Unformatted text preview: e G ∈ A , we have θ ( e G ) ∈ θ ( A ) , so θ ( A ) 6 = /0. (ii) Let x , y ∈ θ ( A ) . Then we may write x = θ a and y = θ b for some a , b ∈ A . Then xy = θ a θ b = θ ( ab ) . Since A is a subgroup, ab ∈ A and we deduce that xy ∈ θ A . (iii) Let x ∈ θ ( A ) . Then we may write x = θ a for some a ∈ A . Then x1 = ( θ a )1 = θ ( a1 ) . Since A is a subgroup, a1 ∈ A and we deduce that x1 ∈ θ A ....
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This note was uploaded on 01/02/2012 for the course MATH 3124 taught by Professor Parry during the Fall '08 term at Virginia Tech.
 Fall '08
 PARRY
 Math, Algebra

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