Abstract Algebra Theory and Applications - Computer Science ...

162 CHAPTER 9 HOMOMORPHISMS AND FACTOR GROUPS9.3 The Isomorphism TheoremsThough at first it is not evident that factor groups correspond exactly tohomomorphic images, we can use factor groups to study homomorphisms.We already know that with every group homomorphism φ : G → H we canassociate a normal subgroup of G, ker φ; the converse is also true. Everynormal subgroup of a group G gives rise to homomorphism of groups.Let H be a normal subgroup of G. Define the natural or canonicalhomomorphismφ : G → G/Hbyφ(g) = gH.This is indeed a homomorphism, sinceφ(g 1 g 2 ) = g 1 g 2 H = g 1 Hg 2 H = φ(g 1 )φ(g 2 ).The kernel of this homomorphism is H. The following theorems describe therelationships among group homomorphisms, normal subgroups, and factorgroups.Theorem 9.9 (First Isomorphism Theorem) If ψ : G → H is a grouphomomorphism with K = ker ψ, then K is normal in G. Let φ : G → G/Kbe the canonical homomorphism. Then there exists a unique isomorphismη : G/K → ψ(G) such that ψ = ηφ.Proof. We already know that K is normal in G. Define η : G/K → ψ(G)by η(gK) = ψ(g). We must first show that this is a well-defined map.Suppose that g 1 K = g 2 K. For some k ∈ K, g 1 k = g 2 ; consequently,η(g 1 K) = ψ(g 1 ) = ψ(g 1 )ψ(k) = ψ(g 1 k) = ψ(g 2 ) = η(g 2 K).Since η(g 1 K) = η(g 2 K), η does not depend on the choice of coset representative.Clearly η is onto ψ(G). To show that η is one-to-one, suppose thatη(g 1 K) = η(g 2 K). Then ψ(g 1 ) = ψ(g 2 ). This implies that ψ(g1 −1 g 2) = e,or g1 −1 g 2 is in the kernel of ψ; hence, g1 −1 g 2K = K; that is, g 1 K = g 2 K.Finally, we must show that η is a homomorphism, butη(g 1 Kg 2 K) = η(g 1 g 2 K)= ψ(g 1 g 2 )= ψ(g 1 )ψ(g 2 )= η(g 1 K)η(g 2 K).