solutions2

# solutions2 - Spring 2010 MATH 4320 Solutions to Prelim 2...

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

Spring 2010 MATH 4320: Solutions to Prelim 2 Instructor: Yuri Berest Problem 1. ( a ) This is straightforward: for example, we have π 1 [( x 1 , x 2 ) · ( y 1 , y 2 ))] = π 1 [( x 1 y 1 , x 2 y 2 )] = x 1 y 1 = π 1 [( x 1 , x 2 )] π 1 [( y 1 , y 2 )] , and similarly for π 2 . Both π 1 and π 2 are surjective, because given any x 1 H 1 and any x 2 H 2 , we have π 1 [( x 1 , x 2 )] = x 1 and π 2 [( x 1 , x 2 )] = x 2 . Finally, Ker( π 1 ) = ( e 1 , H 2 ) and Ker( π 2 ) = ( H 1 , e 2 ) . ( b ) Given any group G with two homomorphisms f 1 : G H 1 and f 2 : G H 2 , we define ϕ : G H 1 × H 2 by ϕ ( x ) := ( f 1 ( x ) , f 2 ( x )) for x G . Then, we see at once that ( π 1 ϕ )( x ) = π 1 [ ϕ ( x )] = π 1 [( f 1 ( x ) , f 2 ( x ))] = f 1 ( x ) and ( π 2 ϕ )( x ) = π 2 [ ϕ ( x )] = π 2 [( f 1 ( x ) , f 2 ( x ))] = f 2 ( x ) for all x G . In other words, f 1 = π 1 ϕ and f 2 = π 2 ϕ as required. Now, suppose there is another homomorphism, say ψ : G H 1 × H 2 , satisfying f 1 = π 1 ψ and f 2 = π 2 ψ . By definition of H 1 × H 2 , the elements ψ ( x ) in the image of ψ can be written as ψ ( x ) = ( g 1 ( x ) , g 2 ( x )) , where g 1 ( x ) H 1 and g 2 ( x ) H 2 . Since f 1 = π 1 ψ and f 2 = π 2 ψ , we have f 1 ( x ) = ( π 1 ψ )( x ) = π 1 [ ψ ( x )] = π 1 [( g 1 ( x ) , g 2 ( x ))] = g 1 ( x ) and f 2 ( x ) = ( π 2 ψ )( x ) = π 2 [ ψ ( x )] = π 2 [( g 1 ( x ) , g 2 ( x ))] = g 2 ( x ) for all x G . Thus, g 1 ( x ) = f 1 ( x ) and g 2 ( x ) = f 2 ( x ) , and therefore ψ ( x ) = ϕ ( x ) for all x G . This proves that ψ = ϕ , which means the uniqueness of ϕ . Problem 2. ( a ) For each h H and for all x 1 and x 2 in N , we have f ( h )( x 1 x 2 ) := h ( x 1 x 2 ) h - 1 = h x 1 ( h - 1 h ) x 2 h - 1 = ( h x 1 h - 1 ) ( h x 2 h - 1 ) . So f ( h )( x 1 x 2 ) = f ( h )( x 1 ) f ( h )( x 2 ) , which means that f ( h ) is a group ho- momorphism N N . Moreover, f ( h ) is bijective for each h H , because it has an inverse (namely f ( h - 1 )). Thus, f : H Aut( N ) is a well-defined map, assigning to elements of H the group automorphisms of N . It remains to show that this map is a homomorphism of groups.

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

View Full Document
Fix any h 1 and h 2 in H . Then f ( h 1 h 2 ) : N
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### 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