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Unformatted text preview: MODEL ANSWERS TO THE EIGHTH HOMEWORK 1. For Chapter 3, Section 3: 1. (a) 1 2 3 4 5 6 7 8 9 2 4 5 1 3 7 8 9 6 = (1 , 2 , 4)(3 , 5)(6 , 7 , 8 , 9) , a product of 2 + 1 + 3 = 6 transpositions. Even. (b) 5 + 2 = 7, odd. (c) 5 + 5 = 10, even. (d) 1 + 2 + 1 + 2 + 2 = 8, even. 1. For Chapter 3, Section 3: 2. We have already seen that a kcycle is a product of k 1 transpositions. Thus the result is clear. 1. For Chapter 3, Section 3: 3. Clear, since they have the same cycle type. 1. For Chapter 3, Section 3: 5. First of the bottom row is missing 4 and 5. Let us try 1 2 3 4 5 6 7 8 9 3 1 2 4 5 7 8 9 6 = (1 , 3 , 2)(6 , 7 , 8 , 9) . This is a product of 5 = 2 + 3 tranpositions. Thus this permutation is odd. Now consider 1 2 3 4 5 6 7 8 9 3 1 2 5 4 7 8 9 6 . This differs from the first permutation by a transposition. Hence it is even, as the other is odd. 1. For Chapter 3, Section 3: 6. Since every element of A n is a product of an even number of transpositions, pairing off the transpositions ar...
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This note was uploaded on 07/15/2008 for the course MATH 111 taught by Professor Long during the Spring '08 term at UCSB.
 Spring '08
 LONG
 Math, Algebra

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