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 = S 3 , 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 = S 3 , 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 = S 3 , 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 . Closed: Let h 1 n 1 , h 2 n 2 ∈ HN . Since N is a normal 1 subgroup, h − 1 2 n 1 h 2 ∈ N , i.e., n 1 h 2 = h 2 n 3 for some n 3 ∈ N . Then h 1 n 1 h 2 n 2 = h 1 h 2 n 3 n 2 ∈ HN . Identity: This is obvious. 2 Inverse: Let hn ∈ HN . We have ( hn ) − 1 = n − 1 h − 1 . Again, 3 since n is normal, n − 1 h − 1 = h − 1 n 1 for some n 1 ∈ 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 . Closed: Let h 1 n 1 , h 2 n 2 ∈ HN . Since N is a normal 1 subgroup, h − 1 2 n 1 h 2 ∈ N , i.e., n 1 h 2 = h 2 n 3 for some n 3 ∈ N . Then h 1 n 1 h 2 n 2 = h 1 h 2 n 3 n 2 ∈ HN . Identity: This is obvious. 2 Inverse: Let hn ∈ HN . We have ( hn ) − 1 = n − 1 h − 1 . Again, 3 since n is normal, n − 1 h − 1 = h − 1 n 1 for some n 1 ∈ 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 . Closed: Let h 1 n 1 , h 2 n 2 ∈ HN . Since N is a normal 1 subgroup, h − 1 2 n 1 h 2 ∈ N , i.e., n 1 h 2 = h 2 n 3 for some n 3 ∈ N . Then h 1 n 1 h 2 n 2 = h 1 h 2 n 3 n 2 ∈ HN . Identity: This is obvious. 2 Inverse: Let hn ∈ HN . We have ( hn ) − 1 = n − 1 h − 1 . Again, 3 since n is normal, n − 1 h − 1 = h − 1 n 1 for some n 1 ∈ 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 . Closed: Let h 1 n 1 , h 2 n 2 ∈ HN . Since N is a normal 1 subgroup, h − 1 2 n 1 h 2 ∈ N , i.e., n 1 h 2 = h 2 n 3 for some n 3 ∈ N . Then h 1 n 1 h 2 n 2 = h 1 h 2 n 3 n 2 ∈ HN . Identity: This is obvious. 2 Inverse: Let hn ∈ HN . We have ( hn ) − 1 = n − 1 h − 1 . Again, 3 since n is normal, n − 1 h − 1 = h − 1 n 1 for some n 1 ∈ 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 . Closed: Let h 1 n 1 , h 2 n 2 ∈ HN . Since N is a normal 1 subgroup, h − 1 2 n 1 h 2 ∈ N , i.e., n 1 h 2 = h 2 n 3 for some n 3 ∈ N . Then h 1 n 1 h 2 n 2 = h 1 h 2 n 3 n 2 ∈ HN . Identity: This is obvious. 2 Inverse: Let hn ∈ HN . We have ( hn ) − 1 = n − 1 h − 1 . Again, 3 since n is normal, n − 1 h − 1 = h − 1 n 1 for some n 1 ∈ 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 . Closed: Let h 1 n 1 , h 2 n 2 ∈ HN . Since N is a normal 1 subgroup, h − 1 2 n 1 h 2 ∈ N , i.e., n 1 h 2 = h 2 n 3 for some n 3 ∈ N . Then h 1 n 1 h 2 n 2 = h 1 h 2 n 3 n 2 ∈ HN . Identity: This is obvious. 2 Inverse: Let hn ∈ HN . We have ( hn ) − 1 = n − 1 h − 1 . Again, 3 since n is normal, n − 1 h − 1 = h − 1 n 1 for some n 1 ∈ 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 . Closed: Let h 1 n 1 , h 2 n 2 ∈ HN . Since N is a normal 1 subgroup, h − 1 2 n 1 h 2 ∈ N , i.e., n 1 h 2 = h 2 n 3 for some n 3 ∈ N . Then h 1 n 1 h 2 n 2 = h 1 h 2 n 3 n 2 ∈ HN . Identity: This is obvious. 2 Inverse: Let hn ∈ HN . We have ( hn ) − 1 = n − 1 h − 1 . Again, 3 since n is normal, n − 1 h − 1 = h − 1 n 1 for some n 1 ∈ 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 . Closed: Let h 1 n 1 , h 2 n 2 ∈ HN . Since N is a normal 1 subgroup, h − 1 2 n 1 h 2 ∈ N , i.e., n 1 h 2 = h 2 n 3 for some n 3 ∈ N . Then h 1 n 1 h 2 n 2 = h 1 h 2 n 3 n 2 ∈ HN . Identity: This is obvious. 2 Inverse: Let hn ∈ HN . We have ( hn ) − 1 = n − 1 h − 1 . Again, 3 since n is normal, n − 1 h − 1 = h − 1 n 1 for some n 1 ∈ 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