Unformatted text preview: Homework 4 Solutions 1. e 2 = e thus e ∈ H and H 6 = Ø. 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 = Ø. Let a,b ∈ A n thus a and b can be written as a = α 1 .. .α r , b = β 1 .. .β s where α i , β i are transpositions and r and s are even. Since the α i ’s and β i ’s are transpositions: ab 1 = ( α 1 .. .α r )( β 1 .. .β s ) 1 = α 1 .. .α r β 1 s .. .β 1 1 = α 1 .. .α r β s .. .β 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 Lagrange’s Theorem. l 6 = 1...
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 Spring '11
 Johnson
 Algebra, Subgroup, Cyclic group, rational entries, step subgroup test

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