# New Results on Neutrosophic Extended Triplet Groups Equipped with a Partial Order
## 4 Homomorphisms and Quotient Sets of po-NETGs  
**Definition 8**. : _Let \((G,*,\leq_{1})\) and \((T,\cdot,\leq_{2})\) be two po-NETGs. The map \(f:G\to T\) is called a po-NETG homomorphism of po-NETGs, if \(f\) satisfies: \(\forall a\in G,\ \forall b\in G\)_  
(1) \(f(a*b)=f(a)\cdot f(b)\)_; (2) \(a\leq_{1}b\) _implies_ \(f(a)\leq_{2}f(b)\)_._  
**Proposition 11**. : _Let \((G,*,\leq_{1})\) and \((T,\cdot,\leq_{2})\) be two po-NETGs, and let \(f:G\to T\) be a po-NETG homomorphism of po-NETGs. The following properties hold:_  
(1) \(\forall a\in G,\ f(neut(a))=neut(f(a))\)_; (2) \(\forall a\in G,\ \{f(b):b\in\{anti(a)\}\}\subseteq\{anti(f(a))\}\)_, and if_ \(f\) _is bijective, then_ \(\{f(b):b\in\{anti(a)\}\}=\{anti(f(a))\}\)_; (3) \(\forall a\in G,\ [f(a)]^{-1}=f(a^{-1})\)_; (4) \(\forall a\in P_{G},\ f(a)\in P_{T}\); (5) \(\forall a\in N_{G},\ f(a)\in N_{T}\)_._ \begin{table}
\begin{tabular}{|c|c|c|c|} \hline \(*\) & a & b & c \\ \hline a & a & b & c \\ \hline b & b & c & a \\ \hline c & c & a & b \\ \hline \end{tabular}
\end{table} 
Table 2: Multiplication \(*\) on \(G\). **Proof. **  
(1) \(\forall\ a\in G,\ \forall\ b\in\{anti(a)\},\) since  
\[f(a)\cdot f(neut(a))=f(a*neut(a))=f(a)=f(neut(a)*a)=f(neut(a))\cdot f(a),\] \[f(a)\cdot f(b)=f(a*b)=f(neut(a))=f(b*a)=f(b)\cdot f(a),\]  
then we obtain \(f(neut(a))=neut(f(a)).\)  
(2) From the proof of (1), we can get that  
\[\forall\ a\in G,\ \forall\ b\in\{anti(a)\},\ f(b)\in\{anti(f(a))\},\]  
and so  
\[\{f(b):b\in\{anti(a)\}\}\subseteq\{anti(f(a))\}.\]  
If \(f\) is bijective, then \(\forall\ d\in\{anti(f(a))\},\ \exists\ c\in G\) such that \(f(c)=d.\) Since  
\[f(c*a)=f(c)\cdot f(a)=d\cdot f(a)=neut(f(a))=f(neut(a)),\]  
we have \(c*a=neut(a).\) Similarly, we can get \(a*c=neut(a).\) Thus, \(c\in anti(a)\) and so  
\[d=f(c)\in\{f(b):b\in\{anti(a)\}\}.\]  
By the arbitrariness of \(d\), we have  
\[\{anti(f(a))\}\subseteq\{f(b):b\in\{anti(a)\}\}.\]  
Then,  
\[\{f(b):b\in\{anti(a)\}\}=\{anti(f(a))\}.\]  
(3) Let \(a\in G\) and \(b\in\{anti(a)\}.\) By (2), \(f(b)\in\{anti(f(a))\}.\) Then by (1), we have  
\[[f(a)]^{-1}=anti(f(a))\cdot neut(f(a))=f(b)\cdot f(neut(a))=f(b*neut(a))=f(a^{- 1}).\]  
(4) Since \(\forall\ a\in P_{G},\ neut(a)\leq_{1}a,\) we have \(neut(f(a))=f(neut(a))\leq_{2}f(a),\) and so \(f(a)\in P_{T}.\)  
(5) It is similar to (4). **Definition 9**. : _Let \((G,*,\leq)\) be a po-NETG and \(\theta\) be an equivalence relation on \(G\). If \(\theta\) satisfies_  
\[f(a)\cdot f(neut(a))=f(a*neut(a))=f(a)=f(neut(a)*a)=f(neut(a))\cdot f(a),\]0  
_then \(\theta\) is called a congruence on \(G\)._  
_Obviously, \(\theta_{1}=\{(a,a):a\in G\}\) and \(\theta_{2}=\{(a,b):\forall\ a,\ b\in G\}\) are both congruences on \(G\), and they are called identity congruence on \(G\) and pure congruence on \(G\), respectively._  
**Definition 10**. : _Let \((G,*,\leq)\) be a po-NETG and \(\theta\) be a congruence on \(G\). A multiplication \(\circ\) on the quotient set \(G/\theta=\{[a]_{\theta}:a\in G\}\) is defined by_  
\[f(a)\cdot f(neut(a))=f(a*neut(a))=f(a)=f(neut(a)*a)=f(neut(a))\cdot f(a),\]1  
**Proposition 12**. : _Let a relation \(\preceq\) on \((G/\theta,\circ)\) be defined by_  
\[f(a)\cdot f(neut(a))=f(a*neut(a))=f(a)=f(neut(a)*a)=f(neut(a))\cdot f(a),\]2  
_Then, \((G/\theta,\circ,\preceq)\) is a po-NETG._  
Proof. : We can verify that \(\circ\) is associative. Let \([a]_{\theta}\in G/\theta\) (see Definition 10), since  
\[f(a)\cdot f(neut(a))=f(a*neut(a))=f(a)=f(neut(a)*a)=f(neut(a))\cdot f(a),\]3  
and  
\[f(a)\cdot f(neut(a))=f(a*neut(a))=f(a)=f(neut(a)*a)=f(neut(a))\cdot f(a),\]4  
we conclude that \((G/\theta,\circ)\) is a NETG, in which \(\forall\;[a]_{\theta}\in G/\theta,\;neut([a]_{\theta})=[neut(a)]_{\theta}\) and \([anti(a)]_{\theta}\in\{anti([a]_{\theta})\}\). Then it is easy to see that \(\preceq\) is a partial order on \((G/\theta,\circ)\).