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Mostar index of graphs associated to groups

  • Masood Ur Rehman EMAIL logo , Muhammad Salman , Sheraz Khan , Ayse Dilek Maden and Faisal Ali
Published/Copyright: August 23, 2022

Abstract

A bond-additive connectivity index, named as the Mostar index, is used to measure the amount of peripheral edges of a simple connected graph, where a peripheral edge in a graph is an edge whose one end vertex has more number of vertices closer as compared to the other end vertex. In this study, we count the contribution of peripheral edges in commuting, non-commuting, and non-conjugate graphs associated to the dihedral and semi-dihedral groups. In fact, we compute the Mostar index of these graphs.

AMS subject classification numbers: 05C09; 05C12; 20E45

1 Introduction

A connectivity index is a form of molecular attribute whose computation depends upon a chemical graph of a chemical substance in the subject of mathematical chemistry. A large variety of numerical values, also known as topological indices, have been suggested and explored in attempt to distil and collect, or summarie, the content contained in graph connectivity patterns (Todeschini and Conosonni, 2002). Topological indices are numerical quantities that describe the topology of a graph and are generally graph invariant (Qiu and Akl, 1995). For instance, the Wiener index is based on the topological proximity of vertices in a graph, and was defined by Wiener in 1947 to estimate the boiling properties of various alkane isomers. Since then, more than 3,000 topological indices of graphs have been recorded in chemical databases. Another type of topological index aims to measure the nonbalancedness among the bonds of a chemical graph on the base of peripherality of bonds (edges), which is named as the Mostar index (Akhter et al., 2021; Došlić et al., 2018). The Mostar index recently discovered as a bond-additive connectivity index determines the quantity of peripherality of certain edges as well as the graph as a whole (Akhter et al., 2021; Došlić et al., 2018). It is a distinct geometric index which counts the contribution |η λ η μ | of every edge e = λμ in a connected graph, where η λ is the quantity of vertices that are closer to the vertex λ than the vertex μ, and η μ is defined in the same way (Ali and Doslic, 2021). Accordingly, this index indicates the degree of specific edges and the degree of peripherality of the graph as a whole. This index attracted many graph theorists to measure the peripherality of various (chemical) graphs. Tepeh addressed the first conjecture about the Mostar index of bicyclic graphs (Tepeh, 2019), which was proposed by Došlić et al. (2018). Further, remarkable work on the Mostar index of carbon nanostructures, trees, and hexagonal chains has been supplied by Arockiaraj et al. (2019), Hayat and Zhou (2019), and Huang et al. (2020).

A classical study of graphs associated with groups attracted many researchers to explore the various theoretical and topological properties of graph. A large number of interesting work have been published by supplying the articles by Abdollahi et al. (2006), Ali et al. (2016), Alolayan et al. (2019), Bhuniya and Bera (2016), Bunday (2006), Cameron and Ghosh (2011), Chakrabarty et al. (2009), and Rahman (2017). Some of the graphs associated to group are defined as follows: Let a group Γ and the center of group Γ be ζ(Γ) = {λ ∈ Γ: λμ = μλμ ∈ Γ}. Γ G denotes the commuting graph of Γ with the vertex set Γ and two distinct vertices λ and μ ∈ Γ from an edge in Γ G if and only if λμ = μλ in Γ (Ali et al., 2016; Bunday, 2006). The non-commuting graph of Γ is denoted by G Γ with the vertex set Γ − ζ(Γ) and two distinct vertices λ and μ ∈ Γ from an edge in G Γ if and only if λμμλ in Γ (Abdollahi et al., 2006; Moghaddamfar et al., 2005; Wei et al., 2020). If λ = gμg −1 or μ = g −1 λg for g ∈ Γ, then λ and μ are said to be conjugate of each other. This relation between elements of Γ is an equivalence relation and is called the conjugacy relation. Due to this equivalence relation, Γ is partitioned into disjoint classes each of which is called a conjugacy class. Mathematically, the conjugacy class of λ ∈ Γ is Cl(λ) = {gλg −1:g ∈ Γ}. G(Γ) denotes a non-conjugate graph with the vertex set Γ and two different vertices λ and μ ∈ Γ from an edge in G(Γ) if and only if λ and μ belong to different conjugacy classes (Alolayan et al., 2019).

Graphs associated with the dihedral group have been considered by Abbas et al. (2021), Salman et al. (2022), and Wei et al. (2020) to study their topological properties such as the Wiener related indices, Harary index, Randić indices, geometric arithmetic indices, atom bond connectivity indices, harmonic index, Hosoya index, and polynomials. This study is aimed to investigate the Mostar index of graphs (commuting, non-commuting, and non-conjugate graphs) associated with the dihedral and semi-dihedral groups.

2 Preliminaries

This section provides partitions of groups under consideration, some basic terminologies of a graph, and the mechanism to compute the Mostar index.

The generating form D n = 〈a, b|a n = b 2 = e, ab = ba −1〉 represents the dihedral group of order 2n, which is the collection of symmetries of regular n-polygon. The center of D n is:

ζ ( D n ) = { e } , if n is odd { e , a n 2 } , if n is even

Let us partition D n as follows:

Ω1 = {e, a, a 2,…, a n−1}, Ω2 = {b, ab, a 2 b,…, a n−1 b}, and Ω3 = Ω1ζ(D n ), then |Ω1| = |Ω2| = n. Further for even n, let Ω 2 = i = 0 n 2 1 Ω 2 i with Ω 2 i = a i b , a i + n 2 be the subsets according to the commuting elements from Ω2, and conjugacy classes of D n are:

(1) Cl ( D n ) = { e } , for any n a n 2 , when n is even { a i b : 0 i n 1 } , when n is odd { a i , a n i } ; 1 i n 1 2 , when n is odd a 2 i b : 0 i < n 2 , when n is even a 2 i + 1 b : 0 i < n 2 , when n is even { a i , a n i } ; 1 i < n 2 , when n is even

The generating form SD8n = 〈a, b|a 4n = b 2 = e, ba = a 2n−1 b〉 represents the semi-dihedral group of order 8n with:

b a i = a 4 n i b , if  n is even a 2 n i b , if n is odd

and

ζ ( SD 8 n ) = { e , a 2 n } , if n is even { e , a n , a 2 n , a 3 n } , if n is odd

Let us partition SD8n as follows:

Ω1 = {e, a, a 2,…, a 4n−1}, Ω2 = {b, ab, a 2 b,…, a 4n−1 b}, and Ω3 = Ω1ζ(D n ), then |Ω1| = |Ω2| = 4n. For odd n ≥ 3, let Ω 2 = i = 0 n 1 Ω 2 i with Ω 2 i = { a i b , a i + n b , a i + 2 n b , a i + 3 n b } be the subsets according to the commuting elements from Ω2, and conjugacy classes of SD8n are:

(2) Cl ( SD 8 n ) = { e } { a i n } ; 1 i 3 Φ 2 j = { a 4 i + j b : 0 i n 1 } ; 0 j 3 Φ 3 i = { a i , a 2 n i } ; for odd i , 1 i n 1 and   2 n + 1 i 3 n 2 Φ 3 i = { a i , a 4 n i } ; for even i , 2 i 2 n 2

where j = 0 3 Φ 2 j = Ω 2 and odd i = 1 n 1 Φ 3 i odd i = 2 n + 1 3 n 2 Φ 3 i even i = 2 2 n 2 Φ 3 i = Ω 3 .

For even n ≥ 2, let Ω 2 = i = 0 2 n 1 Ω 2 i with Ω 2 i = { a i b , a i + 2 n b } be the subsets according to the commuting elements from Ω2, and conjugacy classes of SD8n are:

(3) Cl ( SD 8 n ) = { e } { a 2 n } Φ 2 2 = { a i b : for even i , 0 i 4 n 2 } Φ 2 1 = { a i b : for odd i , 1 i 4 n 1 } Φ 3 i = { a i , a 2 n i } ; for odd i , 1 i n 1 and   2 n + 1 i 3 n 1 Φ 3 i = { a i , a 4 n i } ; for even i , 2 i 2 n 2

where j = 1 2 Φ 2 j = Ω 2 and odd i = 1 n 1 Φ 3 i odd i = 2 n + 1 3 n 1 Φ 3 i even i = 2 2 n 2 Φ 3 i = Ω 3 .

Let a connected and simple graph G have vertex and edge sets symbolized by V(G) and E(G), respectively. We denote the number of edges(size) of a graph G by S(G). The notation K + H denotes the sum of two graphs K and H with V(K) ∪ V(H) as the vertex set and E(K) ∪ E(H) ∪ {λμ: λV(K) ∧ μV(H)} as the edge set. The number of edges in a shortest path between two distinct vertices λ and μ is defined as the distance across λ and μ, indicated by d(λ, μ). The eccentricity of a vertex μ is the number

ecc ( λ ) = max μ V ( G ) d ( λ , μ )

For an edge e = λμ, peripheral neighborhoods of e according to its end vertices λ and μ are defined, Arockiaraj et al. (2020), as follows:

N λ ( e | Γ G ) = { x V ( Γ G ) : d ( λ , x ) < d ( μ , x ) } , N μ ( e | Γ G ) = { x V ( Γ G ) : d ( μ , x ) < d ( λ , x ) }

Then, η λ (e|G) = |N λ (e G )| and η μ (e|G) = |N μ (e G )| are the peripheral degrees of e. Akhter et al. in 2021 and Došlić et al. in 2018 provided the following formula to compute the Mostar index of a graph G

Mo ( G ) = e = λ μ E ( G ) | η λ ( e | G ) η μ ( e | G ) |

The number of vertices adjacent to a vertex λ in G is called the degree of λ and it is denoted by d(λ). A vertex of degree 1 is known as a leaf in G. Whenever we need to find the size, S(G) of a graph G, we will use the formula λ V ( G ) d ( λ ) = 2 S ( G ) provided by the well-known handshake lemma (Rahman, 2017).

Proposition 1

If λ is a leaf in G, then for an edge e = λμ, N λ (e|G) = {λ} and N μ (e|G) = V(G) {λ}.

Proof

Since λ is a leaf in G, so d(λ, x) ≤ d(μ, x) if and only if x = λ, and d(λ, x) > d(μ, x) for all xV(G) − {λ}. Therefore, the result is as follows.

For a non-negative integer k, the set N k (λ) = {μV(G)|d(λ, μ) = k} is known as the k-distance neighborhood of λ in G, where 0 ≤ k ≤ ecc(λ).□

Proposition 2

Let e = λμ be any edge in G, if x ∈ N k (λ) ∩ N k (μ), then xN λ (e|G) ∪ N μ (e|G).

Proof

As d(λ, x) = k = d(μ, x), so the result followed from the definitions of N λ (e|G) and N μ (e|G).□

Remark 1

For any edge e = λμ in G, let us define a set:

N k ( λ ) = { x V ( G ) | d ( λ , x ) = k < d ( μ , x ) } , for 0 k ecc ( λ ) .

Then, N k(λ) = N k (λ) − (∪ tk N t (μ)). Accordingly, N λ ( e | G ) = k = 0 ecc ( λ ) N k ( λ ) .

A neighbor of λ is a vertex adjacent to it in a graph G. The open neighborhood, N(λ), of λ in G is the set of all the neighbors of λ. The closed neighborhood of λ is N[λ] = N(λ) ∪ {λ}. Two vertices λ and μ are false twins in G whenever N(λ) = N(μ), and are true twins whenever N[λ] = N[μ].

Proposition 3

If λ and μ are true twins and e = λμ ∈ E(G), then N λ (e|G) = {λ} and N μ (e|G) = {μ}.

Proof

Since N[λ] = N[μ], so d(λ, x) = d(μ, x)) for each xV(G) − {λ, μ} and

d ( λ , λ ) = 0 1 = d ( μ , λ ) , d ( λ , μ ) = 1 0 = d ( μ , μ )

Thus, for each xV(G) − {λ, μ}, xN k (λ) ∩ N k (μ) for all k ≠ 0. Hence, by Proposition 2, xN λ (e|G) ∪ N μ (e|G) for each xV(G) − {λ, μ}. In fact, only λN λ (e|G) and only μN μ (e|G).□

3 Commuting graphs

The Mostar index of commuting graphs on the dihedral and semi-dihedral groups is computed in this section.

Theorem 1

For n ≥ 3, let Γ be a dihedral group D n . Then,

Mo ( Γ G ) = 3 n ( n 1 ) , w h e n n i s o d d 6 n ( n 2 ) , w h e n n i s e v e n

Proof

In a study by Ali et al. (2016), the following graph theoretical definition of the commuting graph on D n was provided:

Γ G = K 1 + ( K Ω 3 N Ω 2 ) , if n is odd K 2 + K Ω 3 n 2 K 2 , if n is even

Then,

V ( Γ G ) = ζ ( Γ ) Ω 3 Ω 2 , if n is odd ζ ( Γ ) Ω 3 i = 0 n 1 Ω 2 , if n is even

Now, we discuss the following two cases.□

Case 1 (n is odd):

Let E ( Γ G ) = i = 1 3 E i ( Γ G ) with E i G ) = {eE G )|e is of type T i }.

  • Type T 1: e = λμE 1 G ) such that λ ∈ Ω3 and μ ∈ Ω3.

  • Type T 2: e = λμE 2 G ) such that λ ∈ Ω3 and μζ(Γ).

  • Type T 3: e = λμE 3 G ) such that λζ(Γ) and μ ∈ Ω2.

Let e is of type T 1 : Note that N[λ] = ζ(Γ) ∪ Ω3 = N[μ]. Thus, λ and μ are true twins, so Proposition 3 yields that η λ (e G ) = 1 = η μ (e G ). Let t 1 be the number of type T 1 edges. Since Ω3 induces a complete graph K n−1, so t 1 = S ( K n 1 ) = n 1 2 .

Let e is of type T 2 : Since ecc(λ) = 2 and ecc(μ) = 1, so N 0(λ) = {λ}, N 1(λ) = Ω1, N 2(λ) = Ω2 N 0(μ) = {μ}, N 1(μ) = Ω3 ∪ Ω2. Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = N 2 ( λ ) = , N 0 ( μ ) = { μ } , N 1 ( μ ) = Ω 2

and hence η λ (e G ) = 1 and η μ (e G ) = n + 1. Let t 2 be the number of type T 2 edges, then t 2 = |Ω3| × |ζ(Γ)| = n − 1.

Let e is of type T 3 : Note that μ is a leaf in Γ G , so Proposition 1 yields that η μ (e G ) = 1 and η λ (e G ) = 2n − 1. Let t 3 be the number of type T 3 edges, then t 3 = |ζ(Γ)| × |Ω2| = n. Now, the Mostar index of Γ G is:

Mo ( Γ G ) = i = 1 3 λ μ E i ( Γ G ) | η λ ( e | Γ G ) η μ ( e | Γ G ) | = t 1 | 1 1 | + t 2 | n + 1 1 | + t 3 | 2 n 1 1 | = 3 n ( n 1 )

Case 2 (n is even):

Let E ( Γ G ) = i = 1 5 E i ( Γ G ) with E i G ) = {eE G )|e is of type T i }.

  • Type T 1: e = λμE 1 G ) such that λ ∈ Ω3 and μ ∈ Ω3.

  • Type T 2: e = λμE 2 G ) such that λζ(Γ) and μζ(Γ).

  • Type T 3: e = λμE 3 G ) such that λ Ω 2 i and μ Ω 2 i .

  • Type T 4: e = λμE 4 G ) such that λ ∈ Ω3 and μζ(Γ).

  • Type T 5: e = λμE 5 G ) such that λζ(Γ) and μ Ω 2 i .

Let e is of type T 1 : Note that N[λ] = ζ(Γ) ∪ Ω3 = N[μ]. Thus λ and μ are true twins, so Proposition 3 yields that η λ (e G ) = 1 = η μ (e G ). Let t 1 be the number of type T 1 edges. Since Ω3 induces a complete graph K n−2, so t 1 = S ( K n 2 ) = n 2 2 .

Let e is of type T 2 : Note that N[λ] = ζ(Γ) ∪ Ω3 ∪ Ω2 = N[μ]. Thus, λ and μ are true twins, so Proposition 3 yields that η λ (e G ) = 1 = η μ (e G ). Let t 2 be the number of type T 2 edges. Since ζ(Γ) induces a complete graph K 2, so t 2 = S(K 2) = 1.

Let e is of type T 3 : Note that N [ λ ] = ζ ( Γ ) Ω 2 i = N [ μ ] . Thus, λ and μ are true twins, so Proposition 3 yields that η λ (e G ) = 1 = η μ (e G ). Let t 3 be the number of type T 3 edges. Since each Ω 2 i induces the complete graph K 2 and 0 i n 2 1 , so t 3 = n 2 × S ( K 2 ) = n 2 .

Let e is of type T 4 : Since ecc(λ) = 2 and ecc(μ) = 1, so:

N 0 ( λ ) = { λ } , N 1 ( λ ) = ( Ω 3 { λ } ) ζ ( Γ ) , N 2 ( λ ) = Ω 2 , N 0 ( μ ) = { μ } , N 1 ( μ ) = Ω 3 ζ ( Γ ) { μ } Ω 2

Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = N 2 ( λ ) = , N 0 ( μ ) = { μ } , N 1 ( μ ) = Ω 2

and hence η λ (e G ) = 1 and η μ (e G ) = n + 1. Let t 4 be the number of type T 4 edges, then t 4 = |Ω3| × |ζ(Γ)| = 2(n − 2).

Let e is of type T 5 : As ecc(λ) = 2 and ecc(μ) = 2, so:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Ω 3 ζ ( Γ ) { λ } Ω 2 N 0 ( μ ) = { μ } , N 1 ( μ ) = Ω 2 i { μ } ζ ( Γ ) , N 2 ( μ ) = Ω 3 Ω 2 Ω 2 i

Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Ω 3 ( Ω 2 Ω 2 i ) , N 0 ( μ ) = { μ } , N 1 ( μ ) = N 2 ( μ ) =

and hence η λ (e G ) = 2n − 3 and η μ (e G ) = 1. Let t 5 be the number of type T 5 edges, then t 5 = |Ω2| × |ζ(Γ)| = 2n.

Now, the Mostar index of Γ G is:

Mo ( Γ G ) = i = 1 5 λ μ E i ( Γ G ) | η λ ( e | Γ G ) η μ ( e | Γ G ) | = t 1 | 1 1 | + t 2 | 1 1 | + t 3 | 1 1 | + t 4 | n + 1 1 | + t 5 | 2 n 3 1 | = 6 n ( n 2 )

Theorem 2

For n ≥ 2, let Γ be a semi-dihedral group. Then,

Mo ( Γ G ) = 192 n ( n 1 ) , w h e n n i s o d d 48 n ( 2 n 1 ) , w h e n n i s e v e n

Proof

Mathematically, the commuting graph on SD8n is defined by Kumar et al. (2020) as follows:

Γ G = K 4 + ( K Ω 3 n K 4 ) , if n is odd K 2 + ( K Ω 3 2 n K 2 ) , if n is even

V ( Γ G ) = ζ ( Γ ) Ω 3 i = 0 n 1 Ω 2 i , if n is odd ζ ( Γ ) Ω 3 i = 0 2 n 1 Ω 2 i , if n is even

Now, we discuss the following two cases.□

Case 1 (n is odd):

Let E ( Γ G ) = i = 1 5 E i ( Γ G ) with E i G ) = {eE G )|e is of type T i }.

  • Type T 1: e = λμE 1 G ) such that λ ∈ Ω3 and μ ∈ Ω3.

  • Type T 2: e = λμE 2 G ) such that λζ(Γ) and μζ(Γ).

  • Type T 3: e = λμE 3 G ) such that λ Ω 2 i and μ Ω 2 i .

  • Type T 4: e = λμE 4 G ) such that λ ∈ Ω3 and μζ(Γ).

  • Type T 5: e = λμE 5 G ) such that λζ(Γ) and μ Ω 2 i .

Let e is of type T 1 : Note that N[λ] = ζ(Γ) ∪ Ω3 = N[μ]. Thus, λ and μ are true twins, so Proposition 3 yields that η λ (e G ) = 1 = η μ (e G ). Let t 1 be the number of type T 1 edges. Since Ω3 induces a complete graph K 4n−4, so t 1 = S ( K 4 n 4 ) = 4 n 4 2 .

Let e is of type T 2 : Note that N[λ] = ζ(Γ) ∪ Ω3 ∪ Ω2 = N[μ]. Thus, λ and μ are true twins, so Proposition 3 yields that η λ (e G ) = 1 = η μ (e G ). Let t 2 be the number of type T 2 edges. Since ζ(Γ) induces a complete graph K 4, so t 2 = S ( K 4 ) = 4 2 .

Let e is of type T 3 : Note that N [ λ ] = ζ ( Γ ) Ω 2 i = N [ μ ] . Thus, λ and μ are true twins, so Proposition 3 yields that η λ (e G ) = 1 = η μ (e G ). Let t 3 be the number of type T 3 edges. Since each Ω 2 i induces the complete graph K 4 and 0 ≤ i ≤n − 1, so t 3 = n × S(K 4) = 6n.

Let e is of type T 4 : Since ecc(λ) = 2 and ecc(μ) = 1, so:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Ω 3 { λ } ζ ( Γ ) , N 2 ( λ ) = Ω 2 N 0 ( μ ) = { μ } , N 1 ( μ ) = Ω 3 ζ ( Γ ) { μ } Ω 2

Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = N 2 ( λ ) = N 0 ( μ ) = { μ } , N 1 ( μ ) = Ω 2

and hence η λ (e G ) = 1 and η μ (e G ) = 4n + 1. Let t 4 be the number of type T 4 edges, then t 4 = |Ω3| × |ζ(Γ)| = 4(4n − 4).

Let e is of type T 5 : Since ecc(λ) = 1 and ecc(μ) = 2, so N 0(λ) = {λ}, N 1(λ) = Ω3ζ(Γ) − {λ} ∪ Ω2, N 0(μ) = {μ}, N 1 ( μ ) = Ω 2 i { μ } ζ ( Γ ) , and N 2 ( μ ) = Ω 3 Ω 2 Ω 2 i .

Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = { Ω 3 Ω 2 Ω 2 i } N 0 ( μ ) = { μ } , N 1 ( μ ) = N 2 ( μ ) =

and hence η λ (e G ) = 8n − 7 and η μ (e G ) = 1. Let t 5 be the number of type T 5 edges, then t 5 = |Ω2| × |ζ(Γ)| = 16n.

Now, the Mostar index of Γ G is:

Mo ( Γ G ) = i = 1 5 λ μ E i ( Γ G ) | η λ ( e | Γ G ) η μ ( e | Γ G ) | = t 1 | 1 1 | + t 2 | 1 1 | + t 3 | 1 1 | + t 4 | 4 n + 1 1 | + t 5 | 8 n 7 1 | = 192 n ( n 1 )

Case 2 (n is even):

Let E ( Γ G ) = i = 1 5 E i ( Γ G ) with E i G ) = {eE G )|e is of type T i }.

  • Type T 1: e = λμE 1 G ) such that λ ∈ Ω3 and μ ∈ Ω3.

  • Type T 2: e = λμE 2 G ) such that λζ(Γ) and μζ(Γ).

  • Type T 3: e = λμE 3 G ) such that λ Ω 2 i and μ Ω 2 i .

  • Type T 4: e = λμE 4 G ) such that λ ∈ Ω3 and μζ(Γ).

  • Type T 5: e = λμE 5 G ) such that λζ(Γ) and μ Ω 2 i .

Let e is of type T 1 : Note that N[λ] = ζ(Γ) ∪ Ω3 = N[μ]. Thus, λ and μ are true twins, so Proposition 3 yields that η λ (e G ) = 1 = η μ (e G ). Let t 1 the number of type T 1 edges. Since Ω3 induces a complete graph K 4n−2, so t 1 = S ( K 4 n 2 ) = 4 n 2 2 .

Let e is of type T 2 : Note that N[λ] = V G ) = N[μ]. Thus, λ and μ are true twins, so Proposition 3 yields that η λ (e G ) = 1 = η μ (e G ). Let t 2 be the number of type T 2 edges. Since ζ(Γ) induces a complete graph K 2, so t 2 = S(K 2) = 1.

Let e is of type T 3 : Note that N [ λ ] = ζ ( Γ ) Ω 2 i = N [ μ ] . Thus, λ and μ are true twins, so Proposition 3 yields that η λ (e G ) = 1 = η μ (e G ). Let t 3 be the number of type T 3 edges. Since each Ω 2 i induces the complete graph K 2| and 0 ≤ i ≤ 2n − 1, so t 3 = 2n × S(K 2) = 2n.

Let e is of type T 4 : Since ecc(λ) = 2 and ecc(μ) = 1, so N 0(λ) = {λ}, N 1(λ) = (Ω3 − {λ}) ∪ ζ(Γ), N 2(λ) = Ω2, N 0(μ) = {μ}, and N 1(μ) = V G ) − {μ}. Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = N 2 ( λ ) = N 0 ( μ ) = { μ } , N 1 ( μ ) = Ω 2

and hence n λ (e G ) = 1 and n μ (e G ) = 4n + 1. Let t 4 be the number of type T 4 edges, then t 4 = |Ω3| × |ζ(Γ)| = 2(4n − 2).

Let e is of type T 5 : Since ecc(λ) = 1 and ecc(μ) = 2. So N 0(λ) = {λ}, N 1(λ) = Ω3 ∪ (ζ(Γ) − {λ} ∪ Ω2, N 0(μ) = {μ}, N 1 ( μ ) = ( Ω 2 i { μ } ) ζ ( Γ ) , and N 2 ( μ ) = Ω 3 ( Ω 2 Ω 2 i ) .

Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = { Ω 3 Ω 2 Ω 2 i } N 0 ( μ ) = { μ } , N 1 ( μ ) = N 2 ( μ ) =

and hence n λ (e G ) = 8n − 3 and n μ (e G ) = 1. Let t 5 be the number of type T 5 edges, then t 5 = |Ω2| × |ζ(Γ)| = 8n.

Now, the Mostar index of Γ G is:

Mo ( Γ G ) = i = 1 5 λ μ E i ( Γ G ) | η λ ( e | Γ G ) η μ ( e | Γ G ) | = t 1 1 1 + t 2 1 1 + t 3 1 1 + t 4 4 n + 1 1 + t 5 | 8 n 3 1 | = 48 n ( 2 n 1 )

4 Non-commuting graphs

In this section, by measuring the amount of peripherality of each edge, we investigate the Mostar index of non-commuting graphs associated with D n and SD8n .

Theorem 3

For n ≥ 3, let Γ be a dihedral group. Then,

Mo ( G Γ ) = n ( n 1 ) ( n 2 ) , w h e n n i s o d d n ( n 2 ) ( n 4 ) , w h e n n i s e v e n

Proof

The non-commuting graph on Γ = D n is mathematically defined by Wei et al. (2020) as follows:

G Γ = K Ω 2 + N Ω 3 , if n is odd K 2 , 2 , , 2 n 2 times , Ω 3 , if n is even

Then,

V ( G Γ ) = Ω 2 Ω 3 , if n is odd i = 0 n 2 1 Ω 2 i Ω 3 , if n is even

Next we discuss the following two cases.

Case 1 (n is odd):

Let E ( Γ G ) = i = 1 2 E i ( Γ G ) with E i G ) = {eE G )|e is of type T i }.

  • Type T 1: e = λμE 1(G Γ) such that λ ∈ Ω2 and μ ∈ Ω2.

  • Type T 2: e = λμE 2(G Γ) such that λ ∈ Ω2 and μ ∈ Ω3.

Let e is of type T 1 : Note that N[λ] = V(G Γ) = N[μ]. Thus, λ and μ are true twins, so Proposition 3 yields that η λ (e|G Γ) = 1 = η μ (e|G Γ). Let t 1 be the number of type T 1 edges. Since Ω2 induces a complete graph K n , so t 1 = S ( K n ) = n 2 .

Let e is of type T 2 : Since ecc(λ) = 1 and ecc(μ) = 2, so N 0(λ) = {λ}, N 1(λ) = Ω3 ∪ (Ω2 − {λ}, N 0(μ) = {μ}, N 1(μ) = Ω2, and N 2(μ) = Ω3 − {μ}. Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Ω 3 { μ } N 0 ( μ ) = { μ } , N 1 ( μ ) = N 2 ( μ ) =

and hence η λ (e|G Γ) = n − 1 and η μ (e|G Γ) = 1. Let t 2 be the number of type T 2 edges, then t 2 = |Ω2| × |Ω3| = n(n − 1).

Now, the Mostar index of G Γ is:

Mo ( G Γ ) = i = 1 2 λ μ E i ( G Γ ) | η λ ( e | G Γ ) η μ ( e | G Γ ) | = t 1 | n 1 1 | + t 2 | 1 1 | = n ( n 1 ) ( n 2 )

Case 2 (n is even):

Let E ( Γ G ) = i = 1 2 E i ( Γ G ) with E i G ) = {eE G )|e is of type T i }.

  • Type T 1: e = λμE 1(G Γ) such that λ Ω 2 i and μ Ω 2 j for ij.

  • Type T 2: e = λμE 2(G Γ) such that λ Ω 2 i and μ ∈ Ω3.

Let e is of type T 1 : Since ecc(λ) = 2 and ecc(μ) = 2, so N 0(λ) = {λ}, N 1 ( λ ) = Ω 3 ( Ω 2 Ω 2 i ) , and N 2 ( λ ) = Ω 2 i { λ } ,

N 0 ( μ ) = { μ } , N 1 ( μ ) = Ω 3 ( Ω 2 Ω 2 j ) N 2 ( μ ) = Ω 2 j { μ }

Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Ω 2 j { μ } , N 2 ( λ ) = N 0 ( μ ) = { μ } , N 1 ( μ ) = Ω 2 i { λ } , N 2 ( μ ) =

and hence η λ (e|G Γ) = 2 and η μ (e|G Γ) = 2. Let t 1 be the number of edges of type T 1. Since the partition of Ω2 with parts Ω 2 i induces a complete multipartite graph, so for any λ Ω 2 i , d ( λ ) = 2 n 2 1 = n 2 in the subgraph of G Γ induced by the set i = 0 n 2 1 Ω 2 i . Since | Ω 2 i | = 2 and there are n 2 such sets, so by the formula of handshake lemma, t 1 = n ( n 2 ) 2 .

Let e is of type T 2 : Since ecc(λ) = 2 and ecc(μ) = 2, so N 0(λ) = {λ}, N 1 ( λ ) = Ω 3 ( Ω 2 Ω 2 i ) , N 2 ( λ ) = Ω 2 i { λ } , N 0(μ) = {μ}, N 1(μ) = Ω2, and N 2(μ) = Ω3 − {μ}.

Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Ω 3 { μ } , N 2 ( λ ) = N 0 ( μ ) = { μ } , N 1 ( μ ) = Ω 2 i { λ } , N 2 ( μ ) =

and hence η λ (e|G Γ) = n − 2 and η μ (e|G Γ) = 2. Let t 2 be the number of type T 2 edges, then t 2 = |Ω2| × |Ω3| = n(n− 2).

Now, the Mostar index of G Γ is:

Mo ( G Γ ) = i = 1 2 λ μ E i ( G Γ ) | η λ ( e | G Γ ) η μ ( e | G Γ ) | = t 1 | 2 2 | + t 2 | n 2 2 | = n ( n 2 ) ( n 4 )

Theorem 4

For n ≥2, let Γ be a semi-dihedral group. Then,

Mo ( G Γ ) = 64 n ( n 1 ) ( n 2 ) , w h e n n i s o d d 32 n ( n 1 ) ( 2 n 1 ) , w h e n n i s e v e n

Proof

Mathematically, the non-commuting graph on Γ = SD8n can be expressed as follows:

G Γ = n N 4 + N 4 n 4 , if n is odd 2 n N 2 + N 4 n 2 , if n is even

V ( G Γ ) = i = 0 n 1 Ω 2 i Ω 3 , if n is odd i = 0 2 n 1 Ω 2 i Ω 3 , if n is even

Next we discuss the following two cases.□

Case 1 (n is odd):

Let E ( Γ G ) = i = 1 2 E i ( Γ G ) with E i G ) = {eE G )|e is of type T i }.

  • Type T 1: e = λμE 1(G Γ) such that λ Ω 2 i and μ Ω 2 j for ij.

  • Type T 2: e = λμE 2(G Γ) such that λ Ω 2 i and μ ∈ Ω3.

Let e is of type T 1 : Since ecc(λ) = 2 and ecc(μ) = 2, so N 0(λ) = {λ}, N 1 ( λ ) = Ω 3 ( Ω 2 Ω 2 i ) , N 2 ( λ ) = Ω 2 i { λ } , N 0(μ) = {μ}, N 1 ( μ ) = Ω 3 ( Ω 2 Ω 2 j ) , and N 2 ( μ ) = Ω 2 j { μ } .

Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Ω 2 j { μ } , N 2 ( λ ) = N 0 ( μ ) = { μ } , N 1 ( μ ) = Ω 2 i { λ } , N 2 ( μ ) =

and hence η λ (e|G Γ) = 4 and η μ (e|G Γ) = 4. Let t 1 be the number of edges of type T 1. Since the partition of Ω2 with parts Ω 2 i induces a complete multipartite graph, so for any λ Ω 2 i , d(λ) = 4(n − 1) = 4n − 4 in the subgraph of G Γ induced by the set i = 0 n 1 Ω 2 i . Since | Ω 2 i | = 4 and there are n such sets, so by the formula of handshake lemma, t 1 = 2n(4n − 4).

Let e is of type T 2 : Since ecc(λ) = 2 and ecc(μ) = 2, so N 0(λ) = {λ}, N 1 ( λ ) = Ω 3 ( Ω 2 Ω 2 i ) , N 2 ( λ ) = Ω 2 i { λ } , N 0(μ) = {μ}, N 1(μ) = Ω2, N 2(μ) = Ω3 − {μ}.

Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Ω 3 { μ } , N 2 ( λ ) = N 0 ( μ ) = { μ } , N 1 ( μ ) = Ω 2 i { λ } , N 2 ( μ ) =

and hence η λ (e|G Γ) = 4n − 4 and η μ (e|G Γ) = 4. Let t 2 be the number of type T 2 edges, then t 2 = |Ω2| × |Ω3| = 4n(4n− 4).

Now, the Mostar index of G Γ is:

Mo ( G Γ ) = i = 1 2 λ μ E i ( G Γ ) | η λ ( e | G Γ ) η μ ( e | G Γ ) | = t 1 | 4 4 | + t 2 | 4 n 4 4 | = 64 n ( n 1 ) ( n 2 )

Case 2 (n is even):

Let E ( Γ G ) = i = 1 2 E i ( Γ G ) with E i G ) = {eE G )|e is of type T i }.

  • Type T 1: e = λμE 1(G Γ) such that λ Ω 2 i and μ Ω 2 j for ij.

  • Type T 2: e = λμE 2(G Γ) such that λ Ω 2 i and μ ∈ Ω3.

Let e is of type T 1 : Since ecc(λ) = 2 and ecc(μ) = 2, so N 0(λ) = {λ}, N 1 ( λ ) = Ω 3 ( Ω 2 Ω 2 i ) , N 2 ( λ ) = Ω 2 i { λ } , N 0(μ) = {μ}, N 1 ( μ ) = Ω 3 ( Ω 2 Ω 2 j ) , and N 2 ( μ ) = Ω 2 j { μ } .

Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Ω 2 j { μ } , N 2 ( λ ) = N 0 ( μ ) = { μ } , N 1 ( μ ) = Ω 2 i { λ } , N 2 ( μ ) =

and hence η λ (e|G Γ) = 2 and η μ (e|G Γ) = 2. Let t 1 be the number of edges of type T 1. Since the partition of Ω2 with parts Ω 2 i induces a complete multipartite graph, so for any λ Ω 2 i , d(λ) = 2(2n− 1) = 4n− 2 in the subgraph of G Γ induced by the set i = 0 2 n 1 Ω 2 i . Since | Ω 2 i | = 2 and there are 2n such sets, so by the formula of handshake lemma, t 1 = 2n(4n− 2).

Let e is of type T 2 : Since ecc(λ) = 2 and ecc(μ) = 2, so N 0(λ) = {λ}, N 1 ( λ ) = Ω 3 ( Ω 2 Ω 2 i ) , N 2 ( λ ) = Ω 2 i { λ } , N 0(μ) = {μ}, N 1(μ) = Ω2, and N 2(μ) = Ω3 − {μ}.

Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Ω 3 { μ } , N 2 ( λ ) = N 0 ( μ ) = { μ } , N 1 ( μ ) = Ω 2 i { λ } , N 2 ( μ ) =

and hence n λ (e|G Γ) = 4n− 2 and n μ (e|G Γ) = 2. Let t 2 be the number of type T 2 edges, then t 2 = |Ω2| × |Ω3| = 4n(4n − 2).

Now, the Mostar index of G Γ is:

Mo ( G Γ ) = i = 1 2 λ μ E i ( G Γ ) | η λ ( e | G Γ ) η μ ( e | G Γ ) | = t 1 | 2 2 | + t 2 | 4 n 2 2 | = 32 n ( 2 n 1 ) ( n 1 )

5 Non-conjugate graphs

In this section, non-conjugate graphs associated with D n and SD8n are considered in the context of Mostar index.

Theorem 5

For n ≥ 3, let Γ be a dihedral group. Then,

Mo ( G ( Γ ) ) = ( n 1 ) ( n 2 n + 1 ) , w h e n n i s o d d n 2 2 ( n 2 2 n + 4 ) , w h e n n i s e v e n

Proof

From conjugacy classes, written in Eq. 1, the number of conjugacy classes of order 1, 2, and n is 1, n 1 2 , and 1, respectively, whenever n is odd, and the number of conjugacy classes of order 1, 2, and n 2 is 2, n 2 2 , and 2, respectively, whenever n is even. Thus, the non-conjugate graph G(Γ) on Γ = D n can be described as follows:

G ( Γ ) = K 1 , 2 , 2 , , 2 n 1 2 times , n , if n is odd K 1 , 1 , 2 , 2 , , 2 n 2 2 times , n 2 , n 2 , if n is even

Here K 1 , 2 , 2 , , 2 n 1 2 times , n , is a complete n + 3 2 -partite graph having n 1 2 parts Ω 3 i = { a i , a n i } for 1 i n 1 2 with i = 1 n 1 2 Ω 3 i = Ω 3 ; one part is ζ(Γ); and one part is Ω2. Similarly, K 1 , 1 , 2 , 2 , , 2 n 2 2 times , n 2 , n 2 , is a complete n + 6 2 -partite graph having n 2 2 parts Ω 3 i = { a i , a n i } for 1 i < n 2 with i = 1 n 2 2 Ω 3 i = Ω 3 ; two parts {e}, { a n 2 } ; and two parts Ω 2 j = a 2 i + j b : 0 i < n 2 for i = 0,1 with Ω 2 0 Ω 2 1 = Ω 2 .□

Now, we discuss the following two cases.

Case 1 (n is odd):

Let E ( Γ G ) = i = 1 4 E i ( Γ G ) with E i G ) = {eE G )|e is of type T i }.

  • Type T 1: e = λμE 1(G(Γ)) such that λ Ω 3 i and μ Ω 3 j for ij.

  • Type T 2: e = λμE 2(G(Γ)) such that λ Ω 3 i and μ ∈ Ω2.

  • Type T 3: e = λμE 3(G(Γ)) such that λζ(Γ) and μ Ω 3 i .

  • Type T 4: e = λμE 4(G(Γ)) such that λζ(Γ) and μ ∈ Ω2.

Let e is of type T 1 : Since ecc(λ) = 2 and ecc(μ) = 2, so:

N 0 ( λ ) = { λ } , N 1 ( λ ) = V ( G ( Γ ) ) Ω 3 i , N 2 ( λ ) = Ω 3 i { λ } N 0 ( μ ) = { μ } , N 1 ( μ ) = V ( G ( Γ ) ) Ω 3 j , N 2 ( μ ) = Ω 3 j { μ }

Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Ω 3 j { μ } , N 2 ( λ ) = N 0 ( μ ) = { μ } , N 1 ( μ ) = Ω 3 i { λ } , N 2 ( μ ) =

and hence η λ (e|G Γ) = 2 and η μ (e|G Γ) = 2. Let t 1 be the number of edges of type T 1. Since the partition of Ω3 with parts Ω 3 i induces a complete multipartite graph, so for any λ Ω 3 i , d ( λ ) = 2 ( n 1 ) 2 1 = n 3 in the subgraph of G Γ induced by the set i = 1 n 1 2 Ω 3 i . Since | Ω 3 i | = 2 and there are n 1 2 such sets, therefore by the formula of handshake lemma, t 1 = ( n 1 ) ( n 3 ) 2 .

Let e is of type T 2 : Since ecc(λ) = 2 and ecc(μ) = 2, so:

N 0 ( λ ) = { λ } , N 1 ( λ ) = V ( G ( Γ ) ) Ω 3 i , N 2 ( λ ) = Ω 3 i { λ } N 0 ( μ ) = { μ } , N 1 ( μ ) = ζ ( Γ ) Ω 3 , N 2 ( μ ) = Ω 2 { μ }

Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Ω 2 { μ } , N 2 ( λ ) = N 0 ( μ ) = { μ } , N 1 ( μ ) = Ω 3 i { λ } , N 2 ( μ ) =

and hence η λ (e|G(Γ)) = n and η μ (e|G(Γ)) = 2. Let t 2 be the number of type T 2 edges, then t 2 = |Ω2| × |Ω3| = n(n− 1).

Let e is of type T 3 : Since ecc(λ) = 1 and ecc(μ) = 2, so:

N 0 ( λ ) = { λ } , N 1 ( λ ) = V ( G ( Γ ) ) { λ } N 0 ( μ ) = { μ } , N 1 ( μ ) = V ( G ( Γ ) ) Ω 3 i , N 2 ( μ ) = Ω 3 i { μ }

Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Ω 3 i { μ } , N 0 ( μ ) = { μ } , N 1 ( μ ) = N 2 ( μ ) =

and hence η λ (e|G(Γ)) = n and η μ (e|G(Γ)) = 1. Let t 4 be the number of type T 4 edges, then t 4 = |ζ(Γ)| × |Ω2| = n.

Now, the Mostar index of G(Γ) is:

Mo ( G ( Γ ) ) = i = 1 4 λ μ E i ( G ( Γ ) ) | η λ ( e | G ( Γ ) ) η μ ( e | G ( Γ ) ) | = t 1 | 2 2 | + t 2 | n 2 | + t 3 | 2 1 | + t 4 | n 1 | = ( n 1 ) ( n 2 n + 1 )

Case 2 (n is even):

Let E ( Γ G ) = i = 1 6 E i ( Γ G ) with E i G ) = {eE G )|e is of type T i }.

  • Type T 1: e = λμE 1(G(Γ)) such that λ Ω 3 i and μ Ω 3 k for ik.

  • Type T 2: e = λμE 2(G(Γ)) such that λ Ω 3 i and μ Ω 2 j .

  • Type T 3: e = λμE 3(G(Γ)) such that λζ(Γ) and μ Ω 3 i .

  • Type T 4: e = λμE 4(G(Γ)) such that λζ(Γ) and μ Ω 2 j .

  • Type T 5: e = λμE 5(G(Γ)) such that λζ(Γ) and μζ(Γ).

  • Type T 6: e = λμE 6(G(Γ)) such that λ Ω 2 j and μ Ω 2 k for j = 0,1, k = 0,1, and jk.

Let e is of type T 1 : Since ecc(λ) = 2 and ecc(μ) = 2, so:

N 0 ( λ ) = { λ } , N 1 ( λ ) = V ( G ( Γ ) ) Ω 3 i , N 2 ( λ ) = Ω 3 i { λ } N 0 ( μ ) = { μ } , N 1 ( μ ) = V ( G ( Γ ) ) Ω 3 k , N 2 ( μ ) = Ω 3 k { μ }

Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Ω 3 k { μ } , N 2 ( λ ) = N 0 ( μ ) = { μ } , N 1 ( μ ) = Ω 3 i { λ } , N 2 ( μ ) =

and hence η λ (e|G(Γ)) = 2 and η μ (e|G(Γ)) = 2. Let t 1 be the number of edges of type T 1. Since the partition of Ω3 with parts Ω 3 i induces a complete multipartite graph, so for any λ Ω 3 i , d ( λ ) = 2 ( n 2 ) 2 1 = n 4 in the subgraph of G Γ induced by the set i = 1 n 2 2 Ω 3 i . Since | Ω 3 i | = 2 and there are n 2 2 such sets, so by the formula of handshake lemma, t 1 = ( n 2 ) ( n 4 ) 2 .

Let e is of type T 2 : Since ecc(λ) = 2 and ecc(μ) = 2, so:

N 0 ( λ ) = { λ } , N 1 ( λ ) = V ( G ( Γ ) ) Ω 3 i , N 2 ( λ ) = Ω 3 i { λ } N 0 ( μ ) = { μ } , N 1 ( μ ) = V ( G ( Γ ) ) Ω 2 j , N 2 ( μ ) = Ω 2 j { μ }

Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Ω 2 j { μ } , N 2 ( λ ) = N 0 ( μ ) = { μ } , N 1 ( μ ) = Ω 3 i { λ } , N 2 ( μ ) =

and hence η λ ( e | G ( Γ ) ) = n 2 and η μ (e|G(Γ)) = 2. Let t 2 be the number of type T 2 edges, then t 2 = |Ω2| × |Ω3| = n(n − 2).

Let e is of type T 3 : Since ecc(λ) = 1 and ecc(μ) = 2, so:

N 0 ( λ ) = { λ } , N 1 ( λ ) = V ( G ( Γ ) ) { λ } N 0 ( μ ) = { μ } , N 1 ( μ ) = V ( G ( Γ ) ) Ω 3 i , N 2 ( μ ) = Ω 3 i { μ }

Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Ω 3 i { μ } , N 0 ( μ ) = { μ } N 1 ( μ ) = , N 2 ( μ ) =

and hence η λ (e|G(Γ)) = 2 and η μ (e|G(Γ)) = 1. Let t 3 be the number of type T 3 edges, then t 3 = |ζ(Γ)| × |Ω3| = 2(n − 2).

Let e is of type T 4 : Since ecc(λ) = 1 and ecc(μ) = 2, so:

N 0 ( λ ) = { λ } , N 1 ( λ ) = V ( G ( Γ ) ) { λ } N 0 ( μ ) = { μ } , N 1 ( μ ) = V ( G ( Γ ) ) Ω 2 j , N 2 ( μ ) = Ω 2 j { μ }

Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Ω 2 i { μ } , N 0 ( μ ) = { μ } N 1 ( μ ) = N 2 ( μ ) = and hence η λ ( e | G ( Γ ) ) = n 2 and η μ (e|G(Γ)) = 1. Let t 4 be the number of type T 4 edges, then t 4 = |ζ(Γ)| × |Ω2| = 2n.

Let e is of type T 5 : Note that N[λ] = V(G(Γ)) = N[μ]. Thus, λ and μ are true twins, so Proposition 3 yields that η λ (e|G Γ) = 1 = η μ (e|G Γ). Let t 5 be the number of type T 5 edges. Since ζ(Γ) induces a complete graph K 2, so t 1 = S(K 2) = 1.

Let e is of type T 6 : Since ecc(λ) = 2 and ecc(μ) = 2, so:

N 0 ( λ ) = { λ } , N 1 ( λ ) = V ( G ( Γ ) ) Ω 2 j , N 2 ( λ ) = Ω 2 j { λ } N 0 ( μ ) = { μ } , N 1 ( μ ) = V ( G ( Γ ) ) Ω 2 k , N 2 ( μ ) = Ω 2 k { μ }

Accordingly, Remark 1 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Ω 2 k { μ } , N 2 ( λ ) = N 0 ( μ ) = { μ } , N 1 ( μ ) = Ω 2 j { λ } , N 2 ( μ ) =

and hence η λ ( e | G ( Γ ) ) = n 2 and η μ ( e | G ( Γ ) ) = n 2 . Let t 6 be the number of type T 6 edges, then t 6 = | Ω 2 j | × | Ω 2 k | = n 2 4 .

Now, the Mostar index of G(Γ) is:

Mo ( G ( Γ ) ) = i = 1 6 λ μ E i ( G ( Γ ) ) | η λ ( e | G ( Γ ) ) η μ ( e | G ( Γ ) ) | = t 1 2 2 + t 2 n 2 2 + t 3 2 1 + t 4 n 2 1 + t 5 1 1 + t 6 | n 2 n 2 | = n 2 2 ( n 2 2 n + 4 )

Theorem 6

For n ≥ 2, let Γ be a semi-dihedral group. Then:

Mo ( G ( Γ ) ) = 4 ( 2 n 1 ) ( 4 n 2 2 n + 1 ) , w h e n n i s e v e n 16 ( n 1 ) ( n 2 2 n + 4 ) , w h e n n i s o d d

Proof

From conjugacy classes, written in Eqs. 2 and 3, the number of conjugacy classes of order 1, 2 and, n is 4, 2n − 2, and 4, respectively, whenever n is odd, and the number of conjugacy classes of order 1, 2, and 2n is 2, 2n − 1, and 2, respectively, whenever n is even. Thus, the non-conjugate graph on Γ = SD8n can be expressed as follows:

G ( Γ ) = K 1 , 1 , 1 , 1 , 2 , 2 , , 2 2 n 2 times , n , n , n , n , if n is odd  K 1 , 1 , 2 , 2 , , 2 2 n 1 times , 2 n , 2 n , if n is even

Here K 1 , 1 , 1 , 1 , 2 , 2 , , 2 2 n 2 times , n , n , n , n , is a complete(2n + 6)-partite graph having n − 1 parts Φ 3 i for odd i; n−1 parts Φ 3 i for even i; four parts from ζ(Γ); and four parts Φ 2 j . Similarly, K 1 , 1 , 2 , 2 , , 2 2 n 1 times , 2 n , 2 n , is a complete (2n + 3)-partite graph having n parts Φ 3 i for odd i;

n − 1 parts Φ 3 i for even i; two parts from ζ(Γ), and two parts Φ 2 1 and Φ 2 2 .

Now, we discuss the following two cases.□

Case 1 (n is odd):

Let E ( Γ G ) = i = 1 6 E i ( Γ G ) with E i G ) = {eE G )|e is of type T i }.

  • Type T 1: e = λμE 1(G(Γ)) such that λ Φ 3 i and μ Φ 3 k for ik.

  • Type T 2: e = λμE 2(G(Γ)) such that λ Φ 3 i and μ Φ 2 j .

  • Type T 3: e = λμE 3(G(Γ)) such that λζ(Γ) and μ Φ 3 i .

  • Type T 4: e = λμE 4(G(Γ)) such that λζ(Γ) and μ Φ 2 j .

  • Type T 5: e = λμE 5(G(Γ)) such that λζ(Γ) and μζ(Γ).

  • Type T 6: e = λμE 6(G(Γ)) such that λ Φ 2 j and μ Φ 2 k for jk.

Let e is of type T 1 : Since ecc(λ) = 2 and ecc(μ) = 2, so accordingly, Remark 3 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Φ 3 k { μ } , N 2 ( λ ) = N 0 ( μ ) = { μ } , N 1 ( μ ) = Φ 3 i { λ } , N 2 ( μ ) =

and hence η λ (e|G(Γ)) = 2 and η μ (e|G(Γ)) = 2. Let t 1 be the number of edges of type T 1. Since the partition of Ω3 with parts Φ 3 i induces a complete multipartite graph, so for any λ Φ 3 i , d(λ) = 2(2n − 2 − 1) = 4n − 6 in the subgraph of G(Γ) induced by the set odd i = 1 n 1 Φ 3 i odd i = 2 n + 1 3 n 2 Φ 3 i even i = 2 2 n 2 Φ 3 i . Since | Φ 3 i | = 2 and there are 2n − 2 such sets, so by the formula of handshake lemma, t 1 = (2n − 2)(4n − 6).

Let e is of type T 2 : Since ecc(λ) = 2 and ecc(μ) = 2, so:

N 0 ( λ ) = { λ } , N 1 ( λ ) = V ( G ( Γ ) ) Φ 3 i , N 2 ( λ ) = Φ 3 i { λ } N 0 ( μ ) = { μ } , N 1 ( μ ) = V ( G ( Γ ) ) Φ 2 j , N 2 ( μ ) = Φ 2 j { μ }

Accordingly, Remark 3 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Φ 2 j { μ } , N 2 ( λ ) = N 0 ( μ ) = { μ } , N 1 ( μ ) = Φ 3 i { λ } , N 2 ( μ ) =

and hence η λ (e|G(Γ)) = n and η μ (e|G(Γ)) = 2. Let t 2 be the number of type T 2 edges, then t 2 = |Ω2| × |Ω3| = 4n(n − 4).

Let e is of type T 3 : Since ecc(λ) = 1 and ecc(μ) = 2, so:

N 0 ( λ ) = { λ } , N 1 ( λ ) = V ( G ( Γ ) ) { λ } N 0 ( μ ) = { μ } , N 1 ( μ ) = V ( G ( Γ ) ) Φ 3 i , N 2 ( μ ) = Φ 3 i { μ }

Accordingly, Remark 3 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Φ 3 i { μ } , N 0 ( μ ) = { μ } , N 1 ( μ ) = N 2 ( μ ) =

and hence η λ (e|G(Γ)) = 2 and η μ (e|G(Γ)) = 1. Let t 3 be the number of type T 3 edges, then t 3= |ζ(Γ)| × |Ω3| = 4(4n − 4).

Let e is of type T 4 : Since ecc(λ) = 1 and ecc(μ) = 2, so:

N 0 ( λ ) = { λ } , N 1 ( λ ) = V ( G ( Γ ) ) { λ } N 0 ( μ ) = { μ } , N 1 ( μ ) = V ( G ( Γ ) ) Φ 2 j , N 2 ( μ ) = Φ 2 j { μ }

Accordingly, Remark 3 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Φ 2 j { μ } , N 0 ( μ ) = { μ } , N 1 ( μ ) = N 2 ( μ ) =

and hence η λ (e|G Γ) = n and η μ (e|G(Γ)) = 1. Let t 4 be the number of type T 4 edges, then t 4 = |ζ(Γ)| × |Ω2| = 16n.

Let e is of type T 5 : Note that N[λ] = V(G(Γ)) = N[μ]. Thus, λ and μ are true twins, so Proposition 4 yields that η λ (e|G Γ) = 1 = η μ (e|G Γ). Let t 5 be the number of type T 5 edges. Since ζ(Γ) induces a complete graph K |ζ(Γ)|, so t 1 = S(K 4) = 42.

Let e is of type T 6 : Since ecc(λ) = 2 and ecc(μ) = 2, so:

N 0 ( λ ) = { λ } , N 1 ( λ ) = V ( G ( Γ ) ) Φ 2 j , N 2 ( λ ) = Φ 2 j { λ } N 0 ( μ ) = { μ } , N 1 ( μ ) = V ( G ( Γ ) ) Φ 2 k , N 2 ( μ ) = Φ 2 k { μ }

Accordingly, Remark 3 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Φ 2 k { μ } , N 2 ( λ ) = N 0 ( μ ) = { μ } , N 1 ( μ ) = Φ 2 j { λ } , N 2 ( μ ) =

and hence η λ (e|G(Γ)) = n and η μ (e|G(Γ)) = n. Let t 6 be the number of edges of type T 6. Since the partition of Ω2 with parts Φ 2 j induces a complete multipartite graph, so for any λ Φ 2 j , d(λ) = 3n in the subgraph of G(Γ) induced by the set j = 0 3 Φ 2 j . Since | Φ 2 j | = n and there are 4 such sets, so by the formula of handshake lemma, t 6 = 6n 2.

Now, the Mostar index of G(Γ) is:

Mo(G(Γ)) = i = 1 6 λ μ E i ( G ( Γ ) ) | η λ ( e | G ( Γ ) ) η μ ( e | G ( Γ ) ) | = t 1 2 2 + t 2 n 2 + n ( T 3 ) 2 1 + n ( T 4 ) | n 1 | + t 5 | 1 1 | + t 6 | n n | = 16 ( n 1 ) ( n 2 n + 1 )

Case 2 (n is even):

Let E ( Γ G ) = i = 1 6 E i ( Γ G ) with E i G ) = {eE G )|e is of type T i }.

  • Type T 1: e = λμE 1(G(Γ)) such that λ Φ 3 i and μ Φ 3 k for ik.

  • Type T 2: e = λμE 2(G(Γ)) such that λ Φ 3 i and μ Φ 2 j .

  • Type T 3: e = λμE 3(G(Γ)) such that λζ(Γ) and μ Φ 3 i .

  • Type T 4; e = λμE 4(G(Γ)) such that λζ(Γ) and μ Φ 2 j .

  • Type T 5: e = λμE 5(G(Γ)) such that λζ(Γ) and μζ(Γ).

  • Type T 6: e = λμE 6(G(Γ)) such that λ Φ 2 j and μ Φ 2 k for jk.

Let e is of type T 1 : Since ecc(λ) = 2 and ecc(μ) = 2, so:

N 0 ( λ ) = { λ } , N 1 ( λ ) = V ( G ( Γ ) ) Φ 3 i , N 2 ( λ ) = Φ 3 i { λ } N 0 ( μ ) = { μ } , N 1 ( μ ) = V ( G ( Γ ) ) Φ 3 k , N 2 ( μ ) = Φ 3 k { μ }

Accordingly, Remark 3 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Φ 3 k { μ } , N 2 ( λ ) = N 0 ( μ ) = { μ } , N 1 ( μ ) = Φ 3 i { λ } , N 2 ( μ ) =

and hence η λ (e|G(Γ)) = 2 and η μ (e|G(Γ)) = 2. Let t 1 be the number of edges of type T 1. Since the partition of Ω3 with parts Φ 3 i induces a complete multipartite graph, so for any λ Φ 3 i , d(λ) = 2(2n − 1 − 1) = 4n − 4 in the subgraph of G(Γ) induced by the set odd i = 1 n 1 Φ 3 i odd i = 2 n + 1 3 n 1 Φ 3 i even i = 2 2 n 2 Φ 3 i . Since | Φ 3 i | = 2 and there are 2n − 1 such sets, so by the formula of handshake lemma, t 1 = (2n − 2)(4n − 2).

Let e is of type T 2 : Since ecc(λ) = 2 and ecc(μ) = 2, so:

N 0 ( λ ) = { λ } , N 1 ( λ ) = V ( G ( Γ ) ) Φ 3 i , N 2 ( λ ) = Φ 3 i { λ } N 0 ( μ ) = { μ } , N 1 ( μ ) = V ( G ( Γ ) ) Φ 2 j , N 2 ( μ ) = Φ 2 j { μ }

Accordingly, Remark 3 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Φ 2 j { μ } , N 2 ( λ ) = N 0 ( μ ) = { μ } , N 1 ( μ ) = Φ 3 i { λ } , N 2 ( μ ) =

and hence η λ (e|G(Γ)) = 2n and η μ (e|G(Γ)) = 2. Let t 2 be the number of type T 2 edges, then t 2 = |Ω2| × |Ω3| = 4n(n − 2).

Let e is of type T 3 : Since ecc(λ) = 1 and ecc(μ) = 2, so:

N 0 ( λ ) = { λ } , N 1 ( λ ) = V ( G ( Γ ) ) { λ } N 0 ( μ ) = { μ } , N 1 ( μ ) = V ( G ( Γ ) ) Φ 3 i , N 2 ( μ ) = Φ 3 i { μ }

Accordingly, Remark 3 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Φ 3 i { μ } N 0 ( μ ) = { μ } , N 1 ( μ ) = , N 2 ( μ ) =

and hence η λ (e|G(Γ)) = 2 and η μ (e|G(Γ)) = 1. Let t 3 be the number of type T 3 edges, then t 3= |ζ(Γ)| × |Ω3| = 4(2n − 1).

Let e is of type T 4 : Since ecc(λ) = 1 and ecc(μ) = 2, so

N 0 ( λ ) = { λ } , N 1 ( λ ) = V ( G ( Γ ) ) { λ } N 0 ( μ ) = { μ } , N 1 ( μ ) = V ( G ( Γ ) ) Φ 2 j , N 2 ( μ ) = Φ 2 j { μ }

Accordingly, Remark 3 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Φ 2 j { μ } N 0 ( μ ) = { μ } , N 1 ( μ ) = , N 2 ( μ ) =

and hence η λ (e|G Γ) = 2n and η μ (e|G(Γ)) = 1. Let t 4 be the number of type T 4 edges, then t 4 = |ζ(Γ)| × |Ω2| = 8n.

Let e is of type T 5 : Note that N[λ] = V(G(Γ)) = N[μ]. Thus, λ and μ are true twins, so Proposition 4 yields that η λ (e|G Γ) = 1 = η μ (e|G Γ). Let t 5 be the number of type T 5 edges. Since ζ(Γ) induces a complete graph K 2, so t 1 = S(K 2) = 1.

Let e is of type T 6 : Since ecc(λ) = 2 and ecc(μ) = 2, so:

N 0 ( λ ) = { λ } , N 1 ( λ ) = V ( G ( Γ ) ) Φ 2 j , N 2 ( λ ) = Φ 2 j { λ } N 0 ( μ ) = { μ } , N 1 ( μ ) = V ( G ( Γ ) ) Φ 2 k , N 2 ( μ ) = Φ 2 k { μ }

Accordingly, Remark 3 implies that:

N 0 ( λ ) = { λ } , N 1 ( λ ) = Φ 2 k { μ } , N 2 ( λ ) = N 0 ( μ ) = { μ } , N 1 ( μ ) = Φ 2 j { λ } , N 2 ( μ ) =

and hence η λ (e|G(Γ)) = 2n and η μ (e|G(Γ)) = 2n. Let t 6 be the number of type T 6 edges, then t 6 = | Φ 2 j | × | Φ 2 k | = 4 n 2 .

Now, the Mostar index of G(Γ) is:

Mo ( G ( Γ ) ) = i = 1 6 λ μ E i ( G ( Γ ) ) | η λ ( e | G ( Γ ) ) η μ ( e | G ( Γ ) ) | = t 1 2 2 + t 2 2 n 2 + t 3 2 1 + t 4 2 n 1 + t 5 1 1 + t 6 | 2 n 2 n | = 4 ( 2 n 1 ) ( 4 n 2 2 n + 1 )

6 Conclusion

A classical field of study associating graphs with algebraic structures is extended by exploring graph distance neighborhood-based property (which is also known as a bond-additive property) of commuting, non-commuting, and non-conjugate graphs associated with the group of symmetries of regular polygon and its semi version. In fact, we determined the Mostar index of these graphs. Basic theme of this work was to propose a different technique which is quite easy and interesting as compares with any of the other direct methods to measure the peripherality of edges. It is based upon the distance structure of a graph captured through observing distance degree neighborhoods. Researchers working on the peripherality measurement of edges of various graphs, especially graphs having twins, can get a remarkable help by understanding the proposed technique.

Acknowledgment

All the authors are thankful to anonymous referees for their suggestions, which help us to improve the article. The authors are also thankful to the Scientific and Technological Research Council of Turkey (TUBITAK) for providing funds to publish this research work.

  1. Funding information: This research was supported by the Scientific and Technological Research Council of Turkey (TUBITAK) (Grant number: 2221-(1059B212100358)).

  2. Author contributions: Masood Ur Rehman: writing – review and editing and project administration; Muhammad Salman: idea and conceptualization, methodology, and supervision; Sheraz Khan: evaluation and writing – original draft; Ayse Dilek Maden: visualization and project administration; Faisal Ali: formal analysis and editing.

  3. Conflict of interest: Authors state no conflict of interest.

  4. Data availability statement: Data used to support this study are included within the article.

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Received: 2022-03-18
Accepted: 2022-05-26
Published Online: 2022-08-23

© 2022 Masood Ur Rehman et al., published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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