CLASSIFICATION OF 2F-MODULES, I

Robert M. Guralnick and Gunter Malle

University of Southern California and Universität Kassel

To John Thompson, on the occasion of his 70th birthday

We classify the 2F-modules for nearly simple groups, excluding the case of modules for groups of Lie type in their defining characteristic. We also show that for all such modules there exists an offender with cubic action. These modules play a crucial role in the current revisions of the classification of the finite simple groups.

1. Introduction

In this paper we study the following situation: Let G be a finite group such that F^{∗}(G) is quasi-simple. For an absolutely irreducible faithful F G-modules V over a finite field F of characteristic ℓ and a subgroup A ≤ G of G we let

f(A) := f_{V} (A) := |A|^{2} · |C_{V} (A)|.

We classify those modules V for which G has an elementary abelian ℓ-subgroup A satisfying

(1.1)  |V | ≤ f(A) .

Variations on this theme have been studied for many years. Perhaps the first person to consider this situation was John Thompson [20] in studying the embeddings of a Sylow p-subgroup S in finite simple groups – especially considering the subgroups C(Z(S)) and N(J(S)). Glauberman also studied this – see [7], [8], [9] and [11]. They were interested in obtaining factorizations of groups and the existence of such modules is an obstruction (hence called failure of factorization modules by Glauberman). In [9], Glauberman showed that certain groups had no such modules (for ℓ = 2). In [10], he proved a factorization theorem for certain 2-constrained groups such that no chief factors were modules satisfying related conditions. In most instances, the case ℓ = 2 has been the focus of investigations. Often there are extra or modified conditions on the modules – for example, the case where the exponent 2 above is

Typeset by AMS-TEX

1991 Mathematics Subject Classification. primary: 20C20, secondary: 20D05, 20E32.

We thank G. Stroth for pointing out an omission in a previous version. The first author was partially supported by the National Science Foundation grant DMS-9970305

replaced by 1 and/or additional conditions such as assuming that [[V, A], A] = 0 have been considered.

More recently, these modules (again with some extra conditions) play a crucial role in the revision of the classification of finite simple groups. Aschbacher and Smith [1] will use the results of this paper (for ℓ = 2) and the sequel in the classification of quasithin finite simple groups. Their paper will be part of the revision program as well as provide the first published proof of the classification of quasithin groups. These modules (for all ℓ) are also important in Meierfrankenfeld’s program of classifying simple groups which are of generic characteristic ℓ-type, see [3] for example.

In most of these applications, one has extra conditions on the module – for example in [1], they assume that A has cubic action on V – i.e., [[[V, A], A], A] = 0. In this paper our approach is to use (1.1) only. It is then quite easy to decide which of the few remaining examples satisfy the additional hypotheses. We feel that this is intrinsically more pleasing. Moreover, if one wants to impose weaker conditions, then the theorem will still apply. As we have noted, many people have assumed that A acts quadratically on V (and used this property extensively). We always assume that V is irreducible and faithful – by considering extensions of the modules involved, one could determine the indecomposable examples as well. Note that a special case of these modules is where G contains an element of order ℓ that centralizes a subspace of codimension 2 on V, and V is defined over 𝔽_{ℓ}. The more general case of classifying representations of groups containing ‘bireflections’ (in particular, pseudoreflections and transvections) is dealt with in [13]. We also note that the condition that A is an elementary ℓ-group could be relaxed in a variety of ways.

An elementary lemma gives that under our conditions f(A) ≤ f(G) = |G|^{2}. But now a simple argument shows that F G-modules satisfying (1.1) are rare: Assume for example that G = 𝔖_{n} is the symmetric group and let ℓ be a prime. Then log_{ℓ}(|G|) ∼ n log_{ℓ}(n) by Stirling’s formula. It is well known that for large enough n only the heart of the permutation module and its product with the sign have dimension less than n^{2}/3. In particular, for large enough n only these two modules can arise. (We will see that in fact even ℓ ≤ 3.) A similar reasoning deals with groups of Lie type defined over fields of characteristic prime to ℓ, using the lower bounds of Landazuri-Seitz-type for dimensions of non-trivial irreducible representations.

We close the paper by determining in Section 6 for which of the triples (G, V, A) satisfying condition (1.1) the action of A is cubic, that is, the commutator [[[V, A], A], A] is zero. Given the short list of examples from the first part, this can be done by straightforward verification.

Thus we just have to make precise the rough sketch given above for the various families of quasi-simple groups. At each point we will explicitly state which properties of the known quasi-simple groups we use. In particular, we will only use properties of the group in question and its subgroups, thus making the result usable for being part of an inductive proof of the classification.

The case of groups of Lie type in their defining characteristic will be considered in a sequel to this paper.

At present, we have to leave open the case of the smallest non-trivial module in characteristic 2 for the sporadic Conway groups Co_{1} and Co_{2}, which we hope to solve in due course.

2. Notation and basic observations

Let F be a finite field of characteristic ℓ, G a finite group and V a finite-dimensional F G-module. For a subgroup A of G write

f(A) := f_{V} (A) := |A|^{2} · |C_{V} (A)|.

V is called a 2F-module for G with offender A if dim(V) > 1 and |V | ≤ f(A). The following well-known observation will be useful (see [4]):

Lemma 2.1. Let A, B ≤ G with f(B) maximal among all subgroups of B and let C := ⟨A, B⟩. Then f(C) ≥ f(A).

Proof. Note that |C| ≥ |A||B|/|A ∩ B| and |C_{V} (C)| ≥ |C_{V} (A)||C_{V} (B)|/|C_{V} (A ∩ B)|. Thus we have

\begin{matrix}f(C) = & |C|^{2}|C_{V} (C)| \\ ≥ & |A|^{2}|B|^{2}|C_{V} (A)||C_{V} (B)|/ (|A ∩ B|^{2}|C_{V} (A ∩ B)|) \\ = & f(A)f(B)/f(A ∩ B) ≥ f(A). \end{matrix}

□

In particular, if V is a 2F-module for G with offender A and f(A) ≥ f(B) for all B ≤ A, then f(A) ≤ f(A^{G}). In particular, if G is the normal closure of A then f(A) ≤ f(G), Moreover, if V is irreducible then f(A) ≤ |G|^{2}. Also it follows that:

Lemma 2.2. Suppose C ≤ G is an elementary abelian ℓ-subgroup containing A and B, with f(A) = f(B) maximal among all subgroups of C. Then f(AB) = f(A).

This shows that when searching for examples satisfying (1.1) we may sometimes assume that A is maximal elementary abelian.

The following can be used to obtain upper bounds on the centralizer of an ℓsubgroup:

Lemma 2.3. Let B ≤ A ≤ G with dim C_{V} (A) = dim C_{V} (B). Then

dim C_{V} (⟨A^{g} | g ∈ C_{G}(B)⟩) = dim C_{V} (B)

Proof. Let g ∈ C_{G}(B). Then C_{V} (A) = C_{V} (B) = C_{V} (B^{g}) = C_{V} (A^{g}). □

We will often use this in the following form. Let B be an elementary abelian ℓsubgroup, g an ℓ-element in C_{G}(B) \ B and A := ⟨B, g⟩. If ⟨A^{h} | h ∈ C_{G}(B)⟩ has smaller centralizer in V than B, then A has smaller centralizer than B. For example, if C_{G}(B) is an extension of B by a simple group H, then all classes of ℓ-elements in C_{G}(B) outside B generate H. But H contains ℓ′-elements and their centralizer can be read off from the Brauer character of V.

The following result is very useful for inductive arguments; it will be used in the second part.

Proposition 2.4. Let V be a 2F-module for G. Let H ≤ G such that H contains an offender. The either there exists an offender A ≤ U := O_{ℓ}(H), or C_{V} (U) is a 2F-module for H/U.

Proof. Choose an offender A ≤ H such that all subgroups A′ of A satisfy f_{V} (A′) ≤ f_{V} (A). We may assume that A is not contained in U. Let W := C_{V} (U). Then

C_{V} (A) ∩ C_{V} (U) = C_{C_{V} (U)}(A) = C_{W} (A) ,

so |C_{W} (A)| = |C_{V} (A) ∩ C_{V} (U)| ≥ |C_{V} (A)||C_{V} (U)|/|C_{V} (A ∩ U)|. We obtain

|W | = |C_{V} (U)| ≤ \frac{|C_{W} (A)||C_{V} (A ∩ U)|}{|C_{V} (A)|} = \frac{|C_{W} (A)||C_{V} (A ∩ U)||A|^{2}}{f_{V} (A)} .

Now let Ā := AU/U = A/A ∩ U, so |A| = |Ā||A ∩ U|. Then

|W | ≤ |C_{W} (A)||Ā|^{2} \frac{|A ∩ U|^{2}|C_{V} (A ∩ U)|}{f_{V} (A)} = f_{W} (Ā)\frac{f_{V} (A ∩ U)}{f_{V} (A)} ≤ f_{W} (Ā)

by the choice of A, hence Ā is an offender for H/U on W. □

Let’s make some easy observations. Assume that G has ℓ-rank r, then clearly |A| ≤ ℓ^{r} for any elementary abelian ℓ-subgroup A. Write d := dim(V) for the dimension over the smallest field of definition and let d_{ℓ} be the maximal dimension of the centralizer space in V of a non-trivial ℓ-element. Let f be the degree over 𝔽_{ℓ} of the smallest field over which the representation is defined. Then condition (1.1) certainly implies

(2.5)  f(d − d_{ℓ}) ≤ 2r .

For x ∈ G, let k(x) be the minimal number of conjugates of x which generate ⟨x^{G}⟩. Write

k_{ℓ} := max{k(x) | x ∈ G, x ≠ 1 = x^{ℓ}}.

If we assume that every irreducible constitutent of V is faithful, then ⟨x^{G}⟩ has no fixed points on V and so (d − d_{ℓ})k_{ℓ} ≥ d. Thus (1.1) implies

(2.6)  fd ≤ 2rk_{ℓ} .

This weaker condition will already rule out most possibilities for G.

The next result can be used when treating automorphism groups. The only interesting case for us is the one of outer automorphisms of order ℓ (since for V irreducible and faithful, we can always restrict to ⟨A^{G}⟩).

Proposition 2.7. Let G be a finite group with a normal subgroup H ≤ G of index ℓ. Let W be an absolutely irreducible 𝔽_{ℓ^{f}} H-module of dimension dim W ≥ 2 whose induction V to G is again irreducible. If V is a 2F-module for G with offender A, then one of the following holds:

(i) W is a 2F-module for H with offender A′ := A ∩ H;

(ii) ℓ = f = 2, z := dim C_{W} (A′) = 1, dim W ≤ r where |A| = 2^{r};

(iii) |A| = ℓ = 2, dim W = 2, f ≤ 2.

Proof. If A ≤ H then clearly we arrive at case (i). Hence we may assume that A′ := A ∩ H < A, so [A : A′] = ℓ. Write |A| = ℓ^{r}, so |A′| = ℓ^{r−1}. Furthermore, let z := dim C_{W} (A′). Since V is induced from W and A ≰ H we then have dim V = ℓ dim W and dim C_{V} (A) = z.

First assume that V is not defined over a proper subfield of 𝔽_{ℓ^{f}}. Then the 2Fcondition for V yields f(dℓ − z) ≤ 2r. Hence

f(d − z) ≤ (2r + fz)/ℓ − fz ≤ r − fz/2,

so f(d − z) ≤ r − 1. Thus either A′ is an offender for H on W or r = 1. In the latter case A′ = 1, z = d, and we obtain f(dℓ − d) = fd(ℓ − 1) ≤ 2. Then d ≥ 2 forces f = 1, d = ℓ = 2 as in case (iii).

It remains to consider the case where V is defined over the subfield of index ℓ of 𝔽_{ℓ^{f}}. In particular f = ℓf′ with an integer f′. Here the 2F-property gives f(dℓ−z)/ℓ ≤ 2r, so fd ≤ 2r + fz/ℓ. Hence

f(d − z) ≤ 2r + fz/ℓ − fz = 2r − ℓf′z(1 − 1/ℓ) = 2r − f′z(ℓ − 1).

This shows that A′ is an offender for H on W unless f′z(ℓ − 1) ≤ 1. This can only happen if ℓ = 2, f′ = z = 1. Then the first inequality yields 2d ≤ 2r + 1, so d ≤ r, and we are in case (ii). If r = 1 then f′(dℓ − z) ≤ 2, and again we find d = ℓ = 2, f′ = 1, which is case (iii). □

We fix the following notation. Let G be a finite group such that F^{∗}(G) is quasisimple. Write S for the unique non-abelian simple composition factor of G. Let ℓ be a prime dividing |G|. We’ll consider the different possibilities for S according to the classification of finite simple groups, excluding in this part the groups of Lie-type in their defining characteristic.

3. Alternating groups

To handle the alternating groups we will make use of the following properties: lower bounds for degrees of absolutely irreducible representations due to James [15] and Wagner [21], and the modular character tables of 𝔄_{n} for n ≤ 10 (if ℓ is odd) respectively n ≤ 12 (if ℓ is even) [16].

Proposition 3.1. Let S = 𝔄_{n} with n ≥ 5, (n, ℓ) ≠ (5, 2), (5, 5), (6, 3), (8, 2). Then the absolutely irreducible F G-modules V satisfying (1.1) are given in Table 6.3.

Proof. 1. We first consider the deleted permutation module. The commutator space for an ℓ-element on the permutation module is at least (ℓ − 1)⌊n/ℓ⌋, while the ℓ-rank of 𝔖_{n} is ⌊n/ℓ⌋. By (2.5) this forces ℓ ≤ 3, and it is clear that in this case we get examples. Thus from now on we will assume that V is not the deleted permutation module.

2. We next work out the cases n ≤ 7 from the known subgroup structure and the modular character tables [16]. For n = 5 we have to consider ℓ = 3. Since G = 𝔄_{5} has cyclic Sylow 3-subgroups and is generated by two 3-elements we have d ≤ 4 by (2.6). Clearly d = 3 for G and d = 2 for 2.G give examples, while d = 4 for G has f = 2.

For n = 6 and ℓ = 5 we again have d ≤ 4. The case d = 3 for 3.𝔄_{6} has f = 2, while d = 4 for 2.𝔄_{6} restricts to the irreducible 4-dimensional representation of 2.𝔄_{5} = SL_{2}(5) in which 5-elements act indecomposably. Thus neither leads to an example. For ℓ = 2 the rank is 2 and k_{2} = 3, so df ≤ 12. This leads to d = 3, f = 2 for 3.S which gives an example with |A| ∈ {2, 4} and 2-dimensional centralizer. The group PGL_{2}(9) = 𝔄_{6}.2_{2} also has 2-rank 2, and there is a unique class of four groups containing outer involutions. Two of these generate, forcing d ≤ 8, so d = 8, f = 1 is the only possibility. Outer involutions invert elements of order 5, so have 4-dimensional commutator space. But the centralizer is 2 × D_{5}, and 5-elements have no fixed points, so the centralizer of a four group involving outer elements is too small. The group M_{10} = 𝔄_{6}.2_{3} has no outer involutions. The symmetric group 𝔖_{6} has 2rank 3. Since k_{2} = 5, d ≤ 10 if |A| = 2. If |A| ≥ 4 then it contains inner involutions and k(x) = 3 for an inner involution. Thus d ≤ 12 if |A| = 4. If |A| = 8 then two conjugates of A generate, so d ≤ 12 again. This leaves the deleted permutation modules and d = 6 for 3.𝔖_{6}, which occurs. Finally Aut(𝔄_{6}) cannot be generated by the conjugates of an elementary abelian 2-subgroup, since one of the cosets of 𝔄_{6} does not contain involutions.

For n = 7 and ℓ = 5, 7 we have d ≤ 4 by (2.6). The possibility d = 4 for 2.𝔄_{7} restricts irreducibly to SL_{2}(ℓ), so does not give an example. Also, d = 3 for 3.𝔄_{7} with ℓ = 5 has f = 2, so ℓ = 5, 7 is not possible. If ℓ = 3 then d ≤ 8. Apart from the deleted permutation module we obtain d = 4, 6 for 2.𝔄_{7}, both with f = 2. This actually rules out d = 6. The 4-dimensional representation restricts irreducibly to 2.𝔄_{6} = SL_{2}(9), where the Sylow 3-subgroup acts indecomposably, hence is not an example either. For ℓ = 2 we have rank 2 and three involutions generate, and moreover two four-groups generate, so d ≤ 8 for the simple group. This implies d = 4 for 𝔄_{7}, which occurs with |A| = 4. For 𝔖_{7} and ℓ = 2 either A is cyclic whence d ≤ 12, or it contains an involution from 𝔄_{7}. If A has rank 2 then d ≤ 12 as above. Furthermore, two elementary abelian subgroups of order 8 generate, so d ≤ 12 in the rank-3 case as well. This leads to d ∈ {8, 12}. Explicit computation shows that the 8-dimensional representation gives an example. Since the 12-dimensional representation of G = 3.𝔖_{7} is induced from 3.𝔄_{7}, outer involutions have 6-dimensional commutator space. In particular |A| = 8 and thus A contains transpositions. The centralizer of a transposition in 𝔖_{7} is 2×𝔖_{5}, which contains elements of order 5 with 8-dimensional commutator. Thus A has at most 4-dimensional centralizer. This rules out d = 12.

3. From now on let n ≥ 8. First assume that ℓ is odd. The ℓ-rank of 𝔄_{n} equals ⌊n/ℓ⌋. By an easy induction argument it follows that k_{ℓ} ≤ n/2. We first consider representations of 𝔄_{n}. An application of a result of James [15] shows that for n ≥ 12 any absolutely irreducible F 𝔖_{n}-module either has dimension d ≥ (n − 2)(n − 3) or lies in R_{n}(2) ([18], Prop. 2.2). Here R_{n}(2) denotes the set of those irreducible representations parametrized by an ℓ-regular partition λ such that λ or its dual have a part at least equal to n−2. Hence any absolutely irreducible F 𝔄_{n}-module either has dimension d ≥ (n−2)(n−3)/2 or is a constituent of the restriction of a representation in R_{n}(2). The first possibility violates (2.6). But the non-trivial F 𝔄_{n}-constituents of modules in R_{n}(2) different from the deleted permutation module have dimension d ≥ (n^{2} − 5n + 2)/2 by [15], again contradicting (2.6), except when n = 12, ℓ = 3 and |A| = 3^{4}. But clearly 𝔄_{12} is generated by two suitable conjugates of such a subgroup, which forces d ≤ 16 < 43 = (n^{2} − 5n + 2)/2, a contradiction.

For 8 ≤ n ≤ 11 similar arguments may be used; we just sketch the most difficult case ℓ = 3. If |A| = 3 then d ≤ n, giving the deleted permutation module. If |A| = 9 then d ≤ 2n, which leads to n = 8, d = 13. But for n = 8 two such subgroups generate, thus d ≤ 8 in this case. If |A| = 27 then n ≥ 9. Again two such groups generate if n ≤ 11, hence d ≤ 12 and no new cases arise.

4. We still assume that ℓ is odd, and now consider the faithful representations of the 2-fold cover G = 2 · 𝔄_{n}. According to Wagner [21] the smallest dimension is at least 2^{⌊ \frac{n−s−1}{2} ⌋}, where s denotes the number of 1s in the 2-adic expansion of n. Comparison with the ℓ-rank ⌊n/ℓ⌋ shows that n ≤ 15.

If n < 2ℓ then the ℓ-rank of G is 1 and k_{ℓ} = 2, hence d ≤ 4, violating Wagner’s bound. For ℓ = 7, n = 14, 15, or ℓ = 5, 10 ≤ n ≤ 13, k_{ℓ} = 3, so d ≤ 12, which is only possible if n = 10, ℓ = 5. But then |A| = 25, and two Sylow 5-subgroups generate, whence d = 8. On restriction to 2.𝔄_{5} = SL_{2}(5) we find two copies of the 4-dimensional representation in which elements of order 5 have a 3-dimensional commutator space. Thus this case is out. For ℓ = 5, n = 14, 15, we find d ≤ 24, a contradiction. Finally let ℓ = 3. For n = 10, 11, 12, two conjugates of any 3-subgroups of rank 3 generate, whence d ≤ 12, which is not possible. For n = 13, 14, 15, three conjugates of any rank-3 subgroups generate, giving d ≤ 18, a contradiction. For n = 8, 9 the character tables show that d = 8. In the case n = 8 we need |A| = 9. Restriction to 2.𝔄_{5} yields four 2-dimensional constituents which shows that 3-cycles have at least 4-dimensional commutator spaces. The centralizer of a 3-cycle g ∈ A contains an 𝔄_{5}. If A has 4-dimensional centralizer, then, since 𝔄_{5} is generated by its unique class of 3-elements, 𝔄_{5} has a 4-dimensional centralizer as well. But the character table reveals that elements of order 5 act fixed point freely, so this case is not possible. If n = 9 then restriction to 2.𝔄_{8} (the case just treated) shows that |A| = 27. This gives an example.

5. Now assume ℓ = 2 and G = 𝔄_{n} with n ≥ 9 (since 𝔄_{8} = L_{4}(2)). Then again k_{ℓ} ≤ n/2. For n ≥ 17 again Prop. 2.2 in [18] shows that V ∈ R_{n}(2), which only leaves the deleted permutation module by [15]. So now assume n ≤ 16. If A has rank at most 2 then d ≤ 2n by (2.6). For 9 ≤ n ≤ 11 two suitable maximal rank-3 subgroups generate and the 2-rank is equal to 4. We deduce d ≤ 16 in this case, and by the tables in [16] this leads to the deleted permutation module or (n, d) ∈ {(9, 8), (10, 16), (11, 16)}. The first gives an example with |A| ∈ {8, 16} and centralizer of dimension 2, since it restricts irreducibly to the 8-dimensional representation of 𝔖_{7} considered above. The restriction of the 16-dimensional 𝔄_{10}-module to 𝔄_{9} consists of two copies of the previously considered representation, thus the centralizer space is at most 4dimensional, giving a contradiction. This also rules out n = 11. For 12 ≤ n ≤ 15, two rank-5 subgroups, three rank-4 subgroups and four rank-3 subgroups generate. Thus in any case we get d ≤ 24. Lemma 3 in [15] shows that the only possibility is (n, d) = (12, 16). But this has f = 2, being defined over 𝔽_{4}, which violates the bound. For n = 16 we obtain a contradiction from [15].

It remains to consider ℓ = 2 and G = 𝔖_{n}, n ≥ 9. If A is cyclic we obtain d ≤ 2(n−1) since k_{2} = n − 1. Otherwise A contains at least one element from 𝔄_{n}, hence with commutator space of dimension at least 2d/n. In any case Theorem 7 in [15] shows n ≤ 14 or V restricted to 𝔄_{n} is the deleted permutation module. Assume n ≤ 14. The estimates from the 𝔄_{n}-case plus the tables in [16] show that we are in one of the following cases: (n, d) ∈ {(9, 16), (10, 16), (12, 32)}. Outer involutions invert 3-cycles, so by the character tables have commutator space of dimension at least 8, 8, 16 in the respective cases. This eliminates n = 12 and proves that |A| ≥ 16 in the remaining cases. Suppose n = 9. The centralizer of a transposition is 2 × 𝔖_{7}, whose centralizer is less than 8-dimensional, so this case is out as well. Finally for the 16-dimensional representation of 𝔖_{10} we must have |A| = 32 since the previous argument shows that the centralizer space is at most 7-dimensional. But the representation splits into two 8-dimensional ones upon restriction to 𝔄_{9}. There, the centralizer of a rank-4 subgroup is 2-dimensional (see above), hence it is at most 4-dimensional when d = 16. This rules out the last candidate. □

4. Groups of Lie type in non-defining characteristic

4A. Linear groups.

To deal with the linear groups in non-defining characteristic we make use of the Landazuri-Seitz-Zalesskii lower bounds for irreducible representations and the information in the Atlas and Modular Atlas for some small groups.

Proposition 4.1. Let S = L_{2}(q) with ℓ |̸q, q ≠ 4, 5, 9, (q, ℓ) ≠ (7, 2). Then no absolutely irreducible F G-module satisfies (1.1).

Proof. The minimal degree of a nontrivial projective representation in non-defining characteristic for S := L_{2}(q) is (q − 1)/2 if q is odd, respectively q − 1 if q ≥ 8 is even. First assume that ℓ is odd. Then Aut(S) has ℓ-rank at most 2, and k_{ℓ} ≤ 3 for any subgroup of Aut(S). Thus d ≤ 12, so q ≤ 25 (respectively q ≤ 13 for even q). But for q ≤ 25, q ≠ 8, |Out(S)| is even, hence the ℓ-rank is 1, and k_{ℓ} = 2. This gives the better bound q ≤ 9 (respectively q ≤ 5 or (q, ℓ) = (8, 3) for even q). Thus we are left with q = 7, d ≤ 4, and q = 8, d ≤ 8, both with ℓ = 3. In the first case the relevant 3-modular representations, of degrees 3 and 4, are not defined over 𝔽_{3}, thus no example arises. In the second case S = L_{2}(8) the only possibility is d = 7. Since k(x) = 2 for x an inner 3-element, their centralizer is at most 3-dimensional. Since they are third powers, their Jordan blocks have sizes 3,2,2. Outer 3-elements normalize 7-elements, so have Jordan blocks of sizes 3,3,1. Thus the centralizer is at most 3-dimensional for |A| = 3, at most 2-dimensional for |A| = 9. This gives a contradiction.

Now consider ℓ = 2, so q is odd. The 2-rank of S is 2 and k_{2} = 3, thus in this case d ≤ 12, q ≤ 25. Furthermore there is a single class of involutions. Two of these can be chosen to generate a dihedral group of order q + ϵ, ϵ ∈ {±1}, q ≡ ϵ (mod 4). Since this is maximal for q > 11 we deduce that two suitable conjugates of any four-group in Aut(S) containing inner elements generate at least S. Thus in fact q ≤ 17 in this case. For G = PGL_{2}(q) the minimal degree equals q − 1, the 2-rank is still 2 and four outer involutions generate, forcing d ≤ 8, q ≤ 9, so this case does not arise here. Finally, if G involves field automorphisms, then either A has rank 1, or A contains an involution from PGL_{2}(q). Since again four (conjugate) outer involutions generate, in either case d ≤ 12, so q ≤ 25.

For L_{2}(25) we have d = 12. An inner involution inverts an element of order 13, which acts fixed point freely on the 12-dimensional module. Hence inner involutions have 6-dimensional commutator space, so |A| = 8. The centralizer of such an involution in PGL_{2}(25) is a dihedral group of order 48. If all involutions in this centralizer have the same commutator space, the same is true for the elements of order 3 therein. But their centralizer on the 12-dimensional module is only 4-dimensional. This rules out the case S = L_{2}(25). For q = 19, 23 there are no field automorphisms, so they cannot occur. Next, L_{2}(17) contains a Frobenius group of order 17 · 8, so elements of order 8 have a single Jordan block on the 8-dimensional module. Thus elements of order 4 have two Jordan blocks and inner involutions have 4-dimensional commutator space. The involution centralizer is D_{4}, so is generated by its non-central involutions. If all of these had the same 4-dimensional centralizer, the same would hold for elements of order 4 therein, contradiction. For q = 13 the 6-dimensional representation is not defined over 𝔽_{2}, so it does not occur. For q = 11 the 5- and 12-dimensional representations are not defined over 𝔽_{2}, so df = 10 with f = 2 respectively f = 1. Since elements of order 5 are inverted by involutions, the latter have commutator space of dimension 2 (respectively 4). If a second commuting involution has the same commutator space, then so has an element of order 3 in the centralizer of the first, which is not the case.

This completes the proof of Proposition 4.1. □

Proposition 4.2. Let S = L_{n}(q), n ≥ 3, (n, q) ∉ {(3, 2), (4, 2)}, with ℓ |̸q. Then the absolutely irreducible F G-modules V satisfying (1.1) are given in Table 6.5.

Proof. 1. We proceed as follows. By [19] a lower bound for the minimal degree of projective representations is given by d(n, q) := (q^{n} − 1)/(q − 1) − n (unless n ≤ 4, q ≤ 3). On the other hand f(A) is bounded above by |Aut(G)|^{2}. This severely restricts the possibilities for (n, q, ℓ).

2. Let first n = 3, so q ≠ 2. Then our estimate forces q ≤ 4. First consider S = L_{3}(4). The Sylow ℓ-subgroups are cyclic for ℓ = 5, 7 and two of them generate, so d ≤ 4, which is not possible. For ℓ = 3 the rank of Aut(S) equals 2, three elements generate, so d ≤ 12. This leaves d ∈ {6, 10} for 2.L_{3}(4) and d ∈ {4, 8} for the 4-fold coverings. The cases d = 8, 10 are out since they are not defined over 𝔽_{3}. The 6dimensional representation extends to 2.L_{3}(4).2_{2} from where it restricts irreducibly to 𝔖_{6}. The restriction to 𝔄_{6} = O_{3}(9) is now the direct sum of the natural module and its dual, so we get an example for 2.L_{3}(4). The 4-dimensional representation has f = 2 and embeds 4_{2}.L_{3}(4) into SU_{4}(3). The fusion shows that elements of order 3 have Jordan blocks of sizes 3 and 1, and the Sylow 3-subgroup acts indecomposably. Thus no example arises. The group S.3 = PGL_{3}(4) does not lead to examples.

For S = L_{3}(3) only ℓ = 2, 13 have to be considered. The Sylow 13-subgroups are cyclic and two of them generate, so this cannot occur. Now assume that ℓ = 2. The 2-rank of S is 2, for Aut(S) it equals 3. Since k_{2} = 3, d ≤ 12 respectively d ≤ 18. From the character table we find that d = 12 is the only case. Inner involutions invert 3B-elements, so have at least 4-dimensional commutator. This forces |A| ≥ 4. For a fixed involution g ∈ A we can choose a further involution h in its centralizer such that ⟨A, h⟩ is ⟨g⟩ × 𝔖_{3}, in particular has at most 6-dimensional centralizer. This forces |A| = 8 and hence G = Aut(L_{3}(3)). Let g be an outer involution. Since it inverts an element of order 13, its commutator space is 6-dimensional. The centralizer of g is C = C_{1} × ⟨g⟩ with C_{1} ≅ 𝔖_{4}. The involutions in C_{1} not in the normal Klein four group generate 𝔖_{4}, so any A containing one of these has centralizer of dimension at most 5, which contradicts (1.1). Now assume that A is the normal subgroup of C of order 8. Let g′ be a second outer involution in A. Then C_{1} ∩ A is no longer normal in the centralizer of g′, since C_{1} is the full normalizer of C_{1} ∩ A in S. Thus we are reduced to the previous case.

3. When n ≥ 4 then comparison of our bounds shows that either n = 4, q ≤ 3, or S = L_{5}(2), ℓ = 3. For S = L_{4}(3) and ℓ = 5, 13 the rank is 1 and five Sylow ℓ-subgroups generate, so d ≤ 10, which is impossible. For ℓ = 2 we obtain the bound d ≤ 49, so d ∈ {26, 38}. Involutions from both inner classes invert elements of order 5. From the character table we conclude that the commutator space of any inner involution is at least 10-dimensional for d = 26 (respectively 16-dimensional for d = 38). Thus |A| ≥ 32 for d = 26, while d = 38 is not possible. Since S has 2-rank four, some outer automorphism must be present when d = 26. The 26-dimensional representations only extend to G.2_{2}, thus |A| = 32, and A must have centralizer of dimension 16. But the centralizer of an element g from 2A is of the form D_{4} × 𝔖_{6}. If an elementary abelian subgroup of the 𝔖_{6}-factor of order 8 has the same centralizer as g, then the whole 𝔖_{6} has the same centralizer, which is not the case by the trace of 5-elements. Hence no example arises.

The 3-rank of S = L_{5}(2) is 2, and k_{3} ≤ 6, whence d ≤ 24, a contradiction. □

4B. Unitary groups.

Proposition 4.3. Let S = U_{3}(q) with ℓ |̸q, q ≠ 2. Then the absolutely irreducible F G-modules V satisfying (1.1) are given in Table 6.5.

Proof. We proceed as in the previous proof, using the lower bound q^{2} − q for degrees of nontrivial projective representations from [19]. If ℓ is odd and different from 3 then this yields q ≤ 4. For U_{3}(4) the ℓ-rank is at most 2, and two subgroups of rank 2 generate, thus d ≤ 8, smaller then the lower bound. For U_{3}(3) we have ℓ = 7, so the rank is one and two elements generate, again a contradiction.

For ℓ = 3 we get q ≤ 5. For Aut(U_{3}(5)) the 3-rank is equal to 2, and k_{3} ≤ 4, giving d ≤ 16. For U_{3}(4) the rank is 1, leading to no example.

For ℓ = 2 the bounds force q ≤ 7. For S = U_{3}(7) the 2-rank of Aut(S) is bounded by 3, and k_{2} ≤ 5, thus we get no example. The group S = U_{3}(5) has 2-rank 2 and k_{2} ≤ 4 (since S has a unique class of involutions and three involutions generate the maximal subgroup 𝔄_{7}), so d ≤ 16, impossible. The automorphism group of A has 2-rank 3, and k_{2} ≤ 4 (since there is a unique class of outer involutions and three conjugates of them will generate the maximal subgroup 𝔖_{7}), forcing d ≤ 24, so d ∈ {20, 21}. But outer involutions invert elements of order 7, and these have commutator spaces of dimension 18 for d = 20, 21. Thus outer involutions have commutator space of dimension at least 9, contradicting (1.1).

Finally S = U_{3}(3) has 2-rank 2 and k_{2} = 4, so d ≤ 16 for S. This gives d ∈ {6, 14}. The 6-dimensional representation extends to Aut(S) = G_{2}(2) and the restriction to L_{3}(2).2 remains irreducible. This restricts as the direct sum of the natural representation of L_{3}(2) and its dual, hence we obtain an example. Since inner involutions invert a regular element of order 3, they have at least 5-dimensional commutator space in the 14-dimensional representation, thus this is out. Next, G = Aut(U_{3}(3)) has 2-rank 3, hence d ≤ 30, which implies d ∈ {6, 14}. Clearly the case d = 6 yields an example. The case d = 14 is the adjoint representation of G_{2}(2). Outer involutions have at least 6-dimensional commutator space, since they invert elements of order 7. Assume we have A with 8-dimensional centralizer. Choose a further outer involution g such that H := ⟨A, g⟩ contains elements of order 7. Then [5] shows that H = G. This contradicts the fact that both A and g have 8-dimensional centralizer. □

Proposition 4.4. Let S = U_{n}(q), n ≥ 4, with ℓ |̸q, (S, ℓ) ≠ (U_{4}(2), 3). Then the absolutely irreducible F G-modules V satisfying (1.1) are given in Table 6.5.

Proof. 1. Let first n = 4. The lower bound for degrees of nontrivial projective representations is (q^{2} − 1)(q − 1) when q ≥ 4. Our standard estimate shows that q ≤ 3. Let first ℓ be odd. Then the ℓ-rank of U_{4}(3) is one, which gives no example. For U_{4}(2) we have ℓ = 5, with rank 1 and k_{5} = 2, so d ≤ 4. But the 4-dimensional representation of 2.U_{4}(2) is not defined over 𝔽_{5}.

For ℓ = 2 we only have to treat q = 3. The group S = U_{4}(3) has 2-rank 4 (see [12], 4.10.5). Moreover k_{2} = 3, so d ≤ 24 in this case. This leaves d = 20 for S and (d, f) = (6, 2) for 3.S. The latter gives an example with |A| = 8. Assume d = 20. Inner involutions have at least 8-dimensional commutator since they invert 5-elements, so |A| = 16. Let C be the centralizer of an involution in S. The extra-special group O_{2}(C) ≅ 2^{1+4} does not contain an elementary abelian subgroup of order 2^{4}. Thus inside C there exist two conjugates of A generating a subgroup containing 3-elements. But all 3-elements have at least 12-dimensional commutator, a contradiction. For S.2_{1} the 2-rank of centralizers of outer involutions is at most 4, forcing d ≤ 24. This leads to d = 20 for S.2_{1} and d = 12 for 3_{1}.S.2_{1}. In the first case, since outer involutions from 2C invert 3A-elements, they have at least 9-dimensional commutator, while for 2Belements the centralizer 2×U_{3}(3) has only 2-rank 3, impossible. Since d = 12 for G = 3_{1}.S.2_{1} is induced, outer involutions have 6-dimensional commutator. In particular |A| ≥ 8. The module V splits into a direct sum V_{1} ⊕ V_{2} under G′ = 3_{1}.S. Denote by A′ the intersection of A with G′ = 3_{1}.S. Then clearly dim(C_{V} (A)) = dim(C_{V_{1}}(A′)). Since inner involutions have 4-dimensional centralizer this shows dim(C_{V} (A)) ≤ 4 and hence |A| = 16. In particular |A′| = 8 has centralizer of dimension 4 on V_{1}, as any inner involution. But O_{2} of the centralizer C in G′ of an inner involution does not have a normal subgroup 2^{3}. Thus we obtain a contradiction with Lemma 2.3.

For S.2_{2} the centralizer of an outer involution in 2D is 2 × U_{4}(2), hence S.2_{2} has 2-rank 5. This leads to d = 20 for S.2_{2} and d = 6, 15 with f = 2 for 3_{1}.S.2_{2}. The case d = 6 is the reflection (transvection) representation of 3_{1}.S.2_{2}, so gives an example. When d = 15 estimates for the centralizers of outer involutions show that A has to contain inner involutions. Now 2A-involutions have at most 9-dimensional centralizer. Since f = 2 this forces |A| ≥ 2^{6}, impossible. The 20-dimensional module is the exterior cube of the 6-dimensional representation of S.2_{2} embedding it into U_{6}(2). But the exterior cube of the natural module is not an example for U_{6}(2), see the second part.

For S.2_{3} the outer involution centralizer is 2 × M_{10}, so has 2-rank 3. This gives d ≤ 18, so no example arises.

There are two possible groups S.2^{2}, namely S.(2^{2})_{122} and S.(2^{2})_{133} in Atlas notation. The 2-rank of S.(2^{2})_{122} equals 5, so d ≤ 40. This leaves d = 20, 34 for G = S.(2^{2})_{122}, and d = 12, 30 for G = 3_{1}.S.(2^{2})_{122}. The 12-dimensional representation gives an example. For the 30-dimensional representation we have |A| > 4, so A contains inner involutions. But 2A-involutions have centralizer dimension at most 18, whence |A| ≥ 2^{6}, which is not possible. Similarly, in the 34-dimensional representation, A necessarily contains inner involutions, but these have at most 20-dimensional centralizer. Finally, d = 20 is the restriction of the exterior cube of the natural module for U_{6}(2), so does not give an example. The 2-rank of S.(2^{2})_{133} equals 4, so only d = 20 needs to be considered. Since 2E- and 2F-involutions have centralizer dimension at most 11 on this module, A has to be 2A-pure, hence contained in S, contradiction.

2. If n = 5 we arrive at q ≤ 3. For S = U_{5}(3) we have ℓ = 2, with 2-rank bounded above by 5, and generated by 5 conjugates, so d ≤ 50, impossible. For G = Aut(S) = S.2 there is a single class of outer involutions, with centralizer 2×O_{5}(3) [12], Table 4.5.2. Thus the rank of elementary abelian 2-groups containing outer involutions is at most 5, and again no example arises. For S = U_{5}(2) necessarily ℓ = 3 and d ≤ 48, so either d = 10 or d = 44 with G = S. By [2], E.5.8, any elementary abelian 3-group can be embedded into a maximal torus. By Lemma 2.2 we may assume that A = 3^{4} is maximal, hence equal to a maximal torus. Thus it contains an element g with centralizer 3×U_{4}(2). The 10-dimensional module has two 5-dimensional composition factors for U_{4}(2), so g has 5-dimensional centralizer space. Again by loc. cit. all maximal elementary abelian 3-subgroups of U_{4}(2) are conjugate. The 5-dimensional representation identifies U_{4}(2) with O_{5}(3), and this clearly has an elementary abelian subgroup of order 27 with only 1-dimensional centralizer. Hence A can only have 1-dimensional centralizer, ruling out the case d = 10. The 44dimensional representation is the exterior square of the 10-dimensional one (up to a trivial composition factor), hence it occurs neither.

3. For n ≥ 6 the only possibilities are q = 2, n ≤ 7 and ℓ = 3. For S = U_{6}(2) the 3rank is 4 and six conjugates generate. This leads to d = 21. Since this representation extends to PGU_{6}(2) = S.3 we may argue there. As above we may assume that A is a maximal torus of order 3^{5}. Let g be an element with centralizer 3 × U_{5}(2). On restriction to U_{5}(2) we find two copies of the 10-dimensional representation considered above, and we already saw that any 3^{4} has only 1-dimensional centralizer there. Hence A has at most 3-dimensional centralizer, impossible. For U_{7}(2), ℓ = 3, we can reduce the question inductively to the case considered before. □

4C. Symplectic and orthogonal groups.

Proposition 4.5. Let S be a finite simple symplectic or orthogonal group of Lie type in characteristic different from ℓ and in dimension at least 4, S ≠ S_{4}(2)′, S_{4}(3). Then the absolutely irreducible F G-modules V satisfying (1.1) are given in Table 6.5.

Proof. 1. Let’s first consider the symplectic groups defined over fields of odd order. The minimal degree of non-trivial projective representations of S_{2n}(q), q odd, is given by (q^{n} − 1)/2. Let first S = S_{4}(q). For odd ℓ the ℓ-rank of Aut(S) is bounded above by 3, and at most four conjugates generate, so d ≤ 24. This gives q ≤ 7. The 2-rank is at most 6, and five conjugates generate. Moreover, three conjugates of elementary abelian subgroups of order 8 generate, so d ≤ 36, and again q ≤ 7. For S = S_{4}(7) a more precise study of the ℓ-ranks shows that ℓ = 2, with rank 5. This leads to d = 24. Identify S = O_{5}(7) and write 2A, 2B for the class of involutions with 4 resp. 2 eigenvalues -1 in the natural 5-dimensional representation. From the embedding O_{4}^{−}(7) < O_{5}(7) we see that 2B-involutions invert elements of order 25, so have 12-dimensional commutator. But clearly there are no 2A-pure subgroups of order at least 4, so this case is out. For S = S_{4}(5) and ℓ = 3, 13 we get d ≤ 12, so the only possible degree is d = 12 for 2.S, but in both cases f = 2. When ℓ = 2 then d ≤ 50, so d = 12, f = 2, or d = 40. Write 2A, 2B for the two classes of inner involutions as in the case S_{4}(7). Then 2B-involutions invert 5C-elements, hence have 6-dimensional respectively at least 16-dimensional commutator space when d = 12 resp. d = 40. Since 2A-pure subgroups have order 2 neither gives an example. The group S_{4}(3) is isomorphic to U_{4}(2) and has already been considered in Proposition 4.4.

For S_{6}(q) we arrive at q ≤ 3 for odd ℓ and q ≤ 5 for ℓ = 2. For S_{6}(5), ℓ = 2, the only possibility would be d = 62, but there f = 2. For S_{6}(3) all odd primes ℓ ≠ 3 have rank 1, so ℓ = 2. Then the rank is 7, seven conjugates generate, hence d ≤ 98. This gives d = 13, f = 2, or d = 78. The latter is no example since both inner involutions invert elements from class 3G, hence have at least 26-dimensional commutator space. In the case of d = 13 the same argument shows that A has to be 2A-pure of order |A| ≥ 2^{4}. The centralizer C of a 2A-involution has type 2(𝔄_{4}×U_{4}(2)), so A cannot be contained in its O_{2}. Now U_{4}(2) has three 4-dimensional composition factors, hence any four group containing involutions outside O_{2}(C) has at most 7dimensional centralizer. Thus |A| ≥ 2^{6}, but U_{4}(2) does not contain an elementary abelian 2^{3} with 3-dimensional centralizer on the natural module. This rules out the case d = 13. Outer involutions of S_{6}(3).2 have centralizer 2 × L_{3}(3): 2 of 2-rank 4. This gives d ≤ 32, so d = 26. But this representation is induced, so the commutator space of outer involutions is 13-dimensional, impossible.

For S = S_{8}(q) only q = 3, ℓ = 2 is possible, and leads to d = 40 with f = 2. The 40-dimensional Weil representations of S split into three copies of 13-dimensional Weil representations of S_{6}(3) plus a trivial constituent. From the character values it can be seen that any involution in S has at least 12-dimensional commutator space, but the 2-rank of Aut(S) is not larger than 10, hence we obtain no example.

For S_{2n}(q), n ≥ 5, the generic estimate now gives S_{10}(3), ℓ = 2 as the only other possibility. Here d = 121 with f = 2. Since S has 2-rank 10, this is not possible for S, so G = S.2. But the outer involution has centralizer L_{4}(3) × 2 by [12], Table 4.5.2, so the rank of A containing outer involutions is at most 5, ruling out this case as well.

2. We are left with the orthogonal groups. Lower bounds for minimal degrees of non-trivial projective representations of S are given in [19], and f(A) is at most equal to |G|^{2}. For S_{2n}(q) with q even this leaves only the possibilities S_{4}(4), S_{6}(2) and S_{8}(2). For S_{4}(4) we get d ≤ 26, but k_{ℓ} ≤ 4 and the ℓ-rank is at most 2, hence no example arises. For S = S_{6}(2) the estimate yields d ≤ 27. For ℓ = 5, 7 the rank is 1 and 3 conjugates generate, hence there are no examples. For ℓ = 3 the rank is 3 and k_{3} ≤ 4, so in fact d ≤ 24. This leaves d = 7, 14, 21 for S and d = 8 for 2.S. The first clearly is an example. The (3X, 3X, 5A) structure constant does not vanish for any class of 3-elements 3X, so any of these has commutator of dimension at least 6 resp. 8 when d = 14, 21 respectively. This rules out d = 21 and forces |A| = 3^{3} when d = 14. Assume in this case that A has 8-dimensional centralizer, as all of its non-identity elements. This gives a contradiction to the fact that the centralizer of 3A-elements of type 3×𝔖_{6} contains 5-elements with 2-dimensional centralizer. Explicit computation shows that d = 8 is an example. For S = S_{8}(2) the estimate yields ℓ = 3, d ≤ 44, so d = 35. Here the 3-rank is 4. Since the (3X, 3X, 5A)-structure constants are all non-zero, any 3-element has at least 12-dimensional commutator space, so this is not an example.

3. For the orthogonal groups over fields of odd order and in dimension at least 7 the only cases satisfying our basic inequality are O_{7}(3) and O_{8}^{±}(2). For S = O_{7}(3) we necessarily have ℓ = 2. Then log_{2}(|3.S.2|) ≤ 70, giving d = 27, f = 2 for G = 3.S. Involutions from class 2A have six equal eigenvalues on the natural module for S, so no 2A-pure four groups exist in S. Involutions from the other two classes invert elements from 3G, so have at least 9-dimensional commutator space for d = 27. Thus this cases is out.

For S = O_{8}^{+}(2) we have ℓ = 3, 5. For ℓ = 5 the three classes of 5-elements are fused under the triality automorphism. Three elements from class 5A generate (via the maximal subgroup 𝔄_{9}), so the same is true for the other two classes. Consequently d ≤ 12, so the only possibility is d = 8 for G = 2.S with A a Sylow 5-subgroup. This is the 5-modular reduction of the ordinary reflection representation of the Weyl group W (E_{8})′. Let K be (the intersection with 2.S of) a reflection subgroup A_{4} × A_{4} (visible in the extended Dynkin diagram). Then the restriction to K consists of two copies of the deleted permutation module for the two factors. In particular we see 5-elements with at least 6-dimensional commutator. When ℓ = 3 then we either have G = 2.S, d = 8, or G = S, d = 28. Since S has 3-rank 4, d = 8 gives an example for ℓ = 3. When d = 28 we use that without loss |A| = 3^{4} by Lemma 2.2. In particular A contains elements from all classes of 3-elements. But one of these classes contains elements normalizing 7-elements, thus with at least 16-dimensional commutator. This yields a contradiction.

For S = O_{8}^{−}(2) no examples arise. □

4D. Exceptional groups.

Proposition 4.6. Let S be a finite simple exceptional group of Lie type in characteristic different from ℓ, S ≠ ^{2}G_{2}(3)′, G_{2}(2)′. Then no absolutely irreducible F G-module satisfies (1.1).

Proof. Lower bounds for minimal degrees of non-trivial projective representations of S are given in [19], while and upper bound for f(A) is |Aut(S)|^{2}. The resulting inequality is only satisfied for the groups ^{2}B_{2}(8), G_{2}(3), G_{2}(4), and F_{4}(2). For ^{2}B_{2}(8) the ℓ-rank is 1 and k_{ℓ} = 2, so d ≤ 6, which is impossible. For S = G_{2}(3) and ℓ = 7, 13, the rank is 1, k_{ℓ} ≤ 3, thus d ≤ 6, a contradiction. For ℓ = 2 the rank is 3 (respectively 4 for S.2), and k_{2} ≤ 3 (resp. 4), so d ≤ 18 (resp. d ≤ 32). Hence by [16] the only possibility is d = 14. Inner involutions invert elements from class 3C which have 12-dimensional commutator space, so A has to have rank at least 3. The subgroup U_{3}(3).2 has odd index in S and the 14-dimensional representation restricts irreducibly, so we must have G = S.2 by the proof of Proposition 4.3. Outer involutions have centralizer 2 × L_{2}(8): 3 and commutator space of dimension at least 6 on the 14dimensional module. Under restriction to L_{2}(8): 3 this has two trivial composition factors and two 6-dimensional ones. The latter are induced from the natural module of L_{2}(8), so involutions have 3-dimensional commutator. Clearly the centralizer space of an outer involution contains such a representation, hence any non-inner Klein four group has at least 9-dimensional commutator, which violates the upper bound.

For S = G_{2}(4) and ℓ = 7, 13 the rank is 1 and two ℓ-elements generate, so d ≤ 4, which is not possible. The 5-rank equals 2 and three 5-elements respectively two Sylow 5-subgroups generate, so d ≤ 8, again impossible.

The 3-rank also equals 2 and k_{3} ≤ 3, so d ≤ 12, that is d = 12 and G = 2.S. Elements of class 3B act non-trivially on subgroups of order 13, which have no fixed points, so A cannot contain 3B-elements. But the centralizer of a 3A-element is SL_{3}(4).2_{3} and the fusion shows that it contains only two 3A-elements.

For F_{4}(2) and ℓ = 5, 7, 13, 17 the rank is at most 2 and k_{ℓ} ≤ 4, so d ≤ 16, which is not possible. For ℓ = 3 the rank is 4, and k_{3} ≤ 5, so d ≤ 40, again too small. □

5. Sporadic groups

In the case of sporadic groups at the moment we have to leave open two cases: the 22-dimensional respectively 24-dimensional 𝔽_{2}G-modules for G = Co_{2} respectively G = Co_{1}. In both cases, there exist subgroups A ≤ G with f(A) at most a factor 2 off |V |, e.g. A maximal elementary abelian of order 2^{10} respectively 2^{11}.

Proposition 5.1. Let S be a sporadic simple group. Then the absolutely irreducible F G-modules V satisfying (1.1) are given in Table 6.7, or possibly the 22-dimensional 𝔽_{2}Co_{2}-module or the 24-dimensional 𝔽_{2}Co_{1}-module.

Proof. We refer to the table of ℓ-ranks of sporadic groups in [12], Table 5.6.1, and to the table of lower bounds for degrees of faithful representations in [17], p. 137, plus for some groups like the Conway groups the more detailed information in [14]. The tables show that for the sporadic groups J_{1}, He, ON, J_{4}, F i_{23} F i′_{24}, HN, Ly, T h, B and M no examples can occur. We deal with the remaining 15 groups case by case.

• For S = M_{11} the basic inequality leaves ℓ = 3, d = 5, clearly an example with |A| = 3^{2}, and ℓ = 2, d = 10. In the latter case involutions have at least 4-dimensional commutator, so |A| = 2^{2}, but O_{2} of the involution centralizer is a quaternion group, so the standard argument rules out this possibility.

• For S = M_{12} we get ℓ ≤ 3. If ℓ = 3 then only d = 6 for 2.S meets the bound, and we get an example with |A| = 3^{2}. For ℓ = 2 we have to consider d = 10 for S and for S.2. This is the deleted permutation module for the 5-fold transitive permutation representation of S, and explicit computation shows that this gives an example with |A| ≥ 8.

• For S = M_{22} we necessarily have ℓ ≤ 3. For ℓ = 3 we get d ≤ 12 which is not possible since the 10-dimensional representation of 2.S is not defined over 𝔽_{3}. Thus ℓ = 2. The 2-rank of S is 4 and k_{2} = 3, so d ≤ 24. Thus either d = 10 for S, or d = 6, f = 2 for 3.S. Both give examples with |A| ∈ {8, 16}. The group 3.S has a subgroup 2^{4}.(3.𝔄_{6}), and upon restriction to 3.𝔄_{6} the 6-dimensional module has two 3-dimensional composition factors, hence the 2^{4} has 3-dimensional centralizer. For G = S.2 the 2-rank is 5, and k_{2} ≤ 4, hence d ≤ 40. This gives d = 10, 34 for S.2, d = 12, 30 for 3.S.2. If d = 34 then A contains inner involutions, but this is not possible since already three of these generate. Clearly d = 10 is an example. Since the 30-dimensional representation of 3.S.2 is induced, outer involutions have fixed space of dimension 15, thus cannot lie in A. The 12-dimensional representation of 3.S.2 is induced from S.2. Explicit computation in that representation shows that the centralizer space of an elementary abelian 2-subgroup A involving outer involutions is of dimension at most 4, 3, 3, 1 for A of order 2^{2}, 2^{3}, 2^{4}, 2^{5} respectively. Hence we do not get an example.

• For S = J_{2} it follows that ℓ ≤ 5. For ℓ = 5 we have d = 6 for 2.S. All 5-elements have at least 4-dimensional commutator space, since the restriction to 𝔄_{5} = L_{2}(5) containing 5AB-elements (respectively SL_{2}(5) containing 5CD-elements) has only two constituents. The centralizer in S of a 5A-element is 5 × 𝔄_{5}, thus if a Sylow 5-subgroup had more than 2-dimensional centralizer, the same were true for the 𝔄_{5}. But this contains elements from class 3A without fixed points on the module. Hence this case is out. For ℓ = 3 the rank is 2. Then k_{3} ≤ 3 and two Sylow 3subgroups generate, whence d ≤ 8. Since the representation of 2.S with d = 6 has f = 2 we get no example. For ℓ = 2 the rank of S is 4, k_{2} ≤ 4, so d ≤ 32. This leads to d ∈ {6, 14}, both with f = 2. Since inner involutions invert elements from 3B their commutator space is at least 5-dimensional when d = 14, thus this case is out. On the other hand d = 6 gives an example. For J_{2}.2 we find d = 12. The centralizer of an outer involution g is ⟨g⟩ × PGL_{2}(7), so |A| ≤ 8. On the other hand outer involutions have 6-dimensional centralizer. If A is an example then all conjugates of A by the centralizer of g have to have the same centralizer. But they generate a group containing elements of order 7 with trivial centralizer, contradiction.

• For S = M_{23} the only possibility is ℓ = 2 with d ≤ 24, so d = 11. This yields an example with |A| ≥ 8.

• For S = HS we find ℓ = 2 and d ≤ 24, respectively d ≤ 50 for S.2, so d = 20. Let first G = S. Since 2B-elements invert 5A-elements, their commutator is 10dimensional, so A has to be 2A-pure. Furthermore, 2A-elements invert 5B-elements. Hence their commutator on the 20-dimensional module is at least 8-dimensional and necessarily |A| = 2^{4}. Let C = O_{2}(C).𝔖_{5} with O_{2}(C) = D_{4} ∗ D_{4} ∗ 4 ([12], Table 5.3m) be the centralizer of a 2A-involution. As A cannot be contained in O_{2}(C) and 𝔄_{5} is generated by its class of involutions, we deduce that 𝔄_{5} has 12-dimensional centralizer. But this contradicts the character table. The group G = S.2 has 2-rank 5. Outer 2Dinvolutions have 10-dimensional commutator (inverting 11-elements) and centralizer 2^{4}.𝔖_{5} ([12], Table 5.3m), so A cannot contain 2D by the standard argument. Since A has to contain inner involutions, our considerations for S show that |A| ≥ 2^{4}. Clearly A can be chosen in the centralizer C = 2 × 𝔖_{8} of a fixed 2C-involution g. The constituents of the restriction to 𝔖_{8} are of dimensions 6,6,8: two times the deleted permutation module and the induced of the natural module for L_{4}(2) = 𝔄_{8}. Note that |A| ≥ 2^{4} implies that A ∩ 𝔖_{8} has to contain involutions not of permutation type 2^{4}, hence with 4-dimensional commutator on the 8-dimensional constituent. But g has to centralize this 8-dimensional module and has at least 6-dimensional commutator.

So A has at most 10-dimensional centralizer, forcing |A| = 2^{5}. But in fact A ∩ 𝔖_{8} clearly acts non-trivially on the two 6-dimensional composition factors, thus A has at most 9-dimensional centralizer, contradiction.

• For S = J_{3} either ℓ = 3, d ≤ 18, which is not possible since d = 18 has f = 2, or ℓ = 2, d ≤ 30. This leads to d = 9, f = 2, for 3.S, or d = 18 for 3.S.2. The inner involutions invert elements of order 17, so have 4-dimensional (resp. 8-dimensional) commutator space when d = 9 respectively d = 18. Since the centralizer of an outer involution is 2 × L_{2}(17) with 2-rank 3, this rules out the second case. In the first case we get |A| = 2^{4}. The centralizer C of an inner involution g ∈ A is (D_{4} ∗ Q_{8}).𝔄_{5} (see [12], Table 5.3h). Since A cannot be contained in O_{2}(C) and g already has 4dimensional commutator, the centralizer of A is at most 4-dimensional, so we obtain no example.

• For S = M_{24} we necessarily have ℓ = 2. The rank is 6, and k_{2} ≤ 4, so d ≤ 48. This leaves d = 11 or d = 44. The first clearly is an example since it is one upon restriction to M_{23}. On the other hand inner involutions invert elements of order 5, so have at least 18-dimensional commutator when d = 44, a contradiction.

• For S = M^{c}L we arrive at ℓ ≤ 3. When ℓ = 3 k_{3} ≤ 5, while the 3-rank is 4, so d ≤ 40, which implies d = 21. The (3A, 3A, 10A)- and (3B, 3B, 7A)-structure constants in S are non-zero, so elements from 3A (resp. 3B) have at least 10-dimensional (resp. 12dimensional) commutator. Thus this case does not occur. For ℓ = 2 the rank is 4 and k_{2} ≤ 4, so d = 22. Involutions invert elements from 5B, so have at least 8-dimensional commutator space. Thus, |A| = 16 and C_{V} (A) is 14-dimensional. Thus, A is trivial on the fixed space of any involution in A. The centralizer of an involution is 2.𝔄_{8}. Since this is generated by any class of elementary abelian subgroups of order 16 and contains elements of order 7 (which do not centralize a 14-dimensional subspace) we don’t get an example. The rank of 2-subgroups of Aut(S) involving outer involutions is 3 (since the centralizer is 2 × M_{11}), hence this gives no further example.

• For S = Ru we arrive at ℓ = 2, and since the 2-rank is 6 we have d ≤ 48, so d = 28 has to be considered. Involutions from class 2B have 14-dimensional commutator space, since they invert elements of order 29, so A has to be 2A-pure. Elements from 2A invert elements of order 13, so have at least 12-dimensional commutator. In particular, to get an example we need that 2A-elements have 16-dimensional centralizer, |A| = 2^{6}, and C_{V} (g) = C_{V} (A) for all involutions g ∈ A. Since the center of the centralizer of a 2A-involution only has order 2, there exist involutions in A with distinct centralizer. But these centralizers are maximal subgroups, so two of them generate the group. Since both would also stabilize C_{V} (A), this is a contradiction.

• For S = Suz the Suzuki group the estimates give ℓ ≤ 5. For ℓ = 5 the rank is 2 and k_{5} ≤ 5. This forces d ≤ 20, but d = 12 has f = 2, so no case arises. For ℓ = 3 the rank is 5, thus d ≤ 50, which gives d = 12 for 2.S. First assume that A contains 3Celements. The commutator space of a 3C-element is 8-dimensional, as can be seen by restricting to L_{3}(3): 2, so |A| ≥ 3^{4}. Since the centralizer has structure 3^{2}: 4×𝔄_{6}, hence 3-rank 4, we even have |A| = 3^{4}. This is ruled out by the standard argument from the structure of the centralizer. Thus A is 3C-free. Now any 3B-element g is uniquely expressible as a product of two commuting 3A-elements h_{1}, h_{2}, and thus any other 3element commuting with g also commutes with h_{1}, h_{2}. Hence any elementary abelian 3-subgroup of S not containing 3C-elements is contained in an elementary abelian 3-subgroup generated by 3A-elements, with the same normalizer. So we may assume that A contains 3A-elements. Upon restriction to the centralizer 6_{2}.U_{4}(3) of a 3Aelement the 12-dimensional module has two 6-dimensional composition factors. Thus 3A-elements have 6-dimensional centralizer and |A| ≥ 3^{3}. Any 3-element of 2.U_{4}(3) = SO_{6}^{−}(3) has at least 2-dimensional commutator on this 6-dimensional centralizer, so A has centralizer dimension at most 4, that is |A| ≥ 3^{4}. But subgroups of order 3^{3} of SO_{6}^{−}(3) have at most 3-dimensional centralizer on the natural module, hence |A| = 3^{5}. By [22, 2.2] there exists a unique class of elementary abelian subgroups of that order in G, with normalizer 3^{5}: M_{11}. The complement M_{11} has composition factors of dimensions 1,1,5,5, hence either A has 1-dimensional or 6-dimensional centralizer. The first possibility violates the 2F-condition. In the second case, all 3-elements in A would have the same centralizer. But this is impossible since 3A-centralizers are maximal and do not contain a central 3^{5}.

For ℓ = 2 the rank is 6, so d ≤ 60, which leads to d = 12 with f = 2 for 3.S, or d = 24 for 3.S.2. The two classes 2A, 2B of involutions of S have at least 4respectively 6-dimensional commutator space when d = 12.Let’s first assume that A is 2A-pure. Then |A| ≥ 2^{4}, and in fact A = 2^{4} since by [22, 2.4] the largest 2Apure 2-subgroup of S has order 2^{4}. So C_{V} (A) = C_{V} (g) for all involutions g ∈ A. But 2A-centralizers C ≅ 2^{1+6}.O_{6}^{−}(2) are maximal in S and do not contain a 2^{4} in their center, so this leads to a contradiction. Hence A must contain 2B-elements, and |A| = 2^{6}. Since products of commuting 2A-elements lie again in class 2A, in the second case A has to contain a 2B-pure four-group. According to [12, Table 5.3o] this has centralizer (2^{2} × L_{3}(4)).2. But L_{3}(4) does not have 6-dimensional centralizer in this representation, hence the standard argument rules out this case. For G = 3.S.2 the outer involutions have 12-dimensional commutator since the representation with d = 24 is induced. On the other hand the centralizers are 2 × M_{12}.2 respectively 2 × J_{2}.2, both of 2-rank at most 5. This gives a contradiction.

• For S = Co_{3} we obtain ℓ = 2, 3 and d = 22. For ℓ = 3 the rank is 5. The (3C, 3C, 23A)-structure constant does not vanish, so 3C-elements have at least 11dimensional commutator space on the 22-dimensional module. Thus A cannot contain 3C-elements. The same argument applies to 3B-elements, hence A has to be 3Apure. Restriction to M^{c}L contains the 21-dimensional constituent considered above, so 3A-elements have at least 10-dimensional commutator, which forces |A| = 3^{5}. But by [6], Lemma 5.11, S has a unique class of elementary abelian 3^{5}-subgroups, and these are not 3A-pure. For ℓ = 2 the rank is 4. The character table shows that outer involutions from 2A (resp. 2B) have 8-dimensional (resp. 10-dimensional) commutator space. Thus A has to be 2A-pure. But the centralizer of a 2A-involution is 2 · S_{6}(2), which is generated by any conjugacy class of elementary abelian 2-subgroups of order at least 4. Thus the centralizer space of A must have dimension less than 14.

• For S = Co_{2} either ℓ = 3, d = 23, or ℓ = 2, d = 22. Let first ℓ = 3. The subgroup H = U_{6}(2) contains a full Sylow 3-subgroup of S, and the restriction of the 23-dimensional representation contains the 21-dimensional representation of H. But the latter is not an example by Prop. 4.4. The case ℓ = 2 was excluded in the statement.

• For S = Co_{1} the estimates give ℓ ≤ 5. For ℓ = 5 the rank is 3. From the character table it follows that for X ∈ {A, B, C} the (5X, 5X, 3A)-structure constants are nonzero. Since 3A-elements have trivial centralizer for d = 24 all 5-elements have at least 12-dimensional commutator space, which shows that A cannot exist. For ℓ = 3 the rank is 6, so d ≤ 60, again only giving d = 24 for 2.S. The (3X, 3X, 5A)-structure constant is non-zero for X ∈ {A, B, C, D}, and 5A-elements act fixed point freely, so all 3-elements have at least 12-dimensional commutator. In particular |A| = 3^{6}. The centralizers of 3-elements are given in [12, Table 5.3l] and the standard argument shows that this case cannot occur.

For ℓ = 2 the rank is 11, thus d ≤ 110 whence d = 24 for S, which was excluded.

• For S = F i_{22} we arrive at ℓ = 2, with rank 10, so d ≤ 154. This leaves d = 27 with f = 2 for 3.S and d = 78 for S. Since all involutions invert elements from 3A, they have at least 21-dimensional commutator in the 78-dimensional representation. This violates the required bound. Now consider a 27-dimensional module V for 3.S. Elements from 2C have at least 12-dimensional commutator, so cannot lie in A. Elements from 2B have at least 10-dimensional commutator, so |A| = 2^{10}. But O_{2} of the centralizer (2×2_{+}^{1+8}): O_{6}^{−}(2).2 of a 2B-involution [12, Table 5.3t] does not contain such an A, hence the standard argument forces A to be 2A-pure. But the structure constants show that 2A-pure subgroups have order 2, while 2A-elements have at least 5-dimensional commutator. Thus this case is also impossible. In Aut(S) = S.2 we again arrive at d = 78 for S.2 or d = 54 for 3.S.2. Since all outer involutions invert elements from 3C, their commutator space is at least 24-dimensional when d = 78 while clearly the 2-rank is at most 11. The 54-dimensional representation is induced, thus outer involutions have 27-dimensional commutator space, which is again impossible. □

6. Cubic action

In this section we determine which of the examples (G, V) classified in the previous sections have cubic action, that is, there exists an elementary abelian ℓ-subgroup A ≤ G such that

(6.1)  [[[V, A], A], A] = 0  and  |V/C_{V} (A)| ≤ |A|^{2}.

We call these the 2F-modules with cubic offender.

Theorem 6.2. The 2F-modules with cubic offender of alternating groups S = 𝔄_{n} with n ≥ 5, (n, ℓ) ∉ {(5, 2), (5, 5), (6, 3), (8, 2)}, are given in Table 6.3.

Proof. By the result of Proposition 3.1 we only have to show that all cases in Table 6.3 admit A with cubic action.

The four infinite families corresponding to the deleted permutation module are examples since double-transpositions respectively 3-cycles have centralizer space of codimension 2 and clearly have cubic action. In particular |A| = ℓ in these cases. For 𝔖_{n} on the deleted permutation module in characteristic 2 we can take A generated by a transposition.

The case 3.𝔄_{6} is a transvection group, so |A| = 2 works. The 6-dimensional representation of 3.𝔖_{6} splits into the two 3-dimensional 3.𝔄_{6}-modules considered before.

6.3. 2F-modules for alternating groups.

G Out  d  f  ℓ  |A|
𝔄_{n}  2 n − 2  1  2 2|n  2/2
𝔄_{n}  2 n − 1  1  2 2 |̸n  2/2
3.𝔄_{6}  3  2  2  2
3.𝔖_{6}  6  1  2  4
𝔄_{7}  4  1  2  2
𝔖_{7}  8  1  2  8
𝔄_{9}  8  1  2  8
𝔄_{n}  n − 2  1  3 3|n  3
𝔄_{n}  n − 1  1  3 3 |̸n  3
2.𝔄_{5}  2  2  3  3
2.𝔄_{9}  8  1  3  27

Thus a transvection of 3.𝔄_{6} has centralizer dimension 4 on the 6-dimensional module, and on this a commuting outer involution has 2-dimensional centralizer. This gives an example with |A| = 4.

The group 𝔄_{7} ≤ L_{4}(2) is a bi-transvection group. Explicit calculation with representing matrices shows that the 8-dimensional representation of 𝔖_{7} also gives an example with |A| = 8. The 8-dimensional representation of 𝔄_{9} restricts irreducibly to the subgroup 𝔖_{7}, hence gives an example by the previous case.

The group 2.𝔄_{5} ≤ SL_{2}(9) is a transvection group. In the last case explicit computation shows that A of order 27 has cubic action. □

Theorem 6.4. The 2F-modules with cubic offender of simple groups of Lie type in non-defining characteristic are given in Table 6.5.

6.5. 2F-modules for groups of Lie type in non-defining characteristic.

G  Out d  f  ℓ  |A|
U_{3}(3)  2 6  1  2  2/4
3_{1}.U_{4}(3)  2_{2} 6  2  2  8/2
3_{1}.U_{4}(3).(2^{2})_{122}  12  1  2  32
2.L_{3}(4)  6  1  3  9
S_{6}(2)  7  1  3  3
2.S_{6}(2)  8  1  3  27
2.O_{8}^{+}(2)  8  1  3  3

Proof. We have seen in Section 4 that the modules satisfying (1.1) are precisely those of Table 6.5. It remains to prove that we have cubic action in all these cases. The group U_{3}(3) in its 6-dimensional representation is generated by bi-transvections. The group 3_{1}.U_{4}(3).2_{2} is a transvection group. The derived group S_{6}(2) of the Weyl group of type E_{7} contains 3-elements with a single Jordan block of length 3 inside a reflection subgroup of type A_{2}. The same argument applies to the derived group O_{8}^{+}(2) of the Weyl group of type E_{8}. The remaining cases can be checked by direct computation. □

Theorem 6.6. The 2F-modules with cubic offender of sporadic simple groups, different from the 22- or 24-dimensional 𝔽_{2}-modules of the Conway group Co_{2} respectively Co_{1}, are given in Table 6.7.

6.7. 2F-modules for sporadic groups.

G  Out d  f  ℓ  |A|
M_{12}  2 10  1  2  8/16
M_{22}  2 10  1  2  8/8
3.M_{22}  6  2  2  8
J_{2}  6  2  2  16
M_{23}  11  1  2  8
M_{24}  11  1  2  8
M_{11}  5  1  3  9
2.M_{12}  6  1  3  9

Proof. By the result of Proposition 5.1 all 2F-modules of sporadic groups are listed in Table 6.7. It can now be checked by explicit computation in the corresponding representations that in all cases there exist offenders with cubic action, of the indicated orders. □

References

[1] M. Aschbacher and S. Smith, A classification of quasithin groups, preprint (2001).

[2] A. Borel et al., Seminar on algebraic groups and related finite groups, Lecture notes in Math., vol. 131, Springer, Berlin Heidelberg New York, 1970.

[3] A. Chermak, Small modules for small groups, preprint (2001).

[4] A. Chermak and A. Delgado, A measuring argument for finite groups, Proc. Amer. Math. Soc. 107 (1989), 907–914.

[5] J.H. Conway, R.T. Curtis, S.P. Norton, R.A. Parker, R.A. Wilson, Atlas of Finite Groups, Clarendon Press, Oxford, 1985.

[6] L. Finkelstein, The maximal subgroups of Conway’s group C_{3} and McLaughlin’s group, J. Algebra 25 (1973), 58–89.

[7] G. Glauberman, Failure of factorization in p-solvable groups, Quart. J. Math. Oxford Ser. (2) 24 (1973), 71–77.

[8] G. Glauberman, Failure of factorization in p-solvable groups. II, Quart. J. Math. Oxford Ser. (2) 26 (1975), 257–261.

[9] G. Glauberman, Factorizations in local subgroups of finite groups, Regional Conference Series in Mathematics, No. 33, Amer. Math. Soc., Providence, R.I., 1977.

[10] G. Glauberman, Factorizations for 2-constrained groups, Proc. London Math. Soc. (3) 41 (1980), 385–438.

[11] G. Glauberman and R. Niles, A pair of characteristic subgroups for pushing-up in finite groups, Proc. London Math. Soc. (3) 46 (1983), 411–453.

[12] D. Gorenstein, R. Lyons, R. Solomon, The classification of the finite simple groups, Number 3, Mathematical Surveys and Monographs, Amer. Math. Soc., Providence, 1994.

[13] R. Guralnick and J. Saxl, Generation of finite almost simple groups by conjugates, to appear, J. Algebra (2001).

[14] G. Hiss, G. Malle, Low-dimensional representations of quasi-simple groups, LMS J. of Comput. and Math. 4 (2001), 22–63.

[15] G.D. James, On the minimal dimensions of irreducible representations of symmetric groups, Math. Proc. Cam. Phil. Soc. 94 (1983), 417–424.

[16] C. Jansen, K. Lux, R. Parker, R. Wilson, An Atlas of Brauer Characters, Clarendon Press, Oxford, 1995.

[17] M.W. Liebeck, C.E. Praeger, J. Saxl, The maximal factorizations of the finite simple groups and their automorphism groups, Memoirs of the AMS, Amer. Math. Soc., Providence, 1990.

[18] K. Magaard, G. Malle, Irreducibility of alternating and symmetric squares, Manuscripta Math. 95 (1998), 169–180.

[19] G.M. Seitz, A.E. Zalesskii, On the minimal degrees of projective representations of the finite Chevalley groups, II, J. Algebra 158 (1993), 233–243.

[20] J. Thompson, Factorizations of p-solvable groups, Pacific J. Math. 16 (1966), 371–372.

[21] A. Wagner, An observation on the degrees of projective representations of the symmetric and alternating groups over an arbitrary field, Arch. Math. 29 (1977), 583–589.

[22] R. A. Wilson, The complex Leech lattice and maximal subgroups of the Suzuki group, J. Algebra 84 (1983), 151–188.

R.M. G.: Department of Mathematics, University of Southern California, Los Angeles CA 90089 - 1113, USA

   E-mail address: guralnic@math.usc.edu

G. M.: FB Mathematik/Informatik, Universität Kassel, Heinrich-Plett-Str. 40, D– 34132 Kassel, Germany

   E-mail address: malle@mathematik.uni-kassel.de