Abstract Algebra Theory and Applications - Computer Science ...

9.2 GROUP HOMOMORPHISMS 1574. If H 2 is a subgroup of G 2 , then φ −1 (H 2 ) = {g ∈ G : φ(g) ∈ H 2 } is asubgroup of G 1 . Furthermore, if H 2 is normal in G 2 , then φ −1 (H 2 ) isnormal in G 1 .Proof. (1) Suppose that e and e ′ are the identities of G 1 and G 2 , respectively;thene ′ φ(e) = φ(e) = φ(ee) = φ(e)φ(e).By cancellation, φ(e) = e ′ .(2) This statement follows from the fact thatφ(g −1 )φ(g) = φ(g −1 g) = φ(e) = e.(3) The set φ(H 1 ) is nonempty since the identity of H 2 is in φ(H 1 ).Suppose that H 1 is a subgroup of G 1 and let x and y be in φ(H 1 ). Thereexist elements a, b ∈ H 1 such that φ(a) = x and φ(b) = y. Sincexy −1 = φ(a)[φ(b)] −1 = φ(ab −1 ) ∈ φ(H 1 ),φ(H 1 ) is a subgroup of G 2 by Proposition 2.10.(4) Let H 2 be a subgroup of G 2 and define H 1 to be φ −1 (H 2 ); that is,H 1 is the set of all g ∈ G 1 such that φ(g) ∈ H 2 . The identity is in H 1 sinceφ(e) = e. If a and b are in H 1 , then φ(ab −1 ) = φ(a)[φ(b)] −1 is in H 2 since H 2is a subgroup of G 2 . Therefore, ab −1 ∈ H 1 and H 1 is a subgroup of G 1 . IfH 2 is normal in G 2 , we must show that g −1 hg ∈ H 1 for h ∈ H 1 and g ∈ G 1 .Butφ(g −1 hg) = [φ(g)] −1 φ(h)φ(g) ∈ H 2 ,since H 2 is a normal subgroup of G 2 . Therefore, g −1 hg ∈ H 1 .Let φ : G → H be a group homomorphism and suppose that e is theidentity of H. By Proposition 9.3, φ −1 ({e}) is a subgroup of G. Thissubgroup is called the kernel of φ and will be denoted by ker φ. In fact, thissubgroup is a normal subgroup of G since the trivial subgroup is normal inH. We state this result in the following theorem, which says that with everyhomomorphism of groups we can naturally associate a normal subgroup.Theorem 9.4 Let φ : G → H be a group homomorphism. Then the kernelof φ is a normal subgroup of G.Example 9. Let us examine the homomorphism φ : GL 2 (R) → R ∗ definedby A ↦→ det(A). Since 1 is the identity of R ∗ , the kernel of this homomorphismis all 2×2 matrices having determinant one. That is, ker φ = SL 2 (R).□