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Unformatted text preview: Homework 5 Solutions 1. (a) Note: for an arbitrary commutator from G , x 1 y 1 xy , with x,y G and for g G , gx 1 y 1 xyg 1 = ( gx 1 g 1 )( gy 1 g 1 )( gxg 1 )( gyg 1 ) = ( gxg 1 ) 1 ( gyg 1 ) 1 ( gxg 1 )( gyg 1 ) Thus the conjugate of a commutator is also a commutator. Further note that if a G is a commutator than a 1 is also a commutator, since if a = x 1 y 1 xy , a 1 = ( x 1 y 1 xy ) 1 = y 1 x 1 yx . For h G we can write h = a 1 a k where each a i is a commutator from G (and any product of commutators is an element from G since G is a subgroup containing all the commutators). For g G , ghg 1 = ( ga 1 g 1 ) ( ga k g 1 ) which is also a product of commutators by the previous note. Hence ghg 1 G and thus G C G . (b) Let a,b G . a 1 b 1 abG = G since a 1 b 1 ab G . Thus b 1 abG = aG which implies abG = baG . Thus for any elements aG , bG G/G : ( aG )( bG ) = abG = baG = ( bG )( aG ) so G/G is abelian....
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This note was uploaded on 04/19/2011 for the course MATH 21373 taught by Professor Johnson during the Spring '11 term at Carnegie Mellon.
 Spring '11
 Johnson
 Algebra

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