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Unformatted text preview: 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 | 6 = | H | , we know that H 6 ∼ = 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 ....
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This note was uploaded on 02/29/2012 for the course MATH 3362 taught by Professor Ryandaileda during the Fall '10 term at Trinity University.
- Fall '10