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Unformatted text preview: Homework 4 Solutions 1. e 2 = e thus e H and H 6 = . Now let x,y H , hence a,b G such that x = a 2 , y = b 2 . xy 1 = a 2 ( b 2 ) 1 = ( ab 1 ) 2 since G is abelian. Thus xy 1 H and H G by the one step subgroup test. 2. (1) = (12)(12) thus (1) A n and A n 6 = . Let a,b A n thus a and b can be written as a = 1 .. . r , b = 1 .. . s where i , i are transpositions and r and s are even. Since the i s and i s are transpositions: ab 1 = ( 1 .. . r )( 1 .. . s ) 1 = 1 .. . r  1 s .. . 1 1 = 1 .. . r s .. . 1 Since r + s is even, ab 1 is an even permutation and hence ab 1 A n . Thus A n S n by the one step subgroup test. 3. [ G : H ] =  G  /  H  = (  G  /  K  )(  K  /  H  ) = [ G : K ][ K : H ] 4. BWOC, suppose a H K such that a 6 = e . Let o ( a ) = l 6 = 1. a H and a K implies that l  H  and l  K  by the corollary to Lagranges Theorem. l 6 = 1...
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 Spring '11
 Johnson
 Algebra

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