This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 twosided 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 twosided 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 pcycles. 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 ....
View Full
Document
 Winter '11
 PhanThuongCang

Click to edit the document details