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MAT301H1S Lec5101 Burbulla Chapter 8: External Direct Products MAT301H1S Lec5101 Burbulla Week 7 Lecture Notes Winter 2020 Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 8: External Direct Products Chapter 8: External Direct Products Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 8: External Direct Products What is an External Direct Product of Groups? Definition: Let G1; G2;:::; Gn be a finite collection of groups. The external direct product of G1; G2;:::; Gn is the set G1 ⊕ G2 ⊕ · · · ⊕ Gn = f(g1; g2;:::; gn) j gi 2 Gi ; 1 ≤ i ≤ ng with component-wise `multiplication' 0 0 0 0 0 0 (g1; g2;:::; gn)(g1; g2;:::; gn) = (g1g1; g2g2;:::; gngn): Warning: many other books use the notation G1 × G2 × · · · × Gn instead. (In particular, WeBWorK uses this notation!) Theorem: G1 ⊕ G2 ⊕ · · · ⊕ Gn is a group, its identity element is (eG1 ; eG2 ;:::; eGn ); and if each Gi is finite, then jG1 ⊕ G2 ⊕ · · · ⊕ Gnj = jG1j jG2j · · · jGnj: Proof: exercise left for the reader. Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 8: External Direct Products Example 1 Consider R with addition. In linear algebra, the Euclidean vector spaces 2 3 n R = R ⊕ R; R = R ⊕ R ⊕ R; R = R ⊕ R ⊕ · · · ⊕ R | {z } n times are examples of external direct products you should be familiar with, although in linear algebra we usually write the elements as columns, instead of rows, and call them vectors: 2 3 2 3 2 3 x1 y1 x1 + y1 6 x2 7 6 y2 7 6 x2 + y2 7 ~x + ~y = 6 7 + 6 7 = 6 7 4 ::: 5 4 ::: 5 4 ::: 5 xn yn xn + yn Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 8: External Direct Products Example 2 Z3 ⊕ Z4 = f(0; 0); (1; 0); (2; 0); (0; 1); (1; 1); (2; 1)g [ f(0; 2); (1; 2); (2; 2); (0; 3); (1; 3); (2; 3)g: jZ3 ⊕ Z4j = 12: The order of (1; 0) is 3, since (1; 0) + (1; 0) + (1; 0) = (3; 0); and 3 ≡ 0 mod 3: The order of (0; 2) is 2, since (0; 2) + (0; 2) = (0; 4); and 4 ≡ 0 mod 4: The order of (1; 3) is 12: observe 2(1; 3) = (2; 6) 6= (0; 0); 3(1; 3) = (3; 9) 6= (0; 0); and 4(1; 3) = (4; 12) 6= (0; 0); and 6(1; 3) = (6; 18) 6= (0; 0): But the order of (1; 3) must divide 12, so the only choice left is 12. Thus Z3 ⊕ Z4 ≈ Z12: Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 8: External Direct Products Example 3 U(4) ⊕ U(10) = f(1; 1); (1; 3); (1; 7); (1; 9); (3; 1); (3; 3); (3; 7); (3; 9)g: The product (3; 7)(3; 9) = (9; 63) = (1; 3); since the first component is multiplied mod 4, and the second component is multiplied mod 10. jU(4) ⊕ U(10)j = 8; but it is not cyclic, even though U(4) and U(10) are both cyclic. Its not cyclic because it has at least two elements of order 2: both (1; 9) and (3; 9) have order 2 in U(4) ⊕ U(10): (As we saw in Chapter 4, the number of elements of order 2 in a cyclic group of even order must be φ(2) = 1:) Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 8: External Direct Products Some Properties of External Direct Products Let G1; G2;:::; Gn; G; H be finite groups. 1. The order of (g1; g2;:::; gn) 2 G1 ⊕ G2 ⊕ · · · ⊕ Gn is lcm(jg1j; jg2j;:::; jgnj): 2. Let G and H be cyclic groups. Then G ⊕ H is cyclic if and only if jGj and jHj are relatively prime. 3. The external direct product G1 ⊕ G2 ⊕ · · · ⊕ Gn of a finite number of finite cyclic groups is cyclic if and only if jGi j and jGj j are relatively prime, for each pair i 6= j: 4. Let m = n1n2 ··· nk : Then Zm ≈ Zn1 ⊕ Zn2 ⊕ · · · ⊕ Znk if and only if ni and nj are relatively prime for each pair i 6= j: Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 8: External Direct Products Example 4 jZ6 ⊕ U(14)j = 6 × 6 = 36: In Z6 ⊕ U(14) the order of (3; 3) is 6; because j3j = 2 in Z6; j3j = 6 in U(14); and lcm(2; 6) = 6: jS4 ⊕ U(9)j = 24 × 6 = 144: In S4 ⊕ U(9) the order of ((24)(31); 4) is 6; because j(24)(31)j = 2 in S4; j4j = 3 in U(9); and lcm(2; 3) = 6: cos(2π=3) − sin(2π=3) In SL(2; ) ⊕ the order of ; 1 R Z4 sin(2π=3) cos(2π=3) is 12, because the matrix has order 3 and j1j = 4 in Z4: Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 8: External Direct Products Example 5 Find the number of elements of order 5 in Z25 ⊕ Z5: Solution: j(a; b)j = lcm(jaj; jbj) = 5 ) (jaj; jbj) = (1; 5); (5; 1) or (5; 5): If jaj = 5; then a = 5; 10; 15 or 20: If jbj = 1; then b = 0: So in this case there will be 4 × 1 = 4 elements (a; b) of order 5. If jbj = 5; and jaj = 5; then b = 1; 2; 3 or 4; and a is as in the first case. In this case there are 4 × 4 = 16 elements (a; b) of order 5. If jaj = 1; then a = 0; if jbj = 5; then b = 1; 2; 3 or 4: In this case there are 4 elements (a; b) of order 5. All together, there are 24 elements of order 5 in Z25 ⊕ Z5: Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 8: External Direct Products Example 6 Z30 ≈ Z2 ⊕ Z15 ≈ Z2 ⊕ Z3 ⊕ Z5: Consequently, Z2 ⊕ Z30 ≈ Z2 ⊕ Z2 ⊕ Z3 ⊕ Z5: Similarly, Z30 ≈ Z3 ⊕ Z10 and consequently Z2 ⊕ Z30 ≈ Z2 ⊕ Z3 ⊕ Z10 ≈ Z6 ⊕ Z10: This shows that Z2 ⊕ Z30 ≈ Z6 ⊕ Z10: But Z2 ⊕ Z30 is not isomorphic to Z60; one has an element of order 60, but the other one doesn't. Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 8: External Direct Products U(n) As An External Direct Product Theorem 8.3: if s and t are relatively prime, then U(st) ≈ U(s) ⊕ U(t): Proof: (outline) define f : U(st) −! U(s) ⊕ U(t) by f (x) = (x mod s; x mod t): Check f is a homomorphism; exercise left for the reader. Show f is one-to-one: x 2 ker(f ) ) x ≡ 1 mod s and x ≡ 1 mod t: So sjx − 1 and tjx − 1: Since s and t are relatively prime, this implies stjx − 1: Thus x ≡ 1 mod (st): So f is a one-to-one correspondence between two fintite sets of the same order, jU(st)j = φ(st) = φ(s)φ(t) = jU(s)j jU(t)j; so f must be onto as well. Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 8: External Direct Products Defintion: if k is a divisor of n we define Uk (n) = fx 2 U(n) j x ≡ 1 mod kg: Observation: If s and t are relatively prime and n = st; then Us (n) ≈ U(t) and Ut (n) ≈ U(s): For example: U(21) = f1; 2; 4; 5; 8; 10; 11; 13; 16; 17; 19; 20g; jU(21)j = 12; U(3) = f1; 2g; jU(3)j = 2; and U(7) = f1; 2; 3; 4; 5; 6g; jU(7)j = 6; U7(21) = f1; 8g and U3(21) = f1; 4; 10; 13; 16; 19g: The isomorphisms are f : U7(21) −! U(3) by f (x) = x mod 3; g : U3(21) −! U(7) by g(y) = y mod 7; as you can check. Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 8: External Direct Products Example 7 As a corollary to Theorem 8.3, we have if m = n1n2 ··· nk and each pair ni ; nj is relatively prime for i 6= j; then U(m) ≈ U(n1) ⊕ U(n2) ⊕ · · · ⊕ U(nk ): For example: U(105) ≈ U(7) ⊕ U(15) ≈ U(21) ⊕ U(5) ≈ U(3) ⊕ U(5) ⊕ U(7): Each of these smaller groups in turn are cyclic, U(3) ≈ Z2; U(5) ≈ Z4; U(7) ≈ Z6: So U(105) ≈ Z2 ⊕ Z4 ⊕ Z6 and jU(105)j = 2 × 4 × 6 = 48: Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 8: External Direct Products Example 8 Recall that Aut (Z105) ≈ U(105): The calculations in the previous slide thus give us a description of Aut (Z105): For example, consider the question: how many automorphisms of Z105 are there with order 12? We can answer this question more easily by asking, how many elements of order 12 are there in Z2 ⊕ Z4 ⊕ Z6? If (a; b; c) 2 Z2 ⊕ Z4 ⊕ Z6 has order 12, then lcm(jaj; jbj; jcj) = 12 , (jaj; jbj; jcj) = (1; 4; 3); (2; 4; 3); (1; 4; 6); or (2; 4; 6): There are four possibilities in each case, so Aut (Z105) has 16 automorphisms of order 12. Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla.
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