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Unformatted text preview: Solutions For Homework 3 Ch. 2 2. Show that the set { 5 , 15 , 25 , 35 } is a group under multiplication modulo 40. What is the identity element of this group? Can you see any relationship between this group and U (8)? Answer. We need to follow the definition of a group given on page 43: Notation. Let us denote the operation given in the question, multiplication modulo 40, with · and the usual multiplication of integers a and b with ab or ( a )( b ). So, we have a · b = ab mod 40 Step 1. Check that the operation defined in the question is a binary operation: 5 15 25 35 5 25 35 5 15 15 35 25 15 5 25 5 15 25 35 35 15 5 35 25 Calculations above show that the operation defined in the question assigns to each ordered pair of elements of { 5 , 15 , 25 , 35 } an element in { 5 , 15 , 25 , 35 } . Therefore, it is a binary operation. Step 2. Check associativity of the given operation: Since we have a ( bc ) = a ( bc ) for every a , b , c in Z , we get a ( bc ) = ( ab ) c for every a , b , c in { 5 , 15 , 25 , 35 } . Clearly, a ( bc ) = ( ab ) c implies that a ( bc ) mod 40 = ( ab ) c mod 40 . So, we get a · ( b · c ) = a ( b · c ) mod 40 (def. of · ) = [( a mod 40)(( b · c ) mod 40)] mod 40 (pr. of mod) = [( a mod 40)(( bc mod 40) mod 40)] mod 40 (def. of · ) = [( a mod 40)( bc mod 40)] mod 40 (pr. of mod) = a ( bc ) mod 40 (pr. of mod) = ( ab ) c mod 40 (assoc. of Z) = [(( ab ) mod 40)( c mod 40)] mod 40 (pr. of mod) = [(( ab ) mod 40) mod 40)( c mod 40)] mod 40 (pr. of mod) = [(( a · b ) mod 40)( c mod 40)] mod 40 (def. of · ) = ( a · b ) c mod 40 (pr. of mod) = ( a · b ) · c (def. of · ) ....
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 Spring '08
 STAFF
 Multiplication

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