A 9 -dimensional algebra which is not a block of a finite group
aa r X i v : . [ m a t h . R T ] M a y A -DIMENSIONAL ALGEBRA WHICH IS NOT A BLOCK OF A FINITEGROUP MARKUS LINCKELMANN AND WILLIAM MURPHY
Abstract.
We rule out a certain 9-dimensional algebra over an algebraically closed field tobe the basic algebra of a block of a finite group, thereby completing the classification of basicalgebras of dimension at most 12 of blocks of finite group algebras. Introduction
Basic algebras of block algebras of finite groups over an algebraically closed field of dimensionat most 12 have been classified in [11], except for one 9-dimensional symmetric algebra over analgebraically closed field k of characteristic 3 with two isomorphism classes of simple modules forwhich it is not known whether it actually arises as a basic algebra of a block of a finite groupalgebra. The purpose of this paper is to show that this algebra does not arise in this way. It isshown in [11, Section 2.9] that if A is a 9-dimensional basic algebra over an algebraically closedfield k of prime characteristic p with two isomorphism classes of simple modules such that A isisomorphic to a basic algebra of a block B of kG for some finite group G , then the algebra A hasthe Cartan matrix C = (cid:18) (cid:19) , Since the elementary divisors of C are 9 and 1, it follows that p = 3 and that a defect group P of B is either cyclic (in which case A is a Brauer tree algebra) or P is elementary abelian of order 9.We will show that the second case does not arise. Theorem 1.1.
Let k be an algebraically closed field of prime characteristic p . Let G be a finitegroup and B a block of kG with Cartan matrix C as above. Then p = 3 , the defect groups of B arecyclic of order , and B is Morita equivalent to the Brauer tree algebra of the tree with two edges,exceptional multiplicity and exceptional vertex at the end of the tree. The proof of Theorem 1.1 proceeds in the following stages. We first identify in Theorem 2.1any hypothetical basic algebra A of a block with Cartan matrix C as above and a noncyclic defectgroup. It turns out that there is only one candidate algebra, up to isomorphism. In Section 3 wegive a description of the structure of this candidate A , and we show in Theorem 5.1 that A is notisomorphic to a basic algebra of a block. The proof of Theorem 2.1 amounts essentially to fillingin the details in [11, section 2.9]. For the proof of Theorem 5.1 we combine a stable equivalenceof Puig [15], a result of Brou´e in [3] on the invariance of stable centres under stable equivalencesof Morita type, results of Kiyota [7] on blocks with an elementary abelian defect group of order 9,and properties of blocks with symmetric stable centres from [6]. A slightly different approach to Date : May 6, 2020. proving Theorem 5.1 is outlined in the last section, first showing in Proposition 6.1 a more preciseresult on the stable equivalence class of A , and then using Rouquier’s stable equivalences for blockswith elementary abelian defect groups of rank 2 from [16].2. The basic algebra A of a noncyclic block with Cartan matrix C The following result is stated in [4] without proof; for the convenience of the reader we give adetailed proof, following in part the arguments in [11, Section 2.9].
Theorem 2.1.
Let k be an algebraically closed field of prime characteristic p . Let A be a basicalgebra with Cartan matrix C = (cid:18) (cid:19) , such that A is Morita equivalent to a block B of kG , for some finite group G , with a noncyclicdefect group P . Then p = 3 , we have P ∼ = C × C , and A is isomorphic to the algebra given bythe quiver i j γ δ αβ with relations δ = γ = αβ , δγ = γδ = 0 , δα = γα = 0 , and βδ = βγ = 0 . In particular, we have | Irr( B ) | = dim k ( Z ( A )) = 6 , and the decomposition matrix of B is equal to D = . Proof.
As mentioned above, since the elementary divisors of the Cartan matrix C are 1 and 9, itfollows that p = 3 and that B has a defect group P of order 9. Since P is assumed to be noncyclic,it follows that P ∼ = C × C .Let { i, j } be a primitive decomposition of 1 in A . Set S = Ai/J ( A ) i and T = Aj/J ( A ) j .It follows from the entries of the Cartan matrix that we may choose notation such that Ai hascomposition length 6 and Aj has composition length 3. Since the top and bottom compositionfactor of Aj are both isomorphic to T , it follows that Aj is uniserial, with composition factors T , S , T (from top to bottom).We label the two vertices of the quiver of A by i and j . The quiver of A contains a unique arrowfrom i to j and no loop at j because J ( A ) j/J ( A ) j ∼ = S . Thus there is an A -homomorphism α : Ai → Aj with image Im( α ) = V uniserial of length 2, with composition factors S , T . Since Aj is uniserialof length 3, it follows that V = J ( A ) j is the unique submodule of length 2 in Aj .The symmetry of A implies that the quiver of A contains a path from j to i . This forces thatthe quiver of A has an arrow from j to i . Since the Cartan matrix of A implies that Ai has exactlyone composition factor T , it follows that the quiver of A contains exactly one arrow from j to i .This arrow corresponds to an A -homomorphism β : Aj → Ai which is not injective as Aj is an injective module. Thus U = Im( β ) is a submodule of Ai of lengthat most 2. The length of U cannot be 1, because the top composition factor of U is T , but theunique simple submodule of Ai is isomorphic to S . Thus U is a uniserial submodule of length 2of Ai , with composition factors T , S . It follows that β ◦ α is an endomorphism of Ai with imagesoc( Ai ).Since U = Im( β ) and β corresponds to an arrow in the quiver of A , it follows that U is notcontained in J ( A ) i . Thus the simple submodule U/ soc( Ai ) of J ( A ) i/ soc( Ai ) is not contained inthe radical of J ( A ) i/ soc( Ai ), and therefore must be a direct summand. Let M be a submodule of Ai such that M/ soc( Ai ) is a complement of U/ soc( Ai ) in J ( A ) i/ soc( Ai ). Then J ( A ) i = U + M soc( Ai ) = U ∩ M and, by the Cartan matrix, M has composition length 4, and all composition factors of M are iso-morphic to S , and soc( M ) = soc( Ai ). Equivalently, M/ soc( Ai ) has length 3, with all compositionfactors isomorphic to S . We rule out some cases. (1) M/ soc( Ai ) cannot be semisimple. Indeed, if it were semisimple, then J ( A ) i/ soc( Ai ) = U/ soc( Ai ) ⊕ M/ soc( Ai ) would be semisimple. This would imply that J ( A ) i = { } . Since also J ( A ) j = { } , it would follow that ℓℓ ( A ) = 3. But a result of Okuyama in [14] rules this out.Thus M/ soc( Ai ) is not semisimple. (2) M/ soc( Ai ) cannot be uniserial. Indeed, if it were, then the quiver of A would have a uniqueloop at i , corresponding to an endomorphism γ of Ai mapping Ai onto M (with kernel necessarilyequal to U because M has no composition factor isomorphic to T ). Then γ = 0 and γ has imagesoc( Ai ) ∼ = S .By construction, α maps U to soc( Aj and β maps V to soc( Ai ). Thus β ◦ α sends Ai ontosoc( Ai ), and α ◦ β sends Aj onto soc( Aj ). Thus γ and β ◦ α differ at most by nonzero scalar. Wemay choose α such that γ = β ◦ α .The homomorphism α sends M to zero, because Aj contains no simple submodule isomorphicto S . Thus α ◦ γ = 0. Also, since U is the kernel of γ , we have γ ◦ β = 0.It follows from these relations that A is a Brauer tree algebra, of a tree with two edges, excep-tional multiplicity 4, and exceptional vertex at an end of the Brauer tree. This would force P tobe cyclic, contradicting the current assumption that P ∼ = C × C . (3) M/ soc( Ai ) cannot be indecomposable. Indeed, if it were, then it would have Loewy length 2because it has composition length 3, but is neither of length 1 (because it is not semisimple) norof length 3 (because it is not uniserial). But then either its socle or its top is simple, and thereforeit would have to be either a quotient of Ai , or a submodule of Ai . We rule out both cases.Suppose first that M/ soc( Ai ) is a quotient of Ai . Note that then M itself has a simple top,isomorphic to S , hence is a quotient of Ai because Ai is projective. Comparing composition lengths MARKUS LINCKELMANN AND WILLIAM MURPHY yields M ∼ = Ai/U . But also U + M = J ( A ) i , so the image of M in Ai/U is the unique maximalsubmodule J ( A ) i/U of Ai/U ∼ = M . Thus J ( A ) M is the unique maximal submodule of M , andthat maximal submodule is isomorphic to a quotient of M , hence has itself a unique maximalsubmodule. This however would imply that M/ soc( Ai ) is uniserial of length 3, which was ruledout earlier.Suppose finally that M/ soc( Ai ) is a submodule of Ai . Then it must be a submodule of M ,because it does not have a composition factor T . Moreover, M and the image of M/ soc( Ai ) in M both have the same simple socle soc( Ai ). Thus M/ soc( Ai ) divided by its socle (which is simple)is a submodule of M/ soc( Ai ), which has a simple socle. Thus the first and second socle seriesquotients are both simple, again forcing M/ soc( Ai ) to be uniserial, which is not possible. (4) Combining the above, it follows that M/ soc( Ai ) is a direct sum of S and a uniserial moduleof length 2 with both composition factors S . That is, we have M = M + M for some submodules M i of M with M ∩ M = soc( Ai ) = soc( M ) M / soc( Ai ) ∼ = S and M / soc( A i ) uniserial of length 2. It follows that M and M are uniserial, of lengths 2 and 3,respectively.We choose now M as follows. By construction, we have a direct sum J ( A ) i/ soc( Ai ) = U/ soc( Ai ) ⊕ M / soc( Ai ) ⊕ M / soc( Ai )Thus we have J ( A ) i/ ( U + M ) ∼ = ( J ( A ) i/ soc( Ai )) / ( U/ soc( Ai ) ⊕ M / soc( Ai )) ∼ = M / soc( Ai ) . This is a uniserial module with two composition factors isomorphic to S . Thus Ai/ ( U + M )is uniserial with three composition factors isomorphic to S , because Ai/J ( A ) i ∼ = S . Since inparticular its socle is simple, isomorphic to S , this module is isomorphic to a submodule of Ai .Choose an embedding A/ ( U + M ) → Ai and replace M by the image of this embedding. Thenthe composition of canonical maps γ : Ai → Ai/ ( U + M ) → Ai is an A -endomorphism of Ai with kernel U + M and uniserial image M of length three. Notethat M is uniserial of length two, so both a quotient and a submodule of Ai . Thus there is anendomorphism δ : Ai → Ai with image M . Since M ⊆ ker( γ ), we have γ ◦ δ = 0 . We show next that we also have δ ◦ γ = 0 . One way to see this is to observe that this is a calculation in the split local 5-dimensional symmetricalgebra End A ( Ai ) ∼ = ( iAi ) op , which as a consequence of [8, B. Theorem], is commutative. There is a (slightly more general) argument that works in this case. Since the A -module Ai ,and hence also the image of γ , is generated by i , it suffices to show that δ ( γ ( i )) = 0. Now since γ ◦ δ = 0, we have 0 = γ ( δ ( i )) = γ ( δ ( i ) i ) = δ ( i ) γ ( i )Note that δ ( i ) = δ ( i ) = iδ ( i ) ∈ iAi , and similarly, γ ( i ) ∈ iAi . Since Im( δ ) = M has length 2,wehave Im( δ ) ⊆ soc ( A ). Thus δ ( i ) ∈ soc ( A ) ∩ iAi ⊆ soc ( iAi ), and since iAi is symmetric, we havesoc ( iAi ) ⊆ Z ( iAi ). It follows that δ ( i ) γ ( i ) = γ ( i ) δ ( i ) = δ ( γ ( i ) i ) = δ ( γ ( i ))whence δ ( γ ( i )) = 0, and so δ ◦ γ = 0 by the previous remarks. Thus M ⊆ ker( δ ). Since Im( δ ) = M has no composition factor T , it follows that U ⊆ ker( δ ). Together we get that U + M ⊆ ker( δ ).Comparing composition lengths yields ker( δ ) = U + M . This implies that ker( δ ) ∩ Im( δ ) = soc( Ai )ker( γ ) ∩ Im( γ ) = soc( Ai )and hence the endomorphisms δ and γ both map Ai onto soc( Ai ). Thus they differ by a nonzeroscalar. Up to adjusting δ , β , we may therefore assume that δ = γ = β ◦ α Since ker( α ) contains M + M , it follows that α ◦ δ = α ◦ γ = 0 . By taking these relations into account, it follows that End A ( A ) is spanned k -linearly by the set { i, j, α, β, γ, γ , δ, δ , α ◦ β } so this is a basis of End A ( A ). We have identified here i , j with the canonical projections of A onto Ai and Aj . Note that End A ( A ) is the algebra opposite to A . This accounts for the reverse order inthe relations of the generators in A (denoted abusively by the same letters). This shows that thequiver with relations of A is as stated. The equation C = ( D t ) D implies that the second columnof D has exactly two nonzero entries and that these are equal to 1. The first row has either fiveentries equal to 1, which yields | Irr( B ) | = 6 and the decomposition matrix D as stated. Or thefirst row has one entry 2 and one entry 1. This would lead to a decomposition matrix of the form D = . In particular, this would yield | Irr( B ) | = 3. But this is not possible, since dim k ( Z ( A )) is clearlygreater than 3; indeed, Z ( A ) contains the linearly independent elements 1, δ , γ , γ . This concludesthe proof. (cid:3) MARKUS LINCKELMANN AND WILLIAM MURPHY The structure of the algebra A Let k be an algebraically closed field. Throughout this section we denote by A the k -algebragiven in Theorem 2.1. We keep the notation of this theorem and identify the generators i , j , α , β , γ , δ with their images in A . Lemma 3.1. (i)
The set { i, j, α, β, βα, γ, γ , δ, δ } is a k -basis of A . (ii) The set { α, β, αβ − βα } is a k -basis of [ A, A ] . (iii) The set { , γ, γ , δ, δ , βα } is a k -basis of Z ( A ) . (iv) The set { αβ, βα } is a k -basis of soc( A ) .Proof. This follows immediately from the relations of the quiver of A . (cid:3) Lemma 3.2.
There is a unique symmetrising form s : A → k such that s ( αβ ) = s ( βα ) = 1 and such that s ( i ) = s ( j ) = s ( α ) = s ( β ) = s ( γ ) = s ( γ ) = s ( δ ) = 0 The dual basis with respect to the form s of the basis { i, j, α, β, βα, γ, γ , δ, δ } is, in this order, the basis { αβ, βα, β, α, j, γ , γ, δ, i } Proof.
Straightforward verification. (cid:3)
See [3, § projective ideal Z pr ( A ) in Z ( A ) and the stable centre Z ( A ) = Z ( A ) /Z pr ( A ). Lemma 3.3.
Let char( k ) = 3 . The projective ideal Z pr ( A ) is one-dimensional, with basis { αβ − βα } , we have an isomorphism of k -algebras Z ( A ) ∼ = k [ x, y ] / ( x − y , xy, y ) induced by the map sending x to γ and y to δ , and after identifying x and y with their in thequotient, the following statements hold: (i) The set { , x, x , y, y } is a k -basis of Z ( A ) , and in particular dim k ( Z ( A )) = 5 . (ii) The set { x, x , y, y } is a k -basis of J ( Z ( A )) . (iii) The set { x , y } is a k -basis of J ( Z ( A )) . (iv) The set { y } is a k -basis of soc( Z ( A )) , and J ( Z ( A )) = soc( Z ( A )) . (v) The k -algebra Z ( A ) is a symmetric algebra.Proof. It follows from lemma 3.2 that the relative trace map Tr A from A to Z ( A ) is given byTr A ( u ) = iuαβ + juβα + αuβ + βuα + βαuj + γuγ + γ uγ + δuδ + iuδ for all u ∈ A . One checks, using char( k ) = 3, thatTr A ( i ) = − Tr A ( j ) = βα − αβ and that Tr A vanishes on all basis elements different from i , j . Statement (i) then follows from therelations in the quiver of A and Lemma 3.1. The algebra Z ( A ) is split local, proving statement (ii), whilst a straightforward computation shows both statement (iii) and (iv). Finally, a simpleverification proves that the map s : Z ( A ) → k such that s ( y ) = 1and such that s (1) = s ( x ) = s ( x ) = s ( y ) = 0is a symmetrising form on Z ( A ). One verifies also that the dual basis with respect to the form s of the basis { , x, y, x , y } is, in this order, the basis { y , x , y, x, } . This completes the proof. (cid:3)
Remark 3.4.
Note that by a result of Erdmann [5, I.10.8(i)], A is of wild representation type.4. The stable centre of the group algebra k ( P ⋊ C ) . Let k be a field of characteristic 3. Set P = C × C and E the subgroup of Aut( P ) of order2 such that the nontrivial element t of E acts as inversion on P . Denote by H = P ⋊ E thecorresponding semidirect product; this is a Frobenius group. Denote by r and s a generator of thetwo factors C of P . The following Lemma holds in greater generality (see Remark 4.1 in [6]); westate only what we need in this paper. Lemma 4.1.
The projective ideal Z pr ( kH ) is one-dimensional, with k -basis { P x ∈ P xt } , and wehave an isomorphism of k -algebras Z ( kH ) ∼ = ( kP ) E induced by the map sending x + x − in ( kP ) E to its image in Z ( kH ) . In particular, we have dim k ( Z ( kH )) = 5 , and the image of the set { , r + r , s + s , r s + rs , rs + r s } is a k -basis of Z ( kH ) .Proof. The relative trace map Tr H from kH to Z ( kH ) satisfies Tr H = Tr HP ◦ Tr P . We calculatefor all a ∈ P Tr P ( a ) = X g ∈ P gag − = X | P | a = 9 · a = 0Thus for every c ∈ kP we have Tr H ( c ) = Tr HP (Tr P ( c )) = 0. On the other hand, for every elementof the form at in H , where a ∈ P , we haveTr H ( at ) = X g ∈ P g ( at ) g − + X g ∈ P ( gt )( at )( gt ) − = ( a + a − ) X x ∈ P xt = 2 · (cid:0) X x ∈ P xt (cid:1) The conjugacy classes of G are given by { } , { r, r } , { s, s } , { r s, rs } , { rs, r s } and { xt | x ∈ P } .The last statement follows. (cid:3) MARKUS LINCKELMANN AND WILLIAM MURPHY
Lemma 4.2.
There is an isomorphism of k -algebras Z ( kH ) ∼ = (cid:0) k [ x, y ] / ( x , y ) (cid:1) E with inverse induced by the map sending x to r − and y to s − , where the nontrivial element t of E acts by x t = x + 2 x and y t = y + 2 y . After identifying x and y with their images in Z ( kH ) the following statements hold: (i) The image of the set { , x , y , xy + x y + xy , x y } is a k -basis of Z ( kH ) , and in particularwe have dim k ( Z ( kH )) = 5 . (ii) The set { x , y , xy + x y + xy , x y } is a k -basis of J ( Z ( kH )) . (iii) The set { x y } is a k -basis of soc( Z ( kH )) , and J ( Z ( kH )) = soc( Z ( kH )) . In particular, dim k ( J ( Z ( kH )) ) = 1 . (iv) The k -algebra Z ( kH ) is symmetric.Proof. By Lemma 4.1 we have Z ( kH ) ∼ = ( kP ) E . Since k has characteristic 3, we have an isomor-phism kP ∼ = k [ x, y ] / ( x , y ) induced by the map given in the statement of the lemma. Under thisisomorphism, the action of t on x and y is given by x t = x + 2 x and y t = y + 2 y as stated. It isstraightforward to then verify that this isomorphism gives r + r t x + 2 ,s + s t y + 2 ,rs + ( rs ) t x + y + 2 xy + 2 x y + 2 xy + x y ,r s + ( r s ) t x + y + xy + x y + xy . This proves the statement (i) and (ii). A straightforward computation proves statement (iii).The final statement is given in general in [6, Corollary 1.3], with an explicit symmetrising form s : Z ( kH ) → k given by s ( x y ) = 1 and sending all other basis elements to 0. (cid:3) Proof of Theorem 1.1
Theorem 1.1 will be an immediate consequence of Theorem 2.1 and the following result.
Theorem 5.1.
Let k be an algebraically closed field of prime characteristic p , and let A be thealgebra given in Theorem 2.1. Then A is not isomorphic to a basic algebra of a block of a finitegroup algebra over k .Proof. Arguing by contradiction, suppose that A is isomorphic to a basic algebra of a block B of kG , for some finite group G . Denote by P a defect group of B . By Theorem 2.1 we have p = 3 and P ∼ = C × C . By Lemma 3.3, the stable centre Z ( A ) is symmetric, hence so is Z ( B ), as A and B are Morita equivalent. It follows from [6, Proposition 3.8] that we have an algebra isomorphism Z ( A ) ∼ = ( kP ) E where E is the inertial quotient of the block B . Again by Lemma 3.3, we have dim k (( kP ) E ) = 5,or equivalently, E has five orbits in P . The list of possible inertial quotients in Kiyota’s paper [7]shows that E is isomorphic to one of 1, C , C × C , C , C , D , Q , SD . In all cases exceptfor E ∼ = C is the action of E on P determined, up to equivalence, by the isomorphism class of E . Thus if E contains a cyclic subgroup of order 4, then E has at most 3 orbits, and if E is theKlein four group, then E has 4 orbits. Therefore we have E ∼ = C . If the nontrivial element t of E has a nontrivial fixed point in P (or equivalently, if t centralises one of the factors C of P and acts as inversion on the other), then E has 6 orbits. Thus t has no nontrivial fixed point in P ,and the group H = P ⋊ E is the Frobenius group considered in the previous section. By a resultof Puig [15, 6.8] (also described in [13, Theorem 10.5.1]), there is a stable equivalence of Moritatype between B and kH , hence between A and kH . By a result of Brou´e [3, 5.4] (see also [12,Corollary 2.17.14]), there is an algebra isomorphism Z ( A ) ∼ = Z ( kH ). This, however, contradictsthe calculations in the Lemmas 3.3 and 4.2, which show that the dimension of J ( Z ( A )) and of J ( Z ( kH )) are different. This contradiction completes the proof. (cid:3) Proof of Theorem 1.1.
Arguing by contradiction, if a defect P of B is not cyclic, then P ∼ = C × C because the Cartan matrix of B has elementary divisors 9 and 1. But then B has a basic algebraisomorphic to the algebra A in Theorem 2.1. This, however, is ruled out by Theorem 5.1. (cid:3) Further remarks
Using the arguments of the proof of Theorem 5.1 it is possible to prove some slightly strongerstatements about the stable equivalence class of the algebra A from Theorem 2.1. Proposition 6.1.
Let k be an algebraically closed field of prime characteristic p and let A be thealgebra in Theorem 2.1. Let P be a finite p -group, E a p ′ -subgroup of Aut( P ) , and τ ∈ H ( E ; k × ) .There does not exist a stable equivalence of Morita type between A and the twisted group algebra k τ ( P ⋊ E ) .Proof. Arguing by contradiction, suppose that there is a stable equivalence of Morita type between A and k τ ( P ⋊ E ). Note that k τ ( P ⋊ E ) is a block of a central p ′ -extension of P ⋊ E with defectgroup P , so its Cartan matrix has a determinant divisible by | P | . By [12, Proposition 4.14.13], theCartan matrices of the algebras A and k τ ( P ⋊ E ) have the same determinant, which is 9. Since A isclearly not of finite representation type (cf. Remark 3.4), it follows that P is not cyclic, hence P ∼ = C × C . Using as before Brou´e’s result [3, 5.4], we have an isomorphism Z ( A ) ∼ = Z ( k τ ( P ⋊ E )).Since Z ( A ) is symmetric, so is Z ( k τ ( P ⋊ E )). Since k τ ( P ⋊ E ) is a block of a central p ′ -extensionof P ⋊ E with defect group P and inertial quotient E , it follows again from [6, Proposition 3.8]that Z ( A ) ∼ = ( kP ) E . From this point onward, the rest of the proof follows the proof of Theorem5.1, whence the result. (cid:3) Remark 6.2.
By results of Rouquier [16, 6.3] (see also [10, Theorem A2 ]), for any block B withan elementary abelian defect group of rank 2 there is a stable equivalence of Morita type between B and its Brauer correspondent, which by a result of K¨ulshammer [9], is Morita equivalent to atwisted semidirect product group algebra as in Proposition 6.1. Thus Theorem 5.1 can be obtainedas a consequence of Proposition 6.1 and Rouquier’s stable equivalence. Remark 6.3.
A slightly different proof of Theorem 5.1 makes use of Brou´e’s surjective algebrahomomorphism Z ( B ) → ( kZ ( P )) E from [2, Proposition III (1.1)], induced by the Brauer homo-morphism Br P , where here P is a (not necessarily abelian) defect group of a block B of a finitegroup algebra kG , with k an algebraically closed field of prime characteristic p . If P is normal in G , then it is easy to see that Brou´e’s homomorphism is split surjective, but this is not known ingeneral. If B is a block with P nontrivial such that there exists a stable equivalence of Moritatype between B and its Brauer correspondent, then this implies the existence of at least some splitsurjective algebra homomorphism Z ( B ) → kZ ( P ) E . Kiyota’s list in [7] shows that if A were isomorphic to a basic algebra of a block with defectgroup P ∼ = C × C , then E would be isomorphic to one of C or D (subcase (b) in Kiyota’s list).The case C can be ruled out as above, and the case D can be ruled out by using Rouquier’sstable equivalence, and by showing that if E ∼ = D , then ( kP ) E is uniserial of dimension 3, but Z ( A ) admits no split surjective algebra homomorphism onto a uniserial algebra of dimension 3.Note that Z ( A ) does though admit a surjective algebra homomorphism onto a uniserial algebraof dimension 3, so the splitting is an essential point in this argument, and may warrant furtherinvestigation. References [1] D. J. Benson,
Representations and cohomology, Vol. I: Cohomology of groups and modules.
Cambridgestudies in advanced mathematics , Cambridge University Press (1991).[2] M. Brou´e. Brauer coefficients of p -subgroups associated with a p -block of a finite group. J. Algebra (1979), 365–383.[3] M. Brou´e, Equivalences of Blocks of Group Algebras , in: Finite dimensional algebras and related topics,Kluwer (1994), 1–26.[4] C. Eaton, https://wiki.manchester.ac.uk/blocks/index.php/Blocks with basic algebras of low dimension[5] K. Erdmann,
Blocks of tame representation type and related algebras , Lecture Notes Math. , SpringerVerlag, Berlin Heidelberg (1990).[6] R. Kessar, M. Linckelmann,
On blocks with Frobenius inertial quotient . J. Algebra
249 (1) , (2002), 127–146.[7] M. Kiyota, On -blocks with an elementary abelian defect group of order . J. Fac. Sci. Univ. Tokyo Sect.IA, Math. (1984), 33-58.[8] B. K¨ulshammer, Symmetric local algebras and small blocks of finite groups . J. Algebra (1984), 190–195.[9] B. K¨ulshammer, Crossed products and blocks with normal defect groups , Comm. Algebra (1985), 147–168.[10] M. Linckelmann, Trivial source bimodule rings for blocks and p -permutation equivalences , Trans. Amer.Math. Soc. (2009), 1279–1316.[11] M. Linckelmann, Finite-dimensional algebras arising as blocks of finite group algebras.
Representations ofalgebras, Contemp. Math. , Amer. Math. Soc., Providence, RI, (2018), 155–188.[12] M. Linckelmann,
The Block Theory of Finite Group Algebras, Volume 1 , LMS Student Society Texts, ,(2018)[13] M. Linckelmann, The Block Theory of Finite Group Algebras, Volume 2 , LMS Student Society Texts, ,(2018)[14] T. Okuyama, On blocks of finite groups with radical cube zero.