{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

week03tutsols

# week03tutsols - THE UNIVERSITY OF SYDNEY Math2968...

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}