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Unformatted text preview: Math 150a: Modern Algebra Homework 4 Solutions 2.3.16: Give an example of two isomorphic groups such that there is more than one isomorphism between them. Solution: Well, at first thought, Z / 3 = C 3 , and there are two isomorphisms between them: 1 mapsto x , or 1 mapsto x 2 . Another familiar example of isomorphic groups is D 3 = S 3 . Each symmetry of the triangle permutes its 3 vertices. But we can label these three vertices in many different ways each producing a different isomorphism from D 3 S 3 . 1 2 3 1 2 3 b b b b b b b b b Notice that relabeling the elements being permuted is an automorphism of S 3 . Given any iso morphism : G H , and an automorphism Aut ( H ) ; then ( ) : G H is a different isomorphism from G to H . In fact, any two isomorphisms between G and H are related by an auto morphism of H in this way. For any 1 , 2 : G H , ( 2 1 1 ) Aut ( H ) ; and 2 = ( 2 1 1 ) 1 . G H H G H H ( ) 1 2 ( 2 1 1 ) 2.4.6: Let f : R + C be the map f ( x ) = e ix . Prove that f is a homomorphism, and determine its kernel and image. Solution: The rules of exponents continue to work for complex numbers, so x , y R + f ( x + y ) = e i ( x + y ) = e ix + iy = e ix e iy = f ( x ) f ( y ) . Hence, f is a homomorphism. To find its kernel and image, we will need to use Eulers formula: e ix = cos x + i sin x . In complex analysis, this formula gives insight into the geometry of com plex multiplication; and relates rectangular coordinates to polar coordinates via the exponential. Consider the image of f , im ( f ) = { z C  x R + so that f ( x ) = z } . So, z im ( f ) z = cos x + i sin x for some x . Thus, the image of f is the unit circle: { a + ib C  a 2 + b 2 = 1 } . To find the kernel of f , we need to solve f ( x ) = 1 for x , since ker ( f ) = { x R +  f ( x ) = 1 C } . f ( x ) = e ix = cos x + i sin x = 1 + i , cos x = 1, and sin x = , x = 2 n , n Z . square 2.4.13: (a) Let H be a subgroup of G , and let g G . The conjugate subgroup gHg 1 is defined to be the set of all conjugates ghg 1 , where h H . Prove that gHg 1 is a subgroup of G . Solution: Consider general elements x , y gHg 1 . Then for some h , k H , x = ghg 1 and y = gkg 1 . So, xy = ( ghg 1 )( gkg 1 ) = ghkg 1 = g ( hk ) g 1 gHg 1 since hk H . Hence gHg 1 is closed under multiplication. Additionally, gh 1 g 1 gHg 1 , since h 1 H . So ( ghg 1 )( gh 1 g 1 ) = ( gh 1 g 1 )( ghg 1 ) = e ....
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This homework help was uploaded on 02/01/2008 for the course MATH 150A taught by Professor Kuperberg during the Spring '03 term at UC Davis.
 Spring '03
 Kuperberg
 Algebra

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