{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MATH 111 m5 - MODEL ANSWERS TO THE FIFTH HOMEWORK 1 Chapter...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
MODEL ANSWERS TO THE FIFTH HOMEWORK 1. Chapter 3, Section 5: 1 (a) Yes. Given a and b Z , φ ( ab ) = [ ab ] = [ a ][ b ] = φ ( a ) φ ( b ) . This map is clearly surjective but not injective. Indeed the kernel is easily seen to be n Z . (b) No. Suppose that G is not abelian and that xy = yx . Then x - 1 y - 1 = y - 1 x - 1 . On the other hand φ ( xy ) = ( xy ) - 1 = y - 1 x - 1 = x - 1 y - 1 = φ ( x ) φ ( y ) , and one wrong certainly does not make a right. (c) Yes. Suppose that x and y are in G . As G is abelian φ ( xy ) = ( xy ) - 1 = y - 1 x - 1 = x - 1 y - 1 = φ ( x ) φ ( y ) . Thus φ is a homomorphism. Suppose that a G . Then a is the inverse of b = a - 1 , so that φ ( b ) = a . Thus φ is surjective. Suppose that a is in the kernel of φ . Then a - 1 = e and so a = e . Thus the kernel of φ is trivial and φ is injective. (d) Yes. φ is a homomorphism as the product of two positive numbers is positive, the product of two negative numbers is positive and the product of a negative and a positive number is negative. This map is clearly surjective. The kernel consists of all positive real numbers. Thus φ is far from injective. 1
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
(e) Yes. Suppose that x and y are in G . Then φ ( xy ) = ( xy ) n = x n y n = φ ( x ) φ ( y ) . In general this map is neither injective nor surjective. For example, if G = Z and n = 2 then the image of φ is 2 Z , and for example 1 is not in the image. Now suppose that G = Z 4 and n = 2. Then 2[2] = [4] = [0], so that [2] is in the kernel. 10. We need to check that aHa - 1 = H for all a G . If we pick a H there is nothing to prove. Now a = f i g j . Conjugation by a is the same as conjugation by g j followed by conjugation by f i . So we only need to worry about conjugation by f . Now gf = fg - 1 so that fgf - 1 = g - 1 . Thus conjugation by f leaves H fixed, as it sends a generator to a generator. 12. Let g G . We want to show that gZg - 1 Z . Pick z Z . Then z commutes with g , so that gzg - 1 = zgg - 1 = z Z . Thus Z is normal in G .
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern