Kernel and Image Definitions

Abstract Algebra · Axiom Academy

LESSON Kernel and Image of Group Homomorphisms Understanding two fundamental substructures that arise from group homomorphisms Think of the kernel as measuring "how far" φ is from being injective (one-to-one). If Ker(φ) = e , then φ is injective. The larger the kernel, the more elements "collapse" to the identity. The image tells us which elements of G' are actually "reachable" through φ. If Im(φ) = G', then φ is surjective (onto). The image is always a subset of G'. Ker(φ) = σ ∈ S₃ : φ(σ) = 1 = all even permutations = e, (123), (132) = A₃ (the alternating group) Im(φ) = φ(σ) : σ ∈ S₃ = 1, -1 (both values are achieved) 4. Theorem: Kernel is a Normal Subgroup Ker(φ) is a subgroup: • Identity: φ(e) = e' ⇒ e ∈ Ker(φ) • Closure: If φ(a) = e' and φ(b) = e', then φ(ab) = φ(a)φ(b) = e'e' = e' • Inverses: If φ(a) = e', then φ(a⁻¹) = (φ(a))⁻¹ = (e')⁻¹ = e' Ker(φ) is normal: For any g ∈ G and k ∈ Ker(φ): φ(gkg⁻¹) = φ(g)φ(k)φ(g⁻¹) = φ(g)e'φ(g)⁻¹ = φ(g)φ(g)⁻¹ = e' Therefore gkg⁻¹ ∈ Ker(φ), proving normality. 5. Theorem: Image is a Subgroup Identity: Since φ(e) = e', we have e' ∈ Im(φ). Closure: If a', b' ∈ Im(φ), then there exist a, b ∈ G with φ(a) = a' and φ(b) = b'. Then: a'b' = φ(a)φ(b) = φ(ab) ∈ Im(φ) Inverses: If a' ∈ Im(φ), then φ(a) = a' for some a ∈ G. Then: (a')⁻¹ = (φ(a))⁻¹ = φ(a⁻¹) ∈ Im(φ) 6. Example: Determinant Homomorphism This is SL₂(ℝ), the special linear group—all 2×2 matrices with determinant 1. It's a normal subgroup of GL₂(ℝ).

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