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 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 ∈ 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 = {g ∈ G : θ(g) = eH }.

1 Example 21.3 In Example 21.1, a ∈ Ker θ iff [a] = θ(a) = [0], i.e iff a = nq for some q ∈ ZZ > Thus,Ker θ = {nq : q ∈ ZZ}. 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.

Proof. (i) By Theorem 18.2(i), θ(eG) = eH so that eG ∈ Ker θ. Hence, Ker θ 6= ∅. Now, let x, y ∈ Ker θ. Then

−1 −1 −1 −1 θ(xy ) = θ(x)θ(y ) = θ(x)(θ(y)) = eH eH = eH . 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

−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 ∈ 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 −1 Let G = S3 and H =< (12) >= {(1), (12)}. Since (123)(12)(123) = (23) 6∈ 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−1ng ∈ N for all n ∈ N and g ∈ G;

2 Proof. (i) → (ii): Suppose that gng−1 ∈ N for all n ∈ N and g ∈ G. In particular, g−1n(g−1)−1 ∈ N since g−1 ∈ G. But (g−1)−1 = g so that g−1ng ∈ N. (ii) → (i): Suppose that g−1ng ∈ N for all n ∈ N and g ∈ G. Since (g−1)−1 = g then gng−1 = (g−1)−1ng−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. Proof. 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. Then θ is one-to-one if and only if Ker θ = {eG}. Proof. Suppose first that θ is one-to-one. Let x ∈ Ker θ. Then θ(x) = eH = θ(eG). Hence, x = eG. Thus, Ker θ ⊆ {eG}. Since θ(eG) = eH then {eG} ⊆ Ker θ. It follows that Ker θ = {eG}. Conversely, suppose that Ker θ = {eG}. 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 ∈ 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 β : ZZn → ZZm, 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 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] ∈ ZZ6. 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 An /Sn for all n ≥ 2.

Exercise 21.9 Consider the subgroup H = A3 = {(1), (123), (132)} 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 ∈ 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∈CH 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 = {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 = {eG} then hk = kh for all h ∈ H and k ∈ K.

4 Exercise 21.16 The center of the group G is defined by

Z(G) = {g ∈ G : xg = gx∀x ∈ G}.

Prove that Z(G) / G.

Exercise 21.17 Let G and H be groups. Prove that G × {eH } 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 {eG} 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.

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