# Fundamental Homomorphism Theorems for Neutrosophic Extended Triplet Groups

^{*}

## Abstract

**:**

## 1. Introduction

## 2. Preliminaries

#### 2.1. Neutrosophic Extended Triplet

**Definition**

**1**

**([3]).**The set N is called a neutrosophic extended triplet set if for any x ∈ N there exist e

^{neut(x)}∈ N and e

^{anti(x)}∈ N. Thus, a neutrosophic extended triplet is an object of the form (x, e

^{neut(x)}, e

^{anti(x)}) where e

^{neut(x)}is extended neutral of x, which can be equal or different from the classical algebraic unitary element if any, such that

^{anti(x)}∈ N is the extended opposite of x such that

**Theorem**

**1**

**([11]).**Let (N, ∗) be a commutative NET with respect to ∗ and a, b ∈ N;

- (i)
- $neut\left(a\right)*neut\left(b\right)=neut\left(a*b\right);$
- (ii)
- $anti\left(a\right)*anti\left(b\right)=anti\left(a*b\right);$

**Theorem**

**2**

**([11]).**Let (N, ∗) be a commutative NET with respect to ∗ and a ∈ N;

- (i)
- $neut\left(a\right)*neut\left(a\right)=neut\left(a\right);$
- (ii)
- $anti\left(a\right)*neut\left(a\right)=neut\left(a\right)*anti\left(a\right)=anti\left(a\right)$

#### 2.2. NETG

**Definition**

**2**

**([3]).**Let (N, ∗) be a neutrosophic extended triplet set. Then (N, ∗) is called a NETG, if the following classical axioms are satisfied.

- (a)
- (N, ∗) is well defined, i.e., for any $x,y\in N$ one has $x*y\in N.$
- (b)
- (N, ∗) is associative, i.e., for any$x,y,z\in N$one has$x*\left(y*z\right)=\left(x*y\right)*z.$

#### 2.3. Neutrosophic Extended Triplet Subgroup

**Definition**

**3**

**([26]).**Given a NETG (N, ∗), a neutrosophic triplet subset H is called a neutrosophic extended triplet subgroup of N if it itself forms a neutrosophic extended triplet group under ∗. Explicity this means

- (1)
- The extended neutral element ${e}^{neut\left(x\right)}$ lies in H.
- (2)
- For any $x,y\in H,x\ast y\in H.$
- (3)
- If $x\in H$ then ${e}^{anti\left(x\right)}\in H.$

**Definition**

**4.**

#### 2.4. Neutro-Homomorphism

**Definition**

**5**

**([26]).**Let (N

_{1}, ∗) and (N

_{2}, ∘) be two NETGs. A mapping $f:{N}_{1}\to {N}_{2}$ is called a neutro-homomorphism if

- (a)
- For any $x,y\in N,$ we have$$f\left(x*y\right)=f\left(x\right)\circ f\left(y\right)$$
- (b)
- If $(x,neut\left(x\right),anti\left(x\right)$ is a neutrosophic extended triplet from N
_{1}, then$$f\left(neut\left(x\right)\right)=neut\left(f\left(x\right)\right)$$$$f\left(anti\left(x\right)\right)=anti\left(f\left(x\right)\right).$$

**Definition**

**6**

**([26]).**Let f: N

_{1}→N

_{2}be a neutro-homomorphism from a NETG (N

_{1}, ∗) to a NETG (N

_{2}, ∘). The neutrosophic triplet image of f is

**Definition**

**7**

**([26]).**Let f: N

_{1}→N

_{2}be a neutro-homomorphism from a NETG (N

_{1}, ∗) to a NETG (N

_{2}, ∘) and B ⊆ N

_{2}. Then

**Definition**

**8**

**([26]).**Let $f:{N}_{1}\to {N}_{2}$ be a neutro-homomorphism from a NETG (N

_{1}, ∗) to a NETG (N

_{2}, ∘). The neutrosophic triplet kernel of f is a subset

_{2}.

**Definition**

**9.**

**Definition**

**10**

**([26]).**Let N be a NETG and $H\subseteq N.\forall x\in N,$ the set $xh/h\in H$ is called neutrosophic triplet coset denoted by xH. Analogously,

#### 2.5. Neutrosophic Triplet Normal Subgroup and Quotient Group

**Definition**

**11**

**([26]).**A neutrosophic extended triplet subgroup H of a NETG of N is called a neutrosophic triplet normal subgroup of N if $aH\left(anti\left(a\right)\right)\subseteq H,\forall x\in N$ and we denote it as $H\u22b4N\text{}and\text{}H\u22b2N\text{}if\text{}H\ne N.$

**Example**

**1.**

**Definition**

**12**

**([26]).**If N is a NETG and H ⊴ N is a neutrosophic triplet normal subgroup, then the neutrosophic triplet quotient group N/H has elements $xH:x\in N,$ the neutrosophic triplet cosets of H in N, and operation $\left(xH\right)\left(yH\right)=\left(xy\right)H.$

## 3. Neutro-Monomorphism, Neutro-Epimorphism, Neutro-Isomorphism, Neutro-Automorphism

#### 3.1. Neutro-Monomorphism

**Definition**

**13.**

_{1}, ∗) and (N

_{2}, ∘) be two NETG’s. If a mapping $f:{N}_{1}\to {N}_{2}$ of NETG is only one to one (injective) f is called neutro-monomorphism.

**Theorem**

**3.**

_{1}, ∗) and (N

_{2}, ∘) be two NETG’s. $\phi :{N}_{1}\to {N}_{2}$ is a neutro-monomorphism of NETG if and only if $ker\phi =\left\{neu{t}_{N1}\right\}.$

**Proof.**

_{N}

_{1}. Conversely, assume kerφ = φ(neut

_{N}

_{1}). Let a,b ∈ N

_{1}such that φ(a) = φ(b). We need to show that a = b.

**Definition**

**14.**

_{1}, ∗) and (N

_{2}, ∘) be two NETG’s. If a mapping $f:{N}_{1}\to N$ is only onto (surjective) f is called neutro-epimorphism.

**Theorem**

**4.**

^{−1}: H → N.

**Proof.**

**Theorem**

**5.**

**Proof.**

_{n}is injective. ☐

**Theorem**

**6.**

**Proof.**

- (1)
- Let’s show that ${f}^{-1}\left(h\right)\subseteq x\text{}kerf.$If $x\in {f}^{-1}\left(h\right),$ then $f\left(x\right)=h$ and $b\in {f}^{-1}\left(h\right),$ then $f\left(b\right)=h.$ If $f\left(x\right)=f\left(y\right),$ then:$$anti\left(f\left(x\right)\right)f\left(x\right)=anti\left(f\left(x\right)\right)f\left(b\right)\left(\mathrm{by}\text{}\mathrm{theorem}\text{}1\right)$$$$neu{t}_{H}=f\left(anti\left(x\right)\right)f\left(b\right)\text{}\left(\mathrm{by}\text{}\mathrm{definition}\text{}1\right)$$$$\Rightarrow anti\left(x\right)b\in kerf.$$For at least $k\in kerf,anti\left(x\right)b=k.$ If $b=xk,$ then,$$b\in xkerf\Rightarrow {f}^{-1}\left(h\right)\subseteq xkerf$$
- (2)
- Let’s show that $xkerf\subseteq {f}^{-1}\left(h\right).$ Let $b\in xkerf.$ For at least $k\in kerf,b=xk$$$\Rightarrow f\left(b\right)=f\left(xk\right)=f\left(x\right)f\left(k\right)=h\text{}neu{t}_{H}=h$$If ${f}^{-1}\left(h\right)=b$ and $b\in {f}^{-1}\left(h\right),$ then$$xkerf\subseteq {f}^{-1}\left(h\right)$$

**Theorem**

**7.**

_{1}and N

_{2}.

- (1)
- $If\text{}{H}_{2}\u22b4{N}_{2},\text{}then\text{}{\phi}^{-1}\left({H}_{2}\right)\u22b4{N}_{1}.$
- (2)
- $If{H}_{1}\u22b4{N}_{1}\text{}and\text{}\phi \text{}is\text{}a\text{}neutro-epimorhism\text{}then\text{}\phi \left({H}_{1}\right)\u22b4{N}_{2}.$

**Proof.**

- (1)
- If $x\in {\phi}^{-1}\left({H}_{2}\right)$ and $a\in {N}_{1},$ then $\phi \left(x\right)\in {H}_{2}$ and so $\phi (\left(ax\right)\left(anti\left(a\right)\right)=\phi \left(a\right)\phi \left(x\right)anti\left(\phi \left(a\right)\right)\in {H}_{2}$. Since H
_{2}is neutrosophic triplet normal subgroup. We conclude $ax\left(anti\left(a\right)\right)\in {\phi}^{-1}\left({H}_{2}\right).$ - (2)
- Since H
_{1}is neutrosophic triplet normal subgroup, we have $\phi \left(a\right)\phi \left({H}_{1}\right)anti\left(\phi \left(a\right)\right)\subseteq \phi \left({H}_{1}\right).$ Since we assume φ is surjective, every $b\in {N}_{2}$ can be written as $b=\phi \left(a\right),a\in {N}_{1}.$ Therefore, $b\phi \left({H}_{1}\right)anti\left(b\right)\in \phi \left({H}_{1}\right).$ ☐

**Theorem**

**9.**

**Proof.**

**Theorem**

**10.**

**Proof.**

#### 3.2. Neutro-Isomorphism

**Definition**

**15.**

_{1}, ∗) and (N

_{2}, ∘) be two NETGs. If a mapping $f:{N}_{1}\to {N}_{2}$ neutro-homomorphism is one to one and onto f is called neutro-isomorphism. Here, N

_{1}and N

_{2}are called neutro-isomorphic and denoted as ${N}_{1}\cong {N}_{2}.$

**Theorem**

**11.**

_{1}, ∗) and (N

_{2}, ∘) be two NETG’s. If $f:{N}_{1}\to {N}_{2}$ is a neutro-isomorphism of NETG’s, then so is ${f}^{-1}:{N}_{2}\to {N}_{1}.$

**Proof.**

#### 3.3. Neutro-Automorphism.

**Definition**

**16.**

_{1}, ∗) and (N

_{2}, ∘) be two NETG’S. If a mapping $f:{N}_{1}\to {N}_{2}$ is one to one and onto f is called neutro-automorphism.

**Definition**

**17.**

**Proposition**

**1.**

**Proof.**

- (1)
- $\forall x,y\in N,$ we have to show that$$f\left(x\right)=f\left(y\right)\Rightarrow x=y.ax\left(anti\left(a\right)\right)=ay\left(anti\left(a\right)\right)\phantom{\rule{0ex}{0ex}}\Rightarrow ax\left(anti\left(a\right)\right)a=ay\left(anti\left(a\right)\right)a\Rightarrow ax\left(neut\left(a\right)\right)\phantom{\rule{0ex}{0ex}}=ay\left(neut\left(a\right)\right)\Rightarrow ax=ay\Rightarrow anti\left(a\right)ax=anti\left(a\right)ay\phantom{\rule{0ex}{0ex}}\Rightarrow neut\left(a\right)x=neut\left(a\right)y\Rightarrow x=y.$$Therefore, f is one to one.
- (2)
- $\forall x,y\in N,$ we have to show that$$f\left(x\right)=ax\left(anti\left(a\right)\right)=y.ax\left(anti\left(a\right)\right)a=ya\phantom{\rule{0ex}{0ex}}\Rightarrow ax\left(neut\left(a\right)\right)=ya\Rightarrow ax=ya\Rightarrow anti\left(a\right)ax=anti\left(a\right)ya\phantom{\rule{0ex}{0ex}}\Rightarrow neut\left(a\right)x=anti\left(a\right)ya\Rightarrow x=anti\left(a\right)ya.$$So, f is onto. Therefore, f
_{a}is a neutro-automorphism. ☐

**Lemma**

**1.**

**Proof.**

**Theorem**

**12.**

**Proof.**

_{1}H

_{2}is a neutrosophic extended triplet subgroup. Then, for all $a\in {H}_{1},b\in {H}_{2},$ we have $anti\left(a\right)anti\left(b\right)\in {H}_{1}{H}_{2},i.e.,{H}_{2}{H}_{1}\subseteq {H}_{1}{H}_{2}.$ But also for $h\in {H}_{1}{H}_{2}$ we find $a\in {H}_{1},b\in {H}_{2}$ thereby $anti\left(h\right)=ab,$ and then $h=anti\left(b\right)anti\left(a\right)\in {H}_{2}{H}_{1}.$ So ${H}_{1}{H}_{2}\subseteq {H}_{2}{H}_{1}$, that’s, ${H}_{1}{H}_{2}={H}_{2}{H}_{1}.$ On the other hand, assume that ${H}_{1}{H}_{2}={H}_{2}{H}_{1}.$ Then $\forall a,{a}^{\prime}\in {H}_{1},b,{b}^{\prime}\in {H}_{2}$ we have $ab{a}^{\prime}{b}^{\prime}\in a{H}_{2}{H}_{1}{b}^{\prime}=a{H}_{1}{H}_{2}{b}^{\prime}={H}_{1}{H}_{2}.$ Furthermore, $\forall a\in {H}_{1},b\in {H}_{2}$ we have $anti\left(ab\right)=anti\left(b\right)anti\left(a\right)\in {H}_{2}{H}_{1}={H}_{1}{H}_{2}.$ ☐

## 4. Fundamental Theorem of Neutro-Homomorphism

**Theorem**

**13.**

_{1}, N

_{2}be NETG’s and $\varphi :{N}_{1}\to {N}_{2}$ be a neutro-homomorphism. Then, ${N}_{1}/ker\left(\varphi \right)\cong im\left(\varphi \right).$ Furthermore if ϕ is neutro-epimorphism, then

_{1}/kerϕ ≅ N

_{2}.

**Proof.**

_{1}. Let $K=ker\left(\varphi \right),$ and recall that ${N}_{1}/K=\{aK:a\in {N}_{1}\}.$ Define $i:{N}_{1}/K\to im\left(\varphi \right),i:nK\to \varphi \left(n\right),n\in {N}_{1}.$ Thus, we need to check the following conditions.

- (1)
- i is well defined
- (2)
- i is injective
- (3)
- i is surjective
- (4)
- i is a neutro-homomorphism

- (1)
- We must show that if $aK=bK,$ then $i\left(aK\right)=\left(bK\right).$ Suppose $aK=bK.$ We have $aK=bK\Rightarrow anti\left(b\right)aK=K\Rightarrow anti\left(b\right)a\in K.$Here, $neu{t}_{\left(n2\right)}=\varphi \left(anti\left(b\right)a\right)=\varphi (anti\left(b\right)\varphi \left(a\right)$ $=anti\left(\varphi \left(b\right)\right)\varphi \left(a\right)\Rightarrow \varphi \left(a\right)=\varphi \left(b\right).$ Hence, $i\left(aK\right)=\varphi \left(a\right)=\varphi \left(b\right)=i\left(bK\right).$ Therefore, it is well defined.
- (2)
- We must show that $i\left(aK\right)=i\left(bK\right)\Rightarrow aK=bK.$ Suppose that $i\left(aK\right)=i\left(bK\right).$ Then$$i\left(aK\right)=i\left(bK\right)\Rightarrow aK=bK.\phantom{\rule{0ex}{0ex}}\Rightarrow \varphi \left(anti\left(b\right)\right)\varphi \left(a\right)=neu{t}_{\left(n2\right)}\Rightarrow \varphi \left(anti\left(b\right)a\right)=neu{t}_{\left(n2\right)}\Rightarrow anti\left(b\right)a\in K\phantom{\rule{0ex}{0ex}}\Rightarrow anti\left(b\right)aK=K(a{N}_{2}={N}_{2}\iff a\in {N}_{2}).$$Thus, i is injective.
- (3)
- We must show that for any element in the domain (N
_{1}/K) gets mapped to it by i. let’s pick any element $\varphi \left(a\right)\in im\left(\varphi \right).$ By definition, $i\left(aK\right)=\varphi \left(a\right),$ hence i is surjective. - (4)
- We must show that $i\left(aK\text{}bK\right)=i\left(aK\right)i\left(bK\right).i\left(aK\text{}bK\right)=i\left(abK\right)\left(aK\text{}bK=abK\right)$ $=\varphi \left(ab\right)=\varphi \left(a\right)\varphi \left(b\right)=i\left(aK\text{}bK\right)=i\left(aK\right)i\left(bK\right).$ Thus, i is a neutro-homomorphism.

**Corollary 1**

**(A Few Special Cases of Fundamental Theorem of Neutro-homomorphism).**

- Let N = (1, 1, 1) be a trivial neutrosophic extended triplet. If ϕ: N
_{1}→N_{2}is an embedding, then neutrosophic ker(ϕ) = {neut(1) = 1N_{1}}. The Theorem 12 says that im(ϕ) ≅ {N_{1}/1N_{1}} ≅ N_{1}. - If ϕ: N
_{1}→N_{2}is a map ϕ(n) = neut(1) = 1N_{2}for all n_{2}∈ N_{1}, then neutrosophic ker(ϕ) = N_{1}, so Theorem 13 says that 1N_{2}= im(ϕ) ≅ N_{1}/N_{1}.

**Example**

**2.**

_{n}(the neutrosophic extended triplet subgroup of even permutation in NETG S

_{n}) has index 2 in S

_{n}.

**Solution.**

_{n}:A

_{n}] = 2. We will construct a surjective neutro-homomorphism φ: S

_{n}→Z

_{2}with neutrosophic triplet $ker\phi ={A}_{n}$. Here the neutrosophic extended triplets of Z

_{2}are (0, 0, 0) and (1, 1, 1). If this is achieved, it would follow that ${S}_{n}/{A}_{n}\cong {Z}_{2},$ so |S

_{n}/A

_{n}| = |Z

_{2}| = 2, and therefore [S

_{n}:A

_{n}] = |S

_{n}/A

_{n}| = 2, as desired. Define φ: S

_{n}→Z

_{2}by φ(f) = $\{\begin{array}{c}\left[0\right]\text{}if\text{}f\text{}is\text{}even\\ \left[1\right]\text{}if\text{}f\text{}is\text{}odd\end{array}$

_{n}. Here if x and y are both even or both odd, then xy is even. If x is even and y is odd, or if x is odd and y is even, then xy is odd. Let us see these four different cases as follows:

- (1)
- x and y are both even. Then xy is also even. So, φ(x) = φ(y) = φ(xy) = [0]. Since [0] + [0] = [0] holds.
- (2)
- x is even, and y is odd. Then xy is odd. So, φ(x) + φ(y) = [0] + [1] = [1] = φ(xy).
- (3)
- x is odd, and y is even. This case is analogous to case 2.
- (4)
- x and y are both odd. Then xy is even, so φ(x) + φ(y) = [1] + [1] = [0] = φ(xy). Thus, we verified that φ is a neutro-homomorphism. Finally, neutrosophic trplet kerφ = {x ∈ S
_{n}: φ(x) = [0]_{2}} is the neutrosophic extended triplet set of all even permutations, so neutrosophic triet $ker\phi ={A}_{n}.$

## 5. First Neutro-Isomorphism Theorem

**Theorem**

**14.**

- (a)
- $HK$is neutrosophic triplet subgroup of N.
- (b)
- $H\bigcap K$is neutrosophic triplet normal subgroup in K.
- (c)
- $\frac{HK}{H}\cong \frac{\mathrm{K}}{H\bigcap K}$

**Proof.**

- (a)
- Let $xy\in HK.$ If $x={h}_{1}{k}_{1}$ and $y={h}_{2}{k}_{2},{h}_{1}{h}_{2}\in H$ and ${k}_{1},{k}_{2}\in K.$ Consider$$\begin{array}{cc}\hfill x\left(anti\left(y\right)\right)=\left({h}_{1}{k}_{1}\right)& anti\left({h}_{2}{k}_{2}\right)\hfill \\ & =\left({h}_{1}{k}_{1}\right)anti\left({k}_{2}\right)anti\left({h}_{2}\right)\\ & ={h}_{1}\left({k}_{1}\left(anti\left({k}_{2}\right)\right)\right)anti\left({h}_{2}\right),({k}_{3}={k}_{1}\left(anti\left(k2\right)\right):{k}_{3}\in K\\ & ={h}_{1}{k}_{3}\left(anti\left({h}_{2}\right)\right)\\ & ={h}_{1}{k}_{3}\left(anti\left({h}_{2}\right)\right)anti\left({k}_{3}\right){k}_{3}\\ & ={h}_{1}{k}_{3}\left(anti\left({h}_{2}\right)\right)anti\left({k}_{3}\right){k}_{3}\\ & ={h}_{1}{h}_{2}{k}_{3}\text{}because\text{}H\u22b2kso\text{}{h}_{3}={k}_{3}\left(anti\left({h}_{2}\right)\right)anti\left({k}_{3}\right)\in H\\ & \Rightarrow x(anti\left(y\right)={h}_{4}{k}_{3}\in HK,\left({h}_{4}={h}_{1}{h}_{2}\right)\\ & \Rightarrow HK\text{}\mathrm{is}\text{}\mathrm{NETG}\text{}\mathrm{of}\text{}N.\end{array}$$
- (b)
- We have to prove $H\cap K$ is neutrosophic triplet normal subgroup in k or $H\cap K\u22b2k.$ Let $x\in H\cap K$ and $x\in K.$ If $x\in H$ and $x\in K,$ then $kx\left(anti\left(k\right)\right)\in H$ because $H\u22b2k$ and $kx\left(anti\left(k\right)\right)\in K$ because $xk\in K.$ Thus, $kx\left(anti\left(k\right)\right)\in H\cap K.$ Since $H\cap K\u22b2k.$
- (c)
- $\frac{HK}{H}\cong \frac{K}{H\bigcap K}$. Let H ⋂ K = D, so $\frac{K}{D}$ = $\frac{K}{H\bigcap K}$. Now let’s define a mapping ϕ: HK→$\frac{K}{D}$ by $\varphi \left(hk\right)=KD.$

- ϕ is well defined$${h}_{1}{k}_{1}={h}_{2}{k}_{2},{h}_{1}{h}_{2}\in H\text{}\mathrm{and}\text{}{k}_{1}{k}_{2}\in K$$$$\begin{array}{c}{k}_{1}{h}_{1}^{\prime}={k}_{2}{h}_{2}^{\prime}\\ \Rightarrow anti\left({k}_{2}\right){k}_{1}{h}_{1}^{\prime}={h}_{2}^{\prime}\\ \Rightarrow anti\left({k}_{2}\right){k}_{1}={h}_{2}^{\prime}\left(anti\left({h}_{1}\right)\right),{h}_{2}^{\prime}\left(anti\left({h}_{1}\right)\right)\in H\\ \Rightarrow anti\left({k}_{2}\right){k}_{1}\in H,\text{}\mathrm{but}\text{}anti\left({k}_{2}\right){k}_{1}\in K\\ \Rightarrow anti\left({k}_{2}\right){k}_{1}\in H\cap K=D\\ \Rightarrow anti\left({k}_{2}\right){k}_{1}\in D\\ \Rightarrow anti\left({k}_{2}\right){k}_{1}D=D\\ \begin{array}{l}\Rightarrow {k}_{1}D={k}_{2}D\\ \Rightarrow \varphi \left({h}_{1}{k}_{1}\right)=\varphi \left({h}_{2}{k}_{2}\right).\end{array}\end{array}$$
- ϕ is neutro-homomorphism.$$\begin{array}{cc}\hfill \mathrm{\Phi}\left({h}_{1}{k}_{1}.{h}_{2}{k}_{2}\right)& =\varphi ({h}_{1}\left({k}_{1}{h}_{2}\right){k}_{2}\hfill \\ & =\varphi \left({h}_{1}h{2}^{\prime}{k}_{1}{k}_{2}\right)\\ & ={K}_{1}{k}_{2}D\\ & ={k}_{1}D{k}_{2}D\\ & =\varphi \left({h}_{1}{k}_{1}\right).\varphi \left({h}_{2}{k}_{2}\right)\end{array}$$
- ϕ is onto.

**Example**

**3.**

= |HK/K|

= |H/H ⋂ K|

= |H : H ⋂ K|

= |H|/|H ⋂ K|, that is

|HK| = |H||K|/|H ⋂ K|

## 6. Second Neutro-Isomorphism Theorem

**Theorem**

**15.**

**Proof.**

- Ѱ is well defined. Let $ak=bk.$$$\begin{array}{c}anti\left(b\right)ak=k\\ anti\left(b\right)a\in k\\ \Rightarrow K\u22b2H\\ anti\left(b\right)a\in H\\ aH=bH\left(anti\left(b\right)aH=H\right)\\ \u0470\left(ak\right)=\u0470\left(bk\right)\end{array}$$
- Ѱ is neutro-homomorphism$${a}_{k},{b}_{k}\in N/K\phantom{\rule{0ex}{0ex}}\u0470\left({a}_{k}{b}_{k}\right)=\u0470\left(abk\right)=abH=aHbH=\u0470\left(ak\right)\u0470\left(bk\right).$$
- Ѱ is ontoFor all $y=aH\in N/H,x=ak\in N/K\Rightarrow $Ѱ$\left(x\right)=y.$
- kerѰ = H/KThe neutral element of $N/H$ is H. Therefore$$ker\text{}\u0470:\{xk\in N/K:\u0470\left(xk\right)=H\}\phantom{\rule{0ex}{0ex}}=\{xk\in N/K:\u0470\left(xk\right)=xH=H\}\phantom{\rule{0ex}{0ex}}\begin{array}{l}=\{xk\in N/K:x\in H\}\hfill \\ =\{xk\in H/K\}\hfill \\ =H/K.\hfill \end{array}$$By Theorem 13 $N/KH/K\cong N/H.$ ☐

## 7. Conclusions

## Author Contributions

## Funding

## Conflicts of Interest

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**MDPI and ACS Style**

Çelik, M.; Shalla, M.M.; Olgun, N.
Fundamental Homomorphism Theorems for Neutrosophic Extended Triplet Groups. *Symmetry* **2018**, *10*, 321.
https://doi.org/10.3390/sym10080321

**AMA Style**

Çelik M, Shalla MM, Olgun N.
Fundamental Homomorphism Theorems for Neutrosophic Extended Triplet Groups. *Symmetry*. 2018; 10(8):321.
https://doi.org/10.3390/sym10080321

**Chicago/Turabian Style**

Çelik, Mehmet, Moges Mekonnen Shalla, and Necati Olgun.
2018. "Fundamental Homomorphism Theorems for Neutrosophic Extended Triplet Groups" *Symmetry* 10, no. 8: 321.
https://doi.org/10.3390/sym10080321