Small Rank Exceptional Hurwitz Groups

Gunter Malle *, Mathematisches Institut und IWR, Im Neuenheimer Feld 368, D - 69120 Heidelberg.

A group G is called a (2, 3)-group if it can be generated by an involution and an element of order 3. This is equivalent to saying that G is a factor group of PSL_{2}(ℤ), which is the free product of two cyclic groups of order two and three. Thus every (2, 3)-group corresponds to a normal (in general: non-congruence) subgroup of SL_{2}(ℤ). There has recently been some interest in determining the (2, 3)-generated finite simple groups (see for example [3,13,14]). In this paper we consider the exceptional groups of Lie type ^{3}D_{4}(q) and ^{2}F_{4}(2^{2n+1}).

Theorem 1: The groups ^{3}D_{4}(q) and ^{2}F_{4}(2^{2n+1})′ are factor groups of the modular group PSL_{2}(ℤ). Here q is any prime power.

A finite group G is called a Hurwitz group if it is a factor group of

G_{2,3,7} := ⟨σ_{1}, σ_{2}, σ_{3} | σ_{1}^{2} = σ_{2}^{3} = σ_{3}^{7} = σ_{1}σ_{2}σ_{3} = 1⟩.

The interest in such groups stems from the fact that they can be represented as a group of 84(γ −1) automorphisms of a Riemann surface of genus γ. Hurwitz showed that this is the biggest possible value for the order of a group acting on a surface of genus γ ≥ 2. For an introduction to the present knowledge on Hurwitz groups see the recent overview [1] or also [3]. Clearly a Hurwitz group is (2, 3)-generated. In [8] we showed that the groups G_{2}(q) for q ≥ 5 and the groups ^{2}G_{2}(3^{2n+1}) are Hurwitz. Here we extend this result to the Steinberg triality groups ^{3}D_{4}(q) and to some of the Ree groups ^{2}F_{4}(2^{2n+1}). Together with the results in [8] this completely determines the Hurwitz groups among simple exceptional groups of Lie type of rank at most 2.

Theorem 2:

(a) The group ^{3}D_{4}(q), q = p^{n}, is a Hurwitz group if and only if p ≠ 3, q ≠ 4.

(b) The group ^{2}F_{4}(2^{2n+1})′ is a Hurwitz group if and only if n ≡ 1 (mod 3).

* The author gratefully acknowledges financial support by the Deutsche Forschungsgemein-

schaft

(This in particular implies Theorem 1 for the groups ^{3}D_{4}(q), p ≠ 3, q ≠ 4, and also for ^{2}F_{4}(2^{2n+1}) with n ≡ 1 (mod 3).)

The method of proof is similar to the one employed in [8]. To show (2, 3)- generation (resp. (2, 3, 7)-generation) of G we first calculate a structure constant from the ordinary character table of G. Then information on the maximal subgroups of G is used to rule out the possibility that all such (2, 3)-pairs (resp. (2, 3, 7)-triples) generate proper subgroups. Since this was already described in [8] in some detail, we will not repeat the precise description.

We remark that the calculation of structure constants was carried out with the computer algebra system CHEVIE [5], which contains the generic character table of the groups ^{3}D_{4}(q) (as first published in [4]) and of ^{2}F_{4}(2^{2n+1}) (computed from the results in [7]).

1 The groups ^{3}D_{4}(q)

In this section we prove that the Steinberg triality groups G := ^{3}D_{4}(q), q = p^{n}, p ≠ 3, q ≠ 4, are Hurwitz groups, and that the remaining groups are (2, 3)-generated. For this we first have to know the classes of involutions, of 3elements and of 7-elements in G. The unipotent classes of G were determined by Spaltenstein [12], while the semisimple ones can be found in [4]. With these references one easily checks the following facts:

Lemma 1: (a) If p ≠ 2 then G has a single class of involutions, with centralizer order q^{4}(q^{2} − 1)(q^{6} − 1).

(b) If p ≠ 3 then G has two classes of elements of order three, with centralizer orders q^{3}(q^{2} − 1)(q^{3} − ϵ)(q^{2} + ϵq + 1), resp. q^{3}(q^{6} − 1)(q − ϵ), where q ≡ ϵ (mod 3), ϵ ∈ {1, −1}.

(c) If p ≠ 7 then G has a single class of regular elements of order 7.

Proof : The first two assertions readily follow from the tables in [4]. If q ≡ 2, 3, 4 or 5 (mod 7), then a 7-Sylow subgroup of G is contained in a maximal torus T of G of order (q^{2} + q + 1)^{2} resp. (q^{2} − q + 1)^{2}. These tori have a relative Weyl group of order 24, and from the action given in [4, Table 1.1], it follows that there exists a single regular class in T containing elements of order 7. If q ≡ ±1 (mod 7), the 7-Sylow subgroups lie in tori T of order (q^{3} − 1)(q − 1) resp. (q^{3} + 1)(q + 1). Of the 48 elements of order 7, 36 lie in subgroups centralizing a root subgroup. The remaining 12 elements are fused into a single class of regular elements by the Weyl group W = S_{3}×Z_{2}. Alternatively, this follows from the embedding G_{2}(q) < ^{3}D_{4}(q) and an application of Proposition 1 in [8].

If p ≠ 2 we write 2A for the class of involutions, 3A and 3B for the two classes of 3-elements (where 3A-elements have centralizer order q^{3}(q + ϵ)(q^{3} − ϵ)^{2}) if p ≠ 3, and 7A for the class of regular 7-elements if p ≠ 7. The strategy is now to prove that there exist triples σ_{1} ∈ 2A, σ_{2} ∈ 3A, σ_{3} ∈ 7A with σ_{1}σ_{2}σ_{3} = 1 generating G. For this we need a further preparatory lemma.

Lemma 2: Let (σ_{1}, σ_{2}, σ_{3}) be a (2, 3, 7)-system of ^{3}D_{4}(q) with σ_{1}σ_{2}σ_{3} = 1. If 2 |̸ q, or 2|q and σ_{3} ∈ 7A, then ⟨σ_{1}, σ_{2}, σ_{3}⟩ is not contained in a maximal parabolic subgroup.

Proof : By [6] the maximal parabolic subgroups of ^{3}D_{4}(q) have the structure P_{1} = [q^{9}] : (SL_{2}(q^{3}) ∘ Z_{q−1}).d respectively P_{2} = [q^{11}] : (Z_{q^{3}−1} ∘ SL_{2}(q)).d, where d = gcd(2, q − 1). Here [n] denotes a group of order n of unspecified structure, as in the Atlas notation [2]. Since a Hurwitz group is necessarily perfect, if ⟨σ_{1}, σ_{2}, σ_{3}⟩ is contained in one of those parabolic subgroups, then already in the commutator subgroups P′_{1} = [q^{9}] : SL_{2}(q^{3}) respectively P′_{2} = [q^{11}] : SL_{2}(q), and we have q ≠ 2, 3. Now first assume that 2 |̸ q. Then we use that the factors of the above commutator subgroups by their unipotent radicals would also have to possess a (2, 3, 7)-system. This is a contradiction, as SL_{2}(q^{3}) and SL_{2}(q) both contain a single, central involution, which can thus not lie in a (2, 3, 7)-system.

If p = 2 and σ_{3} ∈ 7A, then we may first exclude P′_{1}, as this is the centralizer of a p-element, hence in particular can not contain regular semisimple elements. In the second case, a 7-Sylow subgroup of P′_{2} is necessarily cyclic, so any 7element is conjugate to one in the complement SL_{2}(q). But this complement centralizes a p-element lying in a short root subgroup, hence its semisimple elements are again not regular.

Let C = (C_{1}, C_{2}, C_{3}) be a class vector of the finite group H. We write n_{H}(C) or simply n(C) for the normalized structure constant of C (see for example [8]), and ℓ(C) for the number of generating triples in C with product 1 modulo conjugation. See [8] for the connection between n(C) and ℓ(C).

Proposition 1: Theorem 2(a) holds for p ≠ 2, 3, 7 and the class vector C = (2A, 3A, 7A).

Proof : The normalized structure constant n(C) of C = (2A, 3A, 7A) can be calculated from the ordinary character table of G. This calculation is most easily done with the computer algebra system CHEVIE [5], which contains the generic ordinary character table of the groups ^{3}D_{4}(q). The results are collected in Table 1. (They depend on the congruence classes of q modulo 3 and 7.)

Table 1: The (2, 3A, 7A) normalized structure constants in ^{3}D_{4}(p^{n}), p ≠ 3, 7.

q ≡  1 (mod 7) −1 (mod 7) 2, 4 (mod 7) 3, 5 (mod 7)
1 (mod 3)  q^{2} + 1  q^{2} − 2q + 1 q^{2} − 3q + 1 q^{2} + q + 1
−1 (mod 3) q^{2} + 2q + 1  q^{2} + 1  q^{2} − q + 1 q^{2} + 3q + 1

Note that this shows \frac{1}{2}q^{2} < n(C) < 2q^{2} for p ≠ 2, 3, 7, q ≠ 5. It now remains to check that ℓ(C) ≠ 0, i.e., not all of the structure constant can be accounted for by the contribution of proper subgroups. For this we make use of the list of maximal subgroups of G in [6]. In Table 2 we list the maximal subgroups of G which might contribute to the (2A, 3A, 7A)-structure constant, i.e., which are non-solvable and have order potentially divisible by 42. (We have omitted the parabolic subgroups, since these can not contain a triple from C by Lemma 2.)

Table 2: Some maximal subgroups of ^{3}D_{4}(q), q = p^{n}, p odd.

K  occurs for  n_{K}(2, 3, 7)
G_{2}(q)  always  0
P GL_{3}(q)  q ≡ 1 (mod 3)  ≤ 6
P GU_{3}(q)  q ≡ −1 (mod 3)  ≤ 6
(SL_{2}(q^{3}) ∘ SL_{2}(q)) · 2  always  ≤ 18
((q^{2} + q + 1) ∘ SL_{3}(q)).f_{+}.2  always  ≤ 3
((q^{2} − q + 1) ∘ SU_{3}(q)).f_{−}.2  always  ≤ 3
3D_{4}(q_{0})  q = q_{0}^{t}, 3 ≠ t ∈ ℙ  < 2q_{0}^{2}

Here f_{ϵ} := gcd(3, q^{2} + ϵq + 1).

The last column gives an upper bound for the contribution to n(C) possibly coming from this class of maximal subgroups. This is obtained as follows.

In G_{2}(q) there are two classes of 3-elements, with centralizers SL_{3}(q) and SL_{2}(q)(q − 1) if q ≡ 1 (mod 3), respectively SU_{3}(q) and SU_{2}(q)(q + 1) if q ≡ −1 (mod 3). Of these, the first one fuses into the class 3A of G. But the structure constant of G_{2} involving this first class is seen to vanish using CHEVIE. The contribution from the remaining groups can also be computed from the generic character tables [5], respectively from Table 1 for ^{3}D_{4}(q_{0}).

First assume q = 5. Then n(C) = 40 by Table 1, while the proper subgroups contribute at most 30 by Table 2. If q = 25 one also checks that ℓ(C) ≥ 1. For q = p ≥ 11 we have n(C) ≥ 1/2q^{2} ≥ 60, which is larger than the structure constants coming from proper subgroups. Finally, if q = p^{n}, n ≥ 2, we write n = ∏_{i=1}^{r} t_{i}^{e_{i}}, so that the subfield subgroups ^{3}D_{4}(q_{0}) contribute at most

r ; ∑ ; i=1 2p^{2n/t_{i}} ≤ r ; ∑ ; i=1 2p^{n} = 2rp^{n}.

Thus there remains at least

ℓ(C) ≥ \frac{1}{2}p^{2n} − 2rp^{n} − 30 = \frac{1}{2}p^{n}(p^{n} − 4r − ϵ) ≥ 1,

with some ϵ < 1 since p^{n} ≥ 121, for the number ℓ(C) of generating Hurwitztriples. This achieves the proof.

For p = 7 let 7U be the class of regular unipotent elements, which by the description of the elementary divisors of unipotent elements in the overgroup D_{n}(q^{3}) = O_{8}^{+}(q^{3}) of ^{3}D_{4}(q) have order 7.

Proposition 2: Theorem 2(a) holds for p = 7 and the class vector C = (2A, 3A, 7U).

Proof : The normalized structure constant of C, computed with CHEVIE, is displayed in Table 3.

Table 3: The (2A, 3A, 7U) normalized structure constants in ^{3}D_{4}(7^{n}).

\begin{aligned}q ≡ & 1 (mod 3) & −1 (mod 3) \\ & q(q − 1) & q(q + 1) \end{aligned}

The maximal subgroups which could possibly contribute to this structure constant are contained in Table 1. Again, the corresponding structure constant in G_{2}(q) vanishes. It is an easy exercise to check that the contribution from subgroups is always strictly smaller than n(C).

Proposition 3: Theorems 2(a) and 1 hold for ^{3}D_{4}(3^{n}).

Proof : We first show that all (2, 3, 7)-triples of G generate proper subgroups. By Lemma 1(a), G has a single class of involutions. The (unipotent) classes containing elements of order 3 are A_{1}, 3A_{1}, A′_{2} and A′′_{2} in the notation of [12]. In Table 4 we tabulate those (2, 3, 7)-structure constants which are at least 1; the remaining ones can not come from generating triples anyway. In particular, it turns out that only the regular class 7A of 7-elements is involved in possible candidates.

Table 4: The (2A, 3, 7A) normalized structure constants in ^{3}D_{4}(3^{n}).

q ≡  1 (mod 7)  −1 (mod 7) 2, 4 (mod 7) 3, 5 (mod 7)
3A_{1}  1  1  1  1
A′_{2}  \frac{1}{2}q(q + 2) − δ_{+}  \frac{1}{2}q(q − 2) \frac{1}{2}q(q − 1) − δ_{+}  \frac{1}{2}q(q + 1)
A′′_{2}  \frac{1}{2}q(q + 2) \frac{1}{2}q(q − 2) − δ_{−}  \frac{1}{2}q(q − 1) \frac{1}{2}q(q + 1) − δ_{−}

Here δ_{+} := (q^{2} + q)/(q^{2} + q + 1) and δ_{−} := (q^{2} − q)/(q^{2} − q + 1). The interesting maximal subgroups are listed in Table 1. It turns out that the corresponding (2, 3, 7)-structure constants in the maximal subgroup G_{2}(q) are the same as those displayed in Table 4, up to the summands −δ_{ϵ}. We now compare the centralizers of (2, 3, 7A)-triples in G_{2}(q) and ^{3}D_{4}(q). If the centralizer increases in ^{3}D_{4}(q), then by Table 2 the Hurwitz triple must lie inside ((q^{2} + ϵq + 1) × L_{3}^{ϵ}(q)).2, where q^{3} ≡ ϵ (mod 7), since 7A-elements are regular. (Here we write L_{3}^{+}(q) := L_{3}(q), L_{3}^{−}(q) := U_{3}(q).) But the (2, 3, 7)- structure constant of L_{3}^{ϵ}(q).2 equals 1 if q^{3} ≡ ϵ (mod 7). Thus passing from G_{2}(q) to ^{3}D_{4}(q), the centralizer increases for precisely one Hurwitz triple, by q^{2} + ϵq + 1, so the contribution to the structure constant decreases by

1 − \frac{1}{q^{2} + ϵq + 1} = δ_{ϵ} .

Hence all Hurwitz triples in Table 4 come from G_{2}(q), and G can not be a Hurwitz group.

To prove (2, 3)-generation, we calculate n(2A, A′_{2}, [s_{14}]) = q(q − 1)/2, where s_{14} is a representative of the regular elements in the maximal torus of order q^{4} − q^{2} + 1. By [6], any such triple generates G.

For p = 2 let 2U be the larger of the two classes of involutions of G, denoted by 3A_{1} in [12].

Proposition 4: Theorems 2(a) and 1 hold for ^{3}D_{4}(2^{n}) and the class vector C = (2U, 3A, 7A).

Proof : The structure constants in this case turn out to be the same as for p ≠ 2, 3, 7, so are also given by Table 1. The maximal subgroups which could possibly contribute to n(C) are as follows (see [6]):

Table 5: Some maximal subgroups of ^{3}D_{4}(2^{n}).

K  occurs for  n_{K}(2, 3, 7)
G_{2}(q)  always  0
P GL_{3}(q)  q ≡ 1 (mod 3)  ≤ 3
P GU_{3}(q)  2 < q ≡ −1 (mod 3)  ≤ 3
L_{2}(q^{3}) × L_{2}(q)  always  ≤ 10
((q^{2} + q + 1) ∘ SL_{3}(q)).f_{+}.2  always  ≤ 3
((q^{2} − q + 1) ∘ SU_{3}(q)).f_{−}.2  always  ≤ 3
3D_{4}(q_{0})  q = q_{0}^{t}, 3 ≠ t ∈ ℙ  < 2q_{0}^{2}

If q = 2 we have n(C) = 3, but none of the maximal subgroups contains elements from 2U, 3A and 7A [2]. For q ≥ 8, the usual estimate works. Now let q = 4. Then n(C) = 5, of which 3 already comes from the subgroup 3D_{4}(2). It is then clear that C contains no generating triples, since the outer automorphism group of order 6 would have to act faithfully on those. (In fact, the remaining two triples lie already inside the maximal subgroup P GL_{3}(4).) The structure constant n(2U, 3B, 7A) with the second class of 3-elements 3B equals 16 and is the only other (2, 3, 7)-constant not smaller than 1. But in G_{2}(4) we have n_{G_{2}}(2B, 3B, 7A) = 16, originating from J_{2} and L_{2}(13) [8, §6]. Since these two subgroups have trivial centralizer in G as well, they account for the total structure constant in G. It follows that ^{3}D_{4}(4) is not a Hurwitz group. But n(2U, 3A, 241A) = 13, where 241A is a class of elements of order 241 = q^{4} − q^{2} + 1, and it is easily verified that this gives a (2, 3)-generation of ^{3}D_{4}(4).

2 The groups ^{2}F_{4}(q^{2})

In this section we study G := ^{2}F_{4}(q^{2}), q^{2} = 2^{2n+1} with n ≥ 0. The order of G is q^{24}(q^{2} − 1)(q^{6} + 1)(q^{8} − 1)(q^{12} + 1) and hence divisible by 7 if and only if 3|(2n + 1), so n ≡ 1 (mod 3). We take the notation for Steinberg generators and for representatives of conjugacy classes of G from [10]. Let 2B be the class of the unipotent element α_{10}(1), with centralizer order q^{20}(q^{4} − 1), 3A the unique class of elements of order three with representative t_{4} and centralizer order q^{6}(q^{4} −1)(q^{6} +1), and 7A a class of elements of order 7 with representative t_{2} and centralizer order q^{2}(q^{2} − 1)(q^{4} − 1).

Proposition 5: Theorem 2(b) holds with the class vector C = (2B, 3A, 7A).

Proof : It was already noted above that |^{2}F_{4}(2^{2n+1})| with n ≢ 1 (mod 3) is not divisible by 7, hence can not be a Hurwitz group. So now let us assume n ≡ 1 (mod 3). From the CHEVIE table [5] of G = ^{2}F_{4}(q^{2}) one calculates the structure constant n(C) = q^{4} + q^{2} + 1 + \frac{1}{q^{4}−1}. Next we have to determine the contribution coming from proper subgroups. The non-solvable maximal subgroups of G with order divisible by 3 are the following [9]:

Table 6: Some maximal subgroups of ^{2}F_{4}(q^{2}), q^{2} = 2^{2n+1}.

K  occurs for  n_{K}(2, 3, 7)
[q^{22}]: (L_{2}(q^{2}) × (q^{2} − 1))  always  q^{2} + 1 + \frac{1}{q^{2}−1}
SU_{3}(q^{2}): 2  always  ≤ 3
P GU_{3}(q^{2}): 2  always  ≤ 3
Sp_{4}(q^{2}): 2  always  1
2F_{4}(q_{0}^{2})  q^{2} = q_{0}^{2t}, t ∈ ℙ  < \frac{4}{3}q_{0}^{4}

The result follows from the upper bounds for n_{K}(2, 3, 7) in this table by a calculation as for ^{3}D_{4}(q). So it remains to deduce these bounds.

For the subgroups SU_{3}(q^{2}): 2, P GU_{3}(q^{2}): 2 and Sp_{4}(q^{2}): 2, this follows from the generic character tables if we keep in mind that a (2, 3, 7)-triple has to lie already in the commutator subgroup. For the subfield groups ^{2}F_{4}(q_{0}^{2}) the upper bound is obtained by induction, since the classes 2B, 3A and 7A of any subfield group fuse into the same classes in G. We are left with the parabolic subgroup P_{a} := [q^{22}]: (L_{2}(q^{2}) × (q^{2} − 1)). Again, a Hurwitz triple must lie already inside P̃_{a} := [q^{22}]: L_{2}(q^{2}). We determine explicitly all Hurwitz triples in that group and thus compute the contribution to n(C).

For this it is necessary to work with the Steinberg generators of G as given for example in [10]. In this notation we have

P̃_{a} = U_{1}U_{2}U_{4} ⋯ U_{12} · ⟨U_{3}, r_{a}⟩,

where the U_{i} := {α_{i}(t) | t ∈ 𝔽_{q^{2}}} are as in [11] and r_{a} is a fundamental generator of the Weyl group. The Levi factor L_{a} = ⟨U_{3}, r_{a}⟩ of P̃_{a} is isomorphic to the factor group of P̃_{a} modulo its unipotent radical R_{u}(P̃_{a}) = U_{1}U_{2}U_{4} ⋯ U_{12}. Clearly if (σ_{1}, σ_{2}, σ_{3}) is a Hurwitz triple of P̃_{a}, then so is its image in P̃_{a}/R_{u}(P̃_{a}). But the Hurwitz triples in L_{a} ≅ L_{2}(q^{2}) can easily be classified: they are all conjugate in L_{a} to a triple of the form

(α_{3}(γ)r_{a}α_{3}(1), α_{3}(1)r_{a}, α_{3}(γ))

with γ^{3} + γ + 1 = 0 (i.e., γ is an element of order 7 in 𝔽_{8} ≤ 𝔽_{q^{2}}). We have to count the number of liftings of this to Hurwitz triples of P̃_{a}. Since the 7-Sylow subgroup of P̃_{a} is cyclic, we may moreover assume that the lifting of σ_{1} is again α_{3}(γ)r_{a}α_{3}(1). So we look for u ∈ R_{u}(P̃_{a}) = U_{1}U_{2}U_{4} ⋯ U_{12} such that

(α_{3}(1)r_{a}u)^{3} = 1 and (α_{3}(γ)u)^{2} = 1 .  (∗)

By the description of the involution classes in P_{a} in [11, (3.3)] we first deduce that u ∈ U_{4} ⋯ U_{12}. We distinguish two cases according to whether u ∈ U_{5} ⋯ U_{12} or not. In the first case, let

u = α_{5}(c_{1})α_{6}(c_{2})α_{7}(c_{3})α_{8}(c_{4})α_{9}(c_{5})α_{10}(c_{6})α_{11}(c_{7})α_{12}(c_{8}) .

The conditions (∗), evaluated modulo the normal subgroup U_{7}U_{9} ⋯ U_{12} yield c_{1} = c_{2} = c_{4} and c_{1}(γ + γ^{2θ}) = 0, where θ is the automorphism of 𝔽_{q^{2}} with 2θ^{2} = 1 (so θ is exponentiation by 2^{n}). But γ + γ^{2θ} = 0 implies γ = 1, which is not the case, so we have c_{1} = c_{2} = c_{4} = 0. Repeating the calculation with these new values then yields

u = α_{9}(c_{5})α_{10}(c_{5}^{2θ})α_{12}(c_{8}) , with c_{5}, c_{8} ∈ 𝔽_{q^{2}} .

Since we may still conjugate this by the centralizer of L_{a} (which is isomorphic to U_{10}.H_{10}, with a subgroup H_{10} of order q^{2} − 1 of the maximally split torus inside P_{a}), this gives a contribution of q^{2}/(q^{2} − 1) to n_{P_{a}}(2, 3, 7).

Now assume

u = α_{4}(c_{0})α_{5}(c_{1})α_{6}(c_{2})α_{7}(c_{3})α_{8}(c_{4})α_{9}(c_{5})α_{10}(c_{6})α_{11}(c_{7})α_{12}(c_{8})

with c_{0} ≠ 0, where after conjugation with h ∈ H_{10} we can take c_{0} = 1. Calculation modulo U_{7}U_{9} ⋯ U_{12} then gives

c_{1} = \frac{1 + γ^{2θ}}{γ + γ^{2θ}} , c_{2} = c_{1} + 1, c_{4} = c_{1}

(note that we already saw that γ + γ^{2θ} ≠ 0). The complete computation shows that for u to satisfy (∗) we need

\begin{aligned}c_{3} & = \frac{(γ + 1)(γ^{2θ} + 1)}{(γ + γ^{2θ})(γ^{2} + γ^{2θ})}, c_{6} = c_{1}^{2} + c_{1} + c_{3}^{2θ} + c_{5} + c_{5}^{2θ}, \\ c_{7} & = γ^{−1}(c_{1}^{2θ+1} + c_{1}c_{3}γ^{2θ} + c_{3}γ^{2θ} + c_{1}c_{3} + c_{1}c_{5} + c_{3}^{2θ} + c_{5}^{2θ}) . \end{aligned}

Thus there exist q^{4} such triples, corresponding to the choices of c_{5}, c_{8} ∈ 𝔽_{q^{2}}. Under conjugation with U_{10}, this leads to a contribution of q^{2} to n(C). This completes the proof of the table and hence of the proposition.

To prove (2, 3)-generation for arbitray n we just alter the third class in the above class vector C. Namely, let C_{T} be the class in G of a regular semisimple element of order q^{4}+\sqrt{2}q^{3}+q^{2}+\sqrt{2}q+1, hence a generator of a cyclic maximal torus T of G of the same order, denoted by t_{17} in [10].

Proposition 6: Theorem 1 holds for ^{2}F_{4}(2^{2n+1})′ with the class vector C = (2B, 3A, C_{T}).

Proof : Again with CHEVIE one computes the structure constant to be

n(C) = q^{3}(q^{5} − \sqrt{2}q^{4} + 3q − \sqrt{2}) .

Among the maximal subgroups of G, only the normalizer of the torus T contains elements with order divisible by a primitive prime divisor of q^{4} + \sqrt{2}q^{3} + q^{2} + \sqrt{2}q + 1 [9]. This normalizer is an extension of T by the cyclic group of order 12 and thus clearly solvable, while the group generated by elements from C must be perfect. Hence G is (2, 3)-generated.

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