Unformatted text preview: Â¶ ? ? Â¸ â‰ Â¶ Â· ? Â¸ , Â¶ 2 Â· 2 ? Â¸ or Â¶ Â¸ . 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 Â± = f g 1 Â± . f g 2 Â± âˆ’ ( âˆ— ) to derive the properties given in the book. For any ? âˆˆ ? , f(g) = f g. e Â± = f g Â± . f e Â± . Multiplying f g Â± âˆ’ 1 on both sides we get our desired result. b) In the case of rings, many of you noticed that f g Â± âˆ’ 1 may not exist when f(g) is a 0divisor. I am expecting a more specific argument, i.e. a counterexample. For instance, Âº = Â» /6 Â» , ? : Âº â†’ Âº is given by ? Â¼Â± = 3 Â¼ ....
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 Spring '08
 BILLERA
 Algebra, Normal subgroup, Column, Ker, Group homomorphism

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