Modular representation theory of finite groups


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Modular representation theory of finite groups

  1. 1. Chapter 2The Cartan–Brauer TriangleLet G be a finite group. Over any commutative ring R we have the group ring R[G] = ag g : ag ∈ R g∈Gwith addition ag g + bg g = (ag + bg )g g∈G g∈G g∈Gand multiplication ag g bg g = ah bh−1 g g. g∈G g∈G g∈G h∈GWe fix an algebraically closed field k of characteristic p > 0. The representationtheory of G over k is the module theory of the group ring k[G]. This is our primaryobject of study in the following.2.1 The SettingThe main technical tool of our investigation will be a (0, p)-ring R for k which is acomplete local commutative integral domain R such that– the maximal ideal mR ⊆ R is principal,– R/mR = k, and– the field of fractions of R has characteristic zero. jExercise The only ideals of R are mR for j ≥ 0 and {0}.P. Schneider, Modular Representation Theory of Finite Groups, 43DOI 10.1007/978-1-4471-4832-6_2, © Springer-Verlag London 2013
  2. 2. 44 2 The Cartan–Brauer Triangle We note that there must exist an integer e ≥ 1—the ramification index of R—such that Rp = me . There is, in fact, a canonical (0, p)-ring W (k) for k—its ring of RWitt vectors—with the additional property that mW (k) = W (k)p. Let K/K0 be anyfinite extension of the field of fractions K0 of W (k). Then R := a ∈ K : NormK/K0 (a) ∈ W (k)is a (0, p)-ring for k with ramification index equal to [K : K0 ]. Proofs for all of thiscan be found in §3–6 of [9]. In the following we fix a (0, p)-ring R for k. We denote by K the field of fractionsof R and by πR a choice of generator of mR , i.e. mR = RπR . The following threegroup rings are now at our disposal: K[G] ⊆ pr R[G] k[G]such that K[G] = K ⊗R R[G] and k[G] = R[G]/πR R[G].As explained before Proposition 1.7.4 there are the corresponding homomorphismsbetween Grothendieck groups K0 (K[G]) [P ]−→[K⊗R P ] κ ρ K0 (R[G]) K0 (k[G]). [P ]−→[P /πR P ]For the vertical arrow observe that, quite generally for any R[G]-module M, wehave K[G] ⊗R[G] M = K ⊗R R[G] ⊗R[G] M = K ⊗R M.We put RK (G) := R K[G] and Rk (G) := R k[G] .Since K has characteristic zero the group ring K[G] is semisimple, and we have RK (G) = K0 K[G]by Remark 1.7.3.iii. On the other hand, as a finite-dimensional k-vector space thegroup ring k[G] of course is left and right artinian. In particular we have the Cartan
  3. 3. 2.1 The Setting 45homomorphism cG : K0 k[G] −→ Rk (G) [P ] −→ [P ].Hence, so far, there is the diagram of homomorphisms RK (G) Rk (G) κ cG ρ K0 (R[G]) K0 (k[G]).Clearly, R[G] is an R-algebra which is finitely generated as an R-module. Let uscollect some of what we know in this situation.– (Proposition 1.3.6) R[G] is left and right noetherian, and any finitely generated R[G]-module is complete as well as R[G]πR -adically complete.– (Theorem 1.4.7) The Krull–Remak–Schmidt theorem holds for any finitely gen- erated R[G]-module.– (Proposition 1.5.5) 1 ∈ R[G] can be written as a sum of pairwise orthogonal primitive idempotents; the set of all central idempotents in R[G] is finite; 1 is equal to the sum of all primitive idempotents in Z(R[G]); any R[G]-module has a block decomposition.– (Proposition 1.5.7) For any idempotent ε ∈ k[G] there is an idempotent e ∈ R[G] such that ε = e + R[G]πR .– (Proposition 1.5.12) The projection map R[G] −→ k[G] restricts to a bijection between the set of all central idempotents in R[G] and the set of all central idem- potents in k[G]; in particular, the block decomposition of an R[G]-module M re- duces modulo R[G]πR to the block decomposition of the k[G]-module M/πR M.– (Proposition 1.6.10) Any finitely generated R[G]-module M has a projective ∼ = cover P −→ M such that P / Jac(R[G])P − M/ Jac(R[G])M is an isomor- → phism. We emphasize that R[G]πR ⊆ Jac(R[G]) by Lemma 1.3.5.iii. Moreover, theideal Jac(k[G]) = Jac(R[G])/R[G]πR in the left artinian ring k[G] is nilpotent byProposition We apply Proposition 1.7.4 to the rings R[G] and k[G] and wesee that the maps {P } −→ {P /πR P } −→ {P / Jac(R[G])P } induce the commuta-tive diagram of bijections between finite sets R[G] k[G] R[G] = k[G]
  4. 4. 46 2 The Cartan–Brauer Triangleas well as the commutative diagram of isomorphisms ∼ = Z[R[G]] Z[k[G]] ∼ = ∼ = ∼ = K0 (R[G]) K0 (k[G]). ρFor purposes of reference we state the last fact as a proposition. ∼ =Proposition 2.1.1 The map ρ : K0 (R[G]) − K0 (k[G]) is an isomorphism; its in- →verse is given by sending [M] to the class of a projective cover of M as an R[G]-module. We define the composed map ρ −1 κ eG : K0 k[G] − → K0 R[G] − K0 K[G] = RK (G). − →Remark 2.1.2 Any finitely generated projective R-module is free.Proof Since R is an integral domain 1 is the only idempotent in R. Hence the freeR-module R is indecomposable by Corollary 1.5.2. On the other hand, according toProposition 1.7.4.i the map ˜ ˆ R −→ k {P } −→ {P /πR P }is bijective. Obviously, k is up to isomorphism the only simple k-module. Hence R isup to isomorphism the only finitely generated indecomposable projective R-module.An arbitrary finitely generated projective R-module P , by Lemma 1.1.6, is a fi-nite direct sum of indecomposable ones. It follows that P must be isomorphic tosome R n .2.2 The TriangleWe already have the two sides RK (G) Rk (G) eG cG K0 (k[G])of the triangle. To construct the third side we first introduce the following notion.
  5. 5. 2.2 The Triangle 47Definition Let V be a finite-dimensional K-vector space; a lattice L in V is anR-submodule L ⊆ V for which there exists a K-basis e1 , . . . , ed of V such that L = Re1 + · · · + Red . Obviously, any lattice is free as an R-module. Furthermore, with L also aL, forany a ∈ K × , is a lattice in V .Lemma 2.2.1 i. Let L be an R-submodule of a K-vector space V ; if L is finitely generated then L is free. ii. Let L ⊆ V be an R-submodule of a finite-dimensional K-vector space V ; if L is finitely generated as an R-module and L generates V as a K-vector space then L is a lattice in V .iii. For any two lattices L and L in V there is an integer m ≥ 0 such that πR L ⊆ L . mProof i. and ii. Let d ≥ 0 be the smallest integer such that the R-module L has dgenerators e1 , . . . , ed . The R-module homomorphism R d −→ L (a1 , . . . , ad ) −→ a1 e1 + · · · + ad edis surjective. Suppose that (a1 , . . . , ad ) = 0 is an element in its kernel. Since at leastone ai is nonzero the integer j := max j ≥ 0 : a1 , . . . , ad ∈ mRis defined. Then ai = πR bi with bi ∈ R, and bi0 ∈ R × for at least one index1 ≤ i0 ≤ d. Computing in the vector space V we have 0 = a1 e1 + · · · + ad ed = πR (b1 e1 + · · · + bd ed )and hence b1 e1 + · · · + bd ed = 0. −1But the latter equation implies ei0 = − i=i0 bi0 bi ei ∈ i=i0 Rei , which is a con-tradiction to the minimality of d. It follows that the above map is an isomorphism.This proves i. and, in particular, that L = Re1 + · · · + Red .For ii. it therefore suffices to show that e1 , . . . , ed , under the additional assumptionthat L generates V , is a K-basis of V . This assumption immediately guarantees
  6. 6. 48 2 The Cartan–Brauer Trianglethat the e1 , . . . , ed generate the K-vector space V . To show that they are K-linearlyindependent let c1 e1 + · · · + cd ed = 0 with c1 , . . . , cd ∈ K. jWe find a sufficiently large j ≥ 0 such that ai := πR ci ∈ R for any 1 ≤ i ≤ d. Then 0 = πR · 0 = πR (c1 e1 + · · · + cd ed ) = a1 e1 + · · · + ad ed .By what we have shown above we must have ai = 0 and hence ci = 0 for any1 ≤ i ≤ d. iii. Let e1 , . . . , ed and f1 , . . . , fd be K-bases of V such that L = Re1 + · · · + Red and L = Rf1 + · · · + Rfd .We write ej = c1j f1 + · · · + cdj fd with cij ∈ K,and we choose an integer m ≥ 0 such that πR cij ∈ R m for any 1 ≤ i, j ≤ d.It follows that πR ej ∈ Rf1 + · · · + Rfd = L for any 1 ≤ j ≤ d and hence that mπRmL ⊆ L . Suppose that V is a finitely generated K[G]-module. Then V is finite-dimension-al as a K-vector space. A lattice L in V is called G-invariant if we have gL ⊆ Lfor any g ∈ G. In particular, L is a finitely generated R[G]-submodule of V , andL/πR L is a k[G]-module of finite length.Lemma 2.2.2 Any finitely generated K[G]-module V contains a lattice which isG-invariant.Proof We choose a basis e1 , . . . , ed of the K-vector space V . Then L := Re1 +· · · + Red is a lattice in V . We define the R[G]-submodule L := gL g∈Gof V . With L also L generates V as a K-vector space. Moreover, L is finitelygenerated by the set {gei : 1 ≤ i ≤ d, g ∈ G} as an R-module. Therefore, byLemma 2.2.1.ii, L is a G-invariant lattice in V . Whereas the K[G]-module V always is projective by Remark 1.7.3 a G-invariantlattice L in V need not to be projective as an R[G]-module. We will encounter anexample of this later on.
  7. 7. 2.2 The Triangle 49Theorem 2.2.3 Let L and L be two G-invariant lattices in the finitely generatedK[G]-module V ; we then have [L/πR L] = L /πR L in Rk (G).Proof We begin by observing that, for any a ∈ K × , the map ∼ = L/πR L −→ (aL)/πR (aL) x + πR L −→ ax + πR (aL)is an isomorphism of k[G]-modules, and hence [L/πR L] = (aL)/πR (aL) in Rk (G). mBy applying Lemma 2.2.1.iii (and replacing L by πR L for some sufficiently largem ≥ 0) we therefore may assume that L ⊆ L . By applying Lemma 2.2.1.iii againto L and L (while interchanging their roles) we find an integer n ≥ 0 such that πR L ⊆ L ⊆ L . nWe now proceed by induction with respect to n. If n = 1 we have the two exactsequences of k[G]-modules 0 −→ L/πR L −→ L /πR L −→ L /L −→ 0and 0 −→ πR L /πR L −→ L/πR L −→ L/πR L −→ 0.It follows that L /πR L = L/πR L + L /L = L/πR L + πR L /πR L = L/πR L + [L/πR L] − L/πR L = [L/πR L]in Rk (G). For n ≥ 2 we consider the R[G]-submodule M := πR L + L. n−1It is a G-invariant lattice in V by Lemma 2.2.1.ii and satisfies πR L ⊆ M ⊆ L n−1 and πR M ⊆ L ⊆ M.Applying the case n = 1 to L and M we obtain [M/πR M] = [L/πR L]. The induc-tion hypothesis for M and L gives [L /πR L ] = [M/πR M].
  8. 8. 50 2 The Cartan–Brauer Triangle The above lemma and theorem imply that Z[MK[G] ] −→ Rk (G) {V } −→ [L/πR L],where L is any G-invariant lattice in V , is a well-defined homomorphism. If V1and V2 are two finitely generated K[G]-modules and L1 ⊆ V1 and L2 ⊆ V2 areG-invariant lattices then L1 ⊕ L2 is a G-invariant lattice in V1 ⊕ V2 and (L1 ⊕ L2 )/πR (L1 ⊕ L2 ) = [L1 /πR L1 ⊕ L2 /πR L2 ] = [L1 /πR L1 ] + [L2 /πR L2 ].It follows that the subgroup Rel ⊆ Z[MK[G] ] lies in the kernel of the above map sothat we obtain the homomorphism dG : RK (G) −→ Rk (G) [V ] −→ [L/πR L].It is called the decomposition homomorphism of G. The Cartan–Brauer triangle isthe diagram dG RK (G) Rk (G) eG cG K0 (k[G]).Lemma 2.2.4 The Cartan–Brauer triangle is commutative.Proof Let P be a finitely generated projective R[G]-module. We have to show that dG κ [P ] = cG ρ [P ]holds true. By definition the right-hand side is equal to [P /πR P ] ∈ Rk (G). More-over, κ([P ]) = [K ⊗R P ] ∈ RK (G). According to Proposition 1.6.4 the R(G)-module P is a direct summand of a free R[G]-module. But R[G] and hence any freeR[G]-module also is free as an R-module. We see that P as an R-module is finitelygenerated projective and hence free by Remark 2.1.2. We conclude that P ∼ R d is a =G-invariant lattice in the K[G]-module K ⊗R P ∼ K ⊗R R d = (K ⊗R R)d = K d , =and we obtain dG (κ([P ])) = dG ([K ⊗R P ]) = [P /πR P ]. Let us consider two “extreme” situations where the maps in the Cartan–Brauertriangle can be determined completely. First we look at the case where p does notdivide the order |G| of the group G. Then k[G] is semisimple. Hence we havek[G] = k[G] and K0 (k[G]) = Rk (G) by Remark 1.7.3. The map cG , in particular,is the identity.
  9. 9. 2.2 The Triangle 51Proposition 2.2.5 If p |G| then any R[G]-module M which is projective as anR-module also is projective as an R[G]-module.Proof We consider any “test diagram” of R[G]-modules M α β L N 0.Viewing this as a “test diagram” of R-modules our second assumption ensures theexistence of an R-module homomorphism α0 : M −→ L such that β ◦ α0 = α. Since|G| is a unit in R by our first assumption, we may define a new R-module homo-morphism α : M −→ L by ˜ α(x) := |G|−1 ˜ gα0 g −1 x for any x ∈ M. g∈G ˜One easily checks that α satisfies α(hx) = hα(x) ˜ ˜ for any h ∈ G and any x ∈ M. ˜This means that α is, in fact, an R[G]-module homomorphism. Moreover, we com-pute β α(x) = |G|−1 ˜ gβ α0 g −1 x = |G|−1 gα g −1 x g∈G g∈G = |G|−1 gg −1 α(x) = α(x). g∈GCorollary 2.2.6 If p |G| then all three maps in the Cartan–Brauer triangle areisomorphisms; more precisely, we have the triangle of bijections K[G] k[G] {P }−→{K⊗R P } {P }−→{P /πR P } R[G].Proof We already have remarked that cG is the identity. Hence it suffices to showthat the map κ : K0 (R[G]) −→ RK (G) is surjective. Let V be any finitely generatedK[G]-module. By Lemma 2.2.2 we find a G-invariant lattice L in V . It satisfiesV = K ⊗R L by definition. Proposition 2.2.5 implies that L is a finitely generated
  10. 10. 52 2 The Cartan–Brauer Triangleprojective R[G]-module. We conclude that [L] ∈ K0 (R[G]) with κ([L]) = [V ].This argument in fact shows that the map MR[G] / ∼ −→ MK[G] / ∼ = = {P } −→ {K ⊗R P }is surjective. Let P and Q be two finitely generated projective R[G]-modules suchthat K ⊗R P ∼ K ⊗R Q as K[G]-modules. The commutativity of the Cartan–Brauer =triangle then implies that [P /πR P ] = dG [K ⊗R P ] = dG [K ⊗R Q] = [Q/πR Q].Let P = P1 ⊕ · · · ⊕ Ps and Q = Q1 ⊕ · · · ⊕ Qt be decompositions into indecom-posable submodules. Then s t P /πR P = Pi /πR Pi and Q/πR Q = Qj /πR Qj i=1 j =1are decompositions into simple submodules. Using Proposition 1.7.1 the identity s t [Pi /πR Pi ] = [P /πR P ] = [Q/πR Q] = [Qj /πR Qj ] i=1 j =1implies that s = t and that there is a permutation σ of {1, . . . , s} such that Qj /πR Qj ∼ Pσ (j ) /πR Pσ (j ) = for any 1 ≤ j ≤ sas k[G]-modules. Applying Proposition 1.7.4.i we obtain Qj ∼ Pσ (j ) = for any 1 ≤ j ≤ sas R[G]-modules. It follows that P ∼ Q as R[G]-modules. Hence the above map =between sets of isomorphism classes is bijective. Obviously, if K ⊗R P is in-decomposable (i.e. simple) then P was indecomposable. Vice versa, if P is in-decomposable then the above reasoning says that P /πR P is simple. Because of[P /πR P ] = dG ([K ⊗R P ]) it follows that K ⊗R P must be indecomposable. The second case is where G is a p-group. For a general group G we have thering homomorphism k[G] −→ k ag g −→ ag g∈G g∈G
  11. 11. 2.2 The Triangle 53which is called the augmentation of k[G]. It makes k into a simple k[G]-modulewhich is called the trivial k[G]-module. Its kernel is the augmentation ideal Ik [G] := ag g ∈ k[G] : ag = 0 . g∈G g∈GProposition 2.2.7 If G is a p-group then we have Jac(k[G]) = Ik [G]; in particular,k[G] is a local ring and the trivial k[G]-module is, up to isomorphism, the onlysimple k[G]-module.Proof We will prove by induction with respect to the order |G| = p n of G that thetrivial module, up to isomorphism, is the only simple k[G]-module. There is nothingto prove if n = 0. We therefore suppose that n ≥ 1. Note that we have n n n gp = 1 and hence (g − 1)p = g p − 1 = 1 − 1 = 0for any g ∈ G. The center of a nontrivial p-group is nontrivial. Let g0 = 1 be acentral element in G. We now consider any simple k[G]-module M and we denoteby π : k[G] −→ Endk (M) the corresponding ring homomorphism. Then pn n n π(g0 ) − idM = π(g0 − 1)p = π (g0 − 1)p = π(0) = 0.Since g0 is central (π(g0 ) − idM )(M) is a k[G]-submodule of M. But M is simple.Hence (π(g0 ) − idM )(M) = {0} or = M. The latter would inductively imply that n(π(g0 ) − idM )p (M) = M = {0} which is contradiction. We obtain π(g0 ) = idMwhich means that the cyclic subgroup g0 is contained in the kernel of the grouphomomorphism π : G −→ Autk (M). Hence we have a commutative diagram ofgroup homomorphisms π G Autk (M) pr π G/ g0 .We conclude that M already is a simple k[G/ g0 ]-module and therefore is thetrivial module by the induction hypothesis. The identity Jac(k[G]) = Ik [G] now follows from the definition of the Jacobsonradical, and Proposition 1.4.1 implies that k[G] is a local ring. Suppose that G is a p-group. Then Propositions 2.2.7 and 1.7.1 together implythat the map ∼ = Z −→ Rk (G) m −→ m[k]
  12. 12. 54 2 The Cartan–Brauer Triangleis an isomorphism. To compute the inverse map let M be a k[G]-module of finitelength, and let {0} = M0 M1 · · · Mt = M be a composition series. We musthave Mi /Mi−1 ∼ k for any 1 ≤ i ≤ t . It follows that = t t [M] = [Mi /Mi−1 ] = t · [k] and dimk M = dimk Mi /Mi−1 = t. i=1 i=1Hence the inverse map is given by [M] −→ dimk M.Furthermore, from Proposition 1.7.4 we obtain that ∼ = Z −→ K0 R[G] m −→ m R[G]is an isomorphism. Because of dimk k[G] = |G| the Cartan homomorphism, underthese identifications, becomes the map cG : Z −→ Z m −→ m · |G|.For any finitely generated K[G]-module V and any G-invariant lattice L in V wehave dimK V = dimk L/πR L. Hence the decomposition homomorphism becomes dG : RK (G) −→ Z [V ] −→ dimK V .Altogether the Cartan–Brauer triangle of a p-group G is of the form [V ]−→dimK V RK (G) Z 1−→[K[G]] ·|G| Z.We also see that the trivial K[G]-module K has the G-invariant lattice R whichcannot be projective as an R[G]-module if G = {1}. Before we can establish the finer properties of the Cartan–Brauer triangle weneed to develop the theory of induction.2.3 The Ring Structure of RF (G), and InductionIn this section we let F be an arbitrary field, and we consider the group ring F [G]and its Grothendieck group RF (G) := R(F [G]).
  13. 13. 2.3 The Ring Structure of RF (G), and Induction 55 Let V and W be two (finitely generated) F [G]-modules. The group G acts onthe tensor product V ⊗F W by g(v ⊗ w) := gv ⊗ gw for v ∈ V and w ∈ W.In this way V ⊗F W becomes a (finitely generated) F [G]-module, and we obtainthe multiplication map Z[MF [G] ] × Z[MF [G] ] −→ Z[MF [G] ] {V }, {W } −→ {V ⊗F W }.Since the tensor product, up to isomorphism, is associative and commutative thismultiplication makes Z[MF [G] ] into a commutative ring. Its unit element is theisomorphism class {F } of the trivial F [G]-module.Remark 2.3.1 The subgroup Rel is an ideal in the ring Z[MF [G] ].Proof Let V be a (finitely generated) F [G]-module and let α β 0 −→ L − M − N −→ 0 → →be a short exact sequence of F [G]-modules. We claim that the sequence idV ⊗α idV ⊗β 0 −→ V ⊗F L − − → V ⊗F M − − → V ⊗F N −→ 0 −− −−is exact as well. This shows that the subgroup Rel is preserved under multiplicationby {V }. The exactness in question is purely a problem about F -vector spaces. But asvector spaces we have M ∼ L ⊕ N and hence V ⊗F M ∼ (V ⊗F L) ⊕ (V ⊗F N ). = = It follows that RF (G) naturally is a commutative ring with unit element [F ] suchthat [V ] · [W ] = [V ⊗F W ]. Let H ⊆ G be a subgroup. Then F [H ] ⊆ F [G] is a subring (with the same unitelement). Any F [G]-module V , by restriction of scalars, can be viewed as an F [H ]-module. If V is finitely generated as an F [G]-module then V is a finite-dimensionalF -vector space and, in particular, is finitely generated as an F [H ]-module. Hencewe have the ring homomorphism resG : RF (G) −→ RF (H ) H [V ] −→ [V ].On the other hand, for any F [H ]-module W we have, by base extension, the F [G]-module F [G] ⊗F [H ] W . Obviously, the latter is finitely generated over F [G] if the
  14. 14. 56 2 The Cartan–Brauer Triangleformer was finitely generated over F [H ]. The first Frobenius reciprocity says that ∼ = HomF [G] F [G] ⊗F [H ] W, V −→ HomF [H ] (W, V ) α −→ w −→ α(1 ⊗ w)is an F -linear isomorphism for any F [H ]-module W and any F [G]-module V . α βRemark 2.3.2 For any short exact sequence 0 −→ L − M − N −→ 0 of F [H ]- → →modules the sequence of F [G]-modules idF [G] ⊗α idF [G] ⊗β 0 → F [G] ⊗F [H ] L − − − → F [G] ⊗F [H ] M − − − → F [G] ⊗F [H ] N → 0 −−− −−−is exact as well.Proof Let g1 , . . . , gr ∈ G be a set of representatives for the left cosets of H in G.Then g1 , . . . , gr also is a basis of the free right F [H ]-module F [G]. It follows that,for any F [H ]-module W , the map ∼ = W r −→ F [G] ⊗F [H ] W (w1 , . . . , wr ) −→ g1 ⊗ w1 + · · · + gr ⊗ wris an F -linear isomorphism. We see that, as a sequence of F -vector spaces, thesequence in question is just the r-fold direct sum of the original exact sequencewith itself.Remark 2.3.3 For any F [H ]-module W , if the F [G]-module F [G] ⊗F [H ] W issimple then W is a simple F [H ]-module.Proof Let W ⊆ W be any F [H ]-submodule. Then F [G] ⊗F [H ] W is an F [G]-submodule of F [G] ⊗F [H ] W by Remark 2.3.2. Since the latter is simple we musthave F [G] ⊗F [H ] W = {0} or = F [G] ⊗F [H ] W . Comparing dimensions using theargument in the proof of Remark 2.3.2 we obtain [G : H ] · dimF W = dimF F [G] ⊗F [H ] W = 0 or = dimF F [G] ⊗F [H ] W = [G : H ] · dimF W.We see that dimF W = 0 or = dimF W and therefore that W = {0} or = W . Thisproves that W is a simple F [H ]-module. It follows that the map Z[MF [H ] ] −→ Z[MF [G] ] {W } −→ F [G] ⊗F [H ] W
  15. 15. 2.3 The Ring Structure of RF (G), and Induction 57preserves the subgroups Rel in both sides and therefore induces an additive homo-morphism indG : H RF (H ) −→ RF (G) [W ] −→ F [G] ⊗F [H ] W(which is not multiplicative!).Proposition 2.3.4 We have indG (y) · x = indG y · resG (x) H H H for any x ∈ RF (G) and y ∈ RF (H ).Proof It suffices to show that, for any F [G]-module V and any F [H ]-module W ,we have an isomorphism of F [G]-modules F [G] ⊗F [H ] W ⊗F V ∼ F [G] ⊗F [H ] (W ⊗F V ). =One checks (exercise!) that such an isomorphism is given by (g ⊗ w) ⊗ v −→ g ⊗ w ⊗ g −1 v .Corollary 2.3.5 The image of indG : RF (H ) −→ RF (G) is an ideal in RF (G). H We also mention the obvious transitivity relations resH ◦ resG = resG H H H and indG ◦ indH = indG H H Hfor any chain of subgroups H ⊆ H ⊆ G. An alternative way to look at induction is the following. Let W be any F [H ]-module. Then IndG (W ) := φ : G −→ W : φ(gh) = h−1 φ(g) for any g ∈ G, h ∈ H Hequipped with the left translation action of G given by g φ g := φ g −1 gis an F [G]-module called the module induced from W . But, in fact, the map ∼ = F [G] ⊗F [H ] W −→ IndG (W ) H ag g ⊗ w −→ φ g := ag h hw g∈G h∈His an isomorphism of F [G]-modules. This leads to the second Frobenius reciprocityisomorphism ∼ = HomF [G] V , IndG (W ) −→ HomF [H ] (V , W ) H α −→ v −→ α(v)(1) .
  16. 16. 58 2 The Cartan–Brauer Triangle We also need to recall the character theory of G in the semisimple case. For thiswe assume for the rest of this section that the order of G is prime to the characteristicof the field F . Any finitely generated F [G]-module V is a finite-dimensional F -vector space. Hence we may introduce the function χV : G −→ F g g −→ tr(g; V ) = trace of V − V →which is called the character of V . It depends only on the isomorphism class of V .Characters are class functions on G, i.e. they are constant on each conjugacy classof G. For any two finitely generated F [G]-modules V1 and V2 we have χV1 ⊕V2 = χV1 + χV2 and χV1 ⊗F V2 = χV1 · χV2 .Let Cl(G, F ) denote the F -vector space of all class functions G −→ F . By point-wise multiplication of functions it is a commutative F -algebra. The above identitiesimply that the map Tr : RF (G) −→ Cl(G, F ) [V ] −→ χVis a ring homomorphism.Definition The field F is called a splitting field for G if, for any simple F [G]-module V , we have EndF [G] (V ) = F . If F is algebraically closed then it is a splitting field for G.Theorem 2.3.6 i. If the field F has characteristic zero then the characters {χV : {V } ∈ F [G]} are F -linearly independent. ii. If F is a splitting field for G then the characters {χV : {V } ∈ F [G]} form a basis of the F -vector space Cl(G, F ).iii. If F has characteristic zero then two finitely generated F [G]-modules V1 and V2 are isomorphic if and only if χV1 = χV2 holds true.Corollary 2.3.7 i. If F has characteristic zero then the map Tr is injective.ii. If F is a splitting field for G then the map Tr induces an isomorphism of F -algebras ∼ = F ⊗Z RF (G) −→ Cl(G, F ).
  17. 17. 2.4 The Burnside Ring 59iii. If F has characteristic zero then the map MF [G] / ∼ −→ Cl(G, F ) = {V } −→ χV is injective.Proof For i. and ii. use Proposition The Burnside RingA G-set X is a set equipped with a G-action G × X −→ X (g, x) −→ gxsuch that 1x = x and g(hx) = (gh)x for any g, h ∈ G and any x ∈ X.Let X and Y be two G-sets. Their disjoint union X ∪ Y is a G-set in an obviousway. But also their cartesian product X × Y is a G-set with respect to g(x, y) := (gx, gy) for (x, y) ∈ X × Y.We will call X and Y isomorphic if there is a bijective map α : X −→ Y such thatα(gx) = gα(x) for any g ∈ G and x ∈ X. Let SG denote the set of all isomorphism classes {X} of finite G-sets X. In thefree abelian group Z[SG ] we consider the subgroup Rel generated by all elementsof the form {X ∪ Y } − {X} − {Y } for any two finite G-sets X and Y .We define the factor group B(G) := Z[SG ]/ Rel,and we let [X] ∈ B(G) denote the image of the isomorphism class {X}. In fact, themap Z[SG ] × Z[SG ] −→ Z[SG ] {X}, {Y } −→ {X × Y }makes Z[SG ] into a commutative ring in which the unit element is the isomorphismclass of the G-set with one point. Because of (X1 ∪ X2 ) × Y = (X1 × Y ) ∪ (X2 × Y )
  18. 18. 60 2 The Cartan–Brauer Trianglethe subgroup Rel is an ideal in Z[SG ]. We see that B(G) is a commutative ring. Itis called the Burnside ring of G. Two elements x, y in a G-set X are called equivalent if there is a g ∈ G suchthat y = gx. This defines an equivalence relation on X. The equivalence classesare called G-orbits. They are of the form Gx = {gx : g ∈ G} for some x ∈ X.A nonempty G-set which consists of a single G-orbit is called simple (or transi-tive or a principal homogeneous space). The decomposition of an arbitrary G-set Xinto its G-orbits is the unique decomposition of X into simple G-sets. In particular,the only G-subsets of a simple G-set Y are Y and ∅. We let SG denote the set ofisomorphism classes of simple G-sets. ∼ =Lemma 2.4.1 Z[SG ] −→ B(G).Proof If X = Y1 ∪ · · · ∪ Yn is the decomposition of X into its G-orbits then we put π {X} := {Y1 } + · · · + {Yn }.This defines an endomorphism π of Z[SG ] which is idempotent with im(π) =Z[SG ]. It is rather clear that Rel ⊆ ker(π). Moreover, using the identities [Y1 ] + [Y2 ] = [Y1 ∪ Y2 ], [Y1 ∪ Y2 ] + [Y3 ] = [Y1 ∪ Y2 ∪ Y3 ], ..., [Y1 ∪ · · · ∪ Yn−1 ] + [Yn ] = [X]we see that [X] = [Y1 ] + · · · + [Yn ]and hence that {X} − π({X}) ∈ Rel. As in the proof of Proposition 1.7.1 these threefacts together imply Z[SG ] = Z[SG ] ⊕ Rel . For any subgroup H ⊆ G the coset space G/H is a simple G-set with respect to G × G/H −→ G/H g, g H −→ gg H.Simple G-sets of this form are called standard G-sets.Remark 2.4.2 Each simple G-set X is isomorphic to some standard G-set G/H .Proof We fix a point x ∈ X. Let Gx := {g ∈ G : gx = x} be the stabilizer of x in G.Then G/Gx −→ X gGx −→ gxis an isomorphism.
  19. 19. 2.4 The Burnside Ring 61 It follows that the set SG is finite.Lemma 2.4.3 Two standard G-sets G/H1 and G/H2 are isomorphic if and only if −1there is a g0 ∈ G such that g0 H1 g0 = H2 . −1Proof If g0 H1 g0 = H2 then G/H1 −→ G/H2 gH1 −→ gg0 H2is an isomorphism of G-sets. Vice versa, let α: G/H1 −→ G/H2be an isomorphism of G-sets. We have α(1H1 ) = g0 H2 for some g0 ∈ G and then g0 H2 = α(1H1 ) = α(h1 H1 ) = h1 α(1H1 ) = h1 g0 H2 −1for any h1 ∈ H1 . This implies g0 H1 g0 ⊆ H2 . On the other hand −1 −1 g0 H1 = α −1 (1H2 ) = α −1 (h2 H2 ) = h2 α −1 (1H2 ) = h2 g0 H1 −1for any h2 ∈ H2 which implies g0 H2 g0 ⊆ H1 .Exercise 2.4.4 Let G/H1 and G/H2 be two standard G-sets; we then have [G/H1 ] · [G/H2 ] = G/H1 ∩ gH2 g −1 in B(G) g∈H1 G/H2where H1 G/H2 denotes the space of double cosets H1 gH2 in G. Let F again be an arbitrary field. For any finite set X we have the finite-dimensional F -vector space F [X] := ax x : ax ∈ F x∈X“with basis X”. Suppose that X is a finite G-set. Then the group G acts on F [X] by g ax x := ax gx = ag −1 x x. x∈X x∈X x∈XIn this way F [X] becomes a finitely generated F [G]-module (called a permutationmodule). If α : X −→ Y is an isomorphism of finite G-sets then ∼ = α: F [X] −→ F [Y ] ˜ ax x −→ ax α(x) = aα −1 (y) y x∈X x∈X y∈Y
  20. 20. 62 2 The Cartan–Brauer Triangleis an isomorphism of F [G]-modules. It follows that the map SG −→ MF [G] / ∼ = {X} −→ F [X]is well defined. We obviously have F [X1 ∪ X2 ] = F [X1 ] ⊕ F [X2 ]for any two finite G-sets X1 and X2 . Hence the above map respects the subgroupsRel in both sides and induces a group homomorphism b: B(G) −→ RF (G) [X] −→ F [X] .Remark There is a third interesting Grothendieck group for the ring F [G] which isthe factor group AF (G) := Z[MF [G] ]/ Rel⊕with respect to the subgroup Rel⊕ generated by all elements of the form {M ⊕ N } − {M} − {N }where M and N are arbitrary finitely generated F [G]-modules. We note that Rel⊕ ⊆Rel. The above map b is the composite of the maps pr B(G) −→ AF (G) − RF (G) → [X] −→ F [X] −→ F [X] .Remark 2.4.5 For any two finite G-sets X1 and X2 we have F [X1 × X2 ] ∼ F [X1 ] ⊗F F [X2 ] =as F [G]-modules.Proof The vectors (x1 , x2 ), resp. x1 ⊗ x2 , for x1 ∈ X1 and x2 ∈ X2 , form an F -basisof the left-, resp. right-, hand side. Hence there is a unique F -linear isomorphismmapping (x1 , x2 ) to x1 ⊗ x2 . Because of g(x1 , x2 ) = (gx1 , gx2 ) −→ gx1 ⊗ gx2 = g(x1 ⊗ x2 ),for any g ∈ G, this map is an F [G]-module isomorphism. It follows that the map b: B(G) −→ RF (G)
  21. 21. 2.4 The Burnside Ring 63is a ring homomorphism. Note that the unit element [G/G] in B(G) is mapped tothe class [F ] of the trivial module F which is the unit element in RF (G).Lemma 2.4.6i. For any standard G-set G/H we have b [G/H ] = indG (1) H where 1 on the right-hand side denotes the unit element of RF (H ).ii. For any finite G-set X we have tr g; F [X] = {x ∈ X : gx = x} ∈ F for any g ∈ G.Proof i. Let F be the trivial F [H ]-module. It suffices to establish an isomorphism F [G/H ] ∼ F [G] ⊗F [H ] F. =For this purpose we consider the F -bilinear map β: F [G] × F −→ F [G/H ] ag g, a −→ aag gH. g∈G g∈GBecause of β(gh, a) = ghH = gH = β(g, a) = β(g, ha)for any h ∈ H the map β is F [H ]-balanced and therefore induces a well-definedF -linear map ˜ β: F [G] ⊗F [H ] F −→ F [G/H ] ag g ⊗ a −→ aag gH. g∈G g∈GAs discussed in the proof of Remark 2.3.2 we have F [G] ⊗F [H ] F ∼ F [G:H ] as F - = ˜vector spaces. Hence β is a map between F -vector spaces of the same dimension.It obviously is surjective and therefore bijective. Finally the identity ˜ β g ag g ⊗ a ˜ =β ag g g ⊗ a = aag g gH g∈G g∈G g∈G =g aag gH ˜ =g β ag g ⊗ a g∈G g∈G ˜for any g ∈ G shows that β is an isomorphism of F [G]-modules.
  22. 22. 64 2 The Cartan–Brauer Triangle g· ii. The matrix (ax,y )x,y of the F -linear map F [X] − F [X] with respect to the →basis X is given by the equations gy = ax,y x. x∈XBut gy ∈ X and hence 1 if x = gy ax,y = 0 otherwise.It follows that tr g; F [X] = ax,x = 1 = {x ∈ X : gx = x} ∈ F. x∈X gx=xRemark1. The map b rarely is injective. Let G = S3 be the symmetric group on three letters. It has four conjugacy classes of subgroups. Using Lemma 2.4.1, Remark 2.4.2, and Lemma 2.4.3 it therefore follows that B(S3 ) ∼ Z4 . On the other hand, S3 = has only three conjugacy classes of elements. Hence Proposition 1.7.1 and The- orem 2.3.6.ii imply that RC (S3 ) ∼ Z3 . =2. In general the map b is not surjective either. But there are many structural results about its cokernel. For example, the Artin induction theorem implies that |G| · RQ (G) ⊆ im(b). We therefore introduce the subring PF (G) := im(b) ⊆ RF (G).Let H be a family of subgroups of G with the property that if H ⊆ H is a subgroupof some H ∈ H then also H ∈ H. We introduce the subgroup B(G, H) ⊆ B(G)generated by all [G/H ] for H ∈ H as well as its image PF (G, H) ⊆ PF (G) underthe map b.Lemma 2.4.7 B(G, H) is an ideal in B(G), and hence PF (G, H) is an ideal inPF (G).Proof We have to show that, for any H1 ∈ H and any subgroup H2 ⊆ G, the element[G/H1 ] · [G/H2 ] lies in B(G, H). This is immediately clear from Exercise 2.4.4.But a less detailed argument suffices. Obviously [G/H1 ] · [G/H2 ] is the sum of the −1classes of the G-orbits in G/H1 × G/H2 . Let G(g1 H1 , g2 H2 ) = G(H1 , g1 g2 H2 ) −1be such a G-orbit. The stabilizer H ⊆ G of the element (H1 , g1 g2 H2 ) is containedin H1 and therefore belongs to H. It follows that G(g1 H1 , g2 H2 ) = G/H ∈ B(G, H).
  23. 23. 2.4 The Burnside Ring 65Definition i. Let be a prime number. A finite group H is called -hyper-elementary if it contains a cyclic normal subgroup C such that |C| and H /C is an -group.ii. A finite group is called hyper-elementary if it is -hyper-elementary for some prime number .Exercise Let H be an -hyper-elementary group. Then: i. Any subgroup of H is -hyper-elementary;ii. let C ⊆ H be a cyclic normal subgroup as in the definition, and let L ⊆ H be any -Sylow subgroup; then the map C × L −→ H sending (c, g) to cg is a bijection of sets. Let Hhe denote the family of hyper-elementary subgroups of G. By the exerciseLemma 2.4.7 is applicable to Hhe .Theorem 2.4.8 (Solomon) Suppose that F has characteristic zero; then PF (G, Hhe ) = PF (G).Proof Because of Lemma 2.4.7 it suffices to show that the unit element 1 ∈ PF (G)already lies in PF (G, Hhe ). According to Lemma 2.4.6.ii the characters χF [X] (g) = {x ∈ X : gx = x} ∈ Z,for any g ∈ G and any finite G-set X, have integral values. Hence we have thewell-defined ring homomorphisms tg : PF (G) −→ Z z −→ Tr(z)(g)for g ∈ G. On the one hand, by Corollary 2.3.7.i, they satisfy ker(tg ) = ker Tr |PF (G) = {0}. (2.4.1) g∈GOn the other hand, we claim that tg PF (G, Hhe ) = Z (2.4.2)holds true for any g ∈ G. We fix a g0 ∈ G in the following. Since the imagetg0 (PF (G, Hhe )) is an additive subgroup of Z and hence is of the form nZ for somen ≥ 0 it suffices to find, for any prime number , an -hyper-elementary subgroupH ⊆ G such that the integer
  24. 24. 66 2 The Cartan–Brauer Triangle tg0 F [G/H ] = χF [G/H ] (g0 ) = {x ∈ G/H : g0 x = x} = gH ∈ G/H : g −1 g0 g ∈ His not contained in Z. We also fix . The wanted -hyper-elementary subgroup H will be found in a chain of sub-groups C ⊆ g0 ⊆ H ⊆ Nwith C being normal in N which is constructed as follows. Let n ≥ 1 be the orderof g0 , and write n = s m with l m. The cyclic subgroup g0 ⊆ G generated by g0then is the direct product s g0 = g0 × g0 m swhere g0 is an -group and C := g0 is a cyclic group of order prime to . mWe define N := {g ∈ G : gCg −1 = C} to be the normalizer of C in G. It contains g0 , of course. Finally, we choose H ⊆ N in such a way that H /C is an -Sylowsubgroup of N/C which contains the -subgroup g0 /C. By construction H is -hyper-elementary. In the next step we study the cardinality of the set {gH ∈ G/H : g0 gH = gH } = gH ∈ G/H : g −1 g0 g ∈ H .Suppose that g −1 g0 g ∈ H . Then g −1 Cg ⊆ g −1 g0 g = g −1 g0 g ⊆ H . But, thetwo sides having coprime orders, the projection map g −1 Cg −→ H /C has to bethe trivial map. It follows that g −1 Cg = C which means that g ∈ N . This showsthat {gH ∈ G/H : g0 gH = gH } = {gH ∈ N/H : g0 gH = gH }.The cardinality of the right-hand side is the number of g0 -orbits in N/H whichconsist of one point only. We note that the subgroup C, being normal in N andcontained in H , acts trivially on N/H . Hence the g0 -orbits coincide with theorbits of the -group g0 /C. But, quite generally, the cardinality of an orbit, beingthe index of the stabilizer of any point in the orbit, divides the order of the actinggroup. It follows that the cardinality of any g0 -orbit in N/H is a power of . Weconclude that {gH ∈ N/H : g0 gH = gH } ≡ |N/H | = [N : H ] mod .But by the choice of H we have [N : H ]. This establishes our claim. By (2.4.2) we now may choose an element zg ∈ PF (G, Hhe ), for any g ∈ G, suchthat tg (zg ) = 1. We then have tg (zg − 1) = 0 for any g ∈ G, g ∈G
  25. 25. 2.5 Clifford Theory 67and (2.4.1) implies that (zg − 1) = 0. g ∈GMultiplying out the left-hand side and using that PF (G, Hhe ) is additively and mul-tiplicatively closed easily shows that 1 ∈ PF (G, Hhe ).2.5 Clifford TheoryAs before F is an arbitrary field. We fix a normal subgroup N in our finite group G.Let W be an F [N]-module. It is given by a homomorphism of F -algebras π: F [N ] −→ EndF (W ).For any g ∈ G we now define a new F [N]-module g ∗ (W ) by the composite homo-morphism π F [N] −→ F [N ] − EndF (W ) → h −→ ghg −1or equivalently by F [N ] × g ∗ (W ) −→ g ∗ (W ) (h, w) −→ ghg −1 w.Remark 2.5.1 i. dimF g ∗ (W ) = dimF W . ii. The map U → g ∗ (U ) is a bijection between the set of F [N ]-submodules of W and the set of F [N]-submodules of g ∗ (W ).iii. W is simple if and only if g ∗ (W ) is simple. ∗ ∗ iv. g1 (g2 (W )) = (g2 g1 )∗ (W ) for any g1 , g2 ∈ G. v. Any F [N]-module homomorphism α : W1 −→ W2 also is a homomorphism of F [N ]-modules α : g ∗ (W1 ) −→ g ∗ (W2 ).Proof Trivial or straightforward. Suppose that g ∈ N . One checks that then ∼ = W −→ g ∗ (W ) w −→ gwis an isomorphism of F [N ]-modules. Together with Remark 2.5.1.iv/v this impliesthat
  26. 26. 68 2 The Cartan–Brauer Triangle G/N × (MF [N ] / ∼) −→ MF [N ] / ∼ = = gN, {W } −→ g ∗ (W ) ∼is a well-defined action of the group G/N on the set MF [N ] / =. By Remark 2.5.1.iiithis action respects the subset F [N ]. For any {W } in F [N ] we put IG (W ) := g ∈ G : g ∗ (W ) = {W } .As a consequence of Remark 2.5.1.iv this is a subgroup of G, which contains N ofcourse.Remark 2.5.2 Let V be an F [G]-module, and let g ∈ G be any element; then themap set of all F [N ]- set of all F [N ]- −→ submodules of V submodules of V W −→ gWis an inclusion preserving bijection; moreover, for any F [N ]-submodule W ⊆ V wehave: i. The map ∼ = g ∗ (W ) −→ g −1 W w −→ g −1 w is an isomorphism of F [N]-modules; ii. gW is a simple F [N]-module if and only if W is a simple F [N ]-module;iii. if W1 ∼ W2 are isomorphic F [N ]-submodules of V then also gW1 ∼ gW2 are = = isomorphic as F [N]-modules.Proof For h ∈ N we have h(gW ) = g g −1 hg W = gWsince g −1 hg ∈ N . Hence gW indeed is an F [N]-submodule, and the map in theassertion is well defined. It obviously is inclusion preserving. Its bijectivity is im-mediate from g −1 (gW ) = W = g(g −1 W ). The assertion ii. is a direct consequence.The map in i. clearly is an F -linear isomorphism. Because of g −1 ghg −1 w = h g −1 w for any h ∈ N and w ∈ Wit is an F [N]-module isomorphism. The last assertion iii. follows from i. and Re-mark 2.5.1.v.
  27. 27. 2.5 Clifford Theory 69Theorem 2.5.3 (Clifford) Let V be a simple F [G]-module; we then have: i. V is semisimple as an F [N ]-module; ˜ii. let W ⊆ V be a simple F [N]-submodule, and let W ⊆ V be the {W }-isotypic component; then ˜ a. W is a simple F [IG (W )]-module, and b. V ∼ IndG (W ) (W ) as F [G]-modules. = IG ˜Proof Since V is of finite length as an F [N ]-module we find a simple F [N ]-submodule W ⊆ V . Then gW , for any g ∈ G, is another simple F [N ]-submoduleby Remark 2.5.2.ii. Therefore V0 := g∈G gW is, on the one hand, a semisimpleF [N ]-module by Proposition 1.1.4. On the other hand it is, by definition, a nonzeroF [G]-submodule of V . Since V is simple we must have V0 = V , which proves the ˜assertion i. As an F [N ]-submodule the {W }-isotypic component W of V is of theform W = W1 ⊕ · · · ⊕ Wm with simple F [N ]-submodules Wi ∼ W . Let first g be ˜ =an element in IG (W ). Using Remark 2.5.2.i/iii we obtain gWi ∼ gW ∼ W for any = = ˜ ˜ ˜1 ≤ i ≤ m. It follows that gWi ⊆ W for any 1 ≤ i ≤ m and hence g W ⊆ W . We see ˜that W is an F [IG (W )]-submodule of V . For a general g ∈ G we conclude from ˜Remark 2.5.2 that g W is the {gW }-isotypic component of V . We certainly have V= ˜ gW . g∈G/IG (W ) ˜ ˜Two such submodules g1 W and g2 W , being isotypic components, either are equal ˜ = g2 W then g −1 g1 W = W , hence g −1 g1 W ∼ W ,or have zero intersection. If g1 W ˜ ˜ ˜ = 2 2 −1and therefore g2 g1 ∈ IG (W ). We see that in fact V= ˜ gW . (2.5.1) g∈G/IG (W ) ˜The inclusion W ⊆ V induces, by the first Frobenius reciprocity, the F [G]-modulehomomorphism IndG (W ) (W ) ∼ F [G] ⊗F [IG (W )] W −→ V IG ˜ = ˜ ag g ⊗ w −→ ˜ ˜ ag g w. g∈G g∈GSince V is simple it must be surjective. But both sides have the same dimension ˜as F -vector spaces [G : IG (W )] · dimF W , the left-hand side by the argument inthe proof of Remark 2.3.2 and the right-hand side by (2.5.1). Hence this map is anisomorphism which proves ii.b. Finally, since F [G] ⊗F [IG (W )] W ∼ V is a sim- ˜ =ple F [G]-module it follows from Remark 2.3.3 that W ˜ is a simple F [IG (W )]-module. In the next section we will need the following particular consequence of thisresult. But first we point out that for an F [N ]-module W of dimension dimF W = 1
  28. 28. 70 2 The Cartan–Brauer Trianglethe describing algebra homomorphism π is of the form π: F [N ] −→ F.The corresponding homomorphism for g ∗ (W ) then is x → π(gxg −1 ). Since anendomorphism of a one-dimensional F -vector space is given by multiplication by ascalar we have g ∈ IG (W ) if and only if there is a scalar a ∈ F × such that aπ(h)w = π ghg −1 aw for any h ∈ N and w ∈ W.It follows that IG (W ) = g ∈ G : π ghg −1 = π(h) for any h ∈ N (2.5.2)if dimF W = 1.Remark 2.5.4 Suppose that N is abelian and that F is a splitting field for N ; thenany simple F [N ]-module W has dimension dimF W = 1.Proof Let π : F [N] −→ EndF (W ) be the algebra homomorphism describing W .Since N is abelian we have im(π) ⊆ EndF [N ] (W ). By our assumption on thefield F the latter is equal to F . This means that any element in F [N ] acts onW by multiplication by a scalar in F . Since W is simple this forces it to be one-dimensional.Proposition 2.5.5 Let H be an -hyper-elementary group with cyclic normal sub-group C such that |C| and H /C is an -group, and let V be a simple F [H ]-module; we suppose thata. F is a splitting field for C,b. V does not contain the trivial F [C]-module, andc. the subgroup C0 := {c ∈ C : cg = gc for any g ∈ H } acts trivially on V ;then there exists a proper subgroup H H and an F [H ]-module V such that ∼V = IndH (V ) as F [H ]-modules. HProof We pick any simple F [C]-submodule W ⊆ V . By applying Clifford’s Theo-rem 2.5.3 to the normal subgroup C and the module W it suffices to show that IH (W ) = H.According to Remark 2.5.4 the module W is one-dimensional and given by an alge-bra homomorphism π : F [C] −→ F , and by (2.5.2) we have IH (W ) = g ∈ H : π gcg −1 = π(c) for any c ∈ C .The assumption c. means that C0 ⊆ C1 := ker(π|C).
  29. 29. 2.6 Brauer’s Induction Theorem 71We immediately note that any subgroup of the cyclic normal subgroup C also isnormal in H . By assumption b. we find an element c2 ∈ C C1 so that π(c2 ) = 1.Let L ⊆ H be an -Sylow subgroup. We claim that we find an element g0 ∈ L such −1that π(g0 c2 g0 ) = π(c2 ). Then g0 ∈ IH (W ) which establishes what we wanted. Wepoint out that, since H = C × L as sets, we have C0 = {c ∈ C : cg = gc for any g ∈ L}.Arguing by contradiction we assume that π(gc2 g −1 ) = π(c2 ) for any g ∈ L. Then gc2 C1 g −1 = gc2 g −1 C1 = c2 C1 for any g ∈ L.This means that we may consider c2 C1 as an L-set with respect to the conjugationaction. Since L is an -group the cardinality of any L-orbit in c2 C1 is a power of .On the other hand we have |C1 | = |c2 C1 |. There must therefore exist an elementc0 ∈ c2 C1 such that gc0 g −1 = c0 for any g ∈ L(i.e. an L-orbit of cardinality one). We conclude that c0 ∈ C0 ⊆ C1 and hencec2 C1 = c0 C1 = C1 . This is in contradiction to c2 ∈ C1 .2.6 Brauer’s Induction TheoremIn this section F is a field of characteristic zero, and G continues to be any finitegroup.Definition i. Let be a prime number. A finite group H is called -elementary if it is a direct product H = C × L of a cyclic group C and an -group L.ii. A finite group is called elementary if it is -elementary for some prime number .Exercise i. If H is -elementary then H = C × L is the direct product of a cyclic group C of order prime to and an -group L. ii. Any -elementary group is -hyper-elementary.iii. Any subgroup of an -elementary group is -elementary. Let He denote the family of elementary subgroups of G.Theorem 2.6.1 (Brauer) Suppose that F is a splitting field for every subgroup of G;then indG RF (H ) = RF (G). H H ∈He
  30. 30. 72 2 The Cartan–Brauer TriangleProof By Corollary 2.3.5 each indG (RF (H )) is an ideal in the ring RF (G). Hence Hthe left-hand side of the asserted identity is an ideal in the right-hand side. To obtainequality we therefore need only to show that the unit element 1G ∈ RF (G) lies in theleft-hand side. According to Solomon’s Theorem 2.4.8 together with Lemma 2.4.6.iwe have 1G ∈ Z F [G/H ] = Zb [G/H ] = Z indG (1H ) H H ∈Hhe H ∈Hhe H ∈Hhe ⊆ indG RF (H ) . H H ∈HheBy the transitivity of induction this reduces us to the case that G is -hyper-elementary for some prime number . We now proceed by induction with respect tothe order of G and assume that our assertion holds for all proper subgroups H G.We also may assume, of course, that G is not elementary. Using the transitivity ofinduction again it then suffices to show that 1G ∈ indG RF H H . H GLet C ⊆ G be the cyclic normal subgroup of order prime to such that G/C is an -group. We fix an -Sylow subgroup L ⊆ G. Then G = C × L as sets. In C wehave the (cyclic) subgroup C0 := {c ∈ C : cg = gc for any g ∈ G}.Then H0 := C0 × Lis an -elementary subgroup of G. Since G is not elementary we must have H0 G.We consider the induction IndG0 (F ) of the trivial F [H0 ]-module F . By semisim- Hplicity it decomposes into a direct sum IndG0 (F ) = V0 ⊕ V1 ⊕ · · · ⊕ Vr Hof simple F [G]-modules Vi . We recall that IndG0 (F ) is the F -vector space of all Hfunctions φ : G/H0 −→ F with the G-action given by g φ g H0 = φ g −1 g H0 .This G-action fixes a function φ if and only if φ(gg H0 ) = φ(g H0 ) for any g, g ∈G, i.e. if and only if φ is constant. It follows that the one-dimensional subspaceof constant functions is the only simple F [G]-submodule in IndG0 (F ) which is Hisomorphic to the trivial module. We may assume that V0 is this trivial submodule.In RF (G) we then have the equation 1G = indG0 (1H0 ) − [V1 ] − · · · − [Vr ]. H
  31. 31. 2.6 Brauer’s Induction Theorem 73This reduces us further to showing that, for any 1 ≤ i ≤ r, we have [Vi ] ∈ indGi RF (Hi ) Hfor some proper subgroup Hi G. This will be achieved by applying the criterion inProposition 2.5.5. By our assumption on F it remains to verify the conditions b. andc. in that proposition for each V1 , . . . , Vr . Since C0 is central in G and is containedin H0 it acts trivially on IndG0 (F ) and a fortiori on any Vi . This is condition c. For Hb. we note that CH0 = G and C ∩ H0 = C0 . Hence the inclusion C ⊆ G induces abijection C/C0 −→ G/H0 . It follows that the map ∼ = IndG0 (F ) −→ IndC0 (F ) H C φ −→ φ|(C/C0 )is an isomorphism of F [C]-modules. It maps constant functions to constantfunctions and hence the unique trivial F [G]-submodule V0 to the unique trivialF [C]-submodule. Therefore Vi , for 1 ≤ i ≤ r, cannot contain any trivial F [C]-submodule.Lemma 2.6.2 Let H be an elementary group, and let N0 ⊆ H be a normal sub-group such that H /N0 is not abelian; then there exists a normal subgroup N0 ⊆N ⊆ H such that N/N0 is abelian but is not contained in the center Z(H /N0 ) ofH /N0 .Proof With H also H /N0 is elementary (if H = C × L with C and L having co-prime orders then H /N0 ∼ C/C ∩ N0 × L/L ∩ N0 ). We therefore may assume =without loss of generality that N0 = {1}. Step 1: We assume that H is an -groupfor some prime number . By assumption we have Z(H ) = H so that H /Z(H ) isan -group ={1}. We pick a cyclic normal subgroup {1} = N = g ⊆ H /Z(H ).Let Z(H ) ⊆ N ⊆ H be the normal subgroup such that N/Z(H ) = N and let g ∈ Nbe a preimage of g. Clearly N = Z(H ), g is abelian. But Z(H ) N since g = 1.Step 2: In general let H = C × L where C is cyclic and L is an -group. With H alsoL is not abelian. Applying Step 1 to L we find a normal abelian subgroup NL ⊆ Lsuch that NL Z(L). Then N := C × NL is a normal abelian subgroup of H suchthat N Z(H ) = C × Z(L).Lemma 2.6.3 Let H be an elementary group, and let W be a simple F [H ]-module;we suppose that F is a splitting field for all subgroups of H ; then there exists asubgroup H ⊆ H and a one-dimensional F [H ]-module W such that W ∼ IndH W = Has F [H ]-modules.
  32. 32. 74 2 The Cartan–Brauer TriangleProof We choose H ⊆ H to be a minimal subgroup (possibly equal to H ) suchthat there exists an F [H ]-module W with W ∼ IndH (W ), and we observe that = HW necessarily is a simple F [H ]-module by Remark 2.3.3. Let π : H −→ EndF Wbe the corresponding algebra homomorphism, and put N0 := ker(π ). We claimthat H /N0 is abelian. Suppose otherwise. Then, by Lemma 2.6.2, there exists anormal subgroup N0 ⊆ N ⊆ H such that N/N0 is abelian but is not contained in ¯Z(H /N0 ). Let W ⊆ W be a simple F [N]-submodule. By Clifford’s Theorem 2.5.3we have W ∼ IndH = ˜ ¯ I H ¯ (W ) (W ) ˜ ¯ ¯where W denotes the {W }-isotypic component of W . Transitivity of induction im-plies W ∼ IndH IndH = ˜ ¯ ∼ IndH = ˜ ¯ H I H ¯ (W ) (W ) I H ¯ (W ). (W ) ¯By the minimality of H we therefore must have IH (W ) = H which means thatW =W ˜ is {W }-isotypic. ¯ ¯ ¯ On the other hand, W is an F [H /N0 ]-module. Hence W is a simple F [N/N0 ]-module for the abelian group N/N0 . Remark 2.5.4 then implies (note that ¯ ¯ ¯EndF [N/N0 ] (W ) = EndF [N ] (W ) = F ) that W is one-dimensional given by an al-gebra homomorphism χ: F [N/N0 ] −→ F. ¯It follows that any h ∈ N acts on the {W }-isotypic module W by multiplication bythe scalar χ(hN0 ). In other words the injective homomorphism H /N0 −→ EndF W hN0 −→ π (h)satisfies π (h) = χ(hN0 ) · idW for any h ∈ N.But χ(hN0 ) · idW lies in the center of EndF (W ). The injectivity of the homomor-phism therefore implies that N/N0 lies in the center of H /N0 . This is a contradic-tion. We thus have established that H /N0 is abelian. Applying Remark 2.5.4 to Wviewed as a simple F [H /N0 ]-module we conclude that W is one-dimensional.Theorem 2.6.4 (Brauer) Suppose that F is a splitting field for any subgroup of G,and let x ∈ RF (G) be any element; then there exist integers m1 , . . . , mr , elementary
  33. 33. 2.7 Splitting Fields 75subgroups H1 , . . . , Hr , and one-dimensional F [Hi ]-modules Wi such that r x= mi indGi [Wi ] . H i=1Proof Combine Theorem 2.6.1, Proposition 1.7.1, Lemma 2.6.3, and the transitivityof induction.2.7 Splitting FieldsAgain F is a field of characteristic zero.Lemma 2.7.1 Let E/F be any extension field, and let V and W be two finitelygenerated F [G]-modules; we then have HomE[G] (E ⊗F V , E ⊗F W ) = E ⊗F HomF [G] (V , W ).Proof First of all we observe, by comparing dimensions, that HomE (E ⊗F V , E ⊗F W ) = E ⊗F HomF (V , W )holds true. We now consider U := HomF (V , W ) as an F [G]-module via G × U −→ U (g, f ) −→ g f := gf g −1 .Then HomF [G] (V , W ) = U G := {f ∈ U : g f = f for any g ∈ G} is the {F }-isotypic component of U for the trivial F [G]-module F . Correspondingly we obtain HomE[G] (E ⊗F V , E ⊗F W ) = HomE (E ⊗F V , E ⊗F W )G = (E ⊗F U )G .This reduces us to proving that (E ⊗F U )G = E ⊗F U Gfor any F [G]-module U . The element 1 εG := g ∈ F [G] ⊆ E[G] |G| g∈Gis an idempotent with the property that U G = εG · U,
  34. 34. 76 2 The Cartan–Brauer Triangleand hence (E ⊗F U )G = εG · (E ⊗F U ) = E ⊗F εG · U = E ⊗F U G .Theorem 2.7.2 (Brauer) Let e be the exponent of G, and suppose that F containsa primitive eth root of unity; then F is a splitting field for any subgroup of G. ¯Proof We fix an algebraic closure F of F . Step 1: We show that, for any finitely ¯ [G]-module V , there is an F [G]-module V such thatgenerated F ¯ ¯∼ ¯ V = F ⊗F V ¯ as F [G]-modules.According to Brauer’s Theorem 2.6.4 we find integers m1 , . . . , mr , subgroups ¯ ¯H1 , . . . , Hr of G, and one-dimensional F [Hi ]-modules Wi such that r ¯ [V ] = ¯ ¯ mi F [G] ⊗F [Hi ] Wi ¯ i=1 = ¯ ¯ m F [G] ⊗F [Hi ] Wi i − ¯ ¯ ¯ −m F [G] ⊗F [Hi ] Wi i ¯ mi >0 mi <0 = ¯ ¯ m F [G] ⊗F [Hi ] Wi i ¯ − ¯ ¯ −m F [G] ⊗F [Hi ] Wi i ¯ . mi >0 mi <0 ¯ ¯ ¯ ¯Let πi : F [Hi ] −→ F denote the F -algebra homomorphism describing Wi . We haveπi (h)e = π (he ) = π (1) = 1 for any h ∈ H . Our assumption on F therefore implies i i ithat πi (F [Hi ]) ⊆ F . Hence the restriction πi |F [Hi ] describes a one-dimensionalF [Hi ]-module Wi such that W i ∼ F ⊗F W i ¯ = ¯ ¯ as F [Hi ]-modules.We define the F [G]-modules V+ := F [G] ⊗F [Hi ] Wimi and V− := F [G] ⊗F [Hi ] Wi−mi . mi >0 mi <0Then ¯ F ⊗F V+ = ¯ F ⊗F F [G] ⊗F [Hi ] Wimi = ¯ F [G] ⊗F [Hi ] Wimi mi >0 mi >0 = ¯ ¯ F [G] ⊗F [Hi ] F [Hi ] ⊗F [Hi ] Wimi ¯ mi >0
  35. 35. 2.7 Splitting Fields 77 = ¯ ¯ F [G] ⊗F [Hi ] F ⊗F Wimi ¯ mi >0 ∼ = ¯ ¯ m F [G] ⊗F [Hi ] Wi i ¯ mi >0and similarly F ⊗F V− ∼ ¯ = ¯ ¯ −m F [G] ⊗F [Hi ] Wi i . ¯ mi <0It follows that ¯ ¯ ¯ [V ] = [F ⊗F V+ ] − [F ⊗F V− ]or, equivalently, that ¯ ¯ ¯ V ⊕ (F ⊗F V− ) = [F ⊗F V+ ].Using Corollary 2.3.7.i/iii we deduce that V ⊕ (F ⊗F V− ) ∼ F ⊗F V+ ¯ ¯ = ¯ ¯ as F [G]-modules.If V− is nonzero then V− = U ⊕ V− is a direct sum of F [G]-modules whereU is simple. On the other hand let V+ = U1 ⊕ · · · ⊕ Um be a decomposition ¯ ¯into simple F [G]-modules. Then F ⊗F U is a direct summand of F ⊗F V− andhence is isomorphic to a direct summand of F ¯ ⊗F V+ . We therefore must have ¯ ¯HomF [G] (F ⊗F U, F ⊗F Uj ) = {0} for some 1 ≤ j ≤ m. Lemma 2.7.1 implies ¯that HomF [G] (U, Uj ) = {0}. Hence U ∼ Uj as F [G]-modules. We conclude that =V+ ∼ U ⊕ V+ with V+ := i=j Ui , and we obtain = V ⊕ F ⊗F V− ⊕ (F ⊗F U ) ∼ F ⊗F V+ ⊕ (F ⊗F U ). ¯ ¯ ¯ = ¯ ¯The Jordan–Hölder Proposition 1.1.2 then implies that V ⊕ F ⊗F V− ∼ F ⊗F V+ . ¯ ¯ = ¯By repeating this argument we arrive after finitely many steps at an F [G]-moduleV such that V ∼ F ⊗F V . ¯= ¯Step 2: Let now V be a simple F [G]-module, and let F ⊗F V = V1 ⊕ · · · ⊕ V¯m be ¯ ¯a decomposition into simple F¯ [G]-modules. By Step 1 we find F [G]-modules Visuch that Vi ∼ F ⊗F Vi . ¯ = ¯For any 1 ≤ i ≤ m, the F [G]-module Vi necessarily is simple, and we have ¯ ¯HomF [G] (F ⊗F Vi , F ⊗F V ) = {0}. Hence HomF [G] (Vi , V ) = {0} by Lemma 2.7.1. ¯ ∼ V for any 1 ≤ i ≤ m. By comparing dimensions we concludeIt follows that Vi =
  36. 36. 78 2 The Cartan–Brauer Triangle ¯ ¯that m = 1. This means that F ⊗F V is a simple F [G]-module. Using Lemma 2.7.1again we obtain ¯ ¯ ¯ F ⊗F EndF [G] (V ) = EndF [G] (F ⊗F V ) = F . ¯Hence EndF [G] (V ) = F must be one-dimensional. This shows that F is a splittingfield for G. Step 3: Let H ⊆ G be any subgroup with exponent eH . Since eH dividese the field F a fortiori contains a primitive eH th root of unity. Hence F also is asplitting field for H .2.8 Properties of the Cartan–Brauer TriangleWe go back to the setting from the beginning of this chapter: k is an algebraicallyclosed field of characteristic p > 0, R is a (0, p)-ring for k with maximal idealmR = RπR , and K denotes the field of fractions of R. The ring R will be called splitting for our finite group G if K contains a primitiveeth root of unity ζ where e is the exponent of G. By Theorem 2.7.2 the field K, inthis case, is a splitting field for any subgroup of G. This additional condition caneasily be achieved by defining K := K(ζ ) and R := {a ∈ K : NormK /K (a) ∈ R};then R is a (0, p)-ring for k which is splitting for G. It is our goal in this section to establish the deeper properties of the Cartan–Brauer triangle dG RK (G) Rk (G) eG cG K0 (k[G]).Lemma 2.8.1 For any subgroup H ⊆ G the diagram dH RK (H ) Rk (H ) indG H indG H dG RK (G) Rk (G)is commutative.Proof Let W be a finitely generated K[H ]-module. We choose a lattice L ⊆ Wwhich is H -invariant. Then indG dH [W ] H = indG [L/πR L] = k[G] ⊗k[H ] (L/πR L) . H
  37. 37. 2.8 Properties of the Cartan–Brauer Triangle 79Moreover, R[G] ⊗R[H ] L ∼ L[G:H ] is a G-invariant lattice in K[G] ⊗K[H ] W ∼ = =W [G:H ] (compare the proof of Remark 2.3.2). Hence dG indG [W ] H = dG K[G] ⊗K[H ] W = R[G] ⊗R[H ] L /πR R[G] ⊗R[H ] L = k[G] ⊗k[H ] (L/πR L) .Lemma 2.8.2 We have Rk (G) = indG Rk (H ) . H H ∈HeProof Since the indG (Rk (H )) are ideals in Rk (G) it suffices to show that the unit Helement 1k[G] ∈ Rk (G) lies in the right-hand side. We choose R to be splitting for G.By Brauer’s induction Theorem 2.6.1 we have 1K[G] ∈ indG RK (H ) H H ∈Hewhere 1K[G] is the unit element in RK (G). Using Lemma 2.8.1 we obtain dG (1K[G] ) ∈ dG indG RK (H ) H = indG dH RK (H ) H H ∈He H ∈He ⊆ indG Rk (H ) . H H ∈HeIt is trivial to see that dG (1K[G] ) = 1k[G] .Theorem 2.8.3 The decomposition homomorphism dG : RK (G) −→ Rk (G)is surjective.Proof By Lemma 2.8.1 we have dG RK (G) ⊇ dG indG RK (H ) H = indG dH RK (H ) . H H ∈He H ∈HeBecause of Lemma 2.8.2 it therefore suffices to show that dH (RK (H )) = Rk (H ) forany H ∈ He . This means we are reduced to proving our assertion in the case wherethe group G is elementary. Then G = H × P is the direct product of a group H oforder prime to p and a p-group P . By Proposition 1.7.1 it suffices to show that theclass [W ] ∈ Rk (G), for any simple k[G]-module W , lies in the image of dG . Viewed
  38. 38. 80 2 The Cartan–Brauer Triangleas a k[P ]-module W must contain the trivial k[P ]-module by Proposition 2.2.7. Wededuce that W P := {w ∈ W : gw = w for any g ∈ P } = {0}.Since P is a normal subgroup of G the k[P ]-submodule W P in fact is a k[G]-submodule of W . But W is simple. Hence W P = W which means that k[G] actson W through the projection map k[G] −→ k[H ]. According to Corollary 2.2.6we find a simple K[H ]-module V together with a G-invariant lattice L ⊆ V suchthat L/πR L ∼ W as k[H ]-modules. Viewing V as a K[G]-module through the =projection map K[G] −→ K[H ] we obtain [V ] ∈ RK (G) and dG ([V ]) = [W ].Theorem 2.8.4 Let p m be the largest power of p which divides the order of G; theCartan homomorphism cG : K0 (k[G]) −→ Rk (G) is injective, its cokernel is finite,and p m Rk (G) ⊆ im(cG ).Proof Step 1: We show that p m Rk (G) ⊆ im(cG ) holds true. It is trivial from thedefinition of the Cartan homomorphism that, for any subgroup H ⊆ G, the diagram cH K0 (k[H ]) Rk (H ) [P ]−→[k[G]⊗k[H ] P ] indG H cG K0 (k[G]) Rk (G)is commutative. It follows that indG im(cH ) ⊆ im(cG ). HLemma 2.8.2 therefore reduces us to the case that G is an elementary group. Let Wbe any simple k[G]-module. With the notations of the proof of Theorem 2.8.3 wehave seen there that k[G] acts on W through the projection map k[G] −→ k[H ].Viewed as a k[H ]-module W is projective by Remark 1.7.3. Hence k[G] ⊗k[H ] Wis a finitely generated projective k[G]-module. We claim that cG k[G] ⊗k[H ] W = |G/H | · [W ]holds true. Using the above commutative diagram as well as Proposition 2.3.4 weobtain cG k[G] ⊗k[H ] W = indG [W ] = k[G] ⊗k[H ] k · [W ]. HIn order to analyze the k[G]-module k[G] ⊗k[H ] k let h, h ∈ H , g ∈ P , and a ∈ k.Then h gh ⊗ a = ghh ⊗ a = g ⊗ a = gh ⊗ a.This shows that H acts trivially on k[G] ⊗k[H ] k. In other words, k[G] acts onk[G] ⊗k[H ] k through the projection map k[G] −→ k[P ]. It then follows fromProposition 2.2.7 that all simple subquotients in a composition series of the k[G]-
  39. 39. 2.8 Properties of the Cartan–Brauer Triangle 81module k[G] ⊗k[H ] k are trivial k[G]-modules. We conclude that k[G] ⊗k[H ] k = dimk k[G] ⊗k[H ] k · 1 = |G/H | · 1(where 1 ∈ Rk (G) is the unit element). Step 2: We know from Proposition 1.7.1 that, as an abelian group, Rk (G) ∼ Zr =for some r ≥ 1. It therefore follows from Step 1 that Rk (G)/ im(cG ) is isomorphicto a factor group of the finite group Zr /p m Zr . Step 3: It is a consequence of Proposition 1.7.4 that K0 (k[G]) and Rk (G) areisomorphic to Zr for the same integer r ≥ 1. Hence id ⊗cG : Q ⊗Z K0 k[G] −→ Q ⊗Z Rk (G)is a linear map between two Q-vector spaces of the same finite dimension r. Itsinjectivity is equivalent to its surjectivity. Let a ∈ Q and x ∈ Rk (G). By Step 1 wefind an element y ∈ K0 (k[G]) such that cG (y) = p m x. Then a a (id ⊗cG ) m ⊗ y = m ⊗ p m x = a ⊗ x. p pThis shows that id ⊗cG and consequently cG are injective. In order to discuss the third homomorphism eG we first introduce two bilinearforms. We start from the maps (MK[G] / ∼ × (MK[G] / ∼ −→ Z =) =) {V }, {W } −→ dimK HomK[G] (V , W )and (Mk[G] / ∼) × (Mk[G] / ∼) −→ Z = = {P }, {V } −→ dimk Homk[G] (P , V ).They extend to Z-bilinear maps Z[MK[G] ] × Z[MK[G] ] −→ Zand Z[Mk[G] ] × Z[Mk[G] ] −→ Z.Since K[G] is semisimple we have, for any exact sequence 0 → V1 → V → V2 → 0in MK[G] , that V ∼ V1 ⊕ V2 and hence that = dimK HomK[G] (V , W ) − dimK HomK[G] (V1 , W ) − dimK HomK[G] (V2 , W ) = 0.
  40. 40. 82 2 The Cartan–Brauer TriangleThe corresponding fact in the “variable” W holds as well, of course. The first maptherefore induces a well-defined Z-bilinear form , K[G] : RK (G) × RK (G) −→ Z [V ], [W ] −→ dimK HomK[G] (V , W ).Even though k[G] might not be semisimple any exact sequence 0 → P1 → P → ∼P2 → 0 in Mk[G] still satisfies P = P1 ⊕ P2 as a consequence of Lemma 1.6.2.ii.Hence we again have dimk Homk[G] (P , V ) − dimk Homk[G] (P1 , V ) − dimk Homk[G] (P2 , V ) = 0for any V in Mk[G] . Furthermore, for any P in Mk[G] and any exact sequence0 → V1 → V → V2 → 0 in Mk[G] we have, by the definition of projective modules,the exact sequence 0 −→ Homk[G] (P , V1 ) −→ Homk[G] (P , V ) −→ Homk[G] (P , V2 ) −→ 0.Hence once more dimk Homk[G] (P , V ) − dimk Homk[G] (P , V1 ) − dimk Homk[G] (P , V2 ) = 0.This shows that the second map also induces a well-defined Z-bilinear form , k[G] : K0 k[G] × Rk (G) −→ Z [P ], [V ] −→ dimk Homk[G] (P , V ).If {V1 }, . . . , {Vr } are the isomorphism classes of the simple K[G]-modules then[V1 ], . . . , [Vr ] is a Z-basis of RK (G) by Proposition 1.7.1. We have dimK EndK[G] (Vi ) if i = j, [Vi ], [Vj ] K[G] = 0 if i = j.In particular, if K is a splitting field for G then 1 if i = j, [Vi ], [Vj ] K[G] = 0 if i = j.Let {P1 }, . . . , {Pt } be the isomorphism classes of finitely generated indecomposableprojective k[G]-modules. By Proposition 1.7.4.iii the [P1 ], . . . , [Pt ] form a Z-basisof K0 (k[G]), and the [P1 / Jac(k[G])P1 ], . . . , [Pt / Jac(k[G])Pt ] form a Z-basis ofRk (G) by Proposition 1.7.4.i. We have
  41. 41. 2.8 Properties of the Cartan–Brauer Triangle 83 Homk[G] Pi , Pj / Jac k[G] Pj = Homk[G] Pi / Jac k[G] Pi , Pj / Jac k[G] Pj Endk[G] (Pi / Jac(k[G])Pi ) if i = j, = {0} if i = j k if i = j, = {0} if i = j,where the latter identity comes from the fact that the algebraically closed field k isa splitting field for G. Hence 1 if i = j, [Pi ], Pj / Jac k[G] Pj = k[G] 0 if i = j.Exercise 2.8.5i. If K is a splitting field for G then the map ∼ = RK (G) −→ HomZ RK (G), Z x −→ x, . K[G] is an isomorphism of abelian groups.ii. The maps ∼ = ∼ = K0 k[G] −→ HomZ Rk (G), Z and Rk (G) −→ HomZ K0 k[G] , Z y −→ y, . k[G] z −→ . , z k[G] are isomorphisms of abelian groups.Lemma 2.8.6 We have y, dG (x) k[G] = eG (y), x K[G]for any y ∈ K0 (k[G]) and x ∈ RK (G).Proof It suffices to consider elements of the form y = [P /πR P ] for some finitelygenerated projective R[G]-module P (see Proposition 2.1.1) and x = [V ] for somefinitely generated K[G]-module V . We pick a G-invariant lattice L ⊆ V . The as-serted identity then reads dimk Homk[G] (P /πR P , L/πR L) = dimK HomK[G] (K ⊗R P , V ).
  42. 42. 84 2 The Cartan–Brauer TriangleWe have Homk[G] (P /πR P , L/πR L) = HomR[G] (P , L/πR L) = HomR[G] (P , L)/ HomR[G] (P , πR L) = HomR[G] (P , L)/πR HomR[G] (P , L) = k ⊗R HomR[G] (P , L) (2.8.1)where the second identity comes from the projectivity of P as an R[G]-module. Onthe other hand HomK[G] (K ⊗R P , V ) = HomR[G] (P , V ) −i = HomR[G] P , πR L i≥0 −i = HomR[G] P , πR L i≥0 −i = πR HomR[G] (P , L) i≥0 = K ⊗R HomR[G] (P , L). (2.8.2)For the third identity one has to observe that, since P is finitely generated as an −iR-module, any R-module homomorphism P −→ V = i≥0 πR L has to have its −iimage inside πR L for some sufficiently large i. Both, P being a direct summand of some R[G]m (Remark 2.1.2) and L by defini-tion are free R-modules. Hence HomR (P , L) is a finitely generated free R-module.The ring R being noetherian the R-submodule HomR[G] (P , L) is finitely generatedas well. Lemma 2.2.1.i then implies that HomR[G] (P , L) ∼ R s is a free R-module. =We now deduce from (2.8.1) and (2.8.2) that Homk[G] (P /πR P , L/πR L) ∼ k s = and HomK[G] (K ⊗R P , V ) ∼ K s , =respectively.Theorem 2.8.7 The homomorphism eG : K0 (k[G]) −