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week03tutsols - THE UNIVERSITY OF SYDNEY Math2968...

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THE UNIVERSITY OF SYDNEY Math2968 Algebra (Advanced) Semester 2 Tutorial Solutions Week 3 2008 1. Write down the inverses of the following permutations as products of disjoint cycles: α = (1 2 3 4 5 6) , β = (1 2)(3 4)(5 6 7 8) , γ = (1 3 5)(2 4 6) Express each of α , β , γ as a product of transpositions. Solution α - 1 = (6 5 4 3 2 1) , β - 1 = (1 2)(3 4)(8 7 6 5) , γ - 1 = (5 3 1)(6 4 2) α = (1 2)(1 3)(1 4)(1 5)(1 6) , β = (1 2)(3 4)(5 6)(5 7)(5 8) , γ = (1 3)(1 5)(2 4)(2 6) 2. Let α and β be permutations of a finite set. Verify that if ( x 1 x 2 . . . x k ) is a cycle in the decomposition of α then ( x 1 β x 2 β . . . x k β ) is a cycle in the decomposition of β - 1 αβ . Solution Observe that, if ( x 1 x 2 . . . x k ) is a cycle in the decomposition of α then, for each i , ( x i β )( β - 1 αβ ) = ( x i α ) β = x i +1 β (where we put k + 1 1), which shows ( x 1 β x 2 β . . . x k β ) is a cycle in the decomposition of β - 1 αβ . 3. Use the result of the previous exercise to write down β - 1 αβ immediately when (a) α = (1 2 3 4 5 6) , β = (1 6)(2 5)(3 4) . (b) α = (1 6)(2 5)(3 4) , β = (1 2 3 4 5 6) . (c) α = (1 9)(4 2 5 6)(8 3 7) , β = (2 3 8 6 5 4)(9 1) . (d) α = (2 3 8 6 5 4)(9 1) , β = (1 9)(4 2 5 6)(8 3 7) . Solution (a) β - 1 αβ = (6 5 4 3 2 1) (b) β - 1 αβ = (2 1)(3 6)(4 5) (c) β - 1 αβ = (1 9)(2 3 4 5)(6 8 7) (d) β - 1 αβ = (5 7 3 4 6 2)(1 9) 4. Which of the following permutations are even? odd? α = (1 2)(3 4) , β = (1 2 3 4 5) , γ = (1 2 3 4 5 6) δ = (1 2 3 4 5)(1 3 5)(2 5 6 7) , ε = (1 2)(1 3)(1 4 5)(4 2) Solution Here α and β are even, whilst γ , δ and ε are odd. 5. Let α be a permutation of { 1 , 2 , . . . , n } . Explain why α - 1 = α k for some positive integer k . Explain briefly why the set of all odd permutations of { 1 , 2 , . . . , n } does not form a permutation group. Solution If α = 1 then α - 1 = α 1 . Suppose α negationslash = 1. Then α = α 1 α 2 . . . α where α 1 , . . . , α are disjoint cycles of length n 1 , . . . , n respectively, and n 1 > 1. Then α n 1 ...n = 1 , giving α - 1 = α k where k = ( n 1 . . . n
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