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 ( a1 ) m = e , we see that am = 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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 Fall '08
 PARRY
 Math, Algebra, Greatest common divisor, Divisor, Least common multiple, Zürich Hauptbahnhof

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