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

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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. ( 1 2 3 ) S 3 , but ( 1 2 )( 1 2 3 )( 1 2 ) - 1 = ( 1 3 2 ) . Problem 21.26 on page 110. Prove that if N 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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