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Homework5 - Homework 5 Solutions 1(a Note for an arbitrary...

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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 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. (c) i. Assume ϕ : G A is a homomorphism and A is abelian. As in part (a), first we will note that every commutator is in kerϕ . For x, y G , ϕ ( x - 1 y - 1 xy
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