# Algebra

• Notes
• davidvictor
• 3

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Math 150a: Modern Algebra Homework 3 2.2.13: Prove that every subgroup of a cyclic group is cyclic. Solution: Let H be a subgroup of the cyclic group C n = < x | x n = 1 > . We want to show that H is cyclic. Let k be the first positive integer such that x k is in H . Then...for any y H , y = x m , for some m . Now, using the Euclidean algorithm, we can write m = kq + r , where 0 r < k . And x m = x kq + r x m x kq = x r . Since H is a subgroup, x r = x m ( x m ) q H . But 0 r < k , and k is the least positive integer such that x k is in H . So r must equal 0, and k must divide m . Therefore, H = < x k > (the cyclic group generated by x k ). square 2.3.4: (a) Let b = aba 1 . Prove that b n = ab n a 1 . Solution: Consider b 2 = aba 1 aba 1 = abba 1 = ab 2 a 1 . Using induction on n , assume b n = ab n a 1 and consider b ( n + 1 ) . Then b ( n + 1 ) = b n b = ab n a 1 aba 1 = ab ( n + 1 ) a 1 . square (b) Prove that if aba 1 = b 2 , then a 3 ba 3 = b 8 . Solution: a 3 ba 3 = a 2 ( aba 1 ) a 2 = a 2 b 2 a 2 = a ( aba 1 )( aba 1 ) a 1 = ab 2 b 2 a 1 = ( aba 1 )( aba 1 )( aba 1 )( aba 1 ) = b 2 b 2 b 2 b 2 = b 8 . square 2.3.14: Determine the group of automorphisms of the following groups. Solution: (a) + = ( , +) = In order to be an isomorphism, any automorphism of a cyclic group must send the generator to another generator of the group. There are only two generators of the cyclic group , 1 and 1. So there can only be two automorphisms, either sending 1 mapsto→ 1 or 1 mapsto→ − 1. There is only one group of order 2, but it is also easy to verify that one of these automorphisms is the identity and one has order 2. The identity element of any automorphism group is the identity automorphism. And the map sending 1 mapsto→ − 1, is the automorphism of the integers usually called “negation”.

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• Spring '03
• Kuperberg
• Algebra, Cyclic group, xm, B2 B2 B2, S3 D3

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