lecture23 - 23. Quotient groups II 23.1. Proof of the...

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23. Quotient groups II 23.1. Proof of the fundamental theorem of homomorphisms (FTH). We start by recalling the statement of FTH introduced last time. Theorem (FTH) . Let G , H be groups and ϕ : G H a homomorphism. Then G/ Ker ϕ = ϕ ( G ) . ( * * * ) Proof. Let K = Ker ϕ and define the map Φ : G/K ϕ ( G ) by Φ( gK ) = ϕ ( g ) for g G. We claim that Φ is a well defined mapping and that Φ is an isomorphism. Thus we need to check the following four conditions: (i) Φ is well defined (ii) Φ is injective (iii) Φ is surjective (iv) Φ is a homomorphism For (i) we need to prove the implication “ g 1 K = g 2 K Φ( g 1 K ) = Φ( g 2 K ).” So, assume that g 1 K = g 2 K for some g 1 ,g 2 G . Then g - 1 1 g 2 K by Theorem 19.2, so ϕ ( g - 1 1 g 2 ) = e H (recall that K = Ker ϕ ). Since ϕ ( g - 1 1 g 2 ) = ϕ ( g 1 ) - 1 ϕ ( g 2 ), we get ϕ ( g 1 ) - 1 ϕ ( g 2 ) = e H . Thus, ϕ ( g 1 ) = ϕ ( g 2 ), and so Φ( g 1 K ) = Φ( g 2 K ), as desired. For (ii) we need to prove that “Φ( g 1 K ) = Φ( g 2 K ) g 1 K = g 2 K .” This is done by taking the argument in the proof of (i) and reversing all the implication arrows. (iii) First note that by construction Codomain(Φ) = ϕ ( G ). Thus, for surjectivity of Φ we need to show that Range(Φ) = Φ( G/K ) is equal to ϕ ( G ). This is clear since Φ( G/K ) = { Φ( gK ) : g G } = { ϕ ( g ) : g G } = ϕ ( G ) . (iv) Finally, for any g 1 ,g 2 G we have Φ( g 1 K · g 2 K ) = Φ( g 1 g 2 K ) = ϕ ( g 1 g 2 ) = ϕ ( g 1 ) ϕ ( g 2 ) = Φ( g 1 K )Φ( g 2 K ) where the first equality holds by the definition of product in quotient groups. Thus, Φ is a homomorphism. So, we constructed an isomorphism Φ : G/ Ker ϕ ϕ ( G ), and thus G/ Ker ϕ is isomorphic to ϕ ( G ). ± 1
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2 23.2. Applications of FTH. In most applications one uses a special case of FTH stated last time as Corollary 22.5: If ϕ : G H is a surjective homomorphism, then G/ Ker ϕ = H . (***) Typically this result is being applied as follows. We are given a group G , a normal subgroup K and another group H (unrelated to G ), and we are asked to prove that G/K = H . By (***) to prove that G/K = H it suffices to find a surjective homomorphism ϕ : G H such that Ker ϕ = K . Example 1:
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This note was uploaded on 01/18/2012 for the course INFORMATIK 2011 taught by Professor Phanthuongcang during the Winter '11 term at Cornell University (Engineering School).

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lecture23 - 23. Quotient groups II 23.1. Proof of the...

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