{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lecture23

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

This preview shows pages 1–3. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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:
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 6

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online