0.1 Algebras and Modules

Transcription

1 0.1. ALGEBRAS AND MODULES Algebras and Modules Let F be a field. An F-algebra A, is a ring with identity which is also a vector space over F and such that the following compatibility relation holds: (cx)y = c(xy) = x(cy), c F, x, y A Note: In what follows we identify c F with c 1 A A, when appropriate. Accordingly, we also identify F with its copy F 1 A in A. Example 1: If V is a vector space over F, then Hom F (V, V ) is an F - algebra, where addition and scalar multiplication of functions are pointwise, and multiplication is function composition. Example 2: (the group algebra) Let G be a group and K a field. Then the set KG = { c g g c g K} of formal sums in the elements of G with coefficients in K, is a K-algebra under g g) + (c g g) = (d ((c g + d g ) g) (c g g) g g) = (d ( c g d h ) x x G g,h G gh=x ( c ) c g g = g g) (cc Note also that the elements of G form a basis of KG as a vector space over K. Let A, B be K-algebras. A map f : A B is an algebra homomorhism of A and B if: f(xy) = f(x)f(y) f(1 A ) = 1 B f(cx + y) = c f(x) + f(y) x, y A, c K Let A be a K-algebra. An A-module is an abelian group M, such that A acts on M (i.e. we have a map: A M M, (a, m) am), and the following hold: g(v + w) = gv + gw (g + h)v = gv + hv h(gv) = (hg)v g(cv) = c(gv) = (cg)v 1 A v = v v, w M, g, h A, c K. In this case M is also a K-vector space, with addition inherited from the group M, and scalar multiplication defined by kv = (k 1 A )v. Note: Throughout this text, we will only consider modules which are finite dimensional as vector spaces.

2 2 Fix g A. Then g M : M M, defined by g M (v) = gv, is an element of End(M). The map g g M is a K-algebra homomorphism of A and End K (M). The image of A under this homomorphism is written as A M. Note: Sometimes we will identify g with g M, which will be clear from the context. Example 3: Let V be a vector space and A a subalgebra of End K (V ), then V is naturally an A-module (under gv = g(v), v V, g A). Example 4: In the previous example, let V = K n (the vector space of n-tuples over K). If A is any subalgebra of M n (K), then K n is an A-module under matrix multiplication. Example 5: Let A be a K-algebra. Then A is a module over itself by multiplication on the left. We denote this module by A (called the left regular representation of A). Example 6: K is a KG-module under gk = k, k K, g G, and extension by linearity (use the KG-module relations to find ( k gg)k = k gk). Let V be an A-module (here A is a K-algebra). A submodule of V, is a vector subspace W of V, which is invariant under the action of A (so it is itself an A-module under the same action as V ) Example 7: The submodules of A are precisely the left ideals of A. Example 8: If W is a submodule of the A-module V, then the quotient vector space V/M is an A-module in a natural way. Exercise: Let I be a proper ideal of the K-algebra A. Explain the algebraic structures A/I, A /I, and (A/I). Are they all different? 0.2 Module Homomoprhisms Let A be a K-algebra. Let V, W be A-modules. An A-module homomorphism is a K-linear map ϕ : V W, such that ϕ(gv) = g(ϕ(v)), v V, g KA. We then define Hom A (V, W ) = {ϕ ϕ : V W is an A-module homomorphism} End A (V ) = Hom A (V, V ) Note that End A (V ) is a K-algebra, with addition and scalar multiplication of homomorphisms defined poinwise, and multiplication defined as function composition. Moreover, End A (V ) is the centralizer of A V in End K (V ). Note: When A is the group algebra KG, we call KG-modules just G-modules and KGmodule homomorphisms just G-homomorphisms. A non-zero A-module M is called irreducible if its only submodules are 0 and M. Theorem 1:(Schur s Lemma) Let V, W be irreducible G-modules and assume ϕ : V W is a G-homomorphism. Then: 1. Either ϕ is an isomorphism, or ϕ 0 2. If V = W and K is algebraically closed, then ϕ = λi V for some λ K.

3 0.2. MODULE HOMOMOPRHISMS 3 Proof: Part 1 follows from the fact that kerϕ and imϕ are G-submodules of the irreducible G-modules V and W, respectively. 2 ϕ is a K-linear map on the vector space V and so, since K is algebraically closed, it has an eigenvector v 0. Thus, vϕ = vλ, for some λ K. Then v ker(ϕ λi V ) and part 1 implies ϕ = λi V Consequences: 1. If V is an irreducible G-module, then Hom G (V, W ) is a division algebra. 2. If K is algebraically closed, then End G (V ) = K I V, is a field isomorphic to K. Example 1: (Representations of Abelian Groups) Let G be a group and V a G-module over an algebraically closed field K. If we identify the element g G with the map g V End K (V ), then by definition it follows that g G is a G-module homomorphism gh = hg, h G g Z(G). Now assume G is abelian. By above, all g G are G-homomorphisms. Let V be an irreducible G-module. By Schur s lemma, we have g = λ g I V for some λ g K. Hence every subspace of V is G-invariant. Since V is irreducible, this forces dimv = 1 so that V can be identified with K. Then g becomes a linear functional on K: g (c K λ g c K) Since λ gh c = (gh)(c) = g(hc) = g(λ h c) = λ g (λ h c) = λ g λ h c, c K, we get λ gh = λ g λ h for all g, h G. Thus g λ g is a homomorphism of G and the multiplicative group K. In particular, g n = 1 (λ g ) n = 1, since 1 G is the identity map on K. In other words, λ g is a root of unity in K, of order dividing the order of g. Example 2: In the previous example let G = Z 2 Z 3 be generated by elements a of order 2 and, and b of order 3. Also let K = C. Then all possible maps g G λ g C can be arranged in the following character table of G: 1 a b b 2 ab ab 2 ρ ρ w w 2 w w 2 ρ w 2 w w 2 w ρ ρ w w 2 w w 2 ρ w 2 w w 2 w where w C, w 3 = 1 Note that any 2 rows (or columns) are orthogonal. Theorem 2:(Maschke) Assume chark. Let V be a G-module and W a submodule of V. Then there exists a submodule W of V such that W W = V. Proof: Let U be a subspace of V such that W U = V. Let P 0 be the projection of V onto W (with respect to U). Define P : V W by P v = gp 0(g 1 v). Note that P is K-linear since P 0 and g are. For all v V and h G, we have: P (hv) = hh 1 gp 0 (g 1 hv) = h(h 1 g)p 0 ((h 1 g) 1 v) = x G hx 1 P 0 (xv) = hp (v). Therefore, P is a G-module homomorphism. Next, observe that P (w) = w, w W. Indeed, P (w) = gp 0(g 1 w) =

4 4 g(g 1 w) = w, since W is G-invariant, and on W, P 0 is just the identity map. Take W = kerp, which is a submodule of V. We claim that this W satisfies W W = V : First, pick an arbitrary u V. Then, since chark, there exists v V such that u = v. Write u = P (v) + (v P (v)) (1) Note that P (v P (v)) = P (v) P (P (v)) = P (v) P (v) = 0, since P (v) W and on W, P is multiplication by. Hence v P (v) W. Then (1) implies that W + W = V. Finally, if w W W then w = P (w) on one hand, and P (w) = 0 on the other. Again since chark, we get w = 0. Therefore, we have a direct sum: W W = V 0.3 Basic G-Modules Let G be any group, and V, W be G-modules over some field K. Then, 1. V W is a G-module under g(v, w) = (gv, gw), g G. 2. Recall the tensor product V K W, which is defined as the free product of V and W modulo the module relations: (v 1 + v 2 ) w = v 1 w + v 2 w v (w 1 + w 2 ) = v w 1 + v w 2 c(v w) = (cv) w = v (cw) for all v, v 1, v 2 V, w, w 1, w 2 W and c K. Then V K W is a G-module under g(v w) = (gv gw), g G, and extension by linearity. 3. Hom(V, W ) is a G-module as follows: for ϕ Hom(V, W ), define gϕ by (gϕ)(v) = g(ϕ(g 1 v)), g G, and extend linearly. 4. Letting W = K in above, we get that the dual V = Hom(V, K) is a G-module under (gϕ)(v) = g (ϕ(g 1 v)) = ϕ(g 1 v), since g G acts on K as the identity map. Lemma 1: Hom(V, W ) = V W as G-modules. Proof: Consider γ : V W Hom(V, W ), defined by ϕ w γ (v ϕ(v) w) (2) and extended linearly. Then γ is injective: Indeed, let ξ kerγ. We may assume ξ = n ϕ i w i where w i W are linearly independent (otherwise ξ = 0). Then γ(ξ) = 0 implies n ϕ i (v) w i = 0, v V.

5 0.4. REPRESENTATIONS AND CHARACTERS 5 Hence ϕ i (v) = 0, v V, i = 1, n, and thus ξ = 0. We conclude that kerγ = 0. Further, γ must also be onto, since Hom G (V, W ) and V W have same finite dimension dim K V dim K W over K (so they are isomorphic as vector spaces). Finally, γ is a G-homomorphism: Let g G. We have γg(ϕ w)(v) = (γ(gϕ gw))v = gϕ(v) gw, v V. We also have gγ(ϕ w)(v) = g(ϕ(v) w) = ϕ(v) gw v V, since ϕ(v) K. But g acts as the identity on K : gϕ(v) = ϕ(v). Therefore, gγ = γg, as desired 0.4 Representations and Characters Let G be a group and V a vector space over some field K. We assume that chark. A linear map (or representation) of G on V, is a group homomorphism ρ : G GL(V ), Given a representation ρ of G on V, its character χ ρ is a map : G K defined by χ ρ (g) = trace(ρ(g)). Thus, the following diagram commutes: ρ G GL(V ) χ ρ trace K Note: If ρ : G GL(V ) is linear, then V is a G-module under gv = ρ(g)(v). Conversely, if we start with a G-module M, the map g g M defined in section 0.1 is a representation of G on M. We denote its character by χ M. Theorem 3: Let ρ be a linear map of G on M, and V, W any G-modules. Assume the ground field K is algebraically closed. Then 1. The linear operator ρ(g) GL(V ) is diagonalizable, with roots of unity as eigenvalues. 2. χ ρ (h 1 gh) = χ ρ (g), g, h G (i.e. χ ρ is a class function on G). 3. χ V W = χ V + χ W 4. χ V W = χ V χ W 5. χ Hom(V,W ) = χ V χ W 6. If K = C, then χ ρ (g 1 ) = χ ρ (g), g G and χ V = χ V Proof: 1 Let n be the order of g, then (ρg) n = ρ(g n ) = ρ(1) = I M. Hence the minimal polynomial of ρg divides x n 1 = n (x ω i), where w 1,..., w n K are the n th roots of unity. Since x n 1 is separable, (because chark ) so is the minimal polynomial of ρg. Therefore the Jordan form of ρg is diagonal and all eigenvalues of ρg are roots of unity of order dividing the order of g. 2 Since ρ is a homomorphism, χ ρ (h 1 gh) and χ ρ g are traces of similar matrices, so they are equal. 3 Let g G. Fix bases v 1,..., v n, w 1,..., w m of V and W, and build the basis {(v i, 0) i = 1, n} {(0, w j ) j = 1, m} of V W. It remains

6 6 to note that, with respect to these bases, g V W is built of 2 diagonal blocks: g V and g W. 4 Let v 1,..., v n, w 1,..., w m be eigenbases of g V GL(V ) and g W GL(W ) respectively, so that g V (v i ) = λ i v i, g W (w j ) = µ j w j, with λ i, µ j K. It follows that the (v i w j ) s form a basis of V W (they certainly span V W, and there are n m of them, which is precisely dim K (V W )). Note that g V W (v i w j ) = (g V (v i ) g W (w j )) = λ i µ j (v i w j ), so that j µ j = the above is actually a g V W -eigenbasis. Hence χ V W (g) = i,j λ iµ j = i λ i χ V (g) χ W (g). 5 follows from Lemma 1 and part 4. 6 First, if λ 1,..., λ n are the eigenvalues of ρ(g) then 1/λ 1,..., 1/λ n are the eigenvalues of ρ(g 1 ) = (ρg) 1, and χ ρ (g 1 ) = χ ρ (g) follows, since λ i = 1 by part 1. For second part, let g G and v V be an eigenvalue of g GL(V ). Then (gϕ)v = ϕ(g 1 v) = 0.5 The Projection formula. Orthogonality of the irreducible characters Throughout the section, G will be a finite group and K an algebraically closed field such that chark. Let V be a G-module. Define V G = {v V gv = v, g G}. Note that V G is a submodule of V. Consider P = 1 g as a linear map on V (we identify g G with g V End(V )). gh = Then P is a G-module homomorphism: Indeed, for h G we have P h = 1 h 1 (h 1 gh) = h 1 x G x (since g h 1 gh is bijective) = hp. Furthemore, P is the projection of V onto V G (i.e. V = V G kerp ): First, note that imp V G, since hp (v) = (h 1 g)v = 1 hgv = 1 x G xv = P (v), for all v V, h G. Also, v V G P v = 1 gv = v. We conclude that imp = V G and P P = P, i.e. P is the projection onto imp = V G. Lemma 2: Let V, W be G-modules. Consider the G-module H = Hom(V, W ). ( 1. dim K V G = trace 2. H G = Hom G (V, W ) 1 g ) 3. dim K H G = 1 χ V (g)χ W(g) Proof: 1 Pick a basis of V G, and extend it to a basis of V. From the preceding argument it follows that, in this basis, the matrix of P will be diagonal, with first k = dim K V G diagonal entries 1 and the rest 0. So dim K V G = trace(p ). 2 We have f H G (gf)(v) = f(v), g G, v V gf(g 1 v) = f(v), g G, v V f(g 1 v) = (g 1 f)(v), g G, v V fg 1 = g 1 f, g G f Hom G (V, W ), as desired. 3 By Lemma 1, H = V W, as G-modules. So, χ H = χ V W = χ V χ W, by Theorem 3.

7 0.5. THE PROJECTION FORMULA. ORTHOGONALITY OF THE IRREDUCIBLE CHARACTERS7 Using this and part 1, we get ( ) dim K H G 1 = trace g = 1 χ H (g) = 1 χ V (g)χ W (g) Lemma 3: Let V, W be irreducible G-modules, and H = Hom(V, W ). Then { 0, if V W dim K H G = 1, if V = W Proof: If V, W are non-isomorphic G-modules, then Hom G (V, W ) = 0 by Schur s Lemma, so Lemma 3 implies H G = 0. Now assume V = W, and let ϕ, ψ Hom G (V, W ) be non-zero. By Schur s Lemma, ψ is invertible and hence ϕψ 1 Hom G (W, W ). Again, Schur s Lemma yields ϕψ 1 = λi W, for some λ K, i.e. ϕ = λψ. This proves that dim K Hom G (V, W ) = 1, and then by lemma 3 we are done For K = C, it follows from Theorem 3 and Lemmas 2,3 that if V, W are irreducible G-modules, then 1 { 0, if V W χ V (g) χ W (g) = 1, if V (3) = W Let class(g) = {α : G C α is constant on conjugacy classes of G} Then class(g) is a complex vector space under pointwise addition and scalar multiplication of functions. Its dimension is the number of conjugacy classes in G, since the class indicator functions form a basis of class(g). For any α, β class(g) we define α, β = 1 α(g)β(g) (4) It s easy to check that, is a complex inner product on class(g). Note that χ V class(g) for any G-module V. Let α : G C and let V be a CG-module. Consider the map P α,v = α(g)g : V V, which is certainly in End C (V ). When is P α,v a G-module homomorphism, regardless of V? If α class(g), then P α,v is a G-module homomorphism. Indeed, for any h G, we have P α,v h = = h α(g)gh = hα(h 1 gh)h 1 gh (since α class(g)) = α(h 1 gh)h 1 gh = h x G α(x)x = hp α,v, as desired. Further, if V is an irreducible G-module and α class(g) then P α,v = dimv α, χ V I V (5)

8 8 Since α class(g), P α,v is a G-module homomorphism. But V is irreducible, so by Schur s Lemma P α,v = λi V for some λ C. The claim follows, since λ = trace(p α,v ) dimv = 1 dimv α(g)trace(g) = 1 dimv α(g)χ V (g) = dimv α, χ V. Theorem 4: Let G be a finite group. Then the irreducible characters form an orthonormal basis of class(g) as a C-vector space. Proof: First, note that (3) shows that the irreducible characters form an orthonormal set for the inner product given by (4), and hence they are linearly independent. In particular, their number cannot exceed n, the number of conjugacy classes of G. So let χ 1,..., χ k be the irreducible characters of G (k n). To prove that χ 1,..., χ k span class(g), we show that (span{χ 1,..., χ k }) = {0}. Suppose β class(g) satisfies β, χ i = 0, i. Then α, χ i = 0, i, where α = β class(g). Hence, if U is any irreducible G-module (so χ U {χ 1,..., χ k }), then (5) implies that P α,u 0. In fact, it follows that P α,v 0 for all G-modules V : Indeed, by Maschke s theorem V = m j=1 U j, where U 1,..., U m are irreducible G-modules. Then P α,v = α(g)g V = m α(g)( g Uj ) = j=1 m P α,uj = j=1 m 0 Uj 0 V. j=1 In particular, for the left regular representation (CG) we have P α,(cg) 0. Hence, 0 = P α,(cg) (1 G ) = α(g)(g 1 G ) = α(g)g. But {g g G} is a basis of CG. Therefore, α(g) = 0, g G, i.e. α 0, as desired Corollary 4 : 1. If V = m m iv i (by m i V i we mean the direct sum of m i copies of V i ), where V 1,..., V m are non-isomorphic irreducible G-modules, then m i = χ V, χ Vi, i. (First, from (3) it follows that 2 irreducible G-modules are isomorphic if and only if they have same characters. Hence χ V, χ Vi = m j=1 χ m j V j, χ Vi = m j=1 m j χ Vj, χ Vi = m i, by the theorem). This implies 2. Let U, V be G-modules, with U irreducible. The number of times (an isomorphic copy of) U occurs in a decomposition of V into irreducible submodules is χ V, χ U (so it doesn t depend on the decomposition). 3. For any G-modules V, W we have V = W χ V = χ W. (Clearly V = W χ V = χ W. Conversely, assume χ V = χ W. Let V 1,..., V k be a complete set of irreducible G-modules. Then parts 1,2 above, along with Maschke s theorem show that V = k a iv i = W, where ai = χ V, χ Vi = χ W, χ Vi ). 4. Let V 1,..., V m be non-isomorphic irreducible G-modules. If V = m m iv i, then χ V, χ V = m m2 i. 5. A G-module V is irreducible if and only if χ V, χ V = 1 (follows from 4).

9 0.6. THE LEFT REGULAR REPRESENTATION (CG) The left regular representation (CG) Let G be a finite group and let χ be the character of the CG-module (CG). Consider the basis B = {g g G} of CG. Then g G acts on B by permuting its elements. This action is fixed-point free for any g 1. Hence in this basis B, the matrix of g (identified with g (CG) GL(CG)) has 0 s along the diagonal for g 1, and 1 s otherwise. This shows that { if g = 1 χ(g) = (6) 0 if g 1 Now let χ 1,..., χ k be the irreducible characters of G, and V 1,..., V k (one collection of) G-modules corresponding to them. By corollary 4, we have (CG) = k m iv i, where m i = χ, χ i = 1 Thus, V i appears dimv i times in (CG). Hence χ(g)χ i (g) = 1 χ(1)dimv i = dimv i. (CG) = k (dimv i )V i (7) Lemma 4: Let G be a finite group and V 1,..., V k G-modules corresponding to all the irreducible characters χ 1,..., χ k of G. Then 1. k (dimv i) 2 =. 2. k (dimv i)χ i (g) = 0, g G, g 1. Proof: The lemma follows by taking characters in (7), and then using (6) 0.7 The Character Table Let χ 1,..., χ r be the irreducible characters of a group G, and let K 1 = {1}, K 2,..., K r be the conjugacy classes of G. Pick representatives g i K i, i and consider the character table T of G: K 1 K 2 K r 1 g 2 g r χ 1 Recall that. χ i (g j ) χ r χ i, χ j = 1 r K t χ i (g t )χ j (g t ) = δ ij t=1

10 10 which gives orthogonality of the matrix its adjoint. Hence we also have column orthogonality in T : r χ t (g i )χ t (g j ) = t=1 Kj χ i(g j ). But if a matrix is orthogonal, so is { K i = C G(g i ) if i = j 0 if i j The following observations are helpful in constructing character tables of groups: (1) The number of (irreducible) degree 1 characters of G is G = G/[G, G] : Assume ρ is a degree 1 representation of G = G/[G, G] on V. Then ρ : G GL(V ) defined by ρ(g) = ρ (g ) is a degree 1 representation of G, where g = g[g, G]. Conversely, assume ρ is a degree 1 representation of G on V. Then GL(V ) = C, so that ρ(g)ρ(h) = ρ(h)ρ(g), g, h G. Hence ρ : G GL(V ) given by ρ (g ) = ρ(g), is a well-defined degree 1 representation of G, because ρ(xyx 1 y 1 ) = ρ(x)ρ(y)ρ(x 1 )ρ(y 1 ) = ρ(x)ρ(x 1 )ρ(y)ρ(y 1 ) = I V, implies that ρ acts trivially on [G, G]. Therefore, the number of degree 1 characters of G is the number of degree 1 characters of G, and the latter is G, because G is abelian. (2) A character χ of G is irreducible χ, χ = 1. (by Corollary 4 ) (3) If V, W are G-modules such that χ V is irreducible and χ W has degree 1, then χ V W = χ V χ W is also irreducible: Since dimw = 1, χ W (g) is a root of unity for any g G and hence χ W (g)χ W (g) = 1. Then we have χ V W, χ V W = 1 χ V (g)χ W (g)χ V (g)χ W (g) = = 1 χ V (g)χ V (g) = χ V, χ V = 1, so χ V W is irreducible. (4) Assume H G. Then any representation ρ of G/H naturally gives a representation ρ of G. Moreover, ρ is irreducible if and only if ρ is irreducible: The map ρ defined by ρ(g) = ρ (gh), is a well-defined representation of G. Then χ ρ, χ ρ = 1 χ ρ (g)χ ρ (g) = 1 χ ρ (gh)χ ρ (gh) = H = χ ρ, χ ρ, and the second asertion follows from 2). g G/H χ ρ (g )χ ρ (g ) It should be noted that the inner product χ ρ, χ ρ is taken in class(g), whereas χ ρ, χ ρ in class(g/h). (5) Assume θ is a character of G (over C). Then θ, defined by θ(g) = θ(g), g G, is also a character of G. Moreover, θ is irreducible θ is irreducible. Let ρ : G GL(V ) be a representation such that θ = χ ρ. Fix a C-basis B = {v 1,..., v n } of V. Define ρ : G GL(V ) by ρ (g) = ρ(g) GL(V ) (the conjugate of the matrix of ρ(g) in the basis B). Clearly ρ (gh) = ρ(gh) = ρ(g)ρ(h) = ρ(g)ρ(h) = ρ (g)ρ (h), g, h

11 0.7. THE CHARACTER TABLE 11 G. This proves that ρ is a representation of G on V. Also, χ ρ (g) = trace(ρ (g)) = trace(ρ(g) = θ(g), g G, which shows that θ = χ ρ is a character of G. The last part of the statement follows from 2 in above, since θ, θ = θ(g)θ(g) = θ, θ. (6) The Symmetric and Alternating Squares Sym 2, Alt 2 : Let V be an n-dimensional G-module over F, where F is an algebraically closed field such that char(f) 2. If {v 1,..., v n } is a basis of V, then {v i v j 1 i, j n} is a basis of V F V. Thus, there is a unique linear map θ on V F V, given by θ(v i v j ) = v j v i. For any a i, b j F, we have = θ(a 1 v a n v n b 1 v b n v n ) = n a i b j θ(v i v j ) i,j=1 n b j a i (v j v i ) = (b 1 v b n v n ) (a 1 v a n v n ). i,j=1 i.e. θ(u v) = v u, u, v V. In particular, it follows that θ does not depend on the chosen basis {v 1,..., v n } of V. Next, θ (and hence θ I V V ) is a G-module homomorphism, since θ(g(v i v j )) = θ(gv i gv j ) = gv j gv i = g(θ(v i v j )), g G. We define the symetric square: Sym 2 (V V ) = ker(θ I V V ) (or just Sym 2, when V is clear from the context), which is a G-submodule of V V. We have u = n a ij (v i v j ) ker(θ I V V ) i,j=1 n (a ij a ji )v i v j = 0 a ij = a ji, i, j i,j=1 u span{v i v j + v j v i 1 i j n} Clearly the (v i v j + v j v i ) s are linearly independent. Thus, they form a basis of Sym 2 so that dim F Sym 2 = n(n+1). To compute χ 2 Sym 2, let g G. Since F is algebraically closed, it follows that g, regarded as a an element of GL(V ), is diagonalizable. Thus, we may choose v 1,..., v n to be eigenvectors of g, with eigenvalues λ 1,..., λ n. Then g(v i v j + v j v i ) = λ i λ j (v i v j + v j v i ) and hence {λ i λ j 1 i j n} is the set of eigenvalues of g as an element of GL(Sym 2 ). Therefore, χ Sym 2(g) = λ i λ j = 1 n 2 (( λ i ) i j n n λ 2 i ) = (χ V (g)) 2 + χ V (g 2 ) 2 (8) Accordingly, we also define the alternating square: Alt 2 (V V ) = ker(θ + I V V ). Similarly, one shows that {v i v j v j v i 1 i < j n} is a basis of Alt 2, so that dim F Alt 2 = n(n 1) 2. Likewise, we get χ Alt 2(g) = λ i λ j = 1 n 2 (( λ i ) 2 1 i<j n n λ 2 i ) = (χ V (g)) 2 χ V (g 2 ) 2 (9)

12 Some Examples Example 1: Let G = S 3 (the symmetric group on 3 symbols). S (12) (123) χ χ χ Since [S n, S n ] = A n in general, we have S n /[S n, S n ] = Z 2 for n 2. Hence S n has 2 irreducible characters of degree 1: the trivial one (χ 1 ), and the sign character (χ 2 (g) = 1 if g A n and χ 2 (g) = 1 otherwise). The third character can be found directly from the orthogonality relations. Alternatively, for n 2 consider the natural action of S n on a basis B = {e 1, e 2,..., e n } of C n, given by g(e i ) = e g(i). This turns C n into an S n -module V. Clearly W = span{e e n } is a submodule of V. The map φ : V W φ(a 1 e a n e n ) = a a n (e e n ) n is an S n -homomorphism. Indeed for all g S n, g(φ(a 1 e a n e n )) = a a n (e e n ) = a a n (e g(1) + + e g (n)) n n = φ(a 1 e g(1) + + a n e g(n) ) = φ(g(a 1 e a n e n )). Since imφ = W, it follows that U = kerφ is a n 1 dimensional submodule of V. If we represent g S n as a linear map on V then, with respect to the basis B, g is a permutation matrix (its ij element is 1 if g(j) = i and 0 otherwise). Hence χ V (g) is the number of fixed points of g. Since χ W (g) = 1 and V = W U, we get χ U (g) = (number of fixed points of g)-1. It can be shown that U is irreducible. We denote the n dimensional module V by P n (so χ U = χ Pn χ 1 is an irreducible character of S n of degree n 1). For n = 3, χ U = χ 3. Example 2: G = S 4. S (12) (123) (1234) (12)(34) χ χ χ χ χ χ 1, χ 2 are the trivial and sign characters; χ 3 = χ P4 χ 1 is irreducible by above. Also χ 4 = χ 3 χ 2 is irreducible by (3), section 0.7. The last character χ 5 has degree = 2. Now χ 5 = χ 5 χ 2, since both are irreducible characters of degree 2. So we get 2 zeroes in row 5. The other 2 entries are found from orthogonality

13 0.8. SOME EXAMPLES 13 relations. Alternatively, note that H = {1, (12)(34), (14)(23), (13)(24)} = [A 4, A 4 ] is a normal subgroup of S 4, and S 4 /H = S 3. If we lift χ 3 from S 3 = S4 /H to S 4 we get χ 5, which is automatically irreducible by (4) of section 0.7. Example 3: G = A 4 (the alternating group on 4 letters). A (123) (132) (12)(34) χ χ 2 1 w w 2 1 χ 3 1 w 2 w 1 χ where w C \ R, w 3 = 1. We have [A 4, A 4 ] = {1, (12)(34), (13)(24), (14)(23)} = V 4, so A 4 /[A 4, A 4 ] = C 3. Hence A 4 has 3 representations of degree 1, involving complex roots of unity of order dividing 3 (χ 1, χ 2, χ 3 ). Recall that χ P4 χ 1 is an irreducible character of S 4. If we restrict to A 4, we obtain χ 4 (which is irreducible, since χ 4, χ 4 = 1/12 ( ( 1) 2 ) = 1). Example 4: G = S 5. S (12) (123) (1234) (12345) (12)(34) (12)(345) χ χ χ χ χ χ χ χ 1, χ 2 are the trivial and sign characters; χ 3 = χ P5 χ 1 is irreducible and hence, so is χ 4 = χ 3 χ 2. The alternating square of V 3 (the representation whose character is χ 3 ) has character χ 5 (recall that χ Alt 2 V (g) = 1 2 [χ V (g) 2 χ V (g 2 )]), and is irreducible since χ 5, χ 5 = 1/120 ( ) = 1. The other 2 characters are found by orthogonality relations of the character table. Example 5: G = A 5. A (123) (12)(34) (12345) (21345) χ χ χ χ χ χ 2 and χ 3 are restrictions from the characters χ 3 and χ 6 of S 5 (and they are irreducible by computation). To obtain χ 4 and χ 5, recall that A 5 is isomorphic to the group H of

14 14 rotations and symmetries of a regular icosahedron. This gives us a degree 3 representation ρ : A 5 GL(R 3 ) (the origin is the center of the icosahedron). Note that ρ is irreducible since A 5 = H acts transitively on the set of vertices of the icosahedron. Since (12345) = 5, it s immediate that ρ(12345) is a rotation by α = 2π 5 or 2α = 4π 5 (depending on the isomorphism A 5 = H). In the first case, under a suitable basis of R 3, we have ρ(12345) = cos α sin α 0 sin α cos α χ ρ [(12345)] = 2 cos α + 1 = 1 + 5, 2 which reveals the character χ 4. The second case yields the character χ 5. Example 6: G = D 8 = r, s r 4 = s 2 = 1, srs = r 1 (rotations and simmetries of a square). D 8 1 r 2 r, r 1 s, r 2 s rs, r 1 s χ χ χ χ χ Since [D 8, D 8 ] = r 2 and D 8 = D 8/[D 8, D 8 ] = Z 2 Z 2, we have 4 representations of degree 1 (with characters χ 1,..., χ 4 ). These characters involve only ±1, since every element of D 8 has order at most 2 (i.e. D 8 has exponent 2). Finally, χ 5 is found by orthogonality relations. Example 7: G = Q 8 = i, j, k i 2 = j 2 = k 2 = 1, ij = k, jk = i (the quaternions). Q ±i ±j ±k χ χ χ χ χ We have [Q 8, Q 8 ] = 1 and Q 8 = Z 2 Z 2. From these we obtain χ 1,..., χ 4. Again, χ 5 is determined by orthogonality. Remark: The groups D 8, Q 8 have the same character table even though they are not isomorphic. This happened since the degree 1 characters (which depend only on the derived groups D 8 = Q 8 ) were enough to determine the whole character table. 0.9 Properties Of The Character Table T Let G be a finite group and assume ρ is a representation of G on V over some algebraically closed field, whose character is χ. We define K χ = {g G g acts on V as the identity map} (i.e. K χ = kerρ).

15 0.10. MORE ON THE REGULAR REPRESENTATION 15 Note that K χ = {g G χ(g) = χ(1)}: Indeed, χ(g) is a sum of n = dimv = χ(1) roots of unity (namely, the eigenvalues of ρ(g)). Hence, χ(g) = n all eigenvalues of ρ(g) are 1 ρ(g) = I V (because ρ(g) is diagonalizable). Theorem 5: Let G be a finite group. Let χ 1,..., χ n be its irreducible characters. 1. T (the character table of G) determines the lattice of normal subroups of G. Specifically, H G H = i I K χi, for some I {1, 2,..., n}. 2. G is simple the only irreducible χ i whose kernel K χi is non-trivial, is the trivial character χ T determines whether G is solvable. 4. In principle, T determines whether G is nilpotent. Proof: 1 Let H G. Consider the left regular representation of G = G/H, denoted as (CG) 0. Lift (CG) 0 to a representation of G (which permutes H-cosets in G by left multiplication). Let ψ denote its character. Then K ψ = H, since g(g H) = g H g G if and only if g H. Hence g H ψ(g) = ψ(1). On the other hand, write ψ = n a iχ i, where a 1,..., a n are non-negative integers. Then n n ψ(g) a i χ i (g) a i χ i (1) = ψ(1), which implies ψ(g) = ψ(1) χ i (g) = χ i (1), i I, where I = {1 i n a i > 0} g K χi, i I. Therefore, H = i I K χi. Conversely, each K χi is the kernel of a group homomorphism of G, so K χi s (and their intersections) are normal in G. 2 Assume G is simple. Then, since K χ2,..., K χn are proper normal subgroups of G, all of them must be trivial (note that K χ = G χ = χ 1 ). Conversely, if G is not simple, then N G, {1} N G. By 1, N is an intersection of K χi s, so {1} K χi G for some i (i 1). 3 By 1, T determines the lattice of normal subgroups of G. In particular, we can determine if G has a normal series whose succesive quotients are p-groups. This is a criterion for solvability of G More On The Regular Representation Theorem 6: Let G be a finite group and ρ i : G V i, i = 1,..., n be a complete set of irreducible representations of G. Then CG = n End(V i ), as C-algebras. (10) Proof: Consider the map φ : CG n End(V i), determined by ρ 1 (g) 0 g G φ....., and C-linear extension. (11) 0 ρ n (g)

16 16 Then φ is a C-algebra homomorphism, since it is C-linear, and ρ i s are group homomorphisms. Injectivity of φ follows because the left regular representation is faithful. Indeed, if α(g) g kerφ, then ρ i( α(g) g) = 0 V i. Hence ( n a i ρ i )( α(g) g) = 0 n a iv i, a i N. In particular, for the regular representation ρ, ρ( α(g) g) = 0 (CG) 0. So 0 = ρ( α(g) g)[1] = α(g) g in CG. Hence α(g) = 0 g G, which proves kerφ = 0. Next, note that dimcg = = n (dimv i) 2 = dim n End(V i), so φ is surjective too. Therefore φ is a C-algebra isomorphism Identifying End(V i ) with it s copy in W = n End(V i), note that End(V 1 ),..., End(V n ) are (2-sided) ideals of W. Hence, by the theorem, CG is the direct sum of the 2-sided ideals φ 1 [End(V 1 )],..., φ 1 [End(V n )], so we have 1 (CG) 0 = e 1 + +e n, where e i φ 1 [End(V i )] are unique. But obviously I End(V1 ) W =.... n. =. I End(Vi ). 0 I End(Vn) 0 0 so we must have e i φ 0 0. I End(Vi ). 0 0, i = 1,..., n. (12) In particular, e j e i = 0, i j and e j = e 2 j, so φ(e i ) is the projection of n V i onto V i. Also note that e i generates the ideal φ 1 [End(V i )]. Therefore, e i CG φ End(V i ), and CG = n e icg. The isomorphism given by (11) is important, as it enables us to view all the irreducible representations of G (and their characters) simultaneously. Specifically, χ ρi (g) is the trace of the restriction of φ(g) on V i, and this naturally extends to CG. Thus, by (12) φ(e i a) Vj = 0 Vj, j i and φ(e i a) Vi = φ(a) Vi, a CG. Hence, { 0 for j i χ ρj (e i a) =, a CG. (13) χ ρi (a) for j = i Lemma 5: In CG we have: e i = dimv i χ ρi (g 1 ) g. (14) Proof: Write e i = α gg, α g C. Fix g G. Note that α g is the coefficient of 1 in e i g 1, so χ (CG) 0(e i g 1 ) = α g (recall (6), section 0.6). But χ (CG) 0 = n (dimv i)χ ρi, so α g = 1 n j=1 (dimv i )χ ρj (e i g 1 ) = dimv i χ ρi (g 1 ), by (13). Thus (14) holds

17 0.11. FOURIER TRANSFORMS AND CLASS FUNCTIONS Fourier Transforms And Class Functions We will often make the identification α class(g) α(g)g CG. The convolution product of α, β class(g) is defined by α β ( α(g)g)( h G β(h)h). Equivalently, (α β)(x) = g,h G α(g)β(h), x G. This turns class(g) into a C-algebra. gh=x Moreover, Lemma 6: Z(CG) = class(g), as C-algebras. Proof: We naturally define φ : class(g) CG by φ(α) = α(g)g CG. This is clearly an injective C-algebra homomorphism. For any h G, h 1 φ(α)h = α(g)h 1 gh = α(h 1 gh)h 1 gh = φ(α) (since α is a class function). Since φ(α) is centralised by G, which spans CG, we have φ(α) Z(CG). Finally, we show that imφ = Z(CG): Assume a = β gg Z(CG). Then h 1 ah = a, h G β g (h 1 gh) = β g g [β hgh 1 β g ]g = 0 CG β hgh 1 = β g, g, h G Hence β : G C defined by β(g) = β g is a class function, so a = φ(β) imφ Let ρ : G GL(V ) be a representation of G, and α : G C. The Fourier transform of α at ρ is defined to be ˆα(ρ) = α(g)ρ(g) End(V ). From the definition, we see that it s enough to consider Fourier transforms at irreducible representations. Let ρ i : G V i, i = 1, n be a complete set of irreducible representations ˆα of G, so that ρ i End(V i ). Lemma 7: α ˆ β = ˆα ˆβ. Proof: For any representation ρ of G, we have: α ˆ β(ρ) = x G (α β)(x)ρ(x) = = [ ] x G g,h G α(g)β(h) ρ(x) = [ ] x G g,h G α(g)ρ(g)α(h)ρ(h) = [ gh=x gh=x ] [ = α(g)ρ(g) h G β(h)ρ(h)] = ˆα(ρ) ˆβ(ρ), as desired Theorem 7: (Fourier Inversion) Let α, β : G C. Then α(g) = 1 n (dimv i ) trace[ρ i (g 1 )ˆα(ρ i )]. (15) Proof: The right term of (15) is 1 n ( (dimvi ) trace[ρ i (g 1 ) ( h G α(h)ρ i(h) ) ] ) = ( ) = 1 n α(h) (dimv i )χ ρi (g 1 h) = 1 ( α(h)χcg (g 1 h) ) = α(g), h G the last equality following by Lemma 4 h G

18 18 Corollary 7: (Plancherel s Formula) α(g)β(g 1 ) = 1 Proof: The right term of (16) is 1 n [ = 1 α(g) n (dimv i ) trace[ˆα(ρ i ) ˆβ(ρ i )]. (16) ( (dimv i ) trace[ α(g)ρ i(g) ˆβ(ρ i )] ) = ] n (dimv i )trace[ρ i (g) ˆβ(ρ i )] = α(g)β(g 1 ), by (15) applied to β Example: Let G = Z n = g g n = 1 and α : G C. For j = 1, 2,..., n, let ρ j be the representation of G determined by ρ j (g) = w j, where w C is a primitive n th root of unity. Then ˆα(ρ j ) = x G α(x)ρ j(x) = n k=1 α(gk )w jk. Also, Fourier inversion gives α(g k ) = 1 n n j=1 ˆα(ρ j)w jk Clifford s Theorem Let V be an irreducible G-module over some field F. Let H G. In general, V is not an irreducible H-module. Clifford s theorem shows exactly how V breaks up as a direct sum of H-submodules. We establish it as a consequence of a series of observations: (1) Let W be an H-invariant subspace of V. For each g G, gw is a vector subspace of V and dim F (gw ) = dim F W (since g GL(V )). Also, gw is an H-submodule of V : For all g G, h H and w W we have h(gw) = g(g 1 hg)w gw, because g 1 hg H and W is H-invariant. (2) W is an irreducible H-module gw is an irreducible H-module. Indeed, if 0 S W is an H-submodule of W, then by (1), 0 gs gw is an H-submodule of gw. Thus, if W is H-reducible, then so is gw. Conversely, if gw is H-reducible then so is W = g 1 (gw ). (3) θ : W U is an H-isomorphism gθg 1 : gw gu is an H-isomorphism. Assume θ : W U is an H-isomorphism. Then gθg 1 [h(gw)] = gθ[g 1 hg]w = g[g 1 hg]θw (since θ is an H-homomorphism) = hgθw = h[gθg 1 ]gw, h G, w W. Thus, gθg 1 : gw gu is an H-homomorphism. It is an isomorphism, since dim F (gw ) = dim F W = dim F U = dim F (gu). Conversely, if gθg 1 is an H-isomorphism, then so is θ = g(g 1 θg)g 1. (4) Consider gw. This is a G-submodule of V (since any g G permutes the components in the sum). Since V is irreducible as a G-module, we have gw = V. Note that this is not a direct sum, since for example hw = W, h H. (5) From now on, assume W is H-irreducible. If g 1 W g 2 W for some g 1, g 2 G, then g 1 W g 2 W = 0, since g 1 W g 2 W is an H-submodule of the irreducible H-module gw. This enables us to throw away the repeating components in (4), to get a direct sum. By classifying the remaining ones according to H-isomorphism types, we get V = k V i, where V i = g i1 W g ini W, and g i jw = g i j W as H-modules if and only if i = i. We may assume W = g 11 W V 1.

19 0.13. INDUCED REPRESENTATIONS 19 (6) If gw V i then gv 1 = V i. Indeed, since W and g 1j W are isomorphic as H-modules, so are gw and g(g 1j W ). Hence, gw V i g(g 1j W ) V i, j = 1, n 1, so gv 1 = n1 j=1 g 1jW V i. Conversely, note that g 1 (gw ) = W V 1. Since gw V i, a similar to above argument implies that g 1 V i V 1 i.e. V i gv 1. Therefore, gv 1 = V i, as claimed. Applying the above for g = g ij we get g ij V 1 = V i, and in particular, dim F V i = dim F V 1 so n i = dimv i = dimv 1 = n dimw dimw 1. (7) By (6), V i gv i defines a transitive action of G on {V 1,..., V k }. The kernel of this action contains HC G (H). That H is in the kernel, follows since gw is H-invariant g G. Now let x C G (H). Fix g G. Regarding x as an element of GL(V ), x : gw (xg)w is an isomorphism of vector spaces. Since x 1 hx = h, h H, we have xhu = hxu, u gw, so x is actually an H-isomorphism of gw and (xg)w. Thus, gw V i (xg)w V i, x fixes each V i. To sum up the above, we have Theorem 8: (Clifford) Let G be a group, H G. Assume V is an irreducible representation of G, over some field F, and W V is an H-invariant subspace of V. Then: 1. V = V 1 V2 Vk, where V i = g i1 W g in W, g ij G, and g ij W = g i j W as H-modules if and only if i = i. Note also that n does not depend on i. Also, for each g G there is an i {1,..., k} such that gw = g i1 W as H-modules. 2. g G : V i gv i, defines a transitive action of G on {V 1, V 2,..., V k }, with HC G (H) in its kernel. Corollary 8: 1. χ gw (h) = χ W (g 1 hg), g G, h H. 2. dim F V = kn dim F W. Proof: 1 follows, since g 1 hgw = λw hgw = g(λw) = λ(gw) (λ F). 2 follows by looking at dimensions in 1. of theorem Induced Representations Let F be a field, G a finite group and H G. Our aim is to generate representations of G, from those of H. Suppose ρ : H GL(W ) is a representation of H, so that W is a left FH-module. (1) Note that FG is naturally a right FH-module (by right multiplication in FG). Consider W G = FG G FH W (thus, we can move elements of FH between left and right). Now W is a left G-module, under y(x w) = yx w (w W, x, y G), and F-linear extension. Accordingly, this gives a representation of G on W G, which we denote by ρ G. (2) Let {g 1, g 2,..., g k } be a complete set of representatives of left cosets of H in G. Since G = k g ih and g i H are F-bases of FG and Fg i H respectively, we have FG = k Fg ih as F-vector spaces. (3) For i = 1,..., k we let W i = g i FH W. Thus, W i is an F-vector space, isomorphic to W. Also, W i W j = {0}. We claim that W G = k W i. First, note that W i =

20 20 Fg i H FH W (since each h H acts faithfully on W ). Then by above, [ k ] [ k W G = Fg i H W = Fg i H ] k k W = W i = W i, as claimed. FH FH In particular, we get dimw G = G : H dimw. (4) Let g G. Then gg i H = g j H for some j. For any w W, g i w W i and g(g i w) = gg i w = g j (g 1 j gg i ) w = g j (g 1 j gg i )w W j, since g 1 j gg i H. Thus, g G acts on W G = k W i by first permuting the W i s in the same way it permutes the left cosets of H in G (under the regular representation), and then acting within W j s with elements of H. To be more precise, let s combine bases of W 1,..., W k to get a basis B of W G. In this basis, the matrix of ρ G (g) GL(W G ) is a block-permutation matrix, where the non-zero blocks describe the action of H on W j s. It is convenient to set ρ(x) = 0 Wi, x G \ H, i = 1, k. Then, in this basis, ρ G (g) has form ρ(g1 1 gg 1 ) ρ(g1 1 gg k ) ρ G (g) =..... (17) ρ(g 1 k gg 1) ρ(g 1 k gg k) In particular, if dim F W = 1 and ρ is the trivial representation of H, then ρ G (g) is just the permutation representation of G on the left cosets of H in G. (5) If W 1, W 2 are H-modules, then (W 1 W2 ) G = W1 G W G 2. We have g (w 1 + w 2 ) = g w 1 + g w 2 W1 G W G 2, g G, w i W i, so (W 1 W2 ) G W1 G + W 2 G. By (3), dim F (W 1 W2 ) G = G : H (dimw 1 + dimw 2 ) = dim F (W1 G W G 2 ), so the two must be equal. (6) Assume K H G, and let W be a K-module. Then (W H ) G = W G as G-modules. Define φ : (W H ) G W G, by φ[g FH (h FK w)] = gh FK w, and F-linear extension. Then φ[x(g (h w))] = φ[xg (h w)] = xgh w = x(φ[g (h w)]), w W, h H, x, g G. Thus, φ is a G-module homomorphism. From its definition, φ is surjective. On the other hand, (3) implies dim F (W H ) G = G : K dim F W H = G : H H : K dim F W = dim F W G. Therefore, φ is a G-isomorphism. (7) Let χ G ρ denote the character of ρ G. Taking traces in (17) yields χ G ρ (g) = k i gg i ), { where χ 0 χρ (x) if x H ρ(x) = 0 if x G \ H we get χ 0 ρ(g 1 i gg i ) = χ 0 ρ(h 1 g 1 i gg i h), h H. Hence, χ G ρ (g) = 1 H k h H χ0 ρ(g 1. Since g 1 i gg i H h 1 gi 1 gg i h H, h H, χ 0 ρ(h 1 g 1 i gg i h) = 1 H χ 0 ρ(x 1 gx), g G. (18) (8) Actually (18) permits us to extend the{ notion of induced characters to arbitrary class θ(u) if u H functions: Let θ class(h), and θ 0 (u) = 0 if u G \ H. Define θ G : G F, by θ G (g) = 1 H x G θ0 (x 1 gx). Then θ G class(g). Indeed, θ G (g 1 xg) = 1 H u G θ0 (u 1 g 1 xgu) = θ G (x), since as u runs through G, so does gu. x G

21 0.13. INDUCED REPRESENTATIONS 21 (9) (Frobenius Reciprocity) Assume F = C, H G as before, and let φ class(h), θ class(g). Then φ, θ H = φ G, θ (19) Proof: We have φ G, θ = 1 φg (g)θ(g) = 1 H 1 = x G φ0 (x 1 gx)θ(x 1 gx) H = 1 H = 1 H x G φ0 (x 1 gx)θ(g) = y G φ0 (y)θ(y) (since {x 1 gx g G} = G, x G) y H φ(y)θ(y) = φ, θ H, as claimed. In particular, if φ and θ are irreducible characters of H and G, then the number of times φ appears in θ H is equal to the number of times θ appears in φ G. Note: The two inner products appearing in (19) are different. They are taken in class(h) and class(g), respectively. x O G (g) θ0 (x), (10) Let θ class(h) and G. From the definition, we have θ G (g) = 1 H where O G (g) denotes the G-conjugacy class of g. Now any H-conjugacy class is contained in some G-conjugacy class, but not conversely. In general O G (g) H will break up into several H-conjugacy classes O H (h 1 ),..., O H (h m ), h i H. Since θ class(h), θ G (g) = 1 m H {x G x 1 gx O H (h i )} θ(h i ). But if y 1 gy = x O G (g), then z 1 gz = x yz 1 gzy 1 = g zy 1 C G (g) (C G (g) is the centralizer of g in G). Thus, {y G y 1 gy = x} = C G (g), x O G (g). Hence θ G (g) = 1 H m C G (g) O H (h i ) θ(h i ) = G : H O G (g) m O H (h i ) θ(h i ). (20) Note also that if O H (g) H = (i.e. m = 0), then θ G (g) = 0 by definition. Example: H = A 4 < G = A 5. Let θ be the 1-dimensional character of A 4, given by A (123) (132) (12)(34) θ 1 w w 2 1 where w C \ R, w 3 = 1. Then θ G is a character of A 5, of dimension A 5 : A 4 θ(1) = 5. By (20), we have: 1. g = (123) O G (g) H = O H (123) O H (132), so θ G (123) = G:H O G (123) (4θ(123) + 4θ(132)) = 5 20 (4w + 4w2 ) = g = (12)(34) O G (g) H = O H [(12)(34)]. Hence, θ G 5 [(12)(34)] = O G 3θ[(12)(34)] = 1. [(12)(34)] 3. g = (12345) (or (13524)). Since H = A 4 doesn t have any 5-cycles, O G (g) H =, so θ G (g) = 0. Note also that θ G is irreducible. A (123) (12)(34) (12345) (13524) θ G

22 Consequences To Permutation Groups Consider a finite group G acting transitively on a finite set S. For α S, the stabilizer of α, G α = {g G gα = α} is a subgroup of G. The rank of G, r S, is the number of orbits of G α acting on S. To show that this is well-defined, consider α, β S. There exists g G such that gα = β. Then xβ = β xgα = gα (g 1 xg)α = α, i.e. g 1 G β g = G α. If h G α, then ha = b (a, b S) ghg 1 (ga) = gb (note that ghg 1 G β ). Therefore, a ga (which is a bijection of S) is also a bijection between orbits of G α and G β. So the rank of G is well-defined. (1) The rank of G is 2 G is doubly transitive. By definition, the rank of G is 2 G α is transitive on S \ {α} ( α S). Let a, b, c, d S, a b, c d. Since G is transitive, g(a, b) = (c, e) for some g G, e S. Since e, d S \ {c}, there exists h G c such that he = d. So hg(a, b) = (c, d), i.e. G is doubly transitive. Let a S. Since G is doubly transitive, for any b, c S \ {a} there is a g G such that g(a, b) = (a, c). In particular, g G a and gb = c. Thus, G a is transitive on S \ {a}. (2) (Burnside) Let θ be the character of the permutation representation of G on CS. Then r S = θ, θ. Let s take a closer look at the action of G on S. Fix α S, and let H = G α. Note that gα = hα (g, h G) gh 1 H. Thus, if g 1 H,..., g m H are the cosets of H in G, then the orbit of α, O G (α) = {gα g G} = {g 1 α,..., g m α}. In particular, O G (α) = G : H = G : G α (this is Burnside s Lemma, and is true regardless of the transitivity of G on S). Since G is transitive on S, S = O G (α) = {g 1 α,..., g m α}, so we have θ(g) = {β S gβ = β} = {i g(g i α) = g i α} = {i g 1 i gg i H} = 1 G H (g), the last equality following by (17) (1 H is the trivial character of H). Finally, let O H (α 1 ),..., O H (α k ) be the orbits of H acting on S. Summing up the above, r S = 1 k H O H(α i ) G αi = 1 H β S G β = 1 1 H h H {β S hβ = β} = H h H θ(h) = θ H, 1 H = θ, 1 G H = θ, θ, as desired (we also used Frobenius reciprocity). (3) Assume S 2. Then G is doubly transitive on S θ 1 G is an irreducible character of G. Let α S. Since G is transitive on S, 1 = 1 O G(α) G α = 1 β S G β = 1 θ(g) = θ, 1 G. So θ 1 G is a character of G (of degree S 1), and θ 1 G, 1 G = 0. Hence, G is doubly transitive on S θ, θ = 2 θ 1 G, θ 1 G + 1 G, 1 G + 2 θ 1 G, 1 G = 2 θ 1 G, θ 1 G = 1 θ 1 G is irreducible. (4) For all n 2, S n and A n (except A 3 ) have irreducible characters of degree n 1. The group S n is n-transitive in its permutation action on S = {1, 2,..., n}, so by (3), the claim is true for S n. It is also true for A 2. Now let n 4. We are done once we show that A n is doubly transitive on S. In fact, A n is (n 2)-transitive on S. To see this, let (a 1,..., a n 2 ) and (b 1,..., b n 2 ) be two (n 2)-tuples of distinct elements of S. Then π S n such that π(a 1,..., a n 2 ) = (b 1,..., b n 2 ). Note that (S n ) a1 (S n ) an 2 = {τ S n τ(a i ) = a i, i = 1,..., n 2} = S 2, so it has both odd and even permutations. Choose σ (S n ) a1 (S n ) an 2 such that signπ = signσ. Then σπ A n, and σπ(a 1,..., a n 2 ) = (b 1,..., b n 2 ).

23 0.15. FROBENIUS GROUPS Frobenius Groups Theorem 9: (Frobenius) Assume a finite group G acts transitively on a finite set S. If every non-identity element of G fixes at most 1 point of S, then the set of fixed-point-free elements together with the identity, forms a normal subgroup of G. Proof: For i = 0, 1, let F i = {g G g fixes i elements of S}. Then G = {1} F 0 F1, and we have to show that N = {1} F 0 is a normal subroup of G. For any g, h G, g fixes a 1,..., a k S hgh 1 fixes ha 1,..., ha k. This shows that F 0, F 1 (and clearly {1}) are normal subsets of G. So N is normal as a subset of G. The strategy of the proof is to construct a representation of G, whose kernel is N. We shall do this by inducing irreducible representations of a stabilizer G x (x S), and then by considering a certain linear combination of them. Fix x S, and let H = G x. Since ghg 1 = G gx, and G is transitive on S, all 1-point stabilizers G z are conjugate to H. Thus, any g F 1 is conjugate to some element of H. This enables us to select a full set of representatives of G-conjugacy classes A = {1, h 2,..., h r, y 1,..., y s }, with h i H and y j F 0 = N \ {1}. Another consequence is that F 1 = z S (G z \ {1}) (the union is disjoint since any g G fixes at most 1 point of S). Thus, G = N [ z S (G z \ {1}) ]. Since G is transitive on S, S = G : G z, z S. So the above decomposition gives = S ( G x 1)+ N N = S = G : H. Note also that H is self-normalizing in G: indeed, if ghg 1 = h (h, h H, g G), then h fixes x and gx so we must have x = gx, i.e. g H. This also means that B = {1, h 2,..., h r } is a full set of representatives of H-conjugacy classes. Let φ 1,..., φ r be the irreducible characters of H. We have φ G k (1) = G : H φ k(1), φ0 (g 1 h i g) = φ k (h i ) (since H is self-normalizing in G) and φ G k (y j) = 0 φ G k (h i) = 1 H (since g 1 y j g F 0, which is disjoint from H). Let χ 1 be the trivial character of G. For i = 2,..., r we define χ i = φ G i φ i (1) φ G 1 + φ i(1) χ 1 (which is also valid for i = 1). Then 1. χ i (1) = φ i (1). Indeed, χ i (1) = φ G i (1) φ i(1) φ G 1 (1) + φ i(1) = G : H φ i (1) φ i (1) G : H 1 + φ i (1) = φ i (1). 2. χ i (h j ) = φ i (h j ) (this follows from φ G i (h j ) = φ i (h j ), i, j). 3. χ i (y k ) = φ i (1) (because φ G i (y k ) = 0, i, k). These, together with the above observations, imply χ i, χ i = 1 = 1 = 1 = χ i (g)χ i (g) [ χ i (g) 2 + ] χ i (g) 2 g N z S 1 z [ ] N φ i (1) 2 + S χ i (h) 2 G : H 1 h H φ i (h) 2 = φ i, φ i = 1. h H

24 24 Now, by definition we can write χ i = r+s j=1 a jψ j, a j Z, where ψ 1,..., ψ r+s are the irreducible characters of G. Then 1 = χ i, χ i = r+s j=1 a2 j χ i = ±ψ j for some j. But χ i (1) = φ i (1) > 0, so χ i = ψ j is an irreducible character of G. Finally, let χ = r χ i(1) χ i, which is a character of G. For any 1 h H, we have χ(h) = r φ i(1)φ i (h) = 0, by lemma 4. Hence, the same holds for any g G \ N = F 1, since such a g is congujate to some 1 h H. For any y N (and in particular y = 1), χ(y) = r φ i(1) 2 = H. Therefore, N = {g G χ(g) = χ(1) = H } = K χ G, as desired This theorem motivates the following definition of an important class of groups. A Frobenius group is a group G, having a subgroup H such that H g 1 Hg = {1}, g G \ H. This implies that the distinct conjugates of H intersect at the identity (since x 1 Hx y 1 Hy is conjugate to (yx 1 ) 1 H(yx 1 ) H via x), and there are G : H of them (because x 1 Hx = y 1 Hy xh = yh). We make the following observations: (1) In the setting of the above definition, G acts transitively and faithfully on S = {gh g G}, by left multiplication. (to see that the action is faithful, let x G be in its kernel. Then xgh = gh, g G g 1 xg H, g G x ghg 1 x = 1). Furthermore, if 1 g G fixes xh and yh, then as before, g xhx 1 yhy 1 xh = yh. Thus, each 1 g G fixes at most one element of S. Therefore, by Frobenius theorem, N = {1} {g G g acts fixed-point-free on S} G. By the proof of the theorem, we also have G = N [ r (g 1 i Hg i \ {1}) ] (g 1,..., g r are the distinct representatives of cosets of H), and N = G : H. Since N H = {1}, N H = and N G, we deduce that G is the semidirect product of N and H. Note: N is called the Frobenius kernel of the Frobenius group G, and H the Frobenius complement of N in G. (2) Consider the action of H as a group of automorphisms of N. Let φ : H Aut(N) be the conjugation homomorphism: φh = [n N hnh 1 N]. If 1 h H, then {n N φh(n) = n} = {n N hnh 1 = n} = {n N h = nhn 1 } H N = {1} (recall that H is self-normalizing in G). Thus, for every 1 h H, φh fixes only 1 in its action of N. We say that H acts on N as a group of fixed-point-free automorphisms. (3) Conversely, assume now that G is the semidirect product of N by H (N G), and H acts on N as a group of fixed-point-free automorphisms. We show that G is a Frobenius group, with kernel N and complement H. Let g G \ H, so that g = hn with 1 n N, h H. Then g 1 Hg = n 1 h 1 Hhn = n 1 Hn. Now we show H n 1 Hn = {1}: Indeed, h 1 = nh 2 n 1, h 1, h 2 H φh 1 (n) = h 1 nh 1 1 = nh 2 h 1 1 N h 2 h 1 1 H N = {1} h 1 = h 2 and φh 1 (n) = n h 1 = 1, since n 1 and H acts fixed-point-free on N. Therefore, H g 1 Hg = {1}, g G \ H, and G is Frobenius with complement H. To prove N is its kernel, note that N G and N H = {1} imply N [ r (g 1 i Hg i \ {1}) ] = (g 1 H,..., g r H are the cosets of H in G). Then = N H forces G = N [ r (g 1 i Hg i \ {1}) ], so N is the Frobenius kernel of G. (4) Let p, q be distinct primes with p 1(modq). Let φ : Z q Aut(Z p ) be a non-trivial homomorphism. Since Z q is generated by any non-identity element, φ must be injective. Hence φg acts fixed-point-free on Z p, 1 g Z q (if 1 h Z p is fixed by an injective α Aut(Z p ) then α is the identity since h generates Z p ). It follows that the unique