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Math505HWsolutions

# Math505HWsolutions - MATH 405/505 ABSTARCT ALGEBRA HOMEWORK...

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MATH 405/505 ABSTARCT ALGEBRA HOMEWORK MAIN IDEAS OF SOLUTIONS 2.65 Prove that U ( I 9 ) is isomrophic to I 6 and that U ( I 15 ) is isomorphic to I 4 × I 2 . Solution: For the first part note U ( I 9 ) contains 6 elements. Now exhibit an element of order 6. For the second part note that U ( I 15 ) contains 8 elements. Now exhibit an element a of order 4 and an element b of order 2 so that h a i ∩ h b i = 1 and h a ih b i = U ( I 15 ). 2.67 Prove that (i) Aut ( V ) = S 3 , (ii) Aut ( S 3 ) = S 3 , and (iii) Aut ( Z ) = I 2 . Solution: (i) Note that V = I 2 × I 2 , so V = { 0 = (0 , 0) , a = (1 , 0) , b = (0 , 1) , c = (1 , 1) } . Every automorphism of V fixes 0 and permutes a , b , c . This shows that Aut ( V ) is a subgroup of S X , where X = { a, b, c } . Now check that every permutation of X does also define an automorphism. (ii) Note that S 3 contains 3 elements of order 2, x = (12), y = (13), and z = (23). An element of Aut ( S 3 ) permutes these 3 elements, so Aut ( S 3 ) is a subgroup of S X , where X = { x, y, z } . Now check that every permutation of X does also define an automorphism. (iii) Every automophism of Z send a generator to a generator. The result follows from the fact that the only generators of Z are ± 1. 2.68 If G is a finite group such that Aut ( G ) is the trivial group, then G is trivial or G = I 2 . Solution: Let γ g be the inner automorphism that is conjugation by g , so γ g ( g 0 ) = gg 0 g - 1 for all g 0 G . Since Aut ( G ) is trivial, i.e. only containes the identity, γ g is the identity.

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Math505HWsolutions - MATH 405/505 ABSTARCT ALGEBRA HOMEWORK...

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