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HW6_solns

# HW6_solns - Â Â¸ â‰  Â Â Â¸ Â 2 Â 2 Â¸ or Â Â¸ So there are...

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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 f g. g ′− 1 = f g . f g 1 = ? ker f={e} g. g ′− 1 = e g = g′ 51. Find the cardinality of ?𝐿 2 ? 3 Solution: ?𝐿 2 ? 3 = ? ? ? ? ? , ? , ? , ? ∈ ? 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: 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 0-divisor. I am expecting a more specific argument, i.e. a counter-example. For instance, Âº = Â» /6 Â» , ? : Âº â†’ Âº is given by ? Â¼Â± = 3 Â¼ ....
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