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# sep15 - are ﬁnished so we may assume that at least one of...

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Math 3124 Thursday, September 15 September 15 Ungraded Homework Problem 14.6 page 79 Find the order of (1 2)(3 4 5) in S 5 . Since (1 2)(3 4 5) is already written as a product of disjoint cycles, this is [2,3] = 6 ([2,3] means least the common multiple of 2 and 3). Problem 14.9 page 79 Let α denote the clockwise rotation of the plane through 90 about a fixed point p . What is the order of α ? The order of α is o ( α ) . The answer is 4. Problem 14.10 page 79 (a) Repeat the previous problem, 14.9, with 40 in place of 90 . (b) What is the order of α if α denotes rotation through ( 360 / k ) ( k N ). (a) The answer is 9. (b) We want to find o ( α ) , that is the smallest positive integer n such that 360 n / k is an integer multiple of 360. This is 360 / ( k , 360 ) (where ( k , 360 ) denotes the greatest common divisor). Problem 14.23 page 80 Prove that a nonidentity element has order 2 iff it is its own inverse. x has order 2 iff x = e and x 2 = e iff x = x - 1 and x = e . Problem 14.30 page 80 Prove that if G is a group and a G , then o ( a - 1 ) = o ( a ) . Let m and n be the orders of a - 1 and a respectively; thus m , n N . If m = n = , then we
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Unformatted text preview: are ﬁnished, so we may assume that at least one of m , n is ﬁnite; without loss of generality assume that m is ﬁnite. Since ( a-1 ) m = e , we see that a-m = e , hence ( a m )-1 = e and we deduce that a m = e . This shows that n ≤ m , in particular n ∈ N . Similarly m ≤ n and we deduce that m = n , as required. Problem 14.37 page 81 Determine the largest order of an element of S n for 1 ≤ n ≤ 10. We use the fact from problem 14.36 that the order of a permutation is the least common multiple of the lengths of the cycles in its cycle decomposition (e.g. least common multiple of 4 and 6 is 12). Thus the maximum orders are given by the table below. 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 12 15 20 30 For S 6 , use ( 1 2 3 )( 4 5 ) ; for S 7 , use ( 1 2 3 4 )( 5 6 7 ) ; for S 8 , use ( 1 2 3 4 5 )( 6 7 8 ) ; for S 9 , use ( 1 2 3 4 5 )( 6 7 8 9 ) ; for S 10 , use ( 1 2 )( 3 4 5 )( 6 7 8 9 10 ) ....
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