oct20 - g-1 ng N and we deduce that x gN . Therefore Ng gN...

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Math 3124 Thursday, October 20 October 20, Ungraded Homework Problem 21.19 on page 109. True or false: if N ± G , then gng - 1 = n for all n N and g G . Justify your answer. False, e.g. h ( 1 2 3 ) i ± S 3 , but ( 1 2 )( 1 2 3 )( 1 2 ) - 1 = ( 1 3 2 ) . Problem 21.26 on page 110. Prove that if N 6 G , then N ± G if and only if Ng = gN for each g G . First suppose N ± G . If x Ng , then x = ng for some n N which yields x = g ( g - 1 ng ) . Since N ± G , we know that
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Unformatted text preview: g-1 ng N and we deduce that x gN . Therefore Ng gN , and similarly gN Ng . Therefore Ng = gN . Conversely suppose Ng = gN . If x N , then gx = yg for some y N and we see that gxg-1 = ygg-1 = y N . We deduce that gNg-1 N and hence N G , as required....
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This note was uploaded on 01/02/2012 for the course MATH 3124 taught by Professor Parry during the Fall '08 term at Virginia Tech.

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