Abstract Algebra, Lecture 9 Jan Snellman Abstract Algebra, Lecture 9 the Class Equation

Abstract Algebra, Lecture 9 Jan Snellman Abstract Algebra, Lecture 9 The Class Equation

Acting by conjugation Jan Snellman1 The class equation 1Matematiska Institutionen Applications of the Link¨opingsUniversitet class equation Sylow’s theorems

Link¨oping,fall 2019

Lecture notes availabe at course homepage http://courses.mai.liu.se/GU/TATA55/ Abstract Algebra, Lecture 9

Jan Snellman Summary

Acting by conjugation The class equation 1 Acting by conjugation Caychy’s theorem Applications of the 2 The class equation Finite p-groups have non-trivial class equation Example center Sylow’s theorems 3 Applications of the class Groups of size p2 are abelian equation 4 Sylow’s theorems Abstract Algebra, Lecture 9

Jan Snellman Summary

Jan Snellman Lemma Let the group G act on itself by conjugation,

g.x = gxg −1 Acting by conjugation Then The class equation − 1 Orb(x) = gxg 1 g ∈ G . We call this the conjugate class Applications of the containing x and denote it by Cl(x). class equation −1 Sylow’s theorems 2 Stab(x) = g ∈ G gxg = x = { g ∈ G gx = xg }. We call this subgroup the centralizer of x and denote it by C (x). G −1 3 Fix(g) = x ∈ G gxg = x = { x ∈ G gx = xg } = CG (g) −1 4 Fix(G) = x ∈ G gxg = x for all x ∈ X = ∩g∈G Fix(g) = ∩ C (g) = { x ∈ G gx = xg for all g ∈ G }. We call this subgroup g∈G G the center of G and denote it Z(G). Abstract Algebra, Lecture 9

Jan Snellman

Acting by Lemma conjugation The center of G is the union of all singleton conjugacy classes. The class equation Applications of the Proof. class equation Sylow’s theorems Cl(g) = {g} if and only if g commutes with everything. Abstract Algebra, Lecture 9

Jan Snellman Theorem (Class equation) If G is ﬁnite, then

r r Acting by |G| conjugation |G| = |Z(G)| + |Cl(xi )| = |Z(G)| + (1) |CG (xi )| i=1 i=1 The class equation X X Example where the xi ’s are a choice of exactly one group element from each Applications of the class equation conjugacy class with more than one element. Sylow’s theorems Proof. The conjugacy classes are equivalence classes of an equivalence realtion on G, thus they partition G. The center is, as stated before, the union of the conjugacy classes that consist of a single element. Abstract Algebra, Lecture 9

Jan Snellman

Acting by Example conjugation The class equation If G is abelian, then the class equation reads Example Applications of the |G| = |Z(G)| class equation Sylow’s theorems Abstract Algebra, Lecture 9

Jan Snellman Example

In S3, there is one conjugacy class containing the transpositions, and one containing the 3-cycles, and a singleton class containing the identity. The

Acting by class equation is thus conjugation The class equation |S3| = |Z(S3)| + |Cl((1, 2))| + |Cl((1, 2, 3))| Example |S | |S | = |h()i| + 3 + 3 Applications of the CS ((1, 2)) CS ((1, 2, 3)) class equation 3 3 = 1 + 3 + 3 Sylow’s theorems |S | |S | = |h()i| + 3 + 3 CS3 ((1, 2)) CS3 ((1, 2, 3))

from which we conclude that CS3 ((1, 2)) = h(1, 2)i,

CS3 ((1, 2, 3)) = h(1, 2, 3)i. Abstract Algebra, Lecture 9 For general n, the conjugacy classes of Sn are easy to describe: Jan Snellman Theorem

1 Two permutations in Sn are conjugate iﬀ they have the same cycle Acting by type. conjugation 2 The number of permutations in §n with cycle type λ is given by The class equation c(λ, n) (hand-in exam batch 2). Example n! Applications of the 3 Thus, there are exactly c(λ,n) permutations in Sn commuting with a class equation given permutation σ with cycle type λ Sylow’s theorems 4 Z(Sn) = h()i

5 The class equation for Sn is

n! = 1 + c(λ, n) (2) λ`n X Abstract Algebra, Lecture 9

Jan Snellman

Example For n = 4, the numerical partitions of 4, and the corresponding conjugacy Acting by classes, are conjugation The class equation λ σ c(λ, 4) Example (4)(1, 2, 3, 4) 6 Applications of the (3, 1)(1, 2, 3)(4) 8 class equation (2, 2)(1, 2)(3, 4) 3 Sylow’s theorems (2, 1, 1)(1, 2)(3)(4) 6 (1, 1, 1, 1)(1)(2)(3)(4) 1 Abstract Algebra, Lecture 9

Jan Snellman

Acting by Theorem (Cauchy) conjugation If G is a ﬁnite group with |G| = n, and p is a prime number dividing n, The class equation then G contains en element of order p. Applications of the class equation Caychy’s theorem We will prove this important result twice, ﬁrst using the class equation, Finite p-groups have then using an action by a cyclic group. non-trivial center Groups of size p2 are abelian Sylow’s theorems Abstract Algebra, Lecture 9

Jan Snellman Proof (by Class Equation)

• Induction over n, assuming n ≥ p. • If n = p then G is cyclic, done. Acting by • conjugation So assume n > p, p|n. The class equation • If H ≤ G proper subgroup, p||H|, then by induction H contains Applications of the element of order p. class equation • So suppose that p 6 ||H| for all proper subgroups H. Caychy’s theorem Finite p-groups have • Class equation is non-trivial center r Groups of size p2 are |G| abelian n = Z(G) + , |CG (xi )| j=1 Sylow’s theorems X where CG (xi ) are proper subgroups, hence their order is not divisible by p, hence each term in the sum is, hence so is Z(G). Abstract Algebra, Lecture 9

Jan Snellman Prrof (contd)

• So p|Z(G). s Acting by • Z(G) ﬁnite abelian, so Z(G) ' rj . j=1 Zp conjugation j The class equation • Some pj = p, say p1 = p. Q • r1 r1 Applications of the The generator a of the factor Zp1 has order p . class equation r −1 p 1 r1 r1 r1−1 Caychy’s theorem • The element a has order p / gcd(p , p ) = p. Finite p-groups have non-trivial center • Inject this element of r1 into the direct product, it will still have Zp1 Groups of size p2 are abelian order p. Sylow’s theorems • Done. Abstract Algebra, Lecture 9 Jan Snellman Proof using group action

• Recall |G| = n, p|n, p prime • Let C = hri, the cyclic group with p elements. Acting by p conjugation • Let X = { (g1,..., gp) gi ∈ G, g1 ··· gp = 1 } The class equation • Clearly |X | = np−1 Applications of the |Cp| • |Orb((g1,..., gp))| = class equation |Stab((g1,...,gp))| Caychy’s theorem Finite p-groups have Cp if g1 = g2 = ··· = gp non-trivial center • Stab((g1,..., gp)) = 2 Groups of size p are {1} if some gi 6= gj abelian

Sylow’s theorems 1 if g1 = g2 = ··· = gp • Thus |Orb((g1,..., gp))| = p if some gi 6= gj Abstract Algebra, Lecture 9

Jan Snellman

Proof (contd

Acting by • Denote by a the number of singleton orbits, b the number of orbits of conjugation size p The class equation • Since (1, 1,..., 1) ∈ X , a > 0 Applications of the p−1 class equation • Orbits partition X , so n = a + bp Caychy’s theorem • p|n, so p|a Finite p-groups have non-trivial center • Groups of size p2 are Thus exist other singleton orbit (g,..., g) ∈ X apart from (1,..., 1) abelian • This means that g p = 1, hence o(g) = p. Sylow’s theorems Abstract Algebra, Lecture 9 Example Jan Snellman Take G = S3. Then |G| = 6, which is divisible by 3. Let us prove that there is some element in S3 of order 3. Put X = (g, h, h−1g −1) g, h ∈ S Acting by 3 conjugation Then |X | = 62 = 36. The class equation Study the sequence v = ((12), (13), (123)) ∈ X . All cyclic permutations Applications of the except the identity change v, so Stab(v) = {1}, and class equation Caychy’s theorem Orb(v) = {((12), (13), (123)), ((13), (123), (12)), ((123), (12), (13))}. Finite p-groups have Study the sequence w = ((123), (123), (123)) ∈ X . All cyclic permutations non-trivial center 2 Groups of size p2 are preserve w, so Stab(w) = 1, r, r , and abelian Orb(w) = {((123), (123), (123))}. Sylow’s theorems There are 3 elements in S3 whose orders are divisible by 3, so orbit partition of X becomes 36 = 3 + 3 ∗ 11 Can you ﬁnd these eleven orbits of size 3? Abstract Algebra, Lecture 9 Theorem Jan Snellman If |G| = pn, with p prime, then Z(G) is non-trivial.

Proof.

Acting by • z = |Z(G)| conjugation k • If a ∈ G then CG (a) ≤ G, so |CG (a)| = p . The class equation • If G abelian, then Z(G) = G, done. Applications of the class equation • If G not abelian then z < pn, and Caychy’s theorem Finite p-groups have n non-trivial center n p Groups of size p2 are p = z + pkj abelian j X Sylow’s theorems n • p|LHS, p| p , so p|z. pkj • Since z > 0, we get that z > 1, so Z(G) non-trivial. Abstract Algebra, Lecture 9 Theorem Jan Snellman Let |G| = p2, where p is a prime. Then G is abelian.

Proof.

2 Acting by • |Z(G)| ∈ p, p since the center is a non-trivial subgroup. conjugation 2 • If |Z(G)| = p then done. The class equation • Assume, towards a contradiction, that |Z(G)| = p. Applications of the class equation • Then Z(G) cyclic, and normal in G, so G/Z(G) also cyclic. Caychy’s theorem Finite p-groups have • Let G/Z(G) = haZ(g)i. non-trivial center m Groups of size p2 are • Take g, h ∈ G, their images in the quotient are gZ(G) = a Z(G) abelian and hZ(G) = anZ(G). Sylow’s theorems • So g = amx, h = any, x, y ∈ Z(G). • So gh = amxany = xamany = xam+ny = xyam+n = yxam+n = yam+nx = yanamx = anyxam = hg. Abstract Algebra, Lecture 9

Jan Snellman

Recall: Acting by conjugation Definition The class equation A group is a p-group if every element has order which is a power of p. Applications of the class equation Lemma Sylow’s theorems A finite group is a p-group iff its size is a power of p. Abstract Algebra, Lecture 9 Definition Jan Snellman Let |G| = n, and suppose that pk |n but pk+1 6 |n. A subgroup H ≤ G with |H| = pk is called a p-Sylow subgroup.

Theorem (First Sylow thm) Acting by conjugation If |G| = n, with p|n, then G has a p- Sylow subgroup. Furthermore, any The class equation p-subgroup of G is contained in some p-Sylow subgroup. Applications of the class equation Proof. Sylow’s theorems Omitted, read the textbook.

Corollary If |G| = n, with pk |n, then G has a subgroup of size pk .

Remark Note that this does not guarantee the existence of elements of order pk . Abstract Algebra, Lecture 9

Jan Snellman

Example Acting by conjugation If |G| = 12, then there are surely subgroups of size 2, 3, 4. In the dihedral The class equation group with 12 elements, there are no elements of order 4, however, there Applications of the are subgroups of size 4. You can take the subgroup generated by a class equation reflection through a line through two vertices, a reflection through a line Sylow’s theorems perpendicular to the first line, and the antipodal map. Abstract Algebra, Lecture 9

Jan Snellman Theorem (Sylow’s second thm) Any two p-Sylow subgroups H, K of G are conjugate, i.e., there exists Acting by g ∈ G such that conjugation K = gHg −1. The class equation Applications of the Proof. class equation Sylow’s theorems Omitted, read the textbook.

Remark If there is a single p-Sylow subgroup H, then Sylows’s second thm shows that H / G. Abstract Algebra, Lecture 9

Jan Snellman

Theorem (Sylow’s third thm) Let G = m, p prime, p|m. Let r denote the number of p-Sylow subgroups. Acting by conjugation Then r|m The class equation Applications of the and class equation r ≡ 1 mod p Sylow’s theorems

Proof. Omitted, read the textbook. Abstract Algebra, Lecture 9 Example (Svensson Ex. 12.57) Jan Snellman • Suppose |G| = 56 = 23 ∗ 7. • Claim: G has (at least) one proper normal subgroup.

Acting by • n2 number 2-Sylow, n7 number 7-Sylow. conjugation • The class equation

Applications of the n2 ≡ 1 mod 2 class equation 3 2 ∗ 7 ≡ 0 mod n2 Sylow’s theorems n7 ≡ 1 mod 7 3 2 ∗ 7 ≡ 0 mod n7

• Soln to above: (n2, n7) ∈ {(1, 1), (1, 8), (7, 1), (7, 8)}

• If we can exclude n2 = 7, n7 = 8 then done, since unique p-Sylow is normal Abstract Algebra, Lecture 9

Jan Snellman

Example (Svensson Ex. 12.57 cont.)

Acting by • Suppose that we have 7 2-Sylow and 8 7-Sylow conjugation • The class equation Each 7-Sylow is cyclic Applications of the • Two such intersect in the identity, only, by Lagrange. class equation • Picture! Sylow’s theorems • So 8 ∗ 6 = 48 elements of order 7 • Only 56 − 48 = 8 elems left, can’t make 7 groups of order 23 = 8.