Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 6s (2024) 592 https://internationalpubls.com Multi-Party Key Agreement Protocols using Conjugacy Search Problem in Five dimensional Discrete Heisenberg Group N. Sundarakannan1,* ,T. Isaiyarasi2, V. Nagarani3, D. Captain Prabakaran4 , M.H.A. Aysha Chitukka5 1, 2, 3,4 Department of Mathematics, SRM Valliammai Engineering College, Kattangulathur, Chennai 603 203 1 sundarakannann.maths@srmvalliammai.ac.in 2 isaiyarasit.maths@srmvalliammai.ac.in 3nagaraniv.maths@srmvalliammai.ac.in 4captainprabakarand.maths@srmvalliammai.ac.in 5ayshachitukkamha.maths@srmvalliammai.ac.in *Corresponding author : N. Sundarakannan Article History: Received: 18-06-2024 Revised: 27-07-2024 Accepted: 08-08-2024 Abstract: The secrete way of communicating information between entities is the science known as cryptography. In order to communicate the message through an insecure medium the communicating entities need a cryptographic keys. The process of getting a common shared key is the Key agreement Protocol. This paper proposes a multi party key agreement protocol which provides a common key between the communicating parties using conjugacy Search Problem (CSP). To implement this protocol a platform group is required. The five dimensional Discrete Heisenberg Group(DHG) is chosen as the platform group and the conjugacy search problem (CSP) taken from combinatorial group theory acts as the one way function. As a particular case a five party KAP with numerical computation is provided. In general, multiparty CSP with k communicating parties can get a common key in k-1 rounds. Keywords: Cryptography, key agreement protocol, one way function, communicating party, discrete heisenberg group, conjugacy search problem. 1. Introduction: Cryptography is the art of achieving security by encrypting the messages from the sender to the receiver which is made non-readable for others. The process of encoding plain text into cipher text message is called encryption. The reverse process of transforming cipher text called decryption. There are two cryptographic mechanism depends on the key used. If the same key is used for encryption and decryption it is called symmetric key cryptography or private key cryptography. If two different keys are used in a cryptographic mechanism where in one key is used for encryption and another key is used is used for decryption then it is called asymmetric key cryptography or public key cryptography. Key exchange protocol enable secure communication over an untrusted network by deriving and distributing shared keys between two or more parties. There are two type of key exchange protocols commonly known as Key transport protocol and Key agreement protocol. Key transport protocol is a key establishment protocol in which one of the principal generates the key and this key then transferred to all protocol users. Key agreement protocol is a key establishment protocol in which the session key is a function of inputs by all protocol users. mailto:sundarakannann.maths@srmvalliammai.ac.in mailto:isaiyarasit.maths@srmvalliammai.ac.in mailto:3nagaraniv.maths@srmvalliammai.ac.in mailto:captainprabakarand.maths@srmvalliammai.ac.in mailto:ayshachitukkamha.maths@srmvalliammai.ac.in Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 6s (2024) 593 https://internationalpubls.com The public key cryptography system was introduced by Diffe at al.[6] in their article ‘A new directions in cryptography’. In this a key agreement protocol was proposed with in which the discrete logarithm problem served as the one way function. In [5] Anshel et al. key agreement protocols were proposed using the search problems as the one way functions. In these protocols the Braid group was chosen as the platform group. In [7] T. Isaiyarasi T et al. proposed ‘A New Multiparty Key Agreement Protocol using tripled decomposition search problem in Discrete Heisenberg Group’ was proposed. T. Isaiyarasi et al.[9], A New Key Agreement Protocol using two layers of Security was proposed .In [11] ‘Authenticated Key Agreement Protocol for HWSN Network’ was proposed using factorization search problem. 2. Definitions: 2.1 Discrete Heisenberg Group: Let ℋ be the set of all 5-tuples(𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑥5) where 𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑥5∈, and ⋅ be the binary operation. The binary operation ⋅ is defined as follows (𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑥5)⋅(𝑦1, 𝑦2, 𝑦3, 𝑦4, 𝑦5) = (𝑥1 + 𝑦1, 𝑥2 + 𝑦2, 𝑥3 + 𝑦3, 𝑥4 + 𝑦4, 𝑥5 + 𝑦5 + 𝑥1𝑦3 + 𝑥2𝑦4) ---(1) It can be easily verified that the set ℋ is a group but not abelian under this binary operation⋅. For the cryptographic purpose we consider a finite group ℋ. When the modular arithmetic is introduced, this group can be made finite as follows: (𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑥5)⋅(𝑦1, 𝑦2, 𝑦3, 𝑦4, 𝑦5) = (𝑥1 + 𝑦1, 𝑥2 + 𝑦2, 𝑥3 + 𝑦3, 𝑥4 + 𝑦4, 𝑥5 + 𝑦5 + 𝑥1𝑦3 + 𝑥2𝑦4)𝑚𝑜𝑑 𝑝 Where𝑝 is a prime number. (i) The number of elements in ℋ = 𝑍𝑝, 𝑖. 𝑒. , (ℋ) = 𝑝5 (ii) Existence of Identity element : For any (𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑥5)∈ℋ (𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑥5)⋅ (0,0,0,0,0) = (𝑥1 + 0, 𝑥2 + 0, 𝑥3 + 0, 𝑥4 + 0, 𝑥5 + 0 + 𝑥1. 0 + 𝑥2. 0) = (𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑥5) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 6s (2024) 594 https://internationalpubls.com Also (0,0,0,0,0) ⋅ (𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑥5) = (0 + 𝑥1, 0 + 𝑥2, 0 + 𝑥3, 0 + 𝑥4, 0 + 𝑥5 + 0. 𝑥3 + 0. 𝑥4) = (𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑥5) Therefore (0,0,0,0,0) is the identity element of ℋ (iii) Existence of Inverse element : For any(𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑥5)∈ℋ , its inverse is computed as follows. (𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑥5)⋅(𝑦1, 𝑦2, 𝑦3, 𝑦4, 𝑦5) = (0,0,0,0,0) ⟹(𝑥1 + 𝑦1, 𝑥2 + 𝑦2, 𝑥3 + 𝑦3, 𝑥4 + 𝑦4, 𝑥5 + 𝑦5 + 𝑥1𝑦3 + 𝑥2𝑦4) = (0,0,0,0,0) ⟹𝑥1 + 𝑦1 = 0 , 𝑥2 + 𝑦2 = 0, 𝑥3 + 𝑦3 = 0 , 𝑥4 + 𝑦4 = 0 , 𝑥5 + 𝑦5 + 𝑥1𝑦3 + 𝑥2𝑦4 = 0 ⟹ 𝑦1 = −𝑥1, 𝑦2 = −𝑥2, 𝑦3 = −𝑥3, 𝑦4 = −𝑥4, 𝑦5 = −𝑥5 + 𝑥1𝑥3 + 𝑥2𝑥4 ∴ (𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑥5)−1 = (−𝑥1, −𝑥2, −𝑥3, −𝑥4, −𝑥5 + 𝑥1𝑥3 + 𝑥2𝑥4)𝑚𝑜𝑑 𝑝 (iv) (𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑥5)𝑛 = (𝑛𝑥1, 𝑛𝑥2, 𝑛𝑥3, 𝑛𝑥4, 𝑛𝑥5 + 𝑛2(𝑥1𝑥3 + 𝑥2𝑥4))𝑚𝑜𝑑 𝑝 This can be proved by mathematical induction 2.2 Conjugacy serach problem : The conjugacy search problem (CSP): Given a recursive presentation of a group G and two conjugate elements 𝑥, 𝑦 ∈ 𝐺, find a particular element 𝑔 ∈ 𝐺 such that 𝑔−1𝑥𝑔 = 𝑦. 2.3 Conjugacy search problem in discrete Heisenberg group: The elements of the conjugacy search problem are chosen from the discrete heisenberg group ℋ with condition be described below for two party protocol and generate the common public key. Two parties 𝐴1 and 𝐴2 agree on a finite non-abelian group, discrete heisenberg group ℋ = 𝑍𝑝 5 and 𝐴1 publishes an element 𝑥 ∈ ℋ such that 𝑥 does not commute with 𝑔1 and 𝑔2. 𝐴1 chooses an element 𝑔1 ∈ 𝐺1, computes 𝑔1 −1𝑥𝑔1 and sends it to 𝐴2 𝐴2 chooses an element 𝑔2 ∈ 𝐺1, computes 𝑔2 −1𝑥𝑔2 and sends it to 𝐴1 On knowing 𝑔1 , 𝐴1 computes 𝐾𝐴 = 𝑔1 −1𝑔2 −1𝑥𝑔2𝑔1. On knowing 𝑔2, 𝐴2 computes 𝐾𝐵 = 𝑔2 −1𝑔1 −1𝑥𝑔1𝑔2. 𝐾𝐴 = 𝐾𝐵 is there common shared key. 2.4 Commutative condition: Let us consider 𝑔1 = (𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑥5) and 𝑔2 = (𝑦1, 𝑦2, 𝑦3, 𝑦4, 𝑦5) 𝑔1. 𝑔2 = (𝑥1 + 𝑦1, 𝑥2 + 𝑦2, 𝑥3 + 𝑦3, 𝑥4 + 𝑦4, 𝑥5 + 𝑦5 + 𝑥1𝑦3 + 𝑥2𝑦4)𝑚𝑜𝑑 𝑝 And 𝑔2. 𝑔1 = (𝑦1 + 𝑥1, 𝑦2 + 𝑥2, 𝑦3 + 𝑥3, 𝑦4 + 𝑥4, 𝑦5 + 𝑥5 + 𝑦1𝑥3 + 𝑦2𝑥4)𝑚𝑜𝑑 𝑝 If 𝑔1. 𝑔2 = 𝑔2. 𝑔1 then for 5-tuple be identically equal. So first four values are exactly equal. But for fifth value is equal if 𝑥1𝑦3 + 𝑥2𝑦4 = 𝑦1𝑥3 + 𝑦2𝑥4. 2.5 Inverse element: The inverse of 𝑔1and 𝑔2are 𝑔1 −1 = (−𝑥1, −𝑥2, −𝑥3, −𝑥4, −𝑥5 + 𝑥1𝑥3 + 𝑥2𝑥4) and 𝑔2 −1 = (−𝑦1, −𝑦2, −𝑦3, −𝑦4, −𝑦5 + 𝑦1𝑦3 + 𝑦2𝑦4) Now 𝑔1 −1. 𝑔2 −1 = (−𝑥1 − 𝑦1, −𝑥2 − 𝑦2, −𝑥3 − 𝑦3, −𝑥4 − 𝑦4, −𝑥5 − 𝑦5 + (−𝑥1)(−𝑦3) + (−𝑥2)(−𝑦4)) = (−(𝑥1 + 𝑦1), −(𝑥2 + 𝑦2), −(𝑥3 + 𝑦3), −(𝑥4 + 𝑦4), −(𝑥5 + 𝑦5) + 𝑥1𝑦3 + 𝑥2𝑦4) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 6s (2024) 595 https://internationalpubls.com And 𝑔2 −1. 𝑔1 −1 = (−𝑦1 − 𝑥1, −𝑦2 − 𝑥2, −𝑦3 − 𝑥3, −𝑦4 − 𝑥4, −𝑦5 − 𝑥5 + (−𝑦1)(−𝑥3) + (−𝑦2(−𝑥4)) = (−(𝑦1 + 𝑥1), −(𝑦2 + 𝑥2), −(𝑦3 + 𝑥3), −(𝑦4 + 𝑥4), −(𝑦5 + 𝑥5) + 𝑦1𝑥3 + 𝑦2𝑥4) If 𝑔1 −1. 𝑔2 −1 = 𝑔2 −1. 𝑔1 −1 then we have 𝑥1𝑦3 + 𝑥2𝑦4 = 𝑦1𝑥3 + 𝑦2𝑥4. Hence, The above result provides that “If 𝑔1and 𝑔2 are commute each other then their inverse also commute each other”. 2.6 N tractability: Given 𝑥, 𝑦 ∈ ℋ, find 𝑔 such that 𝑔−1𝑥 𝑔 = 𝑦. Let 𝑥 = (𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑥5)𝑦 = (𝑦1, 𝑦2, 𝑦3, 𝑦4, 𝑦5) and 𝑔 = (𝑔1, 𝑔2, 𝑔3, 𝑔4, 𝑔5) 𝑔−1𝑥 𝑔 = (−𝑔1, −𝑔2, −𝑔3, −𝑔4, −𝑔5 + 𝑔1𝑔3 + 𝑔2𝑔4)(𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑥5)(𝑔1, 𝑔2, 𝑔3, 𝑔4, 𝑔5) = (−𝑔1 + 𝑥1 + 𝑔1, −𝑔2 + 𝑥2 + 𝑔2, −𝑔3 + 𝑥3 + 𝑔3, −𝑔4 + 𝑥4 + 𝑔4, −𝑔5 + 𝑥5 + 𝑔5 + 𝑔1𝑔3 + 𝑔2𝑔4 − 𝑔1𝑥3 − 𝑥4𝑔4 + (−𝑔1 + 𝑥1)𝑔3 + (−𝑔2 + 𝑥2)𝑔4) 𝑚𝑜𝑑 𝑝 = (𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑥5 − 𝑔1𝑥3 − 𝑔2𝑥4 + 𝑥1𝑔3 + 𝑥2𝑔4) ----(2) Now, 𝑦5 = 𝑥5 − 𝑔1𝑥3 − 𝑔2𝑥4 + 𝑥1𝑔3 + 𝑥2𝑔4 𝑦5 − 𝑥5 = − 𝑔1𝑥3 − 𝑔2𝑥4 + 𝑥1𝑔3 + 𝑥2𝑔4 Suppose that 𝑦5 − 𝑥5 = 0 then 𝑔1𝑥3 + 𝑔2𝑥4 = 𝑥1𝑔3 + 𝑥2𝑔4 To obtain the values of 𝑔1, 𝑔2, 𝑔3, 𝑔4 by using the above single equation is very hard problem in DHG. 3. Five party key agreement protocol using CSP: 3.1 Protocol: Five parties 𝐴1, 𝐴2, 𝐴3, 𝐴4, 𝐴5 agree on a finite non-abelian group, discrete heisenberg group ℋ = 𝑍𝑝 5 One of the five party publishes an element𝑥 ∈ ℋ. 𝐴1 chooses an element 𝑔1 ∈ ℋ, such that 𝑥𝑔1 ≠ 𝑔1𝑥 𝐴2 chooses an element 𝑔2 ∈ ℋ, such that 𝑥𝑔2 ≠ 𝑔2𝑥 𝐴3 chooses an element 𝑔3 ∈ ℋ, such that 𝑥𝑔3 ≠ 𝑔3𝑥 𝐴4 chooses an element 𝑔4 ∈ ℋ, such that 𝑥𝑔4 ≠ 𝑔4𝑥 𝐴5 chooses an element 𝑔5 ∈ ℋ, such that 𝑥𝑔5 ≠ 𝑔5𝑥 Also , 𝑔1, 𝑔2, 𝑔3, 𝑔4, 𝑔5 are commute each other. Round 1: 𝐴1 chooses an element 𝑔1 ∈ ℋ, computes 𝐾11 = 𝑔1 −1𝑥𝑔1 and sends it to 𝐴2 𝐴2 chooses an element 𝑔2 ∈ ℋ, computes 𝐾12 = 𝑔2 −1𝑥𝑔2 and sends it to 𝐴3 𝐴3 chooses an element 𝑔3 ∈ ℋ, computes 𝐾13 = 𝑔3 −1𝑥𝑔3 and sends it to 𝐴4 𝐴4 chooses an element 𝑔4 ∈ ℋ, computes 𝐾14 = 𝑔4 −1𝑥𝑔4 and sends it to 𝐴5 𝐴5 chooses an element 𝑔5 ∈ ℋ, computes 𝐾15 = 𝑔5 −1𝑥𝑔5 and sends it to 𝐴1 Round 2: 𝐴1 computes 𝐾21 = 𝑔1 −1𝐾15𝑔1 and sends it to 𝐴2 𝐴2 computes 𝐾22 = 𝑔2 −1𝐾11𝑔2 and sends it to 𝐴3 Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 6s (2024) 596 https://internationalpubls.com 𝐴3 computes 𝐾23 = 𝑔3 −1𝐾12𝑔3 and sends it to 𝐴4 𝐴4 computes 𝐾24 = 𝑔4 −1𝐾13𝑔4 and sends it to 𝐴5 𝐴5 computes 𝐾25 = 𝑔5 −1𝐾14𝑔5 and sends it to 𝐴1 Round 3: 𝐴1 computes 𝐾31 = 𝑔1 −1𝐾25𝑔1 and sends it to 𝐴2 𝐴2 computes 𝐾32 = 𝑔2 −1𝐾11𝑔2 and sends it to 𝐴3 𝐴3 computes 𝐾33 = 𝑔3 −1𝐾22𝑔3 and sends it to 𝐴4 𝐴4 computes 𝐾34 = 𝑔4 −1𝐾23𝑔4 and sends it to 𝐴5 𝐴5 computes 𝐾35 = 𝑔5 −1𝐾24𝑔5 and sends it to 𝐴1 .Round 4: 𝐴1 computes 𝐾41 = 𝑔1 −1𝐾35𝑔1 and sends it to 𝐴2 𝐴2 computes 𝐾42 = 𝑔2 −1𝐾31𝑔2 and sends it to 𝐴3 𝐴3 computes 𝐾43 = 𝑔3 −1𝐾32𝑔3 and sends it to 𝐴4 𝐴4 computes 𝐾44 = 𝑔4 −1𝐾33𝑔4 and sends it to 𝐴5 𝐴5 computes 𝐾45 = 𝑔5 −1𝐾34𝑔5 and sends it to 𝐴1 Computation of KEY: On knowing 𝑔1 , 𝐴1 computes 𝐾𝐴1 = 𝑔1 −1𝐾45 𝑔1. On knowing 𝑔2 , 𝐴2 computes 𝐾𝐴2 = 𝑔2 −1𝐾41 𝑔2. On knowing 𝑔3 , 𝐴3 computes 𝐾𝐴3 = 𝑔3 −1𝐾42 𝑔3 On knowing 𝑔4 , 𝐴4 computes 𝐾𝐴4 = 𝑔4 −1𝐾43 𝑔4. On knowing 𝑔5 , 𝐴5 computes 𝐾𝐴5 = 𝑔5 −1𝐾44 𝑔3 𝐾𝐴1 = 𝐾𝐴2 = 𝐾𝐴3 = 𝐾𝐴4 = 𝐾𝐴5 is their common shared key. 3.2 Algorithm : SAGE Code for computing binary operation x*y in equation (1) is given below Enter the values : def multiply(x,y): x1,x2,x3,x4,x5 = x y1,y2,y3,y4,y5 = y sum=(mod((x1+y1),7),mod((x2+y2),7),mod((x3+y3),7),mod((x4+y4),7), mod((x5+y5+(x1*y3)+(x2*y4)),7)) return print(sum) multiply((2,2,5,1,3),(1,2,4,4,3)) multiply((1,2,4,4,3),(2,2,5,1,3)). The program is given as follows SAGE code for computing Kij values as given below Enter the values : def kvalue(x,g): x1, x2, x3, x4, x5 = x g1, g2, g3, g4, g5 = g kval=(mod(x1,7), mod(x2,7), mod(x3,7), mod(x4,7), mod((x5-(g1*x3)- (g2*x4)+(x1*g3)+(x2*g4)),7)) return print(kval) The program is given as follows Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 6s (2024) 597 https://internationalpubls.com 3.3 Flow chart : 4. Numerical Example: Consider the five parties 𝐴1, 𝐴2, 𝐴3, 𝐴4, 𝐴5agree on a finite non-abelian group, Discrete Heisenberg group ℋ = 𝑍7 5 and one of the public key is 𝑥 = (2, 2, 5, 1, 3) ∈ ℋ = 𝑍7 5 𝐴1 chooses an element 𝑔1 = (1, 2, 4, 4, 3) ∈ 𝑍7 5, such that 𝑥𝑔1 ≠ 𝑔1𝑥 𝐴2 chooses an element 𝑔2 = (3, 1, 4, 6, 2) ∈ 𝑍7 5, such that 𝑥𝑔2 ≠ 𝑔2𝑥 𝐴3 chooses an element 𝑔3 = (3, 4, 6, 4, 6) ∈ 𝑍7 5, such that 𝑥𝑔3 ≠ 𝑔3𝑥 𝐴4 chooses an element 𝑔4 = (1, 0, 5, 3, 1) ∈ 𝑍7 5, such that 𝑥𝑔4 ≠ 𝑔4𝑥 𝐴5 chooses an element 𝑔5 = (6, 2, 1, 5, 1) ∈ 𝑍7 5, such that 𝑥𝑔5 ≠ 𝑔5𝑥 Also 𝑔1, 𝑔2, 𝑔3, 𝑔4, 𝑔5 are commute each other. Round 1: 𝐴1 chooses an element 𝑔1 = (1, 2, 4, 4, 3) ∈ 𝑍7 5 and computes 𝐾11 = 𝑔1 −1𝑥𝑔1 = (6, 5, 3, 3, 2)(2, 2, 5, 1, 3)(1, 2, 4, 4, 3) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 6s (2024) 598 https://internationalpubls.com = (2, 2, 5, 1, 5)and sends it to 𝐴2 𝐴2 chooses an element 𝑔2 = (3, 1, 4, 6, 2) ∈ 𝑍7 5computes 𝐾12 = 𝑔2 −1𝑥𝑔2 = (4, 6, 3, 1, 2)(2, 2, 5, 1, 3)(3, 1, 4, 6, 2) =(2, 2, 5, 1, 0)and sends it to 𝐴3 𝐴3 chooses an element 𝑔3 = (3, 4, 6, 4, 6) ∈ 𝑍7 5computes 𝐾13 = 𝑔3 −1𝑥𝑔3 = (4, 3, 1, 3, 0)(2, 2, 5, 1, 3)(3, 4, 6, 4, 6) = (2, 2, 5, 1, 4)and sends it to 𝐴4 𝐴4 chooses an element 𝑔4 = (1, 0, 5, 3, 1) ∈ 𝑍7 5computes 𝐾14 = 𝑔4 −1𝑥𝑔4 = (6, , 0, 2, 4, 4)(2, 2, 5, 1, 3)(1, 0, 5, 3, 1) = (2, 2, 5, 1, 0)and sends it to 𝐴5 𝐴5 chooses an element 𝑔5 = (6, 2, 1, 5, 1) ∈ 𝑍7 5computes 𝐾15 = 𝑔5 −1𝑥𝑔5 = (1, 5, 6, 2, 1)(2, 2, 5, 1, 3)(6, 2, 1, 5, 1) = (2, 2, 5, 1, 4)and sends it to 𝐴1 Round 2: 𝐴1 computes 𝐾21 = 𝑔1 −1𝐾15𝑔1 = (6, 5, 3, 3, 2)(2, 2, 5, 1, 4)(1, 2, 4, 4, 3) = (2, 2, 5, 1, 6) and sends it to 𝐴2 𝐴2 computes 𝐾22 = 𝑔2 −1𝐾11𝑔2 = (4, 6, 3, 1, 2)(2, 2, 5, 1, 5)(3, 1, 4, 6, 2) = (2, 2, 5, 1, 2) and sends it to 𝐴3 𝐴3 computes 𝐾23 = 𝑔3 −1𝐾12𝑔3 = (4, 3, 1, 3, 0)(2, 2, 5, 1, 0)(3, 4, 6, 4, 6) = (2, 2, 5, 1, 1) and sends it to 𝐴4 𝐴4 computes 𝐾24 = 𝑔4 −1𝐾13𝑔4 = (6, , 0, 2, 4, 4)(2, 2, 5, 1, 4)(1, 0, 5, 3, 1) = (2, 2, 5, 1, 1) and sends it to 𝐴5 𝐴5 computes 𝐾25 = 𝑔5 −1𝐾14𝑔5 = (1, 5, 6, 2, 1)(2, 2, 5, 1, 0)(6, 2, 1, 5, 1) = (2, 2, 5, 1, 1) and sends it to 𝐴1 Round 3: 𝐴1 computes 𝐾31 = 𝑔1 −1𝐾25𝑔1 = (6, 5, 3, 3, 2)(2, 2, 5, 1, 1)(1, 2, 4, 4, 3) = (2, 2, 5, 1, 3) and sends it to 𝐴2 𝐴2 computes 𝐾32 = 𝑔2 −1𝐾11𝑔2 = (4, 6, 3, 1, 2)(2, 2, 5, 1, 6)(3, 1, 4, 6, 2) = (2, 2, 5, 1, 3) and sends it to 𝐴3 𝐴3 computes 𝐾33 = 𝑔3 −1𝐾22𝑔3 = (4, 3, 1, 3, 0)(2, 2, 5, 1, 2)(3, 4, 6, 4, 6) = (2, 2, 5, 1, 3) and sends it to 𝐴4 𝐴4 computes 𝐾34 = 𝑔4 −1𝐾23𝑔4 = (6, , 0, 2, 4, 4)(2, 2, 5, 1, 1)(1, 0, 5, 3, 1) = (2, 2, 5, 1, 5) and sends it to 𝐴5 𝐴5 computes 𝐾35 = 𝑔5 −1𝐾24𝑔5 = (1, 5, 6, 2, 1)(2, 2, 5, 1, 1)(6, 2, 1, 5, 1) = (2, 2, 5, 1, 2) and sends it to 𝐴1 Round 4: 𝐴1 computes 𝐾41 = 𝑔1 −1𝐾35𝑔1 = (6, 5, 3, 3, 2)(2, 2, 5, 1, 2)(1, 2, 4, 4, 3) = (2, 2, 5, 1, 4) and sends it to 𝐴2 𝐴2 computes 𝐾42 = 𝑔2 −1𝐾31𝑔2 = (4, 6, 3, 1, 2)(2, 2, 5, 1, 3)(3, 1, 4, 6, 2) = (2, 2, 5, 1, 0) and sends it to 𝐴3 𝐴3 computes 𝐾43 = 𝑔3 −1𝐾32𝑔3 = (4, 3, 1, 3, 0)(2, 2, 5, 1, 3)(3, 4, 6, 4, 6) = (2, 2, 5, 1, 4) and sends it to 𝐴4 𝐴4 computes 𝐾44 = 𝑔4 −1𝐾33𝑔4 = (6, , 0, 2, 4, 4)(2, 2, 5, 1, 3)(1, 0, 5, 3, 1) = (2, 2, 5, 1, 0) and sends it to 𝐴5 𝐴5 computes 𝐾45 = 𝑔5 −1𝐾34𝑔5 = (1, 5, 6, 2, 1)(2, 2, 5, 1,5)(6, 2, 1, 5, 1) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 6s (2024) 599 https://internationalpubls.com = (2, 2, 5, 1, 6) and sends it to 𝐴1 Computation of KEY: On knowing 𝑔1 , 𝐴1 computes 𝐾𝐴1 = 𝑔1 −1𝐾45 𝑔1 = (6, 5, 3, 3, 2)(2, 2, 5, 1, 6)(1, 2, 4, 4, 3) = (2, 2, 5, 1, 1). On knowing 𝑔2 , 𝐴2 computes 𝐾𝐴2 = 𝑔2 −1𝐾41 𝑔2 = (4, 6, 3, 1, 2)(2, 2, 5, 1, 4)(3, 1, 4, 6, 2) = (2, 2, 5, 1, 1). On knowing 𝑔3 , 𝐴3 computes 𝐾𝐴3 = 𝑔3 −1𝐾42 𝑔3 = (4, 3, 1, 3, 0)(2, 2, 5, 1, 0)(3, 4, 6, 4, 6) = (2, 2, 5, 1, 1) On knowing 𝑔4 , 𝐴4 computes 𝐾𝐴4 = 𝑔4 −1𝐾43 𝑔4 = (6, , 0, 2, 4, 4)(2, 2, 5, 1, 4)(1, 0, 5, 3, 1) = (2, 2, 5, 1, 1) On knowing 𝑔5 , 𝐴5 computes 𝐾𝐴5 = 𝑔5 −1𝐾44 𝑔3 = (1, 5, 6, 2, 1)(2, 2, 5, 1,0)(6, 2, 1, 5, 1) = (2, 2, 5, 1, 1) 𝐾𝐴1 = 𝐾𝐴2 = 𝐾𝐴3 = 𝐾𝐴4 = 𝐾𝐴5 = (2, 2, 5, 1, 1) is their common shared key. This key has been utilized further for encryption or decryption. Public Key gi K ij of Round 1 K ij of Round 2 K ij of Round 3 K ij of Round 4 Computation key 𝑥 = (2, 2, 5, 1, 3) (1, 2, 4, 4, 3) (2, 2, 5, 1, 5) (2, 2, 5, 1, 6) (2, 2, 5, 1, 3) (2, 2, 5, 1, 4) (2, 2, 5, 1, 1) (3, 1, 4, 6, 2) (2, 2, 5, 1, 0) (2, 2, 5, 1, 2) (2, 2, 5, 1, 3) (2, 2, 5, 1, 0) (2, 2, 5, 1, 1) (3, 4, 6, 4, 6) (2, 2, 5, 1, 4) (2, 2, 5, 1, 1) (2, 2, 5, 1, 3) (2, 2, 5, 1, 4) (2, 2, 5, 1, 1) (1, 0, 5, 3, 1) (2, 2, 5, 1, 0) (2, 2, 5, 1, 1) (2, 2, 5, 1, 5) (2, 2, 5, 1, 0) (2, 2, 5, 1, 1) (6, 2, 1, 5, 1) (2, 2, 5, 1, 4) (2, 2, 5, 1, 1) (2, 2, 5, 1, 2) (2, 2, 5, 1, 6) (2, 2, 5, 1, 1) 5. Perfect forward secrecy in the proposed KAP: The long term key in this protocol is the common key arrived by the communicating entities. The common key is arrived in k-1 rounds for k entities. At each round the entities use their private keys to find the public key. The public key thus obtained is communicated to the next entity. The same procedure is taken over by all the entities. At each round the entities use CSP to arrive their public key. The Intractability of CSP is discussed and it is proven to be secure against the attacks. Thus even if the common key is compromised, the session keys at each round cannot be compromised. The proposed KAP has the perfect forward secrecy. In the proposed KAP, the common key is arrived at the kth round by the k-Communicating parties. At each round the communicating parties use their private keys to compute a public key for that round and sends to the other party. If suppose a passive adversary managed to get the public key of one entity, in a particular round, he cannot take part in the communications between the entities. As at each round the communicating entities use their private keys to take part in the process, the adversary is not a le to join the process. The public key is activated by means of CSP in DHG, which is hard to attack. Thus even if the passive adversary managed to get the public key of any particular entity at any particular round he cannot take part in the future process. Thus this KAP is secure against the known key attack. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 6s (2024) 600 https://internationalpubls.com Unknown key share: An unknown key-share attack on an authenticated key agreement protocol is an Attack where by an entity A ends up believing, it shares a key with another entity B and although this is in fact the case that B mistakenly believes the key is instead shared with an entity C ≠ A. The communicating entities may share the key with same other entity with the belief that he/she is sharing with intended parties. But the adversary has to take part in all rounds to act as the intended party. If it is possible then the KAP is vulnerable unknown key share attack. This may have been overcome by proper authentication. No Key Control: Neither entity should be able to force the session key to a value of his choice. Clearly key Control is not possible in the present KAP. As the common key is arrived with the participation of all the communicating parties at each round. Thus this KAP has no key control to any particular entity. 6. Recent Works: The finite dimensional left translation invariant linear spaces of continuous complex valued functions over the Heisenberg group is discussed by Laszlo Szekelyhidi [12] to improve the frame of reference of group. The problem of finding a secure class of nonabelian groups for use as platforms is open and a subject of active research. A polynomial time solution of the CSP in an important class of non- abelian groups, the extraspecial p-groups and the reduction of the CSP in certain types of central products will be discussed by Simran Tinani [13]. 7. CONCLUSION: In this paper a new multi party key agreement protocol is proposed. The conjugacy search problem taken up the role of one way function . This is implemented in a five dimentional discrete Heinsenberg Group which is a non-abelian platform. The computation of k values and choosing g which is not commutative with x to be done by the SageMath algorithm. The interactability of conjugacy search problems also demonstrated. In this multiparty key agreement protocol entities arrive at a common key in k-1 rounds. Also this protocol satisfies the necessary security attributes for KAP. A numerical example for five party KAP is illustrated in this paper. Other search problem such as decomposition search problem, Triple decomposition search problem and twisted conjugacy search problem etc, may also be used to describe a KAP. Any suitable platform group may also be applied to arrive at a common key. The DHG also considered as extraspecial p-group or complex group be the extension of this work and it will give more significant results. 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