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# HW8_soln - Homework#8 Solutions p 132#10 Suppose that is an...

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Homework #8 Solutions p 132, #10 ( ) Suppose that α is an automorphism of G . Let a, b G . Then b - 1 a - 1 = ( ab ) - 1 = α ( ab ) = α ( a ) α ( b ) = a - 1 b - 1 which implies that ab = ba . Since a, b G were arbitrary we conclude that G is abelian. ( ) Suppose that G is abelian. Let a, b G . If α ( a ) = α ( b ) then we have a - 1 = b - 1 which implies that a = b , proving that α is one-to-one. Given any a G , we know that a - 1 G and α ( a - 1 ) = ( a - 1 ) - 1 = a which proves that α is onto. Finally, if a, b G then, since G is abelian, α ( ab ) = ( ab ) - 1 = b - 1 a - 1 = a - 1 b - 1 = α ( a ) α ( b ) so that α is operation preserving. Therefore, α is an automorphism of G . p 132, #12 Let H = Z 7 and G = Z 9 . Since | G | = | H | , we know that H = G . However Aut( Z 7 ) = U (7) = Z 6 and Aut( Z 9 ) = U (9) = Z 6 so that Aut( H ) = Aut( G ). p 134, #30 Let the mapping in question be denoted by s . That is, s : G G is given by s ( g ) = g 2 . Let g, h G . Since G is abelian we have s ( gh ) = ( gh ) 2 = g 2 h 2 = s ( g ) s ( h ) which shows that s preserves operations. In order to show that s is an automorphism of G it remains to show that s is one-to-one and onto. Since G is finite, it suffices to show that s is one-to-one. So suppose that g, h G and s ( g ) = s ( h ). Then, by the definition of s , we have g 2 = h 2 or g 2 h - 2 = e . Again using the fact that G is abelian we have ( gh - 1 ) 2 = e . Since G has no element of order it must be that gh - 1 = e . This implies that g = h , proving that s is one-to-one, completing the proof that s is an automorphism of G . If we let G = Z then for any n Z we have s ( n ) = 2 n , so that the image of s consists only of even integers. It follows that s is not onto and hence is not an automorphism of Z .

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Automorphism Exercise 1.
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