# New Results on Neutrosophic Extended Triplet Groups Equipped with a Partial Order
## 4 Homomorphisms and Quotient Sets of po-NETGs
Moreover, \(\forall\;[a]_{\theta}\in G/\theta,\;\forall[b]_{\theta}\in G/\theta,\;\forall[ c]_{\theta}\in G/\theta,\;\forall[c]_{\theta}\subseteq[b]_{\theta}\), then \(a\leq b\), so we have \(a*c\leq b*c\), and \(c*a\leq c*b\). Thus,  
\[f(a)\cdot f(neut(a))=f(a*neut(a))=f(a)=f(neut(a)*a)=f(neut(a))\cdot f(a),\]5  
and  
\[f(a)\cdot f(neut(a))=f(a*neut(a))=f(a)=f(neut(a)*a)=f(neut(a))\cdot f(a),\]6  
Thus, \((G/\theta,\circ,\preceq)\) is a po-NETG. In the following, we give an example to illustrate Proposition 12. **Example 3**. : _Consider the po-NETG \((G,*,\leq)\) is given in Example 1. Now we define a relation \(\theta\) on \(G\) by_  
\[f(a)\cdot f(neut(a))=f(a*neut(a))=f(a)=f(neut(a)*a)=f(neut(a))\cdot f(a),\]7  
_Then we can verify that \(\theta\) is a congruence on \(G\) with the following blocks:_  
\[f(a)\cdot f(neut(a))=f(a*neut(a))=f(a)=f(neut(a)*a)=f(neut(a))\cdot f(a),\]8  
_So the quotient set \(G/\theta=\{[0]_{\theta},\;[a]_{\theta},\;[1]_{\theta}\}\). By Proposition 12, we know \((G/\theta,\circ,\preceq)\) is a po-NETG, in which \(neut([0]_{\theta})=[0]_{\theta},\;neut([a]_{\theta})=[c]_{\theta}=[a]_{\theta}, \;neut([1]_{\theta})=[1]_{\theta},\;\{anti([0]_{\theta})\}=\{[0]_{\theta},\;[a ]_{\theta},\;[1]_{\theta}\},\;\{anti([a]_{\theta})\}=\{[a]_{\theta}\},\;\{anti([ 1]_{\theta})\}=\{[a]_{\theta},\;[1]_{\theta}\},\) and then \(G/\theta\) is a chain, because \([0]_{\theta}\preceq[a]_{\theta}\preceq[1]_{\theta}\)._  
**Proposition 13**. : _Let \((G,*,\leq)\) be a po-NETG and \(\theta\) be a congruence on \(G\). Then the natural mapping \(\
atural_{\theta}:(G,*,\leq)\rightarrow(G/_{\theta},\circ,\preceq)\) given by \(\
atural_{\theta}(a)=[a]_{\theta}\) is a po-NETG homomorphism of po-NETGs._  
Proof. : As \(\
atural_{\theta}(a*b)=[a*b]_{\theta}=[a]_{\theta}\circ[b]_{\theta}=\
atural_{ \theta}(a)\circ\
atural_{\theta}(b)\;\forall\;a\in G,\;\forall b\in G.\) If \(a\leq b\), then \([a]_{\theta}\preceq[b]_{\theta}\) which implies \(\
atural_{\theta}(a)\preceq\
atural_{\theta}(b)\). Thus, the natural mapping \(\
atural_{\theta}:(G,*,\leq)\rightarrow(G/_{\theta},\circ,\preceq)\) is a po-NETG homomorphism of po-NETGs. Next, we give an example to explain Proposition 13. **Example 4**. : _From Example 3, we consider the natural mapping \(\
atural_{\theta}:(G,*,\leq)\rightarrow(G/_{\theta},\circ,\preceq)\). Thus, \(\
atural_{\theta}(0)=[0]_{\theta},\;\
atural_{\theta}(a)=\
atural_{\theta}(b)= \
atural_{\theta}(c)=[a]_{\theta},\;\
atural_{\theta}(1)=[1]_{\theta}.\) It is easy to verify that \(\
atural_{\theta}\) is a po-NETG homomorphism of po-NETGs._  
**Proposition 14**. : _Let \((G,*,\leq_{1})\) and \((T,\cdot,\leq_{2})\)be two po-NETGs and \(f:(G,*,\leq_{1})\rightarrow(T,\cdot,\leq_{2})\) be a po-NETG homomorphism of po-NETGs. We shall use the notation_  
\[f(a)\cdot f(neut(a))=f(a*neut(a))=f(a)=f(neut(a)*a)=f(neut(a))\cdot f(a),\]9  
_then we can get the following properties:_  
1. \(\mathit{Kerf}\) _is a congruence on_ \(G\)_;_
2. \(f\) _is a injective po-NETG homomorphism of po-NETGs if and only if_ \(\mathit{kerf}\) _is an identity congruence on_ \(G\)_;_
3. _There exists an injective po-NETG homomorphism of po-NETGs_ \(g:(G/\mathit{Kerf},\circ,\preceq)\rightarrow(T,\cdot,\leq_{2})\) _such that_ \(f=g\circ\
atural_{\mathit{Kerf}}\)_._  
Proof. :
1. Obviously, \(\mathit{Kerf}\) is an equivalence relation on \(G\). Let \(\forall\ a\in G,\forall\ b\in G,\forall\ c\in G,\forall\ d\in G\), if \((a,b)\in\mathit{Kerf}\) and \((c,d)\in\mathit{Kerf}\), then \(f(a)=f(b)\) and \(f(c)=f(d)\). Since \(f\) is a po-NETG homomorphism of po-NETGs, we have \(f(a*c)=f(a)\cdot f(c)=f(b)\cdot f(d)=f(b*d)\), and so \((a*c,b*d)\in\mathit{Kerf}\). Thus, \(\mathit{Kerf}\) is a congruence on \(G\). 2. If \(f\) is an injective po-NETG homomorphism of po-NETGs and if \((a,b)\in\mathit{kerf}\) then \(f(a)=f(b)\). Therefore, we get \(a=b\). Hence, by the arbitrariness of \((a,b)\), we obtain \(\mathit{kerf}\) is an identity congruence on \(G\). Conversely, suppose that \(\mathit{kerf}\) is an identity congruence on \(G\).