gallian_sol.pdf - See discussions stats and author profiles for this publication at https/www.researchgate.net/publication/280733004 Abstract Algebra

# gallian_sol.pdf - See discussions stats and author profiles...

• 36

This preview shows page 1 - 5 out of 36 pages.

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.

Subscribe to view the full document.

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
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

Subscribe to view the full document.

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 .
• Fall '19
• lakshmi burra

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern

Ask Expert Tutors You can ask 0 bonus questions You can ask 0 questions (0 expire soon) You can ask 0 questions (will expire )
Answers in as fast as 15 minutes