homework_1_solutions

# 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 σ ∈ S n has order m when σ m = 1 yet σ k 6 = 1 for 1 ≤ k < m . a. Show that a cycle σ = ( 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 σ = ( a 1 a 2 ··· a m ) on the set Σ = { a 1 , a 2 , ..., a m } . Note that σ acts trivially on { 1 , 2 , ..., n } - Σ. It is easy to see by induction that σ k : a i 7→ a k + i for each positive integer k . Since a i 6 = a j the permutation σ k acts nontrivially on Σ for 1 ≤ k < m . On the other hand, σ m acts trivially on Σ i.e., σ m : a i 7→ a i , so that σ m = 1 . Hence σ has order m ....
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homework_1_solutions - MA 553 HOMEWORK ASSIGNMENT#1...

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