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Homework1Solutions - Homework 1 Solutions Math 332 Spring...

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Homework 1 Solutions Math 332, Spring 2010 Problem 1. Let G be a group. Two elements g, h G are said to be conjugate if there exists an element a G for which h = a - 1 ga . (a) Proposition. Conjugacy is an equivalence relation on G . Proof. We must prove that conjugacy is reflexive, symmetric, and transitive. Reflexive If g G , then g = e - 1 ge , and therefore g is conjugate to g . Symmetric Let g, h G and suppose that g is conjugate to h . Then g = a - 1 ha for some element a G . Solving the above equation for h yields h = aga - 1 = b - 1 gb, where b = a - 1 . This proves that h is conjugate to g , and therefore conjugacy is symmetric. Transitive Let g, h, k G and suppose that g is conjugate to h and h is conjugate to k . Then g = a - 1 ha and h = b - 1 kb for some elements a, b G . Substituting the second equation into the first yields g = a - 1 b - 1 kba = c - 1 kc, where c = ba . This proves that g is conjugate to k , and therefore conjugacy is transitive. 1
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(b) Recall that the Cayley table for D 3 is e r r 2 a b c e e r r 2 a b c r r r 2 e b c a r 2 r 2 e r c a b a a c b e r 2 r b b a c r e r 2 c c b a r 2 r e where r and r 2 are the rotations and a , b , and c are the reflections. Using this table, it is easy to verify that r and r 2 are conjugate: a - 1 ra = ara = r 2 , as are a , b
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