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Unformatted text preview: MODEL ANSWERS TO THE FIFTH HOMEWORK 1. Chapter 3, Section 5: 1 (a) Yes. Given a and b Z , ( ab ) = [ ab ] = [ a ][ b ] = ( a ) ( 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 ( xy ) = ( xy )- 1 = y- 1 x- 1 = x- 1 y- 1 = ( x ) ( y ) , and one wrong certainly does not make a right. (c) Yes. Suppose that x and y are in G . As G is abelian ( xy ) = ( xy )- 1 = y- 1 x- 1 = x- 1 y- 1 = ( x ) ( y ) . Thus is a homomorphism. Suppose that a G . Then a is the inverse of b = a- 1 , so that ( b ) = a . Thus is surjective. Suppose that a is in the kernel of . Then a- 1 = e and so a = e . Thus the kernel of is trivial and is injective. (d) Yes. 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 is far from injective. 1 (e) Yes. Suppose that x and y are in G . Then ( xy ) = ( xy ) n = x n y n = ( x ) ( y ) . In general this map is neither injective nor surjective. For example, if G = Z and n = 2 then the image of is 2 Z , and for example 1 is not in the image. Now suppose that G = Z 4 and n = 2. Then 2 =  = , so that  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 ....
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