Properties of Homomorphisms

Abstract Algebra · Axiom Academy

LESSON Properties of Homomorphisms Fundamental properties that all group homomorphisms satisfy: preservation of identity, inverses, and powers. Proof: Consider φ(e G ). We know that e G · e G = e G in G. Applying φ to both sides: φ(e G · e G ) = φ(e G ) φ(e G ) · φ(e G ) = φ(e G ) Multiplying both sides by φ(e G ) -1 gives φ(e G ) = e H Proof: We need to show that φ(a -1 ) is the inverse of φ(a) in H. Consider their product: φ(a) · φ(a -1 ) = φ(a · a -1 ) = φ(e G ) = e H Similarly, φ(a -1 ) · φ(a) = e H . Therefore, φ(a -1 ) is the unique inverse of φ(a). Proof: We prove this by induction for positive n, then extend to negative n using the inverse property. Base case: n = 1 is trivial. n = 0 gives φ(e G ) = e H (identity property). Inductive step: Assume φ(a k ) = φ(a) k . Then: φ(a k+1 ) = φ(a k · a) = φ(a k ) · φ(a) = φ(a) k · φ(a) = φ(a) k+1 4. Composition of Homomorphisms Proof: We need to verify that (ψ ∘ φ)(ab) = (ψ ∘ φ)(a) · (ψ ∘ φ)(b) for all a, b ∈ G: (ψ ∘ φ)(ab) = ψ(φ(ab)) = ψ(φ(a) · φ(b)) = ψ(φ(a)) · ψ(φ(b)) = (ψ ∘ φ)(a) · (ψ ∘ φ)(b) 5. Inverse Images of Subgroups Proof: We verify the subgroup criteria: 1. Non-empty: Since e H ∈ K and φ(e G ) = e H , we have e G ∈ φ -1 (K). 2. Closure: If a, b ∈ φ -1 (K), then φ(a), φ(b) ∈ K. Since K is a subgroup, φ(a) · φ(b) ∈ K. But φ(a) · φ(b) = φ(ab), so ab ∈ φ -1 (K). 3. Inverses: If a ∈ φ -1 (K), then φ(a) ∈ K. Since K is a subgroup, φ(a) -1 ∈ K. But φ(a) -1 = φ(a -1 ), so a -1 ∈ φ -1 (K).

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