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See discussions, stats, and author profiles for this publication at: Abstract Algebra Solutions Research · August 2015 DOI: 10.13140/RG.2.1.2799.6004 CITATIONS 0 READS 13,965 1 author: Evan Oman Black River Systems Co. 4 PUBLICATIONS 0 CITATIONS SEE PROFILE All content following this page was uploaded by Evan Oman on 06 August 2015. The user has requested enhancement of the downloaded file.
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Evan Oman MATH 5371 Joe Gallian September 16, 2013 Homework 1 Gallian 2.6 : In each case, perform the indicated operation: a) In C * , (7 + 5 i ) ( - 3 + 2 i ) (7 + 5 i ) ( - 3 + 2 i ) = - 21 + 14 i - 15 i + 10 i 2 = - 21 - i + 10 ( - 1) = - ( i + 31) C * b) In GL (2 , Z 13 ), det 7 4 1 5 First recall the definition of the determinant of a 2 × 2 matrix: det a b c d = a · d - b · c So in our case we would have: det 7 4 1 5 = 7 · 5 - 4 · 1 = 31 13 5 Z 13 c) In GL (2 , R ), 6 3 8 2 - 1 Recall the definition of the inverse of a 2 × 2 matrix: A - 1 = a b c d - 1 = 1 det A d - b - c a So if we let B = 6 3 8 2 , we find that det[ B ] = 6 · 2 - 8 · 3 = 12 - 24 = - 12 R 1
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Therefore B - 1 = - 1 12 2 - 3 - 8 6 = - 1 6 1 4 2 3 - 1 2 GL (2 , R ) d) In GL (2 , Z 13 ), 6 3 8 2 - 1 Here we use the same formula but since the entries are elements of Z 13 , we must adjust the values according to this new modulus. So from part c) : det[ B ] = - 12 13 1. So our inverse becomes: B - 1 = 1 1 2 - 3 - 8 6 13 2 10 5 6 GL (2 , Z 13 ) Gallian 2.20 : For any integer n > 2, show that there are at least 2 elements in U ( n ) that satisfy x 2 = 1. Let n Z such that n > 2. Then we know that that U ( n ), the group of integers modulo n under multiplication, contains only the elements which are relatively prime to n . So we can write U ( n ) = { x mod n | gcd( n, x ) = 1 } So clearly 1 U ( n ) because the gcd( n, 1) = 1(which is true for any number) and because 1 = e under multiplication which must be an element of U ( n ) by U ( n ) being a group. Then since 1 is the identity it follows that (1) 2 = 1 giving one element satisfying x 2 = 1. The next element we will consider is n - 1. By Lemma 0.1 we know that n and ( n - 1) are relatively prime so n - 1 U ( n ). Then squaring ( n - 1) 2 yields: ( n - 1) 2 = n 2 - 2 n + 1 = n 2 - 2 n + 1 n 1 Therefore we have found two elements that satisfy x 2 = 1, guaranteeing at least 2 elements U ( n ). Lemma 0.1 Let n Z such that n > 2. Assume by contradiction that n, n - 1 are not relatively prime. 2
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Then a N \{ 1 } such that a | n and a | ( n - 1) which is to say that l, m N { 1 } such that m · a = n and l · a = ( n - 1). Rearranging we have ma = la + 1 a = 1 m - l . However since a, l, m N \{ 1 } , we arrive at a contradiction. Therefore we conclude that n, n - 1 must be relatively prime for n > 2. Gallian 2.34 : Prove that in a group, ( ab ) 2 = a 2 b 2 if and only if ab = ba . ( ) Assume ( ab ) 2 = a 2 b 2 . Then ( ab ) 2 = a 2 b 2 abab = a 2 b 2 a - 1 abab = a - 1 a 2 b 2 bab = ab 2 babb - 1 = ab 2 b - 1 ba = ab Therefore ( ab ) 2 = a 2 b 2 ab = ba . ( ) Assume ab = ba . Then ab = ba a ( ab ) = a ( ba ) a 2 b = aba ( a 2 b ) b = ( aba ) b a 2 b 2 = ( ab ) 2 Therefore ab = ba ( ab ) 2 = a 2 b 2 .
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