Unformatted text preview: G . We must show that af ( N ) a1 ⊆ f ( N ). Recall that af ( N ) a1 = { axa1 : x ∈ f ( N ) } . So, we must show that if x is any element of f ( N ), then axa1 belongs to f ( N ). Since f is onto, we know that a = f ( g ) for some g ∈ G . And since x belongs to f ( N ), we know that x = f ( n ) for some n ∈ N . Since N is normal, gng1 belongs to gNg1 , which is equal to N . Hence gng1 ∈ N , and therefore f ( gng1 ) ∈ f ( N ). Since f is a homomorphism, we therefore have axa1 = f ( g ) f ( n ) f ( g )1 = f ( gng1 ) ∈ f ( N ) . Thus we have shown that every element of af ( N ) a1 belongs to f ( N ), so af ( N ) a1 ⊆ f ( N ). Therefore f ( N ) is normal by de±nition. ± 1...
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 Spring '10
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 Math, Algebra, Group Theory, Normal subgroup, Subgroup

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