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 G. We shall see that an isomorphism 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 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 : Z Z n 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 2n so that θ is not one-to-one. However, θ is onto. Example 21.2 Define θ : Z Z by 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 = {g G : θ(g) = e H }. 1
Example 21.3 In Example 21.1, a Ker θ iff [a] = θ(a) = [0], i.e iff a = nq for some q Z > Thus,Ker θ = {nq : q Z}. In Example 21.2, a Ker θ iff 2a = θ(a) = 0, i.e. iff a = 0. Hence, Ker θ = {0}. 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 Ker θ and g G we have gxg 1 Ker θ. (i) By Theorem 18.2(i), θ(e G ) = e H so that e G Ker θ. Hence, Ker θ. Now, let x, y Ker θ. Then θ(xy 1 ) = θ(x)θ(y 1 ) = θ(x)(θ(y)) 1 = e H e 1 H = e H. Thus, xy 1 Ker θ. By Theorem 7.5, Ker θ is a subgroup of G. (ii) Let x Ker θ and g G. Then θ(gxg 1) = θ(g)θ(x)θ(g 1 ) = θ(g)e H (θ(g)) 1 = θ(g)(θ(g)) 1 = e H. Thus, gxg 1 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 H for all g G and h H. We write H G. Example 21.4 Let H be any subgroup of an Abelian group G. Since hg = gh for all g G and all h H then ghg 1 = h H for all g G and h H. That is, H G. Example 21.5 Let G = S 3 and H =< (12) >= {(1), (12)}. Since (123)(12)(123) 1 = (23) H then H is not a normal subgroup of G. Lemma 21.1 The following statements are equivalent: (i) gng 1 N for all n N and g G; (ii) g 1 ng N for all n N and g G; 2
(i) (ii): Suppose that gng 1 N for all n N and g G. In particular, g 1 n(g 1 ) 1 N since g 1 G. But (g 1 ) 1 = g so that g 1 ng N. (ii) (i): Suppose that g 1 ng N for all n N and g G. Since (g 1 ) 1 = g then gng 1 = (g 1 ) 1 ng 1 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. From Theorem 18.2 (iv), we know that θ(n) is a subgroup of H. Let y θ(n) and h H. Then y = θ(x) θ(n) for some x N and h = θ(g) for some g G (since θ is onto). But N G so that gxg 1 N. Thus, θ(gxg 1 ) θ(n). But θ(gxg 1 ) = θ(g)θ(x)θ(g 1 ) = hyh 1 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. Ker θ = {e G }. Then θ is one-to-one if and only if Suppose first that θ is one-to-one. Let x Ker θ. Then θ(x) = e H = θ(e G ). Hence, x = e G. Thus, Ker θ {e G }. Since θ(e G ) = e H then {e G } Ker θ. It follows that Ker θ = {e G }. Conversely, suppose that Ker θ = {e G }. Suppose that θ(x) = θ(y). Then e H = θ(x)(θ(y)) 1 = θ(x)θ(y 1 ) = θ(xy 1 ). Thus, xy 1 Ker θ. But then xy 1 = e G so that x = y. Review Problems Exercise 21.1 Define θ : Z 6 Z 3 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 : Z 3 Z 3 defined by θ([a] 3 ) = [a] 6 is not well-defined. (b) For which pairs m, n is β : Z n Z m, given by β([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 {1, 1, i, i} of complex numbers (operation multiplication). Define θ : Z G by θ(n) = i n. Show that θ is a homomorphism and determine Ker θ. Exercise 21.5 There is a unique homomorphism θ : Z 6 S 3 such that θ([1]) = (123). Determine θ([k]) for each [k] Z 6. Which elements are in Ker θ? Exercise 21.6 Prove that N G if and only if gn = Ng for all g 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 A n S n for all n 2. Exercise 21.9 Consider the subgroup H = A 3 = {(1), (123), (132)} of S 3. Let x = (12) and h = (123). Show that xh hx and xh = Hx. This shows that the equality xh = Hx does not mean that xh = hx for all x G and h H. Exercise 21.10 Prove that if C denote the collection of all normal subgroups of a group G. prove that N = H C H 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 S 3. Exercise 21.13 Let H be a subgroup of G and K G. (a) Prove that HK is a subgroup of G, where HK = {hk : h H and k K}. (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 = {e G } then hk = kh for all h H and k K. 4
Exercise 21.16 The center of the group G is defined by Prove that Z(G) G. Z(G) = {g G : xg = gx x G}. Exercise 21.17 Let G and H be groups. Prove that G {e H } is a normal subgroup of G H. Exercise 21.18 Let N be a normal subgroup of G, and let a, b, c, d 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 G. Prove that ak = {x G : θ(x) = θ(a)}. Exercise 21.21 Let S be any set, and let B be any proper subset of S. Let H = {θ Sym(S) : θ(b) = B}. Prove that H is a subgroup of Sym(S) that is not normal. Exercise 21.22 A group G is called simple if {e G } and G are the only normal subgroups of G. Prove that a cyclic group of prime order is simple. Exercise 21.23 Let U and V be nonabelian simple groups. Show that G = U V has precisely four different normal subgroups. 5