# HW6_solns - ? ? ? , 2 2 ? or . So there are 6 choices....

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Selected Solutions for HW 6 32. Let f: G G be a group homomorphism. Then ker(f) is a subgroup of G, and f is 1-1 iff ker(f) = {e}. Proof: Let g,g be in ker(G). We need to show g.g , g 1 are both in ker(G), i.e. f(g.g ) = f( g 1 ) = e. But f(g.g ) = f(g).f(g ) = e.e = e, f( g 1 ) = (f(g)) 1 = e 1 = e , we are done. For the second part, f g ± = f g ’± ²³³´ f g. g ′− 1 ± = f g ± . f g ± 1 = ? ker f={e} ²³³³³´ g. g ′− 1 = e ²³³´ g = g′ 51. Find the cardinality of 2 ? 3 ± Solution: 2 ? 3 ± = µ¶ · ? ? ? ¸¹ · , ? , ? , ? ∈ ? 3 ; ·? − ?? 0} If we treat the 2x2 matrix as 2 column vectors, we want the 2 vectors · ? ¸ , ? ? ¸ to be linearly independent. There are 9 choices of the first column · ? ¸ , but · ? ¸ cannot be 0 0 ¸ . So there are 8 choices of the first column. For the second column,
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Unformatted text preview: ? ? ? , 2 2 ? or . So there are 6 choices. Therefore in total there are 48 elements in the group. 4. 53(mod 72) 7. 38(mod 420) Additional Problem 1: a) We can only use the property f g 1 . g 2 = f g 1 . f g 2 ( ) to derive the properties given in the book. For any ? ? , f(g) = f g. e = f g . f e . Multiplying f g 1 on both sides we get our desired result. b) In the case of rings, many of you noticed that f g 1 may not exist when f(g) is a 0-divisor. I am expecting a more specific argument, i.e. a counter-example. For instance, = /6 , ? : is given by ? = 3 ....
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## This note was uploaded on 09/05/2011 for the course MATH 3360 taught by Professor Billera during the Spring '08 term at Cornell University (Engineering School).

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