homework_1_solutions - MA 553 HOMEWORK ASSIGNMENT #1...

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Unformatted text preview: MA 553 HOMEWORK ASSIGNMENT #1 SOLUTIONS Problem 1. Let G be a set satisfying the following four axioms: A1: There exists a binary operation : G G G i.e., a b G for all a, b G . A2: Associativity holds i.e., ( a b ) c = a ( b c ) for all a, b, c G . A3: A left identity exists i.e., there exists at least one element e G such that e a = a for all a G . A4: There exist left inverses i.e., for each a G , there is at least one element b G such that b a = e . Show that ( G, ) is a group. Solution: We must show that ( G, ) satisfies four axioms: G1: a b G for all a, b G . G2: ( a b ) c = a ( b c ) for all a, b, c G . G3: There exists an element e G such that a e = e a = a for all a G . G4: For each a G , there exists an element b G such that a b = b a = e . It suffices then to show that axioms (G3) and (G4) hold. Fix an element a G and a left inverse b corresponding to a left identity e . First we show that b is a two-sided inverse for a . Denote c = a b ; we show that c = e . We have c c = ( a b ) ( a b ) = a (( b a ) b ) = a ( e b ) = a b = c. Denoting d as a left inverse for c also corresponding to e , we have c = e c = ( d c ) c = d ( c c ) = d c = e. Hence a b = b a = e as desired. Now we show that e is a two-sided identity. We have a e = a ( b a ) = ( a b ) a = e a = a. Problem 2. Let n be a positive integer. A permutation S n has order m when m = 1 yet k 6 = 1 for 1 k < m . a. Show that a cycle = ( a 1 a 2 a m ) of length m has order m . b. Prove that the order of a permutation in S n equals the least common multiple of the lengths of the cycles in its cycle decomposition. c. Let p be a prime. Show that a permutation has order p if and only if its cycle decomposition is a product of commuting p-cycles. Solution: (a.) Consider the action of = ( a 1 a 2 a m ) on the set = { a 1 , a 2 , ..., a m } . Note that acts trivially on { 1 , 2 , ..., n } - . It is easy to see by induction that k : a i 7 a k + i for each positive integer k . Since a i 6 = a j the permutation k acts nontrivially on for 1 k < m . On the other hand, m acts trivially on i.e., m : a i 7 a i , so that m = 1 . Hence has order m ....
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homework_1_solutions - MA 553 HOMEWORK ASSIGNMENT #1...

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