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Solutions_MT2s11

# Solutions_MT2s11 - MAT 312/AMS 351 Spring 2011 Solutions to...

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MAT 312/AMS 351, Spring 2011 Solutions to the Midterm II Maximal grade: 100 points 1. (15 points) How many permutations in S 4 have order 2? Answer: 9. Solution: A permutation has order 2, if all non-intersecting cycles in it have length 2. In S 4 , such a permutation can be a transposition or a product of two non-intersecting transpositions. Therefore the list of permutations of order 2 looks as (1 2) , (1 3) , (1 4) , (2 3) , (2 4) , (3 4) , (1 2)(3 4) , (1 3)(2 4) , (1 4)(2 3) . Grading: 5 points for the correct description of possible cycle types, 5 points for the correct list of 6 transpositions, 5 points for the correct list of products of transpositions 2. (15 points) Suppose that a permutation f has odd order. Prove that f is even. Solution: a) We know that f 2 n +1 = e , so 1 = sgn ( e ) = sgn ( f 2 n +1 ) = ( sgn ( f )) 2 n +1 . If sgn ( f ) = - 1, then ( sgn ( f )) 2 n +1 = - 1. Contradiction, there- fore sgn ( f ) = 1. b) The order of f is a least common multiple of the lengths of the non- intersecting cycles in f . Since it is odd, all these lengths should be odd.

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