aa r X i v : . [ m a t h . R T ] D ec ADEQUATE SUBGROUPS II
ROBERT GURALNICK
Abstract.
The notion of adequate subgroups was introduced by Jack Thorne [12].It is a weakening of the notion of big subgroup used by Wiles and Taylor in provingautomorphy lifting theorems for certain Galois representations. Using this idea, Thornewas able to prove some new lifting theorems. It was shown in [6] that certain groups wereadequate. One of the key aspects was the question of whether the span of the semsimpleelements in the group is the full endomorphism ring of an absolutely irreducible module.We show that this is the case in prime characteristic p for p -solvable groups as long thedimension is not divisible by p . We also observe that the condition holds for certaininfinite groups. Finally, we present the first examples showing that this condition neednot hold and give a negative answer to a question of Richard Taylor. Introduction
Let k be a field of characteristic p and let V be a finite dimensional vector space over k . Let ρ : G → GL( V ) be an absolutely irreducible representation. Following [12], we say( G, V ) is adequate if the following conditions hold (we rephrase the conditions slightly):(1) H ( G, k ) = 0;(2) p does not divide dim V ;(3) H ( G, V ⊗ V ∗ ) = 0; and(4) End( V ) is spanned by the elements ρ ( g ) with ρ ( g ) semisimple.If G is a finite group of order prime to p (or G is an algebraic or Lie group in characteris-tic zero), then it is well known that ( G, V ) is adequate. In this case, condition (4) is oftenreferred to as Burnside’s Lemma. It is a trivial consequence of the Artin-WedderburnTheorem.These conditions are a weakening of the conditions used by Wiles and Taylor in studyingthe automorphic lifts of certain Galois representations. Thorne [12] generalized variousresults assuming these hypotheses. We refer the reader to [12] for more references anddetails.In particular, it was shown in [6, Theorem 9] that:
Theorem 1.1.
Let k be a field of characteristic p and G a finite group. Let V be anabsolutely irreducible faithful kG -module. If p ≥ V + 2 , then ( G, V ) is adequate. Mathematics Subject Classification.
Primary 20C20; Secondary 11F80.
Key words and phrases.
Burnside’s Lemma, irreducible representation, p-solvable group, Galoisrepresentations.
The proof depends on the classification of finite simple groups. The main ingredientsinclude a result of the author [5] that reduces to the problem to the case that the subgroupof G generated by elements of order p is a central product of quasisimple finite groupsof Lie type in characteristic p , a result of Serre [10] about complete reducibility of tensorproducts and results on the representation theory of the groups of Lie type in the naturalcharacteristic [9].In this note, we consider (4) and show that this holds under some conditions (none ofthese results depend upon the classification of finite simple groups). We say that ( G, V )is weakly adequate if (4) holds.Recall that a finite group is called p -solvable if every composition factor of G either hasorder p or order prime to p . It is known (cf. [8, Theorem B]) that if G is p -solvable and V is an absolutely irreducible G -module in characteristic p , then G contains an absolutelyirreducible p ′ -subgroup, whence Burnside’s Lemma immediately implies: Theorem 1.2.
Let G be a p -solvable subgroup, k a field of characteristic p and V anabsolutely irreducible kG -module. If p does not divide dim V , then ( G, V ) is weakly ade-quate. This allows us to answer in the affirmative a question of R. Taylor for p -solvable groups. Corollary 1.3.
Let G be a p -solvable subgroup, k a field of characteristic p and V anabsolutely irreducible kG -module. If ( G, V ) satisfies conditions (1), (2) and (3) above,then ( G, V ) is adequate. Recall that a kG -module V is called primitive if G preserves no nontrivial direct sumdecomposition of V .We can also show: Theorem 1.4.
Let G be a p -solvable subgroup, k a field of characteristic p and V anabsolutely irreducible kG -module. If V is primitive, then ( G, V ) is weakly adequate. Note that if dim V is a multiple of p , then no p ′ -subgroup can act irreducibly. We alsocan obtain some results for possibly infinite groups. Theorem 1.5.
Let k be algebraically closed of characteristic p . Let V be finite dimensionalover k . Let Γ be an irreducible subgroup of GL( V ) with Zariski closure G . Let G be theconnected component of G and Γ = G ∩ Γ . Assume that either:(1) [Γ : Γ ] is not a multiple of p ; or(2) dim V is not a multiple of p and G/G is p -solvable.(3) V is primitive and G/G is p -solvable.Then ( G, V ) is weakly adequate. The only condition that is difficult to check for adequacy in the previous results isCondition (3). We do improve Theorem 1.1 for p -solvable groups. We first observe aresult from [5]. DEQUATE SUBGROUPS II 3
Theorem 1.6.
Let k be a field of characteristic p . Let G be a finite subgroup of GL n ( k ) =GL( V ) . Assume that V is a completely reducible kG -module. If p > n and is not a Fermatprime or p > n + 1 , then G has no composition factors of order p . It is not difficult to extend this to the case of Zariski closed subgroups. Also, thecomplete reducibility hypothesis can be relaxed – all we need to assume is that G has nonontrivial normal subgroup consisting of unipotent elements. This result is not explicitlystated in [5] there but it is proved there. The result does depend upon the classificationof finite simple groups (however, for p -solvable groups, it does not).It now easily follows that if G is p -solvable and V is a completely reducible kG -moduleof small dimension, then G is in fact a p ′ -group and this gives: Theorem 1.7.
Let k be an algebraically closed field of characteristic p . Let G be a p -solvable group. Let V be an irreducible kG -module. Then ( G, V ) is adequate if:(1) p > dim V with p not a Fermat prime; or(2) p > dim V + 1 . On the other hand, we present an infinite family of examples of imprimitive absolutelyirreducible G -modules in characteristic p with dim V a multiple of p (including cases where G is p -solvable) wiht ( G, V ) not weakly adequate. These are generalizations of examplesof Capdeboscq and Guralnick.In order for this construction to give such examples where p does not divide dim V , wewere led to prove the following result in [4]: Theorem 1.8.
Let p be a prime. There exists a finite simple group G with a nontrivialSylow p -subgroup P such that some coset of P contains no p ′ -elements. Thompson [11] verified this for p = 2 in response to a question of Paige.Using a variation of the Theorem 1.8, we show that for any prime p , Taylor’s questionfails (i.e. (1), (2) and (3) do not necessarily imply (4)).Note that another way to produce examples with ( G, V ) not weakly adequate is tofind absolutely irreducible G -modules in characteristic p such that (dim V ) is larger thanthe number of p ′ -elements in G . These examples are not so easy come by. The onlyprimitive example we know is with G = F (2) ′ (the Tits group) and V the irreduciblemodule of dimension 2048 in characteristic 2. The number of elements of odd order in G is 3 , , < (2048) . So ( G, V ) is not weakly adequate. It is easy to see that V is aprimitive module (since G contains no proper subgroups of index dividing 2048).This suggests the following variant of the problem: Question 1.9.
Let G be a quasisimple finite group and p a prime. Classify all absolutelyirreducible G -modules in characteristic p such that the number of p ′ -elements in G is lessthan (dim V ) . In particular, (
G, V ) cannot be weakly adequate. We suspect that there are very fewsuch examples.
ROBERT GURALNICK
The paper is organized as follows. In the next section, we discuss p -solvable groupsand prove Theorems 1.2 and 1.4. In the following sections we prove Theorem 1.5 andTheorems 1.6 and 1.7. In the last section, we consider necessary conditions for inducedmodules to be weakly adequate. This allows us to construct many examples that are notweakly adequate including some whose dimension is not a multiple of the characteristic.In particular, this allows us to give a negative answer to Taylor’s question.2. p -solvable Groups We prove Theorems 1.2 and 1.4. As noted above, the first result follows by [8, TheoremB] (see also [3]). We sketch an elementary proof of a slight generalization of what werequire.We first prove a lemma about tensor products. The first statement is well known.
Lemma 2.1.
Let G be a group with a normal subgroup N . Let k be an algebraically closedfield. Let V = U ⊗ k W be a finite dimensional kG -module where U and W are irreducible kG -modules. Assume that N acts irreducibly on U and trivially on W .(1) V is an irreducible kG -module; and(2) If N consists of semisimple elements and ( G, W ) is weakly adequate, then ( G, V ) is weakly adequate.Proof. We prove both statements simultaneously. By assumption, End( U ) ⊗ kI is thelinear span of the images of N in GL( U ) ⊗ kI .Since W is kG -irreducible, we can choose elements g i ∈ G such that g i acts as a i ⊗ b i ∈ GL ( U ) ⊗ GL ( W ) where the b i form a basis for End( W ). If ( G, W ) is weakly adequate,we can furthermore assume that the g i are semisimple elements.Thus, the images of the elements N g i span End( U ) ⊗ End( W ) = End( V ). This showsthat V is an irreducible kG -module and that ( G, V ) is weakly adequate if N consists ofsemisimple elments and the g i are semisimple (because then N g i consists of semisimpleelements). (cid:3) Note that in (2) above, (
G, V ) weakly adequate implies that (
G, W ) is weakly adequate.If p is a prime dividing | G | , a subgroup H is called a p -complement if p does not divide | H | but [ G : H ] is a power of p . It is an easy exercise to see that the following holds (justchoose a minimal normal subgroup and apply the Schur-Zassenhaus result): Lemma 2.2.
Let G be a p -solvable group. Any p ′ -subgroup of G is contained in a p -complement and all p -complements are conjugate. We state the next result for irreducible groups rather than absolutely irreducible groups.Most results in the literature assume the latter.
Lemma 2.3.
Let G be a p -solvable group, k a field of characteristic p and V an irreducible kG -module. Let F = End G ( V ) . Assume that p does not divide dim F V . Then a p ′ -complement H of G acts irreducibly on V and F = End H ( V ) . DEQUATE SUBGROUPS II 5
Proof.
First suppose that k = F (i.e. V is absolutely irreducible). So we may assume that k is algebraically closed. We may also assume that O p ( G ) = 1 (since this acts triviallyon V ). Let N be a minimal normal subgroup of G . So N is a p ′ -group. First supposethat N does not act homogeneously on V (i.e. N has at least two nonisomorphic simplesubmodules on V ). Then we can write V = ⊕ ti =1 V i , where the V i are the homogeneouscomponents of N . Let S be the stabilizer of V . Since dim V = t dim V , p is prime toboth dim V and t . Let K be a p -complement in S and H ≥ K a p -complement of G . Byinduction, K is irreducible on V . Since G = SH (since [ G : S ] = t is prime to p ), H actstransitively on the set of V i .Let W be a nonzero H -submodule of V . Since N ≤ H , W = ⊕ ( W ∩ V i ) and since H is transitive on the V i , we see that W ∩ V = 0. Since K acts irreducibly on V , V ≤ W and since H is transitive on the V i , W = V , whence the result.Suppose that N acts homogeneously. It follows (cf. [2, Theorem 51.7]) that (passingto a p ′ -central cover if necessary), V ∼ = U ⊗ k W where U, W are irreducible kG -moduleswith N irreducible on U and trivial on W . If H is a p -complement, then by induction, U and W are irreducible kH -modules. By Lemma 2.1, this implies that H acts irreduciblyon V .Now suppose that k is not F . Since G is finite, we can assume that k is a finitefield. We can view V as an absolutely irreducible F G -module, By the proof above, V isabsolutely irreducible as an F H -module. Thus, F = End H ( V ). Since V is a semisimple kH -module (by Maschke’s theorem) with endomorphism ring a field, V is an irreducible kH -module. (cid:3) Of course, if p does divide dim F V , then V cannot possibly be irreducible restricted to H , since the dimension of any absolutely irreducible H -module in characteristic p divides | H | . Isaacs [8] proves much more than we do above and in particular studies the restrictionof V to H in all cases. These ideas are related to the Fong-Swan theorem: every absolutelyirreducible G -module is the reduction of a characteristic zero module.Theorem 1.2 now follows by Burnside’s Lemma. Theorem 1.4 now follows from thefollowing observation: Lemma 2.4.
Let G be a p -solvable group with k algebraically closed of characteristic p .If V is a primitive kG -module, then p does not divide dim V .Proof. As above, we may assume that O p ( G ) = 1. Let N be a minimal normal noncentralsubgroup of G . Then N is a p ′ -group and acts homogeneously on V . If N acts irreducibly,then dim V divides | N | and the result holds. Otherwise, V = U ⊗ k W where U and W are primitive kG -modules, whence the result follows by induction on dimension. (cid:3) We now give an example to show that conditions (1), (2) and (4) do not guarantee thatcondition (3) holds (even for solvable groups).Let r = p be an odd prime. Let R be an extraspecial r -group of exponent r and order r a . Let s be a prime distinct from p and r . Let S be an s -group with a faithfulabsolutely irreducible F p S module W . Let X be an irreducible F p S -submodule of the ROBERT GURALNICK semisimple module W ⊗ W ∗ . Set K = XS , a semidirect product. We can choose a sufficiently large so that K embeds in Sp(2 a, r ) and so K acts as a group of automorphismsof R . Then RK ≤ R Sp(2 a, r ) has an irreducible module U over k of dimension p a .Set V = U ⊗ k W (where we extend scalars and view W over k ). Then V ⊗ V ∗ ∼ =( U ⊗ U ∗ ) ⊗ ( W ⊗ W ∗ ). Note that V = V R ⊕ [ R, V ]. and V R ∼ = W ⊗ W ∗ . Thus, H ( G, V ⊗ V ∗ ) = H ( G/R, W ⊗ W ∗ ) ∼ = Hom S ( X, W ⊗ W ∗ ) = 0.3. Infinite Groups
Let k be an algebraically closed field of characteristic p ≥
0. Let Γ be an absolutelyirreducible subgroup of GL d ( k ) = GL( V ). Let G be the Zariski closure of Γ. Let G denote the connected component of 1 in G . Set Γ = Γ ∩ G . Note that G = Γ G , whence G/G ∼ = Γ / Γ .We first note: Lemma 3.1.
Let G be a reductive algebraic group over k (i.e. G is reductive). Let g i ∈ G, ≤ i ≤ r be such that the order of g i G /G is not a multiple of p . Then X := { x ∈ G | g i x is semisimple } contains a Zariski open dense subset of G .Proof. Since the intersection of finitely many open dense sets is open and dense, it sufficesto proves this for r = 1. A straightforward argument reduces this to the case that G is asimple algebraic group and g is either inner or is in the coset of a graph automorphism.If g is inner, the result follows since the set of regular semisimple elements is open anddense. If g is a graph automorphism, the same is true – see [7, Lemma 6.8]. (cid:3) Applying this to the Zariski closure of Γ, we immediately obtain:
Corollary 3.2.
Let g i be a finite set of elements of Γ such that none of the orders of g i Γ in Γ / Γ are a multiple of p . Then X := { x ∈ Γ | g i x is semisimple } is Zariski dense in G . In particular, this implies:
Corollary 3.3. If k is algebraically closed of characteristic and V is an irreduciblefinite dimensional k Γ -module, then (Γ , V ) is weakly adequate. Lemma 3.4.
Suppose that V = U ⊗ k W where U and W are irreducible finite dimensional k Γ -modules and that Γ acts irreducibly on U and trivially on W . If (Γ , W ) is weaklyadequate, then (Γ , V ) is weakly adequate.Proof. If (Γ , W ) is weakly adequate, then we can choose finitely many g i ∈ Γ semsimplewith g i = a i ⊗ b i ∈ GL( U ) ⊗ GL ( W ) where the span of the b i is End( W ). Let X be thesubset of Γ consisting of all elements x such that g i x is semisimple for all g i (take g = 1).By Corollary 3.2, X is Zariski dense in G . Thus, the linear span of X is Zariski dense inthe linear span of G which is precisely End( U ) ⊗ kI . Thus, ∪ g i X consists of semisimpleelements and contains a basis for End( V ) = End( U ) ⊗ End( W ). (cid:3) DEQUATE SUBGROUPS II 7
We now prove Theorem 1.5.
Proof.
First suppose that p does not divide [Γ : Γ ]. It follows by Corollary 3.2 that theset of semisimple elements of Γ contain a Zariski dense subset of G . Thus, the linear spanof the semisimple elements of Γ is Zariski dense in the linear span of G . Since linear spacesare closed, it follows that the two sets have the same linear span, whence the result.Next suppose that p does not divide d and G/G is p -solvable. Let H/G be a p -complement in G/G . The exact same proof as in the previous section shows that H isirreducible on V . Thus, Γ ∩ H (which is Zariski dense in H ) is also irreducible on V . Nowapply (1) to Γ ∩ H .Finally consider (3). Since V is primitive, Γ acts homogeneously on V . Thus, V = U ⊗ k W , where U and W are irreducible k Γ-modules, Γ acts irreducibly on U and triviallyon W . Since Γ / Γ is p -solvable and Γ is trivial on W , (Γ , W ) is weakly adequate byTheorem 1.4. Now apply Lemma 3.4. (cid:3) Composition Factors
We first prove Theorem 1.6. As we noted this is essentially in [5]. We sketch the proofindicating in particular how the classification is not required for the case of p -solvablegroups. Theorem 4.1.
Let G be a completely reducible finite subgroup of GL n ( k ) = GL( V ) with k a field of characteristic p . If H ( G, k ) = 0 , then either n ≥ p or p is a Fermat primeand n = p − .Proof. If p ≤
3, then all we are asserting is that n ≥ p ≥ p > n with H ( G, k ) = 0.Let N be the normal subgroup generated by elements of order p . Then H G, k ) embedsinto H ( N, k ) and so we may assume that N = G . Let A be a minimal normal noncentralsubgroup of G . We consider four cases:Case 1. A is an elementary abelian r -group for some prime r = p .Then G permutes the weight spaces of A and since G is generated by elements of order p , some element of order p does not centralize A , whence it must have an orbit of size p and so n ≥ p .Case 2. A is of symplectic type (i.e A/Z ( A ) is elementary abelian of order r a for someprime r = p with Z ( A ) of order r if r is odd or of order 2 or 4 if r = 2; moreover, A hasexponent r is r is odd and has exponent 4 if r = 2).Again, some element g of order p acts nontrivial on A . Thus, g embeds in Sp(2 a, r ),whence p ≤ r a + 1 with equality if and only r = 2 and p is a Fermat prime. Since theminimal faithful representation of A in characteristic p is r a , the result follows. ROBERT GURALNICK
Case 3. A is a central quotient of a direct product of quaisisimple subgroups and p doesnot divide | A | .Again some element g of order p acts nontrivially on A . If g does not preserve eachquasisimple factor of A , then there are least p such factors and we easily see that n ≥ p .So g normalizes each factor of A . Thus, A is quasisimple. We can assume that A actshomogeneously (and nontrivially) on V (otherwise, we may assume that g permutes thehomogeneous factors and so there would be at least p of them, whence n ≥ p ). Since p does not divide | A | , it follows by Sylow’s theorem, that g will normalize a Sylow r -subgroup of A for each prime r dividing | A | . Thus, g will act nontrivially on some Sylow r -subgroup of G and the result follows from cases 1 and 2.Case 4. A is a central quotient of a direct product of quaisisimple subgroups and p doesdivide | A | .Unfortunately, we do not have a proof without the classification (although we suspectthere is one). We argue as in case 3. Now apply [5, Theorem B] to conclude that A isof Lie type in characteristic p . It follows that g must induce a field automorphism andthis forces n ≥ p (one further possibility is that A = J with p = 11, but then A has noouter automorphisms of odd order). (cid:3) Now Theorem 1.6 follows immediately (if there is a composition factor of order p , therewill be a normal subgroup N of G with H ( N, k ) = 0 and N is still completely reducible).An immediate corollary is: Corollary 4.2.
Let G be a completely reducible p -solvable subgroup of GL n ( k ) = GL( V ) with k a field of characteristic p . If p divides | G | , then either n ≥ p or n = p − with p a Fermat prime. We now prove Theorem 1.7.
Proof.
Assume that p > dim V (or p > dim V + 1 if p is a Fermat prime) and that G isan irreducible subgroup of GL( V ) as in the hypotheses. By the corollary G is in fact a p ′ -group, whence ( G, V ) is adequate. (cid:3) Induced Modules
Let k be an algebraically closed field of characteristic p >
0. Suppose that V =Ind GK ( W ). Let g i be a set of coset representatives for the cosets of K in G . So we canwrite V = W ⊕ . . . ⊕ W m where m = [ G : K ] and W i = g i ⊗ W .So End( V ) = ⊕ ij Hom( W i , W j ). Let π ij be the corresponding projection from End( V )to Hom( W i , W j ). Note that the set of g ∈ G such that π j ( g ) = 0 is g j K . This observationyields: DEQUATE SUBGROUPS II 9
Lemma 5.1. If ( G, V ) is weakly adequate, then π j maps the set of p ′ -elements of g j K toa spanning set of Hom( W , W j ) . In particular, if some coset g j K contains no p ′ -elements,then ( G, V ) is not weakly adequate. Using this criterion, we can produce many examples (
G, V ) which are not weakly ad-equate. Of course, we want V to be irreducible and we also want G to be generated by p ′ -elements.Here is our first family of examples.Let H be any finite group whose order is divisible by p with H generated by its p ′ -elements. Let r be a prime not equal to p and let A be an irreducible H -module suchthat H has a regular orbit on Hom( A, k ∗ ) (this can be easily arranged - if r is sufficientlylarge, then any faithful irreducible module A will have this property). Set G = AH , asemidirect product.Let W be a 1-dimensional kA -module with character λ ∈ Hom(
A, k ∗ ) so that λ isin a regular G -orbit. Set V = W GA . We note that V is an irreducible kG -module ofdimension equal to | H | (since V is a direct sum of 1-dimensional non-isomorphic kA -modules permuted transitively by H ). Clearly, G is generated by its p ′ -elements. If g ∈ G has order divisible by p the coset gA has no p ′ -elements, whence: Theorem 5.2. ( G, V ) is not weakly adequate. In particular, we can take G = AS where A is elementary abelian of order 25 with p = 3 and dim V = 6.In fact, we can generalize these examples. Here is the setup:(1) Let L and T be finite groups each generated by p ′ -elements.(2) Let W be an absolutely irreducible faithful kL -module.(3) Let T be a subgroup of T of index t such that T contains no nontrivial normalsubgroup of T and such that some coset xT of T in T contains no p ′ -elements(eg, if T is a proper subgroup of a Sylow p -subgroup P of T , then let x ∈ P \ T ).Set G = L ≀ T = N T , where N = L × . . . × L m with L i ∼ = L and m = [ T : T ] Then G acts on V := W ⊕ . . . ⊕ W t where W i ∼ = W ( L i acts as L on W i and trivially on W j with j = i and T permutes the W i as it does the coset of T ). We can also describe V as Ind GK ( W ) where K = N T with L acting on W as L does on W and ( L × . . . L t ) T acting trivially on W ). Theorem 5.3.
With notation as above, V is a faithful irreducible kG -module of dimensionequal to m dim W , G is generated by p ′ -elements and ( G, V ) is not weakly adequate.Proof. Since the W i are nonisomorphic irreducible kN -modules and T permutes themtransitively, V is irreducible. Since L and T are generated by p ′ -elements, so is G . Since T contains no nontrivial normal subgroup of T , the kernel of this representation would becontained in N . Clearly N acts faithfully. Since the coset xT N contains no p ′ -elements,the result follows. (cid:3) Using the result of [4] for any odd prime p , we can find a sufficiently large q with p exactly dividing q − T = L ( q ) and T a dihedral subgroup of order 2 p , wecan find a t ∈ T with tT containing no p ′ -elements.This allows us to give a negative answer to Richard Taylor’s question. Theorem 5.4.
Let k be an algebraically closed field of characteristic p . Let T = L ( q ) and let T be a subgroup of T isomorphic to a dihedral group of order p as above. Let L be a cyclic group order and let W be the nontrivial -dimensional kL -module. Set G = L ≀ T T , N = L × . . . × L of order m with m = [ T : T ] . Let T act trivially on W .Set V = Ind GK ( W ) where K = N T Then(1) V is an absolutely irreducible kG -module of dimension m (and so prime to p );(2) G satisfies conditions (1), (2) and (3) of the introduction; and(3) G is not adequate.Proof. As we have seen above, the first condition holds and (
G, V ) is not weakly adequateby the construction Clearly p does not divide m = dim V . By construction G is generatedby p ′ -elements. So it remains to that show H ( G, V ⊗ V ∗ ) = 0.Set U = V ⊗ V ∗ . Since N is a normal p ′ -group, it follows that U = C U ( N ) ⊕ [ N, U ]where C U ( N ) are the fixed points of N on U and [ N, U ] is the submodule generated by allnontrivial irreducible N -submodules. By the inflation restriction sequence, it follows that H ( G, [ N, U ]) = 0. Note that dim C U ( N ) = m and indeed C U ( N ) contains U := W ⊗ W ∗ and the stabilizer of U in G is N T . Thus, C U ( N ) ∼ = Ind GK ( k ). So by Shapiro’s Lemma, H ( G, C U ( N )) ∼ = H ( K, k ) ∼ = H ( T , k ) = 0. (cid:3) One can also produce examples showing that Taylor’s question has a negative answerwith p = 2 as well. For example, we can take T = L (137) and T = A ≤ T and L cyclicof order 3 with W a 1-dimensional nontrivial L -module.Here is a variation of Taylor’s question: Question 5.5.
Let V be an absolutely irreducible primitive kG -module. If ( G, V ) satisfies(1), (2) and (3) of the introduction, is ( G, V ) adequate? Now suppose that G is p -solvable. Let V be an irreducible kG -module. If N is anoncentral normal p ′ -subgroup of G that acts homogeneously, then as usual we can write V = U ⊗ k W . By Lemma 2.1 and the remark following it, ( G, V ) is weakly adequate if andonly if (
G/N, W ) is. Thus, if this is the case, the problem reduces to a smaller module.So we may assume that no noncentral normal p ′ -subgroup acts homogeneously. In thiscase, set N = O p ′ ( G ) (the largest normal p ′ -subgroup). Then V = V ⊕ . . . ⊕ V m with m > V i are the kN -components of V . Thus, V = Ind GK ( V ) where N ≤ K .We ask: Question 5.6. If G is p -solvable and every coset gK of K contains a semisimple element,is ( G, V ) weakly adequate? If the answer is yes, then we have an essentially complete answer as to when an abso-lutely irreducible kG -module V is weakly adequate for G a p -solvable group. DEQUATE SUBGROUPS II 11 Acknowledgments
The author was partially supported by the NSF grant DMS-1001962. Some of thiswork was carried out at the Institute for Advanced Study. The author would like tothank Inna Capdeboscq for her help on constructing the first examples which were notweakly adequate, John Thompson for pointing out his article [11], Simon Guest and DanielGoldstein for help with some MAGMA computations and Richard Taylor for many helpfulcomments and questions. Finally, we would like to thank the referee for their carefulreading and helpful comments.
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