If N E G, then there exists a group H and a homomorphism ϕ : G → H such that N = ker(ϕ) Proof Define ϕ : G/N → H by stipulating ϕ(xN) := ψ(x) (for every x ∈ G) Thus, since N = ker(ψ), ψ(x−1y) = eH and since ψ is a homomorphism we have eH = ψ(x−1y) = ψ(x)−1 ψ(y), which implies ψ(x) = ψ(y)
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[PDF] Lecture 43: The fundamental homomorphism theorem - School of
Proof of the FHT Fundamental homomorphism theorem If φ: G → H is a homomorphism, then Im(φ) ∼= G/ Ker(φ) Proof We will construct an explicit map i : G/
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If N E G, then there exists a group H and a homomorphism ϕ : G → H such that N = ker(ϕ) Proof Define ϕ : G/N → H by stipulating ϕ(xN) := ψ(x) (for every x ∈ G) Thus, since N = ker(ψ), ψ(x−1y) = eH and since ψ is a homomorphism we have eH = ψ(x−1y) = ψ(x)−1 ψ(y), which implies ψ(x) = ψ(y)
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Suppose that φ : G → L is a homomorphism Then 1 φ(eG) = eL 2 ∀g ∈ G, (φ( g))−1 = φ(g−1) 3 ker φ < G 4 Im φ ≤ L Proof 1 Suppose that g ∈ G Then
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Let G and H be groups and ϕ : G → H a homomorphism Then (a) ϕ(G) is a subgroup of H (b) Ker(ϕ) is a subgroup of G Proof (a) First note that by Theorem
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suffices to find a surjective homomorphism ϕ : G → H such that Kerϕ = K Example 1: Let n ≥ 2 be an integer Prove that Z/nZ ∼ = Zn
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Then Ker φ is a subgroup of G Proof We have to show that the kernel is non- empty and closed under products and inverses Note that φ(
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23 sept 2003 · Basically a homomorphism of monoids is a function between them that morphism of groups Then f is injective if and only if ker(f) = {e} Proof
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Proof [We prove by mutual set inclusion ] Suppose x ∈ φ−1(g∨) Then φ(x) =
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