CHAPTER 10 Group Homomorphisms Definitions and Examples Definition (Group Homomorphism). A homomorphism from a group G to a group G is a mapping φ : G G that preserves the group operation: ! φ(ab) = φ(a)φ(b) for all a, b G. 2 Definition (Kernal of a Homomorphism). The kernel of a homomorphism φ : G G is the set Ker φ = x G φ(x) = e ! { 2 | } Example. (1) Every isomorphism is a homomorphism with Ker φ = e . { } (2) Let G = Z under addition and G = 1, 1 under multiplication. Define φ : G G by { − } ! 1, n is even φ(n) = 1, n is odd (− is a homomorphism. For m and n odd: (even–even) φ(2n + 2m) = φ(2(n + m)) = 1 = φ(2n)φ(2m) (even–odd) φ(2n + m) = 1 = 1( 1) = φ(2n)φ(m) − − (odd–odd) φ(n + m) = 1 = ( 1)( 1) = φ(n)φ(m) − − Ker φ = even integers . { } 134 10. GROUP HOMOMORPHISMS 135 (3) φ : GL(2, R) R⇤ (nonzero reals under multiplication) defined by ! φ(A) = det A is a homomorphism. φ(AB) = det AB = (det A)(det B) = φ(A)φ(B). Ker φ = SL(2, R). (4) Let R[x] denote the group of all polynomials with real coefficients under addition. Let φ : R[x] R[x] be defined by φ(f) = f 0. The group operation preservation is simply “the! derivative of a sum is the sum of the derivatives.” φ(f + g) = (f + g)0 = f 0 + g0 = φ(f) + φ(g). Ker φ is the set of all constant polynomials. (5) The natural homomorphism from Z to Zn is defined by φ(m) = m mod n. Ker φ = n . h i (6) Consider φ : R R under addition defined by φ(x) = x2. Since ! φ(x + y) = (x + y)2 = x2 + 2xy + y2 and φ(x + φ(y) = x2 + y2, this is not a homomorphism. (7) Every vector space linear transformation is a group homomorphism and the nullspace is the kernel. 136 10. GROUP HOMOMORPHISMS Properties of Homomorphisms Theorem (10.1 – Properties of Elements Under Homomorphisms). Let φ be a homomorphism from a group G to a group G and let g G. Then: 2 (1) φ(e) = e. n (2) φ(gn) = φ(g) . Proof. (F⇥or (1)⇤ and (2)) Same as in Theorem 6.2. ⇤ (3) If g is finite, then φ(g) g . | | | | | | Proof. Suppose g = n = gn = e. Then | | ) n e = φ(e) = φ(gn) = φ(g) = (Corollary 2 to Theorem 4.1) φ(g) n. ) | | ⇥ ⇤ ⇤ (4) Ker φ G. Proof. By (1), Ker φ = . Suppose a, b Ker φ. Then 6 ; 2 1 1 1 1 φ(ab− ) = φ(a)φ(b− ) = φ(a) φ(b) − = e e− = e e = e, · · 1 so ab− Ker φ. Thus Ker φ G by the one-step test. 2 ⇥ ⇤ ⇤ (5) φ(a) = φ(b) a Ker φ = b Ker φ. () Proof. φ(a) = φ(b) () 1 1 1 1 e = (φ(b))− φ(a) = φ(b− )φ(a) = φ(b− a) b− a Ker φ () 2 () b Ker φ = a Ker φ (by property 6 of the lemma in Chapter 7) ⇤ 10. GROUP HOMOMORPHISMS 137 1 (6) If φ(g) = g0, then φ− (g0) = x G φ(x) = g0 = g Ker φ. { 2 } Proof. 1 [We prove by mutual set inclusion.] Suppose x φ− (g0). Then 2 φ(x) = g0 = φ(g) = x Ker φ = g Ker φ by property (5)= ) ) x g Ker φ (by property 4 of the lemma in Chapter 7). 2 1 Thus φ− (g0) g Ker φ. ✓ Now suppose k Ker φ, so that φ(gk) = φ(g)φ(k) = g0e = g0. Thus, by 2 1 1 definition, gk φ− (g0) = g Ker φ φ− (g0), and so by mutual set inclusion, 2 ) ✓ 1 g Ker φ = φ− (g0). ⇤ Since homomorphisms preserve the group operation, they also preserve many other group properties. Theorem (10.2 – Properties of Subgroups Under Homomorphisms). Let φ : G G be a homomorphism and let H G. Then ! (1) φ(H) = φ(h) h H G. { | 2 } (2) H cyclic = φ(H) cyclic. ) (3) H Abelian = φ(H) Abelian. ) Proof. (For (1), (2), and (3)) Same as in Theorem 6.3. ⇤ (4) H G = φ(H) φ(G). C ) C Proof. Let φ(h) φ(H) and φ(g) φ(G). Then 2 2 1 1 φ(g)φ(h)φ(g)− = φ(ghg− ) φ(H) 2 1 since ghg− H because H G. 2 C ⇤ 138 10. GROUP HOMOMORPHISMS (5) If Ker φ = n, then φ is an n-to-1 mapping from G onto φ(G). | | Proof. Follows directly from property 6 of Theorem 10.1 and the fact that 1 all cosets of Ker φ = φ− (e). ⇤ (6) If H = n, then φ(H) n. | | | | Proof. Let φH denote the restriction of φ to H. Then φH is a homomorphism from H onto φ(H). Suppose Ker φH = t. Then, from (5), φH is a t-to-1 mapping, so φ(H) t = H . | | | | | | ⇤ 1 (7) If K G, then φ− (K) = k G φ(k) K G. { 2 | 2 } Proof. 1 1 1 Clearly e φ− (K), so φ− (K) = . Let k , k φ− (K). Then 2 6 ; 1 2 2 1 φ(k ), φ(k ) K = φ(k )− K. 1 2 2 ) 2 2 Thus 1 1 1 φ(k1k2− ) = φ(k1)φ(k2− ) = φ(k1)φ(k2)− K. 1 1 1 2 By definition, k k− φ− (K) and so φ− (K) G by the one-step test. 1 2 2 ⇤ 1 (8) If K G, then φ− (K) = k G φ(k) K G. C { 2 | 2 } C Proof. 1 1 1 Every element of xφ− (K)x− has the form xkx− where φ(k) K. Since 2 K C G, 1 1 φ(xkx− ) = φ(x)φ(k)φ(x)− K, 1 1 1 1 21 1 and so xkx− φ− (K). Thus xφ− (K)x− φ− (K) and φ− (K) G by 2 ✓ C Theorem 9.1. ⇤ (9) If φ is onto and Ker φ = e , then φ is an isomorphism from G to G. { } Proof. Follows directly from (5). ⇤ 10. GROUP HOMOMORPHISMS 139 Corollary (To (8)). Ker φ C G. Proof. Let K = e in (8). { } ⇤ Example. Consider φ : C? C? defined by φ(x) = x4. φ is a homomor- phism since ! φ(xy) = (xy)4 = x4y4 = φ(x)φ(y). Ker φ = 1, 1, i, i . Thus φ is a 4-to-1 mapping by property (5) of Theorem 10.2. Since{ φ−(31/4)−=}3, again by property (5) of Theorem 10.2, 1 1/4 1/4 1/4 1/4 1/4 φ− (3) = 3 Ker φ = 3 , 3 , 3 i, 3 i { − − } Now let H = cos 60◦ + i sin 60◦ . Since h i 6 (cos 60◦ + i sin 60◦) = cos 360◦ + i sin 360◦ = 1, H = 6. | | 4 φ(H) = (cos 60◦ + i sin 60◦) = cos 240◦ + i sin 240◦ = φ(H) = 3, h i h i ) | | so φ(H) H . | | | | 140 10. GROUP HOMOMORPHISMS Example. Define φ : Z16 Z16 by φ(x) = 4x. φ is a homomorphism ! since, for x, y Z16, 2 φ(x + y) = 4(x + y) = 4x + 4y = φ(x) + φ(y). Ker φ = 0, 4, 8, 12 . Thus φ is 4-to-1. Since φ(3) = 12, by property (5) of Theorem {10.2, } 1 φ− (12) = 3 + Ker φ = 3, 7, 11, 15 . { } Since 3 is cyclic, so is φ( 3 ). h i h i 3 = 3, 6, 9, 12, 15, 2, 5, 8, 11, 14, 1, 4, 7, 10, 13, 0 . h i { } φ( 3 ) = 12, 8, 4, 0 = 4 = 12 . h i { } h i h i Also, φ(3) = 4 and 3 = 16, so φ(3) 3 . | | | | | | | | 1 With K = 0, 4, 8, 12 Z16, φ− (K)= 3 Z16. { } h i Problem (Page 221 # 25). Hom many homomorphisms are there from Z20 onto Z10? How many are there to Z10? Solution. Z20 and Z10 are both cyclic and additive. By property (2) of Theorem 10.1, written additively, φ(ng) = nφ(g). Such homomorphisms are completely deter- mined by φ(1), i.e., if φ(1) = a, φ(x) = φ(x 1) = xφ(1) = xa. By Lagrange, a 10, and by property (3) of Theorem 10.1,· a 1 or a 20. | | | | | | | | Thus a = 1, 5, 10, or 2. | | a = 10 : 1, 3, 7, 9 have order 10 in Z10, so 4 homomorphisms are onto. | | a = 5 : 2, 4, 6, 8. | | a = 2 : 5. | | a = 1 : 0. | | In all cases, φ(x + y) = (x + y)a = xa + ya = φ(x + φ(y). ⇤ 10. GROUP HOMOMORPHISMS 141 The First Isomorphism Theorem Theorem (10.3 — First Isomorphism Theorem). Let φ : G G be a group homomorphism. Then : G/ Ker φ φ(G) defined by (!g Ker φ) = φ(g) is an isomorphism, i.e., G Ker φ φ(!G). ⇡ Proof. That is well-defined, i.e., the correspondence is independent of the particular coset representation chosen, and is 1–1 follows directly from property (5) of Theorem 10.1. is clearly onto. [To show is operation-preserving.] For all x Ker φ, y Ker φ G/ Ker φ, 2 (x Ker φ y Ker φ) = (xy Ker φ) = φ(xy) = φ(x)φ(y) = (x Ker φ) (y Ker φ), so is operation-preserving and this is an isomorphism. ⇤ Corollary. If φ is a homomorphism from a finite group G to G, then φ(G) divides both G and G . | | | | | | Example. The natural homomorphism from Z to Zn is defined by φ(m) = m mod n has Ker φ = n .