“…In an identical way, we find all remaining cycles of the permutations 𝑥 and 𝑦 and given as: 𝑥 = (𝜉 1 , 𝜉 126 )(𝜉 2 , 𝜉 125 )(𝜉 3 , 𝜉 124 )(𝜉 4 , 𝜉 123 )(𝜉 5 , 𝜉 122 )(𝜉 6 , 𝜉 121 )(𝜉 7 , 𝜉 120 )(𝜉 8 , 𝜉 119 )(𝜉 9 , 𝜉 118 ) (𝜉 , 𝜉 117 )(𝜉 11 , 𝜉 116 )(𝜉 12 , 𝜉 115 )(𝜉 13 , 𝜉 114 )(𝜉 14 , 𝜉 113 )(𝜉 15 , 𝜉 112 )(𝜉 16 , 𝜉 111 )(𝜉 17 , 𝜉 110 )(𝜉 18 , 𝜉 109 ) (𝜉 , 𝜉 108 )(𝜉 20 , 𝜉 107 )(𝜉 21 , 𝜉 106 )(𝜉 22 , 𝜉 105 )(𝜉 23 , 𝜉 104 )(𝜉 24 , 𝜉 103 )(𝜉 25 , 𝜉 102 )(𝜉 26 , 𝜉 101 )(𝜉 27 , 𝜉 100 ) (𝜉 28 , 𝜉 99 )(𝜉 29 , 𝜉 98 )(𝜉 30 , 𝜉 97 )(𝜉 31 , 𝜉 96 )(𝜉 32 , 𝜉 95 )(𝜉 33 , 𝜉 94 )( 𝜉 34 , 𝜉 93 )( 𝜉 35 𝜉 92 )(𝜉 36 , 𝜉 91 )(𝜉 37 , 𝜉 90 ) (𝜉 38 , 𝜉 89 )(𝜉 39 , 𝜉 88 )(𝜉 40 , 𝜉 87 )(𝜉 41 , 𝜉 86 )(𝜉 42 , 𝜉 85 )(𝜉 43 , 𝜉 84 )(𝜉 44 64 ). 𝑦 = (0 , ∞, 1)(𝜉 1 , 𝜉 96 , 𝜉 30 ) (𝜉 2 , 𝜉 65 , 𝜉 60 )(𝜉 3 , 𝜉 120 , 𝜉 4 )(𝜉 5 , 𝜉 45 , 𝜉 77 )(𝜉 6 , 𝜉 113 , 𝜉 8 )(𝜉 7 , 𝜉 124 , 𝜉 123 ) (𝜉 , 𝜉 83 , 𝜉 35 )(𝜉 10 , 𝜉 90 , 𝜉 27 )(𝜉 11 , 𝜉 93 , 𝜉 <...…”