10Ma5aHw6sol

10Ma5aHw6sol - M a 5a HOMEWORK 6 SOLUTION FALL 2010 The...

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Ma 5 a HOMEWORK 6 SOLUTION FALL 2010 The exercises are taken from the text, Abstract Algebra (third edi- tion) by Dummit and Foote. Page 138, 11 . Define a homomorphism φ : N S p −→ Aut ( P ) via g 7−→ ( x 7−→ gxg - 1 ). Then φ ( gh )( x ) = ghx ( gh ) - 1 = g ( hxh - 1 ) g - 1 = φ ( g )( φ ( h )( x )) = ( φ ( g ) φ ( h ))( x ), and gxg - 1 P , so φ is well-defined. Clearly, ker ( φ ) = C S p ( P ), and | N S p ( P ) C S p ( P ) | ≤ p 1 = Aut ( P ). Since the conjugacy classes of P all consist of ( p 1) p cycles , there are ( p - 1)! p - 1 in total, so [ S p : N S p ( P )] = ( p - 1)! p - 1 . Therefore, | N S p ( P ) | = p ! ( p - 2)! = p ( p 1). Take any g C S p ( P ), and x = ( x 1 x 2 ...x p ) a p -cycle, then gxg - 1 = ( g ( x 1 ) g ( x 2 ) ...g ( x p )) = ( x 1 ...x p ). Obvious | C S p ( P ) | = | P | = p . Therefore | N S p ( P ) C S p ( P ) | = Aut ( P ), and φ is an isomorphism. Page 138,
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10Ma5aHw6sol - M a 5a HOMEWORK 6 SOLUTION FALL 2010 The...

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