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 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 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 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 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 2 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 2 G gxg = x = f g 2 G gx = xg g. We call this subgroup the centralizer of x and denote it by C (x). G -1 3 Fix(g) = x 2 G gxg = x = f x 2 G gx = xg g = CG (g) -1 4 Fix(G) = x 2 G gxg = x for all x 2 X = \g2G Fix(g) = \ C (g) = f x 2 G gx = xg for all g 2 G g. We call this subgroup g2G 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) = fgg 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 jGj conjugation jGj = jZ(G)j + jCl(xi )j = jZ(G)j + (1) jCG (xi )j 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 jGj = jZ(G)j 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 jS3j = jZ(S3)j + jCl((1; 2))j + jCl((1; 2; 3))j Example jS j jS j = jh()ij + 3 + 3 Applications of the CS ((1; 2)) CS ((1; 2; 3)) class equation 3 3 = 1 + 3 + 3 Sylow's theorems jS j jS j = jh()ij + 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 xn 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 jGj = 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, pjn. The class equation • If H ≤ G proper subgroup, pjjHj, then by induction H contains Applications of the element of order p. class equation • So suppose that p 6 jjHj for all proper subgroups H. Caychy's theorem Finite p-groups have • Class equation is non-trivial center r Groups of size p2 are jGj abelian n = Z(G) + ; jCG (xi )j 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 pjZ(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 jGj = n, pjn, p prime • Let C = hri, the cyclic group with p elements. Acting by p conjugation • Let X = f (g1;:::; gp) gi 2 G; g1 ··· gp = 1 g The class equation • Clearly jX j = np-1 Applications of the jCpj • jOrb((g1;:::; gp))j = class equation jStab((g1;:::;gp))j Caychy's theorem Finite p-groups have Cp if g1 = g2 = ··· = gp non-trivial center • Stab((g1;:::; gp)) = 2 Groups of size p are f1g if some gi 6= gj abelian Sylow's theorems 1 if g1 = g2 = ··· = gp • Thus jOrb((g1;:::; gp))j = 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) 2 X , a > 0 Applications of the p-1 class equation • Orbits partition X , so n = a + bp Caychy's theorem • pjn, so pja Finite p-groups have non-trivial center • Groups of size p2 are Thus exist other singleton orbit (g;:::; g) 2 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 jGj = 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 2 S Acting by 3 conjugation Then jX j = 62 = 36. The class equation Study the sequence v = ((12); (13); (123)) 2 X . All cyclic permutations Applications of the except the identity change v, so Stab(v) = f1g, and class equation Caychy's theorem Orb(v) = f((12); (13); (123)); ((13); (123); (12)); ((123); (12); (13))g. Finite p-groups have Study the sequence w = ((123); (123); (123)) 2 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) = f((123); (123); (123))g. 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 jGj = pn, with p prime, then Z(G) is non-trivial.