On Hurwitz groups of low rank

L. Di Martino

Dipartimento di Matematica e Applicazioni Università degli Studi di Milano-Bicocca Via Bicocca degli Arcimboldi 8, Ed. U7 20126 Milano, Italy. e-mail: dimartino@vmimat.mat.unimi.it

M.C. Tamburini

Dipartimento di Matematica e Fisica Università Cattolica del Sacro Cuore Via Trieste 17 , 25121, Brescia, Italy. e-mail: c.tamburini@dmf.bs.unicatt.it

A.E. Zalesskii

School of Mathematics University of East Anglia Norwich NR4 7TJ, UK. e-mail: a.zalesskii@uea.ac.uk

1. Introduction

Let Δ(2, 3, 7) denote the infinite triangle group, defined by the presentation ⟨X, Y | X^{2} = Y^{3} = (XY)^{7} = 1⟩. A non-trivial group G is said to be (2, 3, 7) -generated (or a Hurwitz group, when finite) if it is an epimorphic image of Δ(2, 3, 7). Hurwitz groups are particularly interesting for the theory of Riemann surfaces. Namely, if H is the automorphism group of a compact Riemann surface of genus g > 1, then H is finite of order not exceeding 84(g − 1), and the Hurwitz groups are exactly those for which |H| = 84(g − 1). As a quotient of a Hurwitz group is again Hurwitz, it is crucial to determine the simple Hurwitz groups. The values of q for which the groups PSL(2, q) are Hurwitz were determined by A. M. Macbeath in 1969 ([Mac]), whereas J. Cohen in 1981 ([Coh]) showed that P SL(3, 2) is the only Hurwitz group in the series PSL(3, q), and none of the groups in the series PSU(3, q^{2}) are Hurwitz. M. Conder in 1980 ([Con]) proved that the alternating groups A_{n} are Hurwitz, provided n > 167. More recently G. Malle has shown that the exceptional simple groups G_{2}(q), q ≥ 5; ^{2}G_{2}(q), q > 3; ^{3}D_{4}(q) for q = p^{m}, p ≠ 3, q ≠ 4, and ^{2}F_{4}(2^{2m+1})′ for m ≡ 1 (3) are Hurwitz ( cfr. [Mal1], [Mal2]). The sporadic Hurwitz groups are also known, with the exception of the Monster. (For a bibliography and related comments, see [Jon]). Recent constructive results of Lucchini, Tamburini and Wilson ([LTW], [Lu], [Wil] ) show that the family of (2, 3, 7) -generated groups, and more generally of (2, 3, k) -generated groups, is very large.

AMS 1991 Subject Classification: 20G40, 20F05.

Key words: Hurwitz groups, classical groups, representations.

In fact, for every k ≥ 7 there are 2^{ℵ_{0}} isomorphism classes of simple (2, 3, k) -generated groups. In particular, most finite classical groups of large rank are Hurwitz ([LT]). In this paper we prove, by contrast, that most quasi-simple classical groups of small rank are not (2, 3, 7) -generated. Note that, whenever the centre of a quasi-simple group is non-trivial, the problem of deciding whether the corresponding simple central quotient is Hurwitz requires further analysis. This is why a general treatment of projective classical groups is not envisaged in this paper. Our first result, however, deals with projective groups. Namely, in Theorem 1 ii) we determine the (2,3,7)-generated subgroups of PSL(3, \overline{F}), where \overline{F} is the algebraic closure of an arbitrary field F. This extends the results of [Mac] and [Coh], which deal with the case F = F_{p}, the Galois field of order p. Our approach essentially amounts to bringing the infinite triangle groups Δ(2, 3, k), k ≥ 7, into the picture. While referring to [Mag] II.5 for details and proofs, we recall here that the group PSL(2, ℝ) over the reals acts naturally on the complex upper half-plane U = {α ∈ ℂ | Im α > 0}, preserving the hyperbolic metric which turns U into a model of the Poincaré plane. In this realization, the subgroup of PSL(2, ℝ) generated by rotations of angles \frac{2π}{2} , \frac{2π}{3} , \frac{2π}{k} around the vertices of a hyperbolic triangle having angles of size \frac{π}{2} , \frac{π}{3} , \frac{π}{k} is isomorphic to Δ(2, 3, k).

Throughout the paper F denotes a field of characteristic p ≥ 0 and \overline{F} denotes the algebraic closure of F. Moreover, we observe explicitly that, for any prime k ≥ 7, a (2, 3, k) -generated group G is a perfect group. It follows that, for any representation f : G → GL(n, F), f(G) is a subgroup of SL(n, F). We start with the following:

Theorem 1 Let k ≥ 7 be a prime, and ϵ ∈ \overline{F} be a primitive k^{th} root of unity ( ϵ = 1 if p = k).

i) Up to isomorphism, a (2, 3, k) -generated subgroup H of PSL(2, \overline{F}) is the projective image of the group

⟨( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) , ( \begin{matrix}ϵ & 1 \\ 0 & ϵ^{−1} \end{matrix} )⟩.

If p = 0, H is isomorphic to Δ(2, 3, k). If p > 0, H is a subgroup of PSL(2, p^{k−1}). In particular H is finite and, for k = 7, by [Mac] H ≃ PSL(2, p) if p ≡ 0, ±1 (mod 7), H ≃ PSL(2, p^{3}) if p ≢ 0, ±1 (mod 7).

ii) If p ≠ 2, the image of H under the orthogonal embedding φ : PSL(2, \overline{F}) → SO(3, \overline{F}) defined by

± ( \begin{matrix}a & b \\ c & d \end{matrix} ) ↦ ( \begin{matrix}a^{2} & 2ab & b^{2} \\ ac & ad + bc & bd \\ c^{2} & 2cd & d^{2} \end{matrix} )

is irreducible for all primes k ≥ 7. Assuming k = 7, let K be an irreducible (2, 3, 7) -generated subgroup of SL(3, \overline{F}). Then either K is isomorphic to PSL(2, 7), or p ≠ 2 and K is isomorphic to φ(H). The irreducible (2, 3, 7) -generated subgroups of PSL(3, \overline{F}) are the projective images of those of SL(3, \overline{F}).

iii) If p = 0 or p ≢ 0, ±1 (mod k), the group H embeds irreducibly into SL(4, \overline{F}), for all primes k ≥ 7. The embedding is orthogonal if p ≠ 2.

iv) If p ≠ 3, the group H embeds irreducibly into a symplectic subgroup of PSL(4, \overline{F}), for all primes k ≥ 7.

We have included items iii) and iv) in order to emphasize the occurrence of orthogonal or symplectic representations of Hurwitz groups in dimension 4 (cfr. Theorem 2). We are aware that they are special cases of much more general embeddings in tensor (or twisted tensor) product spaces.

Our subsequent results only concern linear groups, and are based on the application of a formula due to L.L. Scott (see Section 2) to appropriate representations of a group, and the possible contradictions that one may obtain assuming that the group is (2, 3, 7) -generated. Scott himself observed that his formula could be used for proving that certain linear groups are not (2, 3, 7) -generated (he considered, as examples, the groups SL(6, 3) and SL(9, 3) ). A more systematic application of Scott’s formula in the context of Hurwitz groups appears in [TV], where it is shown that the groups SL(4, q) and SU(4, q^{2}), q odd, are not Hurwitz. Although these techniques work in principle over an arbitrary field, we mainly concentrate on classical groups over finite fields (Theorems 2,3,4), and on the field of rational numbers (Theorem 5). We prove the following “negative” results.

Theorem 2 Let H denote an absolutely irreducible subgroup of SL(n, F), n ∈ {4, 5, 6, 7, 10}. Assume that H is not contained in an orthogonal group if p ≠ 2, and that H is not contained in a symplectic group if p = 2. If n = 6 and p = 2, assume further that F is finite and H = SL(6, F) or SU(6, F). Then H is not (2, 3, 7) -generated. In particular, if n ∈ {4, 5, 6, 7, 10}, then:

i) the groups SL(n, q), Sp(n, q), SU(n, q^{2}) are not Hurwitz, with the only possible exception of Sp(n, 2^{t}) ;

ii) every complex irreducible character of degree n of a (2, 3, 7) -generated group, is real.

Both Theorem 1 and Theorem 2 may be rephrased as statements about representations of a (2,3,7)- generated group G. Clearly, if p ≠ 2, G has no absolutely irreducible representation of degree 2, since the only involution of SL(2, \overline{F}) is −I. In particular G has no complex irreducible representation of degree 2. It follows from Theorem 1 that, if G has a complex irreducible representation of degree 3, then it has a quotient isomorphic to PSL(2, 7). In particular, PSL(2, 7) does have complex irreducible representations of degree 3. The remaining irreducible representations of PSL(2, 7) over ℂ have degrees 6,7 and 8, with real characters. Similarly, the next Hurwitz group SL(2, 8) has non-real irreducible characters of degree 9 and real characters of degree 7 and 8 (e.g. cfr. [Isa], p. 289). Theorem 2 states that, under the given assumptions on F and for the stated values of n, every irreducible representation f : G → SL(n, \overline{F}) is orthogonal (symplectic if p = 2 ). It is worth to recall that, in the natural characteristic, the groups of type G_{2} do have orthogonal representations of degree 7 if p ≠ 2, and symplectic representations of degree 6 if p = 2; whereas the groups of type ^{2}G_{2} and ^{3}D_{4} have orthogonal representations of degree 7 and 8, respectively. Finally, note that Holt, Plesken and Souvignier [HPS] have constructed a 7-dimensional complex representation of Δ(2, 3, 7), whose kernel is generated by the element [X, Y]^{11}.

Theorem 3 Let H be one of the groups:

a) SL(n, q), q ≠ 3^{6} and, if q ≡ 1 (mod 7), q ≠ p, p^{3}; Sp(n, q), q odd; SU(n, q^{2}), q ≠ 3^{3}; n ∈ {8, 9, 11};

b) SL(12, q), q ≢ 1 (mod 7); Sp(12, q), q odd; SU(12, q^{2}), q ≢ −1 (mod 7).

Then H is not Hurwitz .

Theorem 4 Assume 13 ≤ n ≤ 19, and let H be one of the groups SL(n, q), q ≢ 1 (mod 3) ; Sp(n, q), q odd; SU(n, q^{2}), q ≢ −1 (mod 3) . Then H is not Hurwitz, except possibly for the following cases:

a) SL(16, q), q ≡ 8 (mod 21), SL(16, 3^{6h}), SU(16, q^{2}), q ≡ −8 (mod 21), SU(16, 3^{6h+3}).

b) SL(18, q), q odd ≡ 8 (mod 21), SL(18, 3^{6h}); SU(18, q^{2}), q odd ≡ −8 (mod 21), SU(18, 3^{6h+3}).

c) SL(n, q), n ∈ {15, 17, 19}, q ≡ 2, 8, −10 (mod 21) or q ∈ {3^{6h}, 3^{6h+2}, 3^{6h+4}}. d) SU(n, q^{2}) n ∈ {15, 17, 19}, q ≡ 10, −2, −8 (mod 21) or q ∈ {3^{6h+1}, 3^{6h+3}, 3^{6h+5}}.

e) SL(n, 2^{t}), n ∈ {14, 18}, 2^{t} ≡ 2, 8, −10 (mod 21).

Theorem 5 Let ℚ be the field of rational numbers, and let H denote an absolutely irreducible subgroup of SL(n, ℚ), n ≤ 19 or n = 22. If H is not contained in an orthogonal group, then H is not (2, 3, 7) -generated. In particular, for these values of n, the group SL(n, ℤ) is not (2, 3, 7) -generated.

We are grateful to the Referee for the accurate reading of the manuscript.

2. Scott’ s formula

We denote shortly by M the algebra Mat(n, F) of n×n matrices over F, and view M as acting on the left on the vector space V = F^{n} of column vectors. Given a representation f : H → GL(n, F) of a group H over F, for any subset X of H we define d_{V}^{X} as the dimension of the subspace of V fixed pointwise by f(X). In a similar way we define d̂_{V}^{X}, with respect to the dual representation f̂: H → GL(n, F) given by f̂(h) := (f(h^{−1}))^{t}. The following formula, which is a special case of a result of L.L.Scott ([Sc], Theorem 1), will be crucial.

Theorem 2.1 Let f : H → GL(n, F) be a representation of H = ⟨x, y⟩. Then :

d_{V}^{x} + d_{V}^{y} + d_{V}^{xy} ≤ dim V + d_{V}^{H} + d̂_{V}^{H}.

The above formula was conceived by Scott as a generalization of the well-known genus formula [\frac{n}{2}] + 2 [\frac{n}{3}] + 6 [\frac{n}{7}] ≥ 2n − 2 for the index n of a subgroup S of a (2, 3, 7) -generated group G. Indeed, the genus formula is a consequence of Theorem 2.1. For, let G = ⟨x, y⟩ with x^{2} = y^{3} = (xy)^{7} = 1, and consider the transitive permutation representation f : G → GL(n, ℂ) given by the action on the cosets of S. Since the multiplicity in f of the trivial representation is 1, d_{V}^{G} = d̂_{V}^{G} = 1. Scott’ s formula then gives: d_{V}^{x} + d_{V}^{y} + d_{V}^{xy} ≤ n + 2. The genus formula follows easily. Indeed, for every g ∈ G of prime order k, whose cyclic structure admits a cycles of length k, n − d_{V}^{g} = (k − 1)a ≤ (k − 1) [\frac{n}{k}] .

Scott’s formula will be applied essentially to the following representations of a subgroup H of SL(n, F), with generating pair {x, y}.

1) The representation of H by conjugation on M (also called the ’adjoint’ representation of H ) given by f(h) := m ↦ hmh^{−1} ( h ∈ H, m ∈ M ). In this case, the fixed-points subspace of M is the centralizer of H in M, and therefore, if H is supposed to be absolutely irreducible, by Schur’s Lemma it consists of scalar matrices. It follows d_{M}^{H} = d̂_{M}^{H} = 1, hence Scott’s formula reads

(∗) d_{M}^{x} + d_{M}^{y} + d_{M}^{xy} ≤ n^{2} + 2.

Failing of the above inequality means that H is reducible over \overline{F} (and to some extent our problem is reduced to linear groups of smaller degree). Dealing with (∗) requires detailed information on minimal values affordable by d_{M}^{x}, d_{M}^{y} and d_{M}^{xy}. These values will be worked out in Section 3.

2) The representation of H on the symmetric square S of \overline{F}^{n}. If H is absolutely irreducible, then Scott’s formula takes the shape

(∗∗)  d_{S}^{x} + d_{S}^{y} + d_{S}^{xy} ≤ dim S , or  d_{M}^{x} + d_{M}^{y} + d_{M}^{xy} ≤ dim S + 2

(the latter occurring only when H is orthogonal if p ≠ 2 , symplectic if p = 2 ). This will be shown in Section 4.

3. The defect def\overline{M}

We first recall a well-known formula, due to Frobenius (e.g. cfr. [Ja], p. 207). Namely, let B ∈ M := Mat(n, F) and let n_{1} ≤ … ≤ n_{s} be the degrees of the non-trivial invariant factors of λI − B (also called the similarity invariants of B ). The dimension d_{M}^{B} over F of the centralizer of B in M is given by the formula

d_{M}^{B} = ∑_{j=1}^{s}(2s − 2j + 1) n_{j} = (2s + 1)n − 2 ∑_{j=1}^{s} jn_{j}.

In particular d_{M}^{B} ≥ n + s^{2} − s, since

∑_{j=1}^{s} jn_{j} = ∑_{j=1}^{s} n_{j} + ∑_{j=2}^{s} n_{j} + … + (n_{s−1} + n_{s}) + n_{s} ≤

≤ n + (n − 1) + … + (n − s + 2) + (n − s + 1) ≤ sn − \frac{s(s − 1)}{2}.

Moreover, considering the algebraic closure \overline{F} and denoting by dB ; \overline{M} the dimension over \overline{F} of the centralizer of B in \overline{M} = Mat(n, \overline{F}), clearly d_{M}^{B} = dB ; \overline{M}, since the similarity invariants of B remain unchanged in \overline{M}. Now, for a fixed prime k, define dk ; \overline{M} as the minimal dimension of the centralizer of an element of order k in SL(n, \overline{F}), i.e.

dk ; \overline{M} := min \{dA ; \overline{M} | A ∈ SL(n, \overline{F}) , A^{k} = I.\}

Lemma 3.1 Write n = kℓ + r, 0 ≤ r < k.

i) If k = 2 and n ≡ 2 (mod 4), assume p = 2. Let B ∈ SL(n, \overline{F}) have order k, and let n_{1} ≤ … ≤ n_{s} be the degrees the non-trivial invariant factors of λI − B. Then dB ; \overline{M} = dk ; \overline{M} iff n_{i} = k for every i > 1, and dk ; \overline{M} = (2ℓ + 1)r + kℓ^{2}.

ii) If k = 2, n ≡ 2 (mod 4) and p ≠ 2, then d2 ; \overline{M} = 2(ℓ^{2} + 1).

Proof

Let A ∈ GL(n, \overline{F}) be a matrix of order k, such that dA ; \overline{M} ≤ dC ; \overline{M} for all C of order k in GL(n, \overline{F}), and denote by n_{1} ≤ … ≤ n_{s} the degrees of the non-trivial invariant factors of λI − A. Suppose that n_{i} < k for some i > 1 and let m be the maximum of such i ’s. If n_{1} > 1, there exists C ∈ GL(n, \overline{F}) of order k such that λI − C has s invariant factors ≠ 1 of degrees \overline{n}_{1} = n_{1} − 1, \overline{n}_{m} = n_{m} + 1, \overline{n}_{i} = n_{i}, i ≠ 1, m. In this case we get :

dC ; \overline{M} = ∑_{j=1}^{s}(2s − 2j + 1)\overline{n}_{j} = 2(1 − m) + ∑_{j=1}^{s}(2s − 2j + 1)n_{j} < dA ; \overline{M},

a contradiction. On the other hand, if n_{1} = 1, there exists C ∈ GL(n, \overline{F}) of order k such that λI − C has s − 1 invariant factors ≠ 1 with degrees \overline{n}_{i} = n_{i+1} for i ≠ m − 1, \overline{n}_{m−1} = n_{m} + 1. A similar calculation gives again the contradiction dC ; \overline{M} < dA ; \overline{M}. It follows n_{i} = k for all i > 1 and

n_{1} = r if r > 0, n_{1} = k if r = 0. Hence dA ; \overline{M} = (2ℓ + 1)r + kℓ^{2} ≤ dk ; \overline{M}.

i) We need to show that there exists B ∈ SL(n, \overline{F}) such that B^{k} = I and dB ; \overline{M} = dA ; \overline{M}. If k = p or r = 0 we may choose B = A. Otherwise, we write r = 2h + h_{0} (h_{0} ≤ 1) and choose B such that λI − B has non-trivial invariant factors

d_{1}(λ) = (λ ∓ 1)^{h_{0}} ∏_{j=1}^{h}(λ − α^{j})(λ − α^{−j}) , d_{i}(λ) = λ^{k} − 1 , 2 ≤ i ≤ ℓ + 1,

where α is a primitive k -th root of 1 in \overline{F} . We conclude (2ℓ + 1)r + kℓ^{2} = dk ; \overline{M} . Finally, assume dB ; \overline{M} = dk ; \overline{M} . If n_{i} < k for some i > 1, by the above argument there exists C ∈ GL(n, \overline{F}) such that dC ; \overline{M} < dB ; \overline{M} ≤ dC ; \overline{M} , a contradiction.

ii) Let B ∈ SL(n, F) be an involution such that d_{M}^{B} = d2 ; \overline{M} . It is easily seen that the non-trivial invariant factors of λI − B must have degrees n_{1} = n_{2} = 1, n_{3} = … = n_{s} = 2, and the formula can be computed directly. □

From the formulae given in Lemma 3.1, we obtain Table 1. The last two columns list the values of the defect of \overline{M}, defined by

def\overline{M} := n^{2} + 2 − (d2 ; \overline{M} + d3 ; \overline{M} + d7 ; \overline{M}).

4. The symmetric square action

Let \overline{V} = \overline{F}^{n}, with basis \{v_{i} | 1 ≤ i ≤ n\}. We denote by S the symmetric square of \overline{V} , i.e. the subspace of \overline{V} ⊗ \overline{V} generated by the set:

{v_{i} ⊗ v_{i} | 1 ≤ i ≤ n} ∪ {v_{i} ⊗ v_{j} + v_{j} ⊗ v_{i} | 1 ≤ i < j ≤ n},

and denote by E the exterior square, i.e. the subspace generated by the set:

{v_{i} ⊗ v_{j} − v_{j} ⊗ v_{i} | 1 ≤ i < j ≤ n}.

Clearly dim (S) = n(n + 1)/2 and dim (E) = n(n − 1)/2. If p ≠ 2, \overline{V} ⊗ \overline{V} = S ˙+E where each of the summands is invariant under the action g(u ⊗ v) := g(u) ⊗ g(v) for every g ∈ GL(n, \overline{F}) (diagonal action).

Lemma 4.1 Let H be an irreducible subgroup of GL(n, \overline{F}) which fixes a non-zero vector of S. Then:

i) if p ≠ 2, H is contained in an orthogonal group; ii) if p = 2, H is contained in a symplectic group;

iii) the identical representation of H is equivalent to the dual one. In particular d_{S}^{H} = d̂_{S}^{H} = 1.

Table 1: the defect of \overline{M}

n  d2 ; \overline{M}  d2 ; \overline{M}  d3 ; \overline{M}  d7 ; \overline{M}  def\overline{M}  def\overline{M}

p ≠ 2  p = 2  p ≠ 2  p = 2

4  8  8  6  4  0  0

5  13  13  9  5  0  0

6  20  18  12  6  0  2

7  25  25  17  7  2  2

8  32  32  22  10  2  2

9  41  41  27  13  2  2

10  52  50  34  16  0  2

11  61  61  41  19  2  2

12  72  72  48  22  4  4

13  85  85  57  25  4  4

14  100  98  66  28  4  6

15  113  113  75  33  6  6

16  128  128  86  38  6  6

17  145  145  97  43  6  6

18  164  162  108  48  6  8

19  181  181  121  53  8  8

20  200  200  134  58  10  10

21  221  221  147  63  12  12

22  244  242  162  70  10  12

Proof (of Lemma 4.1)

i) and ii) Let \{v_{1}, …, v_{n}\} be the standard basis of \overline{V} = \overline{F}^{n}. The map v_{i} ⊗ v_{j} ↦ v_{i} v_{j}^{t} extends to an isomorphism \overline{V} ⊗ \overline{V} → M(n, \overline{F}) of \overline{F} -vector spaces. Via this isomorphism, the elements of S are identified with the symmetric matrices. Clearly, for every g ∈ GL(n, \overline{F}) and every v_{i}, v_{j} ∈ V,

gv_{i} ⊗ gv_{j} ↦ g v_{i}v_{j}^{t} g^{t}.

Thereby, the diagonal action of GL(n, \overline{F}) on \overline{V} ⊗ \overline{V} is identified with the action of GL(n, \overline{F}) on M(n, \overline{F}) given by g ↦ γ, where γ is the linear transformation B ↦ gBg^{t}, for every B ∈ M(n, \overline{F}). As H fixes a non-zero vector of S, there exists a non-zero symmetric matrix B such that hBh^{t} = B, for every h ∈ H. Note that B\overline{V} is a non-zero H -invariant subspace of \overline{V} . Hence, as H is irreducible, B is non-degenerate. Since (h^{t})^{t}Bh^{t} = B implies (h^{−1})^{t}B^{−1}h^{−1} = B^{−1} for every h ∈ H, it follows that, if p ≠ 2, H is orthogonal. Thus i) holds. If p = 2, ii) will be proved if we show that v^{t}Bv = 0 for every v ∈ \overline{V} . To this purpose let \overline{V}_{0} := \{v ∈ \overline{V} | v^{t}Bv = 0\}. From

(w + αu)^{t}B(w + αu) = w^{t}Bw + 2αw^{t}Bu + u^{t}Bu = w^{t}Bw + u^{t}Bu for α ∈ \overline{F} ,

it follows that \overline{V}_{0} is a subspace of \overline{V} . As it is H -invariant and ≠ \{0\}, we obtain \overline{V}_{0} = \overline{V} . iii) As B^{−1}hB = (h^{−1})^{t} for every h ∈ H, the identical representation of H is equivalent to the dual one. It follows immediately that d_{S}^{H} = d̂_{S}^{H} = 1, since absolutely irreducible groups can fix only one form, up to a scalar multiple (e.g. see [KL], Lemma 2.10.3).

For a matrix A ∈ M(n, \overline{F}), define d_{S}^{A} as the dimension of the subspace of the symmetric square of \overline{V} ⊗\overline{V} fixed pointwise by A⊗A. Define d_{E}^{A} in a similar way. The following result is straightforward.

Lemma 4.2 Let g ∈ GL(n, F) be an element of finite order coprime to p. If ν is an eigenvalue of g, let m_{ν} denote the multiplicity of ν. Then:

d_{S}^{g} = \frac{m_{1}(m_{1} + 1) + m_{−1}(m_{−1} + 1)}{2} + ∑ m_{ν}m_{ν^{−1}}

where the summation runs over all pairs ν, ν^{−1} of eigenvalues of g in \overline{F} , with ν ≠ ν^{−1}. □

In the sequel, for every r ∈ ℕ we denote by π_{r} the companion matrix of λ^{r} − 1 and by J_{r} the Jordan unipotent block of size r. In particular, if r = p, π_{p} and J_{p} are conjugate, and in this case we will usually write π_{p}.

Lemma 4.3

i) Assume p = 2 and let g ∼ ( \underbrace{1, …, 1}_{k_{1}} , \underbrace{π_{2}, …, π_{2}}_{k_{2}} ) ∈ GL(n, F). Then

d_{S}^{g} = \frac{k_{1}^{2} + k_{1}}{2} + k_{2}^{2} + k_{2} + k_{1}k_{2} .

ii) Assume p = 3 and let g ∼ ( \underbrace{1, …, 1}_{k_{1}} , \underbrace{J_{2}, …, J_{2}}_{k_{2}} , \underbrace{π_{3}, …, π_{3}}_{k_{3}} ) ∈ GL(n, F). Then

d_{S}^{g} = \frac{k_{1}^{2} + k_{1}}{2} + k_{2}^{2} + \frac{3k_{3}^{2} + k_{3}}{2} + k_{1}k_{2} + k_{1}k_{3} + 2k_{2}k_{3} .

ii) Assume p = 7 and let g ∈ GL(n, F),

g ∼ ( \underbrace{1, …, 1}_{k_{1}} , \underbrace{J_{2}, …, J_{2}}_{k_{2}} , \underbrace{J_{3}, …, J_{3}}_{k_{3}} , \underbrace{J_{4}, …, J_{4}}_{k_{4}} , \underbrace{J_{5}, …, J_{5}}_{k_{5}} , \underbrace{J_{6}, …, J_{6}}_{k_{6}} , \underbrace{π_{7}, …, π_{7}}_{k_{7}} ) .

Then d_{S}^{g} =

\frac{k_{1}^{2} + k_{1}}{2} + k_{2}^{2} + \frac{3k_{3}^{2} + k_{3}}{2} + 2k_{4}^{2} + \frac{5k_{5}^{2} + k_{5}}{2} + 3k_{6}^{2} + \frac{7k_{7}^{2} + k_{7}}{2} +

k_{1} ∑_{j=2}^{7} k_{j} + 2k_{2} ∑_{j=3}^{7} k_{j} + 3k_{3} ∑_{j=4}^{7} k_{j} + 4k_{4} ∑_{j=5}^{7} k_{j} + 5k_{5} ∑_{j=6}^{7} k_{j} + 6k_{6}k_{7}.

Proof

All these formulae can be computed directly. For p odd, they can also be obtained from Lemma 1.2.3 in [GM], where a corresponding formula is given for d_{S}^{g}, g a unipotent in GL(n, F). □

5. Constraints on (2,3,7)-generating triples

Our aim, in this section, is to provide an explanation for the data collected in the Tables 2, 3, 4, 4′ and 5, which will play a crucial role in the proof of our results. For this purpose we assume that x, y, z = xy are elements of order 2, 3 and 7 respectively, which generate an absolutely irreducible subgroup of SL(n, F). It follows that these generators satisfy condition (∗) of Section 2. In particular Tables 2, 3, 4 and 4′ describe the conjugacy classes to which x, y and z are constrained to belong in order to satisfy the bounds:

d_{M}^{x} ≤ d2 ; \overline{M} + def\overline{M} , d_{M}^{y} ≤ d3 ; \overline{M} + def\overline{M} , d_{M}^{z} ≤ d7 ; \overline{M} + def\overline{M}

which follow from (∗). Note that explicit values for the right-hand sides are deducible from Table 1.

A) Constraints on similarity invariants

In this subsection, as a consequence of the above mentioned bounds, we obtain constraints on the number and the degrees of the similarity invariants of x, y, z.

Lemma 5.1 Let g ∈ GL(n, F) be an involution with s similarity invariants, and set δ := d2 ; \overline{M} + def\overline{M}. Then d_{M}^{g} ≤ δ iff

\frac{n}{2} ≤ s ≤ \frac{n + \sqrt{2δ − n^{2}}}{2} .

Proof

Clearly g has has 2s − n invariants of degree 1 and n − s of degree 2. Hence, by the Frobenius formula, d_{M}^{g} = 2s^{2} − 2ns + n^{2}. Our claim follows immediately. □

From the previous Lemma we obtain Table 2, which lists representatives of all the conjugacy classes to which x may belong, and the corresponding values of d_{S}^{x}, for every n ≤ 22.

Lemma 5.2 Let g ∈ GL(n, F) have order 3, s be the number of its similarity invariants, and denote by k_{1} be the multiplicity of the invariant of degree 1. Setting δ := d3 ; \overline{M} + def\overline{M}, d_{M}^{g} ≤ δ iff the following conditions are satisfied:

k_{1} ≤ min \{\frac{3s − n}{2} , \sqrt{\frac{6δ − 2n^{2}}{3}} \} ; \frac{n}{3} ≤ s ≤ \frac{3k_{1} + 2n + \sqrt{6δ − 2n^{2} − 3k_{1}^{2}}}{6} .

Proof

It is easy to see that g has k_{2} = 3s − 2k_{1} − n invariants of degree 2 and k_{3} = −2s + k_{1} + n of degree 3. Hence, by the Frobenius formula, d_{M}^{g} = 6s^{2} − 6sk_{1} − 4sn + 2k_{1}n + 2k_{1}^{2} + n^{2}. Our claim follows, noting that k_{3} ≤ \frac{n−k_{1}}{3} implies k_{1} ≤ \frac{3s−n}{2}. □

From Lemma 5.2 we obtain Table 3, which lists representatives of all the conjugacy classes to which y may belong, and the corresponding values of d_{S}^{y}, for every n ≤ 22. In this table we set

d_{2} := ( \begin{matrix}0 & −1 \\ 1 & −1 \end{matrix} ) . If p ≠ 3, d_{2} is conjugate to ( ω, ω^{−1} ) , where ω is a primitive cubic root of 1.

If p = 3, d_{2} is conjugate to J_{2}. If k_{ℓ} > 0 set d^{ℓ}(λ) for the similarity invariant of degree ℓ of y and note that, by the condition det y = 1 :

\{ \begin{matrix}k_{1} ≢ 0 (mod 3) \\ k_{2} ≡ 0 (mod 3) \end{matrix} ⇒ d^{1}(λ) = λ − 1 ; k_{2} ≢ 0 (mod 3) ⇒ d^{1}(λ) = λ − ω^{±1};

k_{1} ≢ 0 (mod 3) ⇒ d^{2}(λ) = (λ − 1)(λ − ω^{±1}) ; \{ \begin{matrix}k_{1} ≡ 0 (mod 3) \\ k_{2} ≢ 0 (mod 3) \end{matrix} ⇒ d^{2}(λ) = (λ − ω)(λ − ω^{−1}).

Lemma 5.3 Let g ∈ GL(n, F) be an element of order 7, with s similarity invariants of degrees

n_{1} ≤ … ≤ n_{s} ≤ 7. Set t_{1} := n_{1} − 1 , t_{i} := n_{i} − n_{i−1} (2 ≤ i ≤ s − 1) , δ := \frac{1}{2} (d7 ; \overline{M} + def\overline{M} − n) .

For the stated values of s, d_{M}^{g} ≤ d7 ; \overline{M} + def\overline{M} iff:

s = 3 \{ \begin{matrix}3t_{1} + t_{2} ≤ δ − 3 \\ 2t_{1} + t_{2} ≥ n − 9 \\ 3t_{1} + 2t_{2} ≤ n − 3 \end{matrix} s = 4 \{ \begin{matrix}6t_{1} + 3t_{2} + t_{3} ≤ δ − 6 \\ 3t_{1} + 2t_{2} + t_{3} ≥ n − 10 \\ 4t_{1} + 3t_{2} + 2t_{3} ≤ n − 4 \end{matrix}

s = 5 \{ \begin{matrix}10t_{1} + 6t_{2} + 3t_{3} + t_{4} ≤ δ − 10 \\ 4t_{1} + 3t_{2} + 2t_{3} + t_{4} ≥ n − 11 \\ 5t_{1} + 4t_{2} + 3t_{3} + 2t_{4} ≤ n − 5 \end{matrix} s = 6 \{ \begin{matrix}15t_{1} + 10t_{2} + 6t_{3} + 3t_{4} + t_{5} ≤ δ − 15 \\ 5t_{1} + 4t_{2} + 3t_{3} + 2t_{4} + t_{5} ≥ n − 12 \\ 6t_{1} + 5t_{2} + 4t_{3} + 3t_{4} + 2t_{5} ≤ n − 6. \end{matrix}

Proof

Note that n_{s} = n − ∑_{i<s} n_{i} , n ≤ 7s , n + s^{2} − s ≤ d_{M}^{g}. For each of the stated values of s, the first inequality follows from the Frobenius formula, whereas the second and the third one follow from the conditions 7 ≥ n_{s} ≥ n_{s−1}. □

Using the above Lemma (and ruling out greater values of s via similar mutually incompatible inequalities) we obtain Table 4, which lists representatives of all the conjugacy classes to which z may belong in the modular case, and the corresponding values of d_{S}^{z}, for every n ≤ 22.

B) Constraints on eigenvalue multiplicities

An element g ∈ GL(n, F) of finite order o(g) is called modular if o(g) is a power of p, semisimple if o(g) is coprime to p. In the semisimple case, g is called real if it is conjugate to g^{−1}. Now, assume that g is a semisimple element of order 7. Denote by ϵ a primitive 7-th root of 1 in \overline{F} and let m_{±i} be the multiplicity of ϵ^{±i}, 0 ≤ i ≤ 3, in the Jordan canonical form of g over \overline{F} . We note that g is real whenever m_{1} = m_{−1}, m_{2} = m_{−2}, m_{3} = m_{−3}. Furthermore, we say that g is rational when it is conjugate to all its non-trivial powers, i.e. when m_{1} = m_{2} = m_{3} = m_{−1} = m_{−2} = m_{−3}; and we say that g is semirational when it is conjugate to g^{2}, i.e when m_{1} = m_{2} = m_{−3}, m_{−1} = m_{−2} = m_{3}. Let us remark that, if g has s similarity invariants, then obviously m_{±i} ≤ s for every i, and w.l.o.g. we may assume m_{1} ≥ m_{±i}, for every i > 0. With the above notations, and assuming det g = 1 :

d_{M}^{g} = m_{0}^{2} + ∑_{i>0}(m_{i}^{2} + m_{−i}^{2}) ; m_{1} + 2m_{2} + 3m_{3} ≡ m_{−1} + 2m_{−2} + 3m_{−3} (mod 7).

Using these conditions one may easily obtain representatives of the conjugacy classes to which the generator z may belong, listed in Table 4′.

Lemma 5.4 Let g ∈ GL(n, F) have order 7. If g is real, then 2d_{S}^{g} ≥ d7 ; \overline{M} . If g is semirational, then:

\{ \begin{matrix}m_{0} + 3m_{1} + 3m_{−1} = n \\ m_{0}^{2} + 3m_{1}^{2} + 3m_{−1}^{2} = d_{M}^{g} \\ m_{0}^{2} + m_{0} + 6m_{1}m_{−1} = 2d_{S}^{g} \end{matrix} .

Proof

Assume that g is real. Then d_{M}^{g} = m_{0}^{2} + 2m_{1}^{2} + 2m_{2}^{2} + 2m_{3}^{2} and, by Lemma 4.2 , 2d_{S}^{g} =

m_{0}^{2} + m_{0} + 2m_{1}^{2} + 2m_{2}^{2} + 2m_{3}^{2}. It follows 2d_{S}^{g} − d_{M}^{g} = m_{0} ≥ 0, hence 2d_{S}^{g} ≥ d_{M}^{g} ≥ d7 ; \overline{M} . The

semirationality conditions are obvious. □

The next three lemmas are devoted to obtain Table 5

Lemma 5.5 Let h, g ∈ SL(n, q) be elements of order 3 and 7, respectively.

i) If q ≡ 2 (mod 3), then h is real; ii) if q ≡ 3, 5 (mod 7), then g is rational; iii) if q ≡ 2, 4 (mod 7), then g is semirational; iv) if q ≡ 6 (mod 7), then g is real.

Proof

i) As |F_{q}(ω) : F_{q}| = 2, the minimum polynomial of ω over F_{q} has degree 2. It follows that the similarity invariants of h can only be

λ − 1 , ∏_{k=1}^{2}(λ − ω^{k}) or ∏_{k=0}^{2}(λ − ω^{k}).

ii) As 3 and 5 have order 6 (mod 7), [F_{q}(ϵ) : F_{q}] = 6. Thus the minimum polynomial of ϵ over F_{q} has degree 6. It follows that the similarity invariants of g can only be

λ − 1 , ∏_{k=1}^{6}(λ − ϵ^{k}) or ∏_{k=0}^{6}(λ − ϵ^{k}).

iii) and iv) are proved with similar arguments. □

Remark We recall here that any companion matrix is conjugate to its transpose: in fact they have the same minimum and characteristic polynomial, which coincide. It follows immediately that any matrix in GL(n, F) is conjugate to its transpose.

Lemma 5.6 Let g be a semisimple element of a symplectic or orthogonal group over F.

i) g is real; ii) if F = F_{q} with q ≡ 2, 3, 4, 5 (mod 7) and g has order 7, then g is rational.

Proof

i) Let J be the matrix of the non-degenerate symplectic or orthogonal form defining the relevant group. From g^{t}Jg = J it follows that g is conjugate to (g^{−1})^{t}, hence to g^{−1} by the remark above. ii) q ≡ 3, 5 (mod 7) implies g rational, q ≡ 2, 4 (mod 7) implies g conjugate to g^{2}. On the other hand, by i), g is conjugate to g^{−1}. Then it is readily seen that g is conjugate to all its non-trivial powers, hence is rational.

Lemma 5.7 Let h, g ∈ SU(n, q^{2}) be elements of order 3 and 7, respectively.

i) If q ≡ 1 (mod 3), then h is real; ii) if q ≡ 2, 4 (mod 7), then g is rational; iii) if q ≡ 1 (mod 7), then g is real; iv) if q ≡ 3, 5 (mod 7), then g is semirational.

Proof

Let A ∈ SU(n, q^{2}) be semisimple. W.l.o.g. we may assume that A^{t}\overline{A} = I, where \overline{A} is obtained from A applying the field automorphism α ↦ α^{q} to all its entries. By the above remark, A is conjugate to (\overline{A}^{−1}). Moreover, considering the Jordan canonical form of A over \overline{F} , it is clear that \overline{A} is conjugate to A^{q}. Hence A is conjugate to A^{−q}.

i) h is conjugate to h^{−q} = h^{−1} by the previous observation. ii) g is semirational and conjugate to g^{−2} or to g^{−4}. It follows g rational. iii) g is conjugate to g^{−1}. iv) In this case q^{2} ≡ 2, 4 (mod 7). Therefore g is semirational. □

Table 2: admissible conjugacy classes for x

n  s  Representatives  d_{S}^{x}
4  2  ( (π_{2})^{2} )  6
5  3  ( 1, (π_{2})^{2} )  9
6  4  ( 1^{2}, (π_{2})^{2} )  13
3 , p= 2  ( (π_{2})^{3} ) p= 2  12
7  4  ( −1, (π_{2})^{3} )  16
8  4  ( (π_{2})^{4} )  20
5 , p= 2  ( 1^{2}, (π_{2})^{3} ) p= 2  21
9  5  ( 1, (π_{2})^{4} )  25
10  6  ( 1^{2}, (π_{2})^{4} )  31
5 , p= 2  ( (π_{2})^{5} ) p= 2  30
11  6  ( −1, (π_{2})^{5} )  36
12  6  ( (π_{2})^{6} )  42
7 , p= 2  ( 1^{2}, (π_{2})^{5} ) p= 2  43
13  7,8  ( 1, (π_{2})^{6} ) , ( (−1)^{3}, (π_{2})^{5} )  49, 51
14  8  ( 1^{2}, (π_{2})^{6} )  57
7 , p= 2  ( (π_{2})^{7} ) p= 2  56
15  8,9  ( −1, (π_{2})^{7} ) , ( 1^{3}, (π_{2})^{6} )  64, 66
16  8  ( (π_{2})^{8} )  72
9 , p= 2  ( 1^{2}, (π_{2})^{7} ) p= 2  73
17  9,10  ( 1, (π_{2})^{8} ) , ( (−1)^{3}, (π_{2})^{7} )  81, 83
18  10  ( 1^{2}, (π_{2})^{8} )  91
9,11 , p= 2 ( (π_{2})^{9} ) , ( 1^{4}, (π_{2})^{7} ) p=2  90,94
19  10,11  ( −1, (π_{2})^{9} ) , ( 1^{3}, (π_{2})^{8} )  100,102
20  10,12  ( (π_{2})^{10} ) , ( 1^{4}, (π_{2})^{8} )  110,114
11 , p= 2  ( 1^{2}, (π_{2})^{9} ) p=2  111
21  11,12,13  ( 1, (π_{2})^{10} ) , ( (−1)^{3}, (π_{2})^{9} ) , ( 1^{5}, (π_{2})^{8} )  121,123,127
22  12  ( 1^{2}, (π_{2})^{10} )  133
11,13, p= 2 ( (π_{2})^{11} ) , ( 1^{4}, (π_{2})^{9} ) p=2  132,136

Table 3: admissible conjugacy classes for y

n  s  Representatives  d_{S}^{y}
4  2  ( 1, π_{3} )  4
5  2  ( d_{2}, π_{3} )  5
6  2  ( (π_{3})^{2} )  7
7  3  ( 1, (π_{3})^{2} ) , ( (d_{2})^{2}, π_{3} )  10
8  3, 4  ( d_{2}, (π_{3})^{2} ) , ( (1)^{2}, (π_{3})^{2} )  12,14
9  3,4  ( (π_{3})^{3} )  15
( J_{1}, J_{2}, (π_{3})^{2} ) p = 3  16
10  4  ( 1, (π_{3})^{3} )  19
( (d_{2})^{2}, (π_{3})^{2} ) p = 2  19
11  4,5  ( d_{2}, (π_{3})^{3} ) , ( (1)^{2}, (π_{3})^{3} )  22,24
12  4,5  ( (π_{3})^{4} )  26
( J_{1}, J_{2}, (π_{3})^{3} ) p = 3  27
13  5,6  ( 1, (π_{3})^{4} ) , ( (d_{2})^{2}, (π_{3})^{3} )  31
( (ω)^{2}, (1, ω), (π_{3})^{3} ) p ≠ 3  28
( (J_{1})^{2}, J_{2}, (π_{3})^{3} ) p = 3  33
14  5,6  ( d_{2}, (π_{3})^{4} ) , ( (1)^{2}, (π_{3})^{4} )  35,37
( ω, (1, ω)^{2}, (π_{3})^{3} ) p ≠ 3  33
( J_{1}, (J_{2})^{2}, (π_{3})^{3} ) p = 3  37
15  5,6,7  ( (π_{3})^{5} ) , ( d_{2}^{3}, (π_{3})^{3} ) , ( (1)^{3}, (π_{3})^{4} )  40, 42, 44
( (ω)^{3}, (π_{3})^{4} ) , ( (1, ω)^{3}, (π_{3})^{3} ) p ≠ 3  38, 39
( J_{1}, J_{2}, (π_{3})^{4} ) p = 3  41
16  6,7  ( 1, (π_{3})^{5} ) , ( (d_{2})^{2}, (π_{3})^{4} )  46
( (ω)^{2}, (1, ω), (π_{3})^{4} ) p ≠ 3  43
( (J_{1})^{2}, J_{2}, (π_{3})^{4} ) p = 3  48
17  6,7  ( d_{2}, (π_{3})^{5} ) , ( (1)^{2}, (π_{3})^{5} )  51, 53
( ω, (1, ω)^{2}, (π_{3})^{4} ) p ≠ 3  49
( J_{1}, (J_{2})^{2}, (π_{3})^{4} ) p = 3  53
18  6,7,8  ( (π_{3})^{6} ) , ( (d_{2})^{3}, (π_{3})^{4} ) , ( (1)^{3}, (π_{3})^{5} )  57,59,61
( (ω)^{3}, (π_{3})^{5} ) , ( (1, ω)^{3}, (π_{3})^{4} ) p ≠ 3  55,56
( (J_{1}, J_{2}, (π_{3})^{5} ) p = 3  58
19  7,8  ( 1, (π_{3})^{6} ) , ( d_{2}^{2}, (π_{3})^{5} )  64
( 1, (1, ω)^{3}, (π_{3})^{4} ) , ( (ω)^{2}, (1, ω), (π_{3})^{5} ) p ≠ 3  64,61
( J_{1}, (J_{2})^{3}, (π_{3})^{4} ) , ( (J_{1})^{2}, J_{2}, (π_{3})^{5} ) p = 3  67,66

Table 3 (continuation)

n s  y  d_{S}^{y}
20  7,8,9 ( d_{2}, (π_{3})^{6} ) , ( (d_{2})^{4}, (π_{3})^{4} ) , ( (1)^{2}, (π_{3})^{6} )  70,74,72
( ω, (ω, 1)^{2}, (π_{3})^{5} ) , ( (ω)^{3}, d_{2}, (π_{3})^{5} ) p ≠ 3  68,69
( J_{1}, (J_{2})^{2}, (π_{3})^{5} ) , ( (J_{1})^{3}, J_{2}, (π_{3})^{5} ) p = 3  72,75
21  7,8,9 ( (π_{3})^{7} ) , ( (d_{2})^{3}, (π_{3})^{5} ) , ( (1)^{3}, (π_{3})^{6} )  77,79,81
( (ω)^{3}, (π_{3})^{6} ) , ( (ω, 1)^{3}, (π_{3})^{5} ) p ≠ 3  75,76
( J_{1}, J_{2}, (π_{3})^{6} ) , ( (J_{1})^{2}, (J_{2})^{2}, (π_{3})^{5} ) p = 3  78,81
22  8,9,10 ( 1, (π_{3})^{7} ) , ( (d_{2})^{2}, (π_{3})^{6} ) , ( (1)^{4}, (π_{3})^{6} )  85,91
( (ω)^{2}, (1, ω), (π_{3})^{6} ) , ( 1, (1, ω)^{3}, (π_{3})^{5} ) p ≠ 3  82,85
( (J_{1})^{2}, J_{2}, (π_{3})^{6} ) , ( J_{1}, (J_{2})^{3}, (π_{3})^{5} ) p = 3  87,88
( (ω)^{3}, (d_{2})^{2}, (π_{3})^{5} ) p = 2  85

Table 4 : admissible conjugacy classes for z modular

n  s  Representatives  d_{S}^{z}
4  1  ( J_{4} )  2
5  1  ( J_{5} )  3
6  1  ( J_{6} )  3
7  1,2  ( π_{7} ) , ( J_{1}, J_{6} )  4,5
8  2  ( J_{1}, J_{7} ) , ( J_{2}, J_{6} )  6
9  2,3  ( J_{2}, J_{7} ) , ( J_{3}, J_{6} ) , ( (J_{1})^{2}, J_{7} )  7,8,9
10  2  ( J_{3}, J_{7} )  9
11  2,3  ( J_{4}, J_{7} ) , ( J_{5}, J_{6} ) , ( J_{1}, J_{3}, J_{7} )  10,11,12
12  2,3  ( J_{5}, J_{7} ) , ( (J_{6})^{2} )  12
( J_{1}, J_{4}, J_{7} ) , ( J_{1}, J_{5}, J_{6} ) , ( J_{2}, J_{3}, J_{7} )  13,14
13 2,3 ( J_{6}, J_{7} ) , ( J_{1}, J_{5}, J_{7} ) 13,15
( J_{1}, (J_{6})^{2} ) , ( J_{2}, J_{4}, J_{7} )  15
14  2,3  ( (J_{7})^{2} ) , ( J_{1}, J_{6}, J_{7} ) , ( J_{2}, J_{5}, J_{7} )  15,16,17
15  3,4  ( J_{1}, (J_{7})^{2} ) , ( J_{2}, J_{6}, J_{7} ) , ( J_{3}, J_{5}, J_{7} ) , ( J_{3}, (J_{6})^{2} ) 18, 20
( (J_{4})^{2}, J_{7} ) , ( (J_{1})^{2}, J_{6}, J_{7} ) , ( J_{1}, J_{2}, J_{5}, J_{7} )  20, 21
16  3,4  ( J_{2}, (J_{7})^{2} ) , ( J_{3}, J_{6}, J_{7} ) , ( J_{4}, J_{5}, J_{7} ) , ( J_{4}, (J_{6})^{2} ) 20, 21, 22
( (J_{1})^{2}, (J_{7})^{2} ) , ( J_{1}, J_{2}, J_{6}, J_{7} ) , ( J_{1}, J_{3}, J_{5}, J_{7} )  22,24

Table 4 (continuation)

n s  z  d_{S}^{z}
17 3,4,5 ( J_{3}, (J_{7})^{2} ) , ( J_{4}, J_{6}, J_{7} ) , ( (J_{5})^{2}, J_{7} ) ( J_{5}, (J_{6})^{2} )  23, 25
( J_{1}, J_{2}, (J_{7})^{2} ) , ( J_{1}, J_{3}, J_{6}, J_{7} ) , ( J_{1}, J_{4}, J_{5}, J_{7} )  24,25
( (J_{2})^{2}, J_{6}, J_{7} ) , ( (J_{1})^{3}, (J_{7})^{2} )  26,25,27
18 3,4,5 ( J_{4}, (J_{7})^{2} ) , ( J_{5}, J_{6}, J_{7} ) , ( (J_{6})^{3} ) , ( J_{1}, J_{3}, (J_{7})^{2} )  25,26,27
( J_{1}, (J_{5})^{2}, J_{7} ) , ( (J_{2})^{2}, (J_{7})^{2} ) , ( J_{2}, J_{3}, J_{6}, J_{7} )  29,27,28
( J_{1}, J_{4}, J_{6}, J_{7} ) , ( (J_{1})^{2}, J_{2}, (J_{7})^{2} )  27,29
19 3,4,5 ( J_{5}, (J_{7})^{2} ) , ( (J_{6})^{2}, J_{7} ) , ( J_{1}, J_{4}, (J_{7})^{2} ) , ( J_{1}, J_{5}, J_{6}, J_{7} )  28,29,30
( J_{1}, (J_{6})^{3} ) , ( J_{2}, J_{3}, (J_{7})^{2} ) , ( J_{2}, J_{4}, J_{6}, J_{7} )  31,30
( J_{2}, (J_{5})^{2}, J_{7} ) , ( (J_{3})^{2}, J_{6}, J_{7} ) , ( (J_{1})^{2}, J_{3}, (J_{7})^{2} )  32
( (J_{1})^{2}, J_{4}, J_{6}, J_{7} ) , ( J_{1}, (J_{2})^{2}, (J_{7})^{2} )  32
20 3,4,5 ( J_{6}, (J_{7})^{2} ) , ( J_{1}, (J_{6})^{2}, J_{7} ) , ( J_{1}, J_{5}, (J_{7})^{2} ) , ( J_{2}, J_{4}, (J_{7})^{2} ) 30,32
( J_{2}, (J_{6})^{3} ) , ( (J_{3})^{2}, (J_{7})^{2} ) , ( J_{3}, J_{4}, J_{6}, J_{7} )  34
( J_{2}, J_{5}, J_{6}, J_{7} ) , ( J_{3}, (J_{5})^{2}, J_{7} ) , ( (J_{1})^{2}, J_{4}, (J_{7})^{2} )  33,36,34
( (J_{1})^{2}, J_{5}, J_{6}, J_{7} ) , ( J_{1}, J_{2}, J_{3}, (J_{7})^{2} ) , ( J_{1}, J_{2}, J_{4}, J_{6}, J_{7} )  35
21 3,4,5,6 ( (J_{7})^{3} ) , ( J_{1}, J_{6}, (J_{7})^{2} ) , ( J_{2}, J_{5}, (J_{7})^{2} ) , ( J_{2}, (J_{6})^{2}, J_{7} )  33,34,35
( J_{3}, J_{4}, (J_{7})^{2} ) , ( J_{3}, J_{5}, J_{6}, J_{7} ) , ( (J_{4})^{2}, J_{6}, J_{7} ) , ( J_{3}, (J_{6})^{3} ) 36,37,38
( J_{4}, (J_{5})^{2}, J_{7} ) , ( (J_{1})^{2}, J_{5}, (J_{7})^{2} ) , ( (J_{1})^{2}, (J_{6})^{2}, J_{7} )  39,37
( J_{1}, J_{2}, J_{4}, (J_{7})^{2} ) , ( J_{1}, J_{2}, J_{5}, J_{6}, J_{7} ) , ( J_{1}, (J_{3})^{2}, (J_{7})^{2} )  37,38,39
( J_{1}, J_{3}, J_{4}, J_{6}, J_{7} ) , ( (J_{2})^{2}, J_{3}, (J_{7})^{2} ) , ( (J_{1})^{3}, J_{4}, (J_{7})^{2} )  39,40
22 4,5,6 ( J_{1}, (J_{7})^{3} ) , ( J_{2}, J_{6}, (J_{7})^{2} ) , ( J_{3}, J_{5}, (J_{7})^{2} ) , ( J_{3}, (J_{6})^{2}, J_{7} ) 37,39
( (J_{4})^{2}, (J_{7})^{2} ) , ( (J_{1})^{2}, J_{6}, (J_{7})^{2} ) , ( J_{4}, J_{5}, J_{6}, J_{7} )  39,40
( J_{1}, J_{2}, J_{5}, (J_{7})^{2} ) , ( J_{1}, J_{2}, (J_{6})^{2}, J_{7} ) , ( J_{1}, J_{3}, J_{4}, (J_{7})^{2} ) ,  40,41
( J_{1}, J_{3}, J_{5}, J_{6}, J_{7} ) , ( (J_{2})^{2}, J_{4}, (J_{7})^{2} ) , ( (J_{1})^{3}, J_{5}, (J_{7})^{2} )  42,41,43

In Table 4′ below we set d_{6} := ( ϵ, ϵ^{2}, ϵ^{3}, ϵ^{−3}, ϵ^{−2}, ϵ^{−1} ) .

Table 4′ : admissible conjugacy classes for z semisimple

( p ≠ 2 if n = 14 )

n s Representatives  d_{S}^{z}
4  1 ( ϵ, ϵ^{3}, ϵ^{−3}, ϵ^{−1} ) , ( ϵ, ϵ^{2}, ϵ^{−2}, ϵ^{−1} ) , ( 1, ϵ, ϵ^{2}, ϵ^{−3} )  2,1
5  1 ( 1, ϵ, ϵ^{3}, ϵ^{−3}, ϵ^{−1} ) , ( 1, ϵ, ϵ^{2}, ϵ^{−2}, ϵ^{−1} )  3
6  1 d_{6}  3
2 ( ϵ, 1, ϵ, ϵ^{3}, ϵ^{−3}, ϵ^{−2} ) , ( ϵ, 1, ϵ, ϵ^{2}, ϵ^{−3}, ϵ^{−1} ) p=2  2,3
2 ( 1, 1, ϵ, ϵ^{3}, ϵ^{−3}, ϵ^{−1} ) , ( 1, 1, ϵ, ϵ^{2}, ϵ^{−2}, ϵ^{−1} ) p=2  5
7  1 ( π_{7} )  4
10 2 ( ϵ, ϵ^{2}, ϵ^{−3}, π_{7} ) , ( 1, ϵ, ϵ^{−1}, π_{7} )  7,9
2 ( ϵ, ϵ^{2}, ϵ^{−2}, ϵ^{−1}, d_{6} ) , ( ϵ, ϵ^{3}, ϵ^{−3}, ϵ^{−1}, d_{6} )  9
2 ( ϵ, ϵ^{−3}, ϵ^{−2}, ϵ^{−1}, 1, ϵ, ϵ^{3}, ϵ^{−3}, ϵ^{−2}, ϵ^{−1} ) , ( ϵ, ϵ^{3}, ϵ^{−3}, ϵ^{−2}, 1, ϵ, ϵ^{2}, ϵ^{3}, ϵ^{−3}, ϵ^{−2} ) 7
2 ( ϵ, ϵ^{2}, ϵ^{3}, ϵ^{−2}, 1, ϵ, ϵ^{2}, ϵ^{3}, ϵ^{−2}, ϵ^{−1} ) , ( ϵ, ϵ^{2}, ϵ^{3}, ϵ^{−1}, 1, ϵ, ϵ^{2}, ϵ^{3}, ϵ^{−3}, ϵ^{−1} )  7
3,2 ( ϵ, ϵ, ϵ^{−2}, π_{7} ) , ( 1, ϵ, ϵ^{3}, ϵ^{−2}, 1, ϵ, ϵ^{3}, ϵ^{−3}, ϵ^{−2}, ϵ^{−1} )  7
2 ( 1, ϵ, ϵ^{−3}, ϵ^{−2}, 1, ϵ, ϵ^{2}, ϵ^{−3}, ϵ^{−2}, ϵ^{−1} ) , ( 1, ϵ, ϵ^{2}, ϵ^{3}, 1, ϵ, ϵ^{2}, ϵ^{3}, ϵ^{−3}, ϵ^{−2} )  7
13 2 ( d_{6}, d_{7} )  13
3 ( ϵ, ϵ^{−3}, ϵ, ϵ^{−3}, ϵ^{−2}, ϵ^{−1}, π_{7} ) , ( ϵ, ϵ^{−2}, ϵ, ϵ^{−2}, ϵ^{3}, ϵ^{−1}, π_{7} )  12
3 ( ϵ, ϵ^{−1}, ϵ, ϵ^{−1}, ϵ^{3}, ϵ^{−3}, π_{7} ) , ( ϵ, ϵ^{−1}, ϵ, ϵ^{−1}, ϵ^{2}, ϵ^{−2}, π_{7} )  15
3 ( ϵ, ϵ^{3}, ϵ, ϵ^{3}, ϵ^{2}, ϵ^{−3}, π_{7} ) , ( ϵ, ϵ^{2}, ϵ, ϵ^{2}, ϵ^{3}, ϵ^{−2}, π_{7} )  12
3 ( ϵ, 1, ϵ, ϵ^{3}, ϵ^{−3}, ϵ^{−2}, π_{7} ) , ( ϵ, ϵ^{3}, 1, ϵ, ϵ^{3}, ϵ^{−1}, π_{7} )  12,13
3 ( ϵ, ϵ^{−2}, 1, ϵ, ϵ^{2}, ϵ^{−2}, π_{7} ) , ( ϵ, 1, ϵ, ϵ^{2}, ϵ^{−3}, ϵ^{−1}, π_{7} )  13
3 ( ϵ, 1, ϵ, ϵ^{2}, ϵ^{3}, ϵ^{−2}, ϵ^{−1}, 1, ϵ, ϵ^{2}, ϵ^{3}, ϵ^{−2}, ϵ^{−1} ) , ( 1, ϵ, 1, ϵ, ϵ^{2}, ϵ^{3}, π_{7} )  13
3 ( 1, 1, ϵ, ϵ^{3}, ϵ^{−3}, ϵ^{−1}, π_{7} ) , ( 1, 1, ϵ, ϵ^{2}, ϵ^{−2}, ϵ^{−1}, π_{7} )  15
14 3 ( ϵ, ϵ^{−2}, ϵ, ϵ^{2}, ϵ^{3}, ϵ^{−3}, ϵ^{−2}, π_{7} ) , ( ϵ, ϵ^{3}, ϵ, ϵ^{2}, ϵ^{3}, ϵ^{−2}, ϵ^{−1}, π_{7} )  14
3 ( ϵ, ϵ^{−3}, ϵ, ϵ^{2}, ϵ^{3}, ϵ^{−3}, ϵ^{−1}, π_{7} ) , ( ϵ, ϵ^{2}, ϵ, ϵ^{2}, ϵ^{−3}, ϵ^{−2}, ϵ^{−1}, π_{7} )  15
3 ( (π_{7})^{2} ) , ( ϵ, ϵ^{−3}, 1, ϵ, ϵ^{−3}, ϵ^{−2}, ϵ^{−1}, π_{7} )  15,14
3 ( ϵ, ϵ^{−2}, 1, ϵ, ϵ^{3}, ϵ^{−2}, ϵ^{−1}, π_{7} )  14
3 ( ϵ, ϵ^{−1}, 1, ϵ, ϵ^{3}, ϵ^{−3}, ϵ^{−1}, π_{7} ) , ( ϵ, ϵ^{−1}, 1, ϵ, ϵ^{2}, ϵ^{−2}, ϵ^{−1}, π_{7} )  17
3 ( ϵ, ϵ^{3}, 1, ϵ, ϵ^{2}, ϵ^{3}, ϵ^{−3}, π_{7} ) , ( ϵ, ϵ^{2}, 1, ϵ, ϵ^{2}, ϵ^{3}, ϵ^{−2}, π_{7} )  14
3 ( 1, ϵ, 1, ϵ, ϵ^{3}, ϵ^{−3}, ϵ^{−2}, π_{7} ) , ( 1, ϵ, 1, ϵ, ϵ^{2}, ϵ^{−3}, ϵ^{−1}, π_{7} )  15,16

Table 5 (Lemmas 5.5, 5.6, 5.7)

q  SL(n, q)  Sp(n, q)  SU(n, q^{2})
≡ 1 (21)  y real, z real  y real, z real
≡ 2, −10 (21) y real, z semirat.  y real, z rational  z rational
≡ 4, −5 (21)  z semirat.  y real, z rational  y real, z rational
≡ 5, −4 (21) y real, z rational  y real, z rational  z semirat.
≡ 8 (21)  y real  y real, z real  z real
≡ 10, −2 (21)  z rational  y real, z rational  y real, z semirat.
≡ −8 (21)  z real  y real, z real  y real
≡ −1 (21)  y real, z real  y real, z real
3^{6h}  y modular  y modular, z real  y modular, z real
3^{6h+1}  y modular, z rational  y modular, z rational y modular, z semirat.
3^{6h+2}, 3^{6h+4} y modular, z semirat.  y modular, z rational y modular, z rational
3^{6h+3}  y modular, z real  y modular, z real  y modular
3^{6h+5}  y modular, z rational  y modular, z rational y modular, z semirat.
7^{t}  z modular  y real, z modular  y real, z modular

6. Proof of Theorem 1

We start with an easy but useful remark. Namely, let k ≥ 7 be a prime, G be a (2, 3, k) -generated group, and f : G → SL(n, F) be a representation of degree n ≤ 2. If p ≠ 2, the only involution of SL(2, F) is −I, and therefore f(G) is abelian. Hence f(G) = {1}. This readily implies that a (2, 3, k) -generated subgroup G of SL(3, F) or SL(4, F) which does not fix any non-zero vector is absolutely irreducible. For otherwise, by the above remark, G necessarily admits the eigenvalue 1.

i) Let \overline{H} ≤ SL(2, \overline{F}) be the preimage of a (2, 3, k) -generated subgroup of PSL(2, \overline{F}). Clearly we may assume \overline{H} = ⟨\overline{x}, z⟩ where \overline{x}^{2} = −I, z^{k} = I. Moreover \overline{H} is irreducible, since it is not soluble. Therefore, if u is an eigenvector of z, u and \overline{x}u are linearly independent. It follows that, with respect to the basis \{u, \overline{x}u\},

\overline{x} = ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) , z = z_{i} = ( \begin{matrix}ϵ^{i} & a \\ 0 & ϵ^{−i} \end{matrix} ) 1 ≤ i ≤ k − 1

for a suitable a ∈ \overline{F} . Thus y_{i} := \overline{x}z_{i} has trace a. On the other hand, y_{i} must be conjugate to

± ( \begin{matrix}0 & −1 \\ 1 & −1 \end{matrix} ) , which gives a = ±1. Both choices for z_{i} give rise to conjugate subgroups. This

is clear if p = k, ϵ = 1. On the other hand, if p ≠ k, let t := ( \begin{matrix}1 & ϵ^{−i} − ϵ^{i} \\ −ϵ^{−i} + ϵ^{i} & 1 \end{matrix} ) . t is

non-singular, centralizes \overline{x} and

t^{−1} ( \begin{matrix}ϵ^{i} & −1 \\ 0 & ϵ^{−i} \end{matrix} ) t = ( \begin{matrix}ϵ^{i} & 1 \\ 0 & ϵ^{−i} \end{matrix} )^{−1} .

Hence we may assume a = 1. Let \overline{H}_{i} := ⟨\overline{x}, z_{i}⟩. Note that, for every i, (\overline{x}z_{i})^{3} = −I and \overline{H}_{i} ≤ SL(2, R), where R is isomorphic to ℚ(ϵ) if p = 0, R = F_{p^{k−1}} if p > 0. If p = 0, we may consider the automorphism γ_{i} of GL(n, R) induced by the automorphism of R sending ϵ to ϵ^{i}, 1 ≤ i ≤ k − 1. Since \overline{H}_{i} = γ_{i}(\overline{H}_{1}), the projective image of \overline{H}_{1} is, up to isomorphism, the only (2, 3, k) -generated subgroup of PSL(2, \overline{F}), and therefore (cfr. [Mag]) it is isomorphic to Δ(2, 3, k). Now assume p > 0 and k = 7, hence \overline{H}_{i} ≤ SL(2, F_{p^{6}}). As \overline{H}_{i} is perfect, it follows from the knowledge of the subgroups of PSL(2, p^{6}) that the projective image H_{i} of \overline{H}_{i} is isomorphic to PSL(2, p^{m_{i}}) for some m_{i} ∈ {1, 2, 3, 6}. The results of [Mac] give H_{i} ≃ PSL(2, p) if p ≡ 0, ±1 (mod 7) and H_{i} ≃ PSL(2, p^{3}) if p ≢ 0, ±1 (mod 7). ii) A direct computation shows that the group

φ(H) = ⟨( \begin{matrix}0 & 0 & 1 \\ 0 & −1 & 0 \\ 1 & 0 & 0 \end{matrix} ) , ( \begin{matrix}ϵ^{2} & 2ϵ & 1 \\ 0 & 1 & ϵ^{−1} \\ 0 & 0 & ϵ^{−2} \end{matrix} )⟩

preserves the bilinear symmetric form defined by the matrix antidiag(1, −2, 1); moreover, if p ≠ 2, it does not fix any non-zero vector. Hence φ(H) is an irreducible subgroup of SO(3, \overline{F}), for all k ≥ 7. From now on assume k = 7, and let K = ⟨x, z⟩ be an irreducible subgroup of SL(3, \overline{F}), with x^{2} = z^{7} = (xz)^{3} = 1. Clearly x has a 2-dimensional eigenspace. Therefore, by the irreducibility of K, z cannot have a 2-dimensional eigenspace. In particular, if p ≠ 7, z is either conjugate to z_{1} = diag(η, η^{2}, η^{4}) or to z_{2} = diag(η, 1, η^{−1}), where η = ϵ^{i} for some i with 1 ≤ i ≤ 6. Notice that z leaves invariant a 2-dimensional subspace ⟨u, v⟩. If p ≠ 7, we may assume zu = ηu, zv = η^{2}v if z ∼ z_{1} and zu = ηu, zv = v if z ∼ z_{2}. On the other hand, if p = 7, we may assume zu = u. Since K is irreducible, u, v, xu are linearly independent. Moreover, as det x = 1, xv = au − v + axu for some a ∈ \overline{F} . It follows that, with respect to the basis {u, v − au, xu}

x = ( \begin{matrix}0 & 0 & 1 \\ 0 & −1 & 0 \\ 1 & 0 & 0 \end{matrix} ) , and z = z_{1} = ( \begin{matrix}η & b & c \\ 0 & η^{2} & d \\ 0 & 0 & η^{4} \end{matrix} ) or z = z_{2} = ( \begin{matrix}η & b & c \\ 0 & 1 & d \\ 0 & 0 & η^{−1} \end{matrix} )

where b, c, d are suitable elements of \overline{F} (and η = 1 if p = 7 ). Note that b ≠ 0, again by the irreducibility of K. Thus, conjugating x and z_{i} by diag (1, b^{−1}, 1), we may also assume b = 1.

1) Let K_{1} = ⟨x, z_{1}⟩. As y = xz_{1} must have trace 0, it follows c = η^{2}. Moreover, equating the elements in position (1,1) of y^{2} and y^{−1}, one gets d = η^{4} + η^{5}. This gives

z_{1} = ( \begin{matrix}η & 1 & η^{2} \\ 0 & η^{2} & η^{4} + η^{5} \\ 0 & 0 & η^{4} \end{matrix} ) y = ( \begin{matrix}0 & 0 & η^{4} \\ 0 & −η^{2} & −η^{4} − η^{5} \\ η & 1 & η^{2} \end{matrix} )

and a direct calculation shows that z_{1}^{7} = y^{3} = [x, y]^{4} = 1. But the relations x^{2} = y^{3} = (xy)^{7} = [x, y]^{4} = 1 correspond to a well-known presentation for PSL(2, 7) (see [CM], 7.5, p.96). We conclude K_{1} ≃ PSL(2, 7). Note that K_{1} is irreducible, as it does not fix any non-zero vector.

2) Now let K_{2} := ⟨x, z_{2}⟩, Again y = xz_{2} has trace 0, hence c = 1. Moreover, equating the elements in position (1,1) of y^{2} and y^{−1}, one gets d = 2. This gives

x = ( \begin{matrix}0 & 0 & 1 \\ 0 & −1 & 0 \\ 1 & 0 & 0 \end{matrix} ) , z_{2} = ( \begin{matrix}η & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & η^{−1} \end{matrix} ) .

If p = 2, K_{2} is reducible, whereas if p ≠ 2 K_{2} is irreducible, as it does not fix any non-zero vector. In the latter case, conjugating x and z_{2} by diag (1, 2η^{4}, 1), one gets the matrices

( \begin{matrix}0 & 0 & 1 \\ 0 & −1 & 0 \\ 1 & 0 & 0 \end{matrix} ) = φ( ± ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) ) , ( \begin{matrix}η & 2η^{4} & 1 \\ 0 & 1 & η^{3} \\ 0 & 0 & η^{6} \end{matrix} ) = φ( ± ( \begin{matrix}η^{4} & 1 \\ 0 & η^{−4} \end{matrix} ) )

where φ is the embedding given in the statement. It follows from i) that the projective image of

⟨( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) , ( \begin{matrix}η^{4} & 1 \\ 0 & η^{−4} \end{matrix} )⟩

is isomorphic to H, in other words K_{2} ≃ φ(H).

Finally, let ⟨x, z⟩ ≤ SL(3, \overline{F}) be the preimage of a (2, 3, 7) -generated subgroup of PSL(3, \overline{F}). We may assume p ≠ 3, so that the centre of SL(3, \overline{F}) has order 3 and is generated by the scalar matrix β^{3} I, where β is a primitive 9^{th} root of 1. We may also assume that x and z have respective orders 2 and 7. In particular, x is conjugate to diag(−1, −1, 1) and has an eigenspace of dimension 2. Assume, by way of contradiction, that y = x z has order 9, hence y^{3} = β^{3} I. Since y must have pairwise distinct eigenvalues by the irreducibility of ⟨x, z⟩, y should be conjugate to diag(β, β^{4}, β^{7}), which has determinant ≠ 1, a contradiction. It follows that y has order 3, hence ⟨x, z⟩ is conjugate either to K_{1} or to K_{2}.

iii) Under our assumptions we may consider the automorphism γ_{i} of GL(n, R) induced by ϵ ↦ ϵ^{i}, for some i ≢ 0, ±1 (mod k). The image of H under the map g ↦ g ⊗ γ_{i}(g) is the group

⟨( \begin{matrix}0 & 0 & 0 & 1 \\ 0 & 0 & −1 & 0 \\ 0 & −1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{matrix} ) , ( \begin{matrix}ϵ^{i+1} & ϵ^{i} & ϵ & 1 \\ 0 & ϵ^{i−1} & 0 & ϵ^{−1} \\ 0 & 0 & ϵ^{−i+1} & ϵ^{−i} \\ 0 & 0 & 0 & ϵ^{−i−1} \end{matrix} )⟩.

As 1 is not an eigenvalue of the second generator, it follows, from the remark at the beginning of the proof, that this group is irreducible unless possibly when p = 2 and it leaves invariant a 2dimensional subspace. But this possibility is ruled out, using a direct argument (omitted here, but entirely similar to the one given below, see iv). Note that the above group preserves the symmetric bilinear form defined by the matrix J ⊗ J, where J = antidiag(−1, 1), hence it is orthogonal if p ≠ 2. iv) The group \overline{H} = ⟨( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) , ( \begin{matrix}ϵ & 1 \\ 0 & ϵ^{−1} \end{matrix} )⟩ considered in i) acts on the third symmetric power

of V := \overline{F}^{2}, that is on the space of homogeneous polynomials of degree 3 in two variables t, u. With respect to the basis \{t^{3}, t^{2}u, tu^{2}, u^{3}\}, \overline{H} acts faithfully as

⟨( \begin{matrix}0 & 0 & 0 & −1 \\ 0 & 0 & 1 & 0 \\ 0 & −1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{matrix} ) , ( \begin{matrix}ϵ^{3} & ϵ^{2} & ϵ & 1 \\ 0 & ϵ & 2 & 3ϵ^{−1} \\ 0 & 0 & ϵ^{−1} & 3ϵ^{−2} \\ 0 & 0 & 0 & ϵ^{−3} \end{matrix} )⟩.

\overline{H} preserves the symplectic form defined by the matrix antidiag(−3, 1, −1, 3), and the projective image of \overline{H} is isomorphic to H. If p = 3, this form is degenerate, thus \overline{H} is reducible. Let p ≠ 3. As \overline{H} is perfect and 1 is not a common eigenvalue of \overline{x} and z, it follows easily that \overline{H} cannot leave invariant a subspace of dimension 1 or 3. Assume that \overline{H} leaves invariant a 2-dimensional subspace U = ⟨u_{1}, u_{2}⟩. If p ≠ k, z_{|U} is diagonalizable and must have determinant 1. Thus we may choose as u_{1} an eigenvector of z with corresponding eigenvalue ϵ^{3} or ϵ, and set u_{2} = \overline{x} u_{1}. A direct calculation of the eigenvectors of z gives the two possibilities:

U = ⟨e_{1}, e_{4}⟩  or  U = ⟨ \frac{ϵ}{1 − ϵ^{2}} e_{1} + e_{2} , −e_{3} + \frac{ϵ}{1 − ϵ^{2}} e_{4} ⟩ ,

where \{e_{1}, e_{2}, e_{3}, e_{4}\} is the standard basis of \overline{F}^{4}. In both cases U is not z -invariant. Finally let p = k, hence ϵ = 1. Clearly U contains the unique eigenspace of z, generated by e_{1}. Since \overline{x} e_{1} = e_{4}, we obtain U = ⟨e_{1}, e_{4}⟩. But, as noted above, this subspace is not z -invariant. We conclude that \overline{H} is irreducible. □

7. Case by case analysis

Our aim in this section is to prove Theorems 2,3,4 and 5. In the statements of all these theorems, H is an absolutely irreducible subgroup of SL(n, F). Furthermore, if p ≠ 2, H is not contained in an orthogonal group and, if p = 2, H is not contained in a symplectic group. Hence, by Lemma 4.1, d_{S}^{H} = 0. Now, suppose that H is generated by x, y, z = xy of order 2, 3 and 7 respectively. By applying Scott’ s formula to the adjoint representation of H on M = Mat(n, F) and to the representation of H on the symmetric square S, we obtain the necessary conditions:

(∗) d_{M}^{x} + d_{M}^{y} + d_{M}^{z} ≤ n^{2} + 2  and  (∗∗) d_{S}^{z} ≤ dim S − (d_{S}^{x} + d_{S}^{y}).

We conclude that H is not (2, 3, 7) -generated whenever the assumption that x, y, z satisfy both (∗) (or a refined version of it, cfr. Lemma 7.1 below) and (∗∗) leads to a contradiction.

We note that x, y, z satisfy (∗) precisely when x, y, z belong to the ‘admissible ’ conjugacy classes obtained in Section 5. These classes are listed in Tables 2 and 3 for x and y, and in Table 4 for z modular. For certain values of n they are also listed in Table 4′ for z semisimple. The tables also give the corresponding values of d_{S}^{x}, d_{S}^{y}, d_{S}^{z}.

Proof of Theorem 2

Let H be as stated in the theorem, and assume that it is (2,3,7)-generated. The assumption that x, y, z satisfy (∗) and (∗∗) leads immediately to a contradiction, except in the case n = 6, p = 2, which requires a little more work.

\underline{n = 4} . d_{S}^{x} = 6, d_{S}^{y} = 4 and dim S = 10 imply d_{S}^{z} ≤ 0. A contradiction, as d_{S}^{z} ≥ 1.

\underline{n = 5} . d_{S}^{x} = 9, d_{S}^{y} = 5 and dim S = 15 imply d_{S}^{z} ≤ 1. A contradiction, as d_{S}^{z} = 3.

\underline{n = 6} . Assume first d_{S}^{x} = 13. Then d_{S}^{y} = 7 and dim S = 21 imply d_{S}^{z} ≤ 1. A contradiction, as d_{S}^{z} ≥ 2. We are left with the case d_{S}^{x} = 12, x conjugate to ( (π_{2} )^{3}), p = 2. Now d_{S}^{y} = 7 implies d_{S}^{z} ≤ 2, z conjugate to ( ϵ, ϵ, 1, ϵ^{3}, ϵ^{−3}, ϵ^{−2} ) . The faithful action of H on the exterior square E, of dimension 15, gives rise to an embedding i : H → GL(15, \overline{F}). Recall that in the present case we are assuming H = SL(6, q) or SU(6, q^{2}). It is well-known (e.g. cfr. [Lie]) that i(H) is irreducible. It is also easy to see that i(x), i(y) and i(z) are respectively conjugate to the matrices

( 1^{3}, (π_{2})^{6} ) , ( (π_{3})^{5} ) , ( (1), (ϵ)^{3}, (ϵ^{2})^{2}, (ϵ^{3}), (ϵ^{4})^{3}, (ϵ^{5})^{3}, (ϵ^{6})^{2} ) .

Hence, by applying Scott’ s formula to the action of i(H), by conjugation, on the space M_{E} = Mat(15, \overline{F}), we get the contradiction:

d_{M_{E}}^{i(x)} + d_{M_{E}}^{i(y)} + d_{M_{E}}^{i(z)} = 117 + 75 + 37 = 229 > 227 = dim(M_{E}) + 2.

\underline{n = 7} . d_{S}^{x} = 16, d_{S}^{y} = 10 and dim S = 28 imply d_{S}^{z} ≤ 2. A contradiction, as d_{S}^{z} ≥ 4. \underline{n = 10} . d_{S}^{x} ≥ 30, d_{S}^{y} = 19 and dim S = 55 imply d_{S}^{z} ≤ 6. A contradiction, as d_{S}^{z} ≥ 7. □

The following lemma gives a refined version of (∗), which will be needed in order to prove Theorem 3.

Lemma 7.1 Let H be one of the groups SL(n, q), Sp(n, q) or SU(n, q). Assume either

a) q ≡ 1 (mod 7) and q ∉ {p, p^{3}, 3^{6}}; or

b) q ≡ 3, 5 (mod 7) and q ≠ p . If (x, y, z) is a (2, 3, 7) -generating triple for H and y is real or modular, then:

d_{M}^{x} ≤ d2 ; \overline{M} + def\overline{M} − 2 , d_{M}^{y} ≤ d3 ; \overline{M} + def\overline{M} − 2 , d_{M}^{z} ≤ d7 ; \overline{M} + def\overline{M} − 2.

Proof

a) Let d be the multiplicative order of p (mod 7). Then q = p^{dk}, k ∈ ℕ. Assume first k > 1 and let σ be the automorphism of F_{q} defined by α ↦ α^{p^{d}}. Then σ is non-trivial, as the subfield of the elements fixed by σ has order p^{d} < q. Consider the representation γ_{1} : GL(n, q) → GL(n^{2}, q) such that, for every g ∈ GL(n, q), γ_{1}(g) is the automorphism of Mat(n, q) defined by A ↦ g^{σ}Ag^{−1}. By a well-known theorem of Steinberg, γ_{1}(H) and its dual are irreducible (cfr. [St]). Note that γ_{1} is the representation afforded by the tensor product module V ⊗ V^{∗σ}, where V is the natural module for GL(n, q) and V^{∗σ} denotes the dual of V ‘twisted’ by σ). Scott’ s formula then gives:

d_{M_{1}}^{x} + d_{M_{1}}^{y} + d_{M_{1}}^{z} ≤ n^{2},

where, as usual, d_{M_{1}}^{g} denotes the dimension of the space of fixed points of γ_{1}(g). Under our assumptions, y is conjugate to a matrix over F_{p}, while z is conjugate to a matrix over F_{p^{d}}. As x is clearly conjugate to a matrix over F_{p}, it follows that d_{M_{1}}^{x} = d_{M}^{x}, d_{M_{1}}^{y} = d_{M}^{y}, d_{M_{1}}^{z} = d_{M}^{z} and we are done. Now assume k = 1, i.e. q = p^{d}. From q ∉ {p, p^{3}} it follows d ∈ {2, 6}. In particular p ≠ 2, as 2 has order 3 (mod 7), and p ≠ 3 as we exclude the case q = 3^{6}. Let σ be the automorphism of order 2 of F_{q}, and consider the representation γ_{2} : GL(n, q) → GL(n^{2}, q) such that, for every g ∈ GL(n, q), γ_{2}(g) is the automorphism of Mat(n, q) defined by A ↦ g^{σ}Ag^{t}. Again γ_{2}(H) and its dual are irreducible (cfr. [St]; γ_{2} is the representation afforded by the module V ⊗ V^{σ} ). Hence, with obvious notation, d_{M_{2}}^{x} + d_{M_{2}}^{y} + d_{M_{2}}^{z} ≤ n^{2}. As σ(ϵ) = ϵ^{−1}, σ(ω) = ω^{±1} and x, y, z are diagonalizable, it follows that x^{σ}, y^{σ}, z^{σ} are respectively conjugate to x, y^{±1}, z^{−1}. Our claim follows immediately.

b) In this case q = p^{t}, t > 1. Note that, as q ≡ 3, 5 (mod 7), by Lemma 5.5 z is rational, hence is conjugate to a matrix over F_{p} . Now let σ be the Frobenius automorphism α ↦ α^{p}, and apply the same argument used above for the case k > 1. □

Proof of Theorem 3

Let H be as stated in the theorem, and suppose that it is (2,3,7)-generated. Our strategy is as follows. Using Tables 2,3 and 4 and Lemma 5.4 we show that z cannot be modular, real or semirational. It follows from Lemmas 5.5, 5.6 and 5.7 that H can only be either i) SL(n, q) with q ≡ 1 (mod 7), or ii) SU(n, q) with \sqrt{q} ≡ −1 (mod 7). This proves our statement when H is a symplectic group and when H has degree n = 12. Thus we are left with the cases H = SL(n, q) or SU(n, q), q ≡ 1 (mod 7), n ∈ {8, 9, 11}. We assume, by way of contradiction, q ∉ {p, p^{3}, 3^{6}}. Since y is real or modular (cfr. Table 3) and def\overline{M} = 2, it follows from Lemma 7.1 that z satisfies the condition d_{M}^{z} = d7 ; \overline{M}. Hence, by Lemma 3.1, z has two similarity invariants of respective degrees n − 7 and 7. We reach a contradiction by showing that z does not satisfy the condition (∗∗), i.e. d_{S}^{z} ≤ dim S − (d_{S}^{x} + d_{S}^{y}).

\underline{n = 8} . d_{S}^{x} ≥ 20, d_{S}^{y} ≥ 12 and dim S = 36 imply d_{S}^{z} ≤ 4. z cannot be modular since, for p = 7, d_{S}^{z} = 6. By Lemma 5.4 z cannot be real, since 2d_{S}^{z} < d7 ; \overline{M} = 10. Assume z semirational. Then d_{M}^{z} ≤ 12 implies, by the same Lemma, m_{0} = 2, m_{1} = m_{−1} = 1. But this in turn implies z real, a contradiction. Finally, in cases i) and ii), d_{M}^{z} = d7 ; \overline{M} implies z ∼ ( 1, π_{7} ) , hence d_{S}^{z} = 6, a contradiction.

\underline{n = 9}. d_{S}^{x} = 25, d_{S}^{y} ≥ 15 and dim S = 45 imply d_{S}^{z} ≤ 5. z cannot be modular since, for p = 7, d_{S}^{z} ≥ 7. z cannot be real, as 2d_{S}^{z} < d7 ; \overline{M} = 13. Assume z semirational, and non-real. d_{M}^{z} ≤ d7 ; \overline{M} + def\overline{M} = 15 implies m_{0} = 0, m_{1} = 2, m_{−1} = 1. But, in this case, d_{S}^{z} = 6, against the bound 2d_{S}^{z} ≤ 10. Finally, in cases i) and ii), d_{M}^{z} = d7 ; \overline{M} implies z ∼ ( ϵ, ϵ^{−1}, π_{7} ) , hence d_{S}^{z} = 7, a contradiction.

\underline{n = 11}. d_{S}^{x} = 36, d_{S}^{y} ≥ 22 and dim S = 66 imply d_{S}^{z} ≤ 8. z cannot be modular since, for p = 7, d_{S}^{z} ≥ 10. z cannot be real, as 2d_{S}^{z} < d7 ; \overline{M} = 19. Assume z semirational. d_{M}^{z} ≤ d7 ; \overline{M} + def\overline{M} = 21 implies m_{0} = 2, m_{1} = 2, m_{−1} = 1. But, in this case, d_{S}^{z} = 9, against the bound 2d_{S}^{z} ≤ 16. Finally, in cases i) and ii), d_{M}^{z} = dz ; \overline{M} implies that z is conjugate either to ( ϵ, ϵ^{2}, ϵ^{−1}, ϵ^{−2}, π_{7} ) or to ( 1, ϵ, ϵ^{2}, ϵ^{−3}, π_{7} ) . In both cases d_{S}^{z} ≥ 9, a contradiction.

\underline{n = 12}. d_{S}^{x} ≥ 42, d_{S}^{y} ≥ 26 and dim S = 78 imply d_{S}^{z} ≤ 10. z cannot be modular since, for p = 7, d_{S}^{z} ≥ 12. z cannot be real, as 2d_{S}^{z} < d7 ; \overline{M} = 22. Assume z semirational, and non-real. d_{M}^{z} ≤ d7 ; \overline{M} + def\overline{M} = 26 implies m_{0} = 3, m_{1} = 2, m_{−1} = 1. But, in this case, d_{S}^{z} = 12, against the bound 2d_{S}^{z} ≤ 20. □

Proof of Theorem 4

Let H be as stated in the theorem, and assume that it is (2,3,7)-generated. Note that, under our assumptions, y is real or modular by Lemmas 5.5, 5.6 and 5.7. This assumption is necessary in order to obtain contradictions by playing condition (∗) against (∗∗). Our claims follow from the arguments below and Table 5.

A) x, y, z cannot satisfy (∗) and (∗∗). \underline{n = 13}. d_{S}^{x} ≥ 49, d_{S}^{y} ≥ 31 and dim S = 91 imply d_{S}^{z} ≤ 11, contradicting Tables 4 and 4′. \underline{n = 14}, p ≠ 2. d_{S}^{x} = 57, d_{S}^{y} ≥ 35 and dim S = 105 imply d_{S}^{z} ≤ 13, contradicting Tables 4 and 4′. For the case p = 2, see below.

B) If x, y, z satisfy (∗) and (∗∗), then z is neither modular, nor real, nor semirational. \underline{n = 16}. d_{S}^{x} ≥ 72, d_{S}^{y} ≥ 46 and dim S = 136 imply d_{S}^{z} ≤ 18. Thus z cannot be modular by Table 4. On the other hand, if p ≠ 7, 2d_{S}^{z} ≤ 36 < 38 = d7 ; \overline{M}, hence z is not real. Finally, it is easy to see, using the condition d_{M}^{z} ≤ 44, that z can be semirational (and not real) only if m_{0} = 1, m_{1} = 3, m_{−1} = 2. But in this case d_{S}^{z} = 19, a contradiction. \underline{n = 18}, p ≠ 2. d_{S}^{x} ≥ 91, d_{S}^{y} ≥ 57 and dim S = 171 imply d_{S}^{z} ≤ 23. Using this condition and d_{M}^{z} ≤ 54, it is easy to check that z cannot be semirational. The cases z real or modular are dealt with below.

C) If x, y, z satisfy (∗) and (∗∗), then z is neither modular nor real. \underline{n = 14}, p = 2. d_{S}^{x} ≥ 56 and d_{S}^{y} ≥ 35 imply d_{S}^{z} ≤ 14. Assume z real. Using Lemma 5.4, we get 28 ≥ 2d_{S}^{z} ≥ d_{M}^{z} ≥ d7 ; \overline{M} = 28, whence d_{M}^{z} = d7 ; \overline{M}. By Lemma 3.1 z is conjugate to ( (π_{7})^{2} ) and d_{S}^{z} = 15, a contradiction. \underline{n = 18}, p ≥ 2. d_{S}^{x} ≥ 90 implies d_{S}^{z} ≤ 24. z cannot be modular by Table 4. Assume z real. 48 ≥ 2d_{S}^{z} ≥ d_{M}^{z} ≥ d7 ; \overline{M} = 48 implies d_{M}^{z} = d7 ; \overline{M}. By Lemma 3.1 z is conjugate to ( (ϵ, ϵ^{−1}, (ϵ^{i}, ϵ^{−i}), (π_{7})^{2} ) (i = 2 or 3), and we get d_{S}^{z} = 25, a contradiction. \underline{n = 15}. d_{S}^{x} ≥ 64, d_{S}^{y} ≥ 40 and dim S = 120 imply d_{S}^{z} ≤ 16. Thus z cannot be modular by Table 4. On the other hand, if p ≠ 7, 2d_{S}^{z} ≤ 32 < 33 = d7 ; \overline{M}, hence z is not real.

\underline{n = 17}. d_{S}^{x} ≥ 81, d_{S}^{y} ≥ 51 and dim S = 153 imply d_{S}^{z} ≤ 21. Thus z cannot be modular. On the other hand, if p ≠ 7, 2d_{S}^{z} ≤ 42 < 43 = d7 ; \overline{M}, hence z is not real.

\underline{n = 19}. d_{S}^{x} ≥ 100, d_{S}^{y} ≥ 64 and dim S = 190 imply d_{S}^{z} ≤ 26. z cannot be modular. If p ≠ 7, 2d_{S}^{z} ≤ 52 < 53 = d7 ; \overline{M}, hence z is not real. □

Proof of Theorem 5

Let H be as stated in the theorem, and assume that it is (2,3,7)-generated. As |ℚ(ω) : ℚ| = 2 and |ℚ(ϵ) : ℚ| = 6, y is real and z is rational. Using Lemmas 5.3 and 4.2 it is easy to determine the admissible conjugacy classes of z and, when they exist, to compute d_{S}^{z}. (Alternatively this information can be deduced from Table 4.) Our claim follows since, for the values of n considered in the statement, conditions (∗) and (∗∗) are incompatible. □

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