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Unformatted text preview: 1 Math 417 Exam II Solutions: July 24, 2009 1 . If = 1 t is a product of disjoint r icycles i , prove that has order m = lcm { r 1 , . . . , r t } . This is Proposition 2.55(ii). 2 ( i ) . (10 points) If G is a group with Aut( G ) = { 1 } , prove that g 2 = 1 for every g G . If G is not abelian, then there is a G which in not in the center Z ( G ); that is, there is some g G with aga 1 6 = g . Hence, a : G G , conjugation by a , defined by x 7 axa 1 , is an automorphism of G that is not the identity 1 G because aga 1 6 = g . If G is abelian, then we have proved that : g 7 g 1 is an automorphism of G . Now 6 = 1 G if there exists x G with x 6 = x 1 . Hence, = 1 G if g = g 1 for all g G ; that is, if g 2 = 1 for all g G . Remark: Here is a sketch of how to treat this last case.The hypothesis g 2 = 1 for all g G implies that G can be viewed as a vector space with scalars in I 2 . If  G  > 2, then G has a basis...
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 Spring '08
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 Math, Algebra

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