0.1 Algebras and Modules

0.1. ALGEBRAS AND MODULES 1 0.1 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 1A A, when appropriate. Accordingly, we also identify F with its copy F 1∈ in A. · ∈ · A Example 1: If V is a vector space over F, then HomF(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 = g G cg g cg K of formal sums in the elements of G with coeﬃcients in K, { ∈ · | ∈ } is a K-algebra under P (c g) + (d g) = ((c + d ) g) g · g · g g · g G g G g G X∈ X∈ X∈

(cg g) (dg g) = ( cgdh) x · · · · g G g G x G g,h G X∈ X∈ X∈  gXh=∈x      c c g = (cc g) · g · g · g G ! g G X∈ X∈ 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(1A) = 1B x, y A, c K f(cx + y) = c f(x) + f(y) ∀ ∈ ∈ · 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: × → 7→ g(v + w) = gv + gw (g + h)v = gv + hv h(gv) = (hg)v v, w M, g, h A, c K. g(cv) = c(gv) = (cg)v ∀ ∈ ∈ ∈ 1Av = v 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 )v. · A Note: Throughout this text, we will only consider modules which are ﬁnite dimensional as vector spaces. 2

Fix g A. Then g : M M, deﬁned by g (v) = gv, is an element of End(M). The ∈ M → M map g g is a K-algebra homomorphism of A and EndK(M). The image of A under 7→ M this homomorphism is written as AM . Note: Sometimes we will identify g with gM , which will be clear from the context. Example 3: Let V be a vector space and A a subalgebra of EndK(V ), then V is naturally an A-module (under gv = g(v), v V, g A). ∈ ∈ Example 4: In the previous example, let V = Kn (the vector space of n-tuples over K). n If A is any subalgebra of Mn(K), then K 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 ( g G kgg)k = g G kgk). ∈ ∈ Let V be an A-module (here A is a K-algebra).P A submoP dule 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 (V, W ) = ϕ ϕ : V W is an A-module homomorphism A { | → } EndA(V ) = HomA(V, V )

Note that EndA(V ) is a K-algebra, with addition and scalar multiplication of homomorphisms deﬁned poinwise, and multiplication deﬁned as function composition. Moreover, EndA(V ) is the centralizer of AV in EndK(V ). Note: When A is the group algebra KG, we call KG-modules just G-modules and KG- module 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 for some λ K. V ∈ 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 ) and part 1 implies ϕ = λI6 ∈ ∈ − V V Consequences: 1. If V is an irreducible G-module, then HomG(V, W ) is a division algebra. 2. If K is algebraically closed, then End (V ) = K I , is a field isomorphic to K. G · V 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 gV EndK(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 = λgIV for some λg 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 c K) 7→ ∈ 7→ g ∈ Since λ c = (gh)(c) = g(hc) = g(λ c) = λ (λ c) = λ λ c, c K, we get λ = λ λ for gh h g h g h ∀ ∈ gh g h all g, h G. Thus g λg is a homomorphism of G and the multiplicative group K×. In ∈ n 7→ n particular, g = 1 (λg) = 1, since 1 G is the identity map on K. In other words, λg is a root of unity in⇒K, of order dividing∈the order of g. Example 2: In the previous example let G = Z Z be generated by elements a of 2 × 3 order 2 and, and b of order 3. Also let K = C. Then all possible maps g G λg C can be arranged in the following character table of G: ∈ 7→ ∈ 1 a b b2 ab ab2 ρ1 1 1 1 1 1 1 2 2 ρ2 1 1 w w w w ρ 1 1 w2 w w2 w where w C, w3 = 1 3 ∈ ρ4 1 1 1 1 1 1 ρ 1 −1 w w2 −w −w2 5 − − − ρ 1 1 w2 w w2 w 6 − − − Note that any 2 rows (or columns) are orthogonal.

Theorem 2:(Maschke) Assume charK - G . Let V be a G-module and W – a submodule | | of V . Then there exists a submodule W 0 of V such that W W 0 = V . Proof: Let U be a subspace of V such that W U = V . Let P0 be the projection of V L 1 onto W (with respect to U). Deﬁne P : V W by P v = g G gP0(g− v). Note that P 7→ L ∈ is K-linear since P0 and g are. For all v V and h G, we have: ∈ ∈ P 1 1 1 1 1 1 P (hv) = hh− gP0(g− hv) = h(h− g)P0((h− g)− v) = hx− P0(xv) = hP (v). g G g G x G X∈ X∈ X∈ Therefore, P is a G-module homomorphism. 1 Next, observe that P (w) = G w, w W . Indeed, P (w) = g G gP0(g− w) = | | ∀ ∈ ∈ P 4

1 g G g(g− w) = G w, since W is G-invariant, and on W , P0 is just the identity map. ∈ | | Take W 0 = kerP , which is a submodule of V . We claim that this W 0 satisﬁes W W 0 = VP: First, pick an arbitrary u V . Then, since charK - G , there exists v V sucLh that ∈ | | ∈ u = G v. Write | | u = P (v) + ( G v P (v)) (1) | | − Note that P ( G v P (v)) = P ( G v) P (P (v)) = G P (v) G P (v) = 0, since P (v) W | | − | | − | | −| | ∈ and on W , P is multiplication by G . Hence G v P (v) W 0. | | | | − ∈ Then (1) implies that W + W 0 = V . Finally, if w W W 0 then G w = P (w) on one hand, and P (w) = 0 on the other. ∈ ∩ | | Again since charK - G , we get w = 0. Therefore, we have a direct sum: W W 0 = V | | L 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. RecallL the tensor product V K W , which is defined as the free product of V and W modulo the ”module” relations: N (v1 + v2) w = v1 w + v2 w v (w +⊗w ) = v ⊗w + v ⊗w ⊗ 1 2 ⊗ 1 ⊗ 2 c(v w) = (cv) w = v (cw) ⊗ ⊗ ⊗ for all v, v , v V, w, w , w W and c K. 1 2 ∈ 1 2 ∈ ∈ Then V K W is a G-module under g(v w) = (gv gw), g G, and extension by linearity. ⊗ ⊗ ∀ ∈ N 3. Hom(V, W ) is a G-module as follows: for ϕ Hom(V, W ), define gϕ by (gϕ)(v) = 1 ∈ g(ϕ(g− 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 1 1 under (gϕ)(v) = g (ϕ(g− v)) = ϕ(g− 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 →N γ N ϕ w (v ϕ(v) w) (2) ⊗ 7−→ 7→ · and extended linearly. n Then γ is injective: Indeed, let ξ kerγ. We may assume ξ = i=1 ϕi wi where wi W are linearly independent (otherwise∈ ξ = 0). Then γ(ξ) = 0 implies ⊗ ∈ P n ϕ (v) w = 0, v V. i · i ∀ ∈ i=1 X 0.4. REPRESENTATIONS AND CHARACTERS 5

Hence ϕi(v) = 0, v V, i = 1, n, and thus ξ = 0. We conclude that kerγ = 0. ∀ ∈ ? Further, γ must also be onto, since HomG(V, W ) and V W have same ﬁnite dimension dimKV dimKW over K (so they are isomorphic as vector spaces). Finally,·γ is a G-homomorphism: Let g G. N ∈ We have γg(ϕ w)(v) = (γ(gϕ gw))v = gϕ(v) gw, v V . We also have g⊗γ(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 - G . 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 is a map : G K deﬁned by ρ → χρ(g) = trace(ρ(g)). Thus, the following diagram commutes:

ρ G OO / GL(V ) OOO OOOχρ OOO trace OOO OO ' 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 deﬁned in section 0.1 is a 7→ M representation of G on M. We denote its character by χM .

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. ∈

1 2. χ (h− gh) = χ (g), g, h G (i.e. χ is a class function on G). ρ ρ ∀ ∈ ρ

3. χV W = χV + χW 4. χ L = χ χ V W V · W N 5. χ = χ ? χ Hom(V,W ) V · W 1 6. If K = C, then χ (g− ) = χ (g), g G and χ ? = χ ρ ρ ∀ ∈ V V n n Proof: 1 Let n be the order of g, then (ρg) = ρ(g ) = ρ(1) = IM . Hence the minimal n n th polynomial of ρg divides x 1 = i=1(x ωi), where w1, . . . , wn K are the n roots of unity. Since xn 1 is separable,− (because c−harK - G ) so is the minimal∈ polynomial of ρg. Therefore the Jordan− form of ρg isQdiagonal and all| eigen| values of ρg are roots of unity of 1 order dividing the order of g. 2 Since ρ is a homomorphism, χρ(h− gh) and χρg are traces of similar matrices, so they are equal. 3 Let g G. Fix bases v , . . . , v , w , . . . , w of V ∈ 1 n 1 m and W , and build the basis (v , 0) i = 1, n (0, w ) j = 1, m of V W . It remains { i | } { j | } S L 6

to note that, with respect to these bases, gV W is built of 2 diagonal blocks: gV and gW . 4 Let v , . . . , v , w , . . . , w be eigenbases of g GL(V ) and g GL(W ) respectively, 1 n 1 m L V ∈ W ∈ so that gV (vi) = λivi, gW (wj) = µjwj, with λi, µj K. It follows that the (vi wj)’s form a basis of V W (they certainly span V W , and∈there are n m of them, whic⊗h is precisely · dimK(V W )). Note that gV W (vi wj) = (gV (vi) gW (wj)) = λiµj(vi wj), so that N ⊗N ⊗ ⊗ the above is actually a gV W -eigenbasis. Hence χV W (g) = λiµj = λi µj = N N i,j i j χV (g) χW (g). 5 follows fromN Lemma 1 and part 4. 6 First,N if λ1, . . . , λn are the eigenvalues · 1 P 1 P 1 P of ρ(g) then 1/λ1, . . . , 1/λn are the eigenvalues of ρ(g− ) = (ρg)− , and χρ(g− ) = χρ(g) follows, since λi = 1 by part 1. For second part, let g G and v V be an eigenvalue | | 1 ∈ ∈ of g GL(V ). Then (gϕ)v = ϕ(g− 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 - G . | | 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 { ∈ | ∀ ∈ } 1 P = g G g G | | X∈ as a linear map on V (we identify g G with g End(V )). ∈ V ∈ 1 Then P is a G-module homomorphism: Indeed, for h G we have P h = G g G gh = 1 1 1 1 ∈ | | ∈ h G g G(h− gh) = h G x G x (since g h− gh is bijective) = hP . | | ∈ | | ∈ 7→ P Furthemore, P is the projection of V onto V G (i.e. V = V G kerP ): First, note that P G P 1 1 1 imP V , since hP (v) = (h G g G g)v = G g G hgv = G x G xv = P (v), for all ⊆ G | | ∈ 1 | | ∈ | L| ∈ G v V, h G. Also, v V P v = G g G gv = v. We conclude that imP = V and ∈ ∈ ∈ ⇒ P | | ∈ P P P P = P , i.e. P is the projection onto imP = V G. ◦ P Lemma 2: Let V, W be G-modules. Consider the G-module H = Hom(V, W ).

G 1 K 1. dim V = trace G g G g | | ∈ G P 2. H = HomG(V, W )

G 1 K ? 3. dim H = G g G χV (g)χW (g) | | ∈ P Proof: 1 Pick a basis of V G, and extend it to a basis of V . From the preceding argument G it follows that, in this basis, the matrix of P will be diagonal, with ﬁrst k = dimKV G G diagonal entries 1 and the rest 0. So dimKV = trace(P ). 2 We have f H 1 ∈ 1 ⇔ (gf)(v) = f(v), g G, v V gf(g− v) = f(v), g G, v V f(g− v) = 1 ∀ ∈ ∈ 1 ⇔ 1 ∀ ∈ ∈ ⇔ (g− f)(v), g G, v V fg− = g− f, g G f HomG(V, W ), as desired. 3 By ∀ ∈ ? ∈ ⇔ ∀ ∈ ⇔ ∈ Lemma 1, H = V W , as G-modules. So, χ = χ ? = χ ? χ , by Theorem 3. ∼ H V W V · W N N 0.5. THE PROJECTION FORMULA. ORTHOGONALITY OF THE IRREDUCIBLE CHARACTERS7

Using this and part 1, we get

G 1 1 1 dimKH = trace g = χ (g) = χ ? (g)χ (g) G G H G V W g G ! g G g G | | X∈ | | X∈ | | X∈ Lemma 3: Let V, W be irreducible G-modules, and H = Hom(V, W ). Then

G 0, if V W dimKH = 1, if V = W ∼

Proof: If V, W are non-isomorphic G-modules, then HomG(V, W ) = 0 by Schur’s Lemma, G so Lemma 3 implies H = 0. Now assume V = W , and let ϕ, ψ HomG(V, W ) be ∼ 1 ∈ non-zero. By Schur’s Lemma, ψ is invertible and hence ϕψ− HomG(W, W ). Again, 1 ∈ Schur’s Lemma yields ϕψ− = λI , for some λ K, i.e. ϕ = λψ. This proves that W ∈ dimKHomG(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 (g) χ (g) = (3) G V · W 1, if V = W g G ∼ | | X∈ 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 ∈ 1 α, β = α(g)β(g) (4) h i G g G | | X∈

It’s easy to check that , is a complex inner product on class(G). Note that χV class(G) for any G-moduleh· ·iV . ∈

Let α : G C and let V be a CG-module. Consider the map Pα,V = g G α(g)g : V → ∈ → V , which is certainly in EndC(V ). When is Pα,V a G-module homomorphism, regardless P of V ? – If α class(G), then Pα,V is a G-module homomorphism. Indeed, for an∈y h G, we have ∈ 1 1 P h = α(g)gh = hα(h− gh)h− gh (since α class(G)) = α,V ∈ g G g G X∈ X∈ 1 1 = h α(h− gh)h− gh = h α(x)x = hPα,V , as desired. g G x G X∈ X∈ Further, if V is an irreducible G-module and α class(G) then ∈ G P = | | α, χ I (5) α,V dimV h V i V 8

Since α class(G), P is a G-module homomorphism. But V is irreducible, so by ∈ α,V Schur’s Lemma P = λI for some λ C. The claim follows, since α,V V ∈ trace(P ) 1 1 G λ = α,V = α(g)trace(g) = α(g)χ (g) = | | α, χ . dimV dimV dimV V dimV h V i g G g G X∈ X∈

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 χ1, . . . , χk be the irreducible characters of G (k n). To prove that χ , . . . , χ span class(G), we ≤ 1 k show that (span χ1, . . . , χk )⊥ = 0 . Suppose β class(G) satisﬁes β, χi = 0, i. Then α, χ = 0{, i, where}α = β{ }class(G). Hence,∈ if U is any irreducibleh iG-module∀ h ii ∀ ∈ (so χU χ1, . . . , χk ), then (5) implies that Pα,U 0. In fact, it follows that Pα,V 0 ∈ { } ≡ m ≡ for all G-modules V : Indeed, by Maschke’s theorem V = j=1 Uj, where U1, . . . , Um are irreducible G-modules. Then L m m m P = α(g)g = α(g)( g ) = P = 0 0 . α,V V Uj α,Uj Uj ≡ V g G g G j=1 j=1 j=1 X∈ X∈ M M M

In particular, for the left regular representation (CG)◦ we have P C ◦ 0. Hence, α,( G) ≡

0 = P C ◦ (1 ) = α(g)(g 1 ) = α(g)g. α,( G) G · G g G g G X∈ X∈ But g g G is a basis of CG. Therefore, α(g) = 0, g G, i.e. α 0, as desired { | ∈ } ∀ ∈ ≡ Corollary 4’: m 1. If V ∼= i=1 miVi (by miVi we mean the direct sum of mi copies of Vi), where V1, . . . , Vm are non-isomorphic irreducible G-modules, then mi = χV , χVi , i. (First, fromL(3) it follows that 2 irreducible G-moduleshare isomorphici ∀ if and only if they have same characters. Hence χ , χ = m χ , χ = m m χ , χ = m , by h V Vi i j=1h mj Vj Vi i j=1 jh Vj Vi i i the theorem). This implies 2. Let U, V be G-modules, with U irreducible.P The number ofPtimes (an isomorphic copy of) U occurs in a decomposition of V into irreducible submodules is χV , χU (so it doesn’t depend on the decomposition). h i 3. For any G-modules V, W we have V = W χ = χ . ∼ ⇔ V W (Clearly V ∼= W χV = χW . Conversely, assume χV = χW . Let V1, . . . , Vk be a complete set of irreducible⇒G-modules. Then parts 1,2 above, along with Maschke’s theorem show that V = k a V = W , where a = χ , χ = χ , χ ). ∼ i=1 i i ∼ i h V Vi i h W Vi i 4. Let V1, . . . , Vm be non-isomorphic irreducible G-modules. If V = mL m V , then χ , χ = m m2. ∼ i=1 i i h V V i i=1 i 5. A G-module V is irreducible if and only if χV , χV = 1 (follows from 4). L P h i 0.6. THE LEFT REGULAR REPRESENTATION (CG)◦ 9

0.6 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-p{ oin| t ∈free}for any g = 1. Hence∈ in this basis B, the matrix of g (identified 6 with g(CG)◦ GL(CG)) has 0’s along the diagonal for g = 1, and 1’s otherwise. This shows ∈that 6 G if g = 1 χ(g) = (6) 0| | if g = 1 6 Now let χ1, . . . , χk be the irreducible characters of G, and V1, . . . , Vk – (one collection of) k G-modules corresponding to them. By corollary 4’, we have (CG)◦ ∼= i=1 miVi, where 1 1 L m = χ, χ = χ(g)χ (g) = χ(1)dimV = dimV . i h ii G i G i i g G | | X∈ | |

Thus, Vi appears dimVi times in (CG)◦. Hence

k (CG)◦ ∼= (dimVi)Vi (7) i=1 M

Lemma 4: Let G be a ﬁnite group and V1, . . . , Vk – G-modules corresponding to all the irreducible characters χ1, . . . , χk of G. Then

1. k (dimV )2 = G . i=1 i | | P 2. k (dimV )χ (g) = 0, g G, g = 1. i=1 i i ∀ ∈ 6 Proof:PThe lemma follows by taking characters in (7), and then using (6)

0.7 The Character Table

Let χ1, . . . , χr be the irreducible characters of a group G, and let K1 = 1 , K2, . . . , Kr be the conjugacy classes of G. Pick representatives g K , i and consider{ the} character i ∈ i ∀ table T of G: K1 K2 Kr 1 g · · · g 2 · · · r χ1 . . χi(gj) χr Recall that 1 r χ , χ = K χ (g )χ (g ) = δ h i ji G | t| i t j t ij t=1 | | X 10

Kj | | which gives orthogonality of the matrix G χi(gj). But if a matrix is orthogonal, so is | | its adjoint. Hence we also have column orthogonalitq y in T :

r G | | K = CG(gi) if i = j χt(gi)χt(gj) = i | | 0| | if i = j t=1 ( X 6 The following observations are helpful in constructing character tables of groups:

(1) The number of (irreducible) degree 1 characters of G is G0 = G/[G, G] : | | | | Assume ρ0 is a degree 1 representation of G0 = G/[G, G] on V . Then ρ : G GL(V ) → deﬁned by ρ(g) = ρ0(g0) is a degree 1 representation of G, where g0 = 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 ρ0 : G0 GL(V ) given by ρ0(g0) = ρ(g), is a ∀ ∈ → well-deﬁned degree 1 representation of G0, because

1 1 1 1 1 1 ρ(xyx− y− ) = ρ(x)ρ(y)ρ(x− )ρ(y− ) = ρ(x)ρ(x− )ρ(y)ρ(y− ) = IV , 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 G0, and the latter is G0 , because G0 is abelian. | | | | (2) A character χ of G is irreducible χ, χ = 1. (by Corollary 4’) ⇔ h i (3) If V, W are G-modules such that χV is irreducible and χW has degree 1, then χ = χ χ is also irreducible: V W V · W Since dimWN = 1, χW (g) is a root of unity for any g G and hence χW (g)χW (g) = 1. Then we have ∈ 1 χ , χ = χ (g)χ (g)χ (g)χ (g) = h V W V W i G V W V W g G N N | | X∈ 1 = χ (g)χ (g) = χ , χ = 1, so χ is irreducible. G V V h V V i V W g G | | X∈ N

(4) Assume H / G. Then any representation ρ0 of G/H naturally gives a representation ρ of G. Moreover, ρ is irreducible if and only if ρ0 is irreducible: The map ρ defined by ρ(g) = ρ0(gH), is a well-defined representation of G. Then 1 1 H χ , χ = χ (g)χ (g) = χ 0 (gH)χ 0 (gH) = | | χ 0 (g0)χ 0 (g ) h ρ ρi G ρ ρ G ρ ρ G ρ ρ 0 g G g G g0 G/H | | X∈ | | X∈ | | X∈ = χ 0 , χ 0 , and the second asertion follows from 2). h ρ ρ i It should be noted that the inner product χ , χ is taken in class(G), whereas χ 0 , χ 0 h ρ ρi h ρ ρ i – 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 . Fix⇔ a C-basis B = v , . . . , v → ρ { 1 n} of V . Define ρ0 : G GL(V ) by ρ0(g) = ρ(g) GL(V ) (the conjugate of the matrix of → ∈ ρ(g) in the basis B). Clearly ρ0(gh) = ρ(gh) = ρ(g)ρ(h) = ρ(g)ρ(h) = ρ0(g)ρ0(h), g, h ∀ ∈ 0.7. THE CHARACTER TABLE 11

G. This proves that ρ0 is a representation of G on V . Also, χρ0 (g) = trace(ρ0(g)) = trace(ρ(g) = θ(g), g G, which shows that θ = χ 0 is a character of G. The last part ∀ ∈ ρ of the statement follows from 2 in above, since θ, θ = g G θ(g)θ(g) = θ, θ . h i ∈ h i 2 2 (6) The Symmetric and Alternating Squares SymP, Alt : Let V be an n-dimensional G-module over F, where F is an algebraically closed ﬁeld such that char(F) = 2. If v1, . . . , vn is a basis of V , then vi vj 1 i, j n is a basis of V V . Thus,6 there{is a unique} linear map θ on V { V⊗, giv|en ≤by θ(v≤ }v ) = v v . F F i ⊗ j j ⊗ i For any ai, bj F, we have N ∈ N n θ(a v + + a v b v + + b v ) = a b θ(v v ) 1 1 · · · n n ⊗ 1 1 · · · n n i j i ⊗ j i,j=1 X

n = b a (v v ) = (b v + + b v ) (a v + + a v ). j i j ⊗ i 1 1 · · · n n ⊗ 1 1 · · · n n i,j=1 X i.e. θ(u v) = v u, u, v V . In particular, it follows that θ does not depend ⊗ ⊗ ∀ ∈ on the chosen basis v , . . . , v of V . Next, θ (and hence θ I ) is a G-module { 1 n} − V V homomorphism, since θ(g(vi vj)) = θ(gvi gvj) = gvj gvi = g(θ(Nvi vj)), g G. ⊗ 2 ⊗ ⊗ ⊗ 2∀ ∈ We deﬁne the symetric square: Sym (V V ) = ker(θ IV V ) (or just Sym , when V is clear from the context), which is a G-submodule of V− V . We have N N n n N u = a (v v ) ker(θ I ) (a a )v v = 0 a = a , i, j ij i ⊗ j ∈ − V V ⇔ ij − ji i ⊗ j ⇔ ij ji ∀ i,j=1 i,j=1 X N X u span v v + v v 1 i j n ⇔ ∈ { i ⊗ j j ⊗ i | ≤ ≤ ≤ } Clearly the (vi vj + vj vi)’s are linearly independent. Thus, they form a basis of 2 ⊗ 2 ⊗ n(n+1) Sym so that dimFSym = . To compute χ 2 , let g G. Since F is algebraically 2 Sym ∈ closed, it follows that g, regarded as a an element of GL(V ), is diagonalizable. Thus, we may choose v1, . . . , vn to be eigenvectors of g, with eigenvalues λ1, . . . , λn. Then g(vi vj + vj vi) = λiλj(vi vj + vj vi) and hence λiλj 1 i j n is the set of eigen⊗ values ⊗of g as an elemen⊗t of GL⊗(Sym2). Therefore,{ | ≤ ≤ ≤ }

n n 2 2 1 2 2 (χV (g)) + χV (g ) χ 2 (g) = λ λ = (( λ ) + λ ) = (8) Sym i j 2 i i 2 1 i j n i=1 i=1 ≤X≤ ≤ X X 2 Accordingly, we also deﬁne the alternating square: Alt (V V ) = ker(θ + IV V ). Sim- 2 ilarly, one shows that vi vj vj vi 1 i < j n is a basis of AltN, so that 2 n(n 1) { ⊗ − ⊗ | ≤ ≤ N} F − dim Alt = 2 . Likewise, we get

n n 2 2 1 2 2 (χV (g)) χV (g ) χ 2 (g) = λ λ = (( λ ) λ ) = − (9) Alt i j 2 i − i 2 1 i

Example 1: Let G = S3 (the symmetric group on 3 symbols).

S3 1 2 3 1 (12) (123) χ1 1 1 1 χ 1 1 1 2 − χ3 2 0 1

Since [S , S ] = A in general, we have S /[S , S ] = Z for n 2. Hence S has 2 n n n n n n ∼ 2 ≥ n irreducible characters of degree 1: the trivial one (χ1), and the sign character (χ2(g) = 1 if g A and χ (g) = 1 otherwise). The third character can be found directly from the ∈ n 2 − orthogonality relations. Alternatively, for n 2 consider the natural action of Sn on a basis B = e , e , . . . , e of Cn, given by g(e≥) = e . This turns Cn into an S -module { 1 2 n} i g(i) n V . Clearly W = span e + + e is a submodule of V . The map φ : V W { 1 · · · n} → a + + a φ(a e + + a e ) = 1 · · · n (e + + e ) 1 1 · · · n n n 1 · · · n is an S -homomorphism. Indeed for all g S , g(φ(a e + + a e )) = n ∈ n 1 1 · · · n n a + + a a + + a 1 · · · n (e + + e ) = 1 · · · n (e + + e (n)) n 1 · · · n n g(1) · · · g = φ(a e + + a e ) = φ(g(a e + + a e )). 1 g(1) · · · n g(n) 1 1 · · · n n Since imφ = W , it follows that U = kerφ is a n 1 dimensional submodule of V . If we represent g S as a linear map on V then, with resp− ect to the basis B, g is a permutation ∈ n matrix (its ij element is 1 if g(j) = i and 0 otherwise). Hence χV (g) is the number of ﬁxed points of g. Since χW (g) = 1 and V = W U, we get χU (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 (so χ = χ χ is an irreducibleLcharacter of S of degree n 1). For n = 3, n U Pn − 1 n − χU = χ3.

Example 2: G = S4.

S4 1 6 8 6 3 1 (12) (123) (1234) (12)(34) χ1 1 1 1 1 1 χ 1 1 1 1 1 2 − − χ3 3 1 0 1 1 χ 3 1 0 −1 −1 4 − − χ 2 0 1 0 2 5 − χ , χ are the trivial and sign characters; χ = χ χ is irreducible by above. 1 2 3 P4 − 1 Also χ4 = χ3 χ2 is irreducible by (3), section 0.7. The last character χ5 has degree √24 12 12· 32 32 = 2. Now χ = χ χ , since both are irreducible characters of − − − − 5 5 · 2 degree 2. So we get 2 zeroes in row 5. The other 2 entries are found from orthogonality 0.8. SOME EXAMPLES 13 relations. Alternatively, note that H = 1, (12)(34), (14)(23), (13)(24) = [A , A ] is a { } 4 4 normal subgroup of S4, and S4/H ∼= S3. If we lift χ3 from S3 ∼= S4/H to S4 we get χ5, which is automatically irreducible by (4) of section 0.7.

Example 3: G = A4 (the alternating group on 4 letters).

A4 1 4 4 3 1 (123) (132) (12)(34)

χ1 1 1 1 1 3 2 where w C R, w = 1. χ2 1 w w 1 ∈ \ 2 χ3 1 w w 1 χ 3 0 0 1 4 − We have [A , A ] = 1, (12)(34), (13)(24), (14)(23) = V , so A /[A , A ] = C . Hence 4 4 { } ∼ 4 4 4 4 ∼ 3 A4 has 3 representations of degree 1, involving complex roots of unity of order dividing 3

(χ1, χ2, χ3). Recall that χP4 χ1 is an irreducible character of S4. If we restrict to A4, we obtain χ (which is irreducible,− since χ , χ = 1/12 (32 + 3 ( 1)2) = 1). 4 h 4 4i · · − Example 4: G = S5.

S5 1 10 20 30 24 15 20 1 (12) (123) (1234) (12345) (12)(34) (12)(345) χ1 1 1 1 1 1 1 1 χ2 1 1 1 1 1 1 1 χ 4 −2 1 −0 1 0 −1 3 − − χ 4 2 1 0 1 0 1 4 − − χ5 6 0 0 0 1 2 0 χ 5 1 1 1 0 −1 1 6 − − χ 5 1 1 1 0 1 1 7 − − − χ , χ are the trivial and sign characters; χ = χ χ is irreducible and hence, so is 1 2 3 P5 − 1 χ4 = χ3 χ2. The alternating square of V3 (the representation whose character is χ3) · 2 1 2 2 has character χ5 (recall that χAlt V (g) = 2 [χV (g) χV (g )]), and is irreducible since 2 2 2 − χ5, χ5 = 1/120 (6 + 24 1 + 15 2 ) = 1. The other 2 characters are found by horthogonaliti y relations· of the·character· table.

Example 5: G = A5.

A5 1 20 15 12 12 1 (123) (12)(34) (12345) (21345) χ1 1 1 1 1 1 χ 4 1 0 1 1 2 − − χ3 5 1 1 0 0 − 1+√5 1 √5 χ4 3 0 1 2 −2 − 1 √5 1+√5 χ 3 0 1 − 5 − 2 2

χ2 and χ3 are restrictions from the characters χ3 and χ6 of S5 (and they are irreducible by computation). To obtain χ4 and χ5, recall that A5 is isomorphic to the group H of 14

rotations and symmetries of a regular icosahedron. This gives us a degree 3 representation ρ : A GL(R3) (the origin is the center of the icosahedron). Note that ρ is irreducible 5 → since A5 = H acts transitively on the set of vertices of the icosahedron. Since (12345) = ∼ 2π 4π | | 5, it’s immediate that ρ(12345) is a rotation by α = 5 or 2α = 5 (depending on the H 3 isomorphism A5 ∼= ). In the ﬁrst case, under a suitable basis of R , we have cos α sin α 0 1 + √5 ρ(12345) = sin α cos α 0 χ [(12345)] = 2 cos α + 1 = ,  −  ρ 2 0 0 1 ⇒   which reveals the character χ4. The second case yields the character χ5. 4 2 1 Example 6: G = D = r, s r = s = 1, srs = r− (rotations and simmetries of a 8 h | i square). 2 1 2 1 D8 1 r r, r− s, r s rs, r− s χ1 1 1 1 1 1 χ 1 1 1 1 1 2 − − χ3 1 1 1 1 1 χ 1 1 −1 1 −1 4 − − χ 2 2 0 0 0 5 − 2 Since [D , D ] = r and D0 = D /[D , D ] = Z Z , we have 4 representations of degree 8 8 h i 8 8 8 8 ∼ 2× 2 1 (with characters χ , . . . , χ ). These characters involve only 1, since every element of 1 4 D80 has order at most 2 (i.e. D80 has exponent 2). Finally, χ5 is found by orthogonality relations. Example 7: G = Q = i, j, k i2 = j2 = k2 = 1, ij = k, jk = i (the quaternions). 8 h | − i Q 1 1 i j k 8 − χ1 1 1 1 1 1 χ 1 1 1 1 1 2 − − χ3 1 1 1 1 1 χ 1 1 −1 1 −1 4 − − χ 2 2 0 0 0 5 −

We have [Q8, Q8] = 1 and Q80 ∼= Z2 Z2. From these we obtain χ1, . . . , χ4. Again, χ5 is determined by orthogonalith− i y. ×

Remark: The groups D8, Q8 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 D80 ∼= Q80 ) 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 = g G g acts on V as the identity map (i.e. K = kerρ). χ { ∈ | } χ 0.10. MORE ON THE REGULAR REPRESENTATION 15

Note that K = g G χ(g) = χ(1) : Indeed, χ(g) is a sum of n = dimV = χ(1) roots χ { ∈ | } of unity (namely, the eigenvalues of ρ(g)). Hence, χ(g) = n all eigenvalues of ρ(g) are 1 ρ(g) = I (because ρ(g) is diagonalizable). ⇔ ⇔ V Theorem 5: Let G be a ﬁnite group. Let χ1, . . . , χn be its irreducible characters. 1. T (the character table of G) determines the lattice of normal subroups of G. Specif-

ically, H ¢ G H = i I Kχi , for some I 1, 2, . . . , n . ⇐⇒ ∩ ∈ ⊆ { } 2. G is simple the only irreducible χ whose kernel K is non-trivial, is the trivial ⇐⇒ i χi character χ1. 3. 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 Kψ = H, since g(g0H) = g0H g0 G if and only if g H. Hence g H ψ(g) = ψ(1). On the other hand,∀ write∈ n ∈ ∈ ⇐⇒ ψ = i=1 aiχi, where a1, . . . , an are non-negative integers. Then P n n ψ(g) a χ (g) a χ (1) = ψ(1), | | ≤ i| i | ≤ i| i | i=1 i=1 X X which implies ψ(g) = ψ(1) χ (g) = χ (1), i I, where I = 1 i n a > 0 ⇐⇒ i i ∀ ∈ { ≤ ≤ | i } g Kχi , i I. Therefore, H = i I Kχi . ⇐⇒ ∈ ∀ ∈ ∩ ∈ Conversely, each Kχi is the kernel of a group homomorphism of G, so Kχi ’s (and their

intersections) are normal in G. 2 Assume G is simple. Then, since Kχ2 , . . . , Kχn are proper normal subgroups of G, all of them must be trivial (note that Kχ = G χ = χ1). Conversely, if G is not simple, then N ¢ G, 1 = N = G. By 1, N is an intersection⇔ of K ’s, so 1 = K = G for some i∃(i = 1). {3 }By6 1, T6 determines the lattice of normal χi { } 6 χi 6 6 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. 4

0.10 More On The Regular Representation

Theorem 6: Let G be a ﬁnite group and ρi : G Vi, i = 1, . . . , n be a complete set of irreducible representations of G. Then →

n CG ∼= End(Vi), as C-algebras. (10) i=1 M Proof: Consider the map φ : CG n End(V ), determined by −→ i=1 i

ρ1(g) L 0 φ . ·.· · . g G . .. . , and C-linear extension. (11) ∈ 7−→   0 ρ (g)  · · · n    16

Then φ is a C-algebra homomorphism, since it is C-linear, and ρi’s are group homomorphisms. Injectivity of φ follows because the left regular representation is faithful.

Indeed, if g G α(g) g kerφ, then ρi( g G α(g) g) = 0Vi . Hence ∈ · ∈ ∈ · P n P ( a ρ )( α(g) g) = 0 n , a N∗. i i · i=1 aiVi ∀ i ∈ i=1 g G M X∈ L 0 In particular, for the regular representation ρ, ρ( g G α(g) g) = 0(CG) . ∈ · So 0 = ρ( g G α(g) g)[1] = g G α(g) g in CG. Hence α(g) = 0 g G, which proves ∈ · ∈ · n P ∀ n∈ kerφ = 0. Next, note that dimCG = G = (dimV )2 = dim End(V ), so φ is P P | | i=1 i i=1 i surjective too. Therefore φ is a C-algebra isomorphism nP L Identifying End(Vi) with it’s copy in W = i=1 End(Vi), note that End(V1), . . . , End(Vn) are (2-sided) ideals of W . Hence, by the theorem, CG is the direct sum of the 2-sided ideals 1 1 L 1 φ− [End(V )], . . . , φ− [End(V )], so we have 1 C 0 = e + +e , where e φ− [End(V )] 1 n ( G) 1 · · · n i ∈ i are unique. But obviously

IEnd(V1) 0 n 0 0 . ·.· · . . · · · . 1W = . . . = . .  . . .   . IEnd(Vi) .  0 I i=1 0 0  · · · End(Vn)  X  · · ·      so we must have 0 0 φ . · · · . ei . I . , i = 1, . . . , n. (12) ←→  End(Vi)  ∀ 0 0  · · ·  In particular, e e = 0, i = jand e = e2, so φ(e ) is the projection of n V onto V . j i ∀ 6 j j i i=1 i i 1 φ Also note that ei generates the ideal φ− [End(Vi)]. Therefore, eiCG L End(Vi), and n ←→ CG = i=1 eiCG. The isomorphism given by (11) is important, as it enables us to view all the irreducible L representations of G (and their characters) ”simultaneously”. Speciﬁcally, χρi (g) is the trace of the restriction of φ(g) on Vi, and this naturally extends to CG. Thus, by (12) φ(e a) = 0 , j = i and φ(e a) = φ(a) , a CG. Hence, i |Vj Vj ∀ 6 i |Vi |Vi ∀ ∈ 0 for j = i χρj (eia) = 6 , a CG. (13) χρ (a) for j = i ∀ ∈ i Lemma 5: In CG we have:

dimVi 1 e = χ (g− ) g. (14) i G ρi · g G | | X∈

Proof: Write ei = g G αgg, αg C. Fix g G. Note that αg is the coeﬃcient of 1 in 1 1 ∈ ∈ ∈ n e g− , so χ C 0 (e g− ) = G α (recall (6), section 0.6). But χ C 0 = (dimV )χ , so i ( G) i P | | g ( G) i=1 i ρi n P 1 1 dimVi 1 α = (dimV )χ (e g− ) = χ (g− ), by (13). Thus (14) holds g G i ρj i G ρi j=1 | | X | | 0.11. FOURIER TRANSFORMS AND CLASS FUNCTIONS 17 0.11 Fourier Transforms And Class Functions

We will often make the identification α class(G) g G α(g)g CG. ∈ ↔ ∈ ∈ The convolution product of α, β class(G) is defined by α β ( g G α(g)g)( h G β(h)h). ∈ P· ↔ ∈ ∈ Equivalently, (α β)(x) = g,h G α(g)β(h), x G. This turns class(G) into a C-algebra. · gh=∈x ∀ ∈ P P Moreover, P Lemma 6: Z(CG) ∼= class(G), as C-algebras. Proof: We naturally define φ : class(G) CG by φ(α) = g G α(g)g CG. This is → 1∈ ∈ 1 clearly an injective C-algebra homomorphism. For any h G, h− φ(α)h = g G α(g)h− gh = ∈ P ∈ α(h 1gh)h 1gh = φ(α) (since α is a class function). Since φ(α) is centralised by G, g G − − P whic∈h spans CG, we have φ(α) Z(CG). Finally, we show that imφ = Z(CG): Assume P ∈ 1 a = g G βgg Z(CG). Then h− ah = a, h G ∈ ∈ ∀ ∈ ⇔

P 1 β (h− gh) = β g [β −1 β ]g = 0C β −1 = β , g, h G ⇔ g g ⇔ hgh − g G ⇔ hgh g ∀ ∈ ⇔ g G g G g G X∈ X∈ X∈ Hence β : G C defined by β(g) = β is a class function, so a = φ(β) imφ → g ∈ 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 αˆ(ρ) = α(g)ρ(g) End(V ). ∈ g G X∈ From the definition, we see that it’s enough to consider Fourier transforms at irreducible representations. Let ρ : G V , i = 1, n be a complete set of irreducible representations i → i of G, so that ρ αˆ End(V ). i 7−→ i Lemma 7: αˆβ = αˆβˆ. · ˆ Proof: For any representation ρ of G, we have: α β(ρ) = x G(α β)(x)ρ(x) = · ∈ · P = g,h G α(g)β(h) ρ(x) = g,h G α(g)ρ(g)α(h)ρ(h) = x G ∈ x G ∈ ∈ gh=x ∈ gh=x P P P Pˆ = g G α(g)ρ(g) h G β(h)ρ(h) = αˆ(ρ)β(ρ), as desired ∈ ∈ hP i P Theorem 7: (Fourier Inversion) Let α, β : G C. Then → n 1 1 α(g) = (dimV ) trace[ρ (g− )αˆ(ρ )]. (15) G i · i i i=1 | | X 1 n 1 Proof: The right term of (15) is G i=1 (dimVi) trace[ρi(g− ) h G α(h)ρi(h) ] = | | · ∈ n P P 1 1 1 1 = α(h) (dimV )χ (g− h) = α(h)χC (g− h) = α(g), G · i ρi G G | | h G i=1 ! | | h G X∈ X X∈ the last equality following by Lemma 4 18

Corollary 7: (Plancherel’s Formula)

n 1 1 α(g)β(g− ) = (dimV ) trace[αˆ(ρ )βˆ(ρ )]. (16) G i · i i g G i=1 X∈ | | X 1 n ˆ Proof: The right term of (16) is G i=1 (dimVi) trace[ g G α(g)ρi(g)β(ρi)] = | | · ∈ n P P 1 1 = α(g) (dimV )trace[ρ (g)βˆ(ρ )] = α(g)β(g− ), by (15) applied to β G · i i i g G " i=1 # g G | | X∈ X X∈ n Example: Let G = Zn = g g = 1 and α : G C. For j = 1, 2, . . . , n, let ρj be h | i j → th the representation of G determined by ρj(g) = w , where w C is a primitive n root of n k jk ∈ unity. Then αˆ(ρj) = x G α(x)ρj(x) = k=1 α(g )w . ∈ k 1 n jk Also, Fourier inversion gives α(g ) = αˆ(ρj)w− . P n Pj=1 P 0.12 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 dimF(gW ) = dimFW (since g GL(V )). Also, gW is an H-submodule of V : For ∈ 1 1 all g G, h H and w W we have h(gw) = g(g− hg)w gW , because g− hg H and W is∈H-invarian∈ t. ∈ ∈ ∈ (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 gW6 . Th6 us, if W is H-reducible, then so is gW . Conv6 ersely6, if gW is H-reducible then so is 1 W = g− (gW ). 1 (3) θ : W U is an H-isomorphism gθg− : gW gU is an H-isomorphism. → ⇔ 1→ 1 1 Assume θ : W U is an H-isomorphism. Then gθg− [h(gw)] = gθ[g− hg]w = g[g− hg]θw → 1 (since θ is an H-homomorphism) = hgθw = h[gθg− ]gw, h G, w W . Thus, 1 ∀ ∈ ∈ gθg− : gW gU is an H-homomorphism. It is an isomorphism, since dimF(gW ) = → 1 dimFW = dimFU = dimF(gU). Conversely, if gθg− is an H-isomorphism, then so is 1 1 θ = g(g− θg)g− .

(4) Consider g G 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 g G gW = V . Note that thisPis not a direct sum, since for example hW = W, h H. ∈ ∀ ∈ P (5) From now on, assume W is H-irreducible. If g W = g W for some g , g G, then 1 6 2 1 2 ∈ g1W g2W = 0, since g1W g2W is an H-submodule of the irreducible H-module gW . This ∩enables us to throw awa∩y the repeating components in (4), to get a direct sum. By k classifying the remaining ones according to H-isomorphism types, we get V = i=1 Vi, where Vi = gi1W gin W , and gijW = gi0j0 W as H-modules if and only if i = i0. · · · i ∼ L We may assume W = g11W V1. L L ⊆ 0.13. INDUCED REPRESENTATIONS 19

(6) If gW V then gV = V . Indeed, since W and g W are isomorphic as H-modules, ⊆ i 1 i 1j so are gW and g(g1jW ). Hence, gW Vi g(g1jW ) Vi, j = 1, n1, so gV1 = n1 ⊆ 1 ⇒ ⊆ ∀ j=1 g1jW Vi. Conversely, note that g− (gW ) = W V1. Since gW Vi, a similar to ⊆ 1 ⊆ ⊆ above argument implies that g− V V i.e. V gV . Therefore, gV = V , as claimed. L i ⊆ 1 i ⊆ 1 1 i Applying the above for g = gij we get gijV1 = Vi, and in particular, dimFVi = dimFV1 so dimVi dimV1 ni = dimW = dimW = n1. g G (7) By (6), V ∈ gV defines a transitive action of G on V , . . . , V . The kernel of i 7−→ i { 1 k} this action contains HCG(H). That H is in the kernel, follows since gW is H-invariant g G. Now let x CG(H). Fix g G. Regarding x as an element of GL(V ), ∀ ∈ ∈ ∈ 1 x : gW (xg)W is an isomorphism of vector spaces. Since x− 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 (xg)W V , x fixes each V . ⊆ i ⇒ ⊆ i 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 V V , where V = g W g W, g G, and g W = 1 2 · · · k i i1 · · · in ij ∈ ij ∼ gi0j0 W as H-modules if and only if i = i0. Note also that n does not depend on i. L L L L L Also, for each g G there is an i 1, . . . , k such that gW = g W as H-modules. ∈ ∈ { } ∼ i1 2. g G : V gV , defines a transitive action of G on V , V , . . . , V , with HC (H) ∈ i → i { 1 2 k} G in its kernel.

Corollary 8: 1 1. χ (h) = χ (g− hg), g G, h H. gW W ∀ ∈ ∈ 2. dimFV = kn dimFW . · 1 Proof: 1 follows, since g− hgw = λw hgw = g(λw) = λ(gw) (λ F). 2 follows by ⇔ ∈ looking at dimensions in 1. of theorem 8

0.13 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 G G W = FG 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. AccordinglyN, this gives a represen⊗ tation of⊗G on W∈G, which∈we denote by ρG.

(2) Let g1, g2, . . . , gk be a complete set of representatives of left cosets of H in G. Since k { } k G = i=1 giH and giH are F-bases of FG and FgiH respectively, we have FG = i=1 FgiH as F-vector spaces. U L (3) For i = 1, . . . , k we let Wi = gi FH W . Thus, Wi is an F-vector space, isomorphic to W . Also, W W = 0 . We claim that W G = k W . First, note that W = i ∩ j { } N i=1 i i L 20

Fg H W (since each h H acts faithfully on W ). Then by above, i FH ∈ N k k k k G W = FgiH W = FgiH W = Wi = Wi, as claimed. " i=1 # F i=1 " F # i=1 i=1 M OH X OH X M In particular, we get dimW G = G : H dimW . | | · (4) Let g G. Then ggiH = gjH for some j. For any w W , gi w Wi and ∈ 1 1 ∈ 1⊗ ∈ g(g w) = gg w = g (g− gg ) w = g (g− gg )w W , since g− gg H. Thus, i ⊗ i ⊗ j j i ⊗ j ⊗ j i ∈ j j i ∈ g G acts on W G = k W by first permuting the W ’s in the same way it permutes ∈ i=1 i i the left cosets of H in G (under the regular representation), and then acting within Wj’s L with elements of H. To be more precise, let’s combine bases of W1, . . . , Wk to get a basis B of W G. In this basis, the matrix of ρG(g) GL(W G) is a block-permutation ∈ matrix, where the non-zero blocks describe the action of H on Wj’s. It is convenient to set ρ(x) = 0 , x G H, i = 1, k. Then, in this basis, ρG(g) has form Wi ∀ ∈ \ 1 1 ρ(g− gg ) ρ(g− gg ) 1 1 · · · 1 k G . .. . ρ (g) =  . . .  (17) 1 1 ρ(g− gg ) ρ(g− gg )  k 1 · · · k k    G In particular, if dimFW = 1 and ρ is the trivial representation of H, then ρ (g) is just the permutation representation of G on the left cosets of H in G. G G G (5) If W1, W2 are H-modules, then (W1 W2) = W1 W2 . We have g (w1 + G G G G⊗ G w2) = g w1 + g w2 W1 W2 , g G, wi Wi, so (W1 W2) W1 + W2 . ⊗ ⊗ ∈G ∀ ∈L ∈ L G G⊆ By (3), dimF(W W ) = G : H (dimW + dimW ) = dimF(W W ), so the two 1 2 | | 1 2 1 2 must be equal. L L L L (6) Assume K H G, and let W be a K-module. Then (W H )G = W G as ≤ ≤ H G G ∼ G-modules. Define φ : (W ) W , by φ[g F (h F w)] = gh F w, and F-linear → ⊗ H ⊗ K ⊗ K 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-mo⊗dule⊗ homomorphism.⊗ From⊗its ⊗definition,∀ ∈φ ∈ ∈ H G H is surjective. On the other hand, (3) implies dimF(W ) = G : K dimFW = G : G | | | H H : K dimFW = dimFW . Therefore, φ is a G-isomorphism. | · | G | G G k 0 1 (7) Let χρ denote the character of ρ . Taking traces in (17) yields χρ (g) = i=1 χρ(gi− ggi), 0 χρ(x) if x H 1 1 1 where χ (x) = ∈ . Since g− ggi H h− g− ggih PH, h H, ρ 0 if x G H i ∈ ⇔ i ∈ ∀ ∈ 0 1 0 1 ∈1 \ we get χ (g− gg ) = χ (h− g− gg h), h H. Hence, ρ i i ρ i i ∀ ∈ k G 1 0 1 1 1 0 1 χ (g) = χ (h− g− gg h) = χ (x− gx), g G. (18) ρ H ρ i i H ρ ∀ ∈ i=1 h H x G | | X X∈ | | X∈ (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 θ0(u) = ∈ . ∈ 0 if u G H G G 1 0 1 ∈ \ G Define θ : G F, by θ (g) = H x G θ (x− gx). Then θ class(G). Indeed, G 1 1 → 0 1 1 | | G∈ ∈ θ (g− xg) = H u G θ (u− g− xgu) = θ (x), since as u runs through G, so does gu. | | ∈ P P 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 φG, θ (19) h |H i=h i G 1 G 1 0 1 Proof: We have φ , θ = G g G φ (g)θ(g) = G H g G x G φ (x− gx)θ(g) = h i | | ∈ | |·| | ∈ ∈ 1 P 0 1 1 P P = G H x G g G φ (x− gx)θ(x− gx) | |·| | ∈ ∈ = 1 φ0(y)θ(y) (since x 1gx g G = G, x G) H yPG P − | 1 | ∈ { | ∈ } ∀ ∈ = H y H φ(y)θ(y) = φ, θ H , as claimed. | | P ∈ h | i In particular, if φ andPθ are irreducible characters of H and G, then the number of times G φ appears in θ H is equal to the number of times θ appears in φ . Note: The two| inner products appearing in (19) are different. They are taken in class(H) and class(G), respectively. (10) Let θ class(H) and G. From the definition, we have θG(g) = 1 θ0(x), H x OG(g) ∈ ∈ where O (g) denotes the G-conjugacy class of g. Now any H-conjugacy| class| is contained G P in some G-conjugacy class, but not conversely. In general O (g) H will break up G ∩ into several H-conjugacy classes OH (h1), . . . , OH (hm), hi H. Since θ class(H), G 1 m 1 O ∈ 1 ∈O θ (g) = H i=1 x G x− gx H (hi) θ(hi). But if y− gy = x G(g), then 1 | | 1|{ ∈1 | ∈1 }| · ∈ z− gz = x yz− gzy− = g zy− CG(g) (CG(g) is the centralizer of g in G). Thus, ⇔1P ⇔ ∈ y G y− gy = x = C (g) , x O (g). Hence |{ ∈ | }| | G | ∀ ∈ G 1 m G : H m θG(g) = C (g) O (h ) θ(h ) = | | O (h ) θ(h ). (20) H | G | · | H i | · i O (g) | H i | · i i=1 G i=1 | | X | | X Note also that if O (g) H = ∅ (i.e. m = 0), then θG(g) = 0 by definition. H ∩ Example: H = A4 < G = A5 . Let θ be the 1-dimensional character of A4, given by

A4 1 4 4 3 1 (123) (132) (12)(34) where w C R, w3 = 1. ∈ \ θ 1 w w2 1

Then θG is a character of A , of dimension A : A θ(1) = 5. By (20), we have: 5 | 5 4| · G G:H 1. g = (123) O (g) H = O (123) O (132), so θ (123) = | | (4θ(123) + G H H OG(123) ⇒ ∩ ] | | 4θ(132)) = 5 (4w + 4w2) = 1. 20 − 2. g = (12)(34) O (g) H = O [(12)(34)]. Hence, ⇒ G ∩ H θG[(12)(34)] = 5 3θ[(12)(34)] = 1. OG[(12)(34)] | | ·

3. g = (12345) (or (13524)). Since H = A4 doesn’t have any 5-cycles, OG(g) H = ∅, so θG(g) = 0. ∩

A5 1 20 15 12 12 1 (123) (12)(34) (12345) (13524) θG 5 1 1 0 0 − Note also that θG is irreducible. 22 0.14 Consequences To Permutation Groups

Consider a finite group G acting transitively on a finite set S. For α S, the stabilizer of ∈ α, G = g G gα = α is a subgroup of G. The rank of G, r , is the number of orbits α { ∈ | } S of Gα acting on S. To show that this is well-defined, consider α, β S. There exists 1 ∈ 1 g G such that gα = β. Then xβ = β xgα = gα (g− xg)α = α, i.e. g− Gβg = Gα. ∈ ⇔ 1 ⇔ 1 If h Gα, then ha = b (a, b S) ghg− (ga) = gb (note that ghg− Gβ). Therefore, a ∈ga (which is a bijection ∈of S)⇔is also a bijection between orbits of G∈ and G . So the 7→ α β 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 ⇔ Gα is transitive on S α ( α S). ” ” Let a, b, c, d S, a = b, c = d. Since G is transitiv⇔ e, g(a, b) = (c, e)\for{ }some∀ g∈ G, e⇒ S. Since e, d ∈S c6 , there6 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 and gb = c. Thus,∈ G\ {is}transitive on∈S a . ∈ a a \ { } (2) (Burnside) Let θ be the character of the permutation representation of G on CS. Then r = θ, θ . S h i Let’s take a closer look at the action of G on S. Fix α S, and let H = Gα. Note that 1 ∈ gα = hα (g, h G) gh− H. Thus, if g H, . . . , g H are the cosets of H in G, then ∈ ⇔ ∈ 1 m the orbit of α, OG(α) = gα g G = g1α, . . . , gmα . In particular, OG(α) = G : H = G : G (this is Burnside’s{ | ∈Lemma} ,{and is true regardless} of the transitivit| | y of| G | | α| on S). Since G is transitive on S, S = OG(α) = g1α, . . . , gmα , so we have θ(g) = β 1 { G } |{ ∈ S gβ = β = i g(g α) = g α = i g− gg H = 1 (g), the last equality following | } |{ | i i }| |{ | i i ∈ }| H by (17) (1H is the trivial character of H). Finally, let OH (α1), . . . , OH (αk) be the orbits of 1 k O 1 H acting on S. Summing up the above, rS = H i=1 H (αi) Gαi = H β S Gβ = 1 1 | | | | · | G | | | ∈ | | H h H β S hβ = β = H h H θ(h) = θ H, 1H = θ, 1H = θ, θ , as desired ∈ |{ ∈ | }| ∈ Ph | i h i h Pi (w| |e also used Frobenius reciprocit| y).| P P (3) Assume S 2. Then G is doubly transitive on S θ 1 is an irreducible | | ≥ ⇔ − G character of G. 1 O 1 Let α S. Since G is transitive on S, 1 = G G(α) Gα = G β S Gβ = 1 ∈ | | | | · | | | | ∈ | | G g G θ(g) = θ, 1G . So θ 1G is a character of G (of degree S 1), and θ 1G, 1G = ∈ h i − | |− hP− i 0.| | Hence, G is doubly transitive on S θ, θ = 2 θ 1 , θ 1 + 1 , 1 + 2 θ P ⇔ h i ⇔ h − G − Gi h G Gi h − 1 , 1 = 2 θ 1 , θ 1 = 1 θ 1 is irreducible. G Gi ⇔ h − G − Gi ⇔ − G (4) For all n 2, S and A (except A ) have irreducible characters of degree ≥ n n 3 n 1. The group Sn is n-transitive in its permutation action on S = 1, 2, . . . , n , so by−(3), the claim is true for S . It is also true for A . Now let n 4.{ We are }done n 2 ≥ once we show that A is doubly transitive on S. In fact, A is (n 2)-transitive on n n − S. To see this, let (a1, . . . , an 2) and (b1, . . . , bn 2) be two (n 2)-tuples of distinct − − − elements of S. Then π Sn such that π(a1, . . . , an 2) = (b1, . . . , bn 2). Note that ∃ ∈ − − (S ) 1 (S ) −2 = τ S τ(a ) = a , i = 1, . . . , n 2 = S , so it has both odd n a ∩ · · · n an { ∈ n| i i ∀ − } ∼ 2 and even permutations. Choose σ (S ) (S ) − such that signπ = signσ. Then ∈ n a1 ∩ · · · n an 2 σπ An, and σπ(a1, . . . , an 2) = (b1, . . . , bn 2). ∈ − − 0.15. FROBENIUS GROUPS 23 0.15 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 = g G g fixes i elements of S . Then G = 1 F F , i { ∈ | } { } 0 1 and we have to show that N = 1 F0 is a normal subroup of G. For any g, h G, g 1 { } ∪ U ∈U fixes a , . . . , a S hgh− fixes ha , . . . , ha . This shows that F , F (and clearly 1 ) 1 k ∈ ⇔ 1 k 0 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 S), and then by considering a certain x ∈ linear combination of them. 1 Fix x S, and let H = G . Since gHg− = G , and G is transitive on S, all 1-point ∈ x gx stabilizers Gz are conjugate to H. Thus, any g F1 is conjugate to some element of H. This enables us to select a full set of represen∈ tatives of G-conjugacy classes A = 1, h , . . . , h , y , . . . , y , with h H and y F = N 1 . Another consequence is { 2 r 1 s} i ∈ j ∈ 0 \ { } that F1 = z S(Gz 1 ) (the union is disjoint since any g G fixes at most 1 point of ∈ \ { } ∈ S). Thus, G = N z S(Gz 1 ) . Since G is transitive on S, S = G : Gz , z S. U ∈ \ { } | | | | ∀ ∈ So the above decomposition gives G = S ( Gx 1)+ N N = S = G : H . Note U U | | | |· | |−1 | | ⇒ | | | | | | also that H is self-normalizing in G: indeed, if ghg− = h0 (h, h0 H, g G), then h0 fixes x and gx so we must have x = gx, i.e. g H. This also means∈that B∈= 1, h , . . . , h ∈ { 2 r} is a full set of representatives of H-conjugacy classes. G Let φ1, . . . , φr be the irreducible characters of H. We have φk (1) = G : H φk(1), G 1 0 1 | G| · φk (hi) = H g G φ (g− hig) = φk(hi) (since H is self-normalizing in G) and φk (yj) = 0 ∈ (since g 1|y |g F , which is disjoint from H). Let χ be the trivial character of G. For − j P 0 1 i = 2, . . . , r we∈define χ = φG φ (1) φG + φ (1) χ (which is also valid for i = 1). Then i i − i · 1 i · 1 G G 1. χi(1) = φi(1). Indeed, χi(1) = φi (1) φi(1) φ1 (1) + φi(1) = G : H φi(1) φi(1) G : H 1 + φ (1) = φ (1). − · | | · − · | | · i i 2. χ (h ) = φ (h ) (this follows from φG(h ) = φ (h ), i, j). i j i j i j i j ∀ 3. χ (y ) = φ (1) (because φG(y ) = 0, i, k). i k i i k ∀ These, together with the above observations, imply 1 χ , χ = χ (g)χ (g) h i ii G i i g G | | X∈ 1 = χ (g) 2 + χ (g) 2 G | i | | i | "g N z S 1=g Gz # | | X∈ X∈ 6 X∈ 1 = N φ (1)2 + S χ (h) 2 G | | · i | | | i | " 1=h H # | | 6 X∈ G : H = | | φ (h) 2 = φ , φ = 1. G | i | h i ii h H | | X∈ 24

Now, by definition we can write χ = r+s a ψ , a Z, where ψ , . . . , ψ are the i j=1 j j j ∈ 1 r+s irreducible characters of G. Then 1 = χ , χ = r+s a2 χ = ψ for some j. But hPi ii j=1 j ⇒ i j χi(1) = φi(1) > 0, so χi = ψj is an irreducible character of G. r P Finally, let χ = i=1 χi(1) χi, which is a character of G. For any 1 = h H, we have χ(h) = r φ (1)φ (h) = 0, by· lemma 4. Hence, the same holds for any6 g ∈G N = F , i=1 i i ∈ \ 1 since such a g is congujateP to some 1 = h H. For any y N (and in particular y = 1), χ(y) = Pr φ (1)2 = H . Therefore,6 N =∈ g G χ(g)∈= χ(1) = H = K ¢ G, as i=1 i | | { ∈ | | |} χ desired P This theorem motivates the following definition of an important class of groups. 1 A Frobenius group is a group G, having a subgroup H such that H g− Hg = 1 , g G H. This implies that the distinct conjugates of H intersect at∩ the identit{ y} (since∀ ∈ \1 1 1 1 1 x− Hx y− Hy is conjugate to (yx− )− H(yx− ) H via x), and there are G : H of ∩ 1 1 ∩ | | them (because x− Hx = y− 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 { | ∈ } 1 1 ∈ its kernel. Then xgH = gH, g G g− xg H, g G x g GgHg− x = 1). ∀ ∈ ⇔ ∈ ∀ ∈ ⇔ ∈ ∩ ∈ 1 ⇒ 1 Furthermore, if 1 = g G fixes xH and yH, then as before, g xHx− yHy− xH = yH. Thus,6 eac∈h 1 = g G fixes at most one element∈of S. Therefore,∩ b⇒y Frobenius’ theorem, N = 16 g∈ G g acts fixed-point-free on S ¢ G. By the proof { } ∪ { ∈ | r 1 } of the theorem, we also have G = N i=1(gi− Hgi 1 ) (g1, . . . , gr are the distinct representatives of cosets of H), and N = G : H . Since\ { }N H = 1 , N H = G | | | | ∩ { } | | · | | | | and N ¢ G, we deduce that G is the semidirectU U 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) 1 → be the conjugation homomorphism: φh = [n N hnh− N]. If 1 = h H, then 1 ∈ 7→ ∈ 1 6 ∈ n N φh(n) = n = n N hnh− = n = n N h = nhn− H N = 1 (recall{ ∈ that| H is self-normalizing} { ∈ |in G). Thus,}for {every∈ 1 =| h H, φh}fixes⊆ only∩ 1 in{its} 6 ∈ 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 = 1 1 1 ∈ 1 \ hn with 1 = n N, h H. Then g− Hg = n− h− Hhn = n− Hn. Now we show 1 6 ∈ ∈ 1 1 1 H n− Hn = 1 : Indeed, h1 = nh2n− , h1, h2 H φh1(n) = h1nh1− = nh2h1− ∩ 1 { } ∈ ⇒ ∈ N h2h1− H N = 1 h1 = h2 and φh1(n) = n h1 = 1, since n = 1 and H ⇒ ∈ ∩ { } ⇒ 1 ⇒ 6 acts fixed-point-free on N. Therefore, H g− 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 r 1 ∩ { } imply N i=1(gi− Hgi 1 ) = ∅ (g1H, . . . , grH are the cosets of H in G). Then ∩ \ { } r 1 G = N H forces G = N (g− Hgi 1 ) , so N is the Frobenius kernel of G. | | | | · |U| i=1 i \ { } (4) Let p, q be distinct primes with p 1(modq). Let φ : Z Aut(Z ) be a non-trivial U U ≡ q → p homomorphism. Since Zq is generated by any non-identity element, φ must be injective. Hence φg acts fixed-point-free on Z , 1 = g Z (if 1 = h Z is fixed by an injective p ∀ 6 ∈ q 6 ∈ p α Aut(Z ) then α is the identity since h generates Z ). It follows that the unique ∈ p p 0.16. ALGEBRAIC INTEGERS 25

non-abelian group of order pq, Zp o Zq, is a Frobenius group. (5) (Thompson) The Frobenius kernel of any Frobenius group, is a nilpotent group.

0.16 Algebraic Integers

Let α C be an algebraic number, so that α satisfies some non-zero polynomial with coefficien∈ ts in Q. Denote by m (x) Q[x] the minimal polynomial of α, i.e. the monic α ∈ rational polynomial of minimal degree that α satisfies. Theorem 10: Let α C be algebraic. The following are equivalent: ∈

1. mα(x) actually has coeﬃcients in Z.

2. Z[α] is ﬁnitely generated as an (additive) abelian group.

3. α is a root of some monic polynomial f Z[x]. ∈ k k 1 k k 1 Proof: 1 2 Suppose mα(x) = x + bk 1x − + + b0 Z[x]. Then α = (bk 1α − + ⇒ k m − · · · ∈ − − + b0), so α (and α , m k, by induction) is an integral linear combination of · · · k 1 ∀ ≥ k 1 1, α, . . . , α − . Thus Z[α] is generated as a Z-module by 1, α, . . . , α − 2 3 Assume ⇒ Z[α] = f1(α)Z + + fm(α)Z, where fi Z[x]. Let n = 1 + max1 x m deg fi . Since n n · · · m ∈ ≤ ≤ { } α Z[α], α = i=1 cifi(α) for some c1, . . . , cm Z. Then α is a root of the monic ∈ n m ∈ polynomial f(x) = x cifi(x) Z[x] 3 1 Assume f(α) = 0, where f Z[x] P − i=1 ∈ ⇒ ∈ is monic. Since Q[x] is a Euclidean domain, we may write f(x) = a(x)mα(x) + b(x) in Q[x], with b 0 or degPb

Let G be a ﬁnite group. Recall that if ρ is a representation of G on V (over C), then the minimal polynomial of ρ(g) divides xn 1, where n is the order of g G. Then all the eigenvalues of ρ are roots of unity, hence− algebraic integers. It follows ∈that the character table of G consists of algebraic integers. Our next aim is to show that the degrees of the irreducible characters of G, divide G . | | (1) Let C1, . . . , Cr be the conjugacy classes of G. Abuse notation and write Cj =

g Cj g CG. We also identify f : G C with g G f(g)g CG. Under this iden- ∈ ∈ → ∈ ∈ tiﬁcation, from the proof of lemma 6 we see that class(G) = Z(CG). Also C , . . . , C P P 1 r class(G), and any f class(G) is an integral linear combination of C , . . . , C . Since∈ ∈ 1 r C C class(G), we have i j ∈ r C C = n C , n Z. (21) i j ijk k ijk ∈ Xk=1 (2) Let ρ be an irreducible representation of G on V . Since C Z(CG), ρ(C ) : V V j ∈ j → is a G-module homomorphism. By Schur’s lemma, ρ(Cj) = wρjI, for some wρj C. Applying this to (21), we get ∈

r w w = n w 1, i, j r. ρi ρj ijk ρk ∀ ≤ ≤ Xk=1 Therefore, the ring Z[wρ1, . . . , wρr] is ﬁnitely generated as a Z-module (for example by 1, wρ1, . . . , wρr). It follows that Z[wρj] Z[wρ1, . . . , wρr] is also a ﬁnitely generated Z- module. This implies w Alg. ⊆ ρj ∈ (3) What is w ? Recall that ρ(C ) = w I. Taking traces C χ (C ) = w χ (1). ρj j ρj ⇒ | j| · ρ j ρj · ρ Now since χ is irreducible, r C χ (C )χ (C ) = G χ , χ = G . Whence, ρ k=1 | k| ρ k ρ k | | · h ρ ρi | | P r G wρkχρ(Ck) = | | χρ(1) Xk=1 The left term is an algebraic integer, whereas the right term is rational. So they are both integers and dimρ = χ (1) divides G . ρ | | 0.17 Burnside’s paqb Theorem

Lemma 11: If λ C is an average of roots of unity, and λ Alg, then λ = 0 or λ is a root of unity. ∈ ∈ Proof: Consider the monic irreducible polynomial m (x) Z[x]. The Galois group of λ ∈ the splitting ﬁeld K of mλ over Q, acts transitively on the roots of mλ. Let these roots m m be σ1λ, . . . , σmλ. We have i=1 σiλ = ( 1) mλ(0) Z. We assume without loss that − ∈ 1 σ1, . . . , σm are extended ﬁeld autormorphisms of C. Write λ = n (λ1 + + λn), where λi Q 1 · · · is a root of unity. Then σjλ = n (σjλ1 + + σjλn) and note that σjλi is also a root unity (of same order as λ ). It follows that σ ·λ· · 1 ( σ λ + + σ λ ) = 1, with equality if i | j | ≤ n | j 1| · · · | j n| and only if σ λ = = σ λ λ = = λ (= λ). Multiplying these inequalities over j 1 · · · j n ⇔ 1 · · · n 0.17. BURNSIDE’S P AQB THEOREM 27

1 j m implies m (0) = m σ λ 1. If m (0) = 0, then m (x) = x and λ = 0. ≤ ≤ | λ | j=1 | j | ≤ | λ | λ Otherwise mλ(0) = 1, so we must have σjλ = 1, j = 1, . . . , m, and λ = λ1 = = λn is a root of|unity | Q | | ∀ · · ·

Corollary 9: Let χρ be an irreducible character of a ﬁnite group G, and C1, . . . , Cr be the conjugacy classes of G. If χ (1) and C are relatively prime (for some 1 j r), ρ | j| ≤ ≤ then either χ (C ) = 0 or χρ(Cj ) is a root of unity. ρ j χρ(1) Proof: Since (χ (1), C ) = 1, there exist a, b Z such that a χ (1) + b C = 1. Then ρ | j| ∈ · ρ · | j|

χρ(Cj) Cj χρ(Cj) = aχρ(Cj) + b| | · = aχρ(Cj) + bwρj Alg. χρ(1) χρ(1) ∈

The left term is an average of roots of unity, so the conclusion follows from Lemma 11 Lemma 12: Let G be a ﬁnite group. If there exists a conjugacy class C of G, such that C = pn (p prime and n 1), then G is not simple. | | ≥ Proof: Let 1 = g C. Let χ1, . . . , χm be all the non-trivial irreducible characters of G which don’t vanish6 ∈at g. By column orthogonality in the character table of G,

1 + χi(1)χi(g) = 0. (22) i=1 X

If p χ (1), 1 j m, then 1 = m χi(1) χ (g) Alg, which is impossible, since | i ∀ ≤ ≤ p i=1 p i ∈ 1 Q Z. Therefore, p and χ (g) are relativh ely primei for some j = 1, . . . , m. Since p ∈ \ j P χj(g) = 0, Corollary 9 implies χj(g) = λχj(1), where λ C is a root of unity. Let ρ : G 6 GL(V ) be a representation whose character is χ∈ . Since the eigenvalues of → j ρ(g) are roots of unity whose average is λ, all of them have to equal λ, so ρ(g) = λI. Now assume G is simple. Then G is not abelian (otherwise C = 1), so [G, G] = 1 . It follows that G has only 1 irreducible character of degree |1 (the| trivial one). Hence{ } dimρ = dimCV > 1, which implies kerρ = G. Since G is simple, this forces kerρ = 1 , 6 { } so that G ∼= G/kerρ ∼= ρ(G) as groups. Since ρ(g) Z(ρ(G)), we get 1 = g Z(G), contradicting the simplicity of G ∈ 6 ∈ Theorem 11: (Burnside) Let G be a group. If G = paqb (p, q - primes and a, b 1), | | ≥ then G is solvable. Proof: Assume the statement is false, and let G be a counterexample such that G is minimal. Then G is simple. Indeed, if 1 = H ¡ G then H and G/H are solvable. |Then| { } 6 G is solvable (we get a normal series of G by combining the normal series of H and G/H), contradiction. Let P Sylp(G). Since P is a p-group, we may choose 1 = g Z(P ). Then P C (g), so ∈O (g) = G : C (g) = qm (0 m b). But m =6 0 (otherwise∈ ≤ G | G | | G | ≤ ≤ 6 1 = g Z(G), so the simple G must be abelian of non-prime order), so Lemma 12 implies that6 G∈cannot be simple, contradiction What is striking about this proof is the way that solvability is used. Examining the proof, we see that the theorem is true if we replace solvability by any other property P such that H ¢ G, P (H) P (G/H) P (G). ∀ ∧ ⇒ 28 0.18 Brauer-Tate Theorems

We first introduce some preparatory concepts. Let G be a finite group, and χ1, . . . , χm be its irreducible characters (over C). We call ψ class(G) a generalized character if it is the difference of 2 characters. Equiv- ∈ alently, ψ is a generalized character if it is an integral linear combination of (irreducible) characters of G. Define ch(G) = ψ class(G) ψ is a generalized character of G = { ∈ | } a1χ1 + + amχm a1, . . . , am Z . Since χ1, . . . , χm are C-linearly independent, { · · ·m | ∈ } ch(G) ∼= Z as Z-modules. By 4. of Theorem 3, the product of any 2 characters is a character, so ch(G) is a commutative ring with identity (the trivial character χ1). A finite group E is elementary if E = A P , where P is a p-group and A is a cyclic × p0-group (i.e. ( A , P ) = 1). If we wish to highlight p we call E p-elementary. Define E to be the set of| elemen| | | tary subgroups of G (over all primes). The set of characters of G, induced from (characters of) members of E, generates a Z-submodule of ch(G), which we denote by v(G). Our primary goal is to prove Theorem 12: (Brauer-Tate) ch(G) = v(G). As usual, we make a series of observations which will guide us through the proof of the theorem: (1) Let w C be a primitive root of 1, of order n = G . Let R = Z[w] = Z[x]/mw(x), ∈ n 1 n| 2| ∼ where m (x) = x − + + x + 1. Thus, 1, w, . . . , w − is a basis for R as a Z-module. w · · · Define ch (G) = a χ + + a χ a , . . . , a R = Rm, as R-modules. Accordingly R { 1 1 · · · m m | 1 m ∈ } ∼ we define vR(G) to be the R-submodule of chR(G), generated by characters of G, induced from members of E. From definitions, we have the following diagram of inclusions:

ch(G) / chR(G) O O

? ? v(G) / vR(G)

G G (2) Let H G, φ class(H) and θ class(G). Then (φ θ H ) = φ θ. ≤ ∈ ∈ G 1 · | 0 1 · Proof: For any y G, we have (φ θ H ) (y) = H x G(φ θ H ) (x− yx) = ∈ · | | | ∈ · | P 1 0 1 1 1 0 1 G = φ (x− yx)θ(x− yx) = θ(y) φ (x− yx) = θ(y) φ (y). H · H · x G x G | | X∈ | | X∈ 0 0 0 We used that (φ θ H) = φ θ, which holds since both sides vanish on G H, and θ H = θ on H · | · \ | (3) v(G) is an ideal of ch(G). Let χ ch(G). By deﬁnition, the set A = ψG ψ ch(E) for some E E generates v(G) as∈a Z-module, so it’s enough to check {that |χ ∈ψG v(G), where ψ ∈ch(}E), E E. By (2), χ ψG = (ψ χ )G v(G), since ψ and χ · (and∈ ψ χ ) are characters∈ of E∈. · · |E ∈ |E · |E (4) ch(G) vR(G) = v(G). ∩ n 2 First, note that 1, w, . . . , w − are linearly independent over the ring ch(G). Indeed, n 2 m j assume − ( c χ )w = 0 in class(G), with c C. Interchanging the sums we j=1 i=1 ij i ij ∈ P P 0.18. BRAUER-TATE THEOREMS 29 get

m n 2 n 2 − − ( c wj)χ = 0 c wj = 0, i = 1, m c = 0, i = 1, m, j = 1, n 2. ij i ⇒ ij ∀ ⇒ ij ∀ − i=1 j=1 j=1 X X X

n 2 j In particular, it follows that vR(G) = i=1− v(G)w as R-modules. Now suppose ξ = n 2 ξ0+ξ1w+ +ξn 2w − ch(G) vR(G), ξ0, . . . , ξn 2 v(G). By the linear independence · · · n 2− ∈ ∩ L − ∈ of 1, w, . . . , w − over ch(G), we must have ξ = ξ v(G), which proves the claim. In 0 ∈ particular, if χ1 vR(G) then χ1 v(G) ch(G) = v(G), since v(G) is an ideal of ch(G). ∈ ∈ ⇒

(5) Let p be prime and g G. There exist g1, g2 G such that g1 is a p-element (i.e. α ∈ ∈ g = p , for some α), g is a p0-element (i.e. g and p are coprime), and g = g g = g g . | 1| 2 | 2| 1 2 2 1 Moreover, this representation of g is unique. The components g1, g2 are called the p-part and the p0-part of g, respectively. Also g = g1g2 is the p-decomposition of g. α Proof: Since g1, g2 should commute, it’s reasonable to look at powers of g. Write g = p l, with (p, l) = 1. Then apα + bl = 1 for some a, b Z. Then g = gbl and g| |= gapα ∈ 1 2 satisfy the conditions: indeed g = g /( g , bl) = pα, g = g /( g , apα) = l and | 1| | | | | | 2| | | | | g = g1g2 = g2g1. Now assume g = h1h2 = h2h1 were another such representation. α α Since p l = g = [ h1 , h2 ], the hypothesis implies h1 = p and h2 = l. Then apα+bl | |bl |bl | | | bl | | | | h1 = h1 = h1 = g = g1 (since h2 = 1), and h2 = g/h1 = g/g1 = g2. 1 1 1 (6) Let g = g1g2 be the p-decomposition of g G. Consider x− gx = [x− g1x][x− g2x]. 1 1 ∈ 1 Since x− g1x = g1 , h1 = x− g1x is a p-element. Similarly h2 = x− g2x is a p0- | | 1| | 1 1 1 element. Since x− gx = h1h2 = h2h1, 5 implies that x− gx = [x− g1x][x− g2x] is the 1 p-decomposition of x− gx. For g, h G, write g h if the p0-parts of g and h are conjugate. Note that is an equivalence∈relation. If g∼is conjugate to h, then g h by above. ∼ ∼ It follows that the equivalence classes of are unions of conjugacy classes of G. ∼ (7) Assume G = G G . Then 1 × 2

1. Any conjugacy class of G is the product of a conjugacy class G1 and one of G2.

2. Any irreducible character of G is of form χ1χ2 (which is deﬁned by χ1χ2(g1g2) = χ (g )χ (g ), g G , g G ), where χ , χ are irreducible characters of G and 1 1 2 2 ∀ 1 ∈ 1 2 ∈ 2 1 2 1 G2 respectively.

1 Proof: 1 Let g = g1g2 G with gi Gi. Then OG(g) = (h1h2)− g(h1h2) hi Gi = 1 1 ∈ ∈ { | ∈ } [h− g h ][h− g h ] h G = O (g )O (g ). 2 Assume χ , χ are characters of the { 1 1 1 2 2 2 | i ∈ i} G1 1 G2 2 1 2 irreducible representations ρ1 : G1 GL(V1), and ρ2 : G2 V2 respectively. Then χ χ , is the character of the represen→tation ρ : G = G G→ V V deﬁned by 1 2 1 × 2 → 1 2 ρ(g1g2)(v1 v2) = (ρ1(g1), ρ2(g2)) [recall that if v1, . . . , vk and w1, . . . , wl are bases of ⊗ { } { N } eigenvectors of ρ1(g1) GL(V1) and ρ2(g2) GL(V2) respectively, then vi wj 1 i k, 1 j l is a basis∈of eigenvectors of ρ(∈g g ) GL(V V ), so that{ the⊗ eigen| v≤alues≤ ≤ ≤ } 1 2 ∈ 1 2 of ρ(g1g2) are precisely the pairwise products of eigenvalues of ρ1(g1) and ρ2(g2)]. Let N 30

ψ1, ψ2 be irreducible characters of G1, G2, respectively. We have 1 χ χ , ψ ψ = χ (g )χ (g )ψ (g )ψ (g ) h 1 2 1 2i G 1 1 2 2 1 1 2 2 | | g1 G1 gX2∈G2 ∈ 1 1 = χ1(g1)ψ1(g1) χ2(g2)ψ2(g2) G1 G2 g1 G1 ! g2 G2 ! | | X∈ | | X∈ = χ , ψ χ , ψ . h 1 1i · h 2 2i

For ψ1 = χ1 and ψ2 = χ2, this shows that χ1χ2 is irreducible. Otherwise, χ1χ2, ψ1ψ2 = 0, and in particular χ χ = ψ ψ . Thus, the number of such characters his the numbier of 1 2 6 1 2 irreducible characters of G1 times that of G2, which by 1 is equal to the number of irreducible characters of G. Hence these are all irreducible characters of G. (8) If E = A B E, with A = a , then ξ ch (E) such that ξ(ab) = a , ξ(ajb) = × ∈ h i ∃ ∈ R | | 0, j = 2, 3 . . . , a 1, b B. Pro∀ of: Deﬁne ξ| :|E− C∈as above. Then ξ class(E), because O (ajb) = ajO (b), j = → ∈ E B ∀ 1, a , b B. Let χ1, . . . , χk and ψ1, . . . , ψl be the irreducible characters of A and B | | ∈ j 1 respectively. We may assume that χj(a) = w − for 1 j k, where w C is a primitive root of unity of order k = a . By 7, we may write≤ ≤ ∈ | | ξ = a χ ψ , with a C. ij i j ij ∈ 1 i k 1X≤j≤ l ≤ ≤

Let O1, . . . , Ol be the conjugacy classes of B. From 7 and the deﬁnition of ξ, we get

1 l a = ξ, χ ψ = aO ξ(aO )χ (a)ψ (O ) ij h i ji E | s| · s i j s s=1 | | X 1 l = O ) a wi 1 ψ (O ) a B | s | · | | − · j s s=1 | | · | | X wi 1 l = − O ) ψ (O ) = wi 1 ψ , ψ Z[w] R. B | s | j s − · h 1 ji ∈ ⊆ s=1 | | X Thus a R, i, j, implying ξ ch (E). ij ∈ ∀ ∈ R (9) Let p be a prime and a a p0-element of G. There exists an integer valued function θ vR(G) such that, for g G, θ(g) 1 (mod p) if the p0 part of g is conjugate to a, and∈ θ(g) = 0 otherwise. ∈ ≡ Proof: Consider the centralizer CG(a) of a in G. Deﬁne E = a B, where B j h i × ∈ Sylp(CG(a)). By 8, ξ chR(E) such that ξ(ab) = a , ξ(a b) = 0, j = 2, 3 . . . , a 1, b ∃ ∈ |G | ∀ | |− ∈ B. Now E is an elementary subgroup of G, so ξ vR(G). Since all values of ξ are G G ∈ s 0 1 multiples of a , the same is true for ξ , because ξ (g) = i=1 ξ (xi− gxi), with x1, . . . , xs – distinct coset| | representatives of E in G. If g is not conjugate to some ab, with b B, G 1 0 1P ∈ we have ξ (g) = 0 (in this case x− gx aB, so ξ (x− gx ) = 0, 1 i s). Assume i i 6∈ i i ∀ ≤ ≤ 0.18. BRAUER-TATE THEOREMS 31

1 now that x− gx = ab, where b B. We have ∈ G G 1 1 ξ (g) = ξ (ab) = ξ(x− abx ) = a i 1 i s, x− abx aB . i i | | · |{ | ≤ ≤ i i ∈ }| 1 i s −1≤ ≤ x Xgxi E i ∈

Since b B Syl (C (a)), b is a p-element and ab = ba. But a is a p0-element, so b and ∈ ∈ p G a are the p- and p0- parts of ab, respectively. Therefore, by 5 and 6 we have:

1 1 1 x− abx = ab0, b0 B x− ax = a, x− bx = b0 x C (a) N (B). i i ∈ ⇔ i i i i ⇔ i ∈ G ∩ G Hence ξG(g) = a i 1 i s, x C (a) N (B) = a C (a) N (B) : E , | | · |{ | ≤ ≤ i ∈ G ∩ G }| | | · | G ∩ G | the latter being true since E is a subgroup of C (a) N (B). Further ξG(g) = C (a) G ∩ G | G ∩ NG(B) : B 0 (mod p), because CG(a) NG(B) : B divides CG(a) : B , which is not divisible by| p6≡. Finally, define θ = αξ| G where∩ α Z is suc| h that α| C (a) N| (B) : B 1 ∈ ·| G ∩ G | ≡ (mod p). Then θ vR(G) satisfies the conditions in the hypothesis. Note that θ takes only 2 distinct values∈ by construction (0 and α C (a) N (B) : B ). · | G ∩ G | (10) For a fixed prime p, let L1, . . . , Lt be the equivalence classes of . By 9, there exist θ , . . . , θ v (G) such that, for each i, θ is congruent to 1 (mod ∼p) on L , and equal 1 t ∈ R i i to 0 otherwise. Then ψ = θ1 + + θt vR(G) is integer valued, and satisfies ψ(g) 1 (mod p), g G. · · · ∈ ≡ ∀ ∈ (11) If η is an integer valued class function on G, such that η(g) is divisible by G , g G, then η v (G). | | ∀ ∈ ∈ R Proof: Choose a prime p such that (p, G ) = 1. Then g h g is conjugate to | | ∼ ⇔ h, g, h G, because all elements of G are their own p0-parts. Let O (a ), . . . , O (a ) be ∀ ∈ G 1 G t the conjugacy classes of G. By 9, there exist θ , . . . , θ v (G) such that, for each i, θ 1 t ∈ R i is ai on OG(ai), and 0 otherwise. Since ai divides G, we get that vR(G) contains the functions| | which are G on a conjugacy class| | of G, and 0 outside it. But η is an integral | | linear combination of these, so η v (G). ∈ R α (12) If G = mp with p prime and (p, m) = 1, then mχ1 vR(G). Proof: |By| 10, ψ v (G) such that ψ(g) 1 (mod p),∈ g G. Define θ = ψpα ∃ ∈ R ≡ ∀ ∈ ∈ vR(G) (recall that vR(G) is an ideal of chR(G)). We claim that θ(g) 1 (mod p), g G: α pα≡ p ∀pα ∈k k Indeed, fix g G. We have ψ(g) = 1 + pl, l Z θ(g) = (1 + pl) = 1 + k=1 k p l . ∈α ∈ ⇒ pα k pk pα (pα k+1) For 1 k p , the exponent of p in the numerator of p = · ····· − N is at ≤ ≤ k k! P ∈ least k + α, while the exponent of p in its denominator is at most k. So each term in the above sum is divisible by pα, which yields θ(g) 1 (mod pα). It follows that all values of ≡ χ θ are divisible by pα. Hence all values of m(χ θ) are multiples of G . By 11, we have 1− 1− | | m(χ1 θ) vR(G). Since mθ vR(G), we conclude that mχ1 = m(χ1 θ)+mθ vR(G), as desired.− ∈ ∈ − ∈

α1 αk Finally, write G = p1 . . . pk , where p1, . . . , pk are the distinct primes dividing G . For | | α | | 1 i k, Let m = G /p i . Then m , . . . , m are relatively prime, so 1 = c m + + ≤ ≤ i | | i 1 k 1 1 · · · ckmk, for some c1, . . . , ck Z. By 12, miχ1 vR(G) for each 1 i k. Consequently, χ = c (m χ ) + + c ∈(m χ ) v (G), whic∈ h completes the≤pro≤of of Brauer-Tate’s 1 1 1 1 · · · k k 1 ∈ R theorem 32 0.19 Consequences of Brauer-Tate’s Theorem

In this section, we give some important corollaries and applications of Brauer-Tate’s result. As usual, G denotes a finite group. Lemma 13: Let θ class(G). ∈ Then θ ch(G) θ ch(E), E E. ∈ ⇔ |E ∈ ∀ ∈ Proof: Let u(G) = θ class(G) θ E ch(E), E E . First note that ch(G) u(G). The reason for this {is that,∈ if H | G| and∈ χ is a ∀character∈ } of G, then χ is a character⊆ ≤ |H of H. Next, u(G) is a ring with identity (the trivial character of G, χ1 = 1G). Indeed, if φ, ψ u(G) then φ E, ψ E ch(E), E E (φ ψ) E = φ E ψ E ch(E), (φ ψ) E = φ ∈ψ ch(E), | E | E∈ φ ψ,∀φ ψ∈ ⇒u(G) (we used| the| fact| that∈ ch(E) is·a ring).| |E · |E ∈ ∀ ∈ ⇒ · ∈ By Brauer-Tate’s theorem, χ1 v(G). Hence, if we show that v(G) is an ideal of u(G), then v(G) = u(G) = ch(G), as claimed.∈ Let φ v(G), so that φ = θG for some θ ch(E) ∈ G G∈ and E E. By 2 of section 18, for any ψ u(G) we have φ ψ = θ ψ = (θ ψ E) v(G) since θ,∈φ ch(E) ∈ · · · | ∈ |E ∈ What follows is a nice application of the previous results, which yields a result of practical value about the character table of a group. Theorem 13: Let p be a prime number, and n Z such that pn is the highest power of p dividing G . If χ is an irreducible character of∈G such that pn χ(1), then χ(y) = 0 for | | | any y G whose order is a multiple of p. ∈ χ(y) if (p, y ) = 1 Proof: Define θ : G C by θ(y) = | | . Then θ class(G). We → 0 if p divides y ∈ have to show that θ = χ. We will achiev e this by showing that| | θ ch(G). Indeed, if so, ∈ then θ = a1χ1 + +amχm where a1, . . . , am Z and χ1, . . . , χm are irreducible characters · · · ∈2 2 1 2 of G. Since θ(1) = χ(1) > 0, we get 0 < a1 + + am = θ, θ = G g G θ(g) · · · h i ∈ | | ≤ χ(g) 2 = χ, χ = 1, so all of a , . . . , a are 0, except one which| is| 1 (it cannot be g G 1 m P 1,∈since| θ|(1) >h 0).iHence θ is an irreducible character of G. But θ, χ = θ, θ > 0, − h i h i Pwhich forces θ = χ. Next, by lemma 13 it suffices to show that θ ch(E), E E. |E ∈ ∀ ∈ When E is a p0-group this is trivial, because by definition it follows that θ = χ |E |E ∈ ch(E). Assume now that p E . We have E = A B where A is cyclic, B is of prime power order, and ( A , B )||= |1. If p A , then A×= P Q where P is the Sylow-p | | | | || | × 1 subgroup of A, and Q1 is a p0 group (the product of the other Sylow subgroups of A). So E = (P Q ) B = P (Q B). If p B (so A is a p0-group), then E = A B = B A. × 1 × × 1 × || | × × In any case, we see that E = P Q, where P is a p-group and Q – a p0-group. To establish × θ E ch(E), we show that θ E, ξ (which is the coefficient of ξ in θ E written as a linear com| ∈bination of the irreducibleh | characters)i is an integer for any irreducible| character ξ of E. Let ξ be any irreducible character of E. Since E is the direct product P Q, elements × of P commute with those of Q. Hence the order of gh E (g P, h Q), [ g , h ], is a ∈ ∈ ∈ | | | | multiple of p if and only if g = 1. Therefore θ E = χ on Q, and θ E = 0 otherwise. So we have 6 | | 1 Q θ , ξ = χ(y)ξ(y) = | | χ , ξ Q. h |E i E E h |Q |Qi ∈ y Q | | X∈ | | 1 a It follows that θ E, ξ = P χ Q, ξ Q = b , for some a, b Z, with (a, b) = 1 and b P . h | i | |h | | i ∈ || | 0.19. CONSEQUENCES OF BRAUER-TATE’S THEOREM 33

In particular, we see that b is a power of p. On the other hand, 1 χ(1) O (y) χ(y) θ , ξ = χ(y)ξ(y) = C (y) | G | ξ(y) h |E i E G E | G | · χ(1) y Q y Q | | X∈ | | · | | X∈ χ(1) O (y) χ(y) = C (y) : P | G | ξ(y) (since P C (y)) G Q | G | · χ(1) ≤ G y Q | | · | | X∈ χ(1) = c, for some c Alg, cf. 3 of section 16. G Q · ∈ | | · | | G Q G Therefore, | |·| | θ , ξ Q Alg = Z. But | | (which is an integer by section 16) is χ(1) h |E i ∈ ∩ χ(1) coprime to p by hypothesis. Also ( Q , p) = 1. Since b is a power of p, and it divides G Q | | | |·| |, we must have b = 1. Therefore θ , ξ = a Z, as desired χ(1) h |E i ∈ Lemma 14: (Brauer) Let E be a p-elementary group, and θ an irreducible character of E. Then θ = ξE, for some linear (i.e. degree 1) character ξ of some subgroup of E. Proof: Let E = A P , with A – a cyclic p0-group and P a p-group. By 7 of section × 18, θ = θ1θ2, where θ1 and θ2 are irreducibles of A and P , respectively. Since θ1 is linear (because A is abelian), and the degree of θ is a power of p (since it divides P ), we 2 | | get = degθ = θ(1) = pn, for some n 0. We proceed by induction on n. The theorem statement is true for n = 0 (θ = θE ≥is linear), so we assume n > 0. Let λ be a linear 1 1 if λθ = θ character of E. We have θθ, λ = E g E θ(g)θ(g) λ(g) = θ, λθ = . h i ∈ · h i 0 otherwise | | Note that λθ and θ are irreducible charactersP of E, since θ is (cf. section 8). Let Λ = λ λ is a linear character of E, such that λθ = θ . Note that Λ is a multiplicative group: { | } 1 1 indeed, let λ1, λ2 Λ. Since all values of λ are roots of unity, θ1− = θ1, so λ1− λ2 = λ1λ2 ∈ 1 1 1 is a linear character of E (cf. section 8). Then λ1− λ2θ = λ1− θ = λ1− [λ1θ] = θ, which 1 proves that λ− λ Λ. By deﬁnition of Λ, we may write 1 2 ∈ θθ = λ + φ, (23) λ Λ X∈ where φ is a non-linear character of E. Since p degθθ and p degφ, we get p Λ . By Cauchy’s theorem, there exists η Λ such that η |= p. Now consider| K = kerη||. By| the ∈ | | 1st Isomorphism Theorem, E/K = imη = µ C µp = 1 = Z/pZ. This forces K = ∼ { ∈ | } ∼ A Q, where Q is a subgroup of P , of index p. (To see this, write K = M Q with M A, × × ≤ Q P . Since A is a p0-group, P is a p-group, and K = E /p, we must have M = A). ≤ | | | | Restricting (23) to K, and noting that (1 ) = η = 1 , we ﬁnd that 1 occurs at |E |K |K |K |K least twice in (θθ) K, i.e. 1 K, (θθ) K 2. Next, 2 1 K, θ Kθ K = θ K, θ K , so θ K is a reducible character| ofhK|. Let then| i ≥ψ be an irreducible≤ h | character| | i ofhK|, whic| ih occurs| n in θ K. We have degψ < degθ K = degθ = p . But degψ is a power of p (because K is | n 1 | p-elementary), so degψ p − . On the other hand, applying Frobenius reciprocity yields E ≤ E 1 θ K, ψ = θ, ψ , which implies that θ occurs in ψ (since θ is irreducible). In ≤ h | in h i E n 1 particular, p = degθ degψ = E : K degψ = p degψ p − degψ. The two ≤ n 1 | E| · n · ⇒ ≤ E inequalities force degψ = p − and degψ = p = degθ. Again, since θ occurs ψ , we actually have θ = ψE. Now ψ is an irreducible character of the p-elementary group K, 34

n 1 K of degree p − , so by induction ψ = ξ for some linear character ξ of some subgroup of K. Finally, θ = ψE = (ξK)E = ξE (by 6 of section 13, character induction is transitive), which completes the proof Theorem 14: (Brauer-Tate) Every character of G is an integral linear combination of characters of G, induced from linear characters of elementary subgroups of G. k G Proof: Let θ be a character of G. By theorem 12 (Brauer-Tate), θ = i=1 aiψi where a1, . . . , ak Z and for each i, ψi is an irreducible character of some Ei E. By lemma ∈Ei P∈ 14, ψi = ξi where ξi is a linear character of some subgroup Hi Ei. Hence θ = k Ei G k G E ≤ i=1 ai(ξi ) = i=1 aiξi , and we are done since Hi , 1 i k. (subgroups of elementary groups are elementary) ∈ ∀ ≤ ≤ P P 0.20 Mackey’s Theorems

We consider representations over C. Let G be a ﬁnite group. If we induce a representation of G from a an irrreducible character of a subgroup, it is generally not irreducible anymore. Mackey’s theorems help us understand how it decomposes into irreducibles. (1) Let H G and ρ : H GL(V ) be a representation of H. For x G, we deﬁne ρx : 1 ≤ x → 1 1 x 1 ∈ xHx− GL(V ) by ρ (z) = ρ(x− zx), z xHx− (i.e. ρ (xyx− ) = ρ(y), y H). → ∀ ∈ 1 x ∀ ∈ Since conjugation by x is a group isomorphism of xHx− and H, ρ is a representation of 1 x xHx− on V . In some sense, ρ is really just ρ, which is precisely stated in the following 1 commuting diagram (σ is conjugation by x− ):

ρ H / GL(V )

σ id. ρx 1 xHx− / GL(V )

1 In particular, M is an H-submodule of V if and only if M is an xHx− -submodule of V . 1 We clearly have χρx (xyx− ) = χρ(y), y H. ∀ ∈ x 1 x 1 More generally, if φ class(H), we deﬁne φ : xHx− C by φ (xyx− ) = φ(y), y H. x ∈ 1 → ∀ ∈ Then φ class(xHx− ) (again since conjugation by x is a homomorphism). Moreover, ∈ if φ, ψ class(H) then φx, ψx = φ, ψ . ∈ h i h i (2) (Mackey) Let K, H G and let x1 = 1, x2, . . . , xs be a set of K H double ≤ { s} − coset representatives in G (recall that this means G = i=1 KxiH). For 1 i s, let 1 ≤ ≤ Ki = K xiHx− . If ψ class(H), then ∩ i ∈ U s (ψG) = (ψxi )K . (24) |K |Ki i=1 X Proof: K acts on the left cosets xH x G by left multiplication. The double coset KxH is a union of such cosets, namely{ |the∈ones} in the orbit of xH. The stabilizer of xH 1 1 under this action is K xHx− (indeed k K, kxH = xH k K, x− kx H 1 ∩ ∈ ⇔ ∈ ∈ ⇔ k K xHx− ). Now, for 1 i s let k , . . . , k be left coset representatives of K in ∈ ∩ ≤ ≤ i1 isi i K. Thus, Kx H = k x H k x H (since the action of K on x H can be identiﬁed i i1 i ] · · · ] isi i i 0.20. MACKEY’S THEOREMS 35

with the action of K on the left cosets of the stabilizer Ki of xiH in K). It follows that kijxi 1 i s, 1 j si is a complete set of left coset representatives of H in G. { | ≤ ≤ ≤ ≤ }1 xi 0 xi 0 Next, since Ki = K xiHxi− , we have (ψ Ki ) (z) = (ψ ) (z), z K, 1 i s. Hence, for y K we get∩ | ∀ ∈ ≤ ≤ ∈

s si G 0 1 1 xi 0 1 ψ (y) = ψ (xi− kij− ykijxi) = (ψ ) (kij− ykij) 1 i s i=1 j=1 1 X≤j ≤si X X ≤ ≤ s si s xi 0 1 xi K = (ψ ) (k− yk ) = (ψ ) , as desired |Ki ij ij |Ki i=1 j=1 i=1 X X X

(3) (Mackey) If φ class(K) and ψ class(H) then ∈ ∈ s φG, ψG = φ , ψxi . (25) h i h |Ki |Ki i i=1 X Proof: Using Frobenius reciprocity repeatedly, and (24), we have

s s φG, ψG = φ, ψG = φ, (ψxi )K = φ , ψxi h i h |Ki h |Ki i h |Ki |Ki i i=1 i=1 X X The next result, which is a consequence of the above, in some way fulﬁlles the goal stated in the beginning of the section. (4) Let H G and ψ be an irreducible character of H. ≤ 1. ψG is irreducible for each x G H, there is no irreducible character (of H 1 ⇔ ∈ −1 \ xi −1 ∩ xHx− ) occuring in both ψ H xiHx and ψ H xiHx . | ∩ i | ∩ i

2. If H ¢ G, then ψG is irreducible ψG = ψ for all x G H. ⇔ 6 ∈ \ Proof: 1 Letting K = H and φ = ψ in (25), we get

s s

G G −1 xi −1 −1 xi −1 ψ , ψ = ψ H xiHx , ψ H xiHx = 1 + ψ H xiHx , ψ H xiHx . h i h | ∩ i | ∩ i i h | ∩ i | ∩ i i i=1 i=2 X X s G −1 xi −1 Hence ψ is irreducible i=1 ψ H xiHx , ψ H xiHx = 0 for 2 i s there is ⇔ h | ∩ i | ∩ i i ⇔ ≤ ≤ −1 xi −1 no common irreducible character occuring in both ψ H xiHx and ψ H xiHx . This is P | ∩ i | ∩ i equivalent to what we want, since any x G H can be taken to be one of the double ∈ \1 coset representatives x2, . . . , xs. 2 Since xHx− = H, x G, part 1 above implies: ψG is irreducible ψ and ψx don’t have common irreducible∀ ∈ constituents for all x ⇔ ∈ G H ψ = ψx, x G H, since ψ and ψx are both irreducible characters of H \ ⇔ 6 ∀ ∈ \ 36 0.21 An Application of Mackey’s Results

In this section, our purpose is to describe irreducible characters of a semidirect product, where the normal component is abelian. For a group G, let Irred(G) denote the set of irreducible characters of G. (1) Let G = A o B where A is an abelian group. Thus, A ¢ G, B G, AB = G and 1 ≤ A B = 1. Also (a1b1)(a2b2) = a1(b1a2b1− )b1b2, a1, a2 A, b1, b2 B. Let φ Irred(A) ∩ b∀ ∈ b ∈ 1 ∈ and b B. By previous section, we know that φ , defined by φ (a) = φ(bab− ), a A, is ∈ 1 ∀ ∈ an irreducible character of A = bAb− . This gives a well-defined action of B on Irred(A) b1 b2 1 1 b1b2 (note that (φ ) (a) = φ(b2b1ab1− b2− ) = φ (a), a A, b1, b2 B). Let Bφ be the centralizer in B of φ, under this action. ∀ ∈ ∈ (2) Note that φ is linear, since A is abelian, so that we may identify φ with its character. Now let ρ : B GL(V ) be a representation of B . The map φ ρ : A o B GL(V ), φ → φ × φ → given by φ ρ(ab) = φ(a)ρ(b) (a A, b B) is thus well-defined. Moreover, φ ρ is a representation× of A o B . Indeed,∈for all ∈a , a A, b , b B we have × φ 1 2 ∈ 1 2 ∈ 1 1 φ ρ[(a b )(a b )] = φ ρ[a (b a b− )b b ] = φ(a )φ(b a b− )ρ(b )ρ(b ) × 1 1 2 2 × 1 1 2 1 1 2 1 1 2 1 1 2 = φ(a1)φ(a2)ρ(b1)ρ(b2) (since b1 centralizes φ) = φ(a )ρ(b )φ(a )ρ(b ) (since φ(a ) C) 1 1 2 2 2 ∈ = φ ρ(a b )φ ρ(a b ). × 1 1 × 2 2 Denote the character of φ ρ by φ χ . We have φ χ (ab) = φ(a)χ (b), a A, b B. × × ρ × ρ ρ ∀ ∈ ∈ (3) Let φ Irred(A) and ψ Irred(B ). Then ∈ ∈ φ 1. φ ψ Irred(A o B ). × ∈ φ 2. (φ ψ)G Irred(G). × ∈ 1 Proof: 1 We have φ ψ, φ ψ = a A φ(a)ψ(b)φ(a)ψ(b) A bφ h × × i | |·| | b ∈Bφ ∈ P 1 1 = φ(a)φ(a) ψ(b)ψ(b) = φ, φ ψ, ψ = 1. A  Bφ  h i · h i a A ! b B | | X∈ | | X∈ φ   2 Let H = A o B G, so that φ ψ Irred(H) by part 1. By 4 of previous section, it φ ≤ × ∈ suffices to show that if x = ab G H, a A, b B (i.e. if b Bφ), then (φ ψ) H xHx−1 x ∈ \ ∈ ∈ 6∈ × | ∩ and (φ ψ) H xHx−1 have no common constituents. We show that, a fortiori, (φ ψ) A × x| ∩ 1 ×1 | and (φ ψ) have no common constituents (note that A = xAx− ¢ H xHx− ). We × |A ∩ have (φ ψ) A = ψ(1) φ, so that the constituents of (φ ψ) A are ψ(1) copies of φ. × | x · 1 1 1 ×1 | b 1 Next, (φ ψ) A(a1) = φ ψ(b− a− a1ab) = ψ(1)φ(b− a− a1ab) = ψ(1)φ (a− a1a) = ψ(1)φb(a ×) a | A. This sho× ws that (φ ψ)x = ψ(1)φb, so that its constituents are 1 ∀ 1 ∈ × |A ψ(1) copies of φb. Finally, since b B , we have φ = φb and hence we are done 6∈ φ 6 (4) Let φ, θ Irred(A), ψ Irred(B ) and η Irred(B ). ∈ ∈ φ ∈ θ 1. If φ O (θ) (the orbit of θ under the action of B on Irred(A)), then (φ ψ)G = 6∈ B × 6 (θ η)G. × 0.21. AN APPLICATION OF MACKEY’S RESULTS 37

2. If φ = θ and ψ = η, then (φ ψ)G = (θ η)G. 6 × 6 ×

Proof: We shall use Mackey’s result given by (25). Let K = ABφ and H = ABθ. Let b = 1, b , . . . , b be a set of B -B double coset representatives in B. We claim that { 1 2 s} φ θ b , . . . , b is also a set of K-H double coset representatives in G = A o B. Indeed, { 1 s} assume bi = (a1c1)bj(a2c2) where a1, a2 A, c1 Bφ, c2 Bθ. Since A ¢ G, we can ∈ ∈ ∈ 1 ”move” the a’s to the left to get bi = ac1bjc2 where a = a1(c1bj)a2(c1bj)− A. Hence, a A B = 1 and b = c b c , i.e. B b B = B b B . Conversely, if∈b = c b c ∈ ∩ { } i 1 j 2 φ i θ φ j θ i 1 j 2 with c1 Bφ, c2 Bθ then clearly ABφbiABθ = ABφbjABθ, so the claim holds. Now for ∈ ∈ 1 1 1 1 i s, note that K biHbi− = ABφ A(biBφbi− ) = ABi, where Bi = Bφ biBθbi− . By≤(25),≤ we have ∩ ∩ ∩

s (φ ψ)G, (θ η)G = (φ ψ) , (θ η)bi h × × i h × |ABi × |ABi i i=1 X s 1 = φ(a)ψ(b) θbi (a)ηbi (b)  AB ·  i=1 | i| a A X bX∈Bi  ∈  s   = φ, θbi ψ , ηbi . h i · h |Bi |Bi i i=1 X

In case 1, the last sum is 0, since φ and θbi are distinct irreducible characters of A, 1 i s. Hence (φ ψ)G, (θ η)G = 0 and (φ ψ)G, (θ η)G Irred(G) are distinct.∀ ≤ ≤ h × bi × ibi × × ∈ In case 2, we have φ, θ = φ, φ = 0, unless bi Bφ, i.e. unless i = 1. But for i = 1, h i h i bi ∈ we have b1 = 1, B1 = Bφ and ψ B1 , η Bi = ψ, η = 0, since ψ = η. So, as in 1 above, (φ ψ)G, (θ η)G = 0 and (φh | ψ)G =| (iθ hη)G i 6 h × × i × 6 × We conclude with the following theorem, which characterizes the irreducible character of certain semidirect products.

Theorem 15: Let G = A o B with A ¢ G and A – abelian. Let φ1, . . . , φm be a complete set of orbit representatives for the action of B on Irred(A) described in 1. For 1 i m, let Irred(B ) = ψ 1 j m . Then ≤ ≤ φi { ij | ≤ ≤ i}

Irred(G) = (φ ψ )G 1 i m, 1 j m . { i × ij | ≤ ≤ ≤ ≤ i}

Proof: By 3 and 4, the characters (φ ψ )G are all irreducible and distinct. To prove i × ij they are all irreducibles of G, we verify that the sum of squares of their degrees is G . Indeed, note that G : AB = AB / AB = B : B and (φ ψ )G(1) = |B |: | φi | | | | φi | | φi | i × ij | B (φ ψ )(1) = B : B ψ (1). Note also that B : B = O (φ ) (the size of the φi | · i × ij | φi | · ij | φi | | B i | 38

orbit of φi is the index of its stabilizer in B). Thus, we have

m mi [(φ ψ )G(1)]2 = B : B 2 ψ (1)2 = B : B 2 ψ (1)2 i × ij | φi | · ij | φi | · ij 1 i m 1 i m i=1 j=1 ! 1 ≤Xj ≤mi 1 ≤Xj ≤mi X X ≤ ≤ ≤ ≤ m m = B : B 2 B = B B : B | φi | · | φi | | | · | φi | i=1 i=1 X m X = B O (φ ) = B Irred(A) = B A = G | | | B i | | | · | | | | · | | | | i=1 X