On the characterization of Pythagorean fuzzy subgroups

1.   Introduction

Group theory has various applications in many fields of Mathematics such as Algebraic geometry, Cryptography, Harmonic analysis, Algebraic number theory, etc. Uncertainty is a part of our daily life. There is an uncertainty in almost every problems we face day to day. In 1965, fuzzy set theory was first invented by Zadeh ^[1] to handle uncertainty in real life problems. In 1971 using the concept of fuzzy set, Rosenfeld ^[2] first defined the notion of fuzzy subetaoup. In 1979, using $t$-norm the notion of fuzzy subetaoup was redefined by Anthony and Sherwood ^[3,4]. Das ^[5] introduced the concept of fuzzy level subetaoup. In 1988, Choudhury et al. ^[6] proved various properties of fuzzy subetaoups and fuzzy homomorphism. In 1990, Dixit et al. ^[7] discussed fuzzy level subetaoups and union of fuzzy subetaoups. The notion of anti-fuzzy subetaoups was first proposed by Biswas ^[8]. Ajmal and Prajapati ^[9] gave the idea of fuzzy normal subetaoup, fuzzy coset and fuzzy quotient subetaoup. Chakraborty and Khare ^[10] studied various properties of fuzzy homomorphism. Many more results on fuzzy subetaoup ware introduced by Mukherjee ^[11,12] and Bhattacharya ^[13].

In recent years many researchers studied various properties of fuzzy subetaoups. In 2015, Tarnauceanu ^[14] classified fuzzy normal subetaoup of finite groups. In 2016, Onasanya ^[15] reviewed some anti fuzzy properties of fuzzy subetaoups. Shuaib ^[16] and Shaheryar ^[17] studied the properties of omicron fuzzy subetaoup and omicron anti fuzzy subetaoup. In 2018, Addis ^[18] developed fuzzy homomorphism theorems on groups. In 1986, Atanassov ^[19] invented Intuitionistic fuzzy set. In 1996, Intuitionistic fuzzy subetaoup was first studied by Biswas ^[20]. Zhan and Tan ^[21] introduced intuitionistic fuzzy $M$-group. Furthermore, researchers developed intuitionistic fuzzy subetaoup in many ways ^[22,23,24].

In 2013, Yager ^[25] introduced Pythagorean fuzzy set, where the sum of square of the membership degree and non membership degree lies between $0$ and $1$. Pythagorean fuzzy set is more fruitful in many decision making problems. This concept is perfectly designed to represent vagueness and uncertainty in mathematical way and to produce a formalized tool to handle imprecision to real problems. In 2018, Naz et al. ^[26] proposed a novel approach to decision making problem using Pythagorean fuzzy set. In 2019, Akram and Naz ^[27] applied complex Pythagorean fuzzy set in decision making problems. Ejegwa ^[28] gave an application of Pythagorean fuzzy set in career placements based on academic performance using max-min-max composition. Some results related to it were given by Peng ^[29] and Yang ^[30].

Pythagorean fuzzy set gives a modern way to model vagueness and uncertainty with high precision and accuracy compared to intuitionistic fuzzy set. Group symmetry plays a vital role to analyse molecule structures. Isotope molecules decays with a certain rate, so the fuzzy sense comes into it. If decay rate follows the criteria of Pythagorean fuzzy environment then we can not opted for intuitionistic fuzzy subetaoup to analyse the structure of that isotope at certain time. For this type of situations where we can not opted for intuitionistic fuzzy subetaoup, it is very important to introduce Pythagorean fuzzy subetaoup as a bigger class of intuitionistic fuzzy subetaoup. But till now no algebraic structure is defined on Pythagorean fuzzy environment. In this article, we proved that intuitionistic fuzzy subetaoup is a subclass of Pythagorean fuzzy subetaoup.

This paper is organized as follows: Pythagorean fuzzy subetaoup is described in Section 3. Pythagorean fuzzy coset and Pythagorean fuzzy normal subetaoup are discussed in Section 4. Pythagorean fuzzy level subetaoup and its properties are given in Section 5. Finally, effect of group homomorphism on Pythagorean fuzzy subetaoup is discussed in Section 6 and a conclusion is given in Section 7.

2.   Preliminaries

In this section, we recap some definitions and concepts which are very much important to develop later sections.

Definition 2.1. ^[1] Let $C$ be a crisp set. Then $\mu:C\rightarrow[0, 1]$ is called a fuzzy subset of $C$. Here $\mu(m)$ is called degree of membership of $m\in C$.

Definition 2.2. ^[2] Let $\mu:C\rightarrow[0, 1]$ be a fuzzy subset of a group $(C, \circ)$. Then $\mu$ is said to be a fuzzy subetaoup of $(C, \circ)$ if the following conditions hold:

$(i)$ $\mu(m\circ n)\geq\mu(m)\wedge\mu(n)$ $\forall m, n\in C$,

$(ii)$ $\mu(m^{-1})\geq\mu(m)$ $\forall m\in C$.

Definition 2.3. ^[19] Let $C$ be a crisp set. An intuitionistic fuzzy set (IFS) $I$ on $C$ is defined by $I = \{(m, \mu(m), \nu(m))|m\in C\}$ where $\mu(m)\in[0, 1]$ and $\nu(m)\in[0, 1]$ are the degree of membership and non membership of $m\in C$ respectively, which satisfy the condition $0\leq\mu(m)+\nu(m)\leq 1$ $\forall m\in C$.

Definition 2.4. ^[20] Let $I = \{(m, \mu(m), \nu(m))|m\in C\}$ be a IFS of a group $(C, \circ)$. Then $I$ is said to be an intuitionistic fuzzy subetaoup (IFSG) of $C$ if the following conditions hold:

$(i)$ $\mu(m\circ n)\geq \mu(m)\wedge\mu(n)$ and $\nu(m\circ n)\leq \nu(m)\vee\nu(n)$ $\forall m, n\in C$,

$(ii)$ $\mu(m^{-1})\geq \mu(m)$ and $\nu(m^{-1})\leq\nu(m)$ $\forall m\in C$.

In 2013, Yager ^[25] defined Pythagorean fuzzy subset (PFS) as a generalization of IFS.

Definition 2.5. ^[25] Let $C$ be a crisp set. A Pythagorean fuzzy set (PFS) $\psi$ on $C$ is defined by $\psi = \{(m, \mu(m), \nu(m))|m\in C\}$ where $\mu(m)\in[0, 1]$ and $\nu(m)\in[0, 1]$ are the degree of membership and non membership of $m\in C$ respectively, which satisfy the condition $0\leq\mu^2(m)+\nu^2(m)\leq 1$ $\forall m\in C$.

Some operations on PFSs ^[31] are stated below.

Let $\psi_1 = \{(m, \mu_1(m), \nu_1(m))|m\in C\}$ and $\psi_2 = \{(m, \mu_2(m), \nu_2(m))|m\in C\}$ be two PFSs of $C$. Then the following holds:

● $\psi_1\cup\psi_2 = \left\lbrace\left (m, \mu_1(m)\vee\mu_2(m), \nu_1(m)\wedge\nu_2(m)\right)|m\in C\right\rbrace$.

● $\psi_1\cap\psi_2 = \left\lbrace\left (m, \mu_1(m)\wedge\mu_2(m), \nu_1(m)\vee\nu_2(m)\right)|m\in C\right\rbrace$.

● $\psi^c_1 = \{(m, \nu_1(m), \mu_1(m))|m\in C\}$.

● $\psi_1\subseteq\psi_2$ if $\mu_1(m)\leq\mu_2(m)$ and $\nu_1(m)\geq\nu_2(m)$ for all $m\in C$.

● $\psi_1 = \psi_2$ if $\mu_1(m) = \mu_2(m)$ and $\nu_1(m) = \nu_2(m)$ for all $m\in C$.

Throughout this paper, we will write Pythagorean fuzzy subset as PFS and we will write $\psi = (\mu, \nu)$ instead of $\psi = \{(m, \mu(m), \nu(m))|m\in C\}$.

3.   Pythagorean fuzzy subgroup

In this section, we define Pythagorean fuzzy subetaoup (PFSG) as an extension of fuzzy subetaoup and intuitionistic fuzzy subetaoup (IFSG).

Definition 3.1. Let $(C, \circ)$ be a group and $\psi = (\mu, \nu)$ be a PFS of $C$. Then $\psi$ is said to be a PFSG of $C$ if the following conditions hold:

$(i)$ $\mu^2(m\circ n)\geq \mu^2(m)\wedge\mu^2(n)$ and $\nu^2(m\circ n)\leq \nu^2(m)\vee\nu^2(n)$ $\forall m, n\in C$,

$(ii)$ $\mu^2(m^{-1})\geq \mu^2(m)$ and $\nu^2(m^{-1})\leq\nu^2(m)$ $\forall m\in C$.

Here, $\mu^2(m) = \{\mu(m)\}^2$ and $\nu^2(m) = \{\nu(m)\}^2$ for all $m\in C$.

Example 3.1. Let us take the set $C = \{1, -1, i, -i\}$. Then $(C, .)$ is a group, where '.' is the usual multiplication.

Define a PFS $\psi = (\mu, \nu)$ on $C$ by

Here, $\mu^2(i.-i) = \mu^2(1) = (0.8)^2 = 0.64$ and $\nu^2(i.-i) = \nu^2(1) = (0.1)^2 = 0.01$.

Now, $\mu^2(i)\wedge\mu^2(-i) = \min\{0.25, 0.25\} = 0.25$ and $\nu^2(i)\vee\nu^2(-i) = \max\{0.16, 0.16\} = 0.16$.

So, $\mu^2(i.-i) > \mu^2(i)\wedge\mu^2(-i)$ and $\nu^2(i.-i) < \nu^2(i)\vee\nu^2(-i)$.

Also, $\mu^2(i) = \mu^2(-i)$ and $\nu^2(i) = \nu^2(-i)$.

In the same manner it can be shown that $\mu^2(m.n)\geq \mu^2(m)\wedge\mu^2(n)$, $\nu^2(m.n)\leq \nu^2(m)\vee\nu^2(n)$ for all $m, n\in C$ and $\mu^2(m^{-1})\geq \mu^2(m)$, $\nu^2(m^{-1})\leq\nu^2(m)$ for all $m\in C$. Hence $\psi$ is a PFSG of the group $(C, .)$.

Proposition 3.1. Let $\psi = (\mu, \nu)$ be a PFSG of a group $(C, \circ)$. Then the following holds:

$(i)$ $\mu^2(e)\geq \mu^2(m)$ and $\nu^2(e)\leq\nu^2(m)$ $\forall m\in C$ and

$(ii)$ $\mu^2(m^{-1}) = \mu^2(m)$ and $\nu^2(m^{-1}) = \nu^2(m)$ $\forall m\in C$ where, $e$ is the identity element in $C$.

Proof. Since $\psi = (\mu, \nu)$ is a PFSG of a group $(C, \circ)$, then

$\mu^2(m\circ n)\geq \mu^2(m)\wedge\mu^2(n)$, $\nu^2(m\circ n)\leq \nu^2(m)\vee\nu^2(n)$ and

$\mu^2(m^{-1})\geq \mu^2(m)$, $\nu^2(m^{-1})\leq\nu^2(m)$ for all $m, n\in C$.

$(i)$ Now, $\mu^2(e) = \mu^2(m\circ m^{-1})\geq\mu^2(m)\wedge\mu^2(m^{-1}) = \mu^2(m)$.

Also, $\nu^2(e) = \nu^2(m\circ m^{-1})\leq\nu^2(m)\vee\nu^2(m^{-1}) = \nu^2(m)$ for all $m\in C$.

$(ii)$ We have $\mu^2(m^{-1})\geq \mu^2(m)$ and $\nu^2(m^{-1})\leq\nu^2(m)$ for all $m\in C$.

Putting $m^{-1}$ in place of $m$, we obtain

$\mu^2\left((m^{-1})^{-1}\right)\geq \mu^2(m^{-1})$ $\Rightarrow$ $\mu^2(m)\geq \mu^2(m^{-1})$ for all $m\in C$.

Again, $\nu^2\left((m^{-1})^{-1}\right)\leq \nu^2(m^{-1})$ $\Rightarrow$ $\nu^2(m)\leq \nu^2(m^{-1})$ for all $m\in C$.

Hence combining all the results we get $\mu^2(m^{-1}) = \mu^2(m)$ and $\nu^2(m^{-1}) = \nu^2(m)$ $\forall m\in C$.

Now, we will show that every intuitionistic fuzzy subetaoup (IFSG) of a group $(C, \circ)$ is also a Pythagorean fuzzy subetaoup (PFSG) of the group $(C, \circ)$. But the converse is not true.

Theorem 3.1. If $\psi = (\mu, \nu)$ is a IFSG of a group $(C, \circ)$, then $\psi$ is a PFSG of the group $(C, \circ)$.

Proof. Since $\psi = (\mu, \nu)$ is a IFSG of a group $(C, \circ)$, then

$\mu(m\circ n)\geq \mu(m)\wedge\mu(n)$, $\nu(m\circ n)\leq \nu(m)\vee\nu(n)$ and

$\mu(m^{-1})\geq \mu(m)$, $\nu(m^{-1})\leq\nu(m), \forall m, n\in C$.

Here $\mu(m)\in[0, 1]$ and $\nu(m)\in[0, 1]$ for all $m\in C$. Many case arises.

Case 1: Let $\mu(m) > \mu(n)$ and $\nu(m) > \nu(n)$ for all $m, n\in C$.

So, $\mu^2(m) > \mu^2(n)$ and $\nu^2(m) > \nu^2(n)$ for all $m, n\in C$.

Now, $\mu(m\circ n)\geq \mu(m)\wedge\mu(n) = \mu(n)\Rightarrow \mu^2(m\circ n)\geq \mu^2(n) = \mu^2(m)\wedge\mu^2(n)$

i.e., $\mu^2(m\circ n)\geq \mu^2(m)\wedge\mu^2(n)$.

Again, $\nu(m\circ n)\leq \nu(m)\vee\nu(n) = \nu(m)\Rightarrow \nu^2(m\circ n)\leq \nu^2(m) = \nu^2(m)\vee\nu^2(n)$

i.e., $\nu^2(m\circ n)\leq \nu^2(m)\vee\nu^2(n)$.

Case 2: Let $\mu(m) < \mu(n)$ and $\nu(m) < \nu(n)$ for all $m, n\in C$.

So, $\mu^2(m) < \mu^2(n)$ and $\nu^2(m) < \nu^2(n)$ for all $m, n\in C$.

Now, $\mu(m\circ n)\geq \mu(m)\wedge\mu(n) = \mu(m)\Rightarrow \mu^2(m\circ n)\geq \mu^2(m) = \mu^2(m)\wedge\mu^2(n)$

i.e., $\mu^2(m\circ n)\geq \mu^2(m)\wedge\mu^2(n)$.

Again, $\nu(m\circ n)\leq \nu(m)\vee\nu(n) = \nu(n)\Rightarrow \nu^2(m\circ n)\leq \nu^2(n) = \nu^2(m)\vee\nu^2(n)$

i.e., $\nu^2(m\circ n)\leq \nu^2(m)\vee\nu^2(n)$.

Case 3: Let $\mu(m) = \mu(n)$ and $\nu(m) = \nu(n)$ for all $m, n\in C$.

So, $\mu^2(m) = \mu^2(n)$ and $\nu^2(m) = \nu^2(n)$ for all $m, n\in C$.

Now, $\mu(m\circ n) = \mu(m) = \mu(n)\Rightarrow \mu^2(m\circ n) = \mu^2(m) = \mu^2(n)$

i.e., $\mu^2(m\circ n) = \mu^2(m)\wedge\mu^2(n)$.

Again, $\nu(m\circ n) = \nu(m) = \nu(n)\Rightarrow \nu^2(m\circ n) = \nu^2(m) = \nu^2(n)$

i.e., $\nu^2(m\circ n) = \nu^2(m)\vee\nu^2(n)$.

In this way, considering all the cases we can easily show that

$\mu^2(m\circ n)\geq \mu^2(m)\wedge\mu^2(n)$ and $\nu^2(m\circ n)\leq \nu^2(m)\vee\nu^2(n) \forall m, n\in C$.

Again, $\mu(m^{-1})\geq\mu(m)$ and $\nu(m^{-1})\leq\nu(m)$ for all $m\in C$.

Since $\mu(m)\in[0, 1]$ and $\nu(m)\in[0, 1]$,

$\mu^2(m^{-1})\geq\mu^2(m)$ and $\nu^2(m^{-1})\leq\nu^2(m)$ for all $m\in C$.

Hence $\psi = (\mu, \nu)$ is a PFSG of the group $(C, \circ)$.

Example 3.2. Let us consider the Klein's $4$-group $C = \{e, a, b, c\}$, where $a^2 = b^2 = c^2 = e$ and $ab = c$, $bc = a$, $ca = b$.

Define a PFS $\psi = (\mu, \nu)$ on $C$ by

We can easily verify that $\psi = (\mu, \nu)$ is a PFSG of $C$.

But here $\mu(e)+\nu(e) = 1.1$, which is greater than one. So $\psi$ is not a IFS of $C$. Therefore $\psi$ is not a IFSG of $C$.

This example shows that PFSG may not be an IFSG.

Remark 3.1. Every IFSG of a group $(C, \circ)$ is a PFSG of $(C, \circ)$, but the converse need not be true.

Proposition 3.2. Let $\psi = (\mu, \nu)$ be a PFS of a group $(C, \circ)$. Then $\psi$ is a PFSG of $(C, \circ)$ iff $\mu^2(m\circ n^{-1})\geq \mu^2(m)\wedge\mu^2(n)$ and $\nu^2(m\circ n^{-1})\leq \nu^2(m)\vee\nu^2(n)$ $\forall m, n\in C$.

Proof. Let $\psi = (\mu, \nu)$ be a PFSG of a group $(C, \circ)$.

So, $\mu^2(m\circ n^{-1})\geq \mu^2(m)\wedge\mu^2(n^{-1}) = \mu^2(m)\wedge\mu^2(n)$ and

$\nu^2(m\circ n^{-1})\leq \nu^2(m)\vee\nu^2(n^{-1}) = \nu^2(m)\vee\nu^2(n)$ $\forall m, n\in C$.

Conversely, let us assume that $\mu^2(m\circ n^{-1})\geq \mu^2(m)\wedge\mu^2(n)$ and

$\nu^2(m\circ n^{-1})\leq \nu^2(m)\vee\nu^2(n)$ $\forall m, n\in C$.

Now $\mu^2(e) = \mu^2(m\circ m^{-1})\geq\mu^2(m)\wedge\mu^2(m^{-1}) = \mu^2(m)$ and

$\nu^2(e) = \nu^2(m\circ m^{-1})\leq\nu^2(m)\vee\nu^2(m^{-1}) = \nu^2(m)$ $\forall m\in C$, where $e$ is the identity element of $C$.

Again, $\mu^2(m^{-1}) = \mu^2(e\circ m^{-1})\geq\mu^2(e)\wedge\mu^2(m^{-1}) = \mu^2(m)$ and

$\nu^2(m^{-1}) = \nu^2(e\circ m^{-1})\leq\nu^2(e)\vee\nu^2(m^{-1}) = \nu^2(m)$ $\forall m\in C$.

Therefore, $\mu^2(m\circ n) = \mu^2\left(m\circ(n^{-1})^{-1}\right)\geq\mu^2(m)\wedge\mu^2(n^{-1})\geq\mu^2(m)\wedge\mu^2(n)$ and

$\nu^2(m\circ n) = \nu^2\left(m\circ(n^{-1})^{-1}\right)\leq\nu^2(m)\vee\mu^2(n^{-1})\leq\nu^2(m)\vee\nu^2(n)$ $\forall m, n\in C$.

Hence $\psi = (\mu, \nu)$ is PFSG of the group $(C, \circ)$.

Now we will check whether the union and intersection of two PFSGs of a group $(C, \circ)$ is a PFSG of $C$.

Theorem 3.2. Intersection of two PFSGs of a group $(C, \circ)$ is a PFSG of the group $(C, \circ)$.

Proof. Let $\psi_1 = (\mu_1, \nu_1)$ and $\psi_2 = (\mu_2, \nu_2)$ be two PFSGs of a group $(C, \circ)$.

Then $\psi = \psi_1\cap\psi_2 = (\mu, \nu)$, where $\mu(m) = \mu_1(m)\wedge\mu_2(m)$ and $\nu(m) = \nu_1(m)\vee\nu_2(m)$ for all $m\in C$.

Now for all $m, n\in C$

and

Therefore $\psi = \psi_1\cap\psi_2$ is a PFSG of $(C, \circ)$.

Hence intersection of two PFSGs of a group is also a PFSG of the group.

Corollary 3.1. Intersection of a family of PFSGs of a group $(C, \circ)$ is also a PFSG of the group $(C, \circ)$.

Proof. Let $W = \{\psi_1, \psi_2, \ldots, \psi_p\}$ be a family of PFSGs of $(C, \circ)$.

We have to show that $\psi = \cap_{i = 1}^p\psi_i$ is a PFSG of $(C, \circ)$.

Then $\psi = (\mu, \nu)$ is given by $\mu(m) = \mu_1(m)\wedge\mu_2(m)\wedge\ldots\wedge\mu_p(m)$ and

$\nu(m) = \nu_1(m)\vee\nu_2(m)\vee\ldots\vee\nu_p(m)$ for all $m\in C$.

Now for all $m, n\in C$

and

Therefore $\psi = (\mu, \nu)$ is a PFSG of the group $(C, \circ)$.

Hence intersection of a family of PFSGs of a group is also a PFSG of that group.

Remark 3.2. Union of two PFSGs of a group may not be a PFSG of that group.

Example 3.3. Let us take the group $C$ = ($\mathbb{Z}$, +), the group of integers under usual addition and let $\psi_1 = (\mu_1, \nu_1)$, $\psi_2 = (\mu_2, \nu_2)$ be two PFSGs of $C$ defined by

Let $\psi = \psi_1\cup\psi_2 = (\mu, \nu)$, where

Here, $\mu^2(5+(-3)) = \mu^2(2) = 0$, but $\mu^2(5)\wedge\mu^2(-3) = \min\{0.3^2, 0.15^2\} = 0.15^2$.

So, $\mu^2(5+(-3))\ngeq\mu^2(5)\wedge\mu^2(-3)$.

Again, $\nu^2(5+(-3)) = \nu^2(2) = 0.09$, but $\nu^2(5)\vee\nu^2(-3) = \max\{0, 0.04\} = 0.04$.

So, $\nu^2(5+(-3))\nleq\nu^2(5)\vee\nu^2(-3)$.

Hence $\psi = \psi_1\cup\psi_2 = (\mu, \nu)$ is not a PFSG of $C = (\mathbb{Z}, +)$.

Proposition 3.3. If $\psi = (\mu, \nu)$ is a PFSG of a group $(C, \circ)$. Then $\mu^2(m^k)\geq\mu^2(m)$ and $\nu^2(m^k)\leq\nu^2(m)$ for all $m\in C$ and $k\in \mathbb{N}$. Here $m^k = m\circ m\circ\cdots\circ m$ ($k$ times).

Proof. Since $\psi = (\mu, \nu)$ is a PFSG of a group $(C, \circ)$, then

$\mu^2(m\circ n)\geq \mu^2(m)\wedge\mu^2(n)$ and $\nu^2(m\circ n)\leq \nu^2(m)\vee\nu^2(n)$ for all $m, n\in C$.

So, $\mu^2(m^2) = \mu^2(m\circ m)\geq\mu^2(m)\wedge\mu^2(m) = \mu^2(m)$ and

$\nu^2(m^2) = \nu^2(m\circ m)\leq\nu^2(m)\vee\nu^2(m) = \nu^2(m)$ $\forall m\in C$.

Thus by induction, we can show that

$\mu^2(m^k)\geq\mu^2(m)$ and $\nu^2(m^k)\leq\nu^2(m)$ for all $m\in C$ and $k\in \mathbb{N}$.

The next result produce the condition when equality occurs in the definition of PFSG.

Proposition 3.4. Let $\psi = (\mu, \nu)$ be a PFSG of a group $(C, \circ)$. If $\mu(m)\neq\mu(n)$ and $\nu(m)\neq\nu(n)$, then $\mu^2(m\circ n) = \mu^2(m)\wedge\mu^2(n)$ and $\nu^2(m\circ n) = \nu^2(m)\vee\nu^2(n)$ respectively $\forall m, n\in C$.

Proof. Let us assume that $\mu(m) > \mu(n)$ and $\nu(m) < \nu(n)$.

So, $\mu^2(m) > \mu^2(n)$ and $\nu^2(m) < \nu^2(n)$ for all $m, n\in C$.

Now

This shows that $\mu^2(n)\geq \mu^2(m\circ n)$.

Again $\mu^2(m \circ n)\geq \mu^2(m)\wedge\mu^2(n) = \mu^2(n)$.

Therefore $\mu^2(m\circ n)\geq \mu^2(n)$.

So, $\mu^2(m\circ n) = \mu^2(n) = \mu^2(m)\wedge\mu^2(n)$ for all $m, n\in C$.

Similarly, when $\mu(m) < \mu(n)$ this result also holds.

Also,

Thus $\nu^2(n)\leq \nu^2(m\circ n)$.

Again $\nu^2(m \circ n)\leq \nu^2(m)\vee\nu^2(n) = \nu^2(n)$.

Therefore $\nu^2(m\circ n)\leq \nu^2(n)$.

Therefore $\nu^2(m\circ n) = \nu^2(n) = \nu^2(m)\vee\nu^2(n)$ for all $m, n\in C$.

Similarly, for $\nu(m) > \nu(n)$ this result also holds.

Hence $\mu^2(m\circ n) = \mu^2(m)\wedge\mu^2(n)$ and $\nu^2(m\circ n) = \nu^2(m)\vee\nu^2(n)$, when $\mu(m)\neq\mu(n)$ and $\nu(m)\neq\nu(n)$ respectively for all $m, n\in C$.

Proposition 3.5. Let $\psi = (\mu, \nu)$ be a PFSG of a group $(C, \circ)$ with $e$ as the identity element and $m\in C$. If $\mu^2(m) = \mu^2(e)$ then $\mu^2(m\circ n) = \mu^2(n)$ $\forall n\in C$ and if $\nu^2(m) = \nu^2(e)$ then $\nu^2(m\circ n) = \nu^2(n)$ $\forall n\in C$.

Proof. Let us assume that $\mu^2(m) = \mu^2(e)$ and $\nu^2(m) = \nu^2(e)$.

So, $\mu^2(m\circ n)\geq \mu^2(m)\wedge\mu^2(n) = \mu^2(e)\wedge\mu^2(n) = \mu^2(n)$, since $\mu^2(e)\geq\mu^2(n)\ \forall n\in C$.

Also,

Therefore $\mu^2(m\circ n) = \mu^2(n)$ $\forall n\in C$, when $\mu^2(m) = \mu^2(e)$.

Again, $\nu^2(m\circ n)\leq \nu^2(m)\vee\nu^2(n) = \nu^2(e)\vee\nu^2(n) = \nu^2(n)$, since $\nu^2(e)\leq\nu^2(n)\ \forall n\in C$.

In the same way we can show that $\nu^2(n)\leq\nu^2 (m\circ n)$.

Hence $\nu^2(m\circ n) = \nu^2(n)$ $\forall n\in C$, when $\nu^2(m) = \mu^2(e)$.

Theorem 3.3. Let $\psi = (\mu, \nu)$ be a PFSG of a group $(C, \circ)$. Then the set $N = \{m\in C|\mu^2(e) = \mu^2(m)\ \text{and}\ \nu^2(e) = \nu^2(m)\}$ forms a subetaoup of the group $(C, \circ)$, where $e$ is the identity element in $C$.

Proof. Here $N = \{m\in C|\mu^2(e) = \mu^2(m)\ \text{and}\ \nu^2(e) = \nu^2(m)\}$. Clearly $N$ is non empty, as $e\in N$.

To show $(N, \circ)$ is a subetaoup of $(C, \circ)$, we have to show that $m\circ n^{-1}\in N$ for all $m, n\in C$.

Let $m, n\in N$. Then $\mu^2(m) = \mu^2(e) = \mu^2(n)$, $\nu^2(m) = \nu^2(e) = \nu^2(n)$.

Since $\psi = (\mu, \nu)$ is a PFSG of $(C, \circ)$, then

Similarly, we can show that, $\nu^2(m\circ n^{-1})\leq\nu^2 (e)$.

Again from Proposition 3.1, we have $\mu^2(e)\geq\mu^2(m\circ n^{-1})$ and $\nu^2(e)\leq\nu^2(m\circ n^{-1})$.

Therefore $\mu^2(m\circ n^{-1}) = \mu^2(e)$ and $\nu^2(m\circ n^{-1}) = \nu^2(e)$. So, $m\circ n^{-1}\in N$.

Hence $(N, \circ)$ is a subetaoup of $(C, \circ)$.

4.   Pythagorean fuzzy coset and Pythagorean fuzzy normal subgroup

In this section, we will define Pythagorean fuzzy coset and Pythagorean fuzzy normal subetaoup. Further, we will describe properties related to Pythagorean fuzzy normal subetaoup.

Definition 4.1. Let $\psi = (\mu, \nu)$ be a PFSG of a group $(C, \circ)$. Then for $m\in C$, the Pythagorean fuzzy left coset of $\psi$ is the PFS $m\psi = (m\mu, m\nu)$, defined by $(m\mu)^2(u) = \mu^2(m^{-1}\circ u)$, $(m\nu)^2(u) = \nu^2(m^{-1}\circ u)$ and the Pythagorean fuzzy right coset of $\psi$ is the PFS $\psi m = (\mu m, \nu m)$, defined by $(\mu m)^2(u) = \mu^2(u\circ m^{-1})$, $(\nu m)^2(u) = \nu^2(u\circ m^{-1})$ for all $u \in C$.

Definition 4.2. Let $\psi = (\mu, \nu)$ be a PFSG of a group $(C, \circ)$. Then $\psi$ is a Pythagorean fuzzy normal subetaoup (PFNSG) of the group $(C, \circ)$ if every Pythagorean fuzzy left coset of $\psi$ is also a Pythagorean fuzzy right coset of $\psi$ in $C$.

Equivalently, $m\psi = \psi m$ for all $m\in C$.

Example 4.1. Let us take the group $C = (\mathbb{Z}_3, +_3)$, where '$+_3$' is addition of integers modulo $3$.

Define a PFS $\psi = (\mu, \nu)$ on $C$ by

We can easily verify that $\psi = (\mu, \nu)$ is a PFSG of $C$.

For $m = 1\in C$, the Pythagorean fuzzy left coset of $\psi$ is the PFS $1\psi = (1\mu, 1\nu)$, defined by $(1\mu)^2(u) = \mu^2(1^{-1}+_3 u)$, $(1\nu)^2(u) = \nu^2(1^{-1}+_3 u)$ and the Pythagorean fuzzy right coset of $\psi$ is the PFS $\psi 1 = (\mu 1, \nu 1)$, defined by $(\mu 1)^2(u) = \mu^2(u+_3 1^{-1})$, $(\nu 1)^2(u) = \nu^2(u+_3 1^{-1})$ for all $u \in C$.

When $u = 0$, $(1\mu)^2(0) = \mu^2(1^{-1}+_3 0) = \mu^2(2+_{3}0) = \mu^2(2) = 0.49$ and

$(\mu 1)^2(0) = \mu^2(0+_3 1^{-1}) = \mu^2(0+_{3}2) = \mu^2(2) = 0.49$.

Also, $(1\nu)^2(0) = \nu^2(1^{-1}+_3 0) = \nu^2(2+_{3}0) = \nu^2(2) = 0.04$ and

$(\nu 1)^2(0) = \nu^2(0+_3 1^{-1}) = \nu^2(0+_{3}2) = \nu^2(2) = 0.04$.

So, $(1\mu)^2(0) = (\mu 1)^2(0)$ and $(1\nu)^2(0) = (\nu 1)^2(0)$.

Similarly, we can check that the result holds when $u = 1$ and $2$.

Therefore $(1\mu)^2(u) = (\mu 1)^2(u)$ and $(1\nu)^2(u) = (\nu 1)^2(u)$ for all $u\in C$.

That is $1\psi = \psi 1$.

In the same manner it can be shown that $m\psi = \psi m$ for all $m\in C$.

Hence $\psi = (\mu, \nu)$ is a PFNSG of the group $(\mathbb{Z}_3, +_3)$.

Proposition 4.1. Let $\psi = (\mu, \nu)$ be a PFSG of a group $(C, \circ)$. Then $\psi$ is a PFNSG of $C$ iff $\mu^2(m\circ n) = \mu^2(n\circ m)$ and $\nu^2(m\circ n) = \nu^2(n\circ m)$ for all $m, n\in C$.

Proof. Let $\psi = (\mu, \nu)$ be a PFNSG of a group $(C, \circ)$. Then $m\psi = \psi m$ for all $m\in C$.

That is $(m\mu)^2(u) = (\mu m)^2(u)$ and $(m\nu)^2(u) = (\nu m)^2(u)$ for all $m, u\in C$.

Therefore $\mu^2(m^{-1}\circ u) = \mu^2(u\circ m^{-1})$ and $\nu^2(m^{-1}\circ u) = \nu^2(u\circ m^{-1})$ for all $m, u\in C$.

So, $\mu^2(m\circ n) = \mu^2\left(m\circ (n^{-1})^{-1}\right) = \mu^2\left((n^{-1})^{-1}\circ m\right) = \mu^2(n\circ m)$ and

$\nu^2(m\circ n) = \nu^2\left(m\circ (n^{-1})^{-1}\right) = \nu^2\left((n^{-1})^{-1}\circ m\right) = \nu^2(n\circ m)$ $\forall m, n\in C$.

Conversely, let $\mu^2(m\circ n) = \mu^2(n\circ m)$ and $\nu^2(m\circ n) = \nu^2(n\circ m)$ for all $m, n\in C$.

This gives $\mu^2\left(m\circ (n^{-1})^{-1}\right) = \mu^2\left((n^{-1})^{-1}\circ m\right)$ and $\nu^2\left(m\circ (n^{-1})^{-1}\right) = \nu^2\left((n^{-1})^{-1}\circ m\right)$ $\forall m, n\in C$.

Put $n^{-1} = d$. Then $\mu^2(m\circ d^{-1}) = \mu^2(d^{-1}\circ m)$ and $\nu^2(m\circ d^{-1}) = \nu^2(d^{-1}\circ m)$ $\forall m, d\in C$.

So, $(\mu d)^2(m) = (d\mu)^2(m)$ and $(\nu d)^2(m) = (d\nu)^2(m)$ $\forall m, d\in C$.

This implies that $\mu d = d\mu$ and $\nu d = d\nu$ $\forall d\in C$.

Therefore $\psi d = d\psi$ for all $d\in C$. Hence $\psi$ is a PFNSG of the group $(C, \circ)$.

Proposition 4.2. Let $\psi = (\mu, \nu)$ be a PFSG of a group $(C, \circ)$. Then $\psi$ is a PFNSG of $C$ iff $\mu^2(k\circ u\circ k^{-1}) = \mu^2(u)$ and $\nu^2(k\circ u\circ k^{-1}) = \nu^2(u)$ for all $u, k\in C$.

Proof. Let $\psi = (\mu, \nu)$ be a PFNSG of a group $(C, \circ)$.

Then $\mu^2(m\circ n) = \mu^2(n\circ m)$ and $\nu^2(m\circ n) = \nu^2(n\circ m)$ for all $m, n\in C$.

Now for all $u, k\in C$,

Similarly, we get $\nu^2(k\circ u\circ k^{-1}) = \nu^2(u)$.

Therefore $\mu^2(k\circ u\circ k^{-1}) = \mu^2(u)$ and $\nu^2(k\circ u\circ k^{-1}) = \nu^2(u)$ for all $u, k\in C$.

Conversely, let $\mu^2(k\circ u\circ k^{-1}) = \mu^2(u)$ and $\nu^2(k\circ u\circ k^{-1}) = \nu^2(u)$ for all $u, k\in C$.

Now for all $m, n\in C$,

Similarly, we get $\nu^2(m\circ n) = \nu^2(n\circ m)$. Therefore by Proposition 4.1, $\psi = (\mu, \nu)$ is a PFNSG of the group $(C, \circ)$.

Theorem 4.1. Let $\psi = (\mu, \nu)$ be a PFNSG of a group $(C, \circ)$. Then the set $N = \{m\in C|\mu^2(e) = \mu^2(m)\ \text{and}\ \nu^2(e) = \nu^2(m)\}$ forms a normal subetaoup of the group $(C, \circ)$, where $e$ is the identity element in $C$.

Proof. Clearly $N$ is non empty, as $e\in N$. By Proposition 3.3, we have $N$ is a subetaoup of the group $(C, \circ)$.

Let $k\in C$ and $u\in N$.

As $u\in N$, $\mu^2(e) = \mu^2(u)$ and $\nu^2(e) = \nu^2(u)$.

Since $\psi = (\mu, \nu)$ is a PFNSG of the group $(G, \circ)$, by Proposition 4.2 we have

$\mu^2(k\circ u\circ k^{-1}) = \mu^2(u)$ and $\nu^2(k\circ u\circ k^{-1}) = \nu^2(u)$ for all $u, k\in C$.

Consequently, $\mu^2(k\circ u\circ k^{-1}) = \mu^2(e)$ and $\nu^2(k\circ u\circ k^{-1}) = \nu^2(e)$ for all $u, k\in C$.

Therefore $k\circ u\circ k^{-1}\in N$. Hence $(N, \circ)$ is a normal subetaoup of the group $(C, \circ)$.

5.   Pythagorean fuzzy level subgroup

Definition 5.1. Let $C$ be a crisp set. Let $\psi = (\mu, \nu)$ be a PFS of the set $C$. For $\theta$ and $\tau\in [0, 1]$, the set $\psi_{(\theta, \tau)} = \{m\in C|\mu^2(m)\geq\theta\ \text{and}\ \nu^2(m)\leq\tau\}$ is called a Pythagorean fuzzy level subset (PFLS) of the PFS $\psi$ of $C$, where $0\leq\theta^2+\tau^2\leq 1$.

Proposition 5.1. Let $\psi^{\prime} = (\mu^{\prime}, \nu^{\prime})$ and $\psi^{\prime\prime} = (\mu^{\prime\prime}, \nu^{\prime\prime})$ be two PFSs of the universal set $C$. Then

$(i)$ $\psi^{\prime}_{(\theta, \tau)}\subseteq\psi^{\prime}_{(\epsilon, \delta)}$ if $\epsilon\leq\theta$ and $\tau\leq\delta$ for $\epsilon$, $\tau$, $\theta$ and $\delta$ $\in[0, 1]$,

$(ii)$ $\psi^{\prime}\subseteq\psi^{\prime\prime}\Rightarrow\psi^{'}_{(\theta, \tau)}\subseteq\psi^{\prime\prime}_{(\theta, \tau)}$ for $\theta$ and $\tau$ $\in[0, 1]$.

Proof. $(i)$ Let $m\in \psi^{\prime}_{(\theta, \tau)}\Rightarrow(\mu^{\prime})^2(m)\geq\theta, (\nu^{\prime})^2(m)\leq\tau$.

We have $\epsilon\leq\theta$ and $\tau\leq\delta$. So, $\epsilon\leq\theta\leq(\mu^{\prime})^2(m)$ and $(\nu^{\prime})^2(m)\leq\tau\leq\delta$.

Therefore, $m\in\psi^{\prime}_{(\epsilon, \delta)}$.

Hence $\epsilon\leq\theta$ and $\tau\leq\delta$ $\Rightarrow\psi^{\prime}_{(\theta, \tau)}\subseteq\psi^{\prime}_{(\epsilon, \delta)}$.

$(ii)$ Since $\psi^{\prime}\subseteq\psi^{\prime\prime}$, so $\mu^{\prime}(m)\leq\mu^{\prime\prime}(m)$ and $\nu^{\prime}(m)\geq\nu^{\prime\prime}(m)$ for all $m\in C$.

$\Rightarrow(\mu^{\prime})^2(m)\leq(\mu^{\prime\prime})^2(m)$ and $(\nu^{\prime})^2(m)\geq(\nu^{\prime\prime})^2(m)$ for all $m\in C$.

Let $m\in \psi^{\prime}_{(\theta, \tau)}$. This implies that $(\mu^{\prime})^2(m)\geq\theta$ and $(\nu^{\prime})^2(m)\leq\tau$.

So, $\theta\leq(\mu^{\prime})^2(m)\leq(\mu^{\prime\prime})^2(m)$ and $(\nu^{\prime\prime})^2(m)\leq(\nu^{\prime})^2(m)\leq\tau$.

This shows that $\theta\leq(\mu^{\prime\prime})^2(m)$ and $(\nu^{\prime\prime})^2(m)\leq\tau$.

Therefore $m\in \psi^{\prime\prime}_{(\theta, \tau)}$. Hence $\psi^{'}_{(\theta, \tau)}\subseteq\psi^{\prime\prime}_{(\theta, \tau)}$.

Proposition 5.2. Let $\psi = (\mu, \nu)$ be a PFSG of a group $(C, \circ)$. Then the Pythagorean fuzzy level subset $\psi_{(\theta, \tau)}$ forms a subetaoup of the group $(C, \circ)$, where $\theta\leq\mu^2(e)$ and $\tau\geq\nu^2(e)$, $e$ is the identity element in $C$.

Proof. Here, $\psi_{(\theta, \tau)} = \{m\in C|\mu^2(m)\geq\theta\ \text{and}\ \nu^2(m)\leq\tau\}$.

Clearly $\psi_{(\theta, \tau)}$ is non empty, as $e\in\psi_{(\theta, \tau)}$.

To show that $\psi_{(\theta, \tau)}$ is a subetaoup of $(C, \circ)$, we have to show that for $m, n\in\psi_{(\theta, \tau)}$, $m\circ n^{-1}\in\psi_{(\theta, \tau)}$. Let $m, n\in\psi_{(\theta, \tau)}$.

Then $\mu^2(m)\geq\theta$ & $\nu^2(m)\leq\tau$ and $\mu^2(n)\geq\theta$ & $\nu^2(b)\leq\tau$.

Since $\psi = (\mu, \nu)$ is a PFSG of the group $(C, \circ)$, then

and

Therefore $m\circ n^{-1}\in\psi_{(\theta, \tau)}$. Hence $\psi_{(\theta, \tau)}$ is a subetaoup of the group $(C, \circ)$.

Definition 5.2. The subetaoup $\psi_{(\theta, \tau)}$ of the group $(C, \circ)$ is called Pythagorean fuzzy level subetaoup (PFLSG) of the PFSG $\psi = (\mu, \nu)$.

Proposition 5.3. Let $\psi = (\mu, \nu)$ be a PFS of a group $(C, \circ)$. If the PFLS $\psi_{(\theta, \tau)}$ is a subetaoup of the group $(C, \circ)$, where $\theta\leq\mu^2(e)$ and $\tau\geq\nu^2(e)$ then $\psi = (\mu, \nu)$ is a PFSG of the group $(C, \circ)$.

Proof. Let $m, n\in C$. Given that $\psi = (\mu, \nu)$ is a PFS of $C$ and $\psi_{(\theta, \tau)}$ is a subetaoup of the group $(C, \circ)$.

Let us assume that $\mu^2(m) = \theta_1$, $\mu^2(n) = \theta_2$ with $\theta_1 < \theta_2$ and $\nu^2(m) = \tau_1$, $\nu^2(n) = \tau_2$ with $\tau_1 > \tau_2$.

This implies that $m\in\psi_{(\theta_1, \tau_1)}$ and $n\in\psi_{(\theta_2, \tau_2)}$.

Since, $\theta_1 < \theta_2$ and $\tau_1 > \tau_2$, then $\psi_{(\theta_2, \tau_2)}\subseteq\psi_{(\theta_1, \tau_1)}$. So, $n\in\psi_{(\theta_1, \tau_1)}$.

Now, $m\in\psi_{(\theta_1, \tau_1)}$ and $n\in\psi_{(\theta_1, \tau_1)}$.

So, $m\circ n\in\psi_{(\theta_1, \tau_1)}$, since $\psi_{(\theta_1, \tau_1)}$ is a subetaoup of $(C, \circ)$.

Therefore $\mu^2(m\circ n)\geq\theta_1$ and $\nu^2(m\circ n)\leq\tau_1$

$\Rightarrow \mu^2(m\circ n)\geq\theta_1\wedge\theta_2$ and $\nu^2(m\circ n)\leq\tau_1\vee\tau_2$, since $\theta_1 < \theta_2$ and $\tau_1 > \tau_2$

$\Rightarrow \mu^2(m\circ n)\geq\mu^2(m)\wedge\mu^2(n)$ and $\nu^2(m\circ n)\leq\nu^2(m)\vee\nu^2(n)$.

Again, $m\in\psi_{(\theta_1, \tau_1)}\Rightarrow m^{-1}\in\psi_{(\theta_1, \tau_1)}$, since $\psi_{(\theta_1, \tau_1)}$ is a subetaoup of $(C, \circ)$

$\Rightarrow\mu^2(m^{-1})\geq\theta_1$ and $\nu^2(m^{-1})\leq\tau_1$

$\Rightarrow\mu^2(m^{-1})\geq\mu^2(m)$ and $\nu^2(m^{-1})\leq\nu^2(m)$.

Since $m, n\in C$ are arbitrary,

$\mu^2(m\circ n)\geq\mu^2(m)\wedge\mu^2(n)$, $\nu^2(m\circ n)\leq\nu^2(m)\vee\nu^2(n)$ for all $m, n\in C$ and

$\mu^2(m^{-1})\geq\mu^2(a)$, $\nu^2(m^{-1})\leq\nu^2(m)$ for all $m\in C$.

Hence $\psi = (\mu, \nu)$ is a PFSG of the group $(C, \circ)$.

Proposition 5.4. Let $\psi = (\mu, \nu)$ be a PFNSG of a group $(C, \circ)$. Then the Pythagorean fuzzy level subset $\psi_{(\theta, \tau)}$ forms a normal subetaoup of the group $(C, \circ)$, where $\theta\leq\mu^2(e)$ and $\tau\geq\nu^2(e)$, $e$ is the identity element in $C$.

Proof. Since $\psi = (\mu, \nu)$ is a PFNSG of the group $(C, \circ)$, then for all $u, k\in C$,

$\mu^2(k\circ u\circ k^{-1}) = \mu^2(u)$ and $\nu^2(k\circ u\circ k^{-1}) = \nu^2(u)$.

By Proposition 5.2, $\psi_{(\theta, \tau)} = \{m\in G|\mu^2(m)\geq\theta\ \text{and}\ \nu^2(m)\leq\tau\}$ is a subetaoup of $(C, \circ)$.

Let $k\in G$ and $u\in \psi_{(\theta, \tau)}$. Then $\mu^2(k\circ u\circ k^{-1}) = \mu^2(u)\geq\theta$ and $\nu^2(k\circ u\circ k^{-1}) = \nu^2(u)\leq\tau$.

Therefore $k\circ u\circ k^{-1}\in \psi_{(\theta, \tau)}$. Hence $\psi_{(\theta, \tau)}$ is a normal subetaoup of the group $(C, \circ)$.

6.   Homomorphism on Pythagorean fuzzy subgroup

In this section, we will discuss the effect of group homomorphism on Pythagorean fuzzy subetaoup.

Theorem 6.1. Let $(C_{1}, \circ_{1})$ and $(C_{2}, \circ_{2})$ be two groups. Let $g$ be a surjective homomorphism from $(C_{1}, \circ_{1})$ to $(C_{2}, \circ_{2})$ and $\psi = (\mu, \nu)$ be a PFSG of $(C_{1}, \circ_{1})$. Then $g(\psi) = (g(\mu), g(\nu))$ is a PFSG of $(C_{2}, \circ_{2})$.

Proof. Since $g:C_{1}\rightarrow C_{2}$ is a surjective homomorphism, then $g(C_{1}) = C_{2}$.

Let $m_{2}$ and $n_{2}$ be two elements of $C_{2}$.

Suppose $m_{2} = g(m_{1})$ and $n_{2} = g(n_{1})$ for some $m_{1}$, $n_{1}\in C_{1}$.

We have $g(\psi) = \{(c, g(\mu)(c), g(\nu)(c))|c\in C_{2}\}$. Now

Therefore $(g(\mu))^2(m_{2}\circ_{2} n_{2})\geq (g(\mu))^2(m_{2})\wedge (g(\mu))^2(n_{2})$ for all $m_{2}$ and $n_{2}\in C_{2}$.

Similarly, we can prove that $(g(\nu))^2(m_{2}\circ_{2} n_{2})\leq (g(\nu))^2(m_{2})\vee (g(\nu))^2(n_{2})$ for all $m_{2}$ and $n_{2}\in C_{2}$. Again

Therefore $(g(\mu))^2(m_{2}^{-1}) = (g(\mu))^2(m_{2})$ for all $m_{2}\in C_{2}$.

Similarly, we can show that $(g(\nu))^2(m_{2}^{-1}) = (g^{-1}(\nu))^2(m_{2})$ for all $m_{2}\in C_{2}$. Hence $g(\psi) = (g(\mu), g(\nu))$ is a PFSG of $(C_{2}, \circ_{2})$.

Theorem 6.2. Let $(C_{1}, \circ_{1})$ and $(C_{2}, \circ_{2})$ be two groups. Let $g$ be a bijective homomorphism from $(C_{1}, \circ_{1})$ to $(C_{2}, \circ_{2})$ and $\psi = (\mu, \nu)$ be a PFSG of $(C_{2}, \circ_{2})$. Then $g^{-1}(\psi) = (g^{-1}(\mu), g^{-1}(\nu))$ is a PFSG of $(C_{1}, \circ_{1})$.

Proof. Let $m_{1}$ and $n_{1}$ be two arbitrary elements of $C_{1}$, then $m_{1}\circ_{1} n_{1}\in C_{1}$. We have $g^{-1}(\psi) = \{(c, g^{-1}(\mu)(c), g^{-1}(\nu)(c))|c\in C_{1}\}$. Now

Therefore $(g^{-1}(\mu))^2(m_{1}\circ_{1} n_{1})\geq (g^{-1}(\mu))^2(m_{1})\wedge (g^{-1}(\mu))^2(n_{1})$ for all $m_{1}$ and $n_{1}\in C_{1}$.

Similarly, we can show that $(g^{-1}(\nu))^2(m_{1}\circ_{1} n_{1})\leq (g^{-1}(\nu))^2(m_{1})\vee (g^{-1}(\nu))^2(n_{1})$ for all $m_{1}$ and $n_{1}\in C_{1}$. Again

Therefore $(g^{-1}(\mu))^2(m_{1}^{-1}) = (g^{-1}(\mu))^2(m_{1})$ for all $m_{1}\in C_{1}$.

Similarly, we can show that $(g^{-1}(\nu))^2(m_{1}^{-1}) = (g^{-1}(\nu))^2(m_{1})$ for all $m_{1}\in C_{1}$. Hence $g^{-1}(\psi) = (g^{-1}(\mu), g^{-1}(\nu))$ is a PFSG of $(C_{1}, \circ_{1})$.

Theorem 6.3. Let $(C_{1}, \circ_{1})$ and $(C_{2}, \circ_{2})$ be two groups. Let $g$ be a surjective homomorphism from $(C_{1}, \circ_{1})$ to $(C_{2}, \circ_{2})$ and $\psi = (\mu, \nu)$ be a PFNSG of $(C_{1}, \circ_{1})$. Then $g(\psi) = (g(\mu), g(\nu))$ is a PFNSG of $(C_{2}, \circ_{2})$.

Proof. In view of Theorem 6.1, we can state that $g(\psi) = (g(\mu), g(\nu))$ is a PFSG of $(C_{2}, \circ_{2})$.

Since $\psi = (\mu, \nu)$ is a PFNSG of $(C_{1}, \circ_{1})$, then by Proposition 4.1 $\mu^2(m_{1}\circ_{1} n_{1}) = \mu^2(n_{1}\circ_{1} m_{1})$ and $\nu^2(m_{1}\circ_{1} n_{1}) = \nu^2(n_{1}\circ_{1} m_{1})$ for all $m_{1}$, $n_{1}\in C_{1}$.

Let $m_{2}$ and $n_{2}$ be two elements of $C_{2}$.

Suppose that there exist unique $m_{1}$ and $n_{1}\in C_{1}$, such that $m_{2} = g(m_{1})$ and $n_{2} = g(n_{1})$. Now

Therefore $(g(\mu))^2(m_{2}\circ_{2} n_{2}) = (g(\mu))^2(n_{2}\circ_{2} m_{2})$ for all $m_{2}$ and $n_{2}\in C_{2}$.

Similarly, we can prove that $(g(\nu))^2(m_{2}\circ_{2} n_{2}) = (g(\nu))^2(n_{2}\circ_{2} m_{2})$ for all $m_{2}$ and $n_{2}\in C_{2}$.

Hence by Proposition 4.1, $g(\psi) = (g(\mu), g(\nu))$ is a PFNSG of $(C_{2}, \circ_{2})$.

Theorem 6.4. Let $(C_{1}, \circ_{1})$ and $(C_{2}, \circ_{2})$ be two groups. Let $g$ be a bijective homomorphism from $(C_{1}, \circ_{1})$ to $(C_{2}, \circ_{2})$ and $\psi = (\mu, \nu)$ be a PFNSG of $(C_{2}, \circ_{2})$. Then $g^{-1}(\psi) = (g^{-1}(\mu), g^{-1}(\nu))$ is a PFNSG of $(C_{1}, \circ_{1})$.

Proof. In view of Theorem 6.2, we can state that $g^{-1}(\psi) = (g^{-1}(\mu), g^{-1}(\nu))$ is a PFSG of $(C_{1}, \circ_{1})$.

Since $\psi = (\mu, \nu)$ is a PFNSG of $(C_{2}, \circ_{2})$, then by Proposition 4.1

$\mu^2(m_{2}\circ_{2} n_{2}) = \mu^2(n_{2}\circ_{2} m_{2})$ and $\nu^2(m_{2}\circ_{2} n_{2}) = \nu^2(n_{2}\circ_{2} m_{2})$ for all $m_{2}$, $n_{2}\in C_{2}$.

Let $m_{1}$ and $n_{1}$ be two elements of $C_{1}$. Then

Therefore $(g^{-1}(\mu))^2(m_{1}\circ_{1} n_{1}) = (g^{-1}(\mu))^2(n_{1}\circ_{1} m_{1})$ for all $m_{1}$ and $n_{1}\in C_{1}$.

Similarly, we can prove that $(g^{-1}(\nu))^2(m_{1}\circ_{1} n_{1}) = (g^{-1}(\nu))^2(n_{1}\circ_{1} m_{1})$ for all $m_{1}$ and $n_{1}\in C_{1}$.

Hence by Proposition 4.1, $g^{-1}(\psi) = (g^{-1}(\mu), g^{-1}(\nu))$ is a PFNSG of $(C_{1}, \circ_{1})$.

7.   Conclusion

The purpose of this paper is to initiate the study of Pythagorean fuzzy subetaoup. We have discussed various algebraic attributes of Pythagorean fuzzy subetaoup. We have proved that intuitionistic fuzzy subetaoup of any group is a Pythagorean fuzzy subetaoup of that group. We have introduced the notion of Pythagorean fuzzy coset and Pythagorean fuzzy normal subetaoup. We have presented the necessary and sufficient condition for Pythagorean fuzzy subetaoup to be a Pythagorean fuzzy normal subetaoup. Further, we have proved that Pythagorean fuzzy level subset is a normal subetaoup of the given group. Moreover, we have studied the effect of group homomorphism on Pythagorean fuzzy subetaoup. In our future work, we will work on Pythagorean fuzzy quotient group and order of Pythagorean fuzzy subetaoup. We will also work on the Lagrange theorem in Pythagorean fuzzy subetaoup.

Acknowledgments

This research work of first author is sponsored by Council of Scientific and Industrial Research (CSIR), Human Resource Development Group (HRDG), INDIA. Sanctioned file no. is 09/599(0081)/2018-EMR-I. This work was partially supported by Research Council Faroe Islands and University of the Faroe Islands for the third author. The authors are grateful to the anonymous referees for a careful checking of the details and for helpful comments that improved the overall presentation of this paper.

Conflict of interest

All authors declare that there is no conflict of interest.