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practice2b

practice2b - G We must show that af N a-1 ⊆ f N Recall...

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MATH 4107 PRACTICE PROBLEM WITH SOLUTION February 6, 2007 2.5 #15. Suppose that N is a normal subgroup of a group G , and that f : GtoG 0 is a homomorphism of G onto G 0 . Prove that f ( N ) is a normal subgroup of G 0 . Solution First we must prove that f ( N ) is a subgroup, and then we must prove that it is normal. Remember that f ( N ) is the direct image of N : f ( N ) = { f ( n ) : n N } . Suppose that x , y belong to f ( N ). This means that x = f ( m ) and y = f ( n ) for some m , n N . Since N is a subgroup, we know that mn belongs to N . Hence f ( mn ) f ( N ) by definition of f ( N ). Since f is a homomorphism, we therefore have xy = f ( m ) f ( n ) = f ( mn ) f ( N ) . Hence f ( N ) is closed under compositions. Now suppose that x belongs to f ( N ). This means that x = f ( n ) for some n N . Since N is a subgroup, we have n - 1 N , and therefore f ( n - 1 ) f ( N ). Since f is a homomorphism, we therefore have x - 1 = f ( n ) - 1 = f ( n - 1 ) f ( N ) . Thus f ( N ) is closed under inverses. So, we conclude that f ( N ) is indeed a subgroup of G 0 . Now we show that f ( N ) is normal. Suppose that
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Unformatted text preview: G . We must show that af ( N ) a-1 ⊆ f ( N ). Recall that af ( N ) a-1 = { axa-1 : x ∈ f ( N ) } . So, we must show that if x is any element of f ( N ), then axa-1 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, gng-1 belongs to gNg-1 , which is equal to N . Hence gng-1 ∈ N , and therefore f ( gng-1 ) ∈ f ( N ). Since f is a homomorphism, we therefore have axa-1 = f ( g ) f ( n ) f ( g )-1 = f ( gng-1 ) ∈ f ( N ) . Thus we have shown that every element of af ( N ) a-1 belongs to f ( N ), so af ( N ) a-1 ⊆ f ( N ). Therefore f ( N ) is normal by de±nition. ± 1...
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