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 ¼ θ(G) where K = Ker θ: Moreover, Á ± ´ = θ where ´ 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 Á : G=K ¡! θ(G) by Á(Ka) = θ(a): This is a well-defined mapping. To see this, suppose that Ka = Kb: Since a = ea 2 Ka then a 2 Kb so that a = kb for some k 2 K: Hence, θ(a) = θ(kb) = θ(k)θ(b) = eH θ(b) = θ(b): Thus, Á(Ka) = Á(Kb): Next, we show that Á is a homomorphism. Let Ka; Kb 2 G=K: Then Á(KaKb) = Á(Kab) = θ(ab) = θ(a)θ(b) = Á(Ka)Á(Kb): To show that Á is one-to-one we use Theorem 21.2. That is we show that Ker Á = fKg: If Ka 2 Ker Á then Á(Ka) = eH . Thus, θ(a) = eH so that a 2 K: This implies that Ka = K: Now, if y 2 θ(G) then y = θ(a) = Á(Ka) for some a 2 G: This shows that Á is onto. Hence, Á is an ismomorphism and thus G=K ¼ θ(G): Finally, note that Á±´ and θ have the same domain, codomain and for all g 2 G, (Á ± ´)(g) = Á(´(g)) = Á(Ng) = θ(g): Thus, Á ± ´ = 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 >¼ 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 = fhk : h 2 H and k 2 Kg is a subgroup of G such that K/HK and H=H \ K ¼ HK=K: 1 Proof. First we show that HK is a subgroup of G: Since eG = eGeG and eG 2 H \ K then eG 2 HK so that HK 6= ;: Now, if a; b 2 HK then a = h1k1 and b = h2k2: ¡1 ¡1 ¡1 ¡1 ¡1 Thus, ab = h1k1k2 h2 = h1kh2 ; where k = k1k2 2 K: Since K/G then ¡1 ¡1 0 0 ¡1 ¡1 0 h2kh2 2 K so that h2kh2 = k for some k 2 K: Thus, kh2 = h2 k . This ¡1 ¡1 0 shows that ab = h1h2 k 2 HK: Hence, by Theorem 7.5, HK is a subgroup of G: Next, we show that K / HK: Since k = eH k 2 HK then K ⊆ HK: Consider the product (hk)k0(hk)¡1 = hkk0k¡1h¡1 = hk00h¡1 2 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 2 H and find θ(h1h2) = Kh1h2 = (Kh1)(Kh2) = θ(h1)θ(h2): Finally, we show that Ker θ = H \ K: Indeed, if x 2 H \ K then x 2 K and x 2 H: Thus, Kx = K: That is, θ(x) = K and so x 2 Ker θ: Conversely, if x 2 Ker θ then Kx = K and this implies that x 2 K: Since the kernel is a subset of H then x 2 H: Thus, x 2 H \ K and so Ker θ = H \ K: Applying Theorem 23.1, we obtain H=(H \ K) ¼ 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) ¼ G=H: Proof. First we prove that H=K is a subgroup of G=K: Since K 2 H=K then H=K 6= ;: ¡1 ¡1 ¡1 Since (Kh1)(Kh2) = (Kh1)(Kh2 ) = Kh1h2 2 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 2 H=K and Kg 2 G=K and find that (Kg)(Kh)(Kg)¡1 = KghKg¡1 = Kghg¡1 2 H=K since ghg¡1 2 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 2 K ⊆ H: Hence, g 2 Hg0 and Hg = Hg0: From the definition of θ; we have that θ is onto. θ is a homomorphism: If Kg1; Kg2 2 G=K then θ(Kg1Kg2) = θ(Kg1g2) = Hg1g2 = (Hg1)(Hg2) = θ(Kg1)θ(Kg2): Ker θ = H=K : If Kg 2 Ker θ then θ(Kg) = H; i.e. Hg = H so that g 2 H: Thus, Kg 2 H=K and this shows that Ker θ ⊆ H=K: Now, if Kh 2 H=K then θ(Kh) = Hh = H so that Kh 2 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) ¼ 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=feg ¼ G: Exercise 23.3 For any groups A and B, prove the following. (a) A ¼ A £ feg /A £ B: A£B (b) A£feg ¼ B: Exercise 23.4 Prove that if G is a simple Abelian group then G ¼ ZZp for some prime number p: Exercise 23.5 ZZ18 Prove that <[3]> ¼ ZZ3: Exercise 23.6 Prove that if θ is a homomorphism of G onto H, B / H; and A = fg 2 G : θ(g) 2 Bg, 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 Á defined by Á(S) = S=H; where S 2 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 = feg: Prove that G ¼ 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) = ffa : a 2 Gg is a group under composition. (c) Prove that Inn(G) ¼ GZZ(G); where Z(G) is the center of G: Exercise 23.11 Prove that 6ZZ ¼ 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 jHj 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 jHj jKj jHKj = : jH \ 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 = feGg: Prove that H is isomorphic to a subgroup of G=K: Prove that if G = HK then G=K ¼ H: Exercise 23.17 Suppose that Á : G ! H is an epimorphism and N / G: Prove that there exists a homomorphism from G=N onto H=Á(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 ¼ G=K £ H=K : Exercise 23.19 Let M and n be relatively prime. Show that Á : ZZ £ ZZm £ ZZn, given by Á(a) = ([a]m; [a]n) is a homomorphism, onto, and Ker Á =< mn > : Exercise 23.20 Prove that if m and n are relatively prime then ZZmn ¼ ZZm £ ZZn: 4.