This section introduces the energy of the inverse graph of both the dihedral and symmetric groups. We begin with the following example of a dihedral group.

###
**Example 2**

Let *D*_{n} be the dihedral group of equilateral triangle, with vertices on the unit circle, at angles 0, 2*π*∖3, and 4*π*∖3. The matrix representation according to [16] is given by

$$R_{0} = \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right), R_{1} = \left(\begin{array}{cc} -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{array} \right), R_{2} = \left(\begin{array}{cc} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2} \end{array} \right) $$

$$S_{0} = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right), S_{1} = \left(\begin{array}{cc} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{array} \right), S_{2} = \left(\begin{array}{cc} -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{array} \right) $$

and *D*_{3}={*R*_{0},*R*_{1},*R*_{2},*S*_{0},*S*_{1},*S*_{2}}. By Definition 5, we obtain the compositions of rotations and reflections of the equilateral triangle in the following table;

Also, from Definition 3 and to ease understanding, we rewrite the elements of *D*_{3} = {1,2,3,4,5,6} (where *R*_{0}=1,*R*_{1}=2,*R*_{2}=3,*S*_{0}=4,*S*_{1}=5 and *S*_{2}=6) and *S*={*x*∈*D*_{3}|*x*≠*x*^{−1}} = {2,3}. Then, the associated inverse graph of *D*_{3} are the two disjointed graphs in the figure below.

From Fig. 3, we can draw the adjacency matrix of *Γ*(*D*_{3}) as follows:

$$A = \left(\begin{array}{cccccc} 0 & 1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 1 & 0 \\ \end{array} \right).\ $$

To compute the eigenvalues of *A*, we compute the roots of the characteristics polynomial of *A* which is given as *λ*^{6}−5*λ*^{4}+6*λ*^{2}−2*λ*^{3}+4*λ*. The roots *λ*_{i} (*i*=1,2,...,6) of the polynomial is given as 0, 0, –1, 2, –\(\sqrt {2}\), and \(\sqrt {2}\). Hence, from Definition 6, the energy of the inverse graph of *D*_{3} is \(3+2\sqrt {2}\).

###
**Theorem 5**

Let *G*=*D*_{n} be the dihedral group of order 2*n*, where *n*≥3 and *S*={*x*∈*G*|*x*≠*x*^{−1}} be a subset of *D*_{n}. Then, the energy \(\bar {E}\) of the inverse graph *Γ*(*G*) is given as \(3 + 2\sqrt {2} \leq \bar {E}(\Gamma (G)) < 2m \) where *m* is the number of edges in *Γ*(*G*).

###
*Proof*

Let *G*=*D*_{n} be the dihedral group of order 2*n*, where *n*≥3 and *S*={*x*∈*G*|*x*≠*x*^{−1}}. Suppose *Γ*(*G*) is the inverse graph of the dihedral group with *n* vertices and *m* edges. Let *A* be the adjacency matrix of *Γ* with *λ*_{1},*λ*_{2},⋯,*λ*_{n} as its eigenvalues. Then, to obtain the bounds of the energy of our inverse graphs, we shall adopt the general form of energy of graphs (see Brualdi [20]) given as

$$\begin{array}{*{20}l} \bar{E}(\Gamma(G))\geq \sqrt{2m + n(n-1)\mid det A \mid^{2/n}}. \end{array} $$

(1)

Recall that, the inverse graph *Γ* of finite groups are simple graphs. Furthermore, they have no loops, and thus, the leading diagonals of the adjacency matrix *A* are 0s; therefore, the determinant of *A*; *d**e**t*(*A*)=0 then (1) becomes

$$\begin{array}{*{20}l} \bar{E}(\Gamma(G)) \leq \sqrt{2m}. \end{array} $$

(2)

Also, the energy of a graph is a function of both the vertex set and the edge set (see [20]). Therefore, \( \bar {E}(\Gamma (G)) \in \mathbb {R}^{+} \). And if (2) holds, then it is safe to write the following

$$\begin{array}{*{20}l} \bar{E}(\Gamma(G)) \leq \sqrt{2m} < 2m. \end{array} $$

(3)

And using the inequality property (if *a*≤*b*<*c*, then *a*<*c*) and we can write (3) as

$$\begin{array}{*{20}l} \bar{E}(\Gamma(G)) < 2m. \end{array} $$

(4)

Furthermore, since the least number *n* of *D*_{n} is 3, the lower bound is obvious from Example 2. Then, it is safe to write

$$\begin{array}{*{20}l} \bar{E}(\Gamma(G))\geq 3 + 2\sqrt{2}. \end{array} $$

(5)

(see Example 2), then combining (4) and (5), we can conclude that

$$3 + 2\sqrt{2}\leq \bar{E}(\Gamma(G)) < 2m. $$

□

###
**Theorem 6**

[19] Let *G* be a finite non-abelian group generated by two elements of order 2.Then, *G* is isomorphic to a dihedral group.

###
**Theorem 7**

Let *G*=*D*_{n} be the dihedral group of order 2*n*, where *n*≥3 and *S*={*x*∈*G*|*x*≠*x*^{−1}} be a subset of *D*_{n}. Then, the inverse graph *Γ*(*D*_{n}) is never a complete bipartite graph.

###
*Proof*

Let *G*=*D*_{n} be the dihedral group of order 2*n*, where *n*≥3, *S*={*x*∈*G*|*x*≠*x*^{−1}} and *S*⊆*D*_{n}. Suppose the inverse graph of *D*_{n} is bipartite for all *n*≥3, then the set of vertices *V*(*Γ*(*D*_{n})) can be partitioned into two sets, say *V*_{1} and *V*_{2} (the set of rotation of angle 2*π**k*/*n* and the set of reflections about the line through the origin) and also, there exist two vertices *u*∈*V*_{1} and *v*∈*V*_{2} which are adjacent. But the dihedral groups are non-abelian and these disjoint partitions contain elements which are not commutative. Hence, by Definition 3, it is not possible for a *v*∈*V*_{1} and a *u*∈*V*_{2} to be adjacent. □

###
**Corollary 1**

Let *G* be a finite non-abelian group with exactly two non-self invertible elements. Then, the associated inverse graph *Γ*(*G*) is never a complete bipartite graph.

###
*Proof*

Let *G* be a finite non-abelian group with ∣*S*∣ = 2 and let the partition of the vertex set *V*(*Γ*(*G*))={*V*_{1},*V*_{2}}, where *V*_{1}={*x*∣*x*∈*S*} and *V*_{2}={*y*∣*y*∉*S*}. Suppose *x*,*y*∈*V*(*Γ*(*G*)) and for the pair *x*,*y* to be adjacent, *x*∗*y* will be an element in *S* But *G* is non-abelian and *S* contains only two elements. Then, obviously there is no any other element say *z* such that *z*=*x*∗*y*∈*V*_{1}. □

###
**Remark 2**

Let *G*=*D*_{n} be the dihedral group of order 2*n* with a non bipartite graph, then the adjacency matrix of *G* is non-symmetric and the sum of the absolute value of the eigenvalues is not equal to zero.

###
**Corollary 2**

Let *G* be a finite non-abelian group with an even order *n* and *S*={*x*∈*G*|*x*≠*x*^{−1}}. Then, the associated inverse graph *Γ*(*G*) of the group decomposes into two smaller disjoint subgraphs.

###
*Proof*

Let *G* be a non-abelian group of order *n* where *n* is even, *S*={*x*∈*G*|*x*≠*x*^{−1}}. Suppose *V*_{1} and *V*_{2} are two disjoint vertex sets of *G*, where \(V_{1}(\Gamma (G)) = \{1, 2,..., \frac {n}{2}\}\), \(V_{2}(\Gamma (G)) = \{\frac {n}{2},..., n\}\). Note that, for *x*,*y*∈*G* to be adjacent, then *x*∗*y* or *y*∗*x* must be in *S*. However, *G* is non-abelian and obviously, not all *x* and *y* in *G* are commutative. To clearly show that there are two distinct vertex sets which are disjoint but forms the disjoint subgraphs, we will consider two cases.

Case (I), let *x* and *y* be two vertices in *G*, such that *x*∗*y* and *y*∗*x* are both in *S*, they might not necessarily be the same, but their is a common edge between *x* and *y*. This sets of vertices and edges form a part of the inverse graph *Γ*(*G*) that is connected.

Case (II), let *x* and *y* be two different vertices in *G* such that *x*∗*y* is in *S* but the converse *y*∗*x* is not in *S*, this set of vertices also form the second part of the inverse graph *Γ*(*G*) which is not connected. Thus, clearly, we can conclude that an inverse graph *Γ*(*G*) of a non-abelian group *G* with an even order *n* decomposes into two smaller disjoint subgraphs. □

###
**Example 3**

Let

$$G = S_{3} = \left\{A = \left(\begin{array}{ccc} 1 & 2 & 3 \\ 1 & 2 & 3 \end{array} \right), B = \left(\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 1 & 3 \end{array} \right), C = \left(\begin{array}{ccc} 1 & 2 & 3 \\ 3 & 2 & 1 \end{array} \right),\right.$$

$$\left. D = \left(\begin{array}{ccc} 1 & 2 & 3 \\ 1 & 3 & 2 \end{array} \right), E = \left(\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 3 & 1 \end{array} \right), F = \left(\begin{array}{ccc} 1 & 2 & 3 \\ 3 & 1 & 2 \end{array} \right) \right\}$$

be the symmetric group of order 6. By Definition 3, we obtain that

$$S = \left\{E = \left(\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 3 & 1 \end{array} \right), F = \left(\begin{array}{ccc} 1 & 2 & 3 \\ 3 & 1 & 2 \end{array} \right) \right\}.$$

Since *S*={*x*∈*G*|*x*≠*x*^{−1}}, then the inverse graph *Γ*(*S*_{3}) of *S*_{3} is (Fig. 4)

###
**Theorem 8**

Let *D*_{p}⊆*D*_{n} be a subgroup of the dihedral group of order 2p and *S*_{k}⊆*S*_{n} be a subgroup of the symmetric group of order *k*!; *k*,*p*≥3 are positive integers which are not necessarily equal but 2*p*=*k*! and *S*={*x*∈*D*_{p},*S*_{k}|*x*≠*x*^{−1}}. Then, the inverse graphs of *D*_{p} and *S*_{k} are isomorphic.

###
*Proof*

Let *D*_{p}⊆*D*_{n} be a subgroup of a dihedral group of order 2*p* and *S*_{k}⊆*S*_{n} be a subgroup of a symmetric group of order *k*!; *k*,*p*≥3 are positive integers which are not necessarily equal. Suppose 2*p*=*k*! and *S*={*x*∈*D*_{p} and *S*_{k}|*x*≠*x*^{−1}}, it suffices to show that the inverse graphs of *Γ*(*D*_{p}) and *Γ*(*S*_{k}) are isomorphic then the finite groups *D*_{p} and *S*_{k} have the same adjacency matrix. If the order of ∣*S*∣=2 then by Theorem 6, it is obvious that *D*_{p} and *S*_{k} have the same adjacency matrix, but if ∣*S*∣≠2, then ∣*S*∣ = 1 or ∣*S*∣>2. If ∣*S*∣ = 1, then the relation *Γ*(*D*_{p})≃*Γ*(*S*_{k}) is trivial and we are left to show for when ∣*S*∣≥3. Suppose we claim that *Γ*(*D*_{p})≃*Γ*(*S*_{k}) when ∣*S*∣≥3, then we have to show a map *ψ* which is isomorphic on the two groups, we first show that *ψ* is an injection. So let *x* and *y* be two elements of *D*_{p} and let *a* and *b* be two corresponding vertices of *Γ*(*D*_{p}) if *a* and *b* are equal they must have the same effect on *Γ*(*S*_{k}). Check their effect on the identity vertex, *a*=*a*(*e*)=*ψ*(*e*)=*b*(*e*)=*b* clearly *ψ*(*a*)=*ψ*(*b*) shows the injection of the map *ψ* and *a* = *b* and the bijection is obvious. □

Consequently, we obtain the following remark from Theorems 5 and 8.

###
**Remark 3**

Let *S*_{k} be a subgroup of the symmetric group *S*_{n} (*S*_{k}⊆*S*_{n}), such that the inverse graph of *S*_{k}; *Γ*(*S*_{k})≃*Γ*(*D*_{n}), where *D*_{p} is a subgroup of *D*_{n} and \(k,\ p,\ n\in \mathbb {Z}\). Then, the energies \(\bar {E}(\Gamma (S_{k}))\) and \(\bar {E}(\Gamma (D_{p}))\) are equal.