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 Dn be the dihedral group of equilateral triangle, with vertices on the unit circle, at angles 0, 2π∖3, and 4π∖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 D3={R0,R1,R2,S0,S1,S2}. 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 D3 = {1,2,3,4,5,6} (where R0=1,R1=2,R2=3,S0=4,S1=5 and S2=6) and S={x∈D3|x≠x−1} = {2,3}. Then, the associated inverse graph of D3 are the two disjointed graphs in the figure below.
From Fig. 3, we can draw the adjacency matrix of Γ(D3) 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 λ6−5λ4+6λ2−2λ3+4λ. The roots λi (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 D3 is \(3+2\sqrt {2}\).
Theorem 5
Let G=Dn be the dihedral group of order 2n, where n≥3 and S={x∈G|x≠x−1} be a subset of Dn. 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=Dn be the dihedral group of order 2n, 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 λ1,λ2,⋯,λn 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; det(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 Dn 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=Dn be the dihedral group of order 2n, where n≥3 and S={x∈G|x≠x−1} be a subset of Dn. Then, the inverse graph Γ(Dn) is never a complete bipartite graph.
Proof
Let G=Dn be the dihedral group of order 2n, where n≥3, S={x∈G|x≠x−1} and S⊆Dn. Suppose the inverse graph of Dn is bipartite for all n≥3, then the set of vertices V(Γ(Dn)) can be partitioned into two sets, say V1 and V2 (the set of rotation of angle 2πk/n and the set of reflections about the line through the origin) and also, there exist two vertices u∈V1 and v∈V2 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∈V1 and a u∈V2 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))={V1,V2}, where V1={x∣x∈S} and V2={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∈V1. □
Remark 2
Let G=Dn be the dihedral group of order 2n 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 V1 and V2 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(S3) of S3 is (Fig. 4)
Theorem 8
Let Dp⊆Dn be a subgroup of the dihedral group of order 2p and Sk⊆Sn be a subgroup of the symmetric group of order k!; k,p≥3 are positive integers which are not necessarily equal but 2p=k! and S={x∈Dp,Sk|x≠x−1}. Then, the inverse graphs of Dp and Sk are isomorphic.
Proof
Let Dp⊆Dn be a subgroup of a dihedral group of order 2p and Sk⊆Sn be a subgroup of a symmetric group of order k!; k,p≥3 are positive integers which are not necessarily equal. Suppose 2p=k! and S={x∈Dp and Sk|x≠x−1}, it suffices to show that the inverse graphs of Γ(Dp) and Γ(Sk) are isomorphic then the finite groups Dp and Sk have the same adjacency matrix. If the order of ∣S∣=2 then by Theorem 6, it is obvious that Dp and Sk have the same adjacency matrix, but if ∣S∣≠2, then ∣S∣ = 1 or ∣S∣>2. If ∣S∣ = 1, then the relation Γ(Dp)≃Γ(Sk) is trivial and we are left to show for when ∣S∣≥3. Suppose we claim that Γ(Dp)≃Γ(Sk) 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 Dp and let a and b be two corresponding vertices of Γ(Dp) if a and b are equal they must have the same effect on Γ(Sk). 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 Sk be a subgroup of the symmetric group Sn (Sk⊆Sn), such that the inverse graph of Sk; Γ(Sk)≃Γ(Dn), where Dp is a subgroup of Dn and \(k,\ p,\ n\in \mathbb {Z}\). Then, the energies \(\bar {E}(\Gamma (S_{k}))\) and \(\bar {E}(\Gamma (D_{p}))\) are equal.