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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