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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 0divisor. I am expecting a more specific argument, i.e. a counterexample. 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).
 Spring '08
 BILLERA
 Algebra

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