aa_hw_5_Math 113 - Homework 5 Introduction to Abstract...

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Introduction to Abstract Algebra Homework 5 October 11, 2010 Problem 1 Let G be a group and ( S, · ) be a set with a 2-to-1 operation · : S × S S . Let φ : G S be a surjective map of sets such that φ ( ab ) = φ ( a ) · φ ( b ) for all a, b G . Show that S is a group under · and that φ is a group homomorphism. By definition of · , we have closure. Also, φ ( a ( bc )) = φ ( a ) · φ ( bc ) = φ ( a ) · φ ( b ) · φ ( c ) = φ ( ab ) · φ ( c ) = φ (( ab ) c ) , so we have associativity. Pick s S . Since φ is surjective, there is a g G , s.t. φ ( g ) = s . Also, let e S denote the element φ ( e G ) S , then φ ( e G g ) = φ ( e G ) · φ ( g ) = e S · s = φ ( g ) = s, so we also have an identity element. Finally, we pick on s S and let s = φ ( g ) for some g G . Also, we let s - 1 S be the element φ ( g - 1 ) , and we see that, ss - 1 = φ ( g ) · φ ( g - 1 ) = φ ( gg - 1 ) = φ ( e G ) = e S , so we also have inverses, thus S is a group under · . Now, by definition of homomorphisms (p. 36), φ : G S is a group homomorphism. Problem 2 Let G be a group, and let H be a normal subgroup of G . Let π : G G/H , g gH be the natural projection map. Show that π is surjective, satisfies π ( ab ) = π ( a ) · π ( b ) and that π - 1 ( H ) = H . (a) For every gH G/H , there is at least one g G s.t. gH = g H , namely when g = g . Thus, π is surjective. (b) We have π ( ab ) = ( ab ) H = { g G | ∃ h H : g = ( ab ) h } . Since H G , then H/G is a group, and ( ab ) H = ( aH )( bH ) , thus π ( ab ) = ( ab ) H = ( aH )( bH ) = π ( a ) · π ( b ) . Thus, a, b G : π ( ab ) = π ( a ) · π ( b ) . 1/5
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Introduction to Abstract Algebra Homework 5 October 11, 2010 (c) The statement π - 1 ( H ) = H , says that the set of fibers of H under π is just H itself. Pick some h H , then hH = { g G | ∃ h H : g = h h } , which are just the elements of H .
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