace{0.1cm} g \in G, f(g)=h \rbrace\) and \(f(w_{G})(h)=\inf \lbrace w_{G}(g) \hspace{0.1cm}|\hspace{0.1cm} g \in G, f(g)=h \rbrace.\) Also define \( f^{-1}(\mu_{H}): G \to [0,1]\) as \( f^{-1}(r_{H}e^{iw_{H}})=f^{-1}(r_{H}) e^{if^{-1}(w_{H})}\) such that for all \(g \in G\), we define \(f^{-1}(r_{H}e^{iw_{H}})(g)=r_{H}(f(g))e^{iw_{H}(f(g))}.\)

Proposition 9. Let \(\mu_{G}=r_{G}e^{iw_{G}} \in ACFS(G)\) and \(f: G \to H\) be a group homomorphism, then \(f(\mu_{G})\in ACFS(H).\)

Proof.

• (1) Let \(h_{1} , h_{2}\in H\) and \(g_{1},g_{2} \in G\) such that \( h_{1}=f(g_{1}) \) and \( h_{2}=f(g_{2}).\) Then \begin{eqnarray*} f(r_{G})(h_{1}h_{2})&=&\inf \lbrace r_{G}(g_{1}g_{2}) \hspace{0.1cm}|\hspace{0.1cm} g_{1},g_{2} \in G, f(g_{1})=h_{1}, f(g_{2})=h_{2} \rbrace\\ &\leq& \inf \lbrace S( r_{G}(g_{1}),r_{G}(g_{2})) \hspace{0.1cm}|\hspace{0.1cm} g_{1},g_{2} \in G, f(g_{1})=h_{1}, f(g_{2})=h_{2} \rbrace\\ &=&S(\inf \lbrace r_{G}(g_{1}) \hspace{0.1cm}|\hspace{0.1cm} g_{1} \in G, f(g_{1})=h_{1} \rbrace,\inf \lbrace r_{G}(g_{2}) \hspace{0.1cm}|\hspace{0.1cm} g_{2} \in G, f(g_{2})=h_{2} \rbrace)\\ &=&S(f(r_{G})(h_{1}),f(r_{G})(h_{2})),\end{eqnarray*} and so \( f(r_{G})(h_{1}h_{2}) \leq S(f(r_{G})(h_{1}),f(r_{G})(h_{2})). \)
• (2) Let \(h\in H\) and \(g \in G\) such that \( h=f(g).\) Then \begin{eqnarray*} f(r_{G})(h^{-1})&=&\inf \lbrace r_{G}(g^{-1}) \hspace{0.1cm}|\hspace{0.1cm} g^{-1} \in G, f(g^{-1})=h^{-1} \rbrace \\ &\leq& \inf \lbrace r_{G}(g) \hspace{0.1cm}|\hspace{0.1cm} g \in G, f^{-1}(g)=h^{-1} \rbrace \\ &=&\inf \lbrace r_{G}(g) \hspace{0.1cm}|\hspace{0.1cm} g \in G, f(g)=h \rbrace =f(r_{G})(h),\end{eqnarray*} and so \( f(r_{G})(h^{-1}) \leq f(r_{G})(h). \)
• (3) Let \(h_{1} , h_{2}\in H\) and \(g_{1},g_{2} \in G\) such that \( h_{1}=f(g_{1}) \) and \( h_{2}=f(g_{2}).\) Then \begin{eqnarray*} f(w_{G})(h_{1}h_{2})&=&\inf \lbrace w_{G}(g_{1}g_{2}) \hspace{0.1cm}|\hspace{0.1cm} g_{1},g_{2} \in G, f(g_{1})=h_{1}, f(g_{2})=h_{2} \rbrace \\ &\leq& \inf \lbrace \max \lbrace w_{G}(g_{1}),w_{G}(g_{2}) \rbrace \hspace{0.1cm}|\hspace{0.1cm} g_{1},g_{2} \in G, f(g_{1})=h_{1}, f(g_{2})=h_{2} \rbrace \\ &=&\max \lbrace \inf \lbrace w_{G}(g_{1}) \hspace{0.1cm}|\hspace{0.1cm} g_{1} \in G, f(g_{1})=h_{1} \rbrace,\inf \lbrace w_{G}(g_{2}) \hspace{0.1cm}|\hspace{0.1cm} g_{2} \in G, f(g_{2})=h_{2} \rbrace) \\ &=&\max \lbrace f(w_{G})(h_{1}),f(w_{G})(h_{2}) \rbrace,\end{eqnarray*} and thus \( f(w_{G})(h_{1}h_{2}) \geq \max \lbrace f(w_{G})(h_{1}),f(w_{G})(h_{2}) \rbrace. \)
• (4) Let \(h\in H\) and \(g \in G\) such that \( h=f(g).\) Now \begin{eqnarray*} f(w_{G})(h^{-1})&=&\inf \lbrace w_{G}(g^{-1}) \hspace{0.1cm}|\hspace{0.1cm} g^{-1} \in G, f(g^{-1})=h^{-1} \rbrace \\ &\geq& \inf \lbrace w_{G}(g) \hspace{0.1cm}|\hspace{0.1cm} g^{-1} \in G, f^{-1}(g)=h^{-1} \rbrace \\ &=&\inf \lbrace w_{G}(g) \hspace{0.1cm}|\hspace{0.1cm} g \in G, f(g)=h \rbrace =f(w_{G})(h),\end{eqnarray*} and therefore \( f(w_{G})(h^{-1}) \geq f(w_{G})(h).\)

Thus (1) - (4) mean that \(f(\mu_{G})=f(r_{G}e^{iw_{G}})=f(r_{G})e^{if(w_{G})} \in ACFS(H).\)

Proposition 10. Let \(\mu_{H}=r_{H}e^{iw_{H}} \in ACFS(H)\) and \(f: G \to H\) be a group homomorphism, then \(f^{-1}(\mu_{H})\in ACFS(G).\)

Proof.

• (1) Let \( g_{1},g_{2} \in G \), then \( f^{-1}(r_{H})(g_{1}g_{2})=r_{H}(f(g_{1}g_{2}))=r_{H}(f(g_{1})f(g_{2})) \leq S(r_{H}(f(g_{1})) ,r_{H}(f(g_{2})))=S(f^{-1}(r_{H})(g_{1}),f^{-1}(r_{H})(g_{2})),\) and then \( f^{-1}(r_{H})(g_{1}g_{2}) \leq S(f^{-1}(r_{H})(g_{1}),f^{-1}(r_{H})(g_{2})). \)
• (2) Let \( g \in G \), then \(f^{-1}(r_{H})(g^{-1})=r_{H}(f(g^{-1}))=r_{H}(f^{-1}(g)) \leq r_{H}(f(g)) =f^{-1}(r_{H})(g),\) and thus \( f^{-1}(r_{H})(g^{-1}) \leq f^{-1}(r_{H})(g). \)
• (3) Let \( g_{1},g_{2} \in G \), so \( f^{-1}(w_{H})(g_{1}g_{2})=w_{H}(f(g_{1}g_{2}))=w_{H}(f(g_{1})f(g_{2})) \leq \max \lbrace w_{H}(f(g_{1})) ,w_{H}(f(g_{2})) \rbrace = \max \lbrace f^{-1}(w_{H})(g_{1}),f^{-1}(w_{H})(g_{2}) \rbrace,\) and then \( f^{-1}(w_{H})(g_{1}g_{2}) \leq \max \lbrace f^{-1}(w_{H})(g_{1}),f^{-1}(w_{H})(g_{2})) \rbrace. \)
• (4) Let \( g \in G \), then \( f^{-1}(w_{H})(g^{-1})=w_{H}(f^{-1}(g))\leq w_{H}(f(g)) =f^{-1}(w_{H})(g)\) and then \( f^{-1}(w_{H})(g^{-1}) \leq f^{-1}(w_{H})(g). \)

Therefore (1)-(4) give us \(f^{-1}(r_{H}e^{iw_{H}})(g)=r_{H}(f(g))e^{iw_{H}(f(g))} \in ACFS(G).\)

Proposition 11. Let \(\mu_{G}=r_{G}e^{iw_{G}} \in NACFS(G)\) and \(f: G \to H\) be a group homomorphism. Then \(f(\mu_{G})\in NACFS(H).\)

Proof. From Proposition 9, we have \(f(\mu_{G})\in ACFS(H).\) Let \(g_{1},g_{2} \in G\) and \(h_{1},h_{2} \in H\) such that \( f(g_{1})=h_{1} \) and \( f(g_{2})=h_{2}.\) Now \begin{eqnarray*} f(r_{G})(h_{1}h_{2}h^{-1}_{1})&=&\inf \lbrace r_{G}(g_{1}g_{2}g^{-1}_{1}) \hspace{0.1cm}|\hspace{0.1cm} f(g_{1}g_{2}g^{-1}_{1})=h_{1}h_{2}h^{-1}_{1} \rbrace\\ &=&\inf \lbrace r_{G}(g_{2}) \hspace{0.1cm}|\hspace{0.1cm} f(g_{1})f(g_{2})f(g^{-1}_{1})=h_{1}h_{2}h^{-1}_{1} \rbrace \\ &=&\inf \lbrace r_{G}(g_{2}) \hspace{0.1cm}|\hspace{0.1cm} f(g_{1})f(g_{2})f^{-1}(g_{1})=h_{1}h_{2}h^{-1}_{1} \rbrace \\ &=&\inf \lbrace r_{G}(g_{2}) \hspace{0.1cm}|\hspace{0.1cm} f(g_{2})=h_{2} \rbrace =f(r_{G})(h_{2}).\end{eqnarray*} Also\begin{eqnarray*} f(w_{G})(h_{1}h_{2}h^{-1}_{1})&=&\inf \lbrace w_{G}(g_{1}g_{2}g^{-1}_{1}) \hspace{0.1cm}|\hspace{0.1cm} f(g_{1}g_{2}g^{-1}_{1})=h_{1}h_{2}h^{-1}_{1} \rbrace \\ &=&\inf \lbrace w_{G}(g_{2}) \hspace{0.1cm}|\hspace{0.1cm} f(g_{1})f(g_{2})f(g^{-1}_{1})=h_{1}h_{2}h^{-1}_{1} \rbrace \\ &=&\inf \lbrace w_{G}(g_{2}) \hspace{0.1cm}|\hspace{0.1cm} f(g_{1})f(g_{2})f^{-1}(g_{1})=h_{1}h_{2}h^{-1}_{1} \rbrace \\ &=&\inf \lbrace w_{G}(g_{2}) \hspace{0.1cm}|\hspace{0.1cm} f(g_{2})=h_{2} \rbrace =f(w_{G})(h_{2}).\end{eqnarray*} Then \(f(\mu_{G})(h_{1}h_{2}h^{-1}_{1})=f(r_{G})(h_{1}h_{2}h^{-1}_{1})e^{if(w_{G})(h_{1}h_{2}h^{-1}_{1})}=f(r_{G})(h_{2})e^{if(w_{G})(h_{2})}=f(\mu_{G})(h_{2}).\) and so \(f(\mu_{G})\in NACFS(H).\)

Proposition 12. Let \(\mu_{H}=r_{H}e^{iw_{H}} \in NACFS(H)\) and \(f: G \to H\) be a group homomorphism, then \(f^{-1}(\mu_{H})\in NACFS(G).\)

Proof. Using Proposition 10, we get \(f^{-1}(\mu_{H})\in ACFS(G).\) Let \( g_{1} , g_{2} \in G \), then \( f^{-1}(r_{H})(g_{1}g_{2}g^{-1}_{1})=r_{H}(f(g_{1}g_{2}g^{-1}_{1})) =r_{H}(f(g_{1})f(g_{2})f(g^{-1}_{1}))=r_{H}(f(g_{1})f(g_{2})f^{-1}(g_{1}))=r_{H}(f(g_{2}))=f^{-1}(r_{H})(g_{2}).\) Also \(f^{-1}(w_{H})(g_{1}g_{2}g^{-1}_{1})=w_{H}(f(g_{1}g_{2}g^{-1}_{1}))=w_{H}(f(g_{1})f(g_{2})f(g^{-1}_{1}))=w_{H}(f(g_{1})f(g_{2})f^{-1}(g_{1})) =w_{H}(f(g_{2}))=f^{-1}(w_{H})(g_{2}).\) Thus \(f^{-1}(\mu_{H})(g_{1}g_{2}g^{-1}_{1})=f^{-1}(r_{H})(g_{1}g_{2}g^{-1}_{1})e^{if^{-1}(w_{H})(g_{1}g_{2}g^{-1}_{1})} =f^{-1}(r_{H})(g_{2})e^{if^{-1}(w_{H})(g_{2})} =f^{-1}(\mu_{H})(g_{2}) \) and thus \(f^{-1}(\mu_{H})\in NACFS(G).\)

Proposition 13. Let \(\mu_{1} =r_{1}e^{iw_{1}} \in ACFS(G)\) and \(\mu_{2} =r_{2}e^{iw_{2}} \in ACFS(G)\) and \(f: G \to H\) be a group homomorphism. If \( \mu_{1} \bowtie \mu_{2},\) then \( f(\mu_{1}) \bowtie f(\mu_{2}).\)

Proof. We know that \( f(\mu_{1})=f(r_{1})e^{if(w_{1})}\) and \( f(\mu_{2})=f(r_{2})e^{if(w_{2})}.\) By Proposition 9, we have \( f(\mu_{1}) \in ACFS(H) \) and \( f(\mu_{2}) \in ACFS(H).\) Let \( g_{1},g_{2} \in G\) and \( h_{1},h_{2} \in H\) such that \( f(g_{1})=h_{1} \) and \( f(g_{2})=h_{2}.\) Since \( \mu_{1} \bowtie \mu_{2}\) so \( r_{1}(g_{1}g_{2}g^{-1}_{1}) \leq S(r_{1}(g_{2}),r_{2}(g_{1})) \) and \( w_{1}(g_{1}g_{2}g^{-1}_{1}) \leq \max \lbrace w_{1}(g_{2}),w_{2}(g_{1}) \rbrace. \) Now \begin{eqnarray*} f(r_{1})(h_{1}h_{2}h^{-1}_{1})&=&\inf \lbrace r_{1}(g_{1}g_{2}g^{-1}_{1}) \hspace{0.1cm}|\hspace{0.1cm} f(g_{1}g_{2}g^{-1}_{1})=h_{1}h_{2}h^{-1}_{1} \rbrace \\ &\leq& \inf \lbrace S(r_{1}(g_{2}) ,r_{2}(g_{1}) ) \hspace{0.1cm}|\hspace{0.1cm} f(g_{1})f(g_{2})f(g^{-1}_{1})=h_{1}h_{2}h^{-1}_{1} \rbrace \\ &=&\inf \lbrace T(r_{1}(g_{2}) ,r_{2}(g_{1}) ) \hspace{0.1cm}|\hspace{0.1cm} f(g_{1})f(g_{2})f^{-1}(g_{1})=h_{1}h_{2}h^{-1}_{1} \rbrace \\ &=&S(\inf \lbrace r_{1}(g_{2}) \hspace{0.1cm}|\hspace{0.1cm} f(g_{2})=h_{2} \rbrace,\inf \lbrace r_{2}(g_{1}) \hspace{0.1cm}|\hspace{0.1cm} f(g_{1})=h_{1} \rbrace)\\ &=&S(f(r_{1})(h_{2}),f(r_{2})(h_{1})),\end{eqnarray*} and then \( f(r_{1})(h_{1}h_{2}h^{-1}_{1}) \leq S(f(r_{1})(h_{2}),f(r_{2})(h_{1})).\) Also \begin{eqnarray*} f(w_{1})(h_{1}h_{2}h^{-1}_{1})&=&\inf \lbrace w_{1}(g_{1}g_{2}g^{-1}_{1}) \hspace{0.1cm}|\hspace{0.1cm} f(g_{1}g_{2}g^{-1}_{1})=h_{1}h_{2}h^{-1}_{1} \rbrace \\ &\leq& \inf \lbrace \max \lbrace w_{1}(g_{2}) ,w_{2}(g_{1}) \rbrace \hspace{0.1cm}|\hspace{0.1cm} f(g_{1})f(g_{2})f(g^{-1}_{1})=h_{1}h_{2}h^{-1}_{1} \rbrace \\ &=&\inf \lbrace \max \lbrace w_{1}(g_{2}) ,w_{2}(g_{1}) \rbrace \hspace{0.1cm}|\hspace{0.1cm} f(g_{1})f(g_{2})f^{-1}(g_{1})=h_{1}h_{2}h^{-1}_{1} \rbrace \\ &=&\max \lbrace \inf \lbrace w_{1}(g_{2}) \hspace{0.1cm}|\hspace{0.1cm} f(g_{2})=h_{2} \rbrace,\inf \lbrace w_{2}(g_{1}) \hspace{0.1cm}|\hspace{0.1cm} f(g_{1})=h_{1} \rbrace \rbrace \\ &=&\max \lbrace f(w_{1})(h_{2}),f(w_{2})(h_{1}) \rbrace,\end{eqnarray*} and so \(f(w_{1})(h_{1}h_{2}h^{-1}_{1}) \leq \max \lbrace f(w_{1})(h_{2}),f(w_{2})(h_{1}) \rbrace .\) Hence \( f(\mu_{1}) \bowtie f(\mu_{2}).\)

Proposition 14. Let \(\mu_{1} =r_{1}e^{iw_{1}} \in ACFS(H)\), \(\mu_{2} =r_{2}e^{iw_{2}} \in ACFS(H)\) and \(f: G \to H\) be a group homomorphism. If \( \mu_{1} \bowtie \mu_{2},\) then \( f^{-1}(\mu_{1}) \bowtie f^{-1}(\mu_{2}).\)

Proof. Let \( f^{-1}(\mu_{1})=f^{-1}(r_{1})e^{if^{-1}(w_{1})}\) and \( f^{-1}(\mu_{2})=f^{-1}(r_{2})e^{if^{-1}(w_{2})}\). From Proposition 10, we obtain \( f^{-1}(\mu_{1}) \in ACFS(G) \) and \( f^{-1}(\mu_{2}) \in ACFS(G).\) Let \( g_{1},g_{2} \in G \), then \begin{eqnarray*} f^{-1}(r_{1})(g_{1}g_{2}g^{-1}_{1})&=&r_{1}(f(g_{1}g_{2}g^{-1}_{1}))=r_{1}(f(g_{1})f(g_{2})f(g^{-1}_{1}))=r_{1}(f(g_{1})f(g_{2})f^{-1}(g_{1}))\\ &\leq& S(r_{1}(f(g_{2})),r_{2}(f(g_{1})))=S(f^{-1}(r_{1})(g_{2}),f^{-1}(r_{2})(g_{1}).\end{eqnarray*} Also \begin{eqnarray*}f^{-1}(w_{1})(g_{1}g_{2}g^{-1}_{1})&=&w_{1}(f(g_{1}g_{2}g^{-1}_{1}))=w_{1}(f(g_{1})f(g_{2})f(g^{-1}_{1}))=w_{1}(f(g_{1})f(g_{2})f^{-1}(g_{1}))\\ &\leq& \max \lbrace w_{1}(f(g_{2})),w_{2}(f(g_{1})) \rbrace=\max \lbrace f^{-1}(w_{1})(g_{2}),f^{-1}(w_{2})(g_{1} \rbrace.\end{eqnarray*} Therefore \( f^{-1}(\mu_{1}) \bowtie f^{-1}(\mu_{2}). \)

Acknowledgments

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

Conflict of Interests

The author declares no conflict of interest.