Structure of the Equal Bi-Quadratic Graph of Finite Groups

Introduction

Numerous graphs have been introduced and examined in association with algebraic structures such as square element graph [10,11] , 𝑘 𝑡ℎ -power graph [9] , square power graph [3][4][5] , equalsquare graph [1,2] , cubic power graph [6][7][8] etc. Various characteristics like completeness, traversability, girth, diameter, chromatic number and bipartiteness of line graph are examined in [12] . Authors have examined and introduced graphical representation of Equal-Square graphs having finite groups as their vertex sets in [1] . This research extends the concept of Equal-Square graph to Equal Bi-Quadratic graph of a finite group. Equal Bi-Quadratic of a finite group 𝐺, is an undirected simple graph with different vertices pairs 𝑥 and 𝑦 having edge iff 𝑥 4 = 𝑦 4 . In Section 2, connectedness and structure of Equal Bi-Quadratic graph of cyclic groups are examined where as in section 3, for dihedral group.

Equal Bi-Quadratic Graph of Finite Cyclic Group Theorem 2.1 The Equal Bi-Quadratic graph of finite group 𝐺 of order 𝑚 is an empty graph iff (𝑚, 4) = 1.

Proof. Let 𝐸𝐵𝑄(𝐺) be the Equal Bi-Quadratic graph of finite group 𝐺 of order 𝑚. For (𝑚, 4) = 1, 𝑚 is of form 𝑚 = 4𝑛 + 1 or 4𝑛 + 3. Let us assume that different pair of adjacent vertices 𝑢 and 𝑣 in 𝐸𝐵𝑄(𝐺).

Case (i) 𝑚 = 4𝑛 + 1

As (𝑢, 𝑣) ∈ 𝐸(𝐸𝐵𝑄(𝐺)), so 𝑢 4 = 𝑣 4 → (𝑢 4 ) 3𝑛+1 = (𝑣 4 ) 3𝑛+1 → 𝑢 12𝑛+4 = 𝑣 12𝑛+4 → 𝑢 4𝑛+1 𝑢 4𝑛+1 𝑢 4𝑛+1 𝑢 = 𝑣 4𝑛+1 𝑣 4𝑛+1 𝑣 4𝑛+1 𝑣 → 𝑢 = 𝑣. Thus, there is no pair of distinct vertices adjacent in 𝐸𝐵𝑄(𝐺).

Case (ii) 𝑚 = 4𝑛 + 3

As (𝑢, 𝑣) ∈ 𝐸(𝐸𝐵𝑄(𝐺)), so 𝑢 4 = 𝑣 4 → (𝑢 4 ) 𝑛+1 = (𝑣 4 ) 𝑛+1 → 𝑢 4𝑛+4 = 𝑣 4𝑛+4 → 𝑢 4𝑛+3 𝑢 = 𝑣 4𝑛+3 𝑣 → 𝑢 = 𝑣. Thus, there is no pair of distinct vertices adjacent in 𝐸𝐵𝑄(𝐺). Hence 𝐸𝐵𝑄(𝐺) is an empty graph for (𝑚, 4) = 1. In below figure structure of Equal Bi-Quadratic Graph ofℤ 10 = {𝑒, 𝑥, 𝑥 2 , 𝑥 3 , 𝑥 4 , 𝑥 5 , 𝑥 6 , 𝑥 7 , 𝑥 8 , 𝑥 9 } is shown. ).

Theorem 2.2 Let 𝐸𝐵𝑄(𝐺) be the Equal Bi-Quadratic graph of a finite cyclic group of order

So, there are disjoint In below figure structure of Equal Bi-Quadratic Graph ofℤ 16 = {𝑒, 𝑥, 𝑥 2 , 𝑥 3 , 𝑥 4 , 𝑥 5 , 𝑥 6 , 𝑥 7 , 𝑥 8 , 𝑥 9 , 𝑥 10 , 𝑥 11 , 𝑥 12 , 𝑥 13 , 𝑥 14 , 𝑥 15 } is shown. Proof. Let 𝐸𝐵𝑄(𝐷 𝑚 ) be the Equal Bi-Quadratic graph of a dihedral group 𝐷 𝑚 = {𝑒, 𝑟, 𝑟 2 , ⋯ , 𝑟 𝑚-1 , 𝑠, 𝑟𝑠, 𝑟 2 𝑠, ⋯ , 𝑟 𝑚-1 𝑠} of order 2𝑚. For reflection elements of 𝐷 𝑛 , (𝑟 𝑖 𝑠) 4 = 𝑒 ∀ 0 ≤ ⅈ ≤ 𝑚 -1 and for.

Case (i) (𝑚, 4) = 1

In this case, (𝑟 𝑖 ) 4 ≠ (𝑟 𝑗 ) 4 for all 1 ≤ ⅈ, 𝑗 ≤ 𝑚 -1 and (𝑟 𝑖 𝑠) 4 = 𝑒 for all 0 ≤ ⅈ ≤ 𝑚 -1. Thus, 𝑚 -1 rotation elements of 𝐷 𝑛 forms the (𝑚 -1)𝐾 1 disjoint component of 𝐸𝐵𝑄(𝐷 𝑛 ) and identity element & 𝑚 reflection elements forms the 𝐾 𝑚+1 component of 𝐸𝐵𝑄(𝐷 𝑛 ). Thus, 𝐸𝐵𝑄(𝐷 𝑛 ) = 𝐾 𝑚+1 ∪ (𝑚 -1)𝐾 1 . In below figure structure of Equal Bi-Quadratic Graph of𝐷 5 = {𝑒, 𝑟, 𝑟 2 , 𝑟 3 , 𝑟 4 , 𝑠, 𝑟𝑠, 𝑟 2 𝑠, 𝑟 3 𝑠, 𝑟 4 𝑠} is shown. In below figure structure of Equal Bi-Quadratic Graph of𝐷 6 = {𝑒, 𝑟, 𝑟 2 , 𝑟 3 , 𝑟 4 , 𝑟 5 , 𝑠, 𝑟𝑠, 𝑟 2 𝑠, 𝑟 3 𝑠, 𝑟 4 𝑠, 𝑟 5 𝑠} is shown. In below figure structure of Equal Bi-Quadratic Graph of𝐷 8 = {𝑒, 𝑟, 𝑟 2 , 𝑟 3 , 𝑟 4 , 𝑟 5 , 𝑟 6 , 𝑟 7 , 𝑠, 𝑟𝑠, 𝑟 2 𝑠, 𝑟 3 𝑠, 𝑟 4 𝑠, 𝑟 5 𝑠, 𝑟 6 𝑠, 𝑟 7 𝑠} is shown.

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