# New Results on Neutrosophic Extended Triplet Groups Equipped with a Partial Order
## 3 Partially Ordered NETGs  
An NETG is a special set endowed with a multiplicative operation. Assuming that we introduce a partial order which is compatible with multiplication in a NETG, we will get the definition of partially ordered NETGs as indicated below. **Definition 5**. : _Let \((G,*)\) be a NETG. If there exists a partial order relation \(\leq\) on \(G\) such that \(a\leq b\) implying \(c*a\leq c*b\) and \(a*c\leq b*c\) for all \(a\in G,\ b\in G,\ c\in G\), then \(\leq\) is called a compatible partial order on \(G\), and \((G,*,\leq)\) is called a partially ordered NETG (for short, po-NETG)._  
_Similarly, if \((G,*)\) is a WCNETG and endowed with a compatible partial order, then \((G,*,\leq)\) is called a partially ordered WCNETG (po-WCNETG). Hence, po-WCNETGs must be po-NETGs._  
**Remark 1**. : _Obviously, the properties of NETGs and WCNETGs are holding in po-NETGs and po-WCNETGs, respectively._  
In the following, we give an example of a po-NETG. **Example 1**. : _Let \(G=\{0,\ a,\ b,\ c,\ 1\}\) with the Hasse diagram as shown in Figure 1, in which 0 denotes the bottom element (mean the element is smallest element w.r.t. to partial order) and 1 denotes the top element (mean the element is largest element w.r.t. to partial order) of \(G\). Then \(G\) is a partially ordered set._  
_Define multiplication \(*\) on \(G\) as shown in Table 1, where \(a,b,c\) to label the elements in the po-NETG and the multiplication \(*\) among these elements._We can verify that \((G,*)\) is a WCNETG. Moreover,  
\[neut(0)=0,\ \{anti(0)\}=\{0,\ a,\ b,\ c,\ 1\},\ 0^{-1}=0;\]  
\[neut(a)=c,\ \{anti(a)\}=\{b\},\ a^{-1}=b;\]  
\[neut(b)=c,\ \{anti(b)\}=\{a\},\ b^{-1}=a;\]  
\[neut(c)=c,\ \{anti(c)\}=\{c\},\ c^{-1}=c;\]  
\[neut(1)=1,\ \{anti(1)\}=\{a,\ b,\ c,\ 1\},\ 1^{-1}=1.\]  
It is easy to see that the partial order shown in Fig.1 is compatible with multiplication \(*\). Hence, \((G,*,\leq)\) is a po-WCNETG. Definition 6: If \((G,*,\leq)\) is a po-NETG, then \(a\in G\) is said to be a positive element if \(neut(a)\leq a\); and a negative element if \(a\leq neut(a)\). The subset \(P_{G}\) of all positive elements of \(G\) is called the positive cone of \(G\), and the subset \(N_{G}\) of all negative elements the negative cone. Remark 2: By Proposition 1, \(\forall\ a\in G,\ neut(a)\in P_{G}\cap N_{G}\), so \(P_{G}\cap N_{G}\
eq\emptyset\). Lemma 1: Let \((G,*)\) be an NETG. Then \(\forall\ a\in G\),  
\[[neut(a)]^{-1}=neut(a)=neut(a^{-1}).\]  
Proof: Let \(a\in G\). Then  
\[[neut(a)]^{-1} = anti(neut(a))*neut(neut(a))\] \[= anti(neut(a))*neut(a)\] \[= neut(neut(a))\] \[= neut(a).\]  
\[neut(a)=c,\ \{anti(a)\}=\{b\},\ a^{-1}=b;\]0 Hasse diagram. On the other hand, by Proposition 2(3), we have \(neut(a^{-1})=neut(anti(a)*neut(a))=neut(a)\). **Remark 3**. : _If \(G\) is a po-NETG and \(P\subseteq G\), we shall use the notation_  
\[neut(a)=c,\ \{anti(a)\}=\{b\},\ a^{-1}=b;\]1  
**Proposition 6**. : _Let \((G,*,\leq)\) be a po-NETG. Then \(P_{G}\cap P_{G}^{-1}=\{a\in G:a=neut(a)=a^{-1}\}.\)_  
Proof. : (\(\Longrightarrow\)) Let \(a\in G\). By Proposition 1 and Lemma 1, we have  
\[neut(a)=c,\ \{anti(a)\}=\{b\},\ a^{-1}=b;\]2  
so \(\{a\in G:a=neut(a)=a^{-1}\}\
eq\emptyset\). By Lemma 1, it is clear that  
\[neut(a)=c,\ \{anti(a)\}=\{b\},\ a^{-1}=b;\]3  
(\(\Longleftarrow\)) Let \(b\in P_{G}\cap P_{G}^{-1}\), then \(neut(b)\leq b\) and \(\exists\ c\in P_{G}\) such that \(b=c^{-1}\), so  
\[neut(a)=c,\ \{anti(a)\}=\{b\},\ a^{-1}=b;\]4  
that is, \(b\leq neut(b)\), whence \(b=neut(b)\). Hence,  
\[neut(a)=c,\ \{anti(a)\}=\{b\},\ a^{-1}=b;\]5  
Then we can conclude that \(b\in\{a\in G:a=neut(a)=a^{-1}\}\), and so  
\[neut(a)=c,\ \{anti(a)\}=\{b\},\ a^{-1}=b;\]6  
Thus, \(P_{G}\cap P_{G}^{-1}=\{a\in G:a=neut(a)=a^{-1}\}\). **Remark 4**. : _If \((G,*,\leq)\) is a po-NETG and \(P\subseteq G\), then we shall use the notation_  
\[neut(a)=c,\ \{anti(a)\}=\{b\},\ a^{-1}=b;\]7  
**Proposition 7**.