# GROUP ALGEBRAS. ANDREI YAFAEV

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1 GROUP ALGEBRAS. ANDREI YAFAEV We will associate a certain algebra to a finite group and prove that it is semisimple. Then we will apply Wedderburn s theory to its study. Definition 0.1. Let G be a finite group. We define F[G] as a set of formal sums u = g G λ g g,λ g F endowed with two operations: addition and multiplication defined as follows. and g G λ g g + g G µ g g = g G g G λ g g g G µ h h = g G,h G (λ g + µ g )g (λ h µ h 1 g)g Note that the multiplication is induced by multiplication in G, F and linearity. The following proposition is left to the reader. Proposition 0.1. The set (F[G], +, ) is a ring and is an F-vector space of dimension G with scalar multiplication compatible with group operation. Hence F[G] is an F-algebra. The algebra F[G] is non-commutative unless the group G is comutative. It is clear that the basis elements (elements of G) are invertible in F[G]. Lemma 0.2. The algebra F[G] is a not a division algebra. Proof. It is easy to find zero divisors. Let g G and let m be the order of G (the group G is finite, every element has finite order). Then (1 g)(1 + g + g g m 1 ) = 0 1

2 2 ANDREI YAFAEV In this course we will study F[G]-modules, modules over the algebra F[G]. An important example of a F[G]-module is F[G] itself viewed as a F[G]-module. We leave the verifications to the reader. This module is called a regular F[G]-module and the associated representation a regular representation. Next we introduce the notion of F[G]-homomorphism between F[G]- modules. Definition 0.2. Let V and W be two F[G]-modules. A function φ: V W is called a F[G]-homomorphism if it is a homomorphism from V to W viewed as modules over F[G]. That means that φ is F-linear and satisfies φ(gv) = gφ(v) for all g G and v V. We obviously have the following. Proposition 0.3. Let φ: V W be a F[G]-homomorphism. Then ker(φ) and im(φ) are F[G]-submodules of V and W respectively. Definition 0.3. Let G be a finite group, F a field and V a finite dimensional vector space over F. A representation ρ of G on V is a group homomorphism ρ: G GL(V ) A representation is called faithful if ker(ρ) = {1}. A representation is called irreducible if the only subspaces W of V such that ρ(g)w W are W = {0} and W = V. The following theorem telles us that a representation of G and an F[G]-module are same things. Theorem 0.4. Let G be a finite group and F a field. There is a oneto-one correspondence between representations of G over F and finitely generated left F[G]-modules. Proof. Let V be a (finitely generated) F[G]-module. Then V is a finite dimensional vectore space. Let g be in G, then, by axioms satisfied by a module, the action of g on V defines an invertible linear map which gives an element ρ(g) of GL(V ). It is trivial to check that ρ: G GL(V ) is a group homomorphism i.e. a representation G GL(V ). Let ρ: G GL(V ) be a representation. Let x = g G λ gg be an element of F[G] and let v V. Define xv = g G λ g ρ(g)v

3 GROUP ALGEBRAS. 3 It is easy to check that this defines a structure of an F[G]-module on V. By definition, a morphism between two representation is a morphism of the corresponding F[G]-modules. Two representations are isomorphic (or equivalent) if the corresponding F[G]-modules are isomorphic. Given an F[G]-module V and a basis B of V as F-vector space, for g G, we will denote by [g] B the matrix of the linear transformation defined by g with respect to the basis B. For example : Let D 8 = {a,b : a 4 = b 2 = 1,b 1 ab = a 1 } and define a representation by ρ(a) = ( ) and ρ(b) = ( ) ( ( 1 0 Choose B to be the canonical basis v 1 = and v 0) 2 = of V = F 1) 2. We have: av 1 = v 2 av 2 = v 1 bv 1 = v 1 bv 2 = v 2 This completely determines the structure of V as a F[D 8 ]-module. Conversely, by taking the matrices [a] B and [b] B, we recover our representation ρ. Another example : Let G be the group S n, group of permutations of the set {1,...,n}. Let V be a vector space of dimension n over F (the n here is the same as the one in S n ). Let {v 1,...,v n } be a basis of V. We define gv i = v g(i) The reader will verify that the conditions of the above proposition are verified and hence we construct a F[G]-module called the permutation module. Let n = 4 and let B = {v 1,...,v n } be a basis of F 4. Let g be the prmutation (1, 2). Then The matrix [g] B is gv 1 = v 2,gv 2 = v 1,gv 3 = v 3,gv 4 = v

4 4 ANDREI YAFAEV Lemma 0.5. A representation ρ: G GL n (F) is irreducible (or simple) if and only if the corresponding F[G]-module is simple. Proof. A non-trivial invariant subspace W V is a non-trivial F[G]- submodule, and conversely. Note that ρ being irreducible means that the only ρ(g)-invariant subspaces of V are {0} and V itself. If a representation is reducible i.e. there is a F[G]-submodule W of V, then we can choose a basis B of V (choose a basis B 1 of B and complete it to a basis of B) in such a way that the matrix [g] B for all g is of the form ( Xg Y g 0 Z g ) where X g is a dimw dimw matrix. Clearly, the functions g X g and g Z g are representations of G. Let s look at an example. Take G = C 3 = {a : a 3 = 1} and consider the F[G]-module V (dim V = 3) such that av 1 = v 2,av 2 = v 3,av 3 = v 1 (Easy exercise : check that this indeed defines an F[G]-module) This is a reducible F[G]-module. Indeed, let W = Fw with w = v 1 + v 2 + v 3. Clearly this is a F[G]-submodule. Consider the basis B = {w,v 2,v 3 } of V. Then [I 3 ] B = [a] B = (For the last matrix, note that: aw = w,av 2 = v 3,av 3 = v 1 = w v 2 v 3 ) Given two F[G]-modules, a homomorphism φ: V W of F[G]- modules is what you think it is. It has a kernel and an image that are F[G]-submonules of V and W. Example. Let G be the group S n of permutations. Let V be the permutation module for S n and {v 1,...,v n } a basis for V. Let w = i v i and W = F w. This is a F[G]-module. Define a homomorphism φ: i λ i v i ( i λ i )w This is a F[G]-homomorphism (check!). Clearly, ker(φ) = { λ i v i, λ i = 0} and im(φ) = W

5 GROUP ALGEBRAS. 5 We now get to the first very important result of this chapter. It says that F[G]-modules are semisimple. Theorem 0.6 (Maschke s theorem). Let G be a finite group and F a field such that CharF does not divide G (ex. CharF = 0). Let V be a F[G]-module and U an F[G]-submodule. Then there is an F[G]- submodule W of V such that V = U W In other words, F[G] is a semisimple algebra. Proof. Choose any subspace W 0 of V such that V = U W 0. For any v = u + w, define π: V V by π(v) = u (i.e. π is a projection onto U). We will modify π into an F[G]-homomorphism. Define φ(v) = 1 G gπg 1 (v) g G This clearly is an F-linear morphism V V. Furthermore, im(φ) U (notice that π(g 1 v) U and as U is an F[G]-module, we have gπ(g 1 v) U). Claim 1. : φ is a F[G]-homomorphism. Let x G, we need to show that π(xv) = xπ(v). Let, for g G, h := x 1 g (hence h 1 = g 1 x). Then φ(xv) = 1 G h G x(hπh 1 )(v) = x 1 G (hπh 1 )(v) = xφ(v) This proves the claim. Claim 2. : φ 2 = φ. For u U and g G, we have gu U, therefore φ(gu) = gu. Now φ(u) = 1 (gπg 1 )u = 1 (gπ(g 1 u) = 1 gg 1 u = 1 u = u G G G G Let v V, then φ(u) U and it follows that φ 2 (v) = φ(v), this proves the claim. We let W := ker(φ). Then, as φ is a F[G]-homomorphism, W is a F[G]-module. Now, the minimal polynomial of φ is x 2 x = x(x 1). Hence h G V = ker(φ) ker(φ I) = W U This finishes the proof.

6 6 ANDREI YAFAEV Note that without the assumption that Char(F) does not divide G, the conslusion of Mashke s theorem is wrong. For example let G = C p = {a : a p = 1} over F = F p. Then the function ( ) 1 0 a j j 1 for j = 0,...,p 1 is a representation of G of dimension 2. We have a j v 1 = v 1 a j v 2 = jv 1 + v 2 Then U = Span(v 1 ) is a F[G]-submodule of V. But there is no F[G]-submodule W such that V = U V as (easy) U is the only 1-dimensional F[G]-submodule of V. Similarly, the conclusion of Maschke s theorem fails for infinite groups. Take G = Z and the representation ( ) 1 0 n n 1 The proof of Maschke s theorem gives a procedure to find the complementary subspace. Let G = S 3 and V = {e 1,e 2,e 3 } be the permutation module. Clearly, the submodule U = Span(v 1 + v 2 + v 3 ) is an F[G]-submodule. Let W 0 = Span(v 1,v 2 ). Then V = U W 0 as C-vector spaces. The projection φ onto U is given by φ(v 1 ) = 0, φ(v 2 ) = 0, φ(v 3 ) = v 1 + v 2 + v 3 The F[G]-homomorphism as in the proof of Maschke s theorem is given by Φ(v i ) = 1 3 (v 1 + v 2 + v 3 ) Clearly ker(φ) = Span(v 1 v 2,v 2 v 3 ). The F[G]-submodule is then W = Span(v 1 v 2,v 2 v 3 ) This is the F[G]-submodule such that V = U W. Actually, you may notice that is submodule is W = { λ i v i : λ i = 0} By applying a theorem from the previous chapter. Corollary 0.7. Let G be a finite group and V a F[G] module where F = R or C. There exist simple F[G]-modules U 1,...,U r such that V = U 1 U r In other words, F[G] modules are semisimple. Another corollary:

7 GROUP ALGEBRAS. 7 Corollary 0.8. Let V be an F[G]-module, CharF does not divide G. Let U be an F[G] submodule of V. There is a surjective F[G] homomorphism V U. Proof. By Mascke s theorem there is an F[G]-submodule W such that V = U W. Consider π: u + w u. We can now state Shur s lemma for F[G]-modules: Theorem 0.9 (Schur s lemma). Suppose that F is algebraically closed. Let V and W be simple F[G]-modules. (1) If φ: V W is a F[G]-homomorpism, then either φ is a F[G]-isomorphism or φ(v) = 0 for all v V. (2) If φ: V W is a F[G]-isomorpism, then φ is a scalar multiple of the identity endomorphim I V. This gives a characterisation of simple F[G]-modules and it is also a partial converse to Shur s lemma. Proposition Suppose CharF does not divide G Let V be a nonzero F[G]-module and suppose that every F[G]-homomorphism from V to V is a scalar multiple of I V. Then V is simple. Proof. Suppose that V is reducible, then by Maschke s theorem, we have V = U W where U and W are F[G]-submodules. The projection onto U is a F[G]-homomorphism which is not a scalar multiple of I V (it has a non-trivial kernel!). This contradicts the assumtion. We now apply Shur s lemma to classifying representations of abelian groups. In what follows, the field F is C. Let G be a finite abelian group and V a simple C[G]-module. As G is abelian, we have xgv = g(xv),x,g G Therefore, v xv is a C[G]-homomorphism V V. As V is irreducible, Shur s lemma imples that there exists λ x C such that xv = λ x v for all V. In particular, this implies that every subspace of V is a C[G]-module. The fact that V is simple implies that dim(v ) = 1. We have proved the following: Proposition If G is a finite abelian group, then every simple C[G]-module is of dimension one.

8 8 ANDREI YAFAEV The basic structure theorem for finite abelian groups is the following: Theorem 0.12 (Structure of finite abelian groups). Every finite abelian group G is a direct product of cyclic groups. Let G = C n1 C nr and let c i be a generator for C ni and we write Then g i = (1,...,1,c i, 1,...,1) G =< g 1,...,g r >,g n i = 1,g i g j = g j g i Let ρ: G GL n (C) be an irreducible representation of G. We know that n = 1, hence GL n (C) = C. There exist λ i C such that ρ(g i ) = λ i The fact that g i has order n i implies that λ n i i = 1. This completely determines ρ. Indeed, let g = g i 1 1 g nr r, we get ρ(g) = λ i 1 1 λ ir r As ρ is completely determined by the λ i, we write We have shown: ρ = ρ λ1,...,λ r Theorem Let G = C n1 C nr. The representations ρ λ1,...,λ r constructed above are irreducicible and have degree one. There are exactly G of these representations. Let us look at a few examples. Let G = C n = {a : a n = 1} and let ζ n = e 2πi/n. The n irreducible representations of G are the ρ ζ i n (a k ) = ζ k n where 0 k n 1. Let us classify all irreducible representations of G = C 2 C 2 =< a 1,a 2 >. There are four of them, call them V 1,V 2,V 3,V 4 where V i is a one dimensional vector space with basis v i. We have a 1 v 1 = v 1 a 2 v 1 = v 1 a 1 v 2 = v 2 a 2 v 2 = v 2 a 1 v 3 = v 3 a 2 v 3 = v 3 a 1 v 4 = v 4 a 2 v 4 = v 4

9 GROUP ALGEBRAS. 9 Let us now turn to not necessarily irreducible representations. Let G =< g > be a cyclic group of order n and V a C[G]-module. Then V decomposes as V = U 1 U r into a direct sum of irreducible C[G]-modules. We know that every U i has dimension one and we let u i be a vector spanning U i. As before we let ζ n = e 2πi/n. Then for each i there exists an integer m i such that gu i = ζ m i n u i Let B = {u 1,...,u r } be the basis of V consisting of the u i. Then the matrix [g] B is diagonal with coefficients ζ m i n. As an exercise, the reader will classify representations of arbitrary finite abelian groups (i.e products of cyclic groups). The statement that all irreducible representations of abelian groups have degree one has a converse. Theorem Let G be a finite group such that all irreducible representations of G are of degree one. Then G is abelian. Proof. We can write C[G] = U 1 U n where each U i is simple and hence is of degree one by assumption. Let u i be a generator of U i, then {u 1,...,u n } is a basis of C[G] as a C-vector space. Let g be in C[G], then the matrix of the action of g on C[G] in this basis is diagonal (because U i s are C[G]-modules!). The regular representation of G (action of G on C[G] given by multiplication in G) is faithful. Indeed, suppose g (λ h h) = λ h h for all λ h h C[G]. Then, in particular g 1 = 1 hence g = 1. It follows that the group G is realised as a group of diagonal matrices. Diagonal matrices commute, hence G is abelian. 1. C[G] as a module over itself. In this section we study the structure of C[G] viewed as a module over itself. We know that C[G] decomposes as C[G] = U 1 U r where the U i s are irreducible C[G]-submodules. As by Mashke s theorem C[G] is a semisimple algebra, U i s are the only simple C[G]-modules.

10 10 ANDREI YAFAEV Then we have seen that every irreducible C[G]-module is isomorphic to one of the U i s. In particular there are only finitely many of them. Let s look at examples. Take G = C 3 = {a : a 3 = 1} and let ω = e 2iπ/3. Define v 0 = 1 + a + a 2 v 1 = 1 + ω 2 a + ωa 2 v 3 = 1 + ωa + ω 2 a 2 Let U i = Span(v i ). One checks that av i = ω i v i and U i s are C[G]-submodules. It is not hard to see that v 1,v 2,v 3 form a basis of C[G] and hence C[G] = U 0 U 1 U 2 direct sum of irreducible C[G]-modules. Look at D 6. It contains C 3 =< a >. Define: v 0 = 1 + a + a 2, w 0 = v 0 b v 1 = 1 + ω 2 a + ωa 2, w 1 = v 1 b v 3 = 1 + ωa + ω 2 a 2, w 2 = v 2 b As before, < v i > are < a >-invariant and av 0 = v 0, aw 0 = v 0 bv 0 = w 0, bw 0 = v 0 It follows that Span(u 0,w 0 ) is a C[G] modules. It is not simple, indeed, it is the direct sum U 0 U 1 where U 0 = Span(u 0 + w 0 ) and U 1 = Span(u 0 w 1 ) and they are simple submodules. Notice that the irreducible representation of degree one corresponding to U 0 is the trivial one : sends a and b to 1. The one corresponding to U 1 sends a to 1 and b to 1. Next we get : av 1 = ωw 2, aw 2 = ω 2 w 2 bv 1 = w 2, bw 2 = v 1 Therefore U 2 = Span(v 1,w 2 ) is C[G]-module. It is an easy exercise to show that it is irreducible. The corresponding two-dimensional representation is ( ) ω 0 a 0 ω 1

11 GROUP ALGEBRAS. 11 and Lastly b ( ) av 2 = ω 2 w 1, aw 1 = ωw 1 bv 2 = w 1, bw 1 = v 2 Hence U 3 = Span(v 2,w 1 ) is a C[G]-module and one shows that it is irreducible. In fact the morphism φ that sends v 1 w 1 and w 2 v 2 is C[G]-isomorphism (you need to chack that φ(av) = aφ(v) and φ(bv) = bφ(v) for all v C[G]). Therefore the representations U 2 and U 3 are isomorphic. We have C[G] = U 0 U 1 U 2 U 3 with dimu 0 = dimu 2 = 1 and corresponding representations are nonisomorphic. And dimu 2 = dimu3 = 2 and the corresponding representations are isomorphic. We have completely classified all irreducible representations of C[D 6 ] and realised them explicitly as sumbmodules of C[D 6 ] Wedderburn decomposition revisited. We now apply the results we proved for semisimple modules to the group algebra C[G]. View C[G] as a module over itself (regular module). By Maschke s theorem, this module is semisimple. There exist r distinct simple modules S i and integers n i such that We have C[G] = S n 1 1 S nr r C[G] op = End C[G] (C[G]) = M n1 (End(S 1 )) M nr (End(S r )) As S i is simple and C is algebraically closed, End(S i ) = C. By taking the opposite algebra, we get C[G] = M n1 (C) M nr (C) (note that C op = C because C is commutative.) Each S i becomes a M ni (C)-module. Indeed, S i is a C[G] = M n1 (C) M nr (C)-module and M ni (C) is a subalgebra of C[G]. As a M ni (C)- module, S i is also simple. Indeed, suppose that S i is a non-trivial M ni (C)-submodule of S i. Then, 0 0 S i 0 is a non-trivial C[G]-submodule of S i.

12 12 ANDREI YAFAEV We have seen in the previous chapter that simple M ni (C)-modules are isomorphic to C n i (column vector modules). It follows that dim C (S i ) = n i and as dim C C[G] = G, we get the following very important relation r G = The integers n i s are precisely the degrees of all possible irreducible representations of G. In addition, for any finite group there is always an irreducible one dimensional representation : the trivial one. Therefore we always have n 1 = 1. Using this relation we already can determine the degrees of all irreducible representations of certain groups. For abelian groups they are always one. For D 6 : 6 = We recover what we proved above. For D 8 we have 8 = hence four one-dimensional ones (exercise : determine them) and one two dimensional (determine it!). Same for Q 8. We will now determine the integer r : the number of isomorphism classes of irreducible representations. Definition 1.1. Let G be a finite group. The centre Z(C[G]) of the group algebra C[G] is defined by i=1 Z(C[G]) = {z C[G] : zr = rz for all r C[G]} The centre Z(G) of the group G is defined similarly: We have: Lemma 1.1. Z(G) = {g G : gr = rg for all r G} n 2 i dimz(c[g]) = r Proof. Write C[G] = M n1 (C) M nr (C). Now, the centre of each M ni (C) is C and there are r factors, hence Z(C[G]) = C r. Recall that a conjugacy class of g G is the set {x 1 gx : x G} and G is a disjoint union of conjugacy classes. We show: Theorem 1.2. The number r of irreducible representations is equal to the number of conjugacy classes.

13 GROUP ALGEBRAS. 13 Proof. We calculate the dimension of Z(C[G]) in a different way. Let g G λ gg be an element of Z(C[G]). By definition, for any h G we have h( λ g g)h 1 = λ g g g G g G We have h( g G λ g g)h 1 = g G λ hgh 1g Therefore λ g = λ hgh 1 and therefore the function λ g is constant on conjugacy classes. Hence the centre is generated by the { g K g : K conjugacy class} But this family is also free because conjugacy classes are disjoint hence it is a basis for Z(C[G]). This finishes the proof. For example we recover the fact that irreducible representations of abelian groups are one dimensional : each conjugacy class consists of one element. By what we have seen before, we know that D 6 has three conjugacy classes, D 8 has five Conjugacy classes in dihedral groups. We can in fact determine completely conjugacy classes in dihedral groups. Let G be a finite group and for x G, let us denote by x G the conjugacy class of x. Let C G (x) = {g G : gx = xg} This is a subgroup of G called the centraliser of x. We have x G = G : C G (x) = G C G (x) We have the followin relation (standard result in group theory). Let x 1,...,x m be representatives of conjugacy classes in G. G = Z(G) + x G i Let us now turn to the dihedral group x i / Z(G) G = D 2n = {a,b : a n = b 2 = 1,b 1 ab = a 1 } Suppose that n is odd. Consider a i for 1 i n 1. Then C(a i ) contains the group generated by a: obviously aa i a 1 = a i. It follows that a G = G : C G (a) 2 = G :< a >

14 14 ANDREI YAFAEV On the other hand b 1 a i b = a i so {a i,a i } a ig. As n is odd a i a i (a 2i = 1 implies that n = 2i but n is odd). It follows that a ig 2 hence a ig = 2 C G (a i ) =< a > a ig = {a i,a i } Next C G (b) contains 1 and b. As b 1 a i b = a i and a i a i, therefore a i and a i b do not commute with b. Therefore C G (b) = {1,b}. It follows that b G = n and we have b G = {b,ab,...,a n 1 b} (notice that all elements of G are {1,a,a 2,...,a n 1,b,ab,...,a n 1 b}) We have determined all conjugacy classes in the case n is odd. Proposition 1.3. The dihedral group D 2n with n odd has exactly n+3 2 conjugacy classes and they are {1}, {a,a 1 },...,{a (n 1)/2,a (n 1)/2 }, {b,ab,...,a n 1 b} Suppose n = 2m is even. We have a m = a m such that b 1 a m b = a m = a m and the centraliser of a m contains both a and b, hence C G (a m ) = G The conjugacy class of a m is just a m. As before a ig = {a i,a i } for 1 i m 1. We have a j ba j = a 2j b, a j ba j = a 2j+1 b It follows that b G = {a 2j b : 0 j m 1} and (ab) G = {a 2j+1 b : 0 j m 1} We proved: Proposition 1.4. In D 2n for n = 2m even, there are exactly m + 3 conjugacy classes, they are {1}, {a m }, {a i,a i } for 1 i m 1, {a 2j b : 0 j m 1} and (ab) G = {a 2j+1 b : 0 j m 1} In particular, we know the number of all irreducible representations of D 2n.

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