Proof. Consider a finite solvable group $G=⟨a,b:a^{2^{\beta }}=b^2=e,ab=a^{-1}b⟩$, where $\beta$ is even. Let $\mathrm{\Delta }$ be the subset of $\mathrm{{\rm Y}}\times \mathrm{{\rm Y}}$ and $G$ acts on $\mathrm{\Delta }$ by the action $({\upsilon }_1,{\upsilon }_2{)}^g=({\upsilon }_1^g,{\upsilon }_2^g)$, where ${\upsilon }_1,{\upsilon }_2\in \mathrm{{\rm Y}}$ and $g\in G$. This implies, the number of elements in $\mathrm{\Delta }$ is $2(2^{\beta }{)}^2+2^{\beta }$. Based on regular action, we found three different types of orbitals.

Case I

For orbital of the form $O((e,a^{m}b),(e,a^{n}b))$, $0\le m\le 2^{\beta }-1$ and $1\le n\le 2^{\beta }-1$. We consider two subcases.

1. For $m\le \beta +3$, the vertices in the form of $(e,a^{m}b)$ and
   $(a^{2^{\beta -1}},a^{2^{\beta -1}+i}b)$, $0\le i\le 2^{\beta }-1$ are adjacent to the vertices in the form of $(e,a^{n}b)$ and $(a^{2^{\beta -1}},a^{2^{\beta -1}+i}b)$, $0\le i\le 2^{\beta }-1$. Thus, we have $2^{\beta }-(n+1)$, $1\le n\le 2^{\beta -1}+1$ components of $\bigcup _{i=1}^2{K}_2^i$.

   On the other hand, the vertices in the form of $(e,a^{m}b)$, $(e,a^{n}b)$, $(a^{2^{\beta -1}},a^{m}b)$ and $(a^{2^{\beta -1}},a^{n}b)$ are adjacent to one another to form $2^{\beta -1}$ connected component of four vertices.

   Thus, it follows that ${\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}}=\bigcup _{i=1}^2{K}_2^i\bigcup C_4$.
   

2. For $m\ge \beta +4$, we found that the $O((e,a^{m}b),(e,a^{n}b))$ consist $2^{\beta -1}-(n+1)$, $0\le n\le 2^{\beta -1}-2$ components of $\bigcup _{i=1}^2{K}_2^i$.

   This implies, ${\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}}=\bigcup _{i=1}^2{K}_2^i$.

Case II

For orbital of the type $O((e,a^{m}b),(a^{2^{\beta -1}},a^{n}b))$, $0\le m\le 2^{\beta }-1$ and $1\le n\le 2^{\beta }-1$ there exist ${\upsilon }_1,{\upsilon }_2\in \mathrm{{\rm Y}}$ such that ${\upsilon }_1=(e,a^{m}b)$ is adjacent to ${\upsilon }_2=(a^{2^{\beta -1}},a^{m}b)$ to form one component, $\bigcup _{i=1}^2{K}_2^i$ and one complete component, $K_2$. Hence it follows that ${\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}}=\bigcup _{i=1}^2{K}_2^i\bigcup K_2$.

Case III

(Shows exception) Based on regular action the orbital of the type
$O((e,a^{2^{\beta -1}})(e,a^{n}b))$, $0\le n\le 2^{\beta }-1$ of size two, there is always one common vertex of the form $(e,a^{2^{\beta -1}})$ adjacent to the two vertices of the type $(e,a^{n}b)$ and $(a^{2^{\beta -1}},a^{n}b)$. Hence, there are $2^{\beta }$ connected orbital regular graph of three vertices.

Proof. Consider a finite solvable group $G=⟨a,b:a^{2^n}=b^2=e,ab=b{a}^{2^{n-1}-1}⟩$, $n$ is even. Let $\mathrm{\Delta }$ be the subset of $\mathrm{{\rm Y}}\times \mathrm{{\rm Y}}$ and $G$ acts regularly on $\mathrm{\Delta }$. If ${\mathrm{\Gamma }}^{\ast }$ is an orbital graph then each orbital have distinct orbital graph and two vertices of a graph are linked to each other if ${\upsilon }_1,{\upsilon }_2\in \mathrm{{\rm Y}}$ there exist $g\in G$ such that $({\upsilon }_1,{\upsilon }_2{)}^g=({\upsilon }_1^g,{\upsilon }_2^g)$. To show orbital regular graph of different orbitals and for $i$ even, we have four cases:

Case I

In the orbital $O((e,a^{2^{n-1}})(e,a^{i}b))$, where $0\le i\le 2^n-1$. If $G$ acts regularly on $\mathrm{\Delta }$ then the vertices of the form $(e,a^{2^{n-1}})$ and $(e,a^{i}b)$ are adjacent to the vertices of the form $(a^{2^{n-1}},a^{2^{n-1}}b)$ and $(b,a^{2^{n-1}}b)$. Thus, we found that the orbital $O((e,a^{2^{n-1}})(e,a^{i}b))$ consist of $2^m$, $m\ge 2$ connected orbital regular graph of four vertices.

This implies, ${\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}}=C_4$.

On the other hand, for orbital $O((e,a^{i}b)(a^{i}b,a^{i+2^{n-1}}b))$, $0\le i\le 2^n-1$, the vertex ${\upsilon }_1=(a^{i}b,a^{i+2^{n-1}}b)$ is adjacent to the vertices ${\upsilon }_2=(e,a^{i}b)$ and ${\upsilon }_3=(a^{2^{n-1}},a^{i}b)$, $0\le i\le 2^n-1$. Hence, we have $2^{n-2}$ connected orbital regular graph of three vertices.

Case II

Consider the orbital of the form $O((e,a^{2^{n-1}}),(a^{i}b,a^{i+2^{n-1}}b))$ and $O((a^{i^{\prime }}b,a^{i+2^{n-1}}b),(a^{i^{\prime }+2}b,a^{i+2+2^{n-1}}b))$, $0\le i\le 2^n-1$, $0\le i^{\prime }\le 2^{n-1}.$

If $G$ acts regularly on $\mathrm{\Delta }$, we found that both the orbitals are central orbitals and each having two adjacent vertices of the type $(e,a^{2^{n-1}})$, $(a^{i}b,a^{i+2^{n-1}}b)$ and $(a^{i}b,a^{i+2^{n-1}}b)$, $(a^{i+2}b,a^{i+2+2^{n-1}}b)$. Hence, for both the orbitals we have ${\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}}=K_2$.

Case III

For orbital of the form $O((e,a^{i}b),(e,a^{i+2}b))$ where $0\le i\le 2^n-1$, the orbital regular graph of a particular orbitals are;

Hence, we found that each orbital have ${\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}}=\bigcup _{i=1}^2{K}_2^i\bigcup C_4$ and also there exist one orbital which always have ${\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}}=\bigcup _{i=1}^2{K}_2^i$.

Thus, we can say that the orbital $O((e,a^{i}b),(e,a^{i+2}b))$ have disconnected orbital regular graph.

Case IV

For Orbital $O((e,a^{i}b),(a^{2^{n-1}},a^{i}b))$ where $0\le i\le 2^n-1$, we have;

From above table, we see that each orbital have disconnected orbital regular graph, ${\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}}=\bigcup _{i=1}^2{K}_2^i\bigcup K_2$ except last orbital, ${\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}}=K_2$.

Proof. For $n=2$, there is diagonal orbital of group $G$. This implies that the graph is empty.

For $n=4$, If $G$ acts regularly on $\mathrm{\Delta }$ then ${\upsilon }_1,{\upsilon }_2\in \mathrm{{\rm Y}}$ there exist $g\in G$ such that $({\upsilon }_1,{\upsilon }_2{)}^g=({\upsilon }_1^g,{\upsilon }_2^g)$. The vertex ${\upsilon }_1$ joined the vertex ${\upsilon }_2$ whenever $({\upsilon }_1,{\upsilon }_2{)}^g=({\upsilon }_1^g,{\upsilon }_2^g)$. According to [2] the elements of $\mathrm{\Delta }$ are in the form $$\begin{array}{rl}& ((e,(ab)),(e,(cd))),((e,(ab)),((ab),(cd))),((e,(ab)),(e,(ab)(cd))),\\ & (((e,(ab)),((ab)(cd),(ac)(bd))),((ab)(cd),(e,(ab)(cd))),(((ab)(cd)),((ab),(ab)(cd))),\\ & (((ab)(cd)),((ab)(cd),(ac)(bd))),((e,(ab)(cd)),((ab),(ab)(cd))),\\ & ((e,(ab)),((ab),(ab)(cd))),((e,(ab)(cd)),((ab)(cd),(ac)(bd))),\\ & ((ab),((ab)(cd)),((ab)(cd)),((ac)(bd))).\end{array}$$

Based on regular action, there are $26$ orbitals of size one, five orbitals of size two and four orbitals of size four. Therefor, in the orbital of the form $((e,(ab)),(e,(ac)))$, the vertices of the form $(e,ab)$ are adjacent to the vertices of the form $(e,ac)$. Hence,we found that the orbital regular graph for the orbital is connected graph of two vertices, $K_2$. Similarly, for the remaining $25$ orbitals of size one, we have ${\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}}=K_2$. However, for the orbitals $((e,(ab))(e,(ab)(cd)))$, $((e,(ab)(cd))(e,(ac)(bd)))$ and
$((e,(ab)(cd))(e,(ad)(bc)))$ there are four vertices in each orbital and each vertex is adjacent to their next vertex to form a closed path, that is $$(e,ab)\to (e,(ab)(cd))\to ((ab)(cd),(cd))\to ((ab)(cd))\to (e,ab).$$ This implies, ${\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}}=C_4$.

Next, for the orbital of the form $((e,ab)((ab)(cd),(ad)(bc)))$, $((e,(ab)(cd))((ac)(bd)))$ and $((e,(ab)(cd))((ad)(bc)))$, we found that each orbitals contain connected graph of three vertices;

Exception

Consider an orbitals of the type $((e,ab)(e,(ac)(bd)))$ and $((e,ab)((ab)(cd),(ac)(bd)))$ of size two, we found that there are four vertices in each orbitals. If $G$ acts regularly on the set $\mathrm{\Delta }$ then each orbital form two pairs $(P_1,P_2)$ of connected graph but there is no edge between two pairs which shows that both the pairs are disconnected. This implies, ${\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}}=\bigcup _{i=1}^4{K}_2$. 

Proof. Since $n$ is odd, the number of elements in $\mathrm{\Delta }$ are in the form $$\begin{gathered} ((e,(ab)),(e,(ac))), \\ ((e,(ab)),(e,(bc))),((e,(ac)),(e,(bc))) \end{gathered}$$. Based on regular action on $\mathrm{\Delta }$ there are three central orbitals of size one, that is $$\begin{array}{rl}{\mathrm{\Delta }}_1=& ((e,(ab)),(e,(ac))),\\ {\mathrm{\Delta }}_2=& ((e,(ab)),(e,(bc))),\\ {\mathrm{\Delta }}_3=& ((e,(ac)),(e,(bc))).\end{array}$$

This implies that each orbital have two adjacent vertices which follows that each orbital have ${\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}}=K_2$. 

Proof. Based on Theorem 2, for orbital $O((e,a^{m}b),(e,a^{n}b))$, $m\le \beta +3$ and $1\le n\le 2^{\beta }-1$ there are $2^{\beta }-(n+1)$, $1\le n\le 2^{\beta -1}+1$ complete components of $\bigcup _{i=1}^2{K}_2$ and $2^{\beta -1}$ connected component of four vertices then the number of edges can be computed as follows: $$\begin{array}{rl}|E({\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}})|& =2(2^{\beta }-(n+1))+2^{\beta -1}(\genfrac{}{}{0ex}{}{4}{2})-2\\ & =2(2^{\beta }-(n+1))+2^{\beta -1}(2^2)\\ & =2(2^{\beta }-(n+1))+2^{\beta +1}\\ & =2^{\beta +2}-2(n+1).\end{array}$$ Next, for $m\le \beta +4$ there are $2^{\beta -1}-(n+1)$ complete component of $\bigcup _{i=1}^2{K}_2$, then $$\begin{array}{rl}|E({\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}})|& =2(2^{\beta -1}-(n+1))\\ & =2^{\beta }-2(n+1).\end{array}$$ Now, for orbital $O((e,a^{m}b),(a^{2^{\beta -1}},a^{n}b))$, $0\le m\le 2^{\beta }-1$ and $1\le n\le 2^{\beta }-1$ there are one complete component of $\bigcup _{i=1}^2{K}_2$ and one complete component of $K_2$. Hence it follows that $$\begin{array}{rl}|E({\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}})|& =3\end{array}$$ and for orbital $O((e,a^{2^{\beta -1}})(e,a^{n}b))$, $0\le n\le 2^{\beta }-1$ there are $2^{\beta }$ connected graph of three vertices. Thus $$\begin{array}{rl}|E({\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}})|& =2^{\beta }(\genfrac{}{}{0ex}{}{3}{2})-1\\ & =2^{\beta +1}.\end{array}$$ 

Proof. Based on Theorem 3, we consider each case to compute number of edges of each orbital.

Case I

We found that for the orbital $O((e,a^{2^{n-1}})(e,a^{i}b))$, there are $2^m$, $m\ge 2$ connected graph of four vertices. This implies that $$\begin{array}{rl}|E({\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}})|& =2^m(\genfrac{}{}{0ex}{}{4}{2})-2\\ & =2^{m+2},m\ge 2.\end{array}$$ Also, for $O((e,a^{i}b)(a^{i}b,a^{i+2^{n-1}}b))$ there are $2^{n-2}$ connected graph of three vertices. Thus, $$\begin{array}{rl}|E({\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}})|& =3(2^{n-2}).\end{array}$$

Case II

Consider $O((e,a^{2^{n-1}})(a^{i}b,a^{i+2^{n-1}}b))$ and $O((a^{i^{\prime }}b,a^{i+2^{n-1}}b)(a^{i^{\prime }+2}b,a^{i+2^{n-1}+2}b))$, then ${\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}}=K_2$. This implies $$\begin{array}{rl}|E({\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}})|& =1.\end{array}$$

Case III

For $O((e,a^{i}b)(e,a^{i+2}b))$, we found that each orbital have $(2n-(k+1))$, $k\ge 1$ component of $\bigcup _{i=1}^2{K}_2$ and one connected component of four vertices. Thus, we have $$\begin{array}{rl}|E({\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}})|& =2(2n-(k+1))+4.\end{array}$$ This case follows the exception, i.e., there exist one orbital of the type $((e,a^{2^{n-2}}b)(e,a^{2^{n-2}+2}b)$ which always have one component of $\bigcup _{i=1}^2{K}_2$, then $$\begin{array}{rl}|E({\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}})|& =2.\end{array}$$

Case IV

For orbital $O((e,a^{i}b)(a^{2^{n-1}},a^{i}b))$, we find $$\begin{array}{rl}|E({\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}})|& =2(2n-(k+1))+1,\end{array}$$ and (exception) $O((e,a^{2^{n-2}})(a^{2^{n-1}},a^{2^{n-1}}b))$, we have $$\begin{array}{rl}|E({\mathrm{\Gamma }}_G^{{\ast }^{\mathrm{\Delta }}})|& =1.\end{array}$$