Section 34 Isomorphism theorems Instructor: Yifan Yang Fall 2006 - - PowerPoint PPT Presentation

Section 34 – Isomorphism theorems

Instructor: Yifan Yang Fall 2006

Instructor: Yifan Yang Section 34 – Isomorphism theorems

First isomorphism theorem

Theorem (34.2, First isomorphism theorem) Let φ : G → G′ be a homomorphism with kernel K. Then G/K ≃ φ[G], and a canonical isomorphism is µ : G/K → φ[G] defined by µ(gK) = φ(g). Proof. See Theorem 14.11.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

First isomorphism theorem

Theorem (34.2, First isomorphism theorem) Let φ : G → G′ be a homomorphism with kernel K. Then G/K ≃ φ[G], and a canonical isomorphism is µ : G/K → φ[G] defined by µ(gK) = φ(g). Proof. See Theorem 14.11.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Join of two subgroups

Definition Let H and N be subgroups of a group G. The join H ∨ N of H and N is the smallest subgroup of G that contains both H and

• N. Equivalently, H ∨ N is the intersection of all subgroups of G

containing H and N. Lemma (34.4) If N is a normal subgroup of G, and H is any subgroup of G, then H ∨ N = {hn : h ∈ H, n ∈ N} = {nh : n ∈ N, h ∈ H}. Remark If neither H nor N is normal in G, then HN may not be a subgroup at all. For example, let G = S3, H = {e, (1, 2)}, N = {e, (1, 3)}. Then HN = {e, (1, 2), (1, 3), (1, 3, 2)}, which is not a subgroup.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Join of two subgroups

Definition Let H and N be subgroups of a group G. The join H ∨ N of H and N is the smallest subgroup of G that contains both H and

• N. Equivalently, H ∨ N is the intersection of all subgroups of G

containing H and N. Lemma (34.4) If N is a normal subgroup of G, and H is any subgroup of G, then H ∨ N = {hn : h ∈ H, n ∈ N} = {nh : n ∈ N, h ∈ H}. Remark If neither H nor N is normal in G, then HN may not be a subgroup at all. For example, let G = S3, H = {e, (1, 2)}, N = {e, (1, 3)}. Then HN = {e, (1, 2), (1, 3), (1, 3, 2)}, which is not a subgroup.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Join of two subgroups

Definition Let H and N be subgroups of a group G. The join H ∨ N of H and N is the smallest subgroup of G that contains both H and

• N. Equivalently, H ∨ N is the intersection of all subgroups of G

containing H and N. Lemma (34.4) If N is a normal subgroup of G, and H is any subgroup of G, then H ∨ N = {hn : h ∈ H, n ∈ N} = {nh : n ∈ N, h ∈ H}. Remark If neither H nor N is normal in G, then HN may not be a subgroup at all. For example, let G = S3, H = {e, (1, 2)}, N = {e, (1, 3)}. Then HN = {e, (1, 2), (1, 3), (1, 3, 2)}, which is not a subgroup.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Proof of Lemma 34.4

Let HN denote the set {hn : h ∈ H, n ∈ N}. We first prove that HN is a subgroup of G.

1

Closed: Let h1n1, h2n2 ∈ HN. Since N is a normal subgroup, h−1

2 n1h2 ∈ N, i.e., n1h2 = h2n3 for some n3 ∈ N.

Then h1n1h2n2 = h1h2n3n2 ∈ HN.

2

Identity: This is obvious.

3

Inverse: Let hn ∈ HN. We have (hn)−1 = n−1h−1. Again, since n is normal, n−1h−1 = h−1n1 for some n1 ∈ N. Thus, (hn)−1 ∈ HN. Now HN is a subgroup and clearly contains both H and N. Thus H ∨ N ⊂ HN. On the other hand, any subgroup containing both H and N must also contain all elements of the form hn, h ∈ H, n ∈ N. In particular, we have HN ⊂ H ∨ N. We conclude that H ∨ N = HN.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Proof of Lemma 34.4

Let HN denote the set {hn : h ∈ H, n ∈ N}. We first prove that HN is a subgroup of G.

1

Closed: Let h1n1, h2n2 ∈ HN. Since N is a normal subgroup, h−1

2 n1h2 ∈ N, i.e., n1h2 = h2n3 for some n3 ∈ N.

Then h1n1h2n2 = h1h2n3n2 ∈ HN.

2

Identity: This is obvious.

3

Inverse: Let hn ∈ HN. We have (hn)−1 = n−1h−1. Again, since n is normal, n−1h−1 = h−1n1 for some n1 ∈ N. Thus, (hn)−1 ∈ HN. Now HN is a subgroup and clearly contains both H and N. Thus H ∨ N ⊂ HN. On the other hand, any subgroup containing both H and N must also contain all elements of the form hn, h ∈ H, n ∈ N. In particular, we have HN ⊂ H ∨ N. We conclude that H ∨ N = HN.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Proof of Lemma 34.4

Let HN denote the set {hn : h ∈ H, n ∈ N}. We first prove that HN is a subgroup of G.

1

Closed: Let h1n1, h2n2 ∈ HN. Since N is a normal subgroup, h−1

2 n1h2 ∈ N, i.e., n1h2 = h2n3 for some n3 ∈ N.

Then h1n1h2n2 = h1h2n3n2 ∈ HN.

2

Identity: This is obvious.

3

Inverse: Let hn ∈ HN. We have (hn)−1 = n−1h−1. Again, since n is normal, n−1h−1 = h−1n1 for some n1 ∈ N. Thus, (hn)−1 ∈ HN. Now HN is a subgroup and clearly contains both H and N. Thus H ∨ N ⊂ HN. On the other hand, any subgroup containing both H and N must also contain all elements of the form hn, h ∈ H, n ∈ N. In particular, we have HN ⊂ H ∨ N. We conclude that H ∨ N = HN.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Proof of Lemma 34.4

Let HN denote the set {hn : h ∈ H, n ∈ N}. We first prove that HN is a subgroup of G.

1

Closed: Let h1n1, h2n2 ∈ HN. Since N is a normal subgroup, h−1

2 n1h2 ∈ N, i.e., n1h2 = h2n3 for some n3 ∈ N.

Then h1n1h2n2 = h1h2n3n2 ∈ HN.

2

Identity: This is obvious.

3

Inverse: Let hn ∈ HN. We have (hn)−1 = n−1h−1. Again, since n is normal, n−1h−1 = h−1n1 for some n1 ∈ N. Thus, (hn)−1 ∈ HN. Now HN is a subgroup and clearly contains both H and N. Thus H ∨ N ⊂ HN. On the other hand, any subgroup containing both H and N must also contain all elements of the form hn, h ∈ H, n ∈ N. In particular, we have HN ⊂ H ∨ N. We conclude that H ∨ N = HN.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Proof of Lemma 34.4

Let HN denote the set {hn : h ∈ H, n ∈ N}. We first prove that HN is a subgroup of G.

1

Closed: Let h1n1, h2n2 ∈ HN. Since N is a normal subgroup, h−1

2 n1h2 ∈ N, i.e., n1h2 = h2n3 for some n3 ∈ N.

Then h1n1h2n2 = h1h2n3n2 ∈ HN.

2

Identity: This is obvious.

3

Inverse: Let hn ∈ HN. We have (hn)−1 = n−1h−1. Again, since n is normal, n−1h−1 = h−1n1 for some n1 ∈ N. Thus, (hn)−1 ∈ HN. Now HN is a subgroup and clearly contains both H and N. Thus H ∨ N ⊂ HN. On the other hand, any subgroup containing both H and N must also contain all elements of the form hn, h ∈ H, n ∈ N. In particular, we have HN ⊂ H ∨ N. We conclude that H ∨ N = HN.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Proof of Lemma 34.4

Let HN denote the set {hn : h ∈ H, n ∈ N}. We first prove that HN is a subgroup of G.

1

Closed: Let h1n1, h2n2 ∈ HN. Since N is a normal subgroup, h−1

2 n1h2 ∈ N, i.e., n1h2 = h2n3 for some n3 ∈ N.

Then h1n1h2n2 = h1h2n3n2 ∈ HN.

2

Identity: This is obvious.

3

Inverse: Let hn ∈ HN. We have (hn)−1 = n−1h−1. Again, since n is normal, n−1h−1 = h−1n1 for some n1 ∈ N. Thus, (hn)−1 ∈ HN. Now HN is a subgroup and clearly contains both H and N. Thus H ∨ N ⊂ HN. On the other hand, any subgroup containing both H and N must also contain all elements of the form hn, h ∈ H, n ∈ N. In particular, we have HN ⊂ H ∨ N. We conclude that H ∨ N = HN.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Proof of Lemma 34.4

Let HN denote the set {hn : h ∈ H, n ∈ N}. We first prove that HN is a subgroup of G.

1

Closed: Let h1n1, h2n2 ∈ HN. Since N is a normal subgroup, h−1

2 n1h2 ∈ N, i.e., n1h2 = h2n3 for some n3 ∈ N.

Then h1n1h2n2 = h1h2n3n2 ∈ HN.

2

Identity: This is obvious.

3

Inverse: Let hn ∈ HN. We have (hn)−1 = n−1h−1. Again, since n is normal, n−1h−1 = h−1n1 for some n1 ∈ N. Thus, (hn)−1 ∈ HN. Now HN is a subgroup and clearly contains both H and N. Thus H ∨ N ⊂ HN. On the other hand, any subgroup containing both H and N must also contain all elements of the form hn, h ∈ H, n ∈ N. In particular, we have HN ⊂ H ∨ N. We conclude that H ∨ N = HN.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Proof of Lemma 34.4

Let HN denote the set {hn : h ∈ H, n ∈ N}. We first prove that HN is a subgroup of G.

1

Closed: Let h1n1, h2n2 ∈ HN. Since N is a normal subgroup, h−1

2 n1h2 ∈ N, i.e., n1h2 = h2n3 for some n3 ∈ N.

Then h1n1h2n2 = h1h2n3n2 ∈ HN.

2

Identity: This is obvious.

3

Inverse: Let hn ∈ HN. We have (hn)−1 = n−1h−1. Again, since n is normal, n−1h−1 = h−1n1 for some n1 ∈ N. Thus, (hn)−1 ∈ HN. Now HN is a subgroup and clearly contains both H and N. Thus H ∨ N ⊂ HN. On the other hand, any subgroup containing both H and N must also contain all elements of the form hn, h ∈ H, n ∈ N. In particular, we have HN ⊂ H ∨ N. We conclude that H ∨ N = HN.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Proof of Lemma 34.4

Let HN denote the set {hn : h ∈ H, n ∈ N}. We first prove that HN is a subgroup of G.

1

Closed: Let h1n1, h2n2 ∈ HN. Since N is a normal subgroup, h−1

2 n1h2 ∈ N, i.e., n1h2 = h2n3 for some n3 ∈ N.

Then h1n1h2n2 = h1h2n3n2 ∈ HN.

2

Identity: This is obvious.

3

Inverse: Let hn ∈ HN. We have (hn)−1 = n−1h−1. Again, since n is normal, n−1h−1 = h−1n1 for some n1 ∈ N. Thus, (hn)−1 ∈ HN. Now HN is a subgroup and clearly contains both H and N. Thus H ∨ N ⊂ HN. On the other hand, any subgroup containing both H and N must also contain all elements of the form hn, h ∈ H, n ∈ N. In particular, we have HN ⊂ H ∨ N. We conclude that H ∨ N = HN.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Proof of Lemma 34.4

Let HN denote the set {hn : h ∈ H, n ∈ N}. We first prove that HN is a subgroup of G.

1

Closed: Let h1n1, h2n2 ∈ HN. Since N is a normal subgroup, h−1

2 n1h2 ∈ N, i.e., n1h2 = h2n3 for some n3 ∈ N.

Then h1n1h2n2 = h1h2n3n2 ∈ HN.

2

Identity: This is obvious.

3

Inverse: Let hn ∈ HN. We have (hn)−1 = n−1h−1. Again, since n is normal, n−1h−1 = h−1n1 for some n1 ∈ N. Thus, (hn)−1 ∈ HN. Now HN is a subgroup and clearly contains both H and N. Thus H ∨ N ⊂ HN. On the other hand, any subgroup containing both H and N must also contain all elements of the form hn, h ∈ H, n ∈ N. In particular, we have HN ⊂ H ∨ N. We conclude that H ∨ N = HN.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Proof of Lemma 34.4

Let HN denote the set {hn : h ∈ H, n ∈ N}. We first prove that HN is a subgroup of G.

1

Closed: Let h1n1, h2n2 ∈ HN. Since N is a normal subgroup, h−1

2 n1h2 ∈ N, i.e., n1h2 = h2n3 for some n3 ∈ N.

Then h1n1h2n2 = h1h2n3n2 ∈ HN.

2

Identity: This is obvious.

3

Inverse: Let hn ∈ HN. We have (hn)−1 = n−1h−1. Again, since n is normal, n−1h−1 = h−1n1 for some n1 ∈ N. Thus, (hn)−1 ∈ HN. Now HN is a subgroup and clearly contains both H and N. Thus H ∨ N ⊂ HN. On the other hand, any subgroup containing both H and N must also contain all elements of the form hn, h ∈ H, n ∈ N. In particular, we have HN ⊂ H ∨ N. We conclude that H ∨ N = HN.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Proof of Lemma 34.4

Let HN denote the set {hn : h ∈ H, n ∈ N}. We first prove that HN is a subgroup of G.

1

Closed: Let h1n1, h2n2 ∈ HN. Since N is a normal subgroup, h−1

2 n1h2 ∈ N, i.e., n1h2 = h2n3 for some n3 ∈ N.

Then h1n1h2n2 = h1h2n3n2 ∈ HN.

2

Identity: This is obvious.

3

Inverse: Let hn ∈ HN. We have (hn)−1 = n−1h−1. Again, since n is normal, n−1h−1 = h−1n1 for some n1 ∈ N. Thus, (hn)−1 ∈ HN. Now HN is a subgroup and clearly contains both H and N. Thus H ∨ N ⊂ HN. On the other hand, any subgroup containing both H and N must also contain all elements of the form hn, h ∈ H, n ∈ N. In particular, we have HN ⊂ H ∨ N. We conclude that H ∨ N = HN.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Proof of Lemma 34.4

Let HN denote the set {hn : h ∈ H, n ∈ N}. We first prove that HN is a subgroup of G.

1

Closed: Let h1n1, h2n2 ∈ HN. Since N is a normal subgroup, h−1

2 n1h2 ∈ N, i.e., n1h2 = h2n3 for some n3 ∈ N.

Then h1n1h2n2 = h1h2n3n2 ∈ HN.

2

Identity: This is obvious.

3

Inverse: Let hn ∈ HN. We have (hn)−1 = n−1h−1. Again, since n is normal, n−1h−1 = h−1n1 for some n1 ∈ N. Thus, (hn)−1 ∈ HN. Now HN is a subgroup and clearly contains both H and N. Thus H ∨ N ⊂ HN. On the other hand, any subgroup containing both H and N must also contain all elements of the form hn, h ∈ H, n ∈ N. In particular, we have HN ⊂ H ∨ N. We conclude that H ∨ N = HN.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Second isomorphism theorem

Theorem (Second isomorphism theorem) Let H be a subgroup of G and N be a normal subgroup of G. Then (HN)/N ≃ H/(H ∩ N). Proof. Define a function φ : H → (HN)/N by φ(h) = hN. We claim

1

φ is a homomorphism: This is clear.

2

Ker(φ) = H ∩ N: If h ∈ Ker(φ), then hN = N and thus h ∈ N. Because h is also in H, we find h ∈ H ∩ N. Conversely, every element of H ∩ N is in the kernel. Thus Ker(φ) = H ∩ N.

3

Im(φ) = (HN)/N: Given a = hn ∈ HN, we have aN = (hn)N = hN. Then aN = φ(h). Thus φ is onto. Then by the first isomorphism theorem H/(H ∩ N) ≃ (HN)/N.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Second isomorphism theorem

Theorem (Second isomorphism theorem) Let H be a subgroup of G and N be a normal subgroup of G. Then (HN)/N ≃ H/(H ∩ N). Proof. Define a function φ : H → (HN)/N by φ(h) = hN. We claim

1

φ is a homomorphism: This is clear.

2

Ker(φ) = H ∩ N: If h ∈ Ker(φ), then hN = N and thus h ∈ N. Because h is also in H, we find h ∈ H ∩ N. Conversely, every element of H ∩ N is in the kernel. Thus Ker(φ) = H ∩ N.

3

Im(φ) = (HN)/N: Given a = hn ∈ HN, we have aN = (hn)N = hN. Then aN = φ(h). Thus φ is onto. Then by the first isomorphism theorem H/(H ∩ N) ≃ (HN)/N.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Second isomorphism theorem

Theorem (Second isomorphism theorem) Let H be a subgroup of G and N be a normal subgroup of G. Then (HN)/N ≃ H/(H ∩ N). Proof. Define a function φ : H → (HN)/N by φ(h) = hN. We claim

1

φ is a homomorphism: This is clear.

2

Ker(φ) = H ∩ N: If h ∈ Ker(φ), then hN = N and thus h ∈ N. Because h is also in H, we find h ∈ H ∩ N. Conversely, every element of H ∩ N is in the kernel. Thus Ker(φ) = H ∩ N.

3

Im(φ) = (HN)/N: Given a = hn ∈ HN, we have aN = (hn)N = hN. Then aN = φ(h). Thus φ is onto. Then by the first isomorphism theorem H/(H ∩ N) ≃ (HN)/N.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Second isomorphism theorem

Theorem (Second isomorphism theorem) Let H be a subgroup of G and N be a normal subgroup of G. Then (HN)/N ≃ H/(H ∩ N). Proof. Define a function φ : H → (HN)/N by φ(h) = hN. We claim

1

φ is a homomorphism: This is clear.

2

Ker(φ) = H ∩ N: If h ∈ Ker(φ), then hN = N and thus h ∈ N. Because h is also in H, we find h ∈ H ∩ N. Conversely, every element of H ∩ N is in the kernel. Thus Ker(φ) = H ∩ N.

3

Im(φ) = (HN)/N: Given a = hn ∈ HN, we have aN = (hn)N = hN. Then aN = φ(h). Thus φ is onto. Then by the first isomorphism theorem H/(H ∩ N) ≃ (HN)/N.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Second isomorphism theorem

Theorem (Second isomorphism theorem) Let H be a subgroup of G and N be a normal subgroup of G. Then (HN)/N ≃ H/(H ∩ N). Proof. Define a function φ : H → (HN)/N by φ(h) = hN. We claim

1

φ is a homomorphism: This is clear.

2

Ker(φ) = H ∩ N: If h ∈ Ker(φ), then hN = N and thus h ∈ N. Because h is also in H, we find h ∈ H ∩ N. Conversely, every element of H ∩ N is in the kernel. Thus Ker(φ) = H ∩ N.

3

Im(φ) = (HN)/N: Given a = hn ∈ HN, we have aN = (hn)N = hN. Then aN = φ(h). Thus φ is onto. Then by the first isomorphism theorem H/(H ∩ N) ≃ (HN)/N.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Second isomorphism theorem

Theorem (Second isomorphism theorem) Let H be a subgroup of G and N be a normal subgroup of G. Then (HN)/N ≃ H/(H ∩ N). Proof. Define a function φ : H → (HN)/N by φ(h) = hN. We claim

1

φ is a homomorphism: This is clear.

2

Ker(φ) = H ∩ N: If h ∈ Ker(φ), then hN = N and thus h ∈ N. Because h is also in H, we find h ∈ H ∩ N. Conversely, every element of H ∩ N is in the kernel. Thus Ker(φ) = H ∩ N.

3

Im(φ) = (HN)/N: Given a = hn ∈ HN, we have aN = (hn)N = hN. Then aN = φ(h). Thus φ is onto. Then by the first isomorphism theorem H/(H ∩ N) ≃ (HN)/N.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Second isomorphism theorem

Theorem (Second isomorphism theorem) Let H be a subgroup of G and N be a normal subgroup of G. Then (HN)/N ≃ H/(H ∩ N). Proof. Define a function φ : H → (HN)/N by φ(h) = hN. We claim

1

φ is a homomorphism: This is clear.

2

Ker(φ) = H ∩ N: If h ∈ Ker(φ), then hN = N and thus h ∈ N. Because h is also in H, we find h ∈ H ∩ N. Conversely, every element of H ∩ N is in the kernel. Thus Ker(φ) = H ∩ N.

3

Im(φ) = (HN)/N: Given a = hn ∈ HN, we have aN = (hn)N = hN. Then aN = φ(h). Thus φ is onto. Then by the first isomorphism theorem H/(H ∩ N) ≃ (HN)/N.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Second isomorphism theorem

Theorem (Second isomorphism theorem) Let H be a subgroup of G and N be a normal subgroup of G. Then (HN)/N ≃ H/(H ∩ N). Proof. Define a function φ : H → (HN)/N by φ(h) = hN. We claim

1

φ is a homomorphism: This is clear.

2

Ker(φ) = H ∩ N: If h ∈ Ker(φ), then hN = N and thus h ∈ N. Because h is also in H, we find h ∈ H ∩ N. Conversely, every element of H ∩ N is in the kernel. Thus Ker(φ) = H ∩ N.

3

Im(φ) = (HN)/N: Given a = hn ∈ HN, we have aN = (hn)N = hN. Then aN = φ(h). Thus φ is onto. Then by the first isomorphism theorem H/(H ∩ N) ≃ (HN)/N.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Third isomorphism theorem

Theorem (Third isomorphism theorem) Let H and K be normal subgroup of G with K ≤ H. Then G/H ≃ (G/K)/(H/K). Proof. Define φ : G/K → G/H by φ(gK) = gH. We claim

1

φ is well-defined: Suppose that g1K = g2K. Then g−1

1 g2 ∈ K. Since K ⊂ H, g−1 1 g2 ∈ H and thus g1H = g2H.

2

φ is a homomorphism: Easy.

3

Ker(φ) = H/K: Suppose that gK ∈ Ker(φ). Then gH = H. That is, g ∈ H and gK ∈ H/K.

4

Im(φ) = G/H: For all gH ∈ G/H, we have gH = φ(gK). Thus, φ is onto. Then by the first isomorphism theorem, (G/K)/(H/K) ≃ G/H.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Third isomorphism theorem

Theorem (Third isomorphism theorem) Let H and K be normal subgroup of G with K ≤ H. Then G/H ≃ (G/K)/(H/K). Proof. Define φ : G/K → G/H by φ(gK) = gH. We claim

1

φ is well-defined: Suppose that g1K = g2K. Then g−1

1 g2 ∈ K. Since K ⊂ H, g−1 1 g2 ∈ H and thus g1H = g2H.

2

φ is a homomorphism: Easy.

3

Ker(φ) = H/K: Suppose that gK ∈ Ker(φ). Then gH = H. That is, g ∈ H and gK ∈ H/K.

4

Im(φ) = G/H: For all gH ∈ G/H, we have gH = φ(gK). Thus, φ is onto. Then by the first isomorphism theorem, (G/K)/(H/K) ≃ G/H.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Third isomorphism theorem

Theorem (Third isomorphism theorem) Let H and K be normal subgroup of G with K ≤ H. Then G/H ≃ (G/K)/(H/K). Proof. Define φ : G/K → G/H by φ(gK) = gH. We claim

1

φ is well-defined: Suppose that g1K = g2K. Then g−1

1 g2 ∈ K. Since K ⊂ H, g−1 1 g2 ∈ H and thus g1H = g2H.

2

φ is a homomorphism: Easy.

3

Ker(φ) = H/K: Suppose that gK ∈ Ker(φ). Then gH = H. That is, g ∈ H and gK ∈ H/K.

4

Im(φ) = G/H: For all gH ∈ G/H, we have gH = φ(gK). Thus, φ is onto. Then by the first isomorphism theorem, (G/K)/(H/K) ≃ G/H.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Third isomorphism theorem

Theorem (Third isomorphism theorem) Let H and K be normal subgroup of G with K ≤ H. Then G/H ≃ (G/K)/(H/K). Proof. Define φ : G/K → G/H by φ(gK) = gH. We claim

1

φ is well-defined: Suppose that g1K = g2K. Then g−1

1 g2 ∈ K. Since K ⊂ H, g−1 1 g2 ∈ H and thus g1H = g2H.

2

φ is a homomorphism: Easy.

3

Ker(φ) = H/K: Suppose that gK ∈ Ker(φ). Then gH = H. That is, g ∈ H and gK ∈ H/K.

4

Im(φ) = G/H: For all gH ∈ G/H, we have gH = φ(gK). Thus, φ is onto. Then by the first isomorphism theorem, (G/K)/(H/K) ≃ G/H.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Third isomorphism theorem

Theorem (Third isomorphism theorem) Let H and K be normal subgroup of G with K ≤ H. Then G/H ≃ (G/K)/(H/K). Proof. Define φ : G/K → G/H by φ(gK) = gH. We claim

1

φ is well-defined: Suppose that g1K = g2K. Then g−1

1 g2 ∈ K. Since K ⊂ H, g−1 1 g2 ∈ H and thus g1H = g2H.

2

φ is a homomorphism: Easy.

3

Ker(φ) = H/K: Suppose that gK ∈ Ker(φ). Then gH = H. That is, g ∈ H and gK ∈ H/K.

4

Im(φ) = G/H: For all gH ∈ G/H, we have gH = φ(gK). Thus, φ is onto. Then by the first isomorphism theorem, (G/K)/(H/K) ≃ G/H.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Third isomorphism theorem

Theorem (Third isomorphism theorem) Let H and K be normal subgroup of G with K ≤ H. Then G/H ≃ (G/K)/(H/K). Proof. Define φ : G/K → G/H by φ(gK) = gH. We claim

1

φ is well-defined: Suppose that g1K = g2K. Then g−1

1 g2 ∈ K. Since K ⊂ H, g−1 1 g2 ∈ H and thus g1H = g2H.

2

φ is a homomorphism: Easy.

3

Ker(φ) = H/K: Suppose that gK ∈ Ker(φ). Then gH = H. That is, g ∈ H and gK ∈ H/K.

4

Im(φ) = G/H: For all gH ∈ G/H, we have gH = φ(gK). Thus, φ is onto. Then by the first isomorphism theorem, (G/K)/(H/K) ≃ G/H.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Third isomorphism theorem

Theorem (Third isomorphism theorem) Let H and K be normal subgroup of G with K ≤ H. Then G/H ≃ (G/K)/(H/K). Proof. Define φ : G/K → G/H by φ(gK) = gH. We claim

1

φ is well-defined: Suppose that g1K = g2K. Then g−1

1 g2 ∈ K. Since K ⊂ H, g−1 1 g2 ∈ H and thus g1H = g2H.

2

φ is a homomorphism: Easy.

3

Ker(φ) = H/K: Suppose that gK ∈ Ker(φ). Then gH = H. That is, g ∈ H and gK ∈ H/K.

4

Im(φ) = G/H: For all gH ∈ G/H, we have gH = φ(gK). Thus, φ is onto. Then by the first isomorphism theorem, (G/K)/(H/K) ≃ G/H.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Third isomorphism theorem

Theorem (Third isomorphism theorem) Let H and K be normal subgroup of G with K ≤ H. Then G/H ≃ (G/K)/(H/K). Proof. Define φ : G/K → G/H by φ(gK) = gH. We claim

1

φ is well-defined: Suppose that g1K = g2K. Then g−1

1 g2 ∈ K. Since K ⊂ H, g−1 1 g2 ∈ H and thus g1H = g2H.

2

φ is a homomorphism: Easy.

3

Ker(φ) = H/K: Suppose that gK ∈ Ker(φ). Then gH = H. That is, g ∈ H and gK ∈ H/K.

4

Im(φ) = G/H: For all gH ∈ G/H, we have gH = φ(gK). Thus, φ is onto. Then by the first isomorphism theorem, (G/K)/(H/K) ≃ G/H.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Third isomorphism theorem

Theorem (Third isomorphism theorem) Let H and K be normal subgroup of G with K ≤ H. Then G/H ≃ (G/K)/(H/K). Proof. Define φ : G/K → G/H by φ(gK) = gH. We claim

1

φ is well-defined: Suppose that g1K = g2K. Then g−1

1 g2 ∈ K. Since K ⊂ H, g−1 1 g2 ∈ H and thus g1H = g2H.

2

φ is a homomorphism: Easy.

3

Ker(φ) = H/K: Suppose that gK ∈ Ker(φ). Then gH = H. That is, g ∈ H and gK ∈ H/K.

4

Im(φ) = G/H: For all gH ∈ G/H, we have gH = φ(gK). Thus, φ is onto. Then by the first isomorphism theorem, (G/K)/(H/K) ≃ G/H.

Instructor: Yifan Yang Section 34 – Isomorphism theorems

Homework

Problems 2, 4, 6, 7, 8, 9 of Section 34.

Instructor: Yifan Yang Section 34 – Isomorphism theorems