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jan26 - Math 4124 Wednesday January 26 January 26 Ungraded...

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Math 4124 Wednesday, January 26 January 26, Ungraded Homework Exercise 1.3.1 on page 32 Let σ be the permutation 1 3 2 4 3 5 4 2 5 1 and let τ be the permutation 1 5 2 3 3 2 4 4 5 1 . Find the cycle decomposition of each of the following permutations σ , τ , σ 2 , στ , τσ , and τ 2 σ . First we obtain the cycle decomposition for σ . We have σ 1 = 3, σ 3 = 5 and σ 5 = 1, which yields the cycle (1 3 5); and σ 2 = 4 and σ 4 = 2, which yields the cycle (2 4). Therefore the cycle decomposition of σ (i.e. σ written as a product of disjoint cycles) is (1 3 5)(2 4). Next we find the cycle decomposition of τ . We have τ 1 = 5 and τ 5 = 1, which yields the cycle (1 5), and τ 2 = 3 and τ 3 = 2, which yields the cycle (2 3), and τ 4 = 4, which yields the cycle (4). Normally one omits the 1-cycles. Therefore the cycle decomposition of τ is (1 5)(2 3). The calculation of the cycle decompositions of the other elements can be facilitated by using the cycle decompositions of σ and τ from above. Thus σ 2 = (1 5 3), στ = (2 5 3 4), τσ = (1 2 4 3), and (note τ 2 = e ) τ 2 σ = σ = (1 3 5)(2 4). Exercise 1.3.3 on page 33 Find the orders of the permutations computed in the previous exercise. If a permutation is a product of disjoint cycles of lengths 1 ,..., n , then its order is the lowest common multiple [ 1 ,..., n ] of these lengths. Thus σ has order [ 3 , 2 ] = 6, τ has order [ 2 , 2 ] , σ 2 has order 3, στ has order 4, τσ has order 4, τ 2 σ has order 6. Exercise 1.3.5 on page 33 Find the order of (1 12 8 10 4)(2 13)(5 11 7)(6 9). Since this element is written as a product of disjoint cycles, we may apply Exercise 1.3.15 on page 33. The answer is the lowest common multiple of 5,2,3,2 which is 30. Exercise 1.3.9 on page 33 Let σ = (1 2 3 4 5 6 7 8 9 10 11 12), τ = (1 2 3 4 5 6 7 8), ω = (1 2 3 4 5 6 7 8 9 10 11 12 13 14). For which positive integers i is σ i a 12-cycle, τ i an 8-cycle, ω i a 14-cycle? First consider σ , which we may view as an element of S 12 . Suppose ( i , 12 ) = 1 (that is, i is prime to 12). Then there exists j Z such that i j 1 mod 12, so i j = 1 + 12 r for some r Z . Then ( σ i ) j = σ i j = σ 1 + 12 r = σ ( σ 12 ) r = σ .
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