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Unformatted text preview: e G A , we have ( e G ) ( A ) , so ( A ) 6 = /0. (ii) Let x , y ( A ) . Then we may write x = a and y = b for some a , b A . Then xy = a b = ( ab ) . Since A is a subgroup, ab A and we deduce that xy A . (iii) Let x ( A ) . Then we may write x = a for some a A . Then x1 = ( a )1 = ( a1 ) . Since A is a subgroup, a1 A and we deduce that x1 A ....
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 Fall '08
 PARRY
 Math, Algebra

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