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) ∈ G1 ⊕ G2 ⊕ · · · ⊕ Gn is lcm(|g1|, |g2|,..., |gn|). 2. Let G and H be cyclic groups. Then G ⊕ H is cyclic if and only if |G| and |H| 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 |Gi | and |Gj | 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 = |G|. 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 |G/K|. By induction, G/K has a subgroup H/K with H ≤ G and |H/K| = m/p. Then

|H| m |H| = |K| = · p = m. |K| p

Week 10 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes Example 3

Let |G| = 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 = {1, 8, 12, 14, 18, 21, 27, 31, 34, 38, 44, 47, 51, 53, 57, 64}

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 |G| = 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 |8| = 4. Now find another element x of G such that x ∈/ h8i and G = h8i × hxi. We have h8i = {1, 8, 57, 64}. Try x = 12 : x2 = 144 ≡ 14 mod 65 and 142 = 196 ≡ 1 mod 65. So |12| = 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 ∈ G are conjugate in G if there is an x ∈ 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) = {xax−1 | x ∈ G}. Some observations: 1. If a ∈ Z(G), then cl(a) = {a}. 2. In GL(n, R) two matrices A and B are conjugate if there is an −1 X ∈ 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 |xax−1| = |a|.

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 | a = b = e, bab = a i check that ba = a b and ba2 = ab. Then 1. cl(e) = {e} 2. cl(a) = {a, a2}, since a2 = bab = bab−1 3. cl(a2) = {a, a2}, since a2 = bab ⇒ ba2b = a 4. cl(b) = {ebe, aba−1, a2b(a2)−1, bbb, bab(ba)−1, ba2b(ba2)−1} = {ebe, aba2, a2ba, bbb, bab(ba), ba2b(ba2)} = {b, a2b, ab, b, ba2, ba} = {b, ba, ba2}. That is, the elements of order 2 are all conjugates. Thus 5. cl(ba) = {b, ba, ba2} and 6. cl(ba2) = {b, ba, ba2} 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 ∈ 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 ∈ G is {b ∈ G | a ∼ b} = {xax−1 | x ∈ G} = 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 ∈ G. Then |cl(a)| = [G : C(a)]. Proof: consider the mapping f such that f (xC(a)) = xax−1. Check: 1. f is well-defined: xC(a) = yC(a) ⇔ x−1y ∈ C(a) ⇔ x−1ya = ax−1y ⇔ yay −1 = xax−1 2. f is one-to-one: all the above steps are reversible. 3. f is onto cl(a): follows from definition of f . Thus f is a bijection between the left cosets of C(a) and cl(a): so [G : C(a)] = |cl(a)|. Corollary 1: in a finite group, |cl(a)| divides |G|. Corollary 2: in a finite group, |G| = P[G : C(a)] = P |cl(a)|, where the sum is over one element a from each conjugacy class.

Week 10 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes The Class Equation

The equation from Corollary 2 is known as the class equation of the group G. Since a ∈ Z(G) ⇔ cl(a) = {a}, the class equation is usually written as X |G| = |Z(G)| + |cl(a)|,

where the sum is over representatives of all conjugacy classes with more than one element. From Example 1, the class equation for D3 is 6 = 1 + 2 + 3. Once you’ve found them, class equations always look obvious! In practice, they are useful ‘in reverse,’ as counting tools to limit the number of conjugates of a given order and the number of elements in the center in a group. For example: Theorem 24.2: if |G| = pk , for a prime p, then |Z(G)| > 1. (If the order of a group is a power of a prime p it is known as a p-group. The theorem says every p-group has a non-trivial center.)

Week 10 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes Proof: rewrite the class equation of G as X |G| − |cl(a)| = |Z(G)|.

Since |cl(a)| divides |G| = pk , every term on the left of the equation is a multiple of p, so p divides |Z(G)| and |Z(G)| = pm, for some m ≥ 1. Corollary: if |G| = p2, for p a prime, then G is Abelian. Proof: by Theorem 24.2, Z(G) is also a p-group. By Lagrange’s Theorem, we have |Z(G)| = p or p2. If |Z(G)| = p2, then Z(G) = G, and G is Abelian. If |Z(G)| = p, then G/Z(G) has order p and so is cyclic. But then by Theorem 9.3, G is Abelian. For example, if |G| = 121 = 112, then G is Abelian, and by the Fundamental Theorem of Abelian Groups,

G ≈ Z121 or G ≈ Z11 ⊕ Z11.

Week 10 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes Example 2

What is the class equation of the quaternion group? Q = hi, j, k | i 2 = j2 = k2 = ijk = −1i The quaternions are a p-group. |Q| = 8 = 23. As we showed on Problem Set 1, Z(Q) = {1, −1}, so |Z(Q)| = 2. There are six other elements in Q, i, −i, j, −j, k, −k, all of order 4. Since 6 is not a divisor of 8, the class equation of Q is not 8 = 2 + 6. Nor can the class equation be 8 = 2 + 3 + 3. It must be ether 8 = 2 + 4 + 2 or 8 = 2 + 2 + 2 + 2. But as we also showed on Problem Set 1, |C(i)| = |C(j)| = |C(k)| = 4, so the class equation is 8 = 2 + 2 + 2 + 2. You can confirm that cl(i) = {i, −i}, cl(j) = {j, −j}, cl(k) = {k, −k}.

Week 10 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes

Conjugates in Sn

Generally it is not easy to tell if two elements in a group are conjugates. In one case, though, there is a simple way to determine if two elements are conjugates.

Theorem: α and β in Sn are conjugate if and only if they have the same cycle structure.

Proof: by telling example. In S6, let

α = (231)(45)(6),

β = (462)(31)(5). Let γ = (124365); then β = γαγ−1, as you can check.

Week 10 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes

Example 3: Conjugacy Classes in S5

Cycle structure Number of conjugates Order Parity

(1) 1 1 even

(ab) (5 × 4)/2 = 10 2 odd

(abc) (5 × 4 × 3)/3 = 20 3 even

(abcd) (5 × 4 × 3 × 2)/4 = 30 4 odd

(abcde) (5 × 4 × 3 × 2 × 1)/5 = 24 5 even

1 5×4 3×2
(ab)(cd) 2 2 × 2 = 15 2 even

5×4×3 2×1 (abc)(de) 3 × 2 = 20 6 odd Week 10 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 11: Fundamental Theorem of Abelian Groups Chapter 24: Conjugacy Classes

Since Z(S4) = {(1)}, we have |S5| = 1 + |cl((12))| + |cl((123))| +

|cl((1234))| + |cl((12345)) + |cl((12)(34))| + |cl((123)(45)|

⇔ 120 = 1 + 10 + 20 + 30 + 24 + 15 + 20

Aside: for A5 the class equation is not

60 = 1 + 20 + 24 + 15,

since 24 does not divide 60. All 3-cycles are conjugate in A4, but not all 5-cycles. Why? In S5 the element σ = (12345) has 24 conjugates, so [S5 : C(σ)] = 24; this means |C(σ)| = 5, and C(σ) consists of the five powers of σ. Then in A5, |C(σ)| = 5 and |cl(σ)| = 60/5 = 12. Thus in A5 the elements of order 5 are in two distinct conjugacy classes, and the class equation of A5 is

60 = 1 + 20 + 12 + 12 + 15.

Week 10 Lecture Notes MAT301H1S Lec5101 Burbulla