Modular Representations of the Hall-Janko Group

Peter Sin

University of Florida

Abstract. We study modular representations of the Hall-Janko group and its double cover in characteristics 2, 3 and 5. In particular, we determine all extensions of simple modules. Results on the group G_{2}(2)′ ≅ PSU(3, 3), which is isomorphic to a maximal subgroup of the Hall-Janko group, are also included.

Introduction

In this paper we study the modular representations of the Hall-Janko group G = J_{2}, a sporadic simple group of order 604800 = 2^{7}.3^{3}.5^{2}.7 and of its twofold universal covering group Ĝ. The (Brauer) characters of simple modules have been determined in all characteristics ([Parker], [Kondratev]), as have the decomposition numbers. The blocks with cyclic defect are described in [Feit2] (see also [Hiss-Lux]) and the only non-principal 2-block has been treated in [Landrock]. In these cases it is not difficult to give the Loewy structure of the projective indecomposable modules. The blocks which have not yet been studied are the principal blocks and two other blocks of Ĝ of maximal defect, one for each of the characteristics 3 and 5. We shall determine the extensions between the simple modules in these cases. Characteristics 3 and 5 are relatively easy, and one can try to say more about the projective modules. The characteristic 2 calculations appear to be harder. In our approach, we rely heavily on the representation theory of the algebraic group G_{2}(k), where k is an algebraically closed field of characteristic two which was studied in [Sin] and [Dowd-Sin]. G embeds into this group in such a way that most of the irreducible representaions of G over k extend to G_{2}(k).

It can be expected that computer calculations will eventually furnish the entire Loewy structure of all of the projective indecomposable modules. In this direct approach, it is necessary first to find matrices of group generators acting on suitable projective modules and this first step is not automatic and may be quite challenging. In this paper, we calculate the extensions of simple modules by indirect methods, for which the initial data needed is essentially character-theoretic. Unfortunately, our collection of tricks does not amount to an algorithm. However, we hope that some of them will be useful for investigations of other groups.

§1. Statement of results

Let L be a simple kG-module and P its projective cover. If p ≠ 2, then P is also the projective cover of L when both are regarded as kĜ-modules. For p = 2, the kĜ-projective cover P̂ of L is different from P but the simple module extensions are the same; this follows from 5-term sequences arising from the group extension 1 → Z(Ĝ) → Ĝ → G → 1, using the fact that Z(Ĝ) ≤ [Ĝ, Ĝ].

Theorem. In the tables below the dimension of the space of extensions between two simple modules is the entry of the row labelled by the first module and the column labelled by the second.

p = 2.

There are two conjugacy classes of embeddings of G into G_{2}(4) and the are interchanged by the Frobenius map (See [Wales]). We have chosen our notation (See §2 below) so that (for a fixed embedding of G in G_{2}(k)) the simple modules with subscript 1 are the restrictions of simple G-modules with restricted highest weights. The modules with subscripts 2 are then the restrictions to G of their twists by the Frobenius map. This turns out to be the same as twisting by the outer automorphism of G.

Table 1. The principal 2-block of G.

\begin{matrix}Ext_{kG}^{1} & k & 6_{1} & 6_{2} & 14_{1} & 14_{2} & 36 & 84 \\ k & 0 & 1 & 1 & 0 & 0 & 0 & 1 \\ 6_{1} & 1 & 0 & 0 & 0 & 1 & 1 & 0 \\ 6_{2} & 1 & 0 & 0 & 1 & 0 & 1 & 0 \\ 14_{1} & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ 14_{2} & 0 & 1 & 0 & 0 & 0 & 0 & 1 \\ 36 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\ 84 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \end{matrix}

p = 3. Our notation follows that of [Hiß-Lux]. There is one choice to be made with regard to the pair of dual simple modules of dimension 36 in the nonprincipal block of maximal defect. Since these two are the only simple modules which are not self dual, we can choose either to be the one labelled 36_{1}, since complex conjugation fixes all other simple Brauer characters. We have chosen 36_{1} to be the one for which Ext1 ; _{G}(6_{1}, 36_{1}) ≠ 0 (see below that this condition picks out one of the two 36-dimensional modules.)

Table 2. The principal 3-block of G.

\begin{matrix}Ext_{kG}^{1} & k & 13_{1} & 13_{2} & 21_{1} & 21_{2} & 57_{1} & 57_{2} & 133 \\ k & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 13_{1} & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\ 13_{2} & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ 21_{1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 21_{2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 57_{1} & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 57_{2} & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 133 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \end{matrix}

Table 3. The non-principal 3-block of Ĝ of maximal defect.

\begin{matrix}Ext_{k_{G}}^{1} & 6_{1} & 6_{2} & 14 & 36_{1} & 36_{2} & 50_{1} & 50_{2} & 236 \\ 6_{1} & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 \\ 6_{2} & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 14 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\ 36_{1} & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 36_{2} & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 50_{1} & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 50_{2} & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 236 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \end{matrix}

p = 5. Again our notation matches [Hiß-Lux].

Table 4. The principal 5-block of G.

\begin{matrix}Ext_{kG}^{1} & k & 14 & 21 & 41 & 85 & 189 \\ k & 0 & 0 & 1 & 0 & 0 & 0 \\ 14 & 0 & 1 & 1 & 0 & 0 & 1 \\ 21 & 1 & 1 & 1 & 1 & 0 & 1 \\ 41 & 0 & 0 & 1 & 0 & 1 & 0 \\ 85 & 0 & 0 & 0 & 1 & 0 & 1 \\ 189 & 0 & 1 & 1 & 0 & 1 & 1 \end{matrix}

Table 5. The non-principal 5-block of Ĝ of maximal defect.

\begin{matrix}Ext_{k_{G}}^{1} & 6 & 14 & 56 & 64 & 190 & 202 \\ 6 & 1 & 0 & 1 & 1 & 0 & 0 \\ 14 & 0 & 0 & 0 & 1 & 0 & 1 \\ 56 & 1 & 0 & 1 & 1 & 1 & 1 \\ 64 & 1 & 1 & 1 & 1 & 0 & 0 \\ 190 & 0 & 0 & 1 & 0 & 0 & 0 \\ 202 & 0 & 1 & 1 & 0 & 0 & 0 \end{matrix}

Remark. Notice that there exists a bijection between the sets of simple modules in these two blocks which preserves extensions. The blocks are not Morita equivalent, however, since they have different cartan matrices.

Appendix. Extensions for U_{3}(3) ≅ (G_{2}(2))′

Here we give tables of extensions for the group H = U_{3}(3). The subgroups of G which are isomorphic to H form a unique class of maximal subgroups. These results are obtained by similar methods, except that, as might be expected for this smaller group, the consideration of induced modules is an effective additional technique. The group order is 2^{5}.3^{3}.7, so the characteristic 7 case is covered by the theory of blocks of defect one (see[Hiß-Lux2]). We will treat the other two primes. As far as I am aware, several of the results below appear as isolated calculations, scattered throughout the literature, and there is little doubt that most or all of the others are also known, whether published or not. The main reason for including this appendix is simply to provide a convenient reference.

p = 2.

H has index 2 in G_{2}(2). The latter has three simple modules in the principal block of dimensions 1, 6 and 14. Their restrictions to H are the simple modules in the principal block of H. The other block of G_{2}(2) is of defect zero, containing the 64-dimensional Steinberg module, whose restriction to H splits into two nonisomorphic, 32-dimensional simple projective modules.

Table A1. The principal 2-block of H ≅ (G_{2}(2))′.

\begin{matrix}Ext_{kH}^{1} & k & 6 & 14 \\ k & 0 & 1 & 0 \\ 6 & 1 & 0 & 1 \\ 14 & 0 & 1 & 0 \end{matrix}

Remark. The extensions for the principal block of G_{2}(2) are given by the same table except that the entries on the leading diagonal are 1 instead of 0.

p = 3.

H ≅ U_{3}(3) has two blocks, the principal block and a block of defect 0 containing the 27dimesional Steinberg module. The principal block has 8 simple modules, denoted by their dimensions in the table below. The module 3 affords the natural representation and 6 is its symmetric square. The superscript “^{∗}” denotes duality. The module 7 is the nontrivial composition factor of 3 ⊗ 3^{∗}, that is the quotient of the space of matrices of trace 0 by the scalar matrices, and the module 15 is the kernel of the natural map 3^{∗} ⊗ S^{2}(3) → 3.

Table A2. The principal 3-block of H.

\begin{matrix}Ext_{kH}^{1} & 1 & 3 & 3^{∗} & 6 & 6^{∗} & 7 & 15 & 15^{∗} \\ 1 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 \\ 3 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 \\ 3^{∗} & 0 & 1 & 0 & 0 & 0 & 1 & 1 & 0 \\ 6 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 6^{∗} & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 7 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 15 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\ 15^{∗} & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \end{matrix}

Remark. In this case, a little extra work beyond the calculations needed to complete Table A2 yields the Loewy structure of the projective indecomposable modules.

Preliminary Remark.

In what follows, we will often make use of information drawn from the ATLAS, and the tables of Brauer characters, decomposition matrices, products of characters etc. in [Hiß-Lux], [Parker] and [Kondratev]. We will also use the tables of weights in [Bremner-Moody-Patera] in a few places for p = 2. To avoid undue length, the details of calculations which follow directly from this numerical information, such as the calculation of multiplicities of simple modules in tensor products of simple modules, or in projective modules, will be omitted, but we will try to indicate what information is used in each instance. Some care is needed to find the correct correspondences among the notations of these sources.

§2. Characteristic 2

In §2 we assume k to be an algebraically closed field of characteristic 2.

2.1.

In [Wales] it is shown that G is isomorphic to a maximal subgroup of the Chevalley group G_{2}(4) of rational points over F_{4} inside the algebraic group G_{2}(k). The simple G_{2}(k)- modules whose highest weights are 2-restricted (see [Jantzen]) are the trivial module, a 6-dimensional module, which we denote by 6_{1}, the 14-dimensional adjoint module 14_{1} and the 64-dimensional (first) Steinberg module 64_{1}. These modules remain irreducible upon restriction to G; indeed, all but 64_{1} remain irreducible for the subgroup G_{2}(2)′ of G. We denote the algebraic conjugates under Gal(F_{4}/F_{2}) of these modules with a subscript “2”. The simple modules in the principal block are then k, 6_{1}, 6_{2}, 14_{1}, 14_{2}, 36 := 6_{1} ⊗ 6_{2} and 84 := 6_{1} ⊗ 14_{2}. The module 84 is also isomorphic to 6_{2} ⊗ 14_{1}. (Remark: Since the two tensor products are not isomorphic for G_{2}(4) and G is maximal in G_{2}(4),the last sentence implies that G is the full stabilizer of a certain hermitian form on 6_{1} ⊗ 14_{2}.) We see then that all simple modules in the principal block are restrictions of simple modules for G_{2}(k). There is just one other block, with defect group isomorphic to the fours group, containing 64_{1}, 64_{2} and a simple module 160, which appears in the equation

14_{1} ⊗ 14_{2} ≅ 36 ⊕ 160.  (1)

The structure of the nonprincipal block has been studied in [Landrock] and we will make use of it later. The facts concerning simple modules can all be checked from tables of Brauer characters. Note also that all simple modules are self-dual.

2.2. This subsection contains a general lemma and some immediate applications.

In spite of the section heading, the field in the following lemma is arbitrary. This result is a variation of the main result in [Alperin-Gorenstein].

Lemma 1. Let F be a field, G a group and V a finite-dimensional F G-module, with V^{G} = 0. Let L_{0}, L_{1} and L_{2} be subgroups of G with L_{0} ≤ L_{1} ∩ L_{2} and G = ⟨L_{1}, L_{2}⟩. Set d_{i} = dim_{F} V^{L_{i}}. Suppose

(i) d_{0} ≤ d_{1} + d_{2}; (ii) H^{1}(L_{i}, V) = 0, for i = 1,2.

Then H^{1}(G, V) = 0.

Proof: We will show that every short exact sequence of F G-modules of the form

0 → V → E → F → 0

must split, which is equivalent to E^{G} ≠ 0, since V^{G} = 0. By (ii), the sequence splits upon restriction to L_{1} and L_{2}, hence also to L_{0}. Thus, E^{L_{i}} has dimension d_{i} + 1 for all i. Since E^{L_{0}} ⊇ E^{L_{1}} = E^{L_{2}}, the fact that E^{G} = E^{L_{1}} ∩ E^{L_{2}} is nonzero follows from (i) by counting dimensions.

Remark. Under the given conditions on the subgroups L_{i} (excluding (i) and (ii)), the hypothesis (i) is equivalent to d_{0} = d_{1} + d_{2}, or to M^{L_{0}} = M^{L_{1}} + M^{L_{2}}.

We will now perform some calculations with small modules using Lemma 2.2 In order to do so, we must pick suitable subgroups L_{i} and study their actions on these modules. We will follow the ATLAS notation for conjugacy classes. The notation we are about to introduce will be fixed for the rest of §2. Let x be an element of class 5A. Then its centralizer C(x), is isomorphic to A_{5} × Z_{5}, the second factor being generated by x. This has index 2 in the normalizer N(x) ≅ A_{5} × D_{10} of ⟨x⟩.

Let z be an element of C(x) of order 3 (in class 3A). We have C(z) ≅ 3.PGL(2, 9). From the character table, we see that ⟨xz⟩ is its own centralizer and it follows then from

orders that C(x) and C(z) generate G, in fact we may replace C(z) by its derived subgroup 3.A_{6}. Likewise, if x′ is taken to be an element of C(x) in class 5A but outside ⟨x⟩, then C(x) and C(x′) generate G and have intersection S = ⟨x, x′⟩ ≅ Z_{5} × Z_{5}. (The statements concerning generation can of course be read off from the list of maximal subgroups, but elementary arguments suffice for our needs.) These are typical of the subgroups we will choose for L_{i} in applying Lemma 1. In all cases it will be routine to check that L_{1} and L_{2} generate G and that L_{0} ≤ L_{1} ∩ L_{2}, so we will mainly be concerned with hypotheses (i) and (ii). In verifying these we will need to make use of certain facts about representations of A_{5} = SL(2, 4), A_{6}, etc. These will not be difficult facts and they can be found, for instance, in the examples at the end of [Benson].

The nontrivial simple k(SL(2, 4))-modules are two algebraically conjugate 2-dimensional modules V_{1} and V_{2} and their tensor product V_{12} := V_{1} ⊗ V_{2}. Let ω denote both a primitive 5th root of unity ion k and the character of ⟨x⟩ sending x to that root. As modules for the subgroup C(x)′ ≅ SL(2, 4) of G, we fix V_{1} to be the one on which an element of class 5C has trace ω + ω^{4}. For D_{10} = ⟨x⟩.2, we denote by D_{1} and D_{2} the irreducible 2-dimesional module whose restrictions to ⟨x⟩ are ω + ω^{4} and ω^{2} + ω^{3} respectively.

Lemma 2. As modules for C(x) ≅ SL(2, 4) × D_{10}, we have:

(a) 6_{1}_{|N(x)} ≅ (V_{1} ⊗ k) ⊕ (V_{2} ⊗ D_{2});

(b) 14_{1}_{|N(x)} ≅ (X ⊗ k) ⊕ (k ⊗ D_{1}) ⊕ (V_{12} ⊗ D_{2}), where X ≅ V_{2} ⊗ V_{2}, is uniserial with factors k, V_{1}, k.

Proof: This mostly follows from (Brauer) character calculations. The only point which requires further discussion is the structure of X in (b). It is clear from the characters that 14_{1}_{|N(x)} has a direct summand X ⊗ k with the right composition factors. Since 14_{1} is self-dual, so must X be. Therefore, if X were not uniserial, it would have to be semisimple. Then 14_{1}^{N(x)} would be 2-dimensional. Let S = ⟨x, x′⟩ be as in the discussion before the lemma. It is easy to check that dim_{k}(14_{1}^{S}) = 2, so if X were semisimple, we would have 14_{1}^{N(x)} = 14_{1}^{S}. But then the same would hold with x′ in place of x since they are conjugate in G, leading to the contradiction that 14_{1}^{G} ≠ 0, because N(x) and N(x′) generate G. This proves that X is uniserial. The isomorphism with V_{2} ⊗ V_{2} is a elementary calculation in SL(2, 4).

We next apply these facts in a simple case.

Lemma 3.

(a) H^{1}(G, 36) = 0.

(b) Ext_{G}^{1}(14_{1}, 14_{2}) = 0.

Proof: Equation 2.1(1) shows that (b) follows from (a), as 160 does not belong to the principal block. Let L_{1} = C(x), L_{2} = C(x′) and L_{0} = S. Since 36 ≅ 6_{1} ⊗ 6_{2}, we find using Lemma 2(a) that as modules for SL(2, 4) ≅ L_{1}/⟨x⟩, we have 36^{⟨x⟩} ≅ V_{1} ⊗ V_{2} ≅ V_{12}. Since V_{12} is an injective module, we obtain H^{1}(L_{1}, 36) ≅ H^{1}(L_{1}/⟨x⟩, 36^{⟨x⟩}) = 0a and similarly for L_{2}, which is conjugate to L_{1}. Therefore the hypothesis (ii) of lemma 1 is verified. Hypothesis (i) of Lemma 1 holds because we see from chracters that 36^{S} = 0. So Lemma 1 yields the result.

Pursuing this method further, we have the following.

Lemma 4.

(a) H^{1}(G, 6_{i}) ≅ k.

(b) H^{1}(G, 14_{i}) = 0.

(c) H^{1}(G, 6_{i} ⊗ 6_{i}) = 0.

Proof: (a) It does not matter whether we consider 6_{1} or 6_{2}. We first prove that H^{1}(G, 6_{i}) ≠ 0 by showing that there is a nonsplit extension of k by by some 6_{i}. A Brauer character shows that the component of 6_{1} ⊗ 14_{1} in the principal block has composition factors 6_{1} (twice), k (twice) and 6_{2}. Since Hom_{kG}(k, 6_{1} ⊗ 14_{1}) ≅ Hom_{kG}(6_{1}, 14_{1}) = 0, the existence of the claimed nonsplit extension follows. Let M be such an extension, of k by 6_{1}, say. To prove the lemma it will suffice to show that H^{1}(G, M) = 0. In the notation of the preceding discussion and of Lemma 1, we choose L_{1} = C(x) ≅ SL(2, 4) × ⟨x⟩, L_{2} = C(z)′ and L_{0} = ⟨xz⟩. From characters we can see that 6^{⟨z⟩} = 0, which implies that H^{1}(L_{2}, 6_{1}) = 0, from which it follows (in the notation of Lemma 1) that d_{2} = d_{0} = 1 and also that d_{1} = 0, as M^{G} = 0. We also obtain H^{1}(L_{2}, M) = 0 from the long exact sequence. The only remaining hypothesis to be checked is that H^{1}(L_{1}, M) = H^{1}(L_{1}/⟨x⟩, M^{⟨x⟩}) = 0. By Lemma 2(a), 6_{1}^{⟨x⟩} ≅ V_{1}, a simple 2-dimensional module for L_{1}/⟨x⟩ ≅ SL(2, 4). Since d_{1} = 0 the k(L_{1}/⟨x⟩)-module M^{⟨x⟩} must be a nonsplit extension of k by V_{1}. Now it is a fact about SL(2, 4) that H^{1}(L_{1}/⟨x⟩, V_{1}) ≅ k (see [Benson] for example) and it follows that H^{1}(L_{1}/⟨x⟩, M^{⟨x⟩}) = 0.

(b) We may consider just 14_{1}. We take L_{1} = C(x), L_{2} = C(x′) and L_{0} = S. By Lemma 2(b), we see that d_{1} = d_{2} = 1 and d_{0} = 2, so the condition on fixed points holds. It suffices to check hypothesis (ii) of Lemma 1 for L_{1}, since x and x′ are conjugate in G. We have, using Lemma 2(b),

H^{1}(L_{1}, 14_{1}) ≅ H^{1}(L_{1}/⟨x⟩, 14_{1}^{⟨x⟩}) ≅ H^{1}(SL(2, 4), X).

The last group is easily seen to be zero using H^{1}(SL(2, 4), V_{1}) ≅ k and H^{1}(SL(2, 4), k) = 0. Therfore Lemma 1 applies and (b) is proved.

(c) The composition factors of 6_{1} ⊗ 6_{1} are k (twice), 14_{1} (twice) and 6_{2}. From the natural exact sequences

0 → ∧^{2}(6_{1}) → 6_{1} ⊗ 6_{1} → S^{2}(6_{1}) → 0 and 0 → 6_{2} → S^{2}(6_{1}) → ∧^{2}(6_{1}) → 0, (1)

we see (for example by self-duality) that ∧^{2}(6_{1}) ≅ k ⊕ 14_{1}. Thus H^{1}(G, ∧^{2}(6_{1})) = 0, by (b). It will therefore be enough to show H^{1}(G, S^{2}(6_{1})) = 0. Let E be the preimage in S^{2}(6_{1}) of k ≤ ∧^{2}(6_{1}) under the map in the second sequence of (1). Since S^{2}(6_{1})/E ≅ 14_{1}, we are reduced by (b) to showing H^{1}(G, E) = 0. Now E is an extension of k by 6_{2}, so the result will follow from (a) if we can show that the extension does not split. In our situation this is equivalent to (S^{2}(6_{1}))^{G} = 0. This means we must show that G does not preserve any nonzero quadratic form on 6_{1}. Such a form would be nondegenerate since 6_{1} is simple, so we would obtain an embedding of G into SO(6, k), which is impossible as the latter is isomorphic (as an abstract group) to SL(4, k) and G has no faithful 4-dimensional representation. This completes the proof.

2.3. To make further use of 2.2, Lemma 1 we need to have more detailed knowledge of certain modules.

Lemma 1. We have the following module structures where each row is a socle layer. (a)

6_{1} ⊗ 6_{1} : \begin{aligned}k & & 14_{1} \\ & 6_{2} \\ k & & 14_{1} \end{aligned} .

(b)

6_{1} ⊗ 14_{1} : 64_{1} ⊕ \begin{matrix}6_{1} \\ k \\ 6_{2} \\ k \\ 6_{1} \end{matrix} .

Proof: (a) A partial description of the structure of 6_{1} ⊗ 6_{1} has already been given in the proof of 2.2, Lemma 4(c). To complete the proof of (a), it will be sufficient to show that the subquotient M of 6_{1} ⊗6_{1} obtained by omitting the top and bottom trivial composition factors is uniserial (with factors 14_{1}, 6_{2}, 14_{1}). Inside G is a subgroup L ≅ SL(3, 2) (generated by the root subgroups in G_{2}(2) corresponding to long roots, so L lies inside a maximal subgroup of G isomorphic to G_{2}(2)′). A direct calculation shows that 6_{|L} ≅ V ⊕ V^{∗}, where V is the natural module for L. Since ∧^{2}(6_{1}) ≅ k ⊕ 14_{1}, we have

k ⊕ 14_{1}_{|L} ≅ ∧^{2}(V) ⊕ ∧^{2}(V^{∗}) ⊕ (V ⊗ V^{∗}) ≅ V^{∗} ⊕ V ⊕ W ⊕ k,

where W is the simple 8-dimensional kL-module isomorphic to the space of 3 × 3 matrices of trace zero. Thus, both 6_{1} and 14_{1} are semisimple for L and it follows that if the module M were not uniserial, then its restriction to L would have socle length 2. However, we have (6_{1} ⊗ 6_{1})_{|L} ≅ (V ⊕ V^{∗}) ⊗ (V ⊕ V^{∗}), and it is easily seen that the direct summand V ⊗ V of this module is uniserial with factors V^{∗}, V, V^{∗}. Thus, M_{|L} does not have socle length 2 and (a) is proved.

(b) The composition factors are easily checked. Also, 64_{1} splits off as a direct summand since it belongs to a different block from the other factors. Let U be the complementary summand. We have

Hom_{kG}(k ⊕ 6_{2}, 6_{1} ⊗ 14_{1}) ≅ Hom_{kG}(6_{1} ⊕ 36, 14_{1}) = 0,

so the only simple submodule of U must be 6_{1}. By self-duality, every simple quotient is also isomorphic to 6_{1}. It follows that U has a simple head and a simple socle, both isomorphic to 6_{1}. Then U′ := rad U/ soc U is self dual with composition factors k (twice) and 6_{2}. If U′ were not uniserial it would have to be semisimple and then U would have k ⊕ k in its second socle layer, contradicting H^{1}(G, 6_{1}) ≅ k) (2.2, Lemma 4(a)). Remark. . From the natural split surjection 6_{1} ⊗ ∧^{2}(6_{1}) → ∧^{3}(6_{1}), it is immediate that the module U in (b) is isomorphic to ∧^{3}(6_{1}). Since this module will reappear often we reserve the notation U for it.

We can now establish the triviality of 1-cohomology in many cases.

Lemma 2.

(a) H^{1}(G, U) = 0.

(b) H^{1}(G, 6_{1} ⊗ 14_{1}) = 0.

(c) Let M be a nonsplit extension of k by 6_{1} (unique up to isomorphism by 2.2, Lemma 4(a)). Then Ext_{G}^{1}(6_{1}, M) = 0. In particular there exists no uniserial kG-module with factors 6_{1}, k, 6_{1}.

(d) Ext_{G}^{1}(6_{i}, U) = 0.

(e) Ext_{G}^{1}(6_{1}, 84) = 0.

(f) Ext_{G}^{1}(14_{1}, 36) = 0.

(g) Ext_{G}^{1}(84, 84) = 0.

(h) Ext_{G}^{1}(36, 36) = 0.

(i) Ext_{G}^{1}(14_{1}, 14_{1}) = 0.

Proof: (a) Form characters, we see that U^{⟨z⟩} = U^{⟨xz⟩} and that this is a 2-dimensional subspace of the 8-dimensional space U^{⟨x⟩}. We will apply 2.2 Lemma 1 with L_{1} = C(x) ≅ SL(2, 4) × ⟨x⟩, L_{2} = C(z)′, which is a central extension of A_{6} by ⟨z⟩, and L_{0} = ⟨xz⟩. Since A_{6} has no nontrivial 2-dimensional modules over k, C(z)′ must act trivially on U^{⟨}z⟩, so d_{0} = d_{2} and hypothesis (i) of 2.2 Lemma 1 is satisfied. The triviality of the L_{2}-action on U^{⟨}z⟩ also yields H^{1}(L_{2}, U) = 0. It remains to show H^{1}(L_{1}, U) = 0. We shall prove the stronger statement that H^{1}(L_{1}, 6_{1} ⊗ 14_{1}) ≅ H^{1}(L_{1}/⟨x⟩, (6_{1} ⊗ 14_{1})^{⟨x⟩}) is zero. A straightforward computation using 2.2 Lemma 2 shows that

(6_{1} ⊗ 14_{1})^{⟨x⟩}) ≅ (V_{1} ⊗ X) ⊕ (V_{2} ⊗ V_{12}) ⊕ (V_{2} ⊗ V_{12}).

Now V_{1} ⊗X ≅ V_{1} ⊗V_{2} ⊗V_{2} ≅ V_{12} ⊗V_{2} and V_{12} is an injective module for L_{1}/⟨x⟩ ≅ SL(2, 4), so all summands are injective and the result follows.

(b) This follows from (a) and Lemma 1(b), since 64 is not in the principal block.

(c) We shall apply 2.2, Lemma 1 to the module 6_{1}⊗M. We choose L_{1} = C(x), L_{2} = C(x′), L_{0} = S. First, it is easy to check that M^{S} ≅ k. We claim that M^{L_{1}} = 0. If not, we would have M^{L_{1}} = M^{S}, but then the same would hold with L_{2} in place of L_{1} as they are conjugate and both contain S. This would imply M^{G} ≠ 0, contrary to the definition of M. From the claim it now follows using 2.2, Lemma 2(a) that

M_{|L_{1}} ≅ (Y ⊗ k) ⊕ (V_{2} ⊗ ω^{2}) ⊕ (V_{2} ⊗ ω^{2}),

where Y is a nonsplit extension of k by V_{1}. Using this and 2.2 Lemma 2(a) again one calculates that d_{1} = dim_{k}(6_{1} ⊗ M)^{L_{1}} = dim_{k} Hom_{L_{1}}(6_{1}, M) = 3. Hence d_{2} = 3 as well. Character computations show d_{0} = 6, so hypothesis (i) holds. By conjugacy of L_{1} and L_{2} it suffices to check (ii) for L_{1}, namely that H^{1}(L_{1}, 6_{1} ⊗ M) vanishes. By the above description of M_{|L_{1}} and 2.2, Lemma 2(a),

H^{1}(L_{1}, 6_{1} ⊗ M) ≅ H^{1}(L_{1}/⟨x⟩, (6_{1} ⊗ M)^{⟨x⟩})
≅ H^{1}(SL(2, 4), (V_{1} ⊗ Y) ⊕ (V_{2} ⊗ V_{2}) ⊕ (V_{2} ⊗ V_{2}))
≅ Ext_{SL(2,4)}^{1}(V_{1}, Y) ⊕ Ext_{SL(2,4)}^{1}(V_{2}, V_{2}) ⊕ Ext_{SL(2,4)}^{1}(V_{2}, V_{2}).

This reduces us to facts about k(SL(2, 4)), which are easily read off from the known structure of its projective modules (see [Benson]).

(d) First we show Ext_{G}^{1}(6_{1}, U) = 0. From Lemma 1(b) we see that U has an ascending filtration with factors M, 6_{2}, M^{∗}, with M as in (c). By (c) and 2.2, Lemma 3(a), we are left to show that Ext_{G}^{1}(6_{1}, M^{∗}) = 0, but this follows from 2.2, Lemma 4(a), using the long exact sequence associated to 0 → k → M^{∗} → 6_{1} → 0.

To show Ext_{G}^{1}(6_{2}, U) = 0, we filter U by factors 6_{1}, U′, 6_{1}, where U′ is uniserial with factors k, 6_{2}, k. By 2.2, Lemma 3(a), it suffices to show Ext_{G}^{1}(6_{2}, U′) = 0. Suppose we have an extension 0 → U′ → A → 6_{2} → 0. Since Hom_{KG}(6_{2}, U′) = 0, the sequence splits if and only if Hom_{kG}(6_{2}, A) ≠ 0, so we shall aim to show the latter. First we see that A cannot be uniserial, as this would violate the last statement of (c) (with 1 replaced by 2). Therefore there is a nonzero kG map A → k. The kernel has 2 composition factors 6_{1} and only one k, so we are done by 2.2, Lemma 3(a).

(e) We have

Ext_{G}^{1}(6_{1}, 84) ≅ Ext_{G}^{1}(6_{1}, 6_{2} ⊗ 14_{1}) ≅ Ext_{G}^{1}(6_{2}, 6_{1} ⊗ 14_{1}) ≅ Ext_{G}^{1}(6_{2}, 64_{1} ⊕ U) = 0,

using Lemma 1(b) for the third isomorphism and (d) for the vanishing.

(f) This is immediate from (e) since Ext_{G}^{1}(6_{1}, 6_{2} ⊗ 14_{1}) ≅ Ext_{G}^{1}(14_{1}, 6_{1} ⊗ 6_{2}).

(g) We have by Lemma 1(b),

Ext_{G}^{1}(84, 84) ≅ Ext_{G}^{1}(6_{1}⊗14_{2}, 6_{2}⊗14_{1}) ≅ Ext_{G}^{1}(6_{1}⊗14_{1}, 6_{2}⊗14_{2}) ≅ Ext_{G}^{1}(64_{1}⊕U_{1}, 64_{2}⊕U_{2}),

where U_{1} = U and U_{2} is its galois conjugate. By considering blocks, we see that the last group is isomorphic to Ext_{G}^{1}(64_{1}, 64_{2}) ⊕ Ext_{G}^{1}(U_{1}, U_{2}). The first summand is zero since 64_{1} ⊗ 64_{2}, being the restriction of the Steinberg module for G_{2}(4), is injective. and the second summand can be seen to be zero using the long exact sequence in conjunction with (a) and (d).

(h) This follows from (g) and

Ext_{G}^{1}(84, 84) ≅ Ext_{G}^{1}(6_{1} ⊗ 14_{2}, 6_{2} ⊗ 14_{1}) ≅ Ext_{G}^{1}(6_{1} ⊗ 6_{2}, 14_{2} ⊗ 14_{1})
≅ Ext_{G}^{1}(36, 36 ⊕ 160) ≅ Ext_{G}^{1}(36, 36).

(i) Dualizing the exact sequences in the proof of 2.2, Lemma 4(c) and making use of the self-duality of some modules, we have

0 → S^{2}(6_{1})^{∗} → 6_{1} ⊗ 6_{1} → ∧^{2}(6_{1}) → 0  and  0 → ∧^{2}(6_{1}) → S^{2}(6_{1})^{∗} → 6_{2} → 0. (1)

The structure of all the modules occuring in these sequences can be read off from Lemma 1(a). We claim that Ext_{G}^{1}(14_{1}, S^{2}(6_{1})^{∗}) ≅ k. Since ∧^{2}(6_{1}) ≅ 14_{1} ⊕ k (proof of 2.2 Lemma 4(c)) and Ext_{G}^{1}(14_{1}, 6_{1} ⊗ 6_{1}) ≅ Ext_{G}^{1}(6_{1}, 6_{1} ⊗ 14_{1}) = 0, by (d) and Lemma 1(b), our claim follows from long exact sequence for Ext_{G}(14_{1}, −) applied to the first sequence in (1). Now we apply Ext(14_{1}, −) to the second sequence in (1), resulting in

0 → Ext^{1}(14_{1}, ∧^{2}(6_{1})) → Ext_{G}^{1}(14_{1}, S^{2}(6_{1})^{∗}) → Ext_{G}^{1}(14_{1}, 6_{2})  (2)

Using the structure of 6_{1} ⊗ 6_{1} again, we see that the unique nonsplit extension of 14_{1} by S^{2}(6_{1})^{∗} established by the claim can be realised as the kernel K of any nonzero kG-map 6_{1}⊗6_{1} → k. One also sees from the structure that the quotient of K by ∧^{2}(6_{1}) is a nonsplit extension of 14_{1} by 6_{2}. This shows that the last map in (2) is not zero, hence injective by the claim. Thus, Ext^{1}(14_{1}, ∧^{2}(6_{1})) ≅ Ext^{1}(14_{1}, k ⊕ 14_{1}) = 0.

2.4. By making use of group automorphisms and various natural identities involving tensor products, it is not difficult to check that the results of the last section leave only the following extensions of simple modules to be computed: (i) Ext_{G}^{1}(14_{1}, 6_{2}) ≅ H^{1}(G, 84), (ii) Ext_{G}^{1}(6_{1}, 36), (iii) Ext_{G}^{1}(14_{1}, 84) and (iv) Ext_{G}^{1}(36, 84).

From 2.1 we see

Ext_{G}^{1}(14_{1}, 84) = Ext_{G}^{1}(14_{1}, 6_{1} ⊗ 14_{2}) ≅ Ext_{G}^{1}(6_{1}, 14_{1} ⊗ 14_{2}) ≅ Ext_{G}^{1}(6_{1}, 36 ⊕ 160),

so (ii) and (iii) have the same dimension e. Let d be the dimension of (i). These will turn out to be equal; we prove the easier inequality now and the reverse only after we have found e.

Lemma 1. 1 ≤ d ≤ e.

Proof: The first inequality follows from 2.3, Lemma 1(a). By definition of d, there is a kG-module E with socle isomorphic to 6_{2} and E/ soc E isomorphic to the direct sum of d copies of 14_{1}. Tensoring with 6_{1}, we obtain the short exact sequence

0 → 36 → E ⊗ 6_{1} → (14_{1} ⊗ 6_{1})^{d} → 0.

We now apply Hom_{G}(6_{1}, −) and consider the resulting long exact sequence. We have Hom_{G}(6_{1}, E ⊗ 6) ≅ Hom_{G}(6_{1} ⊗ 6_{1}, E), which can be seen to be zero from the structure of 6_{1} ⊗ 6_{1} (2.3, Lemma 1(a)). By 2.3, Lemma 1(b), we have dim_{k} Hom_{G}(6_{1}, (14_{1} ⊗ 6_{1})^{d} = d. Then the long exact sequence yield d ≤ dim_{k} Ext_{G}^{1}(6_{1}, 36) = e.

We need to know some more about structure of some small modules.

Lemma 2. We have the following module structures where the rows going from top to bottom in each summand are the radical layers.

(a)

6_{2} ⊗ 6_{1} ⊗ 6_{1} : \begin{aligned}6_{2} & & 84 \\ k & & 14_{2} \\ & 6_{1} \\ k & & 14_{2} \\ 6_{2} & & 84 \end{aligned} .

Moreover, the quotient by the unique copy of 84 in the socle does not have a trivial submodule. Neither does the quotient by the copy of 6_{2}.

(b) 6_{2} ⊗ 6_{2} ⊗ 14_{1} ≅ 6_{2} ⊗ 14_{2} ⊗ 6_{1} ≅ (64_{2} ⊗ 6_{1}) ⊕ (U_{2} ⊗ 6_{1}):

\begin{matrix}160 \\ 64_{1} \\ 160 \end{matrix}  ⨁  \begin{matrix}14_{1} & & 36 \\ & 6_{1} \\ & k \\ & 6_{2} \\ k & & 14_{1} \\ & 6_{1} \\ & 36 \end{matrix} .

In particular there is a unique quotient which is a nonsplit extension of 36 by 6_{1}.

(c) We have

(64_{1} ⊗ 14_{2}) ⊕ (U ⊗ 14_{2}) ≅ 6_{1} ⊗ 14_{1} ⊗ 14_{2} ≅ (6_{1} ⊗ 160) ⊕ (6_{1} ⊗ 6_{1} ⊗ 6_{2}).

Moreover, (14_{2} ⊗ 64_{1}) ⊕ 64_{2} ≅ 6_{1} ⊗ 160.

Proof: (a) Since we already know the structure of 6_{1} ⊗ 6_{1} from 2.3, Lemma 1, we see that 6_{2} ⊗ (6_{1} ⊗ 6_{1}) has a filtration with factors 6_{2} ⊕ 84 ≅ 6_{2} ⊗ (k ⊕ 14_{1}), 6_{2} ⊗ 6_{2}, 6_{2} ⊕ 84. The structure of the middle factor is known. Let M be the unique subquotient of 6_{1} ⊗ 6_{1}, obtained by omitting top and bottom trivial factors, which is uniserial with series 14_{1}, 6_{2}, 14_{1} and let N be the uniserial subquotient with series k, 6_{2}, k. Since we know the structure of 6_{2} ⊗ 6_{2} all statements will follow if we show that 6_{2} ⊗ M has no submodules 14_{2} or k (hence also no such quotients by self-duality) and that 6_{2}⊗N has no trivial submodule. We have Hom_{G}(14_{2}, 6_{2} ⊗ M) ≅ Hom_{G}(14_{2} ⊗ 6_{2}, M) = 0, from 2.3, Lemma 1(b) and it is clear that Hom_{G}(k, 6_{2} ⊗ M) ≅ Hom_{G}(6_{2}, M) = 0 and Hom_{G}(k, 6_{2} ⊗ N) ≅ Hom_{G}(6_{2}, N) = 0.

(b) The isomorphisms in the first line follow from 6_{1} ⊗14_{2} ≅ 84 ≅ 6_{2} ⊗14_{1} and 2.3, Lemma 1(b). Using again the structure of 6_{2} ⊗ 6_{2}, we see that (6_{2} ⊗ 6_{2}) ⊗ 14_{1} has a filtration with factors

14_{1} ⊕ (14_{1} ⊗ 14_{2}),  14_{1} ⊗ 6_{1},  14_{1} ⊕ (14_{1} ⊗ 14_{2}).

We recall from 2.1 and 2.3, Lemma 1(b) that 14_{1} ⊗ 14_{2} ≅ 160 ⊕ 36 and 14_{1} ⊗ 6_{1} ≅ 64_{1} ⊕ U.

Let us first consider the summand outside the principal block. The above shows it has a filtration with factors 160, 64_{1}, 160. Since Hom_{G}(64_{1}, 6_{2}⊗6_{2}⊗14_{1}) ≅ Hom_{G}(64_{1}⊗6_{2}, 84) = 0, just from characters, we see that this summand has the claimed structure. It is easy to see from characters that this summand of 6_{2} ⊗6_{2} ⊗14_{1} corresponds under the isomorphism in the statement to 64_{2} ⊗ 6_{1}.

The component in the principal block has a filtration with factors 14_{1} ⊕ 36, U, 14_{1} ⊕ 36. From the structure of U and Hom_{G}(6_{2} ⊗ 6_{2} ⊗ 14_{1}, 6_{1}) ≅ Hom_{G}(84, 36) = 0, we see that the top two radical layers of 6_{2} ⊗ 6_{2} ⊗ 14_{1} are 14_{1} ⊕ 36 and 6_{1} respectively. The last statement of (b) now follows from this, using Ext_{G}^{1}(14_{1}, 6_{1}) = 0 (2.3, Lemma 2(b)). With the structure of U known, the rest of the structure (the position of the second 14_{1} in the radical series) can be easily deduced from 2.2, Lemma 4(b), 2.3, Lemma 2(i) and Hom_{G}(6_{2} ⊗ 6_{2} ⊗ 14_{1}, 14_{1}) ≅ Hom_{G}(84, 84) ≅ k. This proves (b). (c). The main statement is immediate from the isomorphisms (2.1 and 2.3, Lemma 1(b)) 14_{1} ⊗ 14_{2} ≅ 160 ⊕ 36 and 6_{1} ⊗ 14_{1} ≅ 64_{1} ⊕ U. The last statement can be seen from the complementary statement U ⊗ 14_{2} ≅ (6_{1} ⊗ 36) ⊕ 64_{1}. First, these modules have the same composition factors. Also, by (a) 6_{1} ⊗ 36 has a simple quotient 6_{2}. Next, we have by Lemma 1(b),

Hom_{G}(U ⊗ 14_{2}, 6_{2}) ≅ Hom_{G}(14_{1} ⊗ 6_{1} ⊗ 14_{2}, 6_{2}) ≅ Hom_{G}(84, 84) ≅ k,

since Hom_{G}(64_{1} ⊗14_{2} ⊗6_{2}) = 0. This equation shows that both U ⊗14_{2} and 6_{1} ⊗14_{1} ⊗14_{2} have unique summands which have 6_{2} as a quotient, so we are done.

The proof of the key fact Ext_{G}^{1}(36, 84) = 0 will make use of several techniques which we have not discussed yet. First, the fact that various simple modules are restrictions of simple modules for G_{2}(k) will allow us to to apply some fundamental results about induced modules and good filtrations from the representation theory of algebraic groups. These theorems can be found in [Jantzen]. Data such as weights of Weyl modules, composition multiplicities in Weyl modules or tensor products of simple modules are straightforward to compute. They can also be found in [Bremner-Moody-Patera] (for weights), [Sin] and [Dowd-Sin]. Secondly, we need the submodule structure of projective modules in the nonprincipal block. This has been given in [Landrock].

Lemma 3.

(a) The radical series the projective covers of modules in the nonprincipal block are as follows (P (64_{2}) omitted):

P (64_{1}) :  64_{1} 160 64_{2} 160 64_{1} ;  P (160) :  160 64_{1}  64_{2} 160  160 64_{2}  64_{1} 160 .

(b) P (64_{1}) ⊗ 14_{2} ≅ P (36) ⊕ P (84) ⊕ (P (160))^{3}.

(c) P (64_{2}) ⊗ 14_{2} ≅ P (6_{2}) ⊕ (P (64_{2}))^{2} ⊕ (P (160))^{2}.

(d) P (160) ⊗ 14_{2} ≅ P (14_{1} ⊕ (P (36))^{2} ⊕ (P (160))^{6}.

Proof: For (a) we refer to [Landrock]. Part (b) is nothing but character calculations.

Lemma 4. Ext_{G}^{1}(36, 84) = 0.

Proof: From the filtration of P := P (64_{1}):

rad^{2} P ⊂ rad P ⊆ P

we have an induced filtration of P ⊗ 14_{2}. We now consider G-homomorphisms from the factors to 36. Since rad^{2} P is a quotient of P (64_{2}), Lemma 3(c) shows that

Hom_{G}(rad^{2} P ⊗ 14_{2}, 36) = 0.  (1)

Next since rad P/ rad^{2} P ≅ 160, we have

Hom_{G}((rad P/ rad^{2} P) ⊗ 14_{2}, 36) ≅ Hom_{G}(160 ⊗ 14_{2}, 36) ≅ Hom_{G}(160, 14_{2} ⊗ 6_{2} ⊗ 6_{1}) ≅ k,

(2)

by Lemma 2(b). Finally,

Hom_{G}(64_{1} ⊗ 14_{2}, 36) ≅ Hom_{G}(64_{1}, 14_{2} ⊗ 6_{2} ⊗ 6_{1}) = 0,  (3)

again by Lemma 2(b). We also need

Hom_{G}(64_{1} ⊗ 14_{2}, 84) ≅ Hom_{G}(64_{1} ⊗ 6_{2}, 14_{2} ⊗ 14_{1}) ≅ Hom_{G}(64_{2} ⊗ 6_{1}, 36 ⊕ 160) ≅ k, (4)

using Lemma 2(b).

Now Hom_{G}(P ⊗ 14_{2}, 36) ≅ k, by Lemma 3(b). Denote by N the maximal submodule of P ⊗14_{2} which is the kernel of a nonzero G-homomorphism to 84. Then by Lemma 3(b), we see that dim_{k} Ext^{1}(84, 36) = dim_{k} Hom_{G}(N, 36) − 1. By (4) and Lemma 3(b), we see that rad P ⊗ 14_{2} ⊂ N and that the image of N in the top factor (P ⊗ 14_{2})/((rad P) ⊗ 14_{2}) ≅ 64 − 1 ⊗ 14_{2} is the kernel K of a nonzero G-homomorphism 64_{1} ⊗ 14_{2} → 84. Now from (1), (2) and (3) it follows that

dim_{k} Ext^{1}(84, 36) ≤ dim_{k} Hom_{G}(K, 36).  (5)

Now Hom_{G}(K/ rad(64_{1} ⊗ 14_{2}), 36) = 0 by (3), so we may replace K by rad(64_{1} ⊗ 14_{2}) in (5). Then by Lemma 2(c), this may be replaced in turn by rad(6_{1} ⊗ 160). Since 6_{1} ⊗ 160 is a quotient of 6_{1} ⊗ (6_{2} ⊗ 64_{1}), by Lemma 2(b), and Hom_{G}(6_{1} ⊗ 6_{2} ⊗ 64_{1}, 36) ≅ Hom_{G}(6_{2} ⊗ 64_{1}, 6_{1} ⊗ 36) = 0, by Lemma 2, we arrive at the inequality

dim_{k} Ext^{1}(84, 36) ≤ dim_{k} Hom_{G}(rad(6_{2} ⊗ 6_{1} ⊗ 64_{1}), 36).  (6)

Our eventual aim is to show that the right hand side of (6) is zero. Before we can do this, we need to recall some ideas from the representation theory of the algebraic group G = G_{2}(k) (See [Dowd-Sin] II,§3). The simple rational modules L(λ) are paramentrized by non-negative integral combinations λ of the two fundamental dominant weights ω_{1} and ω_{2}. We denote by V (λ) the Weyl module, with simple quotient L(λ) and by H^{0}(λ) the induced module with socle L(λ). In the notation of [Dowd-Sin], we have that the restrictions to G of L(ω_{1}), L(ω_{2}) and L(ω_{1} + ω_{2}) are respectively 6_{1}, 14_{1} and 64_{1}. The Weyl module V (ω_{1}) has radical k, while V (ω_{2}) and V (ω_{1} + ω_{2}) are both simple, hence also isomorphic to the corresponding induced modules.

The module 6_{2} ⊗ (6_{1} ⊗ 64_{1}) is the restriction of the G-module V = L(2ω_{1}) ⊗ (L(ω_{1}) ⊗ L(ω_{1} + ω_{2})). Until further notice, all module structures are for G unless stated otherwise. A fundamental result on algebraic group representations [Jantzen, II, 4.19] states that the tensor product of two Weyl modules has an ascending filtration by Weyl modules, such that the highest weights of the factors appear in descending order with respect to the usual ordering on weights. Moreover, the multiplicities of the factors can be calculated from weight data. For V (ω_{1})⊗V (ω_{1}+ω_{2}), the factors (in ascending order) are: V (2ω_{1}+ω_{2}), V (2ω_{2}), V (3ω_{1}), V (ω_{1} + ω_{2}), V (2ω_{1}), V (ω_{2}).

Since V (ω_{1} + ω_{2}) = H^{0}(ω_{1} + ω_{2}) and Weyl modules can extend induced modules only trivially ([Jantzen], II, 4.13) we see that V (ω_{1})⊗V (ω_{1}+ω_{2}) has a simple quotient L(ω_{1}+ω_{2}). That it also has such a simple submodule is clear from the structure of V (ω_{1}). But L(ω_{1} + ω_{2}) occurs only once as a composition factor in V (ω_{1}) ⊗ V (ω_{1} + ω_{2}), so it must be a direct summand. It now follows that V = L(ω_{1}) ⊗ L(ω_{1} + ω_{2}) has a filtration with the other factors as subquotients:

V_{1} = V (2ω_{1} +ω_{2}), V_{2}/V_{1} = V (2ω_{2}), V_{3}/V_{2} = V (3ω_{1}), V_{4}/V_{3} = V (2ω_{1}), V/V_{4} = V (ω_{2}).

(7)

Since V is self-dual, there is also the dual filtration A_{1} ⊂ … ⊂ A_{5} = V by induced modules H^{0}(λ) (ascending, with the weights in the reverse order to (7)).

The composition factors of each Weyl module are easy to compute. By Considering these two filtrations and the structures of their factors, we shall show that V has a filtration F with the following properties:

(i) Every F-factor with L(ω_{1}) as a composition factor is a nonsplit extension of either k or L(3ω_{1}) by L(ω_{1})

(ii) The unique F-factor with L(2ω_{1}+ω_{2}) as a composition factor is a nonsplit extension

of L(2ω_{2}) by L(2ω_{1} + ω_{2}). By the calculation of G-extensions ([Dowd-Sin]), and a universal property of Weyl mod-

ules, we find that

rad_{G} V_{1}/ rad_{G}^{2} V_{1} ≅ k ⊕ L(ω_{2}) ⊕ L(2ω_{2}),  (8)

so W_{1} = rad_{G} V_{1} has no quotients L(ω_{1}). There is a unique G-homomorphism V → F_{5}/F_{4} ≅ H^{0}(2ω_{1} + ω_{2}) and this factors through V/W_{1}, by the uniqueness of L(2ω_{1} + ω_{2}) as a composition factor in V. Inside H^{0}(2ω_{1} + ω_{2}) is a nonsplit extension D of L(2ω_{2}) by L(2ω_{1} + ω_{2}), from (8). It is clear from considering composition factors that the preimage in V of D V_{2}. Let \overline{V}_{2} be the preimage of D in V/W_{1}. then rad_{G} \overline{V}_{2} ≅ \overline{V}_{1} ⊕ \overline{W}_{2} ≅ L(2ω_{1}+ω_{2})⊕rad_{G} V (2ω_{2}). Since Hom_{G}(rad_{G} V (2ω_{2}), L(ω_{1})) ≅ Ext_{G}^{1}(L(2ω_{2}), L(ω_{1})) = 0, by ([Dowd-Sin]), we see that if W_{2} is the preimage of \overline{W}_{2} in V, then we have a filtration W_{1} ⊂ W_{2} ⊂ V_{2} ⊂ V_{3} ⊂ V_{4} ⊂ V_{5} = V in which no factor has L(ω_{1}) as a homomorphic image and with the unique composition factor L(2ω_{1} + ω_{2}) appearing in the factor V_{2}/W_{2} as required in (ii). This filtration can now be further refined to satisfy (i), using the fact that the only composition factors of V which extend L(ω_{1}) as G-modules are k and L(3ω_{1}). Thus, the filtration F can be constructed.

Then F is of course also a filtration of G-modules. We claim that both (i) and (ii) remain valid for G, namely, that the extensions do not split on restriction. Since the groups of Gextensions between simple modules are at most one-dimensional, it suffices to find nonsplit G-extensions of these modules which do not split on restriction. It is easy to see that L(ω_{1}) ⊗ L(ω_{2}) has the same structure as its restriction 6_{1} ⊗ 14_{1}, given in 2.3, Lemma 1(b). so we have the first extension of (i). The second extension in (i) can be obtained by twisting the dual M^{∗} of the first by Frobenius and tensoring with L(ω_{1}), for Hom_{G}(M ⊗ 6_{1}, 6_{1}) ≅ Hom_{G}(M, 6_{1} ⊗ 6_{1}) = 0, by 2.3 Lemma 1(a). The extension in (ii) represents a nonzero class in Ext_{G}^{1}(L(2ω_{2}), L(2ω_{1}) ⊗ L(ω_{2})) ≅ Ext_{G}^{1}(L(ω_{2}), L(2ω_{1}) ⊗ L(2ω_{2})) The module L(2ω_{1})⊗L(2ω_{2}) has a simple module L(2ω_{1}) and the map Ext_{G}^{1}(L(ω_{2}), L(2ω_{1})) → Ext_{G}^{1}(L(ω_{2}), L(2ω_{1})⊗L(2ω_{2})) is injective (using 2.3, Lemma 1(b)). This proves the claim.

From now on we will consider only G-modules again.

First we observe that all G-composition factors 6_{1} or 84 of 6_{1} ⊗ 64 are restrictions of G-composition factors L(ω_{1}) and L(2ω_{1} + ω_{2}) of L(ω_{1}) ⊗ L(ω_{1} + ω_{2}), and that 6_{1} and 84 are the only composition factors with the property that the tensor product with 6_{2} has 36 as a composition factor. Next, calculations of Hom_{G} spaces shows that 84 is the only simple module in the head of the principal block summand of 6_{2} ⊗ 6_{1} ⊗ 64_{1}, and we have Hom_{G}(6_{2} ⊗ 6_{1} ⊗ 64_{1}, 84) ≅ Hom_{G}(6_{2} ⊗ 64_{1}, 6_{1} ⊗ 6_{1} ⊗ 14_{2}) ≅ k^{2}, by Lemma 2(b). Thus (6_{2} ⊗ 6_{1} ⊗ 64_{1})/ rad 6_{2} ⊗ 6_{1} ⊗ 64_{1} ≅ 84 ⊕ 84. Combining this information we see that the right hand side of (6) will be shown to be zero once we show that for each of the three special F-factors E in (i) and (ii), that E ⊗ 6_{2} does not have a homomorphic image which is 36 or a nonsplit extension of 84 by 36. First, let E be a nonsplit extension of k by 6_{1} or a nonsplit extension of 14_{2} by 84 Hom_{G}(E ⊗ 6_{2}, 36) ≅ Hom_{G}(E, (6_{2} ⊗ 6_{2}) ⊗ 6_{1}) = 0, Lemma 2(a). Also, Hom_{G}(E ⊗ 6_{2}, 84) ≅ Hom_{G}(E, 6_{2} ⊗ 6_{2} ⊗ 14_{1}) = 0, by Lemma 2(b).

Finally, let E be a nonsplit extension of 36 by 6_{1}. As we have mentioned before, we can write E as M ⊗ 6_{1}, where M is a nonsplit extension of 6_{2} by k. So E ⊗ 6_{2} = (M ⊗ 6_{2}) ⊗ 6_{1}. Since Hom_{G}(k, M ⊗ 6_{2}) ≅ Hom_{G}(M, 6_{2}) = 0, we see that M ⊗ 6_{2} has a submodule Y which is a nonsplit extension of k by 6_{2}. Then Y ⊗ 6_{1} is a nonsplit extension of 6_{1} by 36 and this submodule of M ⊗ 6_{2} ⊗ 6_{1} ≅ E ⊗ 6_{2} contains the unique composition factor 36. Therefore the conditions are checked for all three special types of F-factors and the Lemma is proved.

Lemma 5.

(a) (e =) dim_{k} Ext_{G}^{1}(36, 6_{1}) = 1.

(b) (d =) dim_{k} H^{1}(G, 84) = 1.

Proof: By 2.3, Lemma 2(a), we have 0 ≠ Ext_{G}^{1}(14_{1}, 6_{2}) ≅ H^{1}(G, 84), so (b) now follows from (a).

Lemma 2(b) shows that Ext_{G}^{1}(36, 6_{1}) ≠ 0, so we shall aim to prove that if E is any nonsplit extension of 36 by 6_{1}, then it is a homomorphic image of 6_{2} ⊗ 6_{2} ⊗ 14_{1}, for then Lemma 2(b) will imply the uniqueness of E up to isomorphism . We have Hom_{G}(6_{2} ⊗ 6_{2} ⊗ 14_{1}, E) ≅ Hom_{G}(84, 6_{2} ⊗ E). Now 6_{2} ⊗ E has a descending filtration with factors 6_{2} ⊗ 36 ≅ 6_{2} ⊗ 6_{2} ⊗ 6_{1}, 6_{2} ⊗ 6_{1} = 36. By Lemma 2(a), the first factor has a submodule isomorphic to 84, so by Lemma 4, we have the conclusion Hom_{G}(84, 6_{2} ⊗ E) ≠ 0. Finally we observe that any nonzero map in Hom_{G}(6_{2} ⊗ 6_{2} ⊗ 14_{1}, E) must be surjective, since Hom_{G}(6_{2} ⊗ 6_{2} ⊗ 14_{1}, 6_{1}) ≅ Hom_{G}(84, 36) = 0. This completes the proof.

Remark. In the proof of Lemma 4, we saw that the nonzero class in (a) can be realized by the tensor product of 6_{2} with the unique nonsplit extension of 6_{1} by k; the latter is the restriction of the corresponding Weyl module of G_{2}(k). Thus, this nonsplit extension is the restriction of one for the algebraic group. It is not hard to show that all of the nonsplit extensions in the principal block can be obtained similarly from extensions between simple modules for the algebraic group having 4-restricted highest weights, through the processes of restriction and twisting by Gal(F_{4}/F_{2}).

This completes the calculation of simple module extensions in characteristic 2.

§3. Characteristic 3

In §3 we assume k to be an algebraically closed field of characteristic 3. Our notation follows that of [Hiß-Lux], where decomposition matrices, Brauer trees and other charactertheoretic information may be found. Note that there are some differences in labelling of ordinary and 3-modular characters in the tables in [H-L], [Kondratev] and [Parker]/[ATLAS] (no two use the same labelling). The principal block has eight simple modules, as does the unique nonprincipal block of maximal defect of Ĝ. Besides these, there are three blocks of defect zero, with simple modules 189_{1}, 189_{2} and 216, and three blocks of 1.

We label the simple modules by their dimension. The modules with suffix “1” are conjugate to those with “2” by the outer automorphism.

The simple modules in the principal block are k, 13_{1}, 13_{2}, 21_{1}, 21_{2}, 57_{1}, 57_{2} and 133. All these are self-dual. In the other block of maximal defect we have 6_{1}, 6_{2}, 14, 36_{1}, 36_{2}, 50_{1}, 50_{2} and 236. All of these self-dual except for the 36-dimensional ones.

We will give details of the calculations only for the principal block. The same method yields the extensions in the other block of maximal defect; most of them result from a straightforward application of the method and then there are a few more stubborn cases ((236, 236), (236, 50_{1}) and (50_{1}, 50_{2})) where more complicated arguments (but still of the same kind) are needed.

Before going into details, we sketch the method we shall use to compute the extensions. The structures of the projective indecomposable modules in the blocks of defect 1 are easily found from the Brauer trees. By tensoring the projective indecomposable modules in the cyclic blocks with simple modules we obtain a projective module on which we have filtrations induced from any filtration projective in the cyclic block. Moreover, the decomposition of the tensor product into indecomposable summands is a purely character-theoretic question whose solution can be read off from the character table and decomposition matrix. For example, suppose P and Q are uniserial projective indecomposable modules with P the projective cover of rad Q, a common situation in blocks of defect 1. Then if S is any simple module, we can routinely find the decompositions P ⊗ S and Q ⊗ S into their indecomposable summands. From this, we derive information about the head of rad Q⊗S, which must be isomorphic to a submodule of the head of P ⊗S, and hence we obtain information about the indecomposable summands of Q ⊗ S. In practice, if S is small (we shall use modules of dimensions 6 and 13 in our calculations), their may be very few summands of S ⊗ Q in the block of interest and many extensions of simple modules can be deduced immediately using this information, in conjunction with the Cartan matrix and knowledge of the composition factors of tensor products of simple module.

3.1 The principal 3-block. In this subsection, we compute the extensions between simple modules in the principal block. Let us first dispose of some trivial cases:

Ext1 ; _{G}(k, k) = 0,  Ext1 ; _{G}(21_{1}, 57_{1}) = 0,  Ext1 ; _{G}(21_{1}, 21_{2}) = 0,  Ext1 ; _{G}(6_{1}, 21_{2}) = 0.

(1)

The first is obvious, the second is because the corresponding Cartan number is zero and the third is because 21_{1} ⊗ 21_{2} has no component in the principal block.

The next three lemmas simply collect together some relevant character-theoretic information. The computations involved are routine. In order to avoid irrelevant information, we will usually write “≅_{0}” to mean isomorphism of principal block components.

Lemma 1. We have the following uniserial projective modules in blocks of defect 1, with composition series as given:

P (36) : 36, 90, 36;  P (90) : 90, 36, 90.  (a)
P (126_{1}) : 126_{1}, 126_{2}, 126_{1};  P (126_{2}) : 126_{2}, 126_{1}, 126_{2}.  (b)

Lemma 2.

(a) P (126_{1}) ⊗ 6_{1} ≅_{0} P (21_{2}) ⊕ P (57_{1}).

(b) P (126_{2}) ⊗ 6_{1} ≅_{0} P (133).

(c) P (90) ⊗ 13_{1} ≅_{0} P (13_{1}).

(d) P (36) ⊗ 13_{1} ≅_{0} P (13_{2}).

Lemma 3. We have the following composition factors, with multiplicities written in the exponents.

(a) 126_{1} ⊗ 6_{1} ≅_{0} {13_{1}^{2}, 21_{2}^{2}, 57_{1}^{2}, 133}.

(b) 126_{2} ⊗ 6_{1} ≅_{0} {k^{3}, 13_{1}^{2}, 13_{2}^{3}, 21_{1}^{2}, 21_{2}, 57_{2}, 133^{4}}.

(c) 90 ⊗ 13_{1} ≅_{0} {k^{2}, 13_{1}^{4}, 13_{2}, 21_{2}, 57_{1}^{2}, 133^{2}}.

(d) 36 ⊗ 13_{1} ≅_{0} {k^{2}, 13_{1}, 13_{2}^{4}, 21_{1}, 57_{2}^{2}, 133^{2}}.

Proposition 4. The following hold:

(a) Ext1 ; _{G}(21_{2}, k) = 0.

(b) Ext1 ; _{G}(21_{2}, 13_{2}) = 0.

(c) Ext1 ; _{G}(57_{1}, k) = 0.

(d) Ext1 ; _{G}(57_{1}, 13_{2}) = 0.

(e) Ext1 ; _{G}(57_{1}, 57_{2}) = 0.

(f) Ext1 ; _{G}(13_{1}, 13_{1}) = 0.

Proof: This uses only the parts (a) and (b) of Lemmas 2 and 3. Note first that Lemma 2(b) implies that rad P (126_{1}) ⊗ 6_{1} has a simple head isomorphic to 133. It follows that for any simple module S other than 133, we have

Ext1 ; _{G}(21_{2} ⊕ 57_{1}, S) ≅ Hom_{G}(rad(126_{1} ⊗ 6_{1}), S).  (2)

Parts (a)-(e) now follow from the fact that none of the corresponding simple modules S is a composition factor of rad(126_{1} ⊗ 6_{1}), and (f) from (a), (c) and the isomorphism 13_{1} ⊗ 13_{1} ≅ k ⊕ 21_{1} ⊕ 57_{1} ⊕ 90, which is easily verified (using self-duality) from character data.

To go further, we need to look at some small modules. In the following lemma the socle series are described.

Lemma 5.

(a) 36 ≅ 6_{1} ⊗ 6_{2}.

(b) 126_{1} ≅ 6_{1} ⊗ 21_{2}.

(c)

6_{1} ⊗ 6_{1} ≅ \begin{matrix}k \\ 13_{1} \\ k \end{matrix} ⊕ 21_{1}.

(d)

21_{1} ⊗ 21_{1} ≅_{0} \begin{matrix}k \\ 13_{1} \\ k \end{matrix} ⊕ 21_{1}.

(e) soc(21_{2} ⊗ 13_{1}) ≅ 57_{1}.

(f)

6_{1} ⊗ 13_{1} ≅ 14 50_{1} 14 .

(g)

6_{2} ⊗ 14 ≅ \begin{aligned}& 13_{2} \\ k & & 57_{2} \\ & 13_{2} \end{aligned} .

(h)

6_{2} ⊗ 50_{1} ≅ 133 13_{1}  21_{1} 133 .

(i)

6_{1} ⊗ 21_{1} ≅ 50_{1} 14  6_{2} 50_{1}  ⊕ 6_{1}.

(j)

13_{1} ⊗ 21_{1} ≅_{0} \begin{aligned}& 13_{1} \\ k & & 57_{1} \\ & 13_{1} \end{aligned} .

Proof: The composition factors are obtained from the character data, so only the module structure needs proof. There is nothing to prove in (a) and (b). Parts (c) and (d) follow from the decomposition of the tensor square into the symmetric and alternating squares and the fact that the tensor square has one-dimensional fixed points. For (e), we calculate first that 21_{2} ⊗ 13_{1} ≅_{0} {13_{1}^{2}, 57_{1}^{2}, 133}. Since [13_{1} ⊗ 13_{1} : 21_{2}] = 0 and since 133 cannot be a direct summand as its dimension is not divisible by 3 ([Benson-Carlson] Prop. 2.2),

(e) is proved. That the module in (f) must have simple head and socle is clear from the fact that it is a tensor factor of 36 ⊗ 13_{1}, which in turn is a quotient of P (13_{2}), By Lemma 2(d), so (f) holds by self-duality. It then follows from (f) and Lemma 2(d) that 6_{2} ⊗ 14 has simple head and socle 13_{2} so, by self-duality, (g) holds. It is easily checked that [6_{2} ⊗ 13_{1} : 50_{1}] = 0 = [6_{2} ⊗ 21_{1} : 50_{1}] and (h) follows from this. Similar considerations of Hom s establish (i).

Proposition 6.

(a) Ext1 ; _{G}(57_{1}, 57_{1}) = 0.

(b) Ext1 ; _{G}(21_{2}, 57_{1}) = 0.

(c) Ext1 ; _{G}(13_{2}, k) ≅ k.

(d) Ext1 ; _{G}(13_{2}, 57_{2}) ≅ k.

(e) Ext1 ; _{G}(13_{2}, 21_{1}) = 0.

(f) Ext1 ; _{G}(13_{2}, 133) ≅ k.

(g) Ext1 ; _{G}(133, 21_{1}) ≅ k.

(h) Ext1 ; _{G}(133, 57_{2}) = 0.

(i) Ext1 ; _{G}(13_{1}, 13_{2}) = 0.

(j) Ext1 ; _{G}(133, k) = 0.

(k) Ext1 ; _{G}(133, 133) = 0.

(l) Ext1 ; _{G}(21_{1}, 21_{1}) = 0.

Proof: To prove (a) and (b), we return to (2) in the proof of Proposition 4. By selfduality we have soc(126_{1} ⊗ 6_{1}) ≅ 21_{2} ⊕ 57_{1}. We claim that every map in the right hand side of (2) has the simple submodule 57_{1} in its kernel. By Lemma 5 (b) we have 126_{1} ⊗ 6_{1} ≅ 21_{2} ⊗ (6_{1} ⊗ 6_{1}). The only composition factor of 6_{1} ⊗ 6_{1} whose tensor product with 21_{2} has 57_{1} as a composition factor is 13_{1}. The claim now follows from Lemma 5(b). Now (a) and (b) follow from the claim and (2).

Next we consider the module P (13_{2}), which by Lemma 2 (d) has a filtration with factors equal to the principal block parts of 36 ⊗ 13_{1}, rad P (36) ⊗ 13_{1}. since the latter has simple head 13_{1}, by Lemma 2 (c), we see that for any simple module S 13_{1}, we have

Ext1 ; _{G}(13_{2}, S) ≅ Hom_{G}(rad /(36 ⊗ 13_{1}), S).  (3)

Now 36 ⊗ 13_{1} ≅ 6_{2} ⊗ (6_{1} ⊗ 13_{1}) has, by Lemma 5 (f), a filtration with factors 6_{2} ⊗ 14, 6_{2} ⊗ 50_{1}, 6_{2} ⊗ 14. The structures of these factors are given in (g) and (h) of that lemma and (c), (d) (e) and (f) now follow from this information, using (3). We can now obtain (l) from Lemma 5(d) and existing results (Proposition 4 (a) and part (c) of this proposition).

Now we examine P (133) ≅_{0} P (126_{2} ⊗ 6_{1}). From its filtration induced by that of P (126_{2}), we obtain for S 21_{2}, 57_{1},

Ext1 ; _{G}(133, S) ≅ Hom_{G}(rad /(126_{2} ⊗ 6_{1}), S).  (4)

By Lemma 5(b), we have 126_{2} ⊗ 6_{1} ≅ 6_{2} ⊗ (21_{1} ⊗ 6_{1}) and by Lemma 5(i), this is has a filtration with factors 6_{2} ⊗ 50_{1}, 6_{2} ⊗ (14 ⊕ 6_{2}), 6_{2} ⊗ 50_{1}. The structures of the factors are given in Lemma 5 (g) and (h). From this information, we can deduce (g) and (h) from (4). To prove (i), we use the embedding 13_{1} ↪ 13_{1} ⊗ 21_{1} in Lemma 5(i), from which we obtain an embedding Ext1 ; _{G}(13_{2}, 13_{1}) into Ext1 ; _{G}(13_{2}, 13_{1} ≅ 21_{1}) ≅ Ext1 ; _{G}(13_{2} ⊗ 13_{1}, 21_{1}). Since 13_{1} ⊗ 13_{2} ≅_{0} 133, (g) implies (i) and also (i) implies (j). To prove (k) using (4), we need to examine 126_{2} ⊗ 6_{1} ≅ 6_{2}(⊗6_{1} ⊗ 21_{1}) more closely. By Lemma 5 (i), the module 6_{1} ⊗ 21_{1} has a uniserial submodule U with composition factors 14, 50_{1}. and the quotient of rad(6_{2} ⊗ 6_{1} ⊗ 21_{1}) by 6 ⊗ U has no nonzero homomorphisms to 133, so it is enough to prove

Hom_{G}(6_{2} ⊗ U, 133) = 0.  (5)

We claim that U is a qotient of 6_{1} ⊗ 13_{1}. Assuming this, we obtain

Hom_{G}(6_{2} ⊗ U, 133) ≤ Hom_{G}(6_{2} ⊗ 6_{1} ⊗ 13_{1}, 133) ≅ Hom_{G}(36 ⊗ 13_{1}, 133) = 0,

by Lemma 2(d), which proves (5). It remains to verify the claim. By Lemma 5(f) and (i), any nonzero homomorphism from 6_{1} ⊗ 13_{1} to 6_{1} ⊗ 21 maps onto U. We compute:

Hom_{G}(6_{1} ⊗ 13_{1}, 6_{1} ⊗ 21_{1}) ≅ Hom_{G}(6_{1} ⊗ 6_{1}, 13_{1} ⊗ 21_{1}).

The fact that this space is not zero follows from Lemma 5 (c) and (j), using Ext1 ; _{G}(k, 13_{1}) ≅ k (part (c) of this proposition).

Up to automorphisms and duality we have now computed all entries in Table 2.

§4. Characteristic 5

Our notation matches [Hiß-Lux], which is our main source for data concerning Brauer characters, decomposition numbers and products of characters.

In §4 we assume k to be an algebraically closed field of characteristic 5.

4.1. The simple kĜ-modules will be denoted by their dimensions. There are two simple modules of dimension 14 of which one is faithful; this we denote by ˆ14. There are also two mutually dual simple modules of dimension 50, but these play no part in our discussion.

The group G has five 5-blocks. There are three blocks of defect zero containing the simple modules 175, 225 and 300, a block of defect 1 with two simple modules 70 and 90, and the principal block, with Cartan matrix 4.1a) The principal 5-block of G.

k 14 21 41 85 189
k 3 1 2 1 0 0
14 1 6 3 0 1 3
21 2 3 7 2 1 3
41 1 0 2 3 2 1
85 0 1 1 2 3 2
189 0 3 3 1 2 6

The projective indecomposable modules in the block of defect 1 have the following structures:

P (70) : 70 70  90 70 ,  P (90) :  90 70 90 .  (1)

The group Ĝ has four further blocks containing faithful modules. Three, containing 50_{1}, 50_{2} and 350 respectively, are of defect zero, while the fourth whose Cartan matrix is given below has maximal defect.

4.1b) The non-principal 5-block of Ĝ of maximal defect.

6 ˆ14 56 64 190 202
6 6 1 3 3 1 0
ˆ14 1 4 1 2 0 2
56 3 1 7 3 2 2
64 3 2 3 6 0 1
190 1 0 2 0 2 1
202 0 2 2 1 1 3

We now describe some tensor products of simple modules. An expression in braces indicates module having the enclosed composition factors and multiplicities are indicated by an exponent.

Lemma 1.

(a) 14 ⊗ 14 ≅ k ⊕ 14 ⊕ 21 ⊕ 70 ⊕ 90.

(b) 14 ⊗ 21 ≅ 14 ⊕ 21 ⊕ 189 ⊕ 70.

(c) 14 ⊗ 6 ≅ 6 ⊕ ˆ14 ⊕ 64.

(d) 14 ⊗ 90 ≅ {14^{3}, 21, 85, 189^{3}} ⊕ {70^{2}, 90^{2}} ⊕ 225.

(e) 14 ⊗ 70 ≅ {k, 14^{3}, 21^{4}, 41, 85, 189^{3}} ⊕ 70 ⊕ 90.

(f) 14 ⊗ 189 ≅ {14^{3}, 21^{4}, 41^{2}, 85^{3}, 189^{7}} ⊕ 70 ⊕ 90 ⊕ 175 ⊕ 225 ⊕ 300.

(g) 14 ⊗ 41 ≅ 85 ⊕ 189 ⊕ 300.

(h) 21 ⊗ 21 ≅ k ⊕ 14 ⊕ 21 ⊕ 41 ⊕ 85 ⊕ 189 ⊕ 90.

(i) 21 ⊗ 41 ≅ {14, 21^{2}, 41^{2}, 85^{2}, 189^{2}} ⊕ 175.

(j) 6 ⊗ 6 ≅ k ⊕ 14 ⊕ 21.

(k) 6 ⊗ ˆ14 ≅ 14 ⊕ 70.

(l) 6 ⊗ 56 ≅ 21 ⊕ 41 ⊕ 85 ⊕ 189.

(m) 6 ⊗ 64 ≅ 14 ⊕ 21 ⊕ 70 ⊕ 90.

(n) 6 ⊗ 41 ≅ 56 ⊕ 190.

(o) 6 ⊗ 85 ≅ {6, 56^{2}, 190, 202}.

(p) 6 ⊗ 70 ≅ {6, ˆ14^{2}, 56, 64^{2}, 202}.

(q) 6 ⊗ 90 ≅ {6, 56, 64^{2}} ⊕ 350.

(r) ˆ14 ⊗ 70 ≅ {6^{4}, 56^{4}, 64^{3}, 190} ⊕ 350.

(s) ˆ14 ⊗ 90 ≅ { ˆ14^{4}, 56, 64^{3}, 202^{3}} ⊕ 350.

Proof: These results all follow from character computations, block decomposition and self-duality.

Next, we consider tensor products of small projective modules with simple modules. Again, the proof is pure calculation with characters.

Lemma 2.

(a) P (90) ⊗ 14 ≅ P (14) ⊕ P (189) ⊕ P (70) ⊕ P (90)^{2} ⊕ 225.

(b) P (70) ⊗ 14 ≅ P (14) ⊕ P (21) ⊕ P (189) ⊕ P (70) ⊕ P (90)^{2} ⊕ 225.

(c) P (90) ⊗ 6 ≅ P (64) ⊕ 350^{2}.

(d) P (70) ⊗ 6 ≅ P ( ˆ14) ⊕ P (64) ⊕ 350.

(e) P (90) ⊗ ˆ14 ≅ P ( ˆ14) ⊕ P (64) ⊕ P (202) ⊕ 350^{3}.

(f) P (70) ⊗ ˆ14 ≅ P (6) ⊕ P (56) ⊕ P (64) ⊕ 350^{4}.

(g) P (k) ⊗ 14 ≅ P (14) ⊕ P (70) ⊕ 300.

(h) P (k) ⊗ 21 ≅ P (21) ⊕ P (85) ⊕ P (90) ⊕ 175.

(i) P (k) ⊗ 6 ≅ P (6).

From the above two lemmas, we can obtain information about the heads (maximal semisimple quotients) of certain tensor products which will appear as subquotients in filtrations of projective modules in subsequent Ext^{1} calculations. By self-duality, the socles are the same.

Lemma 3.

(a) hd(90 ⊗ 14) ≅ 14 ⊕ 189 ⊕ 70 ⊕ 90 ⊕ 225.

(b) hd(70 ⊗ 14) ≅ 14 ⊕ 21 ⊕ 189 ⊕ 70 ⊕ 90.

(c) hd(90 ⊗ 6) ≅ 64 ⊕ 350.

(d) hd(70 ⊗ 6) ≅ ˆ14 ⊕ 64.

(e) hd(90 ⊗ ˆ14) ≅ ˆ14 ⊕ 64 ⊕ 202 ⊕ 350.

(f) hd(70 ⊗ ˆ14) ≅ 6 ⊕ 56 ⊕ 64.

Proof: Upper bounds on the multiplicities are given by the previous lemma since, for example, 90 ⊗ 14 is a quotient of P (90) ⊗ 14. The exact multiplicities are then easily found by elementary considerations of composition factors, self-duality, etc, using Lemma 1. We give the details for (a) only. The occurence of 225 is clear from Lemma 1(d), that of 14 follows from Lemma 1(a) and that of 189 from Lemma 1(f). By Lemma 1(e), 70 appears, so we need only determine the multiplicity of 90. If it appeared twice then 90 ⊕ 90 would be a direct summand of 90 ⊗ 14. It would then follow from the structure of P (90) that P (90) ⊗ 14 had a subquotient {90^{5}, 70}, which contradicts what we know about P (90).

Remark. A general result ([Benson-Carlson] Prop. 2.2) states that if the field characteristic divides the dimension of an indecomposable module then it divides the dimension of every direct summand of the tensor product of that module with another. We shall use this here and on one later occasion. It has the following consequence: Let M be the component in a block of maximal defect of one of the modules whose heads are given in Lemma 3. Then soc M ⊆ rad M. Here we could also have argued using relative projectivity, but not in our later application.

3.2 Extensions. In several instances, where the Cartan number is zero or is a 2 on the leading diagonal, it is clear that there are only split extensions between the corresponding simple modules, so these entries in the table of extensions can be filled in immediately, and of course, Ext1 ; _{G}(k, k) = 0.

In the first result the filtrations of P (90)⊗14 and P (70)⊗14 induced by certain filtrations of P (90) and P (70) will be used to calculate a number of Ext^{1} groups between simple modules. Further extensions can then be computed from formal manipulations with the equations of Lemma 1. In the succeeding propositions we apply the same method to other tensor products of projective modules with simple modules. Nearly all entries in the tables of extensions for characteristic 5 will be found with little difficulty by this process. The rest require some extra arguments, but still based on the same ideas.

Proposition 1.

(a) Ext1 ; _{G}(14, k) = 0.

(b) Ext1 ; _{G}(189, 41) = 0.

(c) Ext1 ; _{G}(21, k) = k.

(d) Ext1 ; _{G}(21, 41) = k.

(e) Ext1 ; _{G}(14, 14) = k.

(f) Ext1 ; _{G}(14, 85) = 0.

(g) Ext1 ; _{G}(189, 85) = k.

(h) Ext1 ; _{G}(21, 85) = 0.

(i) Ext1 ; _{G}(14, 21) = k.

(j) Ext1 ; _{G}(6, 6) = k.

(k) Ext1 ; _{G}(6, ˆ14) = 0.

(l) Ext1 ; _{G}(6, 56) = k.

(m) Ext1 ; _{G}(6, 64) = k.

Proof: We consider the filtration rad P (90) ⊗ 14 ⊂ P (90) ⊗ 14. Since rad P (90) is a quotient of P (70), it follows that rad P (90) ⊗ 14 is a quotient of P (70) ⊗ 14, whose head is given in 3.1, Lemma 3. In particular, neither k nor 41 appears. By Lemma 2,

P (90) ⊗ 14 ≅ P (14) ⊕ P (189) ⊕ (part outside principal block).  (1)

Since we have composition multiplicities [90 ⊗ 14 : k] = 0 and [90 ⊗ 14 : 41] = 0, we see that k and 41 are not quotients of rad(P (90) ⊗ 14), proving (a) and (b).

Now consider the filtration rad P (70)⊗14 ⊂ P (70)⊗14. Since rad P (70) is a quotient of P (70) ⊕ P (90), it follows that rad P (90) ⊗ 14 is a quotient of (P (70) ⊗ 14) ⊕ (P (90) ⊗ 14), whose head is given in Lemma 3. In particular k and 41 do not appear. By Lemma 2,

P (70) ⊗ 14 ≅ P (14) ⊕ P (21) ⊕ P (189) ⊕ (part outside principal block).  (2)

Since we already know Ext1 ; _{G}(14, k ⊕ 41) = 0 and Ext1 ; _{G}(189, k ⊕ 41) = 0, from (a), (b) and the Cartan matrix, it follows that

Ext1 ; _{G}(21, k) ≅ Hom_{G}(rad / soc(70 ⊗ 14), k),  (3)

and similarly for 41. Now, ignoring parts outside the principal block, rad / soc(70 ⊗ 14) ≅ {21^{2}, k, 41, 14, 189, 85}. We claim (c) and (n) hold. If not, then Ext1 ; _{G}(21, k) = 0 and Ext1 ; _{G}(21, 41) = 0, but then using self-duality, we see that rad / soc(70 ⊗ 14) has a quotient k ⊕ 41, contrary to (3).

Now (e) is immediate from Lemma 1(a).

Let us return to the filtration of P (90) ⊗ 14 at the beginning. We know by Lemma 2(b) that 85 is not a quotient of rad P (90) ⊗ 14 so we have by (1),

Ext1 ; _{G}(14 ⊕ 189, 85) ≅ Hom_{G}(rad / soc(90 ⊗ 14), 85).  (4)

It is clear from self-duality that (in the principal block)

rad / soc(90 ⊗ 14) ≅ 14 ⊕ 21 ⊕ 85 ⊕ 189.  (5)

Thus, Ext1 ; _{G}(14⊕189, 85) ≅ k. From (5), (1) and self-duality it follows that rad / soc(P (14)⊕ P (189) has a unique submodule 85 and unique quotient 85, with the submodule being in the kernel of a quotient map. Therefore by self-duality, the image by projection onto exactly one of rad / soc P (14) or rad / soc P (189) has the same property. Since [P (14) : 85] = 1, it must be the latter. This proves (f) and (g).

Next, returning to the filtration of P (70) ⊗ 14, we see from (2) and Lemma 2 (a), (b) that Ext1 ; _{G}(14⊕21⊕189, 85) ≅ Hom_{G}(rad / soc(70⊗14), 85), so (g) and [rad / soc(70⊗14) : 85] = 1 imply (h).

Parts (i)-(m) are easily derived from the identities in Lemma 1 and the parts already proved.

Next we consider P (k) ⊗ 14 and P (k) ⊗ 21.

Proposition 2.

(a) Ext1 ; _{G}(k, 41) = 0.

(b) Ext1 ; _{G}(6, 190) = 0.

(c) Ext1 ; _{G}(21, 21) = k.

(d) Ext1 ; _{G}(14, 189) = k.

(e) Ext1 ; _{G}(21, 189) = k.

(f) Ext1 ; _{G}(189, 189) = k.

Proof: If Ext1 ; _{G}(k, 41) ≠ 0, then by Lemma 2(g), 14 extends any simple quotient of 41⊗14 nontrivially. Therefore (a) follows from Lemma 1(g) and Proposition 1(f). For (b), we use Lemma 1(h) and (j) to obtain

Ext1 ; _{G}(6, 56 ⊕ 190) ≅ Ext1 ; _{G}(6, 6 ⊗ 41) ≅ Ext1 ; _{G}(6 ⊗ 6, 41)

≅ Ext1 ; _{G}(k ⊕ 14 ⊕ 21, 41),

which gives a reduction to Proposition 1 and (a).

(c) follows from Lemma 1() using (a) and Proposition 1.

Lemma 1(a) and (b) yield

Ext1 ; _{G}(14, 14 ⊕ 21 ⊕ 189) ≅ Ext1 ; _{G}(14, 14 ⊗ 21) ≅ Ext1 ; _{G}(14 ⊗ 14, 21)

≅ Ext1 ; _{G}(k ⊕ 14 ⊕ 21, 21),

and now (d) follows from existing results.

From Lemma 2(h), Proposition 1(c) and Lemma 1(h), we see that Ext1 ; _{G}(21, 189) ≠ 0. Equation (1) of Proposition 1 shows that dim Ext1 ; _{G}(14⊕189, 21) = dim Hom_{G}(rad(P (90)⊗ 14), 21) and from the filtration of P (90) ⊗ 14 and the structure of 90 ⊗ 14 and 70 ⊗ 14 described there, one sees that this dimension is at most 2. Now (e) follows from Proposition 1(i).

To prove (f), we use Lemma 1(b), (a) and (h), to obtain

Ext1 ; _{G}(14 ⊕ 21 ⊕ 189 ⊕ 70, 14 ⊕ 21 ⊕ 189 ⊕ 70)

≅ Ext1 ; _{G}(14 ⊗ 21, 14 ⊗ 21) ≅ Ext1 ; _{G}(14 ⊗ 14, 21 ⊗ 21)

≅ Ext1 ; _{G}(k ⊕ 14 ⊕ 21 ⊕ 70 ⊕ 90, k ⊕ 14 ⊕ 21 ⊕ 41 ⊕ 85 ⊕ 189 ⊕ 90),

and all extensions in the first and last terms other than the one in question are known. Remark. It is now easy to see that P (k) has radical (and socle) series

\begin{aligned}& k \\ & 21 \\ k & 14 & 41 \\ & 21 \\ & k \end{aligned} .

Next we consider the tensor products of P (70) and P (90) with 6 and ˆ14.

Proposition 3.

(a) Ext1 ; _{G}(64, 202) = 0.

(b) Ext1 ; _{G}(64, 56) = k.

(c) Ext1 ; _{G}( ˆ14, 202) = k.

(d) Ext1 ; _{G}( ˆ14, 56) = 0.

(e) Ext1 ; _{G}( ˆ14, 64) = k.

(f) Ext1 ; _{G}(202, 190) = 0.

(g) Ext1 ; _{G}(202, 202) = 0.

(h) Ext1 ; _{G}(202, 56) = k.

(i) Ext1 ; _{G}(56, 190) = k.

(j) Ext1 ; _{G}( ˆ14, ˆ14) = 0.

(k) Ext1 ; _{G}(64, 64) = k.

(l) Ext1 ; _{G}(56, 56) = k.

(m) Ext1 ; _{G}(85, 41) = k.

(n) Ext1 ; _{G}(41, 41) = 0.

Proof: From Lemma 3 (c), (d) and Lemma 1 (p), (q) we see that 70 ⊗ 6 and 90 ⊗ 6 have the folowing structures:

70 ⊗ 6 :  ˆ14  64 6 56 202 ˆ14  64 ,  90 ⊗ 6 : 64 6  56 64 ⊕ 350.  (6)

Since these modules filter P (90) ⊗ 6 ≅ P (64) ⊕ 350^{2} we can deduce (a) and (b). Since they also filter P (70) ⊗ 6 ≅ P (14) ⊕ P (64) ⊕ 350, we also obtain (c) (using (a)) and (d) (using (b)). The module rad P (90) ⊗ 6 has a simple quotient ˆ14 and we have seen that the composition factors 6 and 56 of 90 ⊗ 6 do not extend ˆ14, which yields (e). Next we consider 90 ⊗ ˆ14 and 70 ⊗ ˆ14, which filter P (90) ⊗ ˆ14 ≅ P ( ˆ14) ⊕ P (64) ⊕ P (202) ⊕ 350^{3}. Since rad P (90) ⊗ ˆ14 is a quotient of P (70) ⊗ 14 and since [90 ⊗ ˆ14 : 190] = 0, we deduce (f). Since by Lemma 1(e) and Lemma 3(e), [rad / soc(90 ⊗ ˆ14) : 202] = 1, (g) follows from (c). Also since [rad / soc(90 ⊗ ˆ14) : 56] = 1 and 56 appears only once in the head of rad P (90)⊗ ˆ14, we see that dim Ext1 ; _{G}( ˆ14⊕64⊕202, 56) ≤ 2 from which it follows by (b) that dim Ext1 ; _{G}(202, 56) ≤ 1. To prove (h), we need to find a nonsplit extension of 56 by 202. By Lemma 1, 6 ⊗ 85 ≅ {6, 56^{2}, 202, 190}. We have Hom_{G}(6 ⊗ 85, 56) ≅ Hom_{G}(85, 6 ⊗ 56) ≅ k by Lemma 1(l). The subquotient of 6 ⊗ 85 obtained by omitting the top and bottom factors 56 must be semisimple by self-duality. Since any direct summand of 6 ⊗ 85 must have dimension divisible by 5 (Remark after Lemma 3), we see that 202 is not one and so we have found the desired extension.

To prove (i), we note that the part of P (70) ⊗ ˆ14 in the relevant block is P (6) ⊕ P (56) ⊕ P (64). Neither 90 ⊗ ˆ14 nor 70 ⊗ ˆ14 has 190 as a quotient, so since Ext1 ; _{G}(6 ⊕ 64, 190) = 0 (by Proposition 2(b) and the Cartan matrix),

Ext1 ; _{G}(56, 190) ≅ Ext1 ; _{G}(6 ⊕ 56 ⊕ 64, 190) ≅ Hom_{G}(rad / soc(70 ⊗ ˆ14), 190).  (7)

Since rad / soc(70 ⊗ ˆ14) ≅ {6^{2}, 56^{2}, 64, 190} , the assumption Ext1 ; _{G}(56, 190) = 0 would contradict (7), so (i) holds.

From the filtration of P (90) ⊗ ˆ14 we obtain

Ext1 ; _{G}( ˆ14 ⊕ 64 ⊕ 202, ˆ14) Hom_{G}(rad / soc(90 ⊗ ˆ14), 14),

and this has dimension at most 2. Then (c) and (e) imply (j).

By Lemma 1(c), (a) and (j), we have

Ext1 ; _{G}(6 ⊕ ˆ14 ⊕ 64, 6 ⊕ ˆ14 ⊕ 64) ≅ Ext1 ; _{G}(14 ⊗ 6, 14 ⊗ 6)

≅ Ext1 ; _{G}(14 ⊗ 14, 6 ⊗ 6) ≅ Ext1 ; _{G}(k ⊕ 14 ⊕ 21, k ⊕ 14 ⊕ 21),

and all extensions except Ext1 ; _{G}(64, 64) are already known, whence (k).

We prove (l), (m) and (n) together. First we claim dim Ext1 ; _{G}(56, 56) ≤ 1. From the structure of P (70), we see that P = P (70) ⊗ ˆ14 has a filtration P_{1} ⊂ P_{2} ⊂ P where P_{1} is a quotient of P (90)⊗ ˆ14 while P_{2}/P_{1} and P/P_{2} are both isomorphic to 70⊗ ˆ14. Consideration of the heads of these factors and the composition factors of rad / soc(70⊗ ˆ14 shows that the dim Hom_{G}(rad(P (70) ⊗ ˆ14), 56) ≤ 3. Since the part of P (70) ⊗ ˆ14 in the block of maximal defect is P (6) ⊕ P (56) ⊕ P (64), our claim follows from (b) and Proposition 1 (l). Now using Lemma 1 (l) and (n) and known extensions we have

k ⊕ Ext1 ; _{G}(41, 41) ⊕ Ext1 ; _{G}(85, 41) ≅ Ext1 ; _{G}(21 ⊕ 41 ⊕ 85 ⊕ 189, 41)

≅ Ext1 ; _{G}(6 ⊗ 56, 41) ≅ Ext1 ; _{G}(56, 6 ⊗ 41)

≅ Ext1 ; _{G}(56, 56 ⊕ 190) ≅ k ⊕ Ext1 ; _{G}(56, 56).  (8)

By our claim, the dimension of the last member is at most 2. On the other hand we have Ext1 ; _{G}(85, 41) ≠ 0, for example from the fact that the reduction mod 5 of the complex character of degree 126 has 41 and 85 as composition factors. Now (8) implies (l), (m) and (n).

Finally, we turn to the last missing entry.

Proposition 4. Ext1 ; _{G}(85, 85) = 0.

Proof: Ignoring components outside the principal block, we have P (k) ⊗ 21 ≅ P (21) ⊕ P (85). We define the filtration by setting F_{i} to be the principal block part of soc^{i} P (k)⊗21, i = 1, … 5. Thus F_{5}/F_{4} ≅ 21 and F_{4}/F_{3} ≅ k ⊕ 14 ⊕ 12 ⊕ 41 ⊕ 85 ⊕ 189. From existing results on extensions, it now follows that

F_{3} ≅ rad^{2} P (21) ⊕ rad P (85).  (9)

We will prove that dim Hom_{G}(F_{3}, 85) ≤ 1, while Hom_{G}(rad^{2}(P (21), 85) ≠ 0. The lemma will then follow from (9). Dual to (9), we have F_{2} ≅ soc^{2} P (21)⊕soc P (85), so soc^{2} P (21) ⊆ rad^{2} P (21). Since rad^{2} / soc^{2}(P (21) ≅ {14, 21^{3}, 85, 189} is self-dual and 21 does not extend 85, we have Hom_{G}(rad^{2}(P (21), 85) ≠ 0. Now consider any nonzero ϕ ∈ Hom_{G}(F_{3}, 85). We use the identification (9). Since soc P (85) is the unique minimum submodule of rad P (85) and the latter is not simple, we see that soc P (85) is in the kernel of ϕ. Let \overline{F}_{i} denote the images of F_{i} in F_{3}/ soc P (85) (i=1, 2,3) then \overline{F}_{3}/\overline{F}_{2} ≅ F_{3}/F_{2} is the only factor of this filtration of \overline{F}_{3} which has 85 as a composition factor. Also, from the structure of P (k), we have \overline{F}_{3}/\overline{F}_{2} ≅ (41 ⊗ 21) ⊕ (k ⊗ 21) ⊕ (14 ⊗ 21) and all the composition factors 85 belong to 41 ⊗ 21. Thus, we have shown Hom_{G}(F_{3}, 85) ≅ Hom_{G}(21 ⊗ 41, 85) and it remains to bound the dimension of the second space by 1. Using Lemma 1(i),(j) and (n), we have

k ⊕ Hom_{G}(21 ⊗ 41, 85) ≅ Hom_{G}((k ⊕ 14 ⊕ 21) ⊗ 41, 85)

≅ Hom_{G}((6 ⊗ 6) ⊗ 41, 85) ≅ Hom_{G}(6 ⊗ 41, 6 ⊗ 85)

≅ Hom_{G}(56, 6 ⊗ 85) ⊕ Hom_{G}(190, 6 ⊗ 85)

≅ k ⊕ Hom_{G}(190, 6 ⊗ 85).  (10)

Since [6 ⊗ 85 : 190] = 1, we obtain our estimate and the lemma is proved.

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