Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan 21 Homomorphisms and Normal Subgroups Recall that an isomorphism is a function θ : G ¡! H such that θ is one-to-one, onto and such that θ(ab) = θ(a)θ(b) for all a; b 2 G: We shall see that an isomor- phism is simply a special type of function called a group homomorphism. We will also see a relationship between group homomorphisms and normal subgroups. Definition 21.1 A function θ from a group G to a group H is said to be a homomorphism provided that for all a; b 2 G we have that θ(ab) = θ(a)θ(b): If θ : G ¡! H is a one-to-one homomorphism, we call θ a monomorphism and if θ : G ¡! H is an onto homomorphism, then we call θ an epimorphism. Of course, a bijective homomorphism is an isomorphism. Example 21.1 Define θ : ZZ ¡! ZZn by θ(a) = [a]: Then θ(a + b) = [a + b] = [a] © [b] = θ(a) © θ(b); so that θ is a homomorphism. Note that θ(n) = θ(2n) with n 6= 2n so that θ is not one-to-one. However, θ is onto. Example 21.2 Define θ : ZZ ¡! ZZby θ(a) = 2a: Then θ(a + b) = 2(a + b) = 2a + 2b = θ(a) + θ(b); so that θ is a homomorphism. Note that θ is not onto since there is no integer n that satisfies θ(n) = 3: However, θ is one-to-one since θ(n) = θ(m) implies 2n = 2m and this in turn implies that n = m. We have seen that the range of a homomorphism is a subgroup of the codomain. (Theorem 18.2(iv)). The following subset determines a subgroup of the domain of a homomorphism. Definition 21.2 Let θ : G ¡! H be a group homomorphism. Then the kernel of θ is the set Ker θ = fg 2 G : θ(g) = eH g: 1 Example 21.3 In Example 21.1, a 2 Ker θ iff [a] = θ(a) = [0], i.e iff a = nq for some q 2 ZZ > Thus,Ker θ = fnq : q 2 ZZg. In Example 21.2, a 2 Ker θ iff 2a = θ(a) = 0; i.e. iff a = 0: Hence, Ker θ = f0g: As point it out earlier the kernel is a subgroup of the domain. Theorem 21.1 Let θ : G ¡! H be a homomorphism. Then (i) Ker θ is a subgroup of G: (ii) For any x 2 Ker θ and g 2 G we have gxg¡1 2 Ker θ. Proof. (i) By Theorem 18.2(i), θ(eG) = eH so that eG 2 Ker θ. Hence, Ker θ 6= ;: Now, let x; y 2 Ker θ: Then ¡1 ¡1 ¡1 ¡1 θ(xy ) = θ(x)θ(y ) = θ(x)(θ(y)) = eH eH = eH : Thus, xy¡1 2 Ker θ: By Theorem 7.5, Ker θ is a subgroup of G: (ii) Let x 2 Ker θ and g 2 G. Then ¡1 ¡1 ¡1 θ(gxg¡1) = θ(g)θ(x)θ(g ) = θ(g)eH (θ(g)) = θ(g)(θ(g)) = eH : Thus, gxg¡1 2 Ker θ: Theorem 21.1(ii) is one of the common properties that kernels share: They are all normals in the sense of the following definition. Definition 21.3 Let H be a subgroup of a group G: Then H is normal iff ghg¡1 2 H for all g 2 G and h 2 H: We write H / G: Example 21.4 Let H be any subgroup of an Abelian group G: Since hg = gh for all g 2 G and all h 2 H then ghg¡1 = h 2 H for all g 2 G and h 2 H: That is, H / G: Example 21.5 ¡1 Let G = S3 and H =< (12) >= f(1); (12)g: Since (123)(12)(123) = (23) 62 H then H is not a normal subgroup of G: Lemma 21.1 The following statements are equivalent: (i) gng¡1 2 N for all n 2 N and g 2 G; (ii) g¡1ng 2 N for all n 2 N and g 2 G; 2 Proof. (i) ! (ii): Suppose that gng¡1 2 N for all n 2 N and g 2 G: In particular, g¡1n(g¡1)¡1 2 N since g¡1 2 G: But (g¡1)¡1 = g so that g¡1ng 2 N: (ii) ! (i): Suppose that g¡1ng 2 N for all n 2 N and g 2 G: Since (g¡1)¡1 = g then gng¡1 = (g¡1)¡1ng¡1 2 N: The following lemma shows that the homomorphic image of a normal subgroup is normal for onto maps. Lemma 21.2 Let θ : G ! H be an epimorphism and N / G: Then θ(N) / H: Proof. From Theorem 18.2 (iv), we know that θ(N) is a subgroup of H: Let y 2 θ(N) and h 2 H: Then y = θ(x) 2 θ(N) for some x 2 N and h = θ(g) for some g 2 G (since θ is onto). But N/G so that gxg¡1 2 N: Thus, θ(gxg¡1) 2 θ(N): But θ(gxg¡1) = θ(g)θ(x)θ(g¡1) = hyh¡1 2 θ(N): Hence, θ(N) / H: The following theorem describes a commonly used way for testing whether a homomorphism is one-to-one or not. Theorem 21.2 Let θ : G ¡! H be a homomorphism. Then θ is one-to-one if and only if Ker θ = feGg: Proof. Suppose first that θ is one-to-one. Let x 2 Ker θ: Then θ(x) = eH = θ(eG): Hence, x = eG: Thus, Ker θ ⊆ feGg: Since θ(eG) = eH then feGg ⊆ Ker θ: It follows that Ker θ = feGg: Conversely, suppose that Ker θ = feGg: Suppose ¡1 ¡1 ¡1 that θ(x) = θ(y): Then eH = θ(x)(θ(y)) = θ(x)θ(y ) = θ(xy ): Thus, ¡1 ¡1 xy 2 Ker θ. But then xy = eG so that x = y: Review Problems Exercise 21.1 Define θ : ZZ6 ! ZZ3 by θ([a]6) = [a]3: (a) Prove that θ is well-defined. (b) Prove that θ is a homomorphism. (c) Find Ker θ: Exercise 21.2 (a) Prove that θ : ZZ3 ! ZZ3 defined by θ([a]3) = [a]6 is not well-defined. (b) For which pairs m; n is ¯ : ZZn ! ZZm, given by ¯([a]n) = [a]m, well-defined? Exercise 21.3 (a) Prove that every homomorphic image of an Abelian group is Abelian. (b) Prove that every homomorphic image of a cyclic group is cyclic. 3 Exercise 21.4 Let G denote the subgroup f1; ¡1; i; ¡ig of complex numbers (operation multi- plication). Define θ : ZZ ! G by θ(n) = in: Show that θ is a homomorphism and determine Ker θ: Exercise 21.5 There is a unique homomorphism θ : ZZ6 ! S3 such that θ([1]) = (123): Deter- mine θ([k]) for each [k] 2 ZZ6: Which elements are in Ker θ? Exercise 21.6 Prove that N/G if and only if gN = Ng for all g 2 G: Exercise 21.7 Prove that if N is a subgroup of G such that [G : N] = 2 then N / G: Exercise 21.8 Prove that An /Sn for all n ¸ 2: Exercise 21.9 Consider the subgroup H = A3 = f(1); (123); (132)g of S3: Let x = (12) and h = (123): Show that xh 6= hx and xH = Hx: This shows that the equality xH = Hx does not mean that xh = hx for all x 2 G and h 2 H: Exercise 21.10 Prove that if C denote the collection of all normal subgroups of a group G: prove that N = \H2CH is also a normal subgroup of G: Exercise 21.11 Prove that if N/G then for any subgroup H of G, we have H \ N / H: Exercise 21.12 Find all normal subgroups of S3: Exercise 21.13 Let H be a subgroup of G and K/G. (a) Prove that HK is a subgroup of G; where HK = fhk : h 2 H and k 2 Kg: (b) Prove that HK = KH: (c) Prove that K / HK: Exercise 21.14 Prove that if H and K are normal subgroups of G then HK is a normal subgroup of G: Exercise 21.15 Prove that if H and K are normal subgroups of G such that H \ K = feGg then hk = kh for all h 2 H and k 2 K: 4 Exercise 21.16 The center of the group G is defined by Z(G) = fg 2 G : xg = gx8x 2 Gg: Prove that Z(G) / G: Exercise 21.17 Let G and H be groups. Prove that G £ feH g is a normal subgroup of G £ H: Exercise 21.18 Let N be a normal subgroup of G; and let a; b; c; d 2 G: prove that if aN = cN and bN = dN then abN = cdN: Exercise 21.19 Let G be a non-abelian group of order 8. Prove that G has at least one element of order 4. Hence prove that G has a normal cyclic subgroup of order 4. Exercise 21.20 Suppose that θ : G ! H is a homomorphism. Let K = Ker θ and a 2 G: Prove that aK = fx 2 G : θ(x) = θ(a)g: Exercise 21.21 Let S be any set, and let B be any proper subset of S: Let H = fθ 2 Sym(S): θ(B) = Bg: Prove that H is a subgroup of Sym(S) that is not normal.