Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan

23 Isomorphism Theorems

Theorem 22.2 shows that each quotient group of a group G is the homomorphic image of G. The theorem below shows that the converse is also true. That is, each homomorphic image is isomorphic to a quotient group. Theorem 23.1 (The Fundamental Homomorphism Theorem) Let θ : G −→ H be a homomorphism. Then G/K ≈ θ(G) where K = Ker θ. Moreover, φ ◦ η = θ where η is the natural homomorphism introduced in Theorem 22.2 Proof. By Theorem 21.1, K = Ker θ / G so that by Theorem 22.1, G/K is a group. Define φ : G/K −→ θ(G) by φ(Ka) = θ(a). This is a well-defined mapping. To see this, suppose that Ka = Kb. Since a = ea ∈ Ka then a ∈ Kb so that a = kb for some k ∈ K. Hence, θ(a) = θ(kb) = θ(k)θ(b) = eH θ(b) = θ(b). Thus, φ(Ka) = φ(Kb). Next, we show that φ is a homomorphism. Let Ka, Kb ∈ G/K. Then φ(KaKb) = φ(Kab) = θ(ab) = θ(a)θ(b) = φ(Ka)φ(Kb). To show that φ is one-to-one we use Theorem 21.2. That is we show that Ker φ = {K}. If Ka ∈ Ker φ then φ(Ka) = eH . Thus, θ(a) = eH so that a ∈ K. This implies that Ka = K. Now, if y ∈ θ(G) then y = θ(a) = φ(Ka) for some a ∈ G. This shows that φ is onto. Hence, φ is an ismomorphism and thus G/K ≈ θ(G). Finally, note that φ◦η and θ have the same domain, codomain and for all g ∈ G, (φ ◦ η)(g) = φ(η(g)) = φ(Ng) = θ(g). Thus, φ ◦ η = θ. This completes a proof of the theorem. Example 23.1 Looking at Example 22.3, we see that the mapping θ : ZZ −→ ZZn defined by θ(a) = [a] satisfies the assumptions of Theorem 23.1. Hence, ZZ/ < n >≈ ZZn As a consequence of Theorem 23.1, we have the following two theorems. Theorem 23.2 (First Isomorphism Theorem) Let H and K be subgroups of a group G such that K / G. Then the set HK = {hk : h ∈ H and k ∈ K} is a subgroup of G such that K/HK and H/H ∩ K ≈ HK/K.

1 Proof. First we show that HK is a subgroup of G. Since eG = eGeG and eG ∈ H ∩ K then eG ∈ HK so that HK 6= ∅. Now, if a, b ∈ HK then a = h1k1 and b = h2k2. −1 −1 −1 −1 −1 Thus, ab = h1k1k2 h2 = h1kh2 , where k = k1k2 ∈ K. Since K/G then −1 −1 0 0 −1 −1 0 h2kh2 ∈ K so that h2kh2 = k for some k ∈ K. Thus, kh2 = h2 k . This −1 −1 0 shows that ab = h1h2 k ∈ HK. Hence, by Theorem 7.5, HK is a subgroup of G. Next, we show that K/HK. Since k = eH k ∈ HK then K ⊆ HK. Consider the product (hk)k0(hk)−1 = hkk0k−1h−1 = hk00h−1 ∈ K since K / G. Hence, K/HK. Now, consider the mapping θ : H −→ HK/K defined by θ(h) = Kh. This is a well-defined mapping: if h1 = h2 then Kh1 = Kh2. This mapping is onto from its definition. To see that θ is a homomorphism, we take h1, h2 ∈ H and find

θ(h1h2) = Kh1h2 = (Kh1)(Kh2) = θ(h1)θ(h2).

Finally, we show that Ker θ = H ∩ K. Indeed, if x ∈ H ∩ K then x ∈ K and x ∈ H. Thus, Kx = K. That is, θ(x) = K and so x ∈ Ker θ. Conversely, if x ∈ Ker θ then Kx = K and this implies that x ∈ K. Since the kernel is a subset of H then x ∈ H. Thus, x ∈ H ∩ K and so Ker θ = H ∩ K. Applying Theorem 23.1, we obtain H/(H ∩ K) ≈ HK/K.

Theorem 23.3 (Second Isomorphism Theorem) Assume that H,K/G with K/H. Then H/K/G/K and (G/K)/(H/K) ≈ G/H.

Proof. First we prove that H/K is a subgroup of G/K. Since K ∈ H/K then H/K 6= ∅. −1 −1 −1 Since (Kh1)(Kh2) = (Kh1)(Kh2 ) = Kh1h2 ∈ H/K then by Theorem 7.5, H/K is a subgroup of G/K. To see that H/K/G/K we pick elements Kh ∈ H/K and Kg ∈ G/K and find that

(Kg)(Kh)(Kg)−1 = KghKg−1 = Kghg−1 ∈ H/K since ghg−1 ∈ H (H / G.) Next, we define θ : G/K −→ G/H by θ(Kg) = Hg. This is well-defined map for if Kg = Kg0 then g = kg0 for some k ∈ K ⊆ H. Hence, g ∈ Hg0 and Hg = Hg0. From the definition of θ, we have that θ is onto. θ is a homomorphism: If Kg1, Kg2 ∈ G/K then θ(Kg1Kg2) = θ(Kg1g2) = Hg1g2 = (Hg1)(Hg2) = θ(Kg1)θ(Kg2). Ker θ = H/K : If Kg ∈ Ker θ then θ(Kg) = H, i.e. Hg = H so that g ∈ H. Thus, Kg ∈ H/K and this shows that Ker θ ⊆ H/K. Now, if Kh ∈ H/K then θ(Kh) = Hh = H so that Kh ∈ Ker θ. Hence, H/K ⊆ Ker θ and this shows that Ker θ = H/K. By Theorem 23.1, we have

(G/K)/(H/K) ≈ G/H.

2 Review Problems

Exercise 23.1 Find all the homomorphic images of S3.

Exercise 23.2 Prove that if G is any group with identity e then G/{e} ≈ G.

Exercise 23.3 For any groups A and B, prove the following.

(a) A ≈ A × {e} /A × B. A×B (b) A×{e} ≈ B.

Exercise 23.4 Prove that if G is a simple Abelian group then G ≈ ZZp for some prime number p.

Exercise 23.5 ZZ18 Prove that <[3]> ≈ ZZ3.

Exercise 23.6 Prove that if θ is a homomorphism of G onto H, B/H, and A = {g ∈ G : θ(g) ∈ B}, then A / G.

Exercise 23.7 Give an example to show that if A and B are subgroups of a group G then AB need not be a subgroup of G.

Exercise 23.8 Suppose that N / G. Let C be the collection of all subgroups of G containing H. Prove that the map φ defined by φ(S) = S/H, where S ∈ C is one-to-one and onto.

Exercise 23.9 Let G be a group with normal subgroups H and K such that G = HK and H ∩ K = {e}. Prove that G ≈ H × K.

Exercise 23.10 Let G be a group.

−1 (a) Prove that fa : G → G, given by fa(x) = axa is an isomorphism. (b) Prove that the set Inn(G) = {fa : a ∈ G} is a group under composition. (c) Prove that Inn(G) ≈ GZZ(G), where Z(G) is the center of G.

Exercise 23.11 Prove that 6ZZ ≈ 2ZZ. Hint:Use First Isomorphism Theorem. 12ZZ 4ZZ

3 Exercise 23.12 Prove that if H is a normal subgroup of G of prime index p then for all subgroups K of G either (i) K is a subgroup of H or (ii) G = HK and [K : K ∩ H] = p.

Exercise 23.13 Let G be a finite group, H and N are subgroups of G such that N / G. Prove that if |H| and [G : N] are relatively prime then H is a subgroup of N.

Exercise 23.14 Let G be a finite group and let H and K be subgroups of G with K / G. Prove that |H| |K| |HK| = . |H ∩ K

Exercise 23.15 Let G be a group and let H and K be normal subgroups of G with K ⊆ H. Suppose that G/K is cyclic. Prove that G/H and H/K are cyclic.

Exercise 23.16 Suppose that K/G and H is a subgroup of G such that H∩K = {eG}. Prove that H is isomorphic to a subgroup of G/K. Prove that if G = HK then G/K ≈ H.

Exercise 23.17 Suppose that φ : G → H is an epimorphism and N / G. Prove that there exists a homomorphism from G/N onto H/φ(N).

Exercise 23.18 Let G and H be groups and let K/G and K0 /H.

(a) Prove that K × K0 /G × H. G×H 0 (b) Prove that K×K0 ≈ G/K × H/K . Exercise 23.19 Let M and n be relatively prime. Show that φ : ZZ × ZZm × ZZn, given by φ(a) = ([a]m, [a]n) is a homomorphism, onto, and Ker φ =< mn > .

Exercise 23.20 Prove that if m and n are relatively prime then ZZmn ≈ ZZm × ZZn.

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