Section 13 Homomorphisms

Transcription

1 Section 13 Homomorphisms Instructor: Yifan Yang Fall 2006

2 Homomorphisms Definition A map φ of a group G into a group G is a homomorphism if for all a, b G. φ(ab) = φ(a)φ(b)

3 Examples 1. Let φ : G G be defined by φ(g) = e for all g G. Then clearly, φ(ab) = e = e e = φ(a)φ(b) for all a, b G. This is called the trivial homomorphism. 2. Let φ : Z Z be defined by φ(n) = 2n for all n Z. Then φ is a homomorphism. 3. Let S n be the symmetric group on n letters, and let φ : S n Z 2 be defined by { 0, if σ is an even permutation, φ(σ) = 1, if σ is an odd permutation. Then φ is a homomorphism. (Check case by case.)

4 example 1. Let GL(n, R) be the set of all invertible n n matrices over R. Define φ : GL(n, R) R by φ(a) = det(a). Then φ is a homomorphism since det(ab) = det(a) det(b). 2. Let F be the additive group of all polynomials with real coefficients. For a given real number a, the function φ a : F R defined by φ(f ) = f (a) is a homomorphism, called an evaluation homomorphism. 3. Let n be a positive integer. Define φ n : Z Z n by φ n (r) = r. Then φ n is a homomorphism. 4. Let G = G 1 G 2... G n be a direct product of groups. The projection map π i : G G i defined by π i (a 1, a 2,..., a i,..., a n ) = a i is a homomorphism.

5 Properties of homomorphisms Definition Let φ be a mapping of a set X into a set Y. Let A X and B Y. The image φ[a] of A under φ is {φ(a) : a A}. The set φ[x] is the range of φ. The inverse image φ 1 [B] of B in X is {x X : φ(x) B}.

6 Properties of homomorphisms Theorem (13.12) Let φ be a homomorphism of a group G into a group G. 1. φ(e) = e. 2. φ(a 1 ) = φ(a) 1 for all a G. 3. If H is a subgroup of G, then φ[h] is a subgroup of G. 4. If K is a subgroup of G, then φ 1 [K ] is a subgroup of G.

7 Proof of Theorem Proof of φ(e) = e. Consider φ(a), where a G. We have φ(a) = φ(ae) = φ(a)φ(e). By the cancellation law, φ(e) must equal to the identity e. Proof of φ(a 1 ) = φ(a) 1. We have φ(a)φ(a 1 ) = φ(aa 1 ) = φ(e) = e. Thus, φ(a 1 ) = φ(a) 1.

8 Proof of Theorem Proof of Theorem 13.12(3). We need to prove 1. Closed: Suppose that a, b φ[h]. Then there exist a, b H such that φ(a) = a and φ(b) = b. Thus, a b = φ(a)φ(b) = φ(ab). Since H is a subgroup, ab H. Therefore, a b is in φ[h]. 2. identity: By Part (1), e = φ(e) φ[h]. 3. inverse: Suppose that a φ[h]. Then a = φ(a) for some a H. By Part (b), (a ) 1 = φ(a) 1 = φ(a 1 ), and thus (a ) 1 φ[h].

9 Proof of Theorem Proof of Theorem 13.12(4). We need to show 1. Closed: Suppose that a, b φ 1 [K ]. We have φ(a), φ(b) K. Then φ(ab) = φ(a)φ(b) K because φ(a), φ(b) K and K is a subgroup of G. 2. Identity: By Part (1), we have φ(e) = e K. It follows that e φ 1 [K ] since K is a subgroup. 3. Inverse: Let a φ 1 [K ]. We have φ(a) K. By Part (2), φ(a 1 ) = φ(a) 1. Since K is a subgroup, φ(a) K implies φ(a) 1 K.

10 Why are group homomorphisms important in group theory? Let G = Z 12 and G = Z 7. How many homomorphisms from G to G are there? Solution. Let φ be a homomorphism from Z 12 to Z 7. On the one hand, we have φ(0) = 0, by Theorem On the other hand, we have 0 = 12 in Z 12, and thus φ(0) = φ(12) = φ(1) + + φ(1) = 12φ(1). Since 12 is relatively prime to 7, φ(1) must be equal to 0 in Z 7. It follows that φ(n) = nφ(1) = 0 mod 7 for all n Z 12. In other words, there is only one homomorphism, the trivial homomorphism, from Z 12 to Z 7.

11 Why are group homomorphisms important in group theory? Now consider G = G = Z 12. It can be shown that for all a Z 12, the function φ a : Z 12 Z 12 defined by φ a (r) = ar mod 12 is a homomorphism. From these two examples, we see that group homomorphisms are closely related to group structures. Thus, group homomorphisms are very important in studying structural properties of groups.

12 Kernel Since {e } is a subgroup of G, Theorem shows that φ 1 [{e }] is a subgroup of G. This subgroup is of great importance. Definition Let φ : G G be a group homomorphism. The subgroup φ 1 [{e }] = {g G : φ(g)} is called the kernel of φ, and denoted by Ker(φ).

13 Kernel Theorem (13.15) Let φ : G G be a group homomorphism, and let H = Ker(φ). Let a G. Then the set {x G : φ(x) = φ(a)} = φ 1 [{φ(a)}] is the left coset ah, and is also the right coset Ha. Consequently, the partition of G into left cosets is the same as the partition into right cosets. Corollary (13.18) A group homomorphism φ : G G is one-to-one if and only if Ker(φ) = {e}.

14 Example Question Are there any non-trivial homomorphisms from A 4 to Z 2? Solution Suppose that φ : A 4 Z 2 is a non-trivial homomorphism. Then there is a σ in A 4 such that φ(σ) = 1. By Theorem 13.15, Ker(φ) and σker(φ) are left cosets, and form a partition of A 4. In other words, (A 4 : Ker(φ)) = 2, and Ker(φ) = 6. However, we see earlier that A 4 does not have a subgroup of order 6. Thus, there is no non-trivial homomorphism from A 4 to Z 2.

15 Proof of Theorem We will show that 1. ah φ 1 [{φ(a)}]: Let x = ag ah. Since φ is a homomorphism, we have φ(x) = φ(a)φ(h). By assumption that H = Ker(φ), it follows that φ(x) = φ(a), and x φ 1 [{φ(a)}]. 2. φ 1 [{φ(a)}] ah: Suppose that x φ 1 [{φ(a)}]. We have φ(x) = φ(a), and thus φ(a) 1 φ(x) = e. By Theorem 13.12(2), φ(a) 1 = φ(a 1 ). It follows that φ(a 1 x) = e, and a 1 x H. We see that x ah. The proof of φ 1 [{φ(a)}] = Ha is similar.

16 Examples 1. Let n be a positive integer. Let φ : Z Z n be defined by φ(a) = ā, the residue class modulo n containing n. Then Ker(φ) = nz. Also, for each b Z n, the set φ 1 [{ b}] = {..., b 2n, b n, b, b + n,...} = b + nz, which is indeed a coset. 2. Let U be the set of all complex number z of unit length, i.e., z = 1. Consider φ : R U defined by φ(x) = e ix. The kernel is Ker(φ) = {x : e ix = 1} = {2nπ : n Z}, and the sets φ 1 [{φ(a)}] are equal to {a + 2nπ : n Z} = a + Ker(φ).

17 Normal subgroups Definition A subgroup H of a group G is normal if its left cosets and right cosets coincide, that is, if gh = Hg for all g G. (Alternatively, ghg 1 = H for all g G, or ghg 1 H for all h H and g G.) Corollary (13.20) If φ : G G is a group homomorphism, then Ker(φ) is a normal subgroup.

18 Example Let G = S 3. There are 6 subgroups, namely, {e}, {e, (1, 2)}, {e, (1, 3)}, {e, (2, 3)}, {e, (1, 2, 3), (1, 3, 2)}, and G. Among them, it is easy to see that {e} and G are normal. The subgroup {e, (1, 2, 3), (1, 3, 2)} is normal since it is of index 2 in S 3. The three subgroups of order 2 are not normal. For example, we have (1, 3)(1, 2)(1, 3) = (2, 3) {e(1, 2)}. Thus, {e, (1, 2)} is not a normal subgroup.

19 Homework Do Problems 18, 24, 27, 38, 40, 44, 47, 50, 51, 53 of Section 13.