Unformatted text preview: g1 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 gxg1 = ygg1 = y ∈ N . We deduce that gNg1 ⊆ N and hence N ± G , as required....
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 Fall '08
 PARRY
 Math, Logic, Algebra, Trigraph, October 20

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