Group Theory, Linear Algebra, Character Theory - Lecture Notes | MTH 912, Study notes of Mathematics

Download Group Theory, Linear Algebra, Character Theory - Lecture Notes | MTH 912 and more Study notes Mathematics in PDF only on Docsity! Group Theory Lecture Notes for MTH 912/913 04/05 Ulrich Meierfrankenfeld April 26, 2007 2 Chapter 4 Linear Algebra 4.1 Bilinear Forms Definition 4.1.1 [def:bilinear form] Let R be a ring, V an R-module and W a right R-module and s : V × W → R, (v, w) → (v | w) a function. Let A ⊆ V and B ⊆ W . Suppose that s is R-bilinear, that is ( ∑n i=1 rivi | ∑m j=1wjsj) = ∑n i=1 ∑m j=1 ri(vi | wj)sj for all vi ∈ V,wj ∈W and ri, sj ∈ R. Then (a) [a] s is called a bilinear form. (b) [b] s is called symmetric if V = W and (v | w) = (w | v) for all v, w ∈ V . (c) [z] s is called symplectic if V = W and (v | v) = 0 for all v ∈ V . (d) [c] Let v ∈ V and w ∈ W we say that v and w are perpendicular and write v ⊥ w if (v | w) = 0. (e) [d] We say that A and B are perpendicular and write A ⊥ B if a ⊥ b for all a ∈ A, b ∈ B. (f) [e] A⊥ = {w ∈W | A ⊥ w} and ⊥B = {v ∈ V | v ⊥ B}. A⊥ is called the right perp of A and ⊥B the left perp of B. (g) [f] If A is an R-submodule of V , define sA : W → A∗ by sA(w)(a) = (a | w) for all a ∈ A,w ∈W . (h) [g] If B is an R-submodule of W , define sB : V → B∗ by sB(v)(b) = (v | b) for all v ∈ V, b ∈ B. (i) [h] s is called non-degenerate if V ⊥ = 0 and ⊥W = 0. (j) [i] If V is free with basis V and W is free with basis W, then the V × W matrix MWV (s) = ( (v | w) )v∈V,w∈W is called the Gram Matrix of s with respect to V and W. Observe that the Gram Matrix is just the restriction of s to V ×W. 91 92 Chapter 4. Linear Algebra Let I be a set, R a ring, W = iI R and V = ⊕I R. Define s : V × W → R, (v | w) = ∑ i∈I viwi. Note that this is well defined since almost all vi are zero. Note also that if we view v and w as I × 1 matrices we have (v | w) = vTw. As a second example let V be any R-module and W = V ∗ and define (v | w) = w(v). If V is a free R-module this example is essentially the same as the previous: Lemma 4.1.2 [dual basis] Let V be a free R module with basis V. For u ∈ V define u∗ ∈ V ∗ by u∗(v) = δuv. Define φV : V → ⊕ V R, v → (w∗(v))w∈V and φV∗ : V ∗ → i V R,α→ (α(v))v∈V (a) [a] Both φV and φV∗ are R-isomorphisms. (b) [b] Let w ∈ V ∗ and v ∈ V and put ṽ = φV(v) and w̃ = φV∗(w). Then w(v) = ṽTw̃. Proof: (a) Since V is free with basis V, the map ⊕VR → V, (rv) → ∑ v∈V rvv is an R- isomorphism. Clearly φV is the inverse of this map and so φV is an R-isomorphism. To check that φV∗ is an R-linear map of right R-modules recall first that V ∗ is a right R-module via (wr)(v) = w(v)r. Also iV R is a right R-module via (rv)vr = (rvr)v. We compute φV∗(wr) = ((wr)(v))v = (w(v)r)v = (w(v))vr and so φV∗ is R-linear. Given (rv)v ∈ iV R, then w : V → R,∑v∈V svv → ∑v∈V svrv is the unique element of V ∗ with w(v) = rw for all v ∈ V, that is with φV∗(w) = (rv)v. So φV∗ is a bijection. (b) For u ∈ V let su = u∗(v) and ru = w(u). Then v = ∑ u∈V suu and so w(v) =∑ u∈V suw(u) = ∑ u∈V suru = ṽ Tw̃.  Definition 4.1.3 [dual map] Let R be a ring and α : V →W an R-linear map. Then the R-linear map α∗ : W ∗ → V ∗, φ→ φ ◦ α is called the dual of α. Lemma 4.1.4 [matrix of dual] Let R be a ring and V and W free R modules with basis V and W, respectively. Let α : V → W be an R-linear map and M its matrix with respect to V and W. Let δ ∈W ∗. Then φV∗(α∗(δ)) = MTφW∗(δ) Proof: Let v ∈ V. Then the v-coordinate of φV∗(α∗(δ)) is α∗(δ)(v) = (δ◦α)(v) = δ(α(v)). By definition of M = (mwv)w∈W,v∈V , α(v) = ∑ w∈W mwvw and so Section 4.1. Bilinear Forms 93 φV∗(α∗(δ)) = (δ(α(v)))v = ( ∑ w∈W mwvδ(w)) = MTφW∗(δ)  Lemma 4.1.5 [associated non-deg form] Let R be a ring and s : V ×W → R an R- bilinear form. Let A be an R-subspace of V and B an R-subspace of W . Then sAB : A/A ∩ ⊥B ×B/B ∩A⊥, (a+ (A ∩ ⊥B), b+ (B ∩A⊥) → (a | b) is a well-defined non-degenerate R-bilinear form. Proof: Readily verified.  Lemma 4.1.6 [basic bilinear] Let R be a ring and let s : V ×W → R be an R-bilinear form. (a) [a] Let A be an R-subspace of V , then A⊥ = ker sA. (b) [b] Let B be an R-subspace of W then ⊥B = ker sB. (c) [c] s is non-degenerate if and only if sV and sW are 1-1. Proof: (a) and (b) are obvious and (c) follows from (a) and (b).  Lemma 4.1.7 [finite dim non-deg] Let F be a division ring and s : V ×W → F a non- degenerate F-bilinear form. Suppose that one of V or W is finite dimensional. Then both V and W are finite dimensional, both sV and sW are isomorphisms and dimF V = dimFW . Proof: Without loss dimF V <∞ and so dimV = dimV ∗. By 4.1.6(c), sV and sW are 1-1 and so dimW ≤ dimV ∗ = dimV . So also dimW is finite and dimV ≤ dimW ∗ = dimW . Hence dimV = dimW = dimW ∗ = dimV ∗. Since sV and sW are 1-1 this implies that sV and sW are isomorphisms.  Corollary 4.1.8 [dual s-basis] Let F be a division ring, s : V ×W → F a non-degenerate F-bilinear form, B a basis for V . Suppose that B is finite. Then for each b ∈ B there exists a unique b̃ ∈ W with s(a, b̃) = δab for all a, b ∈ B. Moreover, (b̃ | b ∈ B) is an F-basis for W . Proof: By 4.1.7 sV : W → V ∗ is an isomorphism. Let b∗ ∈ V ∗ with b∗(a) = δab and define b̃ = s−1V (b ∗).  96 Chapter 4. Linear Algebra (a) [a] (v | w) = ṽTMw̃. (b) [b] φV(V ⊥) = Null(M), the Null space of M . (c) [c] φV(⊥W ) = NullMT (d) [d] φW∗(sW (v)) = MTṽ. (e) [e] φV∗(sV (w)) = Mw̃. Proof: (a) We have v = ∑ a∈V ṽaa, w = ∑ b∈W bw̃b and M = ((a | b))ab. Since s is R-bilinear, (v | w) = ∑ a∈V,b∈W ṽa(a | b)w̃b = ṽTMw̃ (b) By (a) w ∈ V ⊥ iff ṽTMw̃ = 0 for all ṽ, iff Mw̃ = 0 and iff w̃ ∈ Null(M). (c) v ∈ ⊥W iff ṽTM = 0, iff MTṽ = 0 iff ṽ ∈ NullMT. (d) Let u = sW (v) and ũ = ΦW∗(v). Then by “right-module” version of 4.1.2 u(w) = w̃T ·op ũ = ũT · w̃. On the other hand u(w) = sW (v)(w) = (v | w) = ṽTM · w̃ = Thus ũT = ṽTM and so ũ = MTv and (d) holds. (e) Let u = sV (w) and ũ = ΦV∗(u). Then by 4.1.2 u(v) = ṽT · ũ. On the otherhand u(v) = sV (w)(v) = (v | w) = ṽT ·Mw̃. So ũ = Mw̃ and (e) holds.  Lemma 4.1.16 [gram matrix of dual basis] Let F be a division ring and s : V ×W → F a non-degenerate F-bilinear form. Let V and W be F-basis for V and W respectively and Ṽ and W̃, the corresponding dual basis for W and V . Let M be the Gram matrix for s with respect to V and W. Let N the Gram matrix for s with respect to W̃ and Ṽ. Then (a) [a] MT is the matrix for idV with respect to V and W̃. (b) [b] N is the matrix for idW with respect to W and Ṽ (c) [c] M and N are inverse to each other. Section 4.1. Bilinear Forms 97 Proof: (a) We have idV : V sW→ W ∗ s−1W→ V . By 4.1.15(d), the matrix of sW with respect to V and W∗ is M . By definiton of W̃ the matrix of s−1W with respect to W∗ and W̃ is the identity matrix. So (a) holds. (b) Similar to (a), use sV and 4.1.15(e). (c) By (b) N−1 is the matrix of idW with respect to Ṽ and W. Note that idV is the adjoint of idW . So by (a) and 4.1.13(b), N−1 = MTT = M .  Lemma 4.1.17 [circ and bilinear] Let R be a commutative ring, G a group and let V and W be RG-modules. Let s : V ×W → R be R-bilinear form. (a) [a] s is G-invariant iff (a◦v | w) = (v | aw) for all a ∈ inRG. (b) [b] Let a ∈ RG. Then AW (a) ≤ (a◦V )⊥ with equality if V ⊥ = 0. Proof: (a) Recall first for a = ∑ g∈G agg ∈ Rg, a◦ = ∑ g∈G agg −1. Thus s is G invariant ⇐⇒ (gu | gw) = (u | w) ∀g ∈ G, u ∈ V,w ∈W (u→ v = gu is a bijection) ⇐⇒ (v | gw) = (g−1v | w) ∀g ∈ G, v ∈ V,w ∈W (s is R bilinear) ⇐⇒ (v | aw) = (a◦v | w) ∀a ∈ RG, v ∈ V,w ∈W (b) By (a) a and a◦ are adjoints. So (b) follows from 4.1.12  Lemma 4.1.18 [extending scalars and bilinear] Let R ≤ R̃ be an extensions of rings and s : V ×W → R an R-bilinear form. There exists a unique R̃-bilinear form s̃ : R̃⊗R V ×W ⊗R R̃→ R̃, (a⊗ v, w ⊗ b) = a((| v), w)b for all a, b ∈ R̃, v ∈ V,w ∈ V . Proof: Observe that the map R̃× V ×W × R̃ toR̃, (a, v, b, w) → a((| v), w)b is R-balanced in (a, v) and (b, w). The universal property of the tensor product now shows the existence of the map s̃. A simple calculation shows that s̃ is R̃-bilinear.  Lemma 4.1.19 [extending scalars and intersections] Let F ≤ K be an extension of division rings and V an F space. (a) [a] Let W be a set of F-subspaces of V . Then ⋂ W∈W K⊗W = K⊗ ⋂ W∈W W 98 Chapter 4. Linear Algebra (b) [b] Let s : V ⊗W → F be an F-bilinear form and extend s to a bilinear form s̃ : K⊗F V ×W ⊗F K → K (see 4.1.18). Let X an F-subspace of V . Then K⊗FX⊥ = (K⊗X)⊥. Proof: (a) Suppose first thatW = {W1,W2}. Then there exists F-subspacesXi ofWi with Wi = Xi⊕(W1∩W2). Observe thatW1+W2 = (W1∩W2)⊕X1⊕X2. ForX an F-subspace of V let X = K⊗FX ≤ K⊗FV . Then Wi = W1 ∩W2⊕Xi and W1 +W2 = W1 ∩W2⊕X1⊕X2 and so W1 ∩W2 = W1 ∩W2. So (a) holds if |W| = 2. By induction it holds if W is finite. In the general case let v ∈ V . Then there exists a finite dimensional U ≤ V with v ∈ U Moreover, there exists a finite subset X of W with U ∩ ⋂ X∈X X = U ∩ ⋂ X∈W X. By the finite case, U ∩ ⋂ X∈X X = U ∩ ⋂ X∈X X and so (a) is proved. (b) Note that X⊥ = ⋂ x∈X x ⊥. So by (a) we may assume that X = Fx for some x ∈ X. If X ⊥ V , then also X ⊥ V and we are done. Otherwise dimV/X⊥ = 1 and so also dimV /X⊥ = 1. From X⊥ ≤ X⊥ < V we conclude that X⊥ = X⊥.  Lemma 4.1.20 [symmetric form for p=2] Let F be a field with char F = 2. Define σ : F → F, f → f2 and let Fσ by the F-space with Fσ = F as abelian group scalar multiplication f ·σ k = f2l. Let s a symmetric form on V and define α : V → Fσ : v → (v | v). Then α is F-linear, W := kerα = {v ∈ V | (v | v) = 0} is an F-subspace, s |W is a symplectic form and dimF V/W ≤ dimF Fσ = dimF2 F. Proof: Since (v+w | v+w) = (v | v)+(v | w)+(w | v)+(w | w) = (v | v)+2(v | w)+(w | w) = (v | v) + (w | w) and (fv | fv) = f2(v | v) = f ·σ (v | v) conclude that α is F-linear. Thus W = kerα is an F-subspace of V and V/W ∼= Imα. Also dimF Imα ≤ dimF Fσ. The map (σ, idF : F× Fσ → F2 × F, (f, k) → (f2, k) provides an isomorphism of the F space Fσ and the F2-space F. So dimF Fσ = dimF2 F. Cleary s |W is a symplectic form.  Lemma 4.1.21 [symplectic forms are even dimensional] Let F be a field, V a finite dimensional F-space and s a non-degenerate symplectic F-form on V . Then there exists an F-basis vi, i ∈ {±1,±2, . . .± n} for V with (vi | vj) = δi,−j · sgn(i). In particular dimF V is even. Proof: Let 0 6= v1 ∈ V . Since v1 /∈ 0 = V ⊥, there exists v ∈ V with (v1 | v) 6= 0 . Let v−1 = (v1 | v)−1v. Then (v1 | v−1) = 1 = −(v−1 | v1). Let W = F〈v1, v−1〉. The Gram Matrix of s on W with respect to (v1, v−1) is ( 0 1 −1 0 ) . So the Gram matrix has determinant 1 6= 0. Thus W is non-degenerate and so V = W fW⊥. Hence also W⊥ is non-degenerate and the theorem follows by induction on dimF V .  Lemma 4.1.22 [selfdual and forms] Let F be field, G a group and V simple FG module. Suppose that V is self-dual (that is V ∗ ∼= V as FG-module). Chapter 5 Representations of the Symmetric Groups 5.1 The Symmetric Groups For n ∈ Z+ let Ωn = {1, 2, 3 . . . , n} and Sym(n) = Sym(Ωn). Let g ∈ Sym(n) and let O(g) = {O1, . . . 0k} be the sets of orbits for g on Ωn. Let |Oi| = ni and choose notation such that n1 ≥ n2 ≥ n3 ≥ . . . nk. Define ni = 0 for all i > 1. Then the sequence (ni)∞i=1 is called the cycle type of g. Pick ai0 ∈ Oi and define aij = gj(ai0) for all j ∈ Z. Then aij = aik if and only if j ≡ k (mod n)i. The denote the element g by g = (a11, a12, . . . a1n1)(a21, a22, . . . , a2n2) . . . (ak1, ak2, . . . aknk). Lemma 5.1.1 [conjugacy classes in sym(n)] Two elements in Sym(n) are conjugate if and only if they have the same cycle type. Proof: Let g be as above and h ∈ Sym(n). Then hgh−1 = (h(a11), h(a12), . . . h(a1n1))(h(a21), h(a22), . . . , h(a2n2)) . . . (h(ak1), h(ak2), . . . h(aknk)) and the lemma is now easily proved.  Definition 5.1.2 [def:partition of n] A partition of n ∈ N is a non decreasing sequence λ = (λi)∞i=1 of non-negative intergers with n = ∑∞ i=1 λi. Note that if λ is a partion of n the necessarily λi = 0 for almost all i. For example (4, 4, 4, 3, 3, 1, 1, 1, 1, 0, 0, 0, . . .) is a partition of 22. We denote such a partition by (43, 32, 14). Observe that the cycle type of g ∈ Sym(n) is a partition of n. Together with 3.1.3(f) we conclude 101 102 Chapter 5. Representations of the Symmetric Groups Lemma 5.1.3 [number of partitions] Let n ∈ Z+. The follwing numbers are equal: (a) [a] The numbers of partitions of n. (b) [b] The numbers of conjugacy classes of Sym(n). (c) [c] The number of isomorphism classes of simple CSym(n)-modules.  Our goal now is to find an explicit 1-1 correspondence between the set of partions of n and the simple CSym(n)-modules. We start by associating a Sym(n)-module Mλ to each partition λ of n. But this modules is not simple. In later section we will determine a simple section of Mλ. Definition 5.1.4 [def:lambda partition] Let I be a set of size n and λ a partition of n. A λ-partition of I is a sequence ∆ = (∆i)∞i=1 of subsets of ∆ such that (a) [a] I = ⋃∞ i=1 ∆i (b) [b] ∆i ∩∆j = ∅ for all 1 ≤ i < j <∞. (c) [c] |∆i| = λi. For example ({1, 3, 5}, {2, 4}, {6}, ∅, ∅, . . .) is a (3, 2, 1) partition of I6 where In = {1, 2, 3, . . . n}. we will write such a partition as 1 3 5 2 4 1 The lines in this array are a remainder that the order of the elements in the row does not matter. On the otherhand since sequences are ordered 1 3 5 2 4 6 6= 2 4 6 1 3 5 Let Mλ be the set of all λ-partions of In. Note that Sym(n) acts on λ via π∆ = (π(∆i))∞i=1). Let F be a fixed field and let Mλ = MλF = FM(λ). Then Mλ is an FSym(n)- module. Note that for M (n−1,1) ∼= FIn. Let (· | ·) the unique bilinear form on Mλ with orthonormal basis Mλ. Then by (· | ·) is Sym(n)-invariant and non-degenerate. 5.2 Diagrams,Tableaux and Tabloids Definition 5.2.1 [def:diagram] Let D ⊆ Z+ × Z+ (a) [z] Let (i, j), (k, l) ∈ Z+ × Z+. Then (i, j) ≤ (k, l) provided that i ≤ k and j ≤ l Section 5.2. Diagrams,Tableaux and Tabloids 103 (b) [a] D is called a diagram i if for all d ∈ D and e ∈ Z+×Z+ with e ≤ d one has e ∈ D. (c) [b] The elements of diagram are called the nodes of the diagram. (d) [c] r : Z+ × Z+ × (i, j) → i and c : Z+ × Z+ × (i, j) → j. (e) [e] The i-th row of D is Di := D∩{i}×Z+ and the j-column of D is Dj := Z+×{j}. (f) [d] λ(D) = (|Di|)∞i=1 and λ′(D) = (|Dj |)∞j Definition 5.2.2 [def:diagram2] λ ∈ Z∞+ define [λ] = {(i, j) ∈ Z+ × Z+ | 1 ≤ j ≤ λi}. Lemma 5.2.3 [basic diagram] Let n ∈ N. Then the map D → λD is a bijection between the Diagram of size n and the partitions of n. The inverse is is by λ→ [λ]. Proof: Let D be a diagram of size n and put λ = λ(D). Let i ∈ N and let j be maximal with (i, j) ∈ D. By maximality of j and the definition of a diagram, (i, k) ∈ D iff k ≤ j. Thus j = |Di| = λi and D = [λ]. Let k ≤ i. Since (i, λi) ∈ D, the defintion of a diagram implies (k, λi) and so λi ≤ λk. Thus λ is non-increasing. Clearly ∑∞ i=1 λi = |D| = n and so λ is a partition of n. Conversely suppose that λ is a partition of n. Let (i, j) ∈ D and (a, b) ∈ Z+ × Z+ with a ≤ i and b ≤ j. Then a ≤ i ≤ λj ≤ λb and so (a, b) ∈ [λ]. Thus [λ] is a diagram. Clearly |[λ]i| = λi, that is λ([λ]) = λ.  We draw diagams as in the following example: [5, 33, 22, 1] = x x x x x x x x x x x x x x x x x x x Definition 5.2.4 [def:dominates] Let λ and µ be partitions of n ∈ Z+. We say that λ dominates µ and write λD µ if j∑ i=1 λi ≥ j∑ i=1 µi for all j ∈ Z+. 106 Chapter 5. Representations of the Symmetric Groups Definition 5.2.9 [def:tableau] Let λ be a partition of n. A λ-tableau is a function t : [λ] → In. We denote tableaux as in the following example 5 1 4 2 3 denotes the [3, 2]-tableau t : (1, 1) → 4, (1, 2) → 1, (1, 3) → 4, (2, 1) → 2, (2, 2) → 3. Definition 5.2.10 [def:partition of tableau] Let t : D → In be a tableau. Then ∆(t) = (t(Di))∞i=1) and ∆ ′(t) = (t(Di))∞i=1. ∆(t) is called the row partition of t and ∆ ′(t) the column partition of t. Note that if t is a λ-tableau, then ∆(t) is a λ partition of In and ∆′(t) is a λ-partition of In. For example if t = 2 4 3 6 1 5 then ∆(t) = 2 4 3 6 1 5 Definition 5.2.11 [def:tabloids] Let s, t be λ-tableaux. (a) [a] s and t are called row-equivalent if ∆(t) = ∆(s). An equivalence class of this relations is called a tabloid and the tabloid containing t is denoted by t. (b) [b] s and t are called column-equivalent if ∆′(t) = ∆′(s). The equivalence class of this relations containing t is denoted by |t|. For example if t = 1 4 2 3 then t = { 1 4 2 3 , 4 1 2 3 , 1 4 3 2 , 4 1 3 2 } Lemma 5.2.12 [action on tableaux] Let λ be partition of n. Let π ∈ Sym(n) and s, t be λ tableaux. (a) [a] Sym(n) acts transitively on the set of λ-tableaux via πt = π ◦ t. (b) [b] π∆(t) = ∆(πt)). (c) [c] s and t are row equivalent iff πs and πt are row equivalent. In particular, Sym(n) acts on the set of λ-tabloids via πt = πt. Section 5.2. Diagrams,Tableaux and Tabloids 107 Proof: (a) Clearly πt = π ◦ t defines an action of Sym(n) on the set of λ tableaux. Since s, t a bijections from [λ] → In, ρ := s ◦ t−1 ∈ Sym(n). Then ρ ◦ t = s and so the action is transitive. (b) Let D = [λ]. Then ∆(t) = (Di)∞i=1) and so π∆(t) = π(t(Di)∞i=1) = (π(t(Di) ∞ i=1) = ((πt)(Di)) ∞ i=1 = ∆(πt) (c) s is row-equivalent to t iff ∆(s) = ∆(t) and so iff π∆(s) = π∆(t). So by (b) iff ∆(πs) = ∆(πt) and iff πt and πs are row-equivalent.  Let ∆ = (∆i)∞i=1 be λ-partition of In. Let π ∈ Sym(n). Recall that π ∈ CG(∆) means π∆ = ∆ and so π(∆i) = ∆i for all i. CSym(n)(∆) = ⋂∞ i=1NSym(n)(∆i)) = i∞ i=1 Sym(∆i). So CSym(n)(∆) has order λ! :=∏∞ i=1 λi!. Definition 5.2.13 [def: row stabilizer] Let t be a tableau. The Rt = CSym(n)(∆(t) and Ct = CSym(t)(∆′(t). Rt is called the row stabilzer and Ct the column stablizer of t. Lemma 5.2.14 [char row equiv] Let s and t be λ-tableaux. The s and t are row equiva- lent iff s = πt for some π ∈ Rt. Proof: Then by 5.2.12(a), s = πt for some π ∈ Sym(n). Then s is row-equivalent to t if and only if ∆(t) = ∆(πt). By 5.2.12(b), ∆(π)t) = π∆(t) and so s and t are row equivalent iff π ∈ Rt.  Lemma 5.2.15 [basic combinatorical lemma] Let λ and µ be partions of n, t a λ- tableau and s a µ-tableau. Suppose that for all i, j, |∆(t)i ∩ |∆′(s)j | ≤ 1 ( That is no two entrees from the same row of t lie in the same column of s). Then λEµ. Moreover if λ = µ, then there exists λ-tableau r such that r is row equivalent to t and r is column equivalent to s. Proof: Fix a column C of Changing the order the entrees of C neither effects the assump- tions nor the conclusions of the lemma. So we may assume that if i appears before j in C, then i also lies earlier row than j in the tableau t. We do this for all the columns of s. It follows that an entree in the k-row of t must lie in one of the first k-rows of s. Thus∑k r=1 λi ≤ ∑l r=1 µi and µ dominates λ. Suppose now that λ = µ. Since λ1 = µ1 and the firs row of t is contained in the first row of s, the first row of ∆(t)1 = ∆(s)1. Proceeding by induction we see that ∆(t)k = ∆(s)t for all s and t. So s and t are row equivalent.  108 Chapter 5. Representations of the Symmetric Groups 5.3 The Specht Module Definition 5.3.1 [def:fh] Let G be a group, H ⊆ G, R a ring and f ∈ RG. Then fH =∑ h∈H fhh. Lemma 5.3.2 [basic fh] Let G be a group, R a ring and f ∈ RG. Suppose that f view as a function is a multiplicative homomorphism. (a) [a] Let A,B ⊆ G such that the maps A×B → G, (a, b) → G is 1−1, then fAB = fAfB. (b) [b] Let A ≤ B ≤ G and T a left-transversal to A in B. Then fB = fT fA. (c) [c] Let A1, A2, An ≤ G and A = 〈Ai | 1 ≤ i ≤ n〉 Suppose A = ini=1Ai, then fA = fA1fA2 . . . fAn. (d) [d] Suppose f is a class function, then for all g ∈ G and H ⊆ G, gfHg−1 = fgHg−1. Proof: (a) Since the map (a, b) → ab is 1−1, every element in AB can be uniquely written has ab with a ∈ A and b ∈ B. Thus fAfB = ∑ a∈A faa · ∑ b∈B fbb = ∑ a ∈ A, b ∈ Bfafbab = ∑ a∈A,b∈B fabab = ∑ c∈AB fcc = fAB (b) is a special case of (a). (c) follows from (a) and induction on n. (d) Readily verified. Since the map t→ ∆(t) is a well defined bijection between the λ tabloids and the the λ partitions of In we will often identify t with ∆(t). In particular, we have t ∈Mλ. Definition 5.3.3 [polytabloid] Let t be λ-tableau. (a) [a] kt = sgnCt = ∑ π∈Ct sgnππ ∈ FSym(n). (b) [b] et = ktt = ∑ π∈Ct sgnππt ∈M λ. et is called a polytabloid. (c) [c] Sλ is the F -subspace of Mλ spanned by the λ-polytabloids. Sλ is called a Specht module. (d) [d] F λ is the left ideal in FSym(n) generated by the kt, t a λ-tableau. As a first example consider t = 3 2 5 1 4 . The Ct = Sym({1, 3})× Sym({{2, 4}, kt = (1− (13) · (1− (24)) = 1− (13)− (24) + (13)(24) and et = 3 2 5 1 4 − 1 2 5 3 4 − 3 4 5 1 2 + 1 4 5 3 2 Section 5.3. The Specht Module 111 Proof: This follows immediately from 5.3.6(b) and 5.3.6(c).  Lemma 5.3.8 [submodules of ml] Supp F is a field and let λ be a partition of n and V be an FSym(n)-submodule of Mλ. Then either F λV = Sµ and Sµ ≤ V orF λV = 0 and Sλ ≤ V . Proof: If F λV = 0, then by 5.3.7, V ≤ Sλ⊥. So suppose F λV 6= 0. Then ksV 6= 0 for some λ-tableau s. So 5.3.6 implies ksV = Fes = ksM λ. Since by 5.3.4(a) implies ksV = ksMλ for all λ-tableaux s. Thus F λV = F λMλ = Sλ and Sλ ≤ V .  If F ≤ K is a field extensions we view Mλ = MλF has a subset of Sµ. Note also that MλK is canonically isomorphic to K⊗F Mλ. Put Dλ = Sλ/(Sλ ∩ Sλ⊥). Lemma 5.3.9 [dl=fldl] Let λ be a partition of n. If F is a field then F λDλ = Dλ. Proof: By 5.3.8 either F λSλ = Sλ or Sλ ≤ Sλ⊥. In the first case F λDλ = Dλ and in the second Dλ = 0 and again F λDλ = Dλ. Proposition 5.3.10 [dl=du] Let λ and µ be partitions of n with Dλ = 0. Suppose F is a field. If Dλ is isomorphic to an FSym(n)-section of Mµ, then λ E µ. In particular, Dλ ∼= Dµ then λ = µ. Proof: By 5.3.9 F λDλ = Dλ 6= 0. Hence also F λDµ 6= 0 and F λMµ 6= 0. So by 5.3.5(a), λE µ. If Dλ ∼= Dµ, the Dµ is a section of Mλ and so µE λ and µ = λ.  Lemma 5.3.11 [scalar extensions of ml] Let λ be a partition of n and F ≤ K a field extension. (a) [a] SλK = KSλ ∼= K ⊗F Sλ. (b) [b] Sλ⊥K = K(Sλ⊥) ∼= K⊗F Sλ⊥. (c) [d] SλK ∩ Sλ⊥K = K(Sλ ∩ Sλ⊥) = K⊗F Sλ ∩ Sλ⊥). (d) [c] DλK ∼= K⊗F Dλ. Proof: (a) is obvious. (b) follows from (a) and 4.1.19(b) (a) follows from (a), (b) and 4.1.19(a). (d) follows from (a) and (c).  Lemma 5.3.12 [dl absolutely simple] Let λ be a partition of n and suppose Dλ 6= 0. Then Dλ is an absolutely simple FSym(n)-module. 112 Chapter 5. Representations of the Symmetric Groups Proof: By 5.3.11(d) it suffices to show that Dλ is simple. So let V be an FSym(n)- submodule of Sλ with Sλ ∩Sλ⊥ ≤ V . By 5.3.8 either Sλ ≤ V or V ≤ Sλ⊥. In the first case V = Sλ and in the second V ≤ Sλ ∩ Sλ⊥ and V = S ∩ Sλ⊥. Thus Dλ = Sλ/(Sλ ∩ Sλ⊥) is simple.  5.4 Standard basis for the Specht module Proposition 5.4.1 [garnir relations] Let t be a λ-tableau, i < j ∈ Z+, X ⊆ ∆′(t)i and Y ⊆ ∆′(t)j. Let T be any transversal to Sym(X)× Sym(Y ) in Sym(X ∪ Y ). (a) [a] sgnT et is independent from the choice of the tranversal T . (b) [b] If |X ∪ Y | > λ′i. Then sgnT et = 0 Proof: (a) Let π ∈ Sym(X ∪ Y ) and ρ ∈ Sym(X)× Sym(Y ) ≤ Ct. Then sgn(πρ) · πρ · et = sgn(π)π · sgn(ρ)ρet 5.3.4(e) = sgn(π)πet and so (a) holds. (b) Since |X ∩ Y | > λ′i ≥ λ′j , there exists i ∈ X and j in Y such that i and j are in the same row of t. So (1 − (ij))πt = 0. If π ∈ Sym(X ∪ Y ), then π and π · (ij) lie in differen cosets of Sym(X) × Sym(Y ). Hence we can choose R ⊆ Sym(X ∪ Y ) such that R ∩ R · (i, j) = ∅ and R ∪ R · (ij) is a transversal to Sym(X) ∪ Sym(Y ). By (a) we may assume T = R∪R · (ij) and so sgnT = sgnRsgn{1,(ij)} = sgnR · (1− (ij)) and sgnT et = sgnR · (1− (ij))et = 0.  Definition 5.4.2 [def:garnir] Let t be a λ-tableau, i < j ∈ Z+, X ⊆ ∆′(t)i and Y ⊆ ∆′(t)j. (a) [a] TXY is the set of all π ∈ Sym(X ∪ SymY ) such that the restrictions of π ◦ t to π−1(X) and π−1(Y ) are increasing. (b) [b] GXY t = sgnTXY . GXY t is called a Garnir element in FSym(n). Lemma 5.4.3 [basic garnir] Let t be a λ-tableau, i < j ∈ Z+, X ⊆ ∆′(t)i and Y ⊆ ∆′(t)j. (a) [a] TXY is a transvsersal to Sym(X)× Sym(Y ) in Sym(X ∪ Y ). Section 5.4. Standard basis for the Specht module 113 (b) [b] If |X ∪ Y | > λ′i. Then GXY tet = 0. Proof: (a) Just observe that if π ∈ Sym(X ∪ Sym(Y ), then there exists a unique element ρ ∈ Sym(X)∪Sym(Y ) such that the restriction of πρ to t−1(X) and to t−1(Y ) are increasing. (b) follows from (a) and 5.4.1(b).  Consider n = 5, λ = (3, 2), t = 1 2 3 4 5 , X = {2, 5}, Y = {3} Then GXY et = 0 gives 1 2 3 4 5 − 1 3 2 4 5 − 1 2 5 4 3 = 0 Definition 5.4.4 [def:increasing tableau] Let λ be a partion of n and t a λ-tableau. (a) [a] rt = r ◦ t−1 and ct = s ◦ t−1. So i ∈ In lies in row rt(i) and column ct(i) of t. (b) [b] We say that t is row-increasing ct is increasing on each row ∆i(t) of t (c) [c] We say that t is column-increasing if rt is increasing on column ∆′i(t). Note that rt only depends on T and so we will also write rt for rt. Indeed r = s iff rt = rs. Lemma 5.4.5 [basic increasing] Let λ be a partion of n and t a λ-tableau. (a) [a] t contains a unique row-increasing tableau. (b) [b] |t| contains a unique column-increasing tableau. (c) [c] Let π ∈ Sym(n) and i ∈ I. Then rt(i)) = rπt(πi). Proof: (a) and (b) are readily verfied. (c) rπt ◦ π = r ◦ (π ◦ t)−1 ◦ π = r ◦ t−1 = rt.  Definition 5.4.6 [def:standart tableau] Let λ be a partition of n and t a λ-tableau. A standard tableau is row- and column-increasing tableau. A tabloid is called standard if it contains a standard tableau. If t is a standard tableau, then et is called standard polytabloid. By 5.4.5(a), a standard tabloid contains a unique standard tableau. We will show that the standard polytabloids form a basis of Sλ for any ring F . For this we need to introduce a total order on the tabloids Definition 5.4.7 [def:order tabloids] Let t and s be the distinct λ-tabloids. Let i ∈ In be maximal with rt(i) 6= rs(i). Then t < s provided that rt(i) < rs(i). 116 Chapter 5. Representations of the Symmetric Groups Theorem 5.4.13 [standard basis] Let F be a ring and λ a partition of n. The standard polytabloids form a basis of Sλ. Moreover, Sλ⊥⊥ = Sλ and there exists an R-basis for Sλ indexed by the nonstandard λ-polytabloids. By 5.4.10(a) and 5.4.12 the standard polytabloids are linearly independent. Let t be λ- tableau. Let |t| be the column equivalence class of t. Total order the column euqivalence classes analog to 5.4.7 We show by downwards induction that et is a F -linear combination of the standard polytableaux. Since et = ±es for any s column-equivalent to t we may assume that t is column increasing. If t is also row-increasing, t is standard tableaux and we are done. So suppose t is not row-increasing so there exists (i, j) ∈ Z+× such that t(i, j) > t(i, j + 1). Let X = {t(k, j) | i ≤ k ≤ λ′i and Y = {t(k, j + 1) | 1 ≤ k ≤ j. Then |X ∪ Y | = λ′j + 1 and so by 5.4.1 ∑ π∈TXY sgnπeπt = 0 Since ct is increasing on X and on Y and since t(i, j) > t(i, j + 1), rt is non-increasing on X ∪Y . So by 5.4.10 |πt| > |— for all 1 6= π ∈ Sym(X∪). Thus by downwards induction eπt is an R-linear combination of the standard polytabloids. Hence the same is true for et = − ∑ 1 6=πT sgnπeπt. The remaining statements now follow from 5.4.12.  5.5 The number of simple modules Definition 5.5.1 [def:p-regular class] Let p be an integer. An element g in a group G is called p-singular if p divides |g|. Otherwise g is called p-regular. A conjugacy class is called p-regular if its elements are p-regular. The goal of this section is to show that if K is an algebraicly closed field, G is a finite group and p = charK then the number of isomorpism classes of simle KG-modules equals the number of p-regular conjugacy classes. Lemma 5.5.2 [cyclic permutation] (a) [a] Let G be a group, n ∈ Z+ and a1, . . . an ∈ G. Then for all i ∈ N ai+1ai+2 . . . ai+n is conjugate a1a2 . . . an in G. (b) [b] Let R be a group, n ∈ Z+ and a1, . . . an ∈ R. Then for all i ∈ N, ai+1ai+2 . . . ai+n ≡ a1a2 . . . an (mod )S(R) Proof: (a) We have a−11 · a1a2 . . . an . . . a1 = a2 . . . ana1. So (a) follows by induction on n. (b) a1 · a2 . . . an − a2 . . . an · a1 ∈ S(R) So (b) follows by induction on n.  Section 5.5. The number of simple modules 117 Definition 5.5.3 [def: sr] Let R be ring and p = charR. Then S(R) = 〈xy − yx | x, y ∈ R〉Z. Let p̃ = p if p 6= 0 and p̃ = 1 if p = 0. T(R) = {r ∈ R | rp̃ m ∈ S(R) for some m ∈ N}. Lemma 5.5.4 [sr for group rings] Let R be a commutative ring and G a group. Then S(RG) consists of all a = ∑ rgg ∈ RG with ∑ g∈C rg = 0 for all conjugacy classes C of G. Proof: Let U consists of a = ∑ rgg ∈ RG with ∑ g∈C rg = 0 for all conjuagacy classes C of G. Note that both S(R) and U are R-submodules. As an R-modules S(R) is spaned by the gh− hg wth g, h ∈ G. By 5.5.2 gh and hg are conjugate in G. Thus gh = hg ∈ U and S(R) ⊆ U . U is spanned by the g − h where g, h in G are conjuagte. Then h = aga−1 and g − h = a−1 · ag = ag · a−1 and so g − h ∈ S(R) and U ⊆ S(R).  Lemma 5.5.5 [basic sr] Let R be a ring with p := charR a prime. (a) [a] (a+ b)p m ≡ apm + bpm mod S(R) for all a, b ∈ R and m ∈ N. (b) [b] T(R) is an additive subgroup of R. (c) [c] Suppose that R = ⊕s i=1Ri. Then S(R) = ⊕r i=1 Si and T (R) = ⊕ T (Ri). (d) [d] Let I be an ideal in R. Then S(R/I) = S(R) + I/I. (e) [e] Let I be a nilpotent ideal in R. Then I ≤ T (R), T (R/I) = T (R)/I and R/T (R) ∼= (R/I)/T (R/I). Proof: (a) Let A = {a, b}p and let H = 〈h〉 be a cyclic group of order p acting on A via h(ai) = (ai+1). Then H has two fixed points on A namely the constant sequence (a) and (b). Since the length of any orbit of H divises |H|, all other orbits have lenghth p. Let C be an orbit of length p for H on A. For a = (a1, a2, . . . ap) ∈ A puy ∏ a = a1a2 . . . ap/ Then by 5.5.2 ∏ a ≡ ∏ b (mod )S(R) for all a, b ∈ C and so ∑ b∈C ∏ b ≡ p ∏ a = 0 mod S(R). Hence for (a+ b)p = ∑ αinA ∏ a ≡ ap + bp mod S(R). (a) now follows by induction on m. (b) Follows from (a). (c) Obvious. (d) Obvious. (e) Since I is nilpotent, Ik = 0 for some integer k. Choose m with pm ≥ k. Then for all i ∈ I, ipm = 0 ∈ S(R) and so i ∈ T(R). Thus I ≤ T(R). Since S(R) + I/I = S(T/I) we have T (R)/I ≤ T (R/I). Conversely if t+ I ∈ T (R/I), then tpl ∈ S(R) + I. Since bith S(R) and I are in T (R), (b) implies tp l ∈ T (R) and so also t ∈ T (R).  Lemma 5.5.6 [tr for group rings] Let F be an integral domain with char F = p. Let G be a periodic group and let Cp be the set of p-regular conjugacy classes of G. For C ∈ Cp let gC ∈ C. Then (gC + S(FG) | C ∈ Cp) is a F -basis for FG/ S(FG). 118 Chapter 5. Representations of the Symmetric Groups Proof: Let g ∈ G and write g = ab with [a, b] = 1, apm = 1 and b, p-regular. Then gp m − bpm = 0 and so by 5.5.5(b), g ≡ mod T(FG). Also by 5.5.4 b ≡ gC where C = Gb. (gC + ( FG) | C ∈ Cp) is a spanning set for FG/ S(FG). Now let rC ∈ R with∑ C∈Cr rcgC ∈ T(FG) Then there exists m ∈ N with ( ∑ C∈Cp rcgC) pm ∈ S(FG). Since gC is p-regular, p - gC and so p is invertible in Z/|gC |Z. Hence there exists mC ∈ Z with |gC | | pmC − 1. Put k = m ∏ C∈Cp mC . Then g pk C = gC and ( ∑ C∈Cp rcgC) pk ∈ S(FG). By 5.5.5(b),∑ C∈Cp rp k C gC = ∑ C∈Cp rp k C g p C ∈ S(FG) Thus 5.5.4 shows that rp k C = 0 for all C ∈ Cp. So also rC = 0 and (gC + ( FG) | C ∈ Cp) is a linearly independent.  Lemma 5.5.7 [sr for matrix ring] Let R be a commutative ring and p = charR. (a) [a] S(Mn(R)) consists of the trace zero matrices and Mn(R)/S(Mn(R)) ∼= R. (b) [b] p = char K is a prime, then T(Mn(R)) = {a ∈ Mn(R) | tr(a)p̃ m = 0for somem ∈ N}}. (c) [c] If R is a field, then S(Mn(R)) = T(Mn(R)) and Mn(R)/T(Mn(R)) ∼= R. Proof: Since tr(xy) = tr(yx) and so S(Mn(R)) ≤ ker tr. ker tr is generted by the matrices Eij and Eii − Ejj with i 6= j. Eij = EiiEij − EijEii and so Eij ∈ S(Mn(R). Eii − Ejj = EijEji − EjiEij and so Eii − Ejj ∈ ker tr. Suppose now that p is a prime and let a ∈ Mn(R). Let b = tr(a)E11 and c = a − b. Then trc = 0, c ∈ S(Mn(R)) and so by 5.5.5 a ∈ T (Mn(R)0 if and only if b ∈ T(Mn(R)). Since tr(bp m ) = tr(a)p m the lemma is proved.  Theorem 5.5.8 [pmodular simple] Let G be a finite group, F an algebraicly closed field and p = charF . Then the number of isomorphism classes of simple FG-modules equals the number of p-regular conjugacy classes. Proof: By 5.5.6 the number of p′ conjugacy classes is dimF FG/T(FG). Let A = FG/J(FG). By 6.3.4 J(FG) is nilpotent and so by 5.5.5(e), FG/T(FG) ∼= A/T(A). By 2.5.24 R ∼= ⊕n i=1 Mdi(F), where n is the number of isomorphism classes of simple FG-modules. Thus by 5.5.5(c) and 5.5.7(c), R/T (R) ∼= Fn. So dimF FG/T(FG) is the number of isomorphism classes of simple FG-modules.  Section 5.6. p-regular partitions 121 Let t, u be λ tableaux. Let A be an equivalence class of tabloids and r ∈ A. Let s ∈ A and choose  as in (2◦). Then t(s)u(s) =  · t(s) ·  · s(r) = t(r)t(s) Thus ∑ s∈A t(s)u(s) = |A|t(r)u(r) By (1◦), |A| = z!. Summing over all the A’s we conclude that z! divides (et | es). Thus (3◦) holds. Let t be λ-tableau. Define σ ∈ Sym(n) by σ(t(i, j)) = t(i, λi + 1 − j) and put t̃ = σt. So t̃ is the tableaux obtained by reversing the rows of t. We will show that (et | () | et̃) =∏∞ i=1(zi!) j . Put Ui := Ui(t) := ⋃ k∈Zi ∆k(t), the union of the rows of t of size i. Note that Ui = Ui(t̃) and U = (Ui) is partion of In. Also put U j i := U j i (t) = Ui ∩∆′j , the part of column j of t lying in Ui. Then U j i (t̃) = U i+1−j i = σ(U j i ). Let P = (U j i ) | i, j ∈ Z). Then P is a partition of In refining both U and column partition. ∆′(t). Hence Sym(U) ≤ Ct. Also σ permutes the Uij and so σ normalizes Sym(U) and so Sym(U) ≤ σCtσ−1 = Ct̃. Observe |U j i (t)| = zj if j ≤ i and U ji (t) = ∅ otherwise. Thus 4◦ [4] |Sym(U)| = ∏ i,j |U j i (t)|! = ∏∞ i=1(zi!) i. We show next 5◦ [5] Let π ∈ Sym(U). Then t(πt) = t̃(πt) = sgnπ. Since π ∈ Ct we have t(πt) = sgnπ. Since π ∈ Ct̃ we have t(πt̃) = sgnπ. Since σ fixes the rows of t, πσπ−1 fixes the rows of πt. Thus πt = πσπ−1πt = πσt = πt̃ and so (5◦) holds. 6◦ [6] Let π ∈ Ct such that πt is involved in et̃. Then π ∈ Sym(U). Since πt is involved in et̃ there exists π̃ ∈ Ct̃ with πt = π̃t̃. Hence for all k ∈ In, rπt)(k) = rπ̃t̃(k) and so rt(π −1k) = rt̃(π̃−1k). Put α = π−1 and α̃ = π∗−1. Then for all k ∈ I. (∗) α ∈ Ct, α̃ ∈ Ct̃ and rt(α(k)) = rt̃(α̃(k)) We need to show that α(U ji ) = U j i = α̃(U j i ) for all i, j. The proof uses double induction. First on j and then downwards on i. 122 Chapter 5. Representations of the Symmetric Groups For I, J ⊂ Z+ let UJI = ⋃ {U ji | i ∈ I, j ∈ J}. If I = Z+ or J = Z+ we drop the subscript I, respectively superscript. For example U≤j = ⋃ Uki | i, k ∈ Z+ | k ≤ j} consists ofthe first j columns of t. Suppose that α(U lk) = U l k = α̃(U l k) whenever l < j or l = j and k > i. Then α(U j >i) = α(U j>i) and α(U j) = U j implies α(U ji ) ⊆ U j ≤i. Hence by (*) also (∗∗) ãlpha(U ji ) ⊆ U≤i Let c = i+ 1− j. Then U ji = Ũ ci and Ũ c<i = ⋃ k<i U c+1−kk and so by induction α̃Ũ c<i = U c <i. Hence α̃(U j i ) ⊆ α̃(Ũ c≥i) = Ũ c≥i ⊆ Ũ≥i = U≥i. So by (**) α̃(U ji ) ⊆ Ui ∩ Ũ c = Ũ ci = U j i and ã(U j i ) = Uij . Hence by (*) also α(U J i ≤ Ui ∩U j = U j i and α(U ji ) = U j i . So (6◦) is proved. From (5◦) and (6◦) we conclude that (et | et̃) = |Sym(U)| = ∏∞ i=1(zi!) i. Since gλ divides (et | et̃) the lemma is proved.  Proposition 5.6.6 [dlambda not zero] Suppose F is an integral domain and λ is a par- tition of n. Let p = charF . Then Dλ 6= 0 iff λ is p-regular. Proof: Since F is an integral domain, p = 0 or p is a prime. Let λ = (izi )i=1. Then p | ∏ i zi! iff p ≤ zi for some i, iff p | ∏ i(zi!) i and iff λ is p-singular. So 5.6.5 implies that p | gλ iff λ is p-singular. And so by 5.6.4, Dλ = 0 iff λ is p-singular.  Theorem 5.6.7 [all simple sym(n)-modules] Let F be a field, n a postive integer and p = charF . (a) [a] Let λ be a p-regular partition of n. Then Dλ is an absolutely simple, selfdual FSym(n)-module. (b) [b] Let I be a simple FSym(n)-module. Then there exists a unique p-regular partition λ of n with I ∼= Dλ. Proof: (a) By 5.6.6 Dλ 6= 0. By 4.1.5, s induces a non-degenerate G-invariant form on Dλ and so by 4.1.6(c), Dλ is isomorphic to its dual. By 5.3.12, Dλ is absolutely simple. (b) If λ and µ are distinct p-regular partition then by 5.3.10 and (a), Dλ and Dµ are non-isomorphic simple FSym(n)-modules. The number of simple FSym(n)-modules is less or equal to the number simple Sym(n)-modules over the algebraic closure of F. The latter number is by 5.5.8 equal to to the number of p′-conjuagacy classes and so by 5.6.2 equal to the number of p-regular partitions of n. So (b) holds.  Section 5.7. Series of R-modules 123 5.7 Series of R-modules Definition 5.7.1 [def:series] Let R be a ring and M and R-module. Let S be a set of R-submodules of M . Then S is called an R-series on M provided that: (a) [a] 0 ∈ S and M ∈ S. (b) [b] S is totally ordered with respect to inclusion. (c) [c] For all ∅ 6= T ⊂ S, ⋂ T ∈ S and ⋃ T ∈ S. For example Z > 2Z > 6Z > 30Z > 210Z > . . . > 0 is an Z-series on Z. Definition 5.7.2 [def:jumps] Let R be a ring, M an R-module and S an R-series on M . For 0 6= A ∈ S put A− = ⋃ {B ∈ S | B ⊂ A}. If A 6= A− then (A−, A) is called a jump of S and A/A− a factor of S. S is called a composition series for R on S provided that all its factors are simple R-modules. The above example is composition series and its sets of factors is isomorphic to Z/pZ, p a prime. Lemma 5.7.3 [basic series] Let R be a ring, M an R-module, S an R-series on M . (a) [a] Let A,B ∈ S with B ⊂ A. Then (B,A) is a jump iff A = C or B = C for all C ∈ S with B ⊆ C ⊆ A. (b) [b] Let U ⊂M . Then there exists a unique A ∈ U minimal with U ⊆ A. If U is finite and contains a non-zero element then A− 6= A and A ∪ U * A−. (c) [c] Let 0 6= m ∈ M . Then there exists a unique jump (B,A) if S with v ∈ A and v 6∈ B. Proof: (a) Suppose first that (B,A) is a jump. Then B = A−. Let C ∈ S with B ⊆ C ⊆ A Suppose C ⊂ A. Then C ⊆ A− = B and C = B. Suppose next that A = C or B = C for all C ∈ S with B ⊆ C ⊆ A. Since B ⊆ A, B ⊆ A−. Let C ∈ S with C ⊂ A. Since S is totally ordered, C ⊆ B or B ⊆ C. In the latter case, B ⊆ C ⊂ A and so by assumption B = C. So in any case C ⊆ B and thus A− ⊆ B. We conclude that B = A− and so (B,A) is a jump. (b) Put A = ⋃ {S ∈ S | U ⊆ S}. By A ∈ S and so clearly is minimal with respect to U ⊆ A and is unique with respect to this property. Suppose now that U is finite and contains a non-zero element. Then A 6= 0. Suppose that A = A−. Then for each u ∈ U we can choose Bu ∈ S with u ∈ Bu and Bu ⊂ A. Since U is finite {Bu, u ∈ U} has a maximal elemeent B. Then U ⊆ B ⊂ A, contradicting the minimality of A Thus A 6= A− and by minimality of A, U * A. (c) Follows from (b) applied to U = {m}.  126 Chapter 5. Representations of the Symmetric Groups (2) θi(et) = { et↓ if j = i 0 if j < i If s is a λ(i)-tableau, then s = t ↓ for a (unique) restrictable λ tableau t with n in row ri. Hence (3) Vi−1 ≤ Vi ∩ ker θi and Vi/Vi ∩ ker θi ∼= Im θi = Sλ (i) Let B be the set of standard λ-polytabloids and Bi the et with t standard and n in row ri. Then by (1) θi(Bi) is the standard basis for Sλ (i) and so is linear independently. Thus also the image of Bi in Vi/Vi ker θi is linearly independent. Consider the series of F -modules 0 = V0 ≤ V1 ∩ ker θ1 ≤ V1 ≤ V2 ∩ ker θ2 < V2 < . . . < Vk−1 ≤ Vk ∩ ker θk < Vk < Sλ Each et ∈ B lies in a unique Bi and so in Vi \ (Vi ∩ kerπi). Thus B ∩ Vi ∩ ker θi ⊆ Vi−1. So we can apply 5.7.4 to the series of F -modules and conlcude that Vi ∩ ker θi/Vi−1 is as the emptyset as an R-basis. Hence Vi−1 = Vi ∩ ker θi. For the same reason Vk = Sλ and theorem now follows from (3).  Theorem 5.8.6 (Branching Theorem) [branching theorem] Let F be a field with charF = 0 and λ a partition of n. (a) [a] Sλ ↓Sym(n−1)= ⊕ µ∈λ↓ Sµ (b) [b] Sλ ↑Sym(n−1)= ⊕ µ∈λ↑ Sµ Proof: (a) Follows from 5.8.5 and Maschke’s Theorem 2.3.2 (b) Follows from (a) and Frobenius Reprocity 2.7.4. 5.9 S(n−2,2) In this section we investigate the Specht modules S(n), S(n−1,1) and Sn−2,2. Lemma 5.9.1 [s(n)] M (n) = S(n) ∼= D(n) ∼= F . Proof: There there a unique (n)-tabloid t and πt = t for all π ∈ Sym(n). Moreover et = t and so S(n) = M (n). Also S(n)⊥ = 0 and the lemma is proved.  Section 5.9. S(n−2,2) 127 Lemma 5.9.2 [s(n-1)] Let xi the unique (n−1, 1)-tabloid with i in row 2. Let z = ∑n i=1 xi be the sum of all λ-tabloids. Then (a) [a] S(n−1,1) = { ∑n i=1 fixi | fi ∈ F, ∑n i=1 fi = 0. (b) [b] S(n−1,1)⊥ = Fz. (c) [c] S(n−1,1)⊥ ∩ S(n−1,1) = {fx | f ∈ F, nf = 0}. Proof: (a) If t is tableau with t(1, 1) = i and t(2, 1) = j, then et = xi − xj . This easily implies (a). (b) ∑ fizi ⊥ xi − xj iff fi = fj . (c) Follows from (a) and (b).  Corollary 5.9.3 [dim d(n-1)] Let F be a field and p = char F. (a) [a] If p - n, then S(n−1,1) ∼= D(n−1,1) has dimension n− 1 over D. (b) [b] If p | n, then D(n−1,1) has dimension n− 2 over F . Proof: Follows immediately from 5.9.2.  To analyze S(n − 2, 2) we introduce the follwing notation: Let n ∈ N with n ≥ 4 and λ = (n − 2, 2). Let P be the set for subsets of size two in In. For P ∈ Pn let xP be the λ-partition (P, In \ P ). Then (xP | P ∈ P) is an F -basis for Mλ. For a, b, c, d pairwise distinct elements in In put eab|cd = xac + xbd − xad − xbc. So eab|cd = et for any λ tableau of the form a c . . . b d . For i ∈ In define xi := ∑ i∈P∈P xP and yi = ∑ i/∈P∈P xP . Also let z = ∑ P∈P xP and observe that xi + yi = z for all i ∈ I. Lemma 5.9.4 [basis for s(n-2,2)perp] (a) [a] x1, x2, . . . xn−1, yn is an F -basis for Sλ⊥. (b) [b] x1, x2, . . . xn−1, z is an F -basis for Sλ⊥. (c) [c] y1, y2, . . . yn−1, z is an F -basis for Sλ⊥. (d) [d] If 2 is invertible in F then x1, x2, . . . xn is an F -basis for Sλ⊥. (e) [e] If n− 2 is invertible in F , then y1, y2, . . . yn is an F -basis for Sλ⊥. 128 Chapter 5. Representations of the Symmetric Groups Proof: (a) We will first show that xi ⊥ eab|cd for all appropriate i, a, b, c, d. If i /∈ {a, b, c, d}, xi and eab|cd have do not share a tabloid and so (xi | eab|cd) = 0. So suppose i = a, then xi and eab|cd share xac and xad with opposite signs and so again xi ⊥ eab|cd. Clearly z ⊥ eab|cd and so also yi ⊥ eab|cd. Thus xi, yi and z are all contained in Sλ⊥. Now let a = ∑ P∈P rPxP ∈ Sλ⊥. We need to show that a is a unique F -linear combi- nation of x1, x2, . . . xn−1, yn. For n 6= i ∈ In, xi is the only one involving xin. So replacing a by a − ∑n−1 i=1 rinxi we assume that rin = 0 for all i 6= n. And we need to show that a is scalar multiple of yn. That is we need to show that rij = rkl whenever {i, j}, {k, l} ∈ P with n /∈ {i, j, k, l}. Suppose first that P ∩ Q 6= ∅ and say i = k and withoutloss j 6= l. Since a ∈ Sλ⊥, a ⊥ ein|jl. Thus rij + rnl − ril − rnj = 0. By assumption rnl = rnj = 0 and so rij = ril = rkl. In the geneal case we conclude rij = rik = rkl and (a) is proved. (b) Observe that z = ∑n−1 i=1 xi − yn. Thus (b) follows from (a). (c) Since yi = z − xi this follows from (b). (d) Observe that ∑n i=1 xi = 2z and so xn = − ∑n−1 i=1 xi + 2z. So (d) follows from (b). (e) We have ∑n i=1 yi = ∑n i=1(z − xi) = nz − ∑n i=1 xi = (n− 2)z. So yn = − ∑n−1 i=1 yi + (n− 2)z and (e) follows from (c).  It might be interesting to observe that y1, . . . , yn−1, xn is only a basis if n−2 is invertible. Indeed xn = − ∑n−1 i=1 xi + 2z = ∑n−1 i=1 (yi − z) + 2z = ∑ i=1 yi + (n− 2)z. We know proceed to compute Sλ ∩ Sλ⊥ if F is a field. Lemma 5.9.5 [s(n-2) cap s(n-2)perp] Suppose F is field and put p = charF . (a) [a] Suppose p = 0 or p is odd and n 6≡ 1, 2 mod p or p = 2 and n ≡ 3 mod 4. Then n Sλ ∩ Sλ⊥ = 0. (b) [b] Suppose p is odd and n ≡ 1 mod p or p = 2, n ≡ 1 mod 4. Then Sλ ∩Sλ⊥ = Fz. (c) [c] Suppose p is odd and n ≡ 2 mod p or p = 2 and n ≡ 2 mod 4, then Sλ ∩ Sλ⊥ = 〈Fyi | 1 ≤ i ≤ n〉 and ∑n i=1 yi = 0. (d) [d] Suppose p = 2 and n ≡ 0 mod 4. Then Sλ ∩ Sλ⊥ = 〈Fyiyj | 1 ≤ i < j ≤ n〉 and∑n i=1 yn = 0. Proof: Since F is a field and (· | ·) is non-degenerate, Sλ⊥⊥ = Sλ and so Sλ ∩ Sλ⊥ = Sλ⊥⊥ ∩ Sλ⊥ is the radical of the restriction of (· | ·) to Sλ. By 5.9.4 y1, y2 . . . yn−1z is basis for Sλ⊥. Let a = r0z + ∑n−1 i=1 riyi. Then Observe that (yi | yi) = ( n−1 2 ) (yi | yj) = ( n−2 2 ) i 6= j (yi | z) = ( n−1 2 ) (z | z) = ( n 2 ) Section 5.9. S(n−2,2) 131 For example if F is a field with charF = p then by 5.9.2 M (n−1,1) ∼ D(n) ⊕D(n−1,1) if p - n and M (n−1,1)∼D(n | D(n−1,1) | D(n) if p | n. If might also be worthwhile to define the following binary operation on classes of R- modules. If A,B are classes of R-modules, then A⊕B denotes the set of all R-modules M such that M ∼= X ⊕ Y with X ∈ A and Y ∈ B. A | B is the class of all R-modules M such that M has an R-submodule X with X ∈ A and M/X ∈ B. A shape of M then can be interpreted as a class of R-modules containing M obtained form the isomorphism classes of R modules and repeated application of the operations ⊕ and |. To improve readabilty we write D(a, b, c . . .) for D(a,b,c,...) in the next lemma. Corollary 5.9.8 [shape of m(n-2,2)] Suppose F is a field. Then D(n−2,2) has simply shapes as follows: (a) [a] Suppose p = 0 or p is odd and n 6≡ 0, 1, 2 mod p or p = 2 and n ≡ 3 mod 4. Then M (n−2,2) ∼ D(n− 2, 2)⊕D(n− 1, 1)⊕D(n) (b) [b] Supose p 6= 0, 2 and n ≡ 0 mod p. Then M (n−2,2) ∼ D(n− 2, 2) ⊕ D(n) D(n− 1, 1) D(n) (c) [c] Suppose p is odd and n ≡ 1 mod p or p = 2, n ≡ 1 mod 4. Then M (n−2,2) ∼ D(n) D(n− 2, 2) D(n) ⊕ D(n− 1, 1) (d) [d] Suppose p is odd and n ≡ 2 mod p. Then M (n−2,2) ∼ D(n− 1, 1) D(n− 2, 2) D(n− 1, 1) ⊕ D(1) (e) [e] Suppose p = 2 and n ≡ 2 mod 4. Then M (n−2,2) ∼ D(n− 1, 1) D(n) D(n− 2, 2) D(n) D(n− 1, 1) ⊕ D(1) 132 Chapter 5. Representations of the Symmetric Groups (f) [f] Suppose p = 2 and n ≡ 0 mod 4. Then M (n−2,2) ∼ D(n− 1, 1)⊕D(n) D(n− 2, 2) D(n− 1, 1)⊕D(n) Proof: This is straighforward from 5.9.5. As an example we consider the case p = 2 and n ≡ 2 (mod 4). Observe that (z | z) = ( n 2 ) 6= 0 and so Mλ = Fz. Thus Mλ ∼ D(n)⊕ z ⊥, and the restrition of (· | ·) to z⊥ is a non-degenerate. 5.9.5 B := Sλ ∩ Sλ⊥ = 〈yi | 1 ≤ 1 ≤ n〉. So B has the submodule, A = 〈yiyj | 1 ≤ u < j ≤ n〉. Since ∑n i=1 yi = 0, B ∼= D(n − 1, 1). Since n is even, A/B 6= 1 and A/B ∼= D(n). Sλ/A = Dλ = D(n−2, 2). Since Sλ⊥ = A+Fz, Sλ = z⊥∩A⊥. So z⊥∩B⊥/Sλ ∼= (A/B)∗ ∼= D(n)∗ ∼= D(n). Moreover, z⊥/z⊥ ∩A⊥ ∼= A∗ ∼= D(n− 1, 1)∗ ∼= D(n− 1, 1). Thus (e) holds.  5.10 The dual of a Specht module Definition 5.10.1 [def:twisted module] Let R be a ring, G a group , M an RG-module and  : G → Z(R)] a multiplicative homomoprhism. Then M is the RG-module which is equal to M as an R-module and g · m = (g)gm for all g ∈ G,m ∈M . Note that this definition is consistent with our definition of the RG-module R. Proposition 5.10.2 [slambdaprime] Let λ be a partion of n. Then Sλ∗ ∼= Mλ/Sλ⊥ ∼= Sλ ′ sgn as FSym(n)-module. Proof: Fix a λ tableau s. Let π ∈ Rs = CG(s). Since Rs = Cs′ , 5.3.4(e) gives πes′ = sgnπes′ = π ·sgn es′ . Hence there exists a unique FSym(n)-linear homorphism (1) αs : Mλ →Mλ ′ with s→ es′ Let t be any λ-tabloids. Then the exists π ∈ Symn with πs = t (namely π = ts−1) and so αs(t)αs(πs) = π ·sgn es′ = sgn(π)eπs′ = sgn(ts−1)et′ that is (2) αs(t) = sgn(ts−1)et′ Section 5.10. The dual of a Specht module 133 Observe that (2) implies (3) Imαs = Sλ ′ Since λ′′ = λ we also obtain a unique FSym(n− 1) linear map (4) αs′ : Mλ →Mλ, t′ → sgn(ts−1)et Then (5) Imαs′ = Sλ We claim that αs′ is the adjoint of αs. That is (6) (αs(t) | r′) = (t | αs′(t))r for all λ-tableaux t, r. Indeed suppose that r′ is involved in involved in αs(t) = sgnts−1et′ . Then there exists β ∈ Ct′ with r′ = βt′ and so there exists δ ∈ Rr′ with δr′ = βt′. Moreover (αs(t) | r′) = sgn(ts−1)sgnβ Observe that δ ∈ Cr and β ∈ Rt. Thus t = βt = δr and so t is involved in er and (t | αs′(r′)) = sgn(rs−1)sgnδ δr = βt implies δrs−1 = βts−1 and so sgn(rs−1)sgnδ = sgn(ts−1)sgnβ and so (6) holds. Let m ∈ Mλ. (· | ·) is non-degenereate, (6) implies αs(m) = 0 iff (αs(m) | m′) = 0 for all m′ ∈Mλ′ iff (m | αs′(m′)) = 0 and iff m ∈ (Imαs′)⊥. So by (5) kerαs = Sλ ⊥ and so Mλ/Sλ⊥ ∼= Mλ/ kerαs ∼= Imαs = Sλ  Lemma 5.10.3 [tensor and twist] Let R be a ring, G a group , M an RG-module and  : G→ Z(R)] a multiplicative homomoprhism. Then M ∼= R ⊗RM as an RG-module. 136 Chapter 6. Brauer Characters If a ∈ A then f(a) = 0 for some monic f ∈ Z[x]. Then also f∗(a) = 0 and f∗ 6= 0. So a∗ is algebraic over Z∗. Let K be an algebraic closure of F and so of Z∗. Let 0 6= k ∈ K. Then km = 1 where m = |Z∗[k]| − 1 is coprime to p. Since U∗ contains all m roots of xm − 1 we get k ∈ U∗. Thus K∗ ⊆ U∗ ⊆ F∗ ⊆ K∗ and the lemma is proved.  Definition 6.1.2 [def:brauer character] Let G be a finite group and M an FG-module. G̃ is the set of p-regular elements in G. Let g ∈ G̃ and choose ξ1, . . . ξn ∈ U such that ηM (g) = ∏n i=1(x − ξ∗i ), where ηM (g) is the characteristic polynomial of g on M . Put φM (g) = ∑n i=1 ξi. Then the function φM : G̃→ A, g → φM (g) is called the Brauer character of G with respect to M . Recall that if H ⊆ G then we view RH as R an an R-submodule of RG. Also note that φM = ∑ g∈G̃ φM (g)g ∈ AG̃ ⊆ AG. Observe also that 1G◦ is the Brauer character of the trivial module FG. Lemma 6.1.3 [basic brauer] Let M be a G-module. (a) [a] φM is a class function. (b) [b] φM (g) = φM (g−1). (c) [c] φM = φM∗. (d) [d] If H ≤ G then φ |H= φM |H . (e) [e] F be the sets of factors of some FG-series on M . Then φM = ∑ F∈F φF Proof: Readily verified. See 3.2.8.  Definition 6.1.4 [def tilde a] (a) [a] For g ∈ G let gp, gp′ be defined by gp, gp′ ∈ 〈g〉, g = gpgp′, gp is a p- and gp′ is a p′-element. (b) [b] For a = ∑ g∈G agg ∈ CG, ã = a |G̃= ∑ g∈G̃ agg. (c) [c] For a = CG̃ define ǎ ∈ CG by ǎ(g) = a(gp′. Recall that χM (g) = trM (g) is the trace of g on M . Section 6.2. Algebraic integers 137 Lemma 6.1.5 [brauer and trace] Let M be a FG-module. Then (φ̌M )∗ = χM . Proof: Let Wi, 1 ≤ i ≤ n be the factors of an F〈g〉 composition series on M . Then since F is algebraically closed, Wi is 1-dimensionaly and g acts as a scalar µi on Wi. Since F contains no non-trivially p-root of unity gp acts trivially on Wi and so also gp′ acts as µi on Wi. Pick ξi ∈ U with ξ∗i = µi. Then φ̌M (g) = φM (gp′) = n∑ i=1 ξi and so (φ̌M (g))∗ = n∑ i=1 µi = χM (g)  Let Sp be a set of representatives for the simple FG-modules. 6.2 Algebraic integers Definition 6.2.1 [def:tracekf] Let F : K be a finite separable field extension and E a splitting field of F over K. Let Σ be set of F-linear monomorphism from F to K. tr = trFK : F → K | f → ∑ σ∈Σ σ(f) Lemma 6.2.2 [basic tracekf] Let F : K be a finite separable field extension. Then s : F× F → K, (a, b) → tr(ab) is a non-degenerate symmetric K-bilinear form. Proof: Clearly s isK-bilinear and symmetric. Suppose that a 6= f ∈ F⊥. Then tr(ab) = 0 for all b ∈ F and since a 6= o, tr(f) = 0 for all f ∈ F . Thus ∑ σ∈Σ σ, contradiction the linear idependence of filed monomorphism [Gr, III.2.4]. Corollary 6.2.3 [trace dual basis] Let F : K be a finite separable field extension and B a K basis for F. Then b ∈ B there exists a unique b̃ ∈ F with tr(ab̃) = δab for all ab ∈ F. Proof: 6.2.2 and 4.1.8.  Definition 6.2.4 [def:integral] Let S be a commutative ring and R a subring. (a) [a] a ∈ R is called integral over S if there exists a monic f ∈ S[x] with f(a) = 0. (b) [b] IntS(R) is the set of elements in S intgeral over R. 138 Chapter 6. Brauer Characters (c) [c] R is integrally closed in S if IntR(S). (d) [d] If Ris an integral domain, then R is called integrall closed if R is integraly closed in its field of fractions FR. Lemma 6.2.5 [basic integral] Let S be a commutative ring, R a subring and a ∈ S. Then the following are equivalent: (a) [a] a is integral over S. (b) [b] R[a] is finitely generated S-submodule of R. (c) [c] There exists a faithful, finitely R-generated R[a] module M Proof: (a)=⇒ (b): Let f ∈ R[x] be monic with f(a) = 0. Then an ∈ R〈1, . . . , an−1〉 and so R[a] = R〈1, a, . . . , an−1〉 is finitely R-generated. (a)=⇒ (b): Take M = R[a]. (b)=⇒ (c): Let B ⊆ M be finite with M = RB. Choose a matrix D = (dij) ∈ MB(R) with ai = ∑ i∈B dijj for all i ∈ B. Let f be the characteristic polynomial of D. Then f ∈ R[x] and f is monic. By Cayley-Hamilton [La, XV Theorem 8] f(D) = 0. Since f(a)i = ∑ j∈B f(D)ijj for all i ∈ I we get f(a)M = 0. Since AR(M) = 0 we have f(a) = 0.  Lemma 6.2.6 [integral closure] Let S be a commutative ring and R a subring of S. (a) [a] Let a ∈ S. If a is integral over R, then also R[a] is integral over R. (b) [b] Let T be a subring of S with R ⊆ T . Then S is integral over R iff T is integral over R and S is integral over T . (c) [c] IntS(R) is a subring of R and IntR(S) is integrally closed in S. Proof: (a) Let b ∈ R[a]. By 6.2.5(b), R[a] is finitely R-generated. Since R[a] is a faithful R[b]-module, 6.2.5(c) implies that b is integral over R. (b) One direction is obvious. So suppose S : T and T : R are integral and let a ∈ S. Let f = sumni=1tix i ∈ T [x] be monic with f(a) = 0. Put R0 = R and inductively Ri = Ri−1[ai]. Then ai is integral over Ri−1, Ri is finitely Ri−1-generated. Also f ∈ Rn[x] and so Rn[a] is finitely Rn-generated. It follows that Rn[a] is finitely R-generated and so by 6.2.5(c), a is integral over R. (c) Let a, b ∈ IntS(R). By (a) R[a] : R and R[a, b] : R[a] are integral. So by (b) R[a, b] : R is integral and so R[a, b] ⊆ IntS(R) and IntS(R) is a subring. Since both IntS(IntS(R) : IntS(R) and IntS(R) are integral, (b) implies that IntS(R) is integrally closed in R.  Section 6.2. Algebraic integers 141 Definition 6.2.17 [def:division] Let M be an R module and N ⊆ M and J ⊆ R. Then N ÷M J =: {m ∈M | Jm ⊆ N} . For example 0 ÷M J = AM (J) and if N is an R-submodule of M , then N ≤ N ÷M J and N ÷M J/N = AM/N (J). If R is an integral domain with field of fraction K and a, b ∈ K with b 6= 0, then Ra÷K Rb = Rab . Definition 6.2.18 [def:fractional ideal] Let R be a integral domain with field of fraction K. A fractional ideal of R is a non-zero R-submodule J of R such that kJ ⊆ R for some k ∈ K]. FI(R) is the set of fractional ideals of R. Observe that FI(R) is an abelian monoid under multiplication with identity element R. A fractional ideal is called invertible if its invertible in the monoid FI(R). FI∗(R) is the group of invertible elements in FI(R). Lemma 6.2.19 [basic monoid] Let H be a monoid. (a) [a] Every h has at most one inverse. (b) [b] Let a, b ∈ H. If H is abelian and ab is invertible, then a and b are invertible. invertible. Proof: (a) If ah = 1 and hb = 1, then b = (ah)b = a(hb) = a. (b) Let h be an inverse of a. Then 1 = h(ab) = (ha)b and so since H is abelian, ha is an inverse of b. By symmetry hb is an inverse for a.  Lemma 6.2.20 [basic invertible] Let R be a integral domain with field of fraction K and let J be a fractional ideal of R. (a) [a] If T 6= 0 is an R-submodule of J , then T is a fraction ideal of R and R÷KJ ⊆ R÷KT . (b) [b] R÷K J is a fractional ideal of I. (c) [c] J is invertible iff and only if (R÷K J)J = R. In this case its inverse is (R÷K J)J . Proof: By defintion of a fractiona ideal there exists k ∈ K] with kJ ⊆ R. (a) Note that kT ⊆ R and so T is a fractional ideal. If lK ⊆ R then also lT ⊆ R and (a) is proved. (b) Since k ∈ R÷K J , R÷K J 6= 0. Let t ∈ J ]. Then by (a) applied to T = Rt, R÷K J ⊆ R÷K Rrt = R 1 t and so t(R÷K J) ⊆ R and R÷K J is a fractional ideal. (c) If (R ÷K J)J = R, then R ÷K J is an inverse for J in FI(R). Suppose now that T ∈ FI(R) with TJ = R. Then clearly T ⊆ R÷F J . Thus R = TJ ⊆ (R÷F J)J ⊆ R Thus both T and R÷K F are inverse of J and so T = R÷K F .  142 Chapter 6. Brauer Characters Lemma 6.2.21 [partial inverse] Let R be an Dedekind domain with field of fraction K and J proper ideal in R. Then R < R÷K J . Proof: Let P be a maximal ideal in R with J ≤ P . Let a ∈ J ]. By 6.2.16 there exists non-zero prime ideals P1, P2, . . . Pn with ∏n i=1 Pi ≤ Ra. We also assume that n is minimal with with property. Since Ra ≤ P and P is a prime ideal we must have Pi ≤ P for some i. By definition of a Dekind domain, Pi is a maximal ideal and so Pi = P . Let Q = ∏n i6=j=1 Pj . Then PQ ≤ Ra and by minimality of n, Q  Ra. Thus Ja−1Q ≤ PQa−1 ≤ R and and a−1Q  R. So a−1Q ≤ R÷K J and hence R÷K J  R. Clearly R ≤ R÷K J and the lemma is proved. Proposition 6.2.22 [fi for dekind] et R be an Dedekind domain with field of fraction K. Let P be a nonzero prime ideal in the Dedekind domain R and J a non-zero ideal with J ⊆ P . Then P invertible and J < JP−1 ≤ R. Proof: Put Q := R÷K. Then R ≤ Q and J ⊆ JQ ⊆ R. Suppose that J = JQ. Since R is Noetherian, J is finitely R-generated. Since K is an integral domain and J 6= 0, J is a faithful Q-module. Thus 6.2.5(c) implies that Q is integral over R. By defintition of a Dekind domain, R is integrally closed in K and so Q ≤ R. But this contradicts 6.2.21 Thus J < JQ−1 and inparticular P < PQ ≤ R. By definition of a Dekind Domain P is a maximal ideal in R and so PQ = P . Thus Q = P−1 and the proposition is proved.  Theorem 6.2.23 [structure of dedekind] Let R be a Dedekind domain and let P be the set of non-zero prime ideals in R. Then the map τ : ⊕PZ → FI(R) | (zP ) → ∏ P∈P P zP is an isomorphism of monoids. In particular, FI(R) is a group. Moreover τ(z) ≤ R if and only if z ∈ ⊕PN. Proof: Clearly τ is an homomorphism. Suppose there exists 0 6= z ∈ ker τ . Let X = {P ∈ P | zP < 0 and Y = {P ∈ P | zP > 00. Then X ∩ Y = ∅ and X ∪ Y 6= ∅. Moreover, τ(z) = R implies ∏ P∈X P−zp = ∏ P∈Y P zP In particular both X and not empty. Let Q ∈ X. Then∏ P∈Y P zP ≤ Q a contrdiction since P  Q for all P ∈ Y and sinceR/Q is a prime ideal. Thus τ is 1− 1. Section 6.3. The Jacobson Radical II 143 Next let J be a proper ideal in R and P a maximal ideal in R with J ≤ P . By 6.2.22 J < JP−1 ≤ R. By Noetherian induction JP−1 = P1 . . . Pn for some prime ideals P1, . . . Pn and so J = PP1 . . . Pn, that is J = τ(z) for some z ∈ ⊕PN. Finally let J be an arbitray fraction ideal in K. Then by definition ther exists kJ ⊆ R for some k ∈ K]. Then k = rs with r, s ∈ R ] and so rJ = skJ ⊆ R. Let u, v ∈ ⊕ P N with τ(u) = Rr and τ(v) = rJ . Then τ(v − u) = (Rr)−1(rJ) = Rr−1rJ = J and so τ is onto.  The next proposition shows that Dedekind domains are not far away from being principal domains. Proposition 6.2.24 [nearly principal] Let R be a Dedekind domain. (a) [a] Let A and B be a fractional ideals of R with B ≤ A. Then A/B is a cyclic R-module. (b) [b] Let A be a fractional ideal of R. Then there exists a, b ∈ A with A = Ra+Rb. Proof: (a) Replacing A and B by kA and kB for a suitable k ∈ R we may assume that B ≤ A ≤ R, Let Q be a finite set of prime ideals in R with A = ∏ P∈Q P aP and B = ∏ P∈Q P bP for some ap, bP ∈ N. Choose xP ∈ P ap \ P ap+1. Observe that P ap+1 + QaQ+1 = R for disctinct P,Q ∈ Q. So by the Chinese Remainder Theorem 2.5.15(e) the exists x ∈ R with x + P ap+1 = xp + P ap+1 for all P ∈ Q. Thus x ∈ ⋂ P∈Q P ap = A and x /∈ P aP +1. Since B ≤ Rx + B, Rx + B = ∏ P∈Q P cP for some cP ∈ N. Since Rx + B ≤ A, cP ≥ aP . Since x /∈ P aP +1, cP ≤ ap. Thus aP = cP for all P ∈ Q and so A = Rx+B. (b) Let 0 6= b ∈ A and put B = Ra. By (a) A/B = Ra + B/B for some a ∈ A. Thus A = Ra+Rb.  6.3 The Jacobson Radical II Lemma 6.3.1 (Nakayama) [nakayama] Let R be a ring and M a non zero finitely gen- erated R-module then J(R)M 6= 0. Let B ⊆ M be minimal with RB = M . Let b ∈ B, then M 6= R(B \ {b} and repplacing M be M/R(B \ {b} we mau assume that M = Rb. Then M ∼= R/AR(b). Let J be maximal left ideal of R with AR(b) ≤ J . Then J(R) +AR(b) ≤ J < R and so also J(R) < M .  Lemma 6.3.2 [jr and inverses] Let R be a ring and x ∈ R. (a) [a] x ∈ J(R) iff rx− 1 has a left inverse for all x ∈ R. (b) [b] x is left invertible in R iff x+ J(R) is left invertible in R/J(R). (c) [c] The J(R) is equal to the right Jacobson radical J(Rop. 146 Chapter 6. Brauer Characters Lemma 6.4.3 [linear independence of characters] (a) [a] (χM |M ∈ Sp) is F-linear independent in FG. (b) [b] (φM |M ∈ Sp) is C-linearly independent in CG̃. Proof: (a) Let fM ∈ F with ∑ fMχM = 0. Pick eM ∈ EndF(M) with trM (eM ) = 1. 2.5.18 there exists aM ∈ FG such that aM acts as eM on N and trivially on N for all M 6= N ∈ Sp. Then 0 = ∑ N∈Sp fNχN (eM ) = fM and so (a) holds. (b) Since all coefficents of φM are in A, φM | M ∈ Sp) is C-linearly independent iff (φM | M ∈ Sp) is Q-linearly independent and iff (φ̌M | M ∈ Sp) is Q-linearly independent. By 6.1.5 (φ̌M )∗ = χM and so by (a) (φ̌M )∗ | M ∈ Sp) is F-linearly independent. So (b) follows from 6.4.2.  Lemma 6.4.4 [existence of a lattice] Let V be an oQ-space and W a finitely generated AI submodule of V with V = QW . Then W is an AI-lattice in V . Proof: Note that W/IIW is a finite dimensional vector space over AI/II = F and so has a basis ui + IIW, 1 ≤ i ≤ n. By 6.4.2 (ui)ni=1 is linearly independent over Q and so also over AI . Let U = Ai〈ui od1 ≤ i ≤ n. Then W = U + IIW . Since II is the unique maximal ideal in AI , II = ( AI). Thus by the Nakayama Lemma 6.3.1 applied to W/U gives W = U . Hence also V = QW = QV 〈ui | 1 ≤ i ≤ n〉  Lemma 6.4.5 [existence of oq lattice] Let E : K be a field extension and M a simple KG-module. If K is algebraicly closed then there exists an G-invarinant K lattice L is M . For any such L, L is a simple KG-module and M ∼= E⊗K L. Proof: Since G is finite there exists a simple KG-submodule L in M . Moreover there is a non-zero EG-linear map α : E⊗K L→M, e⊗ l→ el. Since K is algebraicly closed, E⊗K L is a simple EG-module. The same is true for M and so α is an isomorphism. In particular, any K basis for L is also a E-basis for M and so L is a K-lattice in M . Now let L is any K-lattice in G. If ) 6= N ≤ L is a KG-submodule then EN is a EG-submodule of M . Thus EN = M and dimKN = dimE EN = dimEM = dimK L and so N = L and L is a simple KG-module.  Lemma 6.4.6 [existence of ai lattice] Let M be an CG-module. Then there exists a G-invariant AI-lattice L in M . Section 6.4. A basis for CG̃ 147 Proof: By 6.4.5 there exists a G-invariant Q-lattice V in M . Let X be a Q-basis for V and put L = AIGX. Since G and X are finite, L is finitely AI -generated. Thus by 6.4.4, L is an AI -lattice in V and so also in M .  Lemma 6.4.7 [characters are brauer characters] Let M be an CG-module and L a G-invariant AI-lattice in M . Let M◦ be the FG-module, L/IIL. Then χ∗M = χM◦ and χ̃M = φM◦ Proof: Let B be an AI basis for L, g ∈ G and D the marix for g with respect to B. Then D∗ is the matrix for g with respect to the basis (b+ ILL)b∈B for M◦. Since ηM (g) = det(xI dn −D) we conclude that ηM (g)∗ = ηM◦(g). In particular χM (g)∗ = χM◦(g) and if ηM (g) = ∏n i=1(x− ξi) then ηM◦(g) = ∏n i=1(x− ξ∗i ). So if g ∈ G◦, then χM (g) = φM◦(g). Definition 6.4.8 [def:Irr G] (a) [a] Irr(G) = {χM |M ∈ S} is the set of simple characters of G. (b) [b] IBr(G) = {φM |M ∈ Sp} is the set of simple Brauer characters of G. (c) [c] ZCG̃ := CG̃ ∩ Z(CG) is the set of complex valued class function on G̃. (d) [d] If M be an CG-module and L an G invariant C : AI lattice in M , then M◦ = L/IIL is called a reduction modulo p of M . Theorem 6.4.9 [ibr basis] (a) [a] ZC(G̃) is the C-span of the Brauer characters. (b) [b] IBr(G) is a C-basis forZC(G̃) (c) [c] |S|p = |IBr(G) is the number of p′-conjugacy classes. Proof: (a) Observe that the map˜: Z(CG) → ZC(G̃) is an orthogonal projection and so onto. On the otherhand since Z(CG) is an C -span of the G-characters we conclude from 6.4.7 that the image of˜is conatained in C-span of the Brauer characters. So (a) holds. (b) By 6.1.3(e) every Brauer chacter is a sum of simple Brauet charcters. So by (a), IBr(G) spans ZC(G̃) By 6.4.3(b) IBr(G) is linearly independent over C and so (b) holds. (c) Both IBr(G) and (aC | Cap′ conjugacy class} are bases for ZC(G̃)  Definition 6.4.10 [def:decomposition matrix] (a) [a] D = D(G) = (dphiχ) is the matrix of˜: ZCG → ZCG̃ with respect to Irr(G) and IBr(G). D is called the decompositon matrix of G. 148 Chapter 6. Brauer Characters (b) [b] C = C(G) = (cφψ) is the inverse of Gram matrix of (· | ·) with respect to IBr(G). C is called the Cartan matrix of G. (c) [c] For φ ∈ IBr(G), Φφ = ∑ χ∈Irr(G) dφχχ is called the projective indecomposable character associated to φ. For M ∈ Sp put ΦM = ΦφM . Lemma 6.4.11 [basic decomposition] (a) [a] Let χ ∈ Irr(G). Then χ̃ = ∑ φ∈IBr(G) dφχφ. (b) [z] Let M ∈ S(G), M◦ a p-reduction of M , N ∈ Sp(G) and F a FG-composition series on M . Then dφNχM is the number of factors of |caF isomorphic to N . (c) [b] Let φ, ψ ∈ IBr(G). Then Φφ ∈ ZCG̃ and (Φφ | ψ) = δφψ. So (Φφ | φ ∈ Irr(G)) is the dual basis for ZCG̃. (d) [c] C−1 = ((φ | ψ))φψ (e) [d] C = ((Φφ | Φψ)) is Gram matrix of (cot | ·) with respect to (Φφ | φ ∈ IBr(G). (f) [e] Let φ ∈ Ψ. Then Φφ = Φ̃φ = ∑ ψ∈IBr(G) cφψψ. (g) [f] C = DDT. Proof: (a) Immediate from the definition of D. (b) For N ∈ Sp(G) Let aN be the number of compostion factors of G isomorphic to N . Then by 6.1.3(e), φM◦ = ∑ N∈Sp(G) aNφN . By 6.4.7 φM◦ = χ̃M . So (a) and the linearly independence of IBr(G) implies dφNχM = aN . (c) Follows from 4.1.14 (d) Immediate from the definition of C. (e) and (f) follows from 4.1.16 (g) From (d) and the definition of Φπ: cφψ = ( ∑ χ∈Irr(G) dφχχ | ∑ χ∈Irr(G) dψχχ) = ∑ χ∈Irr(G) dφχdψχ and so (g) holds. Corollary 6.4.12 [dphichi not zero] For each φ ∈ IBr(G), there exists χ ∈ Irr(G) with dφχ6=0. In otherwords, for each M ∈ Sp there exists a M̌ ∈ S such that M is isomorphic to a composition factor of nay p-reduction of M̌ . Proof: Follows from the fact that˜: Z(CG) → ZCG̃ is onto.  Section 6.5. Blocks 151 (a) [a] Let χ ∈ Irr(G) and φ ∈ IBr(G). If dφχ 6= 0 then λφ = λχ. (b) [b] Let B be a block of G then IBr(B) = Irr(B)† and Irr(B) = IBr(B)†. Proof: (a) Let M ∈ S with χ = χM and N ∈ Sp with φ = φN . Let L be an G-invariant AI -lattice in M . Since dφχ6=0, N is isomorphic to FG composition factor of M◦ = L/IIL. Let a ∈ Z(AG). Then a acts as the scalar ωχ(a) on M and on L. Thus a acts as the scalar ωχ(a)∗ = λχ(a∗) on M◦ and on N . Thus λχ(a∗) = λφ(a∗) and (a) holds. (b) φ ∈ IBr(G) with dφχ for some χ ∈ Irr(B) then by (a) φ ∈ B. Thus Irr(B)† ⊆ IBr(B). Conversely if phi ∈ IBr(B) we can choose (by 6.4.12) χ ∈ IBr(G) with dφχ 6= 0. Then by (a) χ ∈ B and so IBr(B) ⊆ Irr(B)†. Thus IBr(B) = Irr(B)†. Similary Irr(B) = IBr(B)†.  Let χ ∈ Irr(G) and φ ∈ IBr(G). Then λχ is defined by ??(??) and λφ by ??(??). If λ = φ then 6.5.3(a) shows that λχ = λφ. Definition 6.5.4 [brauer graph] Let χ, ψ ∈ Irr(G). We say that φ and ψ are linked if there exists φ ∈ IBr(G) with dφχ 6= 0 6= dφψ. The graph on IBr(G) with edges the linked pairs is called the Brauer graph of G. We say χ and ψ are connected if φ and ψ lie in the same connected component of the Brauer graph. Corollary 6.5.5 [blocks and connected component] (a) [a] Let A ⊆ Irr(G). Then A†† consist of all simple characters linked to some element of A. (b) [b] Let A ⊆ Irr(G). Then A is union of connected components of the Brauer graph iff and only if A = A††. (c) [c] If B is a block then Irr(B) is a union of connected components of the Brauer Graph. Proof: (a) Let ψ ∈ Irr(G). Then ψ is linked to some element of A iff there exists χ ∈ A and φ ∈ IBr(G) with dφχ 6= 0 6= dφψ iff there exists φ ∈ A† with dφψ 6= 0 iff ψ ∈ A†† So (a) holds. (b) follows immediately from (a). (c) By 6.5.3 Irr(B)†† = IBr(B)† = Irr(B). 152 Chapter 6. Brauer Characters Proposition 6.5.6 [osima] Let A ⊆ Irr(G) with A = A††. Let x ∈ G̃ and y ∈ G. Then∑ χ∈A χ(x)χ(y) = ∑ φ∈A† φ(x)Φφ(y) Proof: We compute ∑ χ∈A χ(x)χ(y) = ∑ χ∈A  ∑ φ∈IBr(G) dφχφ(x) χ(y) = ∑ χ∈A  ∑ φ∈A† dφχφ(x) χ(y) = ∑ χ∈A† ∑ φ∈A dφχχ(y) φ(x) = ∑ χ∈A†  ∑ φ∈Irr(G) dφχχ(y) φ(x) = ∑ χ∈A† Φφ(y)φ(x)  Corollary 6.5.7 (Weak Block Orthogonality) [weak block orthogonality] Let B be block of G, x ∈ G̃ and y ∈ G \ G̃. Then∑ χ∈Irr(B) χ(x)χ(y) = 0 Since Irr(G)†† = Irr(G) we can apply 6.5.6:∑ χ∈Irr(B) χ(x)χ(y) = ∑ χ∈Irr(B) χ(x)χ(y−1) = ∑ φ∈A† φ(x)Φφ(y−1) Since y−1 6 G̃ 6.4.11(c) implies Φφ(y−1 = 0 and so the Corollary is proved.  Definition 6.5.8 [def:ea] (a) [a] For M ∈ S and χ = χM put eχ = eM ( see 3.1.3(d). (b) [b] For A ⊆ Irr(G), put eA = ∑ χ∈A eχ. Corollary 6.5.9 [ea in ai(tilde g)] Let A ⊆ Irr(G) with A = A††. Then eA ∈ ZAIG̃. Proof: Let χ ∈ A and g ∈ G. By 3.2.12(a), g coefficents of eχ is 1|G|χ(1)χ(x) Let fg be the g-coefficent of eA. Then by 6.5.6 fg = 1 |G| ∑ χ∈A χ(1)χ(x−1) = 1 |G| ∑ φ∈A† φ(1)Φφ(g−1) Section 6.5. Blocks 153 If g /∈ G̃ we conclude that fg = 0 and so (∗) eA ∈ CG̃ Suppose now that g ∈ G̃. Then using 6.5.6 one more time: fg = 1 |G| ∑ χ∈A χ(g−1)χ(1) = 1 |G| ∑ φ∈A† φ(g−1)Φφ(1) = ∑ φ∈A† φ(g−1) Φφ(1) |G| By 6.4.13 Φφ(1)|G| ∈ AI . Also φ(g −1 ∈ A ∈ AI and so fg ∈ Ai. Thus eA ∈ AG. Together with (*) and the fact that eχ is class function we see that the Corollary holds.  Lemma 6.5.10 [unions of blocks] Let A ⊆ Irr(G) with eA ∈ Z(AI(G)). Then A =⋃k i=1 Irr(Bi) for some blocks B1, . . . Bk. Proof: Let χ, ψ ∈ Irr(G). Then ωχ(eψ) = δχψ and so ωχ(eA) = 1 if χ ∈ A and ωχ(eA) = 0 otherwise. By assumption eA ∈ Z(AI(G)) and so λχ(e∗A) = ωχ(eA) and so (∗) χ ∈ A iff λχ(e∗A) = 1 Let B be the block containg χ and ψ ∈ Irr(B). Then λχ(e∗A) = λψ(e∗A) and so by (*), χ ∈ A iff ψ ∈ A.  Theorem 6.5.11 [block=connected components] If B is block, then Irr(B) is con- nected in the Brauer Graph. So the connected components of the Brauer graph are exactly the Irr(B), B a block. Proof: If B is a block then by 6.5.5(c), Irr(B) is the union of connected components. Connversely if A is a connected component then by 6.5.9 eA ∈ Z(AIG) and so by 6.5.10 A is a union of blocks.  Definition 6.5.12 [def:fb] (a) [a] Let B be a block. Then eB = e∗Irr(B) and fB = eIrr(B). (b) [b] Let A be set of blocks. Then eA = ∑ B∈A eB and fA = ∑ BinB fB (c) [c] Let B be block, then FB := FGeB. (d) [d] If A is a set of blocks, then FA = FGeA. (e) [e] Let B be a block then λB = λφ for any φ ∈ IBr(G). 156 Chapter 6. Brauer Characters and so by (1) (3) φFG = ∑ B∈Bl(G) ∑ φ∈IBr(B) Φφ(1)φ Now let B a block. If M is composition factor for FG of FB then eB acts identity on M . So by 6.5.14 φM ∈ B. It follows that (4) φFB = ∑ φ∈IBr(G) dφφ for some dφ ∈ N. Since FG = ∑ B∈Bl(G) FB we conclude (5) φFG = ∑ B∈Bl(G) ∑ φ∈IBr(B) dφφ From (3) and (5) and the linear independence of IBr(G) we get dφ = Φφ(1) for all φ ∈ IBr(G). The lemma now follows from (4) and (2).  6.6 Brauer’s Frist Main Theorem Definition 6.6.1 [def:defect group c] Let C be a conjugacy class of G. (a) [z] A defect group of C is a Sylow p-subgroup of CG(x) for some x ∈ C. (b) [a] Syl(C) is the set of all defect groups of G. (c) [b] We fix gC ∈ C and DC ∈ Sylp(CG(gC)). (d) [d] Let A and B be set of subgroups of G. We write A ≺ B if for all A ∈ A there exists B ∈ B with A ≤ B. (e) [e] Let A be a set subgroups of G. Then CA = {C ∈ C | Syl(C) ≺ A}} and ZA(FG) = F〈aC | C ∈ CA〉. (f) [f] For A ⊆ Z(FG) set CA = {C ∈ C(G) | a(gC) 6= 0 for some a ∈ A}. (g) [g] For A,B,C ∈ C put KABC = {(a, b) ∈ A×B | ab = gC}. Lemma 6.6.2 [trivial zdfg] Let z ∈ Z(FG) and D a set of subgroups of G. Then z ∈ ZD(FG) iff aC ∈ ZD(FG) for all C ∈ Cz and iff Syl(C) ≺ D for all C ∈ Cz. Section 6.6. Brauer’s Frist Main Theorem 157 Proof: Since z = ∑ C∈C(G) z(gC)aC and (aC | C ∈ C(G)) is linearly independent this follows immediately from the definition of ZD(FG).  Lemma 6.6.3 [syl c prec syl a] Let A,B,C ∈ C (a) [a] |KABC | ≡ |{(a, b) ∈ A× B | a, b ∈ CG(DC), ab = gC}| (mod p). (b) [b] If p - |KABC | then Syl(C) ≺ Syl(A). Proof: (a) Observe that CG(gC) acts on KABC by coordinate wise conjugation. All non- trivial orbits of DC on KABC have length divisble by p and so (a) holds. (b) By (a) there exists a ∈ A withDC ∈ CG(a) and soDC ≤ D for someD ∈ Sylp(CG(a). Since G acts transitively on Syl(C), Syl(C) ≺ Syl(A).  Proposition 6.6.4 [zdfg ideal] Let D be set of subgroups of G. Then ZD(FG) is an ideal in G. Proof: Let A,B ∈ C with Syl(A) ≺ D. Then in FG: aAaB = ∑ C∈C |KABC |aC = ∑ C∈C,φ-|KABC | |KABCaC By 6.6.3 Syl(C) ≺ Syl(A) ≺ D whenever p - |KABC |. Then aC ∈ ZD(FG) and so aAaB ∈ ZD(FG).  Definition 6.6.5 [def:fa] (a) [a] G be the set of sets of of subgroups of G. G◦ consist of all A ∈ G such that A,B ∈ A with A ⊆ B implies A = B. (b) [b] If A ∈ G, then max(A) is the set maximal elements of A with respect to inclusion. (c) [c] Let A,B ∈ G. Then A ∧ B := max({A ∩B | A ∈ A, B ∈ B}). (d) [d] Let A, αB ∈ G. The A ∨ B = max(A ∪ B). Lemma 6.6.6 [basis fa] Let A,B,D ∈ G. (a) [a] ≺ is reflexive and transitive. (b) [b] A ≺ maxA and maxA ≺ A. (c) [c] max(A) ∈ G◦ and if A is G-invariant so is maxA. (d) [d] A ≺ B iff max(A) ≺ max(B). 158 Chapter 6. Brauer Characters (e) [e] If all elements in A have the same size, A ∈ G◦. (f) [f] If A is conjugacy class of subgroups of G, then A ∈ G◦. (g) [g] CA = Cmax(A) and ZA(FG) = Zmax(A)(FG). (h) [h] Restricted to G◦, ≺ is a partial ordering. (i) [i] (A ∨ B) ≺ D iff A ≺ D and B ≺ D. (j) [j] D ≺ (A ∧ B) iff D ≺ A and D ≺ B. Proof: (a) Obvious. (b) Clearly maxA ≺ A. Let A ∈ A since G is finite we can choose B ∈ A of maxial size with A ⊆ B. Then B ∈ max(A0 and so A ≺ maxA. (c) If A,B ∈ max(A) with A ⊆ B, then A = B by maximalty of A. (d) Follows from (a) and (b). (e) is obvious. (f) follows from (e). (g) The first statement follows from (d) and the second from the first. (h) Let A,B ∈ A(G) with A ≺ B. Let A ∈ A and choose B ∈ B with A ≤ B. Then choose D ∈ A with B ≤ D. Then A ≤ D and so A = D and A = B. Thus A ⊆ B. By symmetry B ⊆ A. So A = B. (h) now follows from (a). (i) By (d) (A ∨ B) ≺ D iff (A ∪ B) ≺ D and so iff A ≺ D and B ≺ D. (j) By (d) D ≺ (A∧B) iff D ≺ {A∩B | A ∈ A, B ∈ B} and so iff D ≺ A and D ≺ B.  Lemma 6.6.7 [basic zdfg] Let D, E ∈ D◦. (a) [a] If D ≺ E, then CD ⊆ CE and ZD(FG) ≤ ZE(FG). (b) [b] (D ∧ E) ≺ D. (c) [c] CD ∩ CE = CD∧E and ZD(FG) ∩ ZE(FG) = ZD∧E(FG) (d) [d] Let A ⊆ Z(F(G)). Let G◦(A) := {A ∈ G◦ | ZD(FG). Then there exists a unique E ∈ G◦(A) with E ≺ D for all D ∈ G◦(A). We denote this E by Syl(A). (e) [e] If A ⊆ B ⊆ Z(F(G)), then Syl(A) ≺ Syl(B). (f) [f] For all C ∈ C, Syl(aC) = Syl(C) (g) [g] Syl(Z(FG)) = Syl(G) (h) [h] For all A ⊆ Z(F(G)), Syl(A) ≺ Syl(G), that is Syl(A) is a set of p subgroups of G. (i) [i] Let A,B ⊆ Z(FG). Then Syl(A ∪B) = Syl(A) ∨ Syl(B). Section 6.6. Brauer’s Frist Main Theorem 161 Proof: Let C be a defect class of the block B. Then λB(aC) 6= 0 and so aC /∈ J(Z(FB)). Thus by 6.6.14 C ∩ CG(Op(G)) 6= ∅. Since G is transitive on C, C ⊆ CG(Op(G)).  Proposition 6.6.16 [opg in defect group] (a) [a] Op(G) is contained in any defect group of any block of G. (b) [b] If P is a defect group of some block of G and P EG then P = Op(G) (a)Let B be a block, C a defect class of B. By 6.6.15 Op(G) ≤ CG(gC) and so Op(G) ≤ DC . (b) Follows immediateley from (a)  Definition 6.6.17 [def:brauer map] Let P be a p-subgroup. Then BrP : Z(FG) → Z(FCG(P )), a→ a |CG(P ) is called the Brauer map of P . Proposition 6.6.18 [basic brauer map] (a) [a] Let K ⊆ G. Then BrP (aK) = aK∩CG(P ). (b) [b] BrP is an algebra homomophism. (c) [c] If CG(P ) ≤ H ≤ NG(P ) then Im BrP ≤ Z(FH) and so we obtain algebra homomor- phism BrHP : Z(FG) → Z(FH), a ∈ BrP (H) Proof: (a) is obvious. (b) Let A,B ∈ C(G). We need to show that BrP (aAaB) = BrP (aA)BrP (aB). Let g ∈ CG(P ). Then the coeficient of g in BrP (aAaB) is the order of the set {(a, b) ∈ A×B | ab = g} The coefficient of g in BrP (aAaB) is the order of {(a, b) ∈ A×B | a ∈ CG(P ), b ∈ CG(P ), ab = g} Since P centralizes g, P acts on the first set and the second set consists of the fixedpoints of P . So the size of the two sets are equal modulo p and (b) holds. (c) Let α : FG → FCG(P ) be the restriction map. Since CG(P ) E H, α(hah−1) = α(hah−1) for all a ∈ G and all h ∈ H. Hence the same is true for all a ∈ FG, h ∈ H. Thus ImBrP = α(Z(FG)) ≤ Z(FH).  Lemma 6.6.19 [kernel of brauer map] Let P be a p-subgroup of G. 162 Chapter 6. Brauer Characters (a) [a] Let C ∈ C(G). Then C ∩ CG(P ) 6= ∅ iff P ≺ Syl(C). (b) [b] ker BrP = F〈aC | C ∈ C(G), P ⊀ Syl(C)〉 Proof: (a) C ∩ CG(P ) 6= ∅ iff P ≤ CG(g) for some g ∈ C and so iff P ≤ D for some D ∈ Syl(C), that is iff P ≺ Syl(C). (b) Let z = ∑ g∈G z(g)g = ∑ C∈C(G) z(gc)aC ∈ Z(F(G)). Then BrP (z) = 0 iff z(g) = 0 for all g ∈ P , iff z(gc) = 0 for all C ∈ C with C ∩ P 6= ∅ and iff z ∈ F〈aC | C ∩ P = ∅〉. So (a) implies (b).  Proposition 6.6.20 [defect and brauer map] Let B be a block of G and P be a p- subgroup of G. (a) [a] BrP (eB) 6= 0 iff P ≺ Syl(B). (b) [b] P ∈ Syl(B) iff P is p-subgroup maximal with respect to BrP (eB) 6= 0. Proof: (a) By 6.6.19(b), BrP (eP ) 6= 0 iff eB /∈ F〈aC | C ∈ C(G), P ⊀ Syl(C)〉 and so iff P ≺ Syl(C) for some C ∈ C(G) with eB(gC) 6= 0. If P ≺ Syl(B), then by 6.6.13(c), P ≺ Syl(C) for amy defect class C of B. Thus BrP (eB) 6= 0. Conversely suppose BrP (eP ) 6= 0 and let C ∈ C(G) with eB(gC) 6= 0 and P ≺ Syl(C). By 6.6.13(b), Syl(C) ≺ Syl(B) and so (a) is proved. (b) follows immediately from (a).  Definition 6.6.21 [def:lbg] Let H ≤ G and b a block of H. (a) [a] λGb : Z(FG) → F, a→ λb(a |H). (b) [b] If λGb is an algebra homomorphsim, the b G is the unique block of G with λbG = λGb . Lemma 6.6.22 [syl(b) in syl(bg)] Let b be a block of H ≤ G. If bG is defined then Syl(b) ≺ Syl(bG). Proof: Let C be a defect class of B. Then 0 6= λbG(aC) = λGb (aC) = λb(aC∩H). Ot follows that there exists c ∈ C(H) with c ⊆ C and λb(ac) 6= 0. Hence by 6.6.13(a), Syl(b) ≺ Syl(c). Clearly Syl(c) ≺ Syl(C) = Syl(B) and the lemma is proved.  Proposition 6.6.23 [lbg=brplb] Suppose that P is a p-subgroup of G and PCG(P ) ≤ H ≤ NG(P ). (a) [a] λGb = λb ◦ BrP for all blocks b of H. Section 6.6. Brauer’s Frist Main Theorem 163 (b) [b] bG is defined for all blocks b of H. (c) [c] Let B be a block if G and b a block of H. Then B = bG iff λb(BrP (eB)) = 1. (d) [d] Let B be a block. Then BrP (eB) = ∑ {eb | b ∈ Bl(H), bG = B}. (e) [e] Let B be a block of G. Then B = bG for some block b of H iff P ≺ Syl(B). Proof: (a) Let C ∈ (G) we have to show that (∗) λb(aC∩H) = λb(aC∩CG(P )) Since H nomrmalizes C ∩H and C ∩CG(P ). C ∩H \CG(P ) is a union of conjugacy classes of H. Let c ∈ C(H) with c ⊆ C and c ∩ CG(P )∅. Since P ≤ Op(H), CH(Op(H)) ≤ CG(P ) and thus c∩CH(Op(H)) = 1. 6.6.14 implies ac ∈ J(Z(FH)) and so λb(ac) = 0. This implies (*) and so (a) holds. (b) Since both BrP and λb are homomorphism this follows from (a). (c) By (b) λb(BrB(eB) = λbG(eB) = δB,bG . (d) Since BrP is a homomorphism, BrP (eB) is either zero or an idempotent in Z(FH). Hence by 6.5.16(b) ( applied to H Br(eB) = eT for some (possible empty) T ⊆ Bl(H). Let b ∈ Bl(H). The λb(eT ) = 1 if b ∈ T and 0 otherwise. So by (c), T = {b ∈ Bl(G) | B = bG}. (e) By (d) BrP (eB) 6= 0 iff ther exists b ∈ Bl(G) with B = bG. Thus (e) follows from 6.6.20(a).  Definition 6.6.24 [def:G—P] Let P be a p-sugbroups of G. Then C(G|P ) = {C ∈ C(G) | P ∈ Syl(C)} and Bl(G|P ) = {B ∈ Bl(G)midP ∈ Syl(G)}. Proposition 6.6.25 [defect opg] Let B be a block of G with defect group Op(G). Then Syl(C) = {Op(G)} for all C ∈ C(G) with eB(gC) 6= 0 and so eB ∈ C〈aC | C ∈ C(G|Op(G))〉 Proof: Let C ∈ C(G) with eB(gC) 6= 0. Then by 6.6.13(b), Syl(C) ≺ Syl(B) = {Op(G)}. On the otherhand b = B is the unique block of G with B = bG and so by 6.6.23(d), BrOp(G) = eB. It follows that C ≤ CG(Op(G)) and so Op(G) ≺ Syl(C).  Lemma 6.6.26 [first for classes] Let P be a p-subgroup of G. Then the map C(G|P ) → C(NG(P )|P ), C → C ∩ CG(P ) is a well defined bijection. 166 Chapter 6. Brauer Characters Lemma 6.7.3 [p partition] (a) [a] Let 〈h〉 be a finite cyclic group acting on a set Ω. Suppose hp acts fixed-point freely on Ω. Then there exists there exists an < h >-invariant partion of (Ωi)i∈Fp of Ω with hΩi = Ωi+1. (b) [b] If h ≤ H ≤ G with CH(hp) ≤ H, S a ring and w ∈ S[G \H]. If h centralizes w, then there exists wi ∈ S[G \H], i ∈ Fp with hwih−1 = wi+1 and ∑ i∈Fp wi = w. (a) Put H = 〈h〉 act transitively on Ω. Let Ω0 be an orbit for Hp on Ω. Suppose that Ω0 = Ω. Then by the Frattinargument, H = HpCH(ω) and so H/CH(ω) is a p′ group. Thus hp ∈ CH(ω) contrary to the assumptions. Thus Ω0 6= Ω Since Hp EH, H/Hp ∼= Cp acts tranistively on the set of orbits of Hp on Ω. So (a) holds with Ωi = hiΩ0, for i ∈ Fp. (b) Since CG(hp) ≤ H, hp acts fixed-point freely on G \H via conjuagtion. Let Ωi be as in (a) with Ω = G \H and put wi = w |Ωi . Then clearly w = ∑ i∈Fp wi. Now hwi = h(w | Ωi) = hw |hΩi= w |Ωi+1= wi+1 and (b) is proved. Lemma 6.7.4 [eigenvector for h] Let H ≤ G and b a block for G. Suppose that B = bG us defined and that h ∈ H with CG(hp) ∈ H. (a) [a] Let ω ∈ C with ωp = 1. If fB′fb 6= 0, then the exists a unit t in the ring fB′fb · AIG · fB′fb with ht = ωt. (b) [b] If χ ∈ Irr(G) with χ /∈ B. Then χ(hfb) = 0. Proof: (a) Let w be a as in 6.7.2. By 6.7.3(b) theer exists wi ∈ AIG with w = s ∑ i∈Fp wi and hwi = wi+1. By 6.7.2(b), w = fbwfb and so replacing wi by fbwifb we may assume that wi ∈ fb · AIG · fb. Put s = ∑ i∈Fp ω iwi. Then clearly hs = ωs and s ∈ fb · AIG · fb. Put t = fB′s. fB′ ∈ Z(AIG) is a central idempotent, t ∈ fB′fb · AIG · fB′fb and ht = ωt. To complete the proof of (a) we need to show that t is unit in the ring fB′fb · AIG · fB′fb. Since F has no element of multiplicative order p, ω∗ = 1 and so s∗ = ∑ i∈Fp w ∗ i = w ∗ and so by 6.7.2(a), fB′fb)∗ = (fB′w)∗ = (fB′s)∗ = t∗ So 6.3.5 applied with the idempotent f = fB′fb yields that t is a unit in fB′fb·AIG·fB′fb. (b) Let M ∈ S(G) with χ = χM . Put V = fbM . Observe that V that CH submodule of M . Moreover, M = AM (fb)⊕V and fb acts as idV on V . Thus χM (hfb) = χV (fb). Since χ 6∈ B, fBM = 0 and so fB′ act as identity on M and on V . So also fB′fb acts as indentity on V . The V = fB′fbM is a module for the ring fB′fb · AIG · fB′fb If V = 0 clearly (b) holds. So suppose V 6= 0 and so also fB′fb 6= 0. Section 6.7. Brauer’s Second Main Theorem 167 For L be the set of eigenvalues for h on V and for l ∈ L let Vl be the corresponding eigenspace. Then V = ⊕ l∈L Vl. Let ω be a primitive p-root of unity in U and choose t as in (a). Then t is invertible on V . Moreover, if l ∈ L and v ∈ Vl, then htv = hth−1hv = ωtlv = (ωl)tv. Thus tVl ≤ Vtl. In particular tpVl = VtpL = Vl and since tp is invertible, tpVl = Vl and so also tVl = Vtl. T Inparticular < ω〉 acts an L be left multiplication and dimVl = dimVωl. Let L0 be a set of representatoves for the orbits of 〈ω〉 in L. Then χV (h) = ∑ l∈L χVl(h) = ∑ l∈L l dimVl = ∑ l∈L0 ∑p−1 i=0 ω il dimVωil = ∑ l∈L0 (∑p−1 i=0 ω i ) l dimVl = 0  Definition 6.7.5 [def:p-section] Let x ∈ G be a p-element. Then SG(x) = S(x) = {y ∈ G | yp ∈ Gx} is called the p-section if x in G. Lemma 6.7.6 [basic p-section] Let x ∈ G be a p-elemenent and Y a set of representatives for the p′-conjugact classes in CG(x). Then {xy | y ∈ Y } is a set of representaives for the conjugacy classes of G in S(x). Proof: Any s ∈ S(x) is uniquely determined by the pair (sp, sp′). So the lemma follows from 1.1.10  Definition 6.7.7 [def:bx] Let x ∈ G be a p-element and B a block p-block and θ ∈ CG). (a) [a] Let T a block or a set of blocks. Then θT : G→ C | g → θ(fT g). (b) [b] θx : G→ C, x→ θ(xh). (c) [c] Bx = {b ∈ Bl(CG(x))} | bG = B}. Lemma 6.7.8 [fchi selfadjoint] Let T ⊆ Irr(G). Then (a) [a] fT ◦ = fT (b) [b] (afT | b) = (a | bfT ) for all a, b ∈ CG. Proof: By linearity we may assume T = {χ} for some χ ∈ Irr(G). (a) Since χ◦ = chi and fχ = χ(1) |G| χ we have fχ◦ = fχ. (b) By (a) f◦χ = fχ and 3.4.2(c) implies (afχ | b) = (a | bfχ). Lemma 6.7.9 [dual of a block] Let B be a block. (a) [a] B = {ψ | ψ ∈ B} is a block. 168 Chapter 6. Brauer Characters (b) [b] λB(a) = λB(a ◦). (c) [c] fB = fB = f ◦ B. (d) [d] eB = e ◦ B. Proof: (a) and (b): Let ψ ∈ B and M the correspoding module. Then ψ corresponse to M∗. By the definition of the action of a group ring on the dual ρM∗(a) = ρM (a◦)dual. It follows that λψ(a) = λψ(a ◦). Thus λα = λβ iff λα = λb and so (a) and (b) hold. (c): Clearly fB = fB. By 6.7.8, fB = f ◦ T and so (c) holds. (d): Apply ∗ to (c).  Lemma 6.7.10 [theta b] Let T be a block or or a set of blocks and θ ∈ CG. Then θB = θfB. Proof: Let b ∈ G. Then by 6.7.8 θT (b) = θ(fBb) = |G|(θ | fT b) = |G|(θfT | b) = (θfB)(b).  Lemma 6.7.11 [theta fb] Let B be a block. (a) [a] Irr(B) is a basis for CB := CGfB. (b) [b] Both IBr(G) and (Φφ | φ ∈ IBr(G) are a basis for CB̃, where CB̃ := CG̃ ∩ CB. (c) [c] If χ ∈ Irr(B), then χ̃ ∈ FB. (d) [d] For all θ ∈ Z(CG), θ̃fB = θ̃fB and θ̃B = θ̃B. (e) [e] Let θ ∈ Z(CG) and B a block of G. Then θfB = ∑ χ∈Irr(B)(θ | χ)χ. Proof: (a): Let χ ∈ Irr(B). Then χ = |G|φ(1)fχ ∈ CGB and so (a) holds. (b) Let φ ∈ IBr(B). Then by (a) Φψ = ∑ χ∈Irr(B) dφχχ ∈ CB and so (Φφ | φ ∈ IBr(G) is a basis for CB̃. Moreover, φ = ∑ ψ∈IBr(B) (φ | ψ)Φψ ∈ CB and so (b) holds. Section 6.7. Brauer’s Second Main Theorem 171 Theorem 6.7.15 (Brauer’s Second Main Theorem) [second] Let x be a p-element in G and b ∈ Bl(CG(x)). If χ ∈ Irr(G) but χ /∈ Irr(bG), then dxφχ = 0 for all φ ∈ IBr(G). Proof: Follows from 6.7.12(a). Corollary 6.7.16 [chixy] Let x be a p-element in G, y ∈ CG(x) a p′-element, B a block of B and χ ∈ Irr(B). Then χ(xy) = ∑ {dxφχ | b ∈ Bl(CG(x)), B = bG} Proof: This just rephrases 6.7.12(a). Corollary 6.7.17 [gp in defect group] Let B be a block of G, χ ∈ Irr(B) and g ∈ G. If χ(g) 6= 0 then gp is contained in a defect group of B, Proof: Let x = gp, y = gp′ . Since χ(g) = χ(xy) 6= 0, 6.7.16 implies tat there exists b ∈ IBr(G) with B = bG. Since x ∈ Op(CG(x) is contained in any defect group of b, 6.6.22 implies that x is contained a defect group of B.  172 Chapter 6. Brauer Characters Bibliography [Co] M.J. Collins, Representations and characters of finite groups, Cambridge studies in advanced mathematics 22, Cambridge University Press, New York (1990) [Go] D.M. Goldschmidt, Group Characters, Symmetric Functions and the Hecke Algebra, University Lectures Series Volume 4, American Mathematical Society, Providence (1993) [Gr] L.C. Grove, Algebra, Academic Press, New York (1983) [Is] I.M. 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