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MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms MAT301H1S Lec5101 Burbulla Week 5 Lecture Notes Winter 2020 Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms What is a Homomorphism? Definition: Let G and H be groups and suppose f : G −! H is a function such that for all x; y 2 G; f (xy) = f (x)f (y): Then f is called a group homomorphism, or just homomorphism for short. In words, we say \f preserves products." Note: to calculate xy we are using the operation in G; but to calculate f (x)f (y) we are using the operation in H: Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Examples ∗ ∗ 1. f : R −! R by f (x) = jxj f (xy) = jxyj = jxjjyj = f (x)f (y): ∗ ∗ 2 2. f : R −! R by f (x) = x f (xy) = (xy)2 = x2y 2 = f (x)f (y): 3. f : Z −! Zn by f (m) = m mod n; with addition. Use the division algorithm to write m1 = p1n + q1; m2 = p2n + q2: Then m1 + m2 = (p1 + p2)n + q1 + q2; so f (m1+m2) = (q1+q2) mod n ≡ q1 mod n+q2 mod n = f (m1)+f (m2) Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Not All Functions are Homomorphisms 2 Let f : R −! R by f (x) = x : The operation is addition. Then f (x + y) = (x + y)2 = x2 + 2xy + y 2 but f (x) + f (y) = x2 + y 2: So it is not true that f (x + y) = f (x) + f (y) for all x; y 2 R; so f is not a homomorphism. Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Some More Interesting Examples ∗ The determinant function, det : GL(n; R) −! R ; is a homomorphism, since det(AB) = det(A) det(B): Define sgn : Sn −! f1; −1g by 1 if σ is an even permutaton sgn (σ) = −1 if σ is an odd permutaton You have to check that for all α; β 2 Sn; sgn (αβ) = sgn (α) sgn (β); which boils down to checking four cases, namely even plus even is even, even plus odd is odd, odd plus even is odd, and odd plus odd is even. Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Properties of Homomorphisms: Th 10.1 First Three Parts Let f : G −! H be a homomorphism. Then 1. f (eG ) = eH Proof: x = xeG ) f (x) = f (xeG ) = f (x)f (eG ) ) eH = f (eG ): 2. f (xn) = (f (x))n; for all x 2 G and all n 2 Z Proof: f (x2) = f (x x) = f (x) f (x) = (f (x))2; etc for positive n: If n = −1; then −1 −1 −1 xx = eG ) f (xx ) = f (eG ) ) f (x)f (x ) = eH ) f (x−1) = (f (x))−1: 3. If jxj is finite, then jf (x)j divides jxj Proof: let jxj = n: Then n n n x = eG ) f (x ) = f (eG ) ) (f (x)) = eH ) jf (x)j divides n: Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Kernels and Images If f : G −! H is a homomorphism, define 1. ker(f ) = fx 2 G j f (x) = eH g 2. im (f ) = ff (x) j x 2 Gg = f (G) Then ker(f ) ≤ G and im (f ) ≤ H: Proof: (for kernel) I f (eG ) = eH ) eG 2 ker(f ): I x; y 2 ker(f ) −1 −1 −1 ) f (xy ) = f (x)f (y ) = f (x)(f (y)) = eH eH = eH ; so xy −1 2 ker(f ): I Proof that im (f ) ≤ H is left as an exercise. Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Examples ∗ 1. For det : GL(n; R) −! R ; ker(det) = fA 2 GL(n; R) j det(A) = 1g = SL(n; R): 2. For sgn : Sn −! f1; −1g ker(sgn ) = fσ 2 Sn j sgn (σ) = 1g = An: 3. For f : Z −! Zn by f (m) = m mod n; with addition, ker(f ) = fm j m ≡ 0 mod ng = < n > and im (f ) = fm mod n j m 2 Zg = Zn: Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Proper Definition of Determinant Let A = (aij ) be an n × n matrix. Here is the \proper" definition of det(A): X det(A) = sgn (σ) a1σ(1)a2σ(2) ····· anσ(n): σ2Sn For example, if n = 3; the even permutations are (1); (123); (132) and the odd permutations are (12); (23); (31): Then 2 3 a11 a12 a13 det 4 a21 a22 a23 5 = a31 a32 a33 a11a22a33+a12a23a31+a13a21a32−a12a21a33−a11a23a32−a13a22a31: Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Properties of Subgroups Under Homomorphisms First three parts of Theorem 10.2: let f : G −! H be a homomorphism and let K be a subgroup of G: Then 1. f (K) = ff (k) j k 2 Kg is a subgroup of H: 2. If K is cyclic then f (K) is cyclic. 3. If K is Abelian then f (K) is Abelian. Proof: 1. left as an exercise. 2. Suppose K = hki: Then x 2 K ) x = kn: Then f (x) = f (kn) = (f (k))n; which means that f (K) = hf (k)i: 3. Suppose xy = yx for all x; y 2 K: Then f (xy) = f (yx) , f (x)f (y) = f (y)f (x); which means all elements in f (K) commute. Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Example cos θ − sin θ Let f : −! GL(2; ) by f (θ) = : f is a R R sin θ cos θ homomorphism because f (α)f (β) = f (α + β): cos α − sin α cos β − sin β cos(α + β) − sin(α + β) = ; sin α cos α sin β cos β sin(α + β) cos(α + β using appropriate trig identities. Check that ker(f ) = h2πi and im (f ) is the group of 2 × 2 rotation matrices. In particular, R is Abelian, so im (f ) is Abelian and im (f ) is an Abelian subgroup of GL(2; R): If the positive integer n is fixed and 2π 2πk K = = j k 2 ; n n Z then f (K) = hf (2π=n)i ≤ Dn; consisting of all rotations of a regular n-gon. Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms One-to-one and Onto Homomorphisms Let f : G −! H be a group homomorphism. Definition: f is called one-to-one if f (x1) = f (x2) ) x1 = x2: Defintion: f is called onto if every element in H is in im (f ): Theorem: let f : G −! H be a group homomorphism. I f is one-to-one if and only if ker(f ) = feG g I f is onto if and only if im (f ) = H Proof: (for one-to-one) suppose f is one-to-one: Let x 2 ker(f ): Then f (x) = eH and f (eG ) = eH : Therefore x = eG : Now suppose ker(f ) = feG g, and let f (x1) = f (x2): Then −1 −1 −1 −1 f (x1x2 ) = f (x1)f (x2 ) = f (x1)(f (x2)) = f (x1)(f (x1)) = eH : −1 −1 Thus x1x2 2 ker(f ) ) x1x2 = eG ) x1 = x2: Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms What is an Isomorphism? Let f : G −! H be a group homomorphism. Definition: f is called an isomorphism if f is one-to-one and onto. Thus f is an isomorphism if ker(f ) = feG g and im (f ) = H: Note: every isomorphism f : G −! H has an inverse f −1 : H −! G; defined by f −1(x) = y , x = f (y); which is also an isomorphism. Definition: if there is an isomorphism f : G −! H; then we say the groups G and H are isomorphic, and we write G ≈ H: Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Example 1 1 m Let H = 2 GL(2; ) j m 2 and define f : −! H 0 1 R Z Z by 1 m f (m) = : 0 1 Then f is an isomorphism: 1. f is a homomorphism since 1 m + n 1 m 1 n f (m+n) = = = f (m)f (n) 0 1 0 1 0 1 1 0 2. f is one-to-one since f (m) = ) m = 0 0 1 3. f is onto since obviously im (f ) = H: Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Example 2 Let G = hai be a cyclic group of order n: Then G ≈ Zn: k Proof: for k 2 Z; define f : G −! Zn by f (a ) = k mod n: j k I f is well-defined: a = a ) n divides j − k ) j ≡ k mod n: j k j+k I f is a homomorphism: f (a a ) = f (a ) = (j + k) mod n; and f (aj ) + f (ak ) = j mod n + k mod n = (j + k) mod n: I f is one-to-one: k k n q f (a ) = 0 ) k ≡ 0 mod n ) k = qn ) a = (a ) = eG : I f is onto: this is obvious since k 2 Z; so im (f ) = Zn: m Alternate Proof: use g : Zn −! G defined by g(m) = a : (g = f −1:) This makes the algebra easier since g is obviously well-defined.
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