CHARACTER THEORY OF FINITE GROUPS. Chapter 1: REPRESENTATIONS

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1 CHARACTER THEORY OF FINITE GROUPS Chapter 1: REPRESENTATIONS

2 G is a finite group and K is a field. A K-representation of G is a homomorphism X : G! GL(n, K), where GL(n, K) is the group of invertible n n matrices over K. The positive integer n is the degree of X.

3 A K-representation X of G of degree n is irreducible if there does NOT exist a nonzero proper subspace U of the n-dimensional row space over K such that UX (g) U for all g 2 G. To say that X is irreducible thus means that it is NOT similar to a representation Y of the form Y(g) = apple U(g) 0 V(g) Note: Here, U and V are representations.

4 The group algebra K[G] is the vector space over K with basis G. Multiplication is defined by the group product together with the distributive law. If X is a K-representation of G with degree n, then X can be extended linearly to K[G]. The result is an algebra homomorphism X : K[G]! M n (K), where M n (K) is the algebra of n n matrices.

5 Fact: Assuming that K is algebraically closed, the K-representation X of G is irreducible i X maps K[G] onto the full algebra M n (K). Note: The if direction is obvious but only if requires algebraically closed. If X is irreducible and K is algebraically closed, it follows that the only matrices that centralize X (G) are the scalar matrices. This is called Schur s lemma.

6 EXERCISE (1.1): Let K be an arbitrary field, and let z be the sum of all elements of G in the group algebra K[G]. Show that the onedimensional subspace Kz of K[G] is an ideal. Also, prove that z is nilpotent i the characteristic of K divides G. EXERCISE (1.2): With the notation as above, show that K[G] has a proper ideal I such that Kz + I = K[G] i the characteristic of K does not divide G.

7 From now on, we take K = C, though any algebraically closed characteristic zero field will work as well. Fact: C[G] is a direct sum of full matrix algebras: Note: kx C[G] = M ni (C). i=1 kx G =dim(c[g]) = (n i ) 2. i=1

8 What is k? Since each matrix algebra has center of dimension 1, we have k =dim(z(c[g])). Also, the sums in C[G] of the elements of each conjugacy class of G form a basis for Z(C[G]), and thus k = number of classes of G.

9 Observe that the composite map G! C[G]! M ni (C) is an irreducible C-representation of G. Here, the second map is the projection to the i th summand. Write X i to denote this representation. Fact: Every irreducible C-representation of G is similar to one of the X i.

10 Chapter 2: CHARACTERS

11 Given a C-representation X of G, the associated character is the function G! C given by: We say X a ords. (g) = trace(x (g)). Note: Similar representations a ord equal characters. Note: IfX a ords then (1) is the degree of X. Thus (1) is a positive integer; it is called the degree of.

12 Note: (x 1 gx) = trace(x (x 1 gx)) = trace(x (x) 1 X (g)x (x)) = trace(x (g)) = (g). so characters are constant on conjugacy classes. The characters of G thus lie in the C-space cf(g) of class functions of G. Note: Wehavedim(cf(G)) = k, thenumber of classes of G.

13 If and are characters of G a orded by U and V respectively, then the representation X given by apple U(g) 0 X (g) = 0 V(g) a ords the character = +. Thus sums of characters are characters. A character that is not a sum of two characters is said to be irreducible.

14 Note: Every character is a sum of irreducible characters. This follows since character degrees are positive integers, and thus one cannot keep decomposing indefinitely. We will see that a character is uniquely a sum of irreducible characters. These are called the irreducible constituents of. Notation: Irr(G) is the set of irreducible characters of G.

15 Recall: The C-representations X i for 1 apple i apple k come from the decomposition of the group algebra C[G] as a direct sum of matrix rings. Also recall: Up to similarity, these X i are all of the irreducible C-representations of G. THEOREM: The characters a orded by the X i are distinct and linearly independent, and they form the set Irr(G).

16 Proof: Since the X i come from projections into direct summands of C[G], we can find a j 2 C[G] such that X i (a j )=0ifi 6= j, butx i (a i ) is an arbitrary matrix of the appropriate size. Let i be the character a orded by X i, and view i as being defined on the whole of C[G]. Then we can choose the a i such that i(a j ) = 0, but i(a i ) = 1. It follows that the i are distinct and linearly independent.

17 Now let 2 Irr(G) be arbitrary, and let X be a C-representation a ording. We argue that X is irreducible. Otherwise, X is similar to a representation Y such that Y(g) = apple U(g) 0 V(g) and is a orded by Y. Then = +,whereu afords and V a ords. This contradicts the irreducibility of.

18 We now know that each character 2 Irr(G) is a orded by an irreducible representation, and hence is a orded by some X i. It follows that = i for some i. What remains is to show that each character is irreducible. Otherwise, since j is a sum of irreducible characters, it is a sum of characters i with i 6= j, and this contradicts the linear independence. j

19 COROLLARY: X 2Irr(G) (1) 2 = G. COROLLARY: Irr(G) = k, the number of conjugacy classes of G. COROLLARY: G is abelian i every member of Irr(G) has degree 1. Proof: G is abelian i k = G. COROLLARY: ThesetIrr(G) is a basis for the space cf(g) of class functions.

20 A linear character is a character degree (1) = 1. such that the Note: The linear characters of G are exactly the homomorphisms from G into the group C. The principal character 1 G of G is the trivial homomorphism, with constant value 1. Note: The set of linear characters of G forms a group under pointwise multiplication. The identity is 1 G and 1 =, the complex conjugate of.

21 COROLLARY: The number of linear characters of G is G : G 0. Proof: Since a linear character of G is a homomorphism of G into the abelian group C,the derived subgroup G 0 is contained in the kernel of every such character. The linear characters of G, therefore, are exactly the linear characters of the abelian group G/G 0, and the number of these is G/G 0. Fact: The group of linear characters of G is isomorphic to G/G 0.

22 EXERCISE (2.1): Compute the degrees of the irreducible characters of S 3, A 4, S 4 and A 5. EXERCISE (2.2) :If is a character of G and is a linear character, define to be the function defined by ( )(g) = (g) (g). Show that is a character, and that it is irreducible i is irreducible.

23 In fact, if and are arbitrary characters of G then the function defined by ( )(g) = (g) (g) is a character. We sketch a proof. Let A be an m m matrix with entries a i,j, and let B be an n n matrix with entries b k,l.the Kronecker product of A and B, denoted A B, is an mn mn matrix, defined as follows. The rows and columns of A B are indexed by the set of ordered pairs (u, v) with1apple u apple m and 1 apple v apple n. The((i, k), (j, l))-entry of A B is a i,j b k,l.

24 Note: To write A B explicitly as a matrix, one must define some specific order on the index set. The choice of order is essentially irrelevant, however. It is easy to compute that trace(a B) = trace(a)trace(b). Also, if C and D are m m and n n matrices, respectively, it follows from the definition of matrix multiplication that (A B)(C D) =AC BD.

25 Now let X and Y be representations of G, affording characters and, respectively.define Z by Z(g) =X (g) Y(g). It follows from the above remarks that Z is a representation, and that the character a orded by Z is the product. This shows that products of characters are characters. Note: Ifeither or is reducible, it is easy to see that is reducible. Even if and are irreducible, however, their product is usually not irreducible.

26 EXERCISE (2.3): Suppose is the unique member of Irr(G) having degree d for some integer d. Show that (x) = 0 for all elements x 2 G G 0. EXERCISE (2.4): Let be a character of G, and let X a ord. Define the function on G by (g) = det(x (g)). Show that is a linear character of G and that it does not depend on the choice of the representation X a ording. Notation: = det( ). Also, the determinantal order o( ) is the order of det( ) in the group of linear characters of G.

27 Chapter 3: CHARACTER VALUES

28 Let X be a representation of G with degree d and let g 2 G have order n. Then X (g) n = I, the d d identity matrix. It follows by linear algebra that the matrix X (g) is similar to a diagonal matrix whose diagonal entries are n th roots of unity.

29 If X a ords the character,then (g) = trace(x (g)) = d, where the i for 1 apple i apple d are n th roots of unity. Also (g 1 ) = trace(x (g) 1 )= d = (g), where the overbar is complex conjugation.

30 THEOREM: Let be a character of a group G, where is a orded by a representation X, and let g 2 G. Then: (a) (g) apple (1). (b) (g) = (1) i X (g) is a scalar matrix. (c) (g) = (1) i X (g) is the identity matrix.

31 COROLLARY: Let be a character of G. Then {g 2 G (g) = (1)} is the kernel of every representation a ording. In particular, this subset is a normal subgroup of G. Notation: Wewrite ker( )={g 2 G (g) = (1)}. This normal subgroup is called the kernel of, and is faithful if ker( ) = 1.

32 EXERCISE (3.1): Let N / G and 2 Irr(G/N). Define : G! C by setting (g) = (Ng). Show that 2 Irr(G) and that N ker( ). Also, show that the map 7! from Irr(G/N) into Irr(G) defines a bijection from Irr(G/N) to the set { 2 Irr(G) N ker( )}. Note: It is customary to identify with we usually pretend that,so Irr(G/N) ={ 2 Irr(G) N ker( )}.

33 EXERCISE (3.2): Show that \ {ker( ) 2 Irr(G)} =1, the trivial subgroup of G. EXERCISE (3.3): Let N / G. Show that N = \ {ker( ) N ker( )}. Thus, knowing the irreducible characters of G determines the set of all normal subgroups of G.

34 COROLLARY: Let be a character of G afforded by X. Then {g 2 G (g) = (1)} is the preimage in G of the normal subgroup of X (G) consisting of scalar matrices. In particular, this set is a normal subgroup of G. Notation: Wewrite Z( )={g 2 G (g) = (1)}. This normal subgroup is the center of.

35 EXERCISE (3.4): Let be a character of G. Prove that Z( )/ker( ) is a cyclic subgroup of Z(G/ker( )). EXERCISE (3.5): If 2 Irr(G), show that Z( )/ker( )=Z(G/ker( )). EXERCISE (3.6): Let P be a p-group for some prime p. Show that P has a faithful irreducible character if and only if Z(P )iscyclic.

36 EXERCISE (3.7): Let 2 Irr(G), where G is a simple group of even order. Show that (1) 6= 2. HINT: Let g 2 G have order 2. values of (g) and det( )(g). Consider the EXERCISE (3.8): Let G be a simple group and suppose 2 Irr(G) has degree 3. Compute (g) for an element g 2 G of order 2. has de- EXERCISE (3.9): Repeat the above if gree 4.

37 EXERCISE (3.10): IfH G and is a character of G, it is easy to see that the restriction of to H, denoted H, is a character of H. Show that H Z( )i H has the form e,wheree is a positive integer and is a linear character of H. EXERCISE (3.11): Let H G and let be a character of G such that the restriction H is irreducible. Show that C G (H) Z( ).

38 Chapter 4: ORTHOGONALITY

39 G acts by right multiplication on C[G], so each element g 2 G determines a linear transformation on C[G]. We get a C-representation of G by choosing a basis for C[G]. The character a orded by this representation is the regular character of G, denoted or G.

40 To compute, we can use the basis G for C[G]. If R is the corresponding representation, R(g) is an G G matrix with rows and columns indexed by the elements of G. If x, y 2 G, then the entry of R(g) at position (x, y) is1ifxg = y and 0 otherwise. If g 6= 1, therefore, all diagonal entries of R(g) are 0.

41 We thus have (g) = 0 if g 6= 1 G if g =1. Next, we wish to express the character in terms of Irr(G).

42 Fact: = X 2Irr(G) (1). This is proved using a di erent basis for C[G], one compatible with the decomposition as a direct sum of matrix rings. (We omit the proof.) As a check (but not a proof) observe that G = (1) = X 2Irr(G) (1) 2.

43 Recall: C[G] is a direct sum of matrix rings. The summands correspond to members of Irr(G). For 2 Irr(G), let e be the unit element of the direct summand of C[G] corresponding to. For, 2 Irr(G), we have e if = e e = 0 if 6=. The e Note: are the central idempotents of C[G]. P e = 1. 2Irr(G)

44 Let X be the representation obtained by the projection of C[G] onto the direct summand corresponding to, and note that X a ords. Then X (e ) is the (1) (1) identity matrix and X (e )=0if 6=. For arbitrary u 2 C[G], we have X (u) if = X (ue )=X (u)x (e )= 0 if 6=.

45 Let X be the representation obtained by the projection of C[G] onto the direct summand corresponding to, and note that X a ords. Then X (e ) is the (1) (1) identity matrix and X (e )=0if 6=. For arbitrary u 2 C[G], we have X (u) if = X (ue )=X (u)x (e )= 0 if 6=. Thus (ue )= (u) if = 0 if 6=.

46 Write e = P g2g a g g with complex coe cients a g. Then a g is the coe cient of 1 in g 1 e. We have Thus G a g = (g 1 e )= a g = X 2Irr(G) = (1) (g 1 ). (1) (g) G. (1) (g 1 e )

47 Now compare the coe cients of 1 in e =(e ) 2. (1) 2 G Thus 1 = 1 G = X g2g(a g 1)(a g ) = X g2g = (1)2 G 2 X g2g (1) (g) X g2g G (g) (g). (g) (g) = 1 G (1) (g) G! X (g) 2. g2g

48 Now suppose, 2 Irr(G) are di erent, and compare the coe cients of 1 in 0 = e e. Writing e = P a g g and e = P b g g, we get 0= g2g(a X g 1)(b g ) = X g2g (1) (g) G (1) (g) G!. Thus X g2g (g) (g) = 0.

49 We now have the first orthogonality relation: 1 G X g2g (g) (g) = 1 if = 0 if 6=, for, 2 Irr(G).

50 Notation: For class functions and write [, ]= 1 X (x) (x). G g2g on G, we This is an inner product on the space cf(g) of class functions, and Irr(G) is an orthonormal basis for this space. Note: The form [.,.] is linear in the first variable and conjugate-linear in the second. Also [, ] =[, ] and [, ]=[, ].

51 If is an arbitrary class function on G, weknow that is a linear combination of Irr(G). The coe cient of in this linear combination is [, ], so we can write = X [, ]. 2Irr(G) Observe that [, ] = (1) since vanishes on nonidentity group elements and (1) = G. Thus = P (1), which is a fact we stated previously, but did not prove. Of course, this is not a proof.

52 COROLLARY: Let be a nonzero class function on G. Then is a character i [, ] is a nonnegative integer for all 2 Irr(G). COROLLARY: Let be a character of G. Then is irreducible i [, ]=1.

53 EXERCISE (4.1): Let H G and 2 Irr(G), and recall that H is the character of H obtained by restricting to H. Show that [ H, H] apple G : H with equality i has the value 0 on all elements of G H. EXERCISE (4.2): If G is abelian, show that [, ] (1) for all (not necessarily irreducible) characters of G. EXERCISE (4.3): Let H G, withh abelian, and let 2 Irr(G). Show that (1) apple G : H.

54 EXERCISE (4.4): Let 2 Irr(G), where G is a nontrivial group. Show that there is a nonidentity element g 2 G such that (g) 6= 0. EXERCISE (4.5): Let H < G with H is abelian. If 2 Irr(G) with G : H = (1), show that H contains a nonidentity normal subgroup of G. EXERCISE (4.6): Let G act on a set. Let be the corresponding permutation character, so (g) = {t 2 t.g = t}. Show that is a character of G and [, 1 G ] is the number of orbits of G on.

55 EXERCISE (4.7): Let G = A 5, the alternating group, and let be the permutation character of the natural action of G on five points. Show that = 1 G is an irreducible character of G having degree 4. EXERCISE (4.8): Again let G = A 5 and observe that G acts by conjugation on the set of six Sylow 5-subgroups of G. Let be the permutation character of this action. Show that = 1 G is an irreducible character of G having degree 5.

56 It is customary to display the irreducible characters of a group G in a character table. This is a square array of complex numbers with rows indexed by Irr(G) and columns indexed by the set of classes of G. The entry in a given row and column is the common value of the given character on the elements of the given class.

57 Usually, the first column of the character table corresponds to the class of the identity in G, and thus we see the irreducible character degrees of G by reading down this column. It is also customary for the first row of the character table to correspond to the principal character 1 G, and thus the first row consists of 1s. EXERCISE (4.9): Show that knowing the character table of G determines whether or not G is solvable or G is nilpotent.

58 We introduce some temporary notation now. Write Irr(G) ={ 1, 2,..., k} and let g j be a representative of the j th conjugacy class of G. The character table of G is thus the matrix X with (i, j)-entry equal to i (g j ). Write d t to denote the size of the t th conjugacy class of G. Then by the first orthogonality relation, we have:

59 i,j =[ i, j] = 1 G kx t=1 i(g t )d t j (g t ). We can rewrite this in matrix notation: I = 1 G XDXt, where D = diag(d 1,d 2,...,d k ). Then I = 1 G Xt XD.

60 We have so G i,j = G I = X t XD kx t=1 t(g i ) t (g j )d j. Now G /d j = C G (g j ), and thus for x, y 2 G, we have X CG (x) if x y (x) (y) = 0 otherwise, 2Irr(G) where x y means that x and y are conjugate.

61 This is the second orthogonality relation, also called column orthogonality. Since it is a consequence of the first orthogonality relation, or row orthogonality, it gives no additional information about the characters of G. It is,however, useful, especially when constructing character tables. A consequence is that no two columns of the character table can be identical. If x, y 2 G are not conjugate, therefore, there exists 2 Irr(G) such that (x) 6= (y).

62 Chapter 5: INTEGRALITY

63 An algebraic integer is a complex number that is a root of a monic polynomial in Z[x]. Examples: Elements of Z, roots of unity and things like np m where m, n 2 Z. Fact: The algebraic integers are a subring of C. COROLLARY: Character values are algebraic integers.

64 Fact: Rational algebraic integers are in Z. Fact: Suppose u i 2 C for 1 apple i apple n, and let R be the set of Z-linear combinations if the u i. If R is closed under multiplication, then all u i are algebraic integers.

65 Now let 2 Irr(G) and let X a ord. Let z 2 Z(C[G]) so X (z) commutes with all matrices in X (C[G]). Since this is a full matrix ring, we see that X (z) is a scalar matrix. Thus X (z) =!I for some complex number!, and we have (z) = trace(x (z)) = trace(!i) =! (1). It follows that! depends on but not on X, and we write! for!.

66 We have! (z) = (z)/ (1). Also X (z) =! (z)i, and it follows that! algebra homomorphism Z(C[G])! C. is an The function! is sometimes referred to as a central character. Recall that the conjugacy class sums in C[G] form a basis for Z(C[G]) so the central characters are determined by their values on class sums.

67 Let K be a class of G and write b K to denote the sum of the elements of K, so b K 2 Z(C[G]). We have ( b K)= K (g), where g 2 K. Then! ( b K)= ( b K) (1) = K (g) (1).

68 Now suppose K and L are classes of G. Then bk b L is a linear combination of class sums, so we can write bk b L = X M a K,L,M c M, where M runs over the classes of G. The coe cient of each group element on the left of the above equation is a nonnegative integer, so the coe cients a K,L,M must also be nonnegative integers.

69 We have! ( b K)! ( b L)= X a K,L,M! ( c M). As K runs over the classes of G, we see that the set of Z-linear combinations of the complex numbers! ( b K) is closed under multiplication. These numbers, therefore, are algebraic integers. THEOREM: Let g 2 G and 2 Irr(G), and let K be the class of g in G. Then! ( K)= b K (g) (1) is an algebraic integer.

70 THEOREM: (1) divides G for all 2 Irr(G). Proof: For each class K of G, letg K be an element of K. By the first orthogonality relation G = X g2g (g) (g) = X K K (g K ) (g K ), where the second sum runs over the classes of G.

71 Since K (g K )=! ( b K) (1), we have G = (1) X K! ( b K) (x K ). so the rational number G / (1) is an algebraic integer, and hence lies in Z. Note: A related argument can be used to prove the following stronger result. Fact: Let 2 Irr(G). Then G : Z(G) is divisible by (1).

72 EXERCISE (5.1): Let G be a p-group, where G 0 = Z(G) has order p. Prove the following. (a) Each noncentral class of G has size p. (b) G has exactly p characters. 1 nonlinear irreducible (c) The average of (1) 2 for nonlinear 2 Irr(G) is G /p. (d) (1) 2 apple G /p for all 2 Irr(G), and thus (1) 2 = G /p for nonlinear 2 Irr(G). Note: Groups of this type are extraspecial.

73 We begin work now toward a proof of Burnside s famous p a q b -theorem. LEMMA: Let be a character of G, where G = n, and let g 2 G. Suppose (g) < (1). Let = Gal(Q n /Q), whereq n is the n th cyclotomic field. Then (g) < (1) for all 2. Proof: We know that (g) is a sum of (1) roots of unity in Q n. These are not all equal since (g) < (1). It follows that (g) is also a sum of (1) roots of unity that are not all equal. The result follows.

74 THEOREM (Burnside): Let 2 Irr(G) and g 2 G, and assume that (1) is relatively prime to the size of the class K containing g. Theneither (g) =0or g 2 Z( ). Proof: Writea (1) + b K = 1 for integers a, b. Now! = Then (g) (1) = (g) (1) (g) K / (1) is an algebraic integer. is an algebraic integer. a (1) + b K = a (g)+b!

75 Let = Gal(Q n /Q), where n = G, and let z = Y 2 (g) (1), so z is an algebraic integer in Q. Thenz 2 Z. Assume now that g 62 Z( ). Then (g) < (1), so by the lemma, (g) < (1) for all 2. Each factor of z has absolute value less than 1, so z < 1. Since z 2 Z, wehavez = 0. Then (g) = 0 for some,so (g) = 0.

76 THEOREM (Burnside): Let G be nonabelian and simple. Then no nontrivial conjugacy class of G has prime power size. Proof: Suppose K is a power of a prime p, where K is a nontrivial class, and let g 2 K, so g 6= 1. If 2 Irr(G), then Z( ) / G. IfZ( )=G, then is linear. But G 0 = G, so is principal. Thus if 2 Irr(G) is nonprincipal, Z( ) = 1.

77 By the previous theorem, (g) = 0 for all nonprincipal where p does not divide (1). We have 0= (g) = X 2Irr(G) (1) (g) Separate =1 G and sum only over for which p divides (1). This yields 0 = 1 + p, where is an algebraic integer. Since = 1/p 2 Q, this is a contradiction.

78 THEOREM (Burnside): If G has at most two prime divisors, then G is solvable. Proof: Let G be a minimal counterexample, so G is nonabelian. If N / G with 1 <N<G,then N and G/N are solvable, so G is solvable, which is a contradiction. Thus G is simple. Let P > 1 be a Sylow subgroup of G. Choose z 2 Z(P )withz 6= 1. Let K be the class of z in G, so K = G : C G (z), and this divides G : P, which is a prime power, by hypothesis. This contradicts the previous theorem.

79 EXERCISE (5.2): Suppose that no prime other than p or q divides the degree of any member of Irr(G). Show that G cannot be a nonabelian simple group.

A PROOF OF BURNSIDE S p a q b THEOREM

A PROOF OF BURNSIDE S p a q b THEOREM OBOB Abstract. We prove that if p and q are prime, then any group of order p a q b is solvable. Throughout this note, denote by A the set of algebraic numbers. We