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Unformatted text preview: 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: = (1 2 3 4 5 6) , = (1 2)(3 4)(5 6 7 8) , = (1 3 5)(2 4 6) Express each of , , as a product of transpositions. Solution  1 = (6 5 4 3 2 1) ,  1 = (1 2)(3 4)(8 7 6 5) ,  1 = (5 3 1)(6 4 2) = (1 2)(1 3)(1 4)(1 5)(1 6) , = (1 2)(3 4)(5 6)(5 7)(5 8) , = (1 3)(1 5)(2 4)(2 6) 2. Let and be permutations of a finite set. Verify that if ( x 1 x 2 ... x k ) is a cycle in the decomposition of then ( x 1 x 2 ... x k ) is a cycle in the decomposition of  1 . Solution Observe that, if ( x 1 x 2 ... x k ) is a cycle in the decomposition of then, for each i , ( x i )(  1 ) = ( x i ) = x i +1 (where we put k + 1 1), which shows ( x 1 x 2 ... x k ) is a cycle in the decomposition of  1 . 3. Use the result of the previous exercise to write down  1 immediately when (a) = (1 2 3 4 5 6) , = (1 6)(2 5)(3 4) . (b) = (1 6)(2 5)(3 4) , = (1 2 3 4 5 6) . (c) = (1 9)(4 2 5 6)(8 3 7) , = (2 3 8 6 5 4)(9 1) . (d) = (2 3 8 6 5 4)(9 1) , = (1 9)(4 2 5 6)(8 3 7) . Solution (a)  1 = (6 5 4 3 2 1) (b)  1 = (2 1)(3 6)(4 5) (c)  1 = (1 9)(2 3 4 5)(6 8 7) (d)  1 = (5 7 3 4 6 2)(1 9) 4. Which of the following permutations are even? odd? = (1 2)(3 4) , = (1 2 3 4 5) , = (1 2 3 4 5 6) = (1 2 3 4 5)(1 3 5)(2 5 6 7) , = (1 2)(1 3)(1 4 5)(4 2) Solution Here and are even, whilst , and are odd. 5. Let be a permutation of { 1 , 2 ,... ,n } . Explain why  1 = 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....
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This note was uploaded on 09/12/2009 for the course MATH 2968 taught by Professor Easdown during the One '09 term at University of Sydney.
 One '09
 easdown
 Algebra, Permutations

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