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Unformatted text preview: Math 3124 Thursday, September 15 Third Homework Solutions 1. First put the position so that there is the space is in the bottom right hand corner; there will be several ways of doing this and it does not matter which you choose. We will choose 12 14 5 7 10 3 9 11 8 2 1 13 6 4 15 . We now obtain the permutation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 12 14 5 7 10 3 9 11 8 2 1 13 6 4 15 We must decide whether this is a product of an even number or an odd number of transpositions (2cycles). Writing the permutation as a product of disjoint cycles, we obtain ( 1 12 13 6 3 5 10 2 14 4 7 9 8 11 ) This is a 14cycle, hence a product of 13 transpositions (cf. problem 6.9 on page 40) which is odd. Therefore the position cannot be reached. 2. Problem 6.4 on page 40 Write each of the given permutations as a product of dis joint cycles. Also determine whether the permutation in S 6 is in A 6 . (a) 1 2 3 4 5 6 1 6 4 5 3 2 This sends 1 → 1, 2 → 6 → 2, 3 → 4 → 5 → 3, so when written as a product of disjoint cycles is (2 6)(3 4 5). A 2cycle is a product of one transposition and a 3cycle is a product of two transpositions. Therefore theone transposition and a 3cycle is a product of two transpositions....
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This note was uploaded on 01/02/2012 for the course MATH 3124 taught by Professor Parry during the Fall '08 term at Virginia Tech.
 Fall '08
 PARRY
 Math, Algebra

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