Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes MAT301H1S Lec5101 Burbulla Week 10 Lecture Notes Winter 2020 Week 10 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes Week 10 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes Properties of External Direct Products Let G1; G2;:::; Gn; G; H be finite groups. Recall (from Chapter 8): 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 10 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes Example 1 (from Chapter 8) 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 10 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes Example 2 How many external direct products of cyclic groups are there with order 1176? 3 2 Solution: 1176 = 2 · 3 · 7 : The part with order 3 must be Z3: There are two possibilities for the part with order 49: Z49 or Z7 ⊕ Z7: There are three possibilities for the part with order 8: Z8 or Z4 ⊕ Z2 or Z2 ⊕ Z2 ⊕ Z2: Thus the six possibilities with order 1176 are: Z8 ⊕ Z3 ⊕ Z49 ; Z8 ⊕ Z3 ⊕ Z7 ⊕ Z7; Z4 ⊕ Z2 ⊕ Z3 ⊕ Z49 ; Z4 ⊕ Z2 ⊕ Z3 ⊕ Z7 ⊕ Z7; Z2 ⊕ Z2 ⊕ Z2 ⊕ Z3 ⊕ Z49 ; Z2 ⊕ Z2 ⊕ Z2 ⊕ Z3 ⊕ Z7 ⊕ Z7 None of these six are isomorphic to each other. (Why?) Week 10 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes The Fundamental Theorem of Finite Abelian Groups The Fundamental Theorem of Finite Abelian Groups says that the approach of Example 2 will find all possible Abelian groups with a given order. To be more precise: Theorem 11.1: if G is an Abelian group with order n, then there are prime numbers p1; p2;:::; pk ; not necessarily distinct, such that n n n G ≈ Zp1 1 ⊕ Zp2 2 ⊕ · · · ⊕ Zpk k n1 n2 nk and n = p1 · p2 ····· pk : Proof: we will not prove this. See the textbook, if you are interested. The proof requires the establishment of four Lemmas first, and uses induction on the order of the group, n: Week 10 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes Converse to Lagrange's Theorem for Finite Abelian Groups Corollary: if m divides the order of a finite Abelian group G then G has a subgroup of order m: Proof: by induction on n = jGj: If n = 1 then m = 1 and result is trivial. If n > 1 and m divides n; then there is a prime p that divides m: By Theorem 11.1, G has a subgroup K of order p; n if p = pi ; then the cyclic group Zpi i has a subgroup of order p: Then G=K is an Abelian group of order n=p and m=p divides jG=Kj: By induction, G=K has a subgroup H=K with H ≤ G and jH=Kj = m=p: Then jHj m jHj = jKj = · p = m: jKj p Week 10 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes Example 3 Let jGj = 72 = 23 · 32: If G is Abelian then G is isomorphic to one of the six following groups: Z8 ⊕ Z9 ; Z8 ⊕ Z3 ⊕ Z3; Z4 ⊕ Z2 ⊕ Z9 ; Z4 ⊕ Z2 ⊕ Z3 ⊕ Z3; Z2 ⊕ Z2 ⊕ Z2 ⊕ Z9 ; Z2 ⊕ Z2 ⊕ Z2 ⊕ Z3 ⊕ Z3 Each of these six possibilities must have a subgroup of order 12. Finding a subgroup of order 12 in the top four possibilities is easy, since we can find an element a of order 4 in Z8 or Z4 and we can find an element b of order 3 in Z9 or Z3 ⊕ Z3: In the bottom two cases there aren't obvious cyclic subgroups of order 12 but we can take an element of order 3 from Z9 or Z3 ⊕ Z3 and combine it with a copy of Z2 ⊕ Z2; resulting in a subgroup of order 12. Week 10 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes Example 4 Let G = f1; 8; 12; 14; 18; 21; 27; 31; 34; 38; 44; 47; 51; 53; 57; 64g which is a group under multiplication modulo 65. (Not obvious; but you can check it out.) Up to isomorphism, which group is it? Solution: since jGj = 16; it must be one of Z16 Z8 ⊕ Z2 Z4 ⊕ Z4 Z4 ⊕ Z2 ⊕ Z2 Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 But which one? Week 10 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes One way to find the answer is to write G as an internal direct product of two subgroups. Observe that 82 = 64 ≡ −1 mod 65; so j8j = 4: Now find another element x of G such that x 2= h8i and G = h8i × hxi: We have h8i = f1; 8; 57; 64g: Try x = 12 : x2 = 144 ≡ 14 mod 65 and 142 = 196 ≡ 1 mod 65: So j12j = 4 and G = h8i × h12i ≈ Z4 ⊕ Z4: Or: calculate the order of every element in G; and compare with five candidates above. After much work, 8; 12; 18; 21; 27; 31; 34; 38; 44; 47; 53 and 57 all have order 4; and 14; 51 and 64 all have order 2. So G has no elements of order 8 or 16, which immediately rules out the top two choices above. And the bottom choice is ruled out since it has no elements of order 4. Since G only has three elements of order 2 and Z4 ⊕ Z2 ⊕ Z2 has seven elements of order 2, we must have G ≈ Z4 ⊕ Z4: Week 10 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes Conjugates and Conjugacy Classes Definition: a; b 2 G are conjugate in G if there is an x 2 G such that b = xax−1: The set of all conjugates of a is called the conjugacy class of a; and is denoted by cl(a): cl(a) = fxax−1 j x 2 Gg: Some observations: 1. If a 2 Z(G); then cl(a) = fag: 2. In GL(n; R) two matrices A and B are conjugate if there is an −1 X 2 GL(n; R) such that B = XAX : A and B are called similar matrices. 3. All elements in the conjugacy class of a have the same order: since (xax−1)k = xak x−1; as you can check, it follows that (xax−1)k = e , xak x−1 = e , ak = e: Thus jxax−1j = jaj: Week 10 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes Example 1 3 2 2 2 In D3 = ha; b j a = b = e; bab = a i check that ba = a b and ba2 = ab: Then 1. cl(e) = feg 2. cl(a) = fa; a2g; since a2 = bab = bab−1 3. cl(a2) = fa; a2g; since a2 = bab ) ba2b = a 4. cl(b) = febe; aba−1; a2b(a2)−1; bbb; bab(ba)−1; ba2b(ba2)−1g = febe; aba2; a2ba; bbb; bab(ba); ba2b(ba2)g = fb; a2b; ab; b; ba2; bag = fb; ba; ba2g: That is, the elements of order 2 are all conjugates. Thus 5. cl(ba) = fb; ba; ba2g and 6. cl(ba2) = fb; ba; ba2g as well. Notice that the conjugacy classes partition D3 into three disjoint classes. Week 10 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes Conjugacy Is An Equivalence Relation More precisely, you can define an equivalence relation of G by a ∼ b if and only if b = xax−1 for some x 2 G: Then, 1. a ∼ a : a = eae−1 2. a ∼ b implies b ∼ a : b = xax−1 implies a = x−1bx = x−1b(x−1)−1 implies b ∼ a 3. a ∼ b and b ∼ c imply a ∼ c : b = xax−1 and c = yby −1 imply c = y(xax−1)y −1 = (yx)a(yx)−1 implying a ∼ c: The equivalence class of each element a 2 G is fb 2 G j a ∼ bg = fxax−1 j x 2 Gg = cl(a): Note: unlike the partition of G determined by the cosets of a subgroup H; the conjugacy classes of a group need not have the same number of elements. Week 10 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes Conjugacy Classes and Centralizers Theorem 24.1: Let G be a finite group and let a 2 G: Then jcl(a)j = [G : C(a)]: Proof: consider the mapping f such that f (xC(a)) = xax−1: Check: 1.