The maximal subgroups of ^{2}F_{4}(q^{2})

Gunter Malle *, Mathematisches Institut der Universität Heidelberg, Im Neuenheimer Feld 288, D - 6900 Heidelberg.

In this paper the maximal subgroups of the Ree groups ^{2}F_{4}(q^{2}) defined in characteristic 2 [14] are studied. The Ree groups are the only finite groups of Lie type of Lie rank at most two for which the maximal subgroups are not known. Only the smallest case ^{2}F_{4}(2), the automorphism group of the Tits group, was treated by Wilson [20] (see also [5], but note that in both references a maximal subgroup SU_{3}(2): 2 of ^{2}F_{4}(2) was omitted). Tchakerian [19] investigated the maximal subgroups of the Tits group ^{2}F_{4}(2)′. For the classical groups of low rank, complete results have now been obtained (a good source of reference is [7], for example). Also, the maximal subgroups of the exceptional groups ^{2}G_{2}(q^{2}) were determined in [11] and [8], of G_{2}(q) by Aschbacher [1], Cooperstein [6] and Kleidman [8], and of the triality groups ^{3}D_{4}(q) again by Kleidman [9]. The Suzuki groups ^{2}B_{2}(q^{2}) were already studied by Suzuki himself [18].

Main Theorem : Every maximal subgroup of G := ^{2}F_{4}(q^{2}), q^{2} = 2^{2n+1}, n ≥ 1, is conjugate to one of the following :

(1) P_{a} = [q^{22}]: (A_{1}(q^{2}) × (q^{2} − 1)),

(2) P_{b} = [q^{20}]: (^{2}B_{2}(q^{2}) × (q^{2} − 1)),
(3) N_{G}(⟨t_{4}⟩) = C_{G}(t_{4}): 2 ≅ SU_{3}(q^{2}): 2,
(4) N_{G}(T_{8}) ≅ (Z_{q^{2}+1} × Z_{q^{2}+1}): GL_{2}(3),
(5) N_{G}(T_{6}) ≅ (Z_{q^{2}−\sqrt{2}q+1} × Z_{q^{2}−\sqrt{2}q+1}): [96] if q^{2} > 8,
(6) N_{G}(T_{7}) ≅ (Z_{q^{2}+\sqrt{2}q+1} × Z_{q^{2}+\sqrt{2}q+1}): [96],
(7) N_{G}(T_{10}) ≅ Z_{q^{4}−\sqrt{2}q^{3}+q^{2}−\sqrt{2}q+1} : 12,
(8) N_{G}(T_{11}) ≅ Z_{q^{4}+\sqrt{2}q^{3}+q^{2}+\sqrt{2}q+1} : 12,
(9) P GU_{3}(q^{2}): 2,
(10) ^{2}B_{2}(q^{2}) ≀ 2,
(11) B_{2}(q^{2}): 2, or
(12) ^{2}F_{4}(q_{0}), if q_{0} = 2^{2m+1} with (2n + 1)/(2m + 1) ∈ ℙ.

Conversely, there is exactly one class of maximal subgroups of G for each entry in the list.

As an easy consequence, the maximal subgroups of all groups H with ^{2}F_{4}(q^{2}) ≤ H ≤ Aut(^{2}F_{4}(q^{2})) — these are extensions of G by field automorphisms — can be determined:

Corollary : The maximal subgroups of ^{2}F_{4}(q^{2}) : f, f|(2n + 1), not containing ^{2}F_{4}(q^{2}), are obtained from the ones in the above list by adjoining the field automorphism in the obvious way. In particular, no novelties arise.

* The author gratefully acknowledges financial support by the Deutsche Forschungsgemeinschaft

The determination of maximal subgroups of a finite simple group G makes use of the following observation: all maximal subgroups are among the normalizers of characteristically simple subgroups of G, i.e., of direct products of isomorphic simple groups. A (maximal) subgroup of G is called local, if it contains a solvable normal subgroup (and hence an elementary abelian normal subgroup), otherwise it is called nonlocal. Thus the problem divides into two parts, the first of which (namely the local subgroups) turns out to be very easy in our case, because most of the information was already obtained by Shinoda [16]. He determined the maximal tori, the conjugacy classes of elements and the structure of their centralizers in 2F_{4}(q^{2}). His results and his terminology for the classes and the tori will be used throughout. 1 The local subgroups

The local subgroups of the exceptional groups of Lie type are quite well understood. In the case of our groups G = ^{2}F_{4}(q^{2}), the analysis is very straightforward, due to the work of Shinoda [16].

1.1 The 2-locals

By the theorem of Borel and Tits [3] the p-local subgroups of groups of Lie type in characteristic p are all contained in parabolic subgroups. The parabolic subgroups are well known, they can be constructed from a Borel subgroup by adding some of the generating reflections of the Weyl group corresponding to simple roots. In the Ree groups, we arrive at the following :

Proposition 1.1: Every 2-local subgroup of ^{2}F_{4}(q^{2}) has a conjugate contained in one of the maximal parabolic subgroups:

(1) P_{a} = q^{6} · q^{4} · q^{8} · q^{4} : (L_{2}(q^{2}) × (q^{2} − 1)) of order q^{24}(q^{2} − 1)(q^{4} − 1) or

(2) P_{b} = q^{2} · q^{8} · q^{10} : (^{2}B_{2}(q^{2}) × (q^{2} − 1)) of order q^{24}(q^{2} − 1)^{2}(q^{4} + 1).

Remark: These subgroups are the groups denoted by P_{a}, P_{b} respectively in [16]. (Unfortunately, in [17] the notation P_{2}, P_{1} is used.) Generators and relations can be found in [14],

[16] and [17].

1.2 The 3-locals

The normalizer in G of the maximal torus T_{8} contains a Sylow 3-subgroup of G. According to the results of [16], there is just one class of elements of order three in G, with representative element t_{4}. We get the following result about the 3-locals of G:

Proposition 1.2: Any 3-local subgroup of ^{2}F_{4}(q^{2}) is conjugate to a subgroup of one of the following two subgroups of G:

(1) N_{G}(⟨t_{4}⟩) = C_{G}(t_{4}): 2 ≅ 3^{.}U_{3}(q^{2}): 2 of order 2q^{6}(q^{4} − 1)(q^{6} + 1) or

(2) N_{G}(T_{8}) ≅ (Z_{q^{2}+1} × Z_{q^{2}+1}): GL_{2}(3) of order 48(q^{2} + 1)^{2}.

Proof : The centralizer of an element of order three was determined in [16], where explicit generators are given for C_{G}(t_{4}) ≅ 3^{.}U_{3}(q^{2}). As elements of order three are rational, the normalizer is an extension of SU_{3}(q^{2}) of degree two which acts nontrivially on the center. This leads to the structure given in the theorem. Now 3-rank(G) = 2 (this can be seen as SL_{3}(q^{4}) ≥ SU_{3}(q^{2}) ≅ C_{G}(t_{4}) has 3-rank two), so the only other case to be considered is the normalizer of an elementary abelian subgroup E of order nine. But by Theorem 5.8(c) of [2], E2, this embedds into the maximal torus T_{8}, so the normalizer over the centralizer is isomorphic to a subgroup of N_{G}(T_{8})/T_{8} (see for example Proposition 3.7.1 in [4]). As E is characteristic in T_{8}, the assertion of the proposition follows. (The extension in (2) is split, because 3^{2}: GL_{2}(3) in L_{3}(3) < ^{2}F_{4}(2) is split [5].)

1.3 The r-locals, r ≥ 5

Here we can invoke powerful theorems about the structure of elementary abelian subgroups of groups of Lie type containing only semisimple elements. Namely no prime r ≥ 5 divides the order of the Weyl group of type F_{4}. This implies that any direct product of cyclic r-groups can be embedded into a maximal torus of G and moreover the normalizer over the centralizer of such a subgroup is isomorphic to a subgroup of the corresponding group for the maximal torus (see [2] or [4]; these properties are collected in Lemma 1.7 of [13].) This leads to:

Proposition 1.3: Any r-local subgroup of ^{2}F_{4}(q^{2}) for r ≥ 5 is conjugate to a subgroup of one of the groups in Proposition 1.2 or of one of the following groups:

(1) N_{G}(⟨t_{1}⟩) ≅ (Z_{q^{2}−1} : 2) × ^{2}B_{2}(q^{2}),

(2) N_{G}(⟨t_{2}⟩) ≅ (Z_{q^{2}−1} : 2) × L_{2}(q^{2}),
(3) N_{G}(⟨t_{5}⟩) ≅ (Z_{q^{2}+1} : 2) × L_{2}(q^{2}),
(4) N_{G}(⟨t_{7}⟩) ≅ (Z_{q^{2}−\sqrt{2}q+1} : 4) × ^{2}B_{2}(q^{2}),
(5) N_{G}(⟨t_{9}⟩) ≅ (Z_{q^{2}+\sqrt{2}q+1} : 4) × ^{2}B_{2}(q^{2}),
(6) N_{G}(T_{1}) ≅ (Z_{q^{2}−1} × Z_{q^{2}−1}): D_{16},
(7) N_{G}(T_{6}) ≅ (Z_{q^{2}−\sqrt{2}q+1} × Z_{q^{2}−\sqrt{2}q+1}): [96],
(8) N_{G}(T_{7}) ≅ (Z_{q^{2}+\sqrt{2}q+1} × Z_{q^{2}+\sqrt{2}q+1}): [96],
(9) N_{G}(T_{10}) ≅ Z_{q^{4}−\sqrt{2}q^{3}+q^{2}−\sqrt{2}q+1} : 12,
(10) N_{G}(T_{11}) ≅ Z_{q^{4}+\sqrt{2}q^{3}+q^{2}+\sqrt{2}q+1} : 12.

Proof : Let E be an elementary abelian r-subgroup. First assume E cyclic. Then either a generating element is conjugate to one of t_{1}, t_{2}, t_{5}, t_{7} or t_{9}, whence the centralizer and normalizer can be found in table IV of [16]. This accounts for the first five cases of the proposition. Or the centralizer of E is a maximal torus T, and as r does not divide the order of the Weyl group, the normalizer of E is contained in the normalizer of T.

The only other case is E ≅ Z_{r} × Z_{r}, because r-rank(G) ≤ 2. Again the normalizer of E is a subgroup of the normalizer of a maximal torus (see Lemma 1.7 in [13]). These are given in Table III of [16]. Now (2) and (3) both contain a subgroup T_{2}: (2 × 2) ≅ N_{G}(T_{2}),

(1) contains subgroups T_{3}: (2 × 4) ≅ N_{G}(T_{3}) and T_{4}: (2 × 4) ≅ N_{G}(T_{4}), and (4) and (5) have subgroups of the form T_{5}: (4 × 4) ≅ N_{G}(T_{5}), so that these normalizers are not maximal in G. The normalizer of T_{8} already occurs in Proposition 1.2, and N_{G}(T_{9}) ≅ T_{9}: 6 is conjugate to a subgroup of the normalizer of an element of order three as in Proposition 1.2(1). This completes the proof of the proposition and the determination of possible maximal local subgroups.

2 The nonlocal subgroups

The case of nonlocal subgroups is much harder to solve. The classification of finite simple groups will be necessary in the approach used here.

2.1 Simple subgroups

As the nonlocal subgroups are normalizers of direct products of nonabelian simple groups, in a first step a list of possible simple subgroups of ^{2}F_{4}(q^{2}) is compiled (in this section, simple will always mean nonabelian simple):

Lemma 2.1: A simple subgroup of ^{2}F_{4}(q^{2}) is either a group of Lie type in characteristic 2 of Lie rank at most two or isomorphic to one of

L_{2}(13), L_{2}(17), L_{2}(19), L_{2}(25), L_{2}(27), L_{2}(37), L_{2}(49), L_{3}(3), L_{4}(3), S_{6}(3), U_{3}(3), U_{4}(3), G_{2}(3), A_{6}, A_{7}, or a sporadic simple group.

Proof : As ^{2}F_{4}(q^{2}) is defined as a subgroup of F_{4}(q^{2}), every simple subgroup of G = ^{2}F_{4}(q^{2}) also occurs in F_{4}(q^{2}). Now F_{4}(q^{2}) has a nontrivial representation of dimension 26 over 𝔽_{q^{2}}, and the main result of [10] can be cited. Namely, a simple subgroup of G is either a group of Lie type in characteristic 2, one of a finite list of groups of Lie type in odd characteristic, an alternating group or a sporadic group. The groups of Lie type in odd characteristic which can occur as subgroups of F_{4}(q^{2}) are listed in Lemma 5.1 of [13], and discarding those in characteristic 2, one arrives at the above list. Also, no alternating group A_{n}, n ≥ 9, can be a subgroup of ^{2}F_{4}(q^{2}), because G has 3-rank two (Section 1.2). If H is a simple subgroup of G of Lie type in characteristic 2, then its Lie rank can be at most two, as is seen from the embedding of maximal parabolic subgroups of H in those of G (see [12]). This proves the lemma.

Many of the ‘exceptional’ groups in the above lemma can be discarded as possibilities for simple subgroups of G by a study of the local subgroups. The results in the first section show that no Frobenius groups 11: 5 or 29: 14 are contained in G, and as 9 does not divide the order of N_{G}(T)/C_{G}(T) for any maximal torus T of G, there are no Frobenius groups 19: 9 or 37: 18 either. But all sporadic simple groups apart from J_{2} and He contain a Frobenius group of one of the above types [5], as do L_{2}(19) and L_{2}(37).

As q^{2} = 2^{2n+1}, we have q^{2} ≡ −1 (mod 3) and q^{2} ≡ 2, 3 (mod 5). Thus 5 does not divide the order q^{6}(q^{4} − 1)(q^{6} + 1) of the centralizer of an element of order three and G can not have elements of order fifteen. This rules out S_{6}(3), J_{2} and He.

As q^{2} = 2^{2n+1} ≡ 1, 2, 4 (mod 7), we get that 7||G| iff 3|2n+1. Then a Sylow 7-subgroup of G is contained in T_{1}, and hence it is abelian of rank two. All elements of order seven are of type t_{1}, and so they are conjugate to at most two of their primitive powers. But in all of the groups L_{2}(49), U_{3}(3), U_{4}(3), G_{2}(3), A_{7}, elements of order seven are conjugate to at least three of their powers. Thus they can not occur in G. Moreover as U_{3}(3) = G_{2}(2)′, none of G_{2}(2^{m}) or ^{3}D_{4}(2^{m}) is a subgroup of G. The same holds true for L_{3}(2^{m}), as L_{3}(2) = L_{2}(7) has a Sylow 7-subgroup of type 7: 3.

The groups L_{2}(27) and L_{4}(3) have 3-rank at least equal to three, which is more than the 3-rank of G. The same holds true for U_{4}(2), excluding the possibilities U_{4}(2^{m}) and U_{5}(2^{m}).

Finally, one checks that |G| is always prime to 17, whence L_{2}(17) can not occur in G.

Proposition 2.2: A simple subgroup of ^{2}F_{4}(q^{2}) is either (1) L_{2}(2^{m}), m ≥ 2,

(2) U_{3}(2^{m}), m ≥ 2,

(3) S_{4}(2^{m}), m ≥ 2,

(4) ^{2}B_{2}(2^{2m+1}), m ≥ 1,

(5) ^{2}F_{4}(2^{2m+1}), m ≥ 1, or one of

(6) L_{2}(13), L_{2}(25), L_{3}(3), A_{6}, ^{2}F_{4}(2)′.

In the rest of this section, the cases of Proposition 2.2 are examined. Apart from L_{2}(13), prototypes of all these groups already occur over the smallest field in ^{2}F_{4}(2)′. No surprise arises, i.e., L_{2}(13) is not a subgroup of G. In what follows, heavy use will be made of the presentation of G with generators and relations (see for example [17]; note, however, that in the formula for the commutator [α_{2}(t), α_{3}(u)], the last factor with α_{11} is wrong !). The existence of subgroups of the types mentioned in Proposition 2.2 will be proved or disproved by explicit constructions inside G. For the notations α_{i}, U_{i}, see Section II of [17]. Define the subgroups H_{i} := H ∩ ⟨U_{i}, U_{i}^{r_{0}}⟩, i ∈ {2, 3, 7, …, 12}, of H, where r_{0} := (r_{a}r_{b})^{4} = (r_{b}r_{a})^{4}. For shortness, let 2A := [u_{1}], 2B := [u_{2}].

2.2 Some generic subgroups

First, some generic subgroups, which can easily be detected from the structure of ^{2}F_{4}(q^{2}), will be described. At the same time all direct products of isomorphic simple subgroups of G will be found. By the local analysis, the only candidates of simple groups K with K ×K ≤ G are those of types A_{1} and ^{2}B_{2}.

It is easily seen that ⟨U_{10}, r_{a}r_{0}⟩ and ⟨U_{3}, r_{a}⟩ are both isomorphic to A_{1}(q^{2}) and centralize each other. Indeed, all subgroups of type A_{1}(q_{0})×A_{1}(q_{0}) are contained in a conjugate of this generic subgroup S := ⟨U_{3}, U_{10}, r_{a}, r_{0}⟩. For assume A_{1}(q_{0}) < G with nontrivial centralizer. Then by local analysis, it is conjugate to an A_{1}(q_{0}) in C_{G}(⟨t_{2}⟩) ≅ Z_{q^{2}−1} × A_{1}(q^{2}). But this centralizer is a subgroup of the generic A_{1}(q^{2}) × A_{1}(q^{2}). By looking at elements of order q_{0} − 1 inside A_{1}(q_{0}) it is immediate that the centralizer of this subgroup is completely contained in S. (We have q_{0} > 2 since A_{1}(2) ≅ S_{3} is not simple.) Hence any direct product of A_{1}(q_{0}) with another simple group is already realized inside S. From the structure of the normalizer of elements t_{2} inside G one sees that the normalizer of such a direct product cannot contain any field automorphisms of one of the factors. Hence it is contained in the extension S: ⟨r_{b}r_{a}r_{b}⟩ ≅ A_{1}(q^{2}) ≀ 2. (Note that the centralizer is trivial.)

In a similar fashion we find R := ⟨U_{5}, r_{b}r_{0}⟩ × ⟨U_{1}, r_{b}⟩ ≅ ^{2}B_{2}(q^{2}) × ^{2}B_{2}(q^{2}). All direct products of subgroups ^{2}B_{2}(q_{0}) with themselves are contained in a conjugate of R by the same arguments as above. Moreover the element r_{a}r_{b}r_{a} normalizes R and ⟨R, r_{a}r_{b}r_{a}⟩ ≅ ^{2}B_{2}(q^{2})≀2 constitutes the full normalizer of R. Thus we have proved the following result:

Proposition 2.3: The normalizer of any direct product of isomorphic simple subgroups of G is conjugate to a subgroup of R: 2 or S: 2.

One other generic subgroup will be studied. Namely by adding U_{7} and U_{9} to S: 2 above we get a subgroup satisfying the relations of the symplectic group B_{2}(q^{2}). Thus S: 2 is not maximal in G. The normalizer of B_{2}(q^{2}) cannot contain any field automorphisms, since those would have to centralize an A_{6}. But the element r_{b} realizes the graph automorphism of the symplectic group, yielding ⟨S, U_{7}, U_{9}, r_{b}⟩ ≅ B_{2}(q^{2}): 2.

2.3 2B-pure subgroups of types A_{4} and S_{4}

In the further investigation, the possible 2B-pure A_{4}’s and S_{4}’s are needed. The alternating group A_{4} is the semidirect product of an elementary abelian 2-group V = {ι, σ_{1}, σ_{2}, σ_{3}} with a group ⟨τ⟩ of order three. Thus, after conjugation, it is contained in a maximal parabolic subgroup of G, and as 3 |̸ |P_{b}|, in P_{a}. All Levi complements L_{a} in P_{a} are conjugate, as are all elements of order three in L_{a}, so without loss we can assume τ = α_{3}(1)r_{a}. The Levi complement itself does not have a subgroup A_{4}, so V has to be contained in R_{u}(P_{a}), the unipotent radical of P_{a}. By easy calculations with the commutator relations one gets that three classes of 2B-involutions of P_{a} intersect R_{u}(P_{a}), with representatives α_{10}(1), α_{7}(1), α_{2}(1)α_{12}(1) (see also (3.3) of [17]).

First assume σ_{1} ∼ α_{10}(1), so σ_{1} = α_{10}(a)α_{11}(b)α_{12}(c). Then σ_{1}^{τ} = α_{10}(a)α_{11}(b + c) × α_{12}(b), but σ_{1}σ_{1}^{τ} = α_{11}(c)α_{12}(b + c) ∈ 2A, which is a contradiction to A_{4} being 2B-pure.

If σ_{1} ∼ α_{2}(1)α_{12}(1), then σ_{1} ∈ α_{2}(a)α_{5}(b)α_{6}(c)α_{8}(d) · N, N = U_{7}U_{9}U_{10}U_{11}U_{12}. Calculation modulo N yields

σ_{1}^{τ} ≡ α_{2}(a + b + c + d)α_{5}(a + c)α_{6}(a + b)α_{8}(a) ≡ σ_{2},
σ_{2}^{τ} ≡ α_{2}(d)α_{5}(c + d)α_{6}(b + d)α_{8}(a + b + c + d) ≡ σ_{3}.

Now as σ_{1}σ_{2} = σ_{3}, this forces c = b, d = a + b. But for σ_{1} = α_{2}(a)α_{5}(b)α_{6}(b)α_{8}(a + b) · N to be an involution, we arrive at a^{2} + ab + b^{2} = 0. So a^{3} = b^{3}, and as 3 |̸ (q^{2} − 1) it follows that a = b, hence a = b = 0. But then σ_{1} ≁ α_{2}(1)α_{12}(1).

Finally, if σ_{1} ∼ α_{7}(1), similar calculations give σ_{1} = α_{7}(a)α_{9}(b)α_{10}(a + b)α_{11}(c)α_{12}(d).

Proposition 2.4: (a) Any 2B-pure subgroup A_{4} of G is conjugate to one of ⟨α_{7}(a) × α_{9}(b)α_{10}(a + b)α_{11}(c)α_{12}(d), α_{3}(1)r_{a}⟩ for suitable a, b, c, d ∈ 𝔽_{q^{2}}, not both a and b equal to zero.

(b) There are exactly q^{2} classes of 2B-pure subgroups S_{4} of G, with representatives ⟨α_{9}(1)α_{10}(1)α_{12}(t), α_{3}(1), r_{a}⟩, t ∈ 𝔽_{q^{2}}. Each of these has centralizer C_{G}(S_{4}) = U_{10}.

Proof : Part (a) has already been proved. As S_{4} = V: S_{3}, it is also a 2-local subgroup, and the same type of arguments as for A_{4} can be applied. Moreover the centralizer of S_{3} = ⟨α_{3}(1), r_{a}⟩ is known to be equal to ⟨U_{10}, r_{b}r_{a}r_{b}r_{a}r_{b}r_{a}r_{b}⟩ by the local analysis, and so the statement (b) is deduced.

2.4 Rank-1 subgroups of types A_{1} and ^{2}B_{2}

As a first step towards the classification of the subgroups of Lie type in characteristic 2 we determine all rank-1 subgroups, that is, all subgroups of types A_{1}, ^{2}A_{2} and ^{2}B_{2} defined over fields of even order. This will be done by finding BN-pairs of rank 1 and characteristic 2 in G.

Each of A_{1}(q_{0}), ^{2}B_{2}(q_{0}) has an elementary abelian subgroup E of order q_{0} normalized by a subgroup F of order q_{0} − 1. Furthermore there is an involution ρ conjugate to the ones in E, inverting all elements in F, and with ρEρ ⊂ EF ρE. All such constellations in G will now be found.

Obviously EF has to be contained in a maximal parabolic subgroup. Elementary number theory then shows (q_{0} − 1)|(q^{4} − 1), i.e., q_{0} = 2^{m}, m|2(2n + 1), and EF is conjugate to a subgroup of P_{a}.

First assume m even, so we are in the case A_{1} (as ^{2}B_{2} is defined for odd powers of 2 only). The elements of order three in A_{1}(q_{0}) are rational, and involutions inverting an element of order three lie in 2B by the structure of SU_{3}(q^{2}): 2, the normalizer of a subgroup of order three. All subgroups A_{4} = ⟨σ, τ⟩ contained in EF must hence be 2B-pure and by Proposition 2.4(a), we may take σ ∈ U_{7}U_{9}U_{10}U_{11}U_{12}, τ = α_{3}(1)r_{a}. The ”structure identity” of L_{2}(q_{0}) stipulates the existence of an involution ρ with τ^{ρ} = τ^{−1}, (σρ)^{3} = ι. Any element inverting τ has the form ρ = z · r_{a}, z ∈ C_{G}(τ) ≅ SU_{3}(q^{2}). Now if z ∈ C_{G}(τ) ∩ P_{a}, then ⟨ρ, σ, τ⟩ ⊂ P_{a}, contradicting ⟨ρ, σ, τ⟩ ≅ L_{2}(4). Thus z is not contained in the Borel subgroup C_{G}(τ) ∩ P_{a} of SU_{3}(q^{2}), and as SU_{3}(q^{2}) is a BN-pair of rank one, z = u_{1}kr_{b}r_{a}r_{b}r_{a}r_{b}r_{a}r_{b}u_{2}, with u_{1}, u_{2} ∈ C_{G}(τ) ∩ U, k ∈ C_{G}(τ) ∩ P_{a} with k^{q^{4}−1} = ι. Before we proceed, a lemma is needed:

Lemma 2.5: Let u ∈ U_{7}U_{9}U_{10}U_{11}U_{12} ∪ U_{8}U_{9}U_{10}U_{11}U_{12}.

(a) (ur_{0})^{3} = ι iff u = α_{7}(1)α_{9}(1).

(b) (ur_{0})^{5} = ι iff u = α_{8}(1)α_{11}(1).

Let moreover r ∈ W, r^{2} = ι, r ≠ r_{0}. Then up to conjugation by elements of W (c) (ur)^{3} = ι iff r = r_{b}r_{a}r_{b} and u = α_{7}(1)α_{9}(a)α_{10}(a^{θ})α_{12}(b) with a, b ∈ 𝔽_{q^{2}},

(d) (ur)^{5} = ι iff r = r_{a}r_{b}r_{a} and u = α_{8}(1)α_{10}(a^{θ})α_{11}(a)α_{12}(b) with a, b ∈ 𝔽_{q^{2}}.

Proof : (a) We write the required identity in the form (ur_{0})^{2} = r_{0}u^{−1}, so that the righthand side is the Bruhat normal form of the lefthand side (∗). Considering first u = α_{7}(a)α_{9}(b)α_{10}(c)α_{11}(d)α_{12}(e) we get

(∗) = ur_{b}r_{a}r_{b}r_{a}r_{b}r_{a}α_{10}(a)α_{3}(b)α_{9}(c)α_{12}(d)α_{8}(e)r_{a}r_{b}r_{a}r_{b}r_{a}r_{b}.

As a first case assume b = 0. Then

(∗) = ur_{b}r_{a}r_{b}r_{a}r_{b}α_{10}(a)α_{7}(c)α_{11}(d)α_{2}(e)r_{b}r_{a}r_{b}r_{a}r_{b}.

Now if also e = 0, then the Weyl element in the normal form of the lefthand side can not be equal to r_{0}, which is a contradiction. Hence we have e ≠ 0, and so

(∗) = ur_{b}r_{a}r_{b}r_{a}α_{9}(a)α_{3}(c)α_{8}(d)α_{2}(e^{−1})hr_{b}α_{2}(e^{−1})r_{a}r_{b}r_{a}r_{b}.

If c ≠ 0 then from the commutator of α_{3}(c), α_{2}(e^{−1}) in the middle we get a factor α_{6}(ce^{−1}), which when shifted to the left yields α_{4}(ce^{−1}) on the left. But this is impossible if the righthand side is to be the Bruhat normal form of the left. By similar reasoning, if a ≠ 0, then by shifting α_{9}(a) from the middle to the right we obtain a factor α_{3}(a) on the right, which is incompatible with the righthand side. Finally assume d ≠ 0. Then the commutator of α_{8}(d), α_{2}(e^{−1}) in the middle yields α_{10}(de^{−1}), and this is conjugated to α_{3}(de^{−1}) when shifted to the left, again in contradiction to the required normal form. But if a = c = d = 0, then it is immediate that the Weyl element of the lefthand side has odd length, while r_{0} on the righthand side has even length.

So we must have b ≠ 0. Then

(∗) = ur_{b}r_{a}r_{b}r_{a}r_{b}α_{10}(a)α_{3}(b^{−1})h_{3}r_{a}α_{3}(b^{−1})α_{7}(c)α_{11}(d)α_{2}(e)r_{b}r_{a}r_{b}r_{a}r_{b} =
uα_{9}(b^{−1})r_{b}r_{a}r_{b}r_{a}r_{b}r_{a}α_{10}(a)h′_{3}α_{3}(b^{−1})α_{7}(c)α_{11}(d)α_{2}(e)r_{b}r_{a}r_{b}r_{a}r_{b}.

If α_{2}(e) is shifted to the left of this term, a factor α_{8}(e) is obtained, forcing e = 0, because α_{8} does not occur in u. But also c = 0, since otherwise a factor α_{3}(c) on the left would result from shifting α_{7}(c) to the left. If d ≠ 0 we get

(∗) = uα_{9}(b^{−1})r_{b}r_{a}r_{b}r_{a}r_{b}r_{a}r_{b}r_{a}α_{7}(a)h′′_{3}α_{2}(d)α_{11}(db^{−1})r_{b}r_{a}r_{b}α_{9}(b^{−1}) =
uα_{9}(b^{−1})r_{a}r_{b}r_{a}r_{b}r_{a}r_{b}r_{a}α_{3}(a)h′′′_{3}α_{2}(d^{−1})h_{2}r_{b}α_{2}(d^{−1})α_{8}(db^{−1})r_{a}r_{b}α_{9}(b^{−1}),

but now a superfluous factor α_{2}(d^{−1}) occurs on the left, proving d = 0. So finally

(∗) = α_{7}(a)α_{9}(b)α_{9}(b^{−1})r_{b}r_{a}r_{b}r_{a}r_{b}r_{a}r_{b}r_{a}α_{7}(a)h′_{3}r_{b}r_{a}r_{b}α_{9}(b^{−1}) =
α_{7}(a)α_{9}(b)α_{9}(b^{−1})α_{7}(a^{−1})r_{0}h′′_{3}α_{7}(a^{−1})h′′′_{3}α_{9}(^{−1}),

and hence b = b^{−1}, a = a^{−1}, i.e., a = b = 1, and the first half of part (a) is proved. The calculations for u ∈ U_{8}U_{9}U_{10}U_{11}U_{12} are easier. Namely we can assume from the beginning that the factor from U_{8} is nontrivial, since otherwise we would be in the first case.

The proof of part (b) is very similar and hence omitted.

(c) Here we are looking for involutions u, r such that they generate an S_{3}. Hence, as already observed above, both must lie in the class 2B. This leaves four possibilities for r. If r = r_{a} then ⟨u, r_{a}⟩ is a 2-group, and so no solution arises. If r = r_{b}r_{a}r_{b} and u ∈ U_{8}U_{9}U_{10}U_{11}U_{12}, again no solution is obtained. So let u = α_{7}(a)α_{9}(b)α_{10}(c)α_{11}(d)α_{12}(e), then

(ur)^{2} = α_{7}(a)α_{9}(b)α_{10}(c)α_{11}(d)α_{12}(e)r_{b}α_{3}(a^{−1})hr_{a}α_{3}(a^{−1})α_{10}(b)α_{7}(c)α_{2}(d)α_{11}(e)r_{b}

(assuming a ≠ 0 since otherwise we would return to the first case). If d ≠ 0, a factor α_{6}(a^{−1}d) would result from the commutator of α_{3}(a^{−1}), α_{2}(d), which does not cancel. Thus d = 0 and

(ur)^{2} = α_{7}(a)α_{9}(b)α_{10}(c)α_{12}(e)α_{7}(a^{−1})r_{b}hr_{a}r_{b}α_{7}(a^{−1})α_{9}(b)α_{3}(c)α_{8}(e),

and the α_{7}-part on the left cancels only if a = 1. In this case,

(ur)^{2} = r_{b}r_{a}r_{b}α_{9}(b)α_{3}(c)α_{8}(e)α_{7}(1)α_{9}(b)α_{3}(c)α_{8}(e) =
r_{b}r_{a}r_{b}α_{7}(1)α_{9}(c^{2θ})α_{10}(c)α_{12}(e),

and comparison with (ur)^{−1} finally shows c^{2θ} = b, leading to the case in the Lemma.

The other cases are treated in the same way, and yield no further solutions. Also part (d) of the Lemma is proved along the same lines, hence the proof will be omitted, too.

Using the lemma we conclude Proposition 2.6: For each m|(2n+1) there exists exactly one conjugacy class of subgroups L_{2}(2^{2m}) in G with normalizer L_{2}(2^{2m}): 2.

Proof : The element τ and the shape of σ are left invariant under conjugation by H_{10} and C_{G}(τ) ∩ P_{a} =: B_{SU}. We now examine the different possibilities for ρ. If k ∈ H then ρ = u_{1}hr_{0}u_{2}, which is conjugate to u_{3}r_{0} via B_{SU}. This is an involution iff u_{3} = 1 by the uniqueness of the Bruhat decomposition. Using Lemma 2.5(a), (σρ)^{3} = ι implies σ = α_{7}(1)α_{9}(1).

If k ∉ H then k = α_{3}(r)hr_{a}α_{3}(s) with [k, τ] = ι. After conjugation with k′ ∈ B_{SU} we may assume σ ∈ U_{7}U_{9}U_{11}U_{12}. But then arguments as above and use of Lemma 2.5(a) lead to a contradiction. So L_{2}(4) = ⟨α_{3}(1)r_{a}, r_{0}, α_{7}(1)α_{9}(1)⟩ is unique up to conjugation.

To construct L_{2}(2^{2m}) ≥ L_{2}(4), an element of order dividing q^{2} −1 in C_{G}(τ)∩P_{a} inverted by r_{0} has to be found. Elements which satisfy these requirements are easily seen to belong to H_{10} ≅ Z_{q^{2}−1}, so there is one cyclic subgroup of the correct type for each divisor m of 2n + 1. This yields the uniqueness of L_{2}(2^{2m}). By the local analysis C_{G}(L_{2}(2^{2m})) = 1, and moreover elements of the torus of order q^{4} − 1 are conjugate to at most four of their powers, hence field automorphisms of order at most two can be contained in N_{G}(L_{2}(2^{2m})). It is well known that B_{2}(q_{0}) contains a (maximal) subgroup L_{2}(q_{0}^{2}): 2, hence from the generic B_{2}(q^{2}) in 2.2 we can derive the existence.

From now on we may assume m odd, m|(2n + 1), m ≥ 3 (as L_{2}(2) and ^{2}B_{2}(2) are not simple). Thus EF already lies in the Borel subgroup B. By (3.5) in [17], the maximal elementary abelian subgroups of U are V_{1} = U_{2}U_{7}U_{10}U_{11}U_{12}, V_{2} = U_{3}U_{8}U_{9}U_{10}U_{12}, V_{3} = U_{7}U_{9}U_{10}U_{11}U_{12} and V_{4} = U_{8}U_{9}U_{10}U_{11}U_{12}. Hence E, being an elementary abelian subgroup of U, is contained in one of those. Moreover V_{1} = V_{4}^{r_{a}} and V_{2} = V_{3}^{r_{b}}, so without loss we may even assume E ≤ V_{3} or E ≤ V_{4}. After conjugation with u ∈ U we also have F ≤ H.

If F contains regular elements of H, then the only involutions ρ of G inverting F are r_{0}h, h ∈ H, by the definition of regular elements (see also the list of normalizers of semisimple elements in Section 1.3). Conjugating with elements of H we get ρ = r_{0}. Now due to E ≤ V_{3} or E ≤ V_{4}, Lemma 2.5(a) can be applied, showing that only σ = α_{7}(1)α_{9}(1) is a candidate for the structure identity of L_{2}(q_{0}). But this means L_{2}(q_{0}) ≤ ⟨U_{7}U_{9}, H, r_{0}⟩, and the latter is a subgroup of the standard B_{2}(q^{2}). Similarly, testing for the structure identity of ^{2}B_{2}(q_{0})

[18] with ρ = r_{0} we get σ = α_{8}(1)α_{11}(1) by Lemma 2.5(b), again leading to a subgroup of B_{2}(q^{2}). Actually the two subgroups just derived have generic descriptions: The Suzuki group lies in the centralizer of the graph automorphism of the B_{2}(q^{2}), and the A_{1}(q^{2}) is conjugate to the diagonal subgroup of A_{1}(q^{2}) ≀ 2 < B_{2}(q^{2}).

Thus we are led to consider the nonregular elements h of H. They are conjugate to either t_{1} or t_{2} [16], and so by Proposition 1.3 they are inverted by elements of the form z · ρ, with z ∈ C_{G}(h) and ρ an involution in the Weyl group. But all involutions of C_{G}(h) are conjugate inside this group, so without loss either z = 1 or z = r_{0}ρ ∈ C_{G}(h). The second case is implicit in the preceeding calculations, hence we are left with the possibilities discussed in Lemma 2.5(c) and (d). Assume we are in case (c), so u = α_{7}(1)α_{9}(a)α_{10}(a^{θ})α_{12}(b), and ρ = r_{b}r_{a}r_{b}. The nonregular elements h of H inverted by ρ then lie in H_{7}. Thus U_{9} centralizes ⟨r, h, u⟩, and a possible A_{1}(q_{0}) of this type has a nontrivial centralizer. Similarly, in case (d), the possible ^{2}B_{2}(q_{0}) are centralized by U_{11}. By the arguments in Section 2.2 they are easily seen to lie inside subgroups A_{1}(q^{2}) ≀ 2 or ^{2}B_{2}(q^{2}) ≀ 2. Also their normalizers are subgroups of those generic groups, because elements of order q_{0} −1 are conjugate to such of type t_{1} or t_{2}, and hence to only two of their primitive powers (excluding the possibility of field automorphisms).

Proposition 2.7: The normalizer of any simple subgroup A_{1}(q_{0}) or ^{2}B_{2}(q_{0}), q_{0} = 2^{2m+1}, is conjugate to a subgroup of the generic groups S: 2 or R: 2 of Proposition 2.3.

Note that we have found two types of subgroups ^{2}B_{2}(q_{0}) in G, one containing involutions from class 2A, the other from class 2B.

2.5 Rank-1 subgroups of type ^{2}A_{2}

We still have to consider the rank-1 groups K = ^{2}A_{2}(q_{0}). First, the existence of U_{3}(q^{2}) < G is shown. We start from the definition of G as the centralizer of a graph automorphism in F_{4}(q^{2}). Consider the subgroups K := (SU_{3}(q^{2})∘SU_{3}(q^{2})): 2, the normalizer of a certain class of subgroups of order three in F_{4}(q^{2}) ([15], Part 3). The graph automorphism σ centralizes some such subgroups, leading to the centralizer of an element τ of order three, SU_{3}(q^{2}), in G. But the product of σ with one of the outer involutions of K inverts τ and it is immediate that it has to centralize a subgroup U_{3}(q^{2}) of K. Looking at F_{4}(2) one sees that the outer involution of F_{4}(q^{2}): 2 centralizing the U_{3}(q^{2}) is conjugate to σ, and hence U_{3}(q^{2}) is contained in G.

To determine the number of different classes and the normalizer, again a construction is needed. We start from the centralizer of an element of order q_{0} + 1 inside K := U_{3}(q_{0}), which is isomorphic to Z_{(q_{0}+1)/3} × A_{1}(q_{0}). Without loss the A_{1} can be assumed to lie inside ⟨U_{10}, r_{0}r_{a}⟩ by the local analysis (and hence q_{0} = 2^{m}, m|(2n + 1)), with α_{10}(1) one of the elements of order two.

Now U_{3}(2) ≅ 3^{2}: Q_{8}, and there is just one class of such groups in G by the structure of the normalizer of T_{8} (Proposition 1.2). Hence ⟨α_{10}(1), r_{0}r_{a}⟩ ≅ S_{3} in A_{1}(q_{0}) may be extended to a unique U_{3}(2) modulo conjugation by C_{G}(S_{3}) = ⟨U_{3}, r_{a}⟩. Calculation shows that one can choose U_{3}(2) = ⟨α_{10}(1), r_{0}r_{a}, σ⟩, with σ = α_{2}(1)α_{8}(1)α_{7}(1)α_{9}(1)α_{11}(1). We have P_{a} ∩ A_{1}(q_{0}) = ⟨α_{10}(1), h_{10}⟩, with h_{10} ∈ H_{10} an element of order q_{0} − 1. Consequently, there is a unique candidate for U_{3}(q_{0}), namely ⟨α_{10}(1), r_{0}r_{a}, σ, h_{10}⟩. We get

U_{U} = U ∩K = {(a, b, c) | a, b, c ∈ 𝔽_{q_{0}}, (a, b, c)·(a′, b′, c′) = (a+a′, b+b′, c+c′+aa′+bb′+a′b)},

with (a, b, c) := α_{2}(a)α_{5}(b)α_{6}(b)α_{8}(a + b)α_{7}(a^{2θ} + b^{2θ})α_{9}(a^{2θ})α_{10}(c)α_{11}(a^{2θ}b + a^{2θ+1} + b^{2θ+1})α_{12}(b^{2θ+1}). Here θ is an automorphism of 𝔽_{q^{2}} with 2θ^{2} = 1 [16]. With U_{SU} = U ∩ C_{G}(α_{3}(1)r_{a}), the Sylow 2-subgroup of SU_{3}(q^{2}) is given by

U_{SU} = {[a, b, c] | a, b, c ∈ 𝔽_{q^{2}}, [a, b, c] · [a′, b′, c′] = [a + a′, b + b′, c + c′ + aa′ + bb′ + ab′]},

where [a, b, c] := α_{2}(a)α_{5}(b)α_{6}(a + b)α_{8}(a)α_{7}(a^{2θ} + b^{2θ})α_{9}(b^{2θ})α_{10}(c)α_{11}(a^{2θ}b + ab^{2θ} + b^{2θ+1})α_{12}(b^{2θ+1}). An isomorphism between the two Sylow 2-subgroups may be defined by (a, b, c) ↦ [b, a, c].

Now r_{a} centralizes ⟨U_{10}, r_{0}r_{a}⟩ and normalizes U_{U} (via (a, b, c)^{r_{a}} = (a + b, b, c + a^{2} + ab + b^{2})), hence it realizes the graph (field) automorphism of K. Moreover the centralizer in ⟨U_{3}, r_{a}⟩ of α_{3}(1)r_{a} (a subgroup of order q^{2}+1) normalizes K, so that N_{G}(K) ≥ P GU_{3}(q^{2}): 2, and equality holds since any further field automorphism inside the normalizer would contradict the fact that elements in T_{9} < P GU_{3}(q^{2}) are conjugate to only six of their powers.

Proposition 2.8: For each m|(2n + 1), m ≥ 3, there exists one class of U_{3}(2^{m}) in G with normalizer P GU_{3}(2^{m}): 2.

2.6 Subgroups of type L_{3}(3)

We will show that there is only one class of L_{3}(3), which is hence already contained in 2F_{4}(2)′ and can not be maximal for q^{2} > 2.

Proposition 2.9: There is just one class of subgroups L_{3}(3) in G, with normalizer L_{3}(3): 2.

Proof : The group K = L_{3}(3) has just one class of involutions, and as their centralizer order is divisible by three, they fuse into 2B = [u_{2}]. Let S be a maximal subgroup of K isomorphic to S_{4}. Then it has to be 2B-pure, so after conjugation it is equal to ⟨α_{3}(1), r_{a}, α_{9}(1)α_{10}(1)α_{12}(t)⟩ by Proposition 2.4(b). In K, an S_{3} ≤ S_{4} is centralized by an element σ of order three. Moreover, K is generated by S_{4} and σ. Now C_{G}(⟨α_{3}(1), r_{a}⟩) = ⟨U_{10}, r_{0}r_{a}⟩ ≅ L_{2}(q^{2}). Elements of order three in this group have the form σ = α_{10}(s)h_{10}r_{0}r_{a}, s ∈ 𝔽_{q^{2}}^{×}, h_{10} ∈ H_{10}, modulo conjugation by u ∈ U_{10} = C_{G}(S_{4}).

Define τ := α_{9}(1)α_{10}(1)α_{12}(t) · σ. Then by easy calculations with matrices one sees that inside K we have to have τ^{4} = α_{3}(1). This imposes conditions on the unknowns t and s.

To simplify the calculations, define τ̃ to be the conjugate of τ by r_{a}r_{b}r_{a}. Then the fourth power of τ̃ = α_{9}(1)α_{3}(1 + s)α_{8}(t)h_{3}r_{a} has to be equal to α_{10}(1). Both elements lie in P_{a}; modulo the normal subgroup R_{u}(P_{a}) we get that α_{3}(1 + s)h_{3}r_{a} has order two or four in the Levi complement ⟨U_{3}, r_{a}⟩ ≅ P_{a}/R_{u}(P_{a}). But this forces s = 1 by the uniqueness of the Bruhat decomposition. Then τ̃^{2} = α_{9}(1)α_{8}(t)h_{3}α_{7}(1)α_{2}(t)h_{3}^{−1}, and τ̃^{4} = α_{10}(1) yields h_{3} = ι and t = 1. So the uniqueness is shown. By the knowledge of the maximal subgroups of ^{2}F_{4}(2), the existence is clear as well. The normalizer L_{3}(3): 2 is already realized inside 2F_{4}(2).

2.7 Subgroups of type A_{6}

Proposition 2.10: There is just one class of subgroups A_{6} in G with normalizer A_{6}^{.}2^{2} = Aut(A_{6}).

Before we can prove this, we need

Lemma 2.11: Let x be an involution with α_{3}(1)^{x} = α_{9}(1)α_{10}(1)α_{12}(t). Then x = r_{b}α_{1}(1)α_{7}(a)α_{6}(t + 1)α_{10}(b)α_{12}(c)α_{11}(c + (t + 1)^{2θ+1})α_{9}(a)α_{5}(t + 1)α_{4}(1)α_{8}(1)r_{a}r_{b}, with a, b, c ∈ 𝔽_{q^{2}}.

Proof : Conjugate the required equality by r_{b}r_{a}r_{b}. Then we look for y with α_{10}(1)^{y} = α_{3}(1)α_{9}(1)α_{8}(t). Now y_{0} = r_{b}r_{a}r_{b}α_{5}(t + 1)α_{6}(1)α_{11}(1) is a possible solution, so the general

y has the shape y = z · y_{0}, z ∈ C_{G}(α_{10}(1)) = U′H_{10} ∪^{·} U′H_{10}r_{a}U_{3}, U′ = U_{2}U_{3}U_{5} ⋯ U_{12}.

If z = u_{1}hr_{a}u_{2}, then y^{2} ∈ UHr_{0}U, and hence can not be equal to ι. If on the other hand z = uh, then y^{2} = ι implies, after conjugating back by r_{b}r_{a}r_{b} the shape of x given in the lemma. Proof of Prop. 2.10: Let K be a subgroup of G isomorphic to the alternating group A_{6}. Then K contains a maximal, 2B-pure subgroup S_{4}. (2B-pure since K has only one class of involutions, normalizing subgroups of order three.) By Proposition 2.4(b) after conjugation we can assume S_{4} = ⟨α_{3}(1), r_{a}, α_{9}(1)α_{10}(1)α_{12}(t)⟩. The two involutions α_{3}(1), α_{9}(1)α_{10}(1)α_{12}(t) fuse into a single class of A_{6}. One can check that they even have to be conjugate by an involution x. Thus x is as in Lemma 2.11. Moreover we may conjugate by u ∈ U_{10} = C_{G}(S_{4}) and hence assume without loss a = 0.

Now the element r_{a} · α_{3}(1) · x has order four in K. This condition forces t = 1, and then c = 0, d = 1, so that the element x finally is uniquely determined, proving the first part of the assertion. This A_{6} has to be contained in the generic ^{2}F_{4}(2), and the latter already contains the full automorphism group of A_{6} as normalizer. As the centralizer is trivial (see local analysis), the proposition follows.

2.8 Subgroups of type L_{2}(25)

Proposition 2.12: There is just one class of subgroups L_{2}(25) in G, with normalizer L_{2}(25)^{.}2_{3}.

Proof : Let K be a subgroup of G isomorphic to L_{2}(25). First we construct one of the maximal subgroups S_{5} of K. By Proposition 2.6, after conjugation this is equal to ⟨α_{3}(1), r_{a}, α_{7}(1)α_{9}(1), r_{0}⟩. As K has just one class of involutions, the elements α_{3}(1) and α_{9}(1)α_{10}(1) of order two have to be conjugate in K. Calculations in K show that they are even conjugate via a third involution x. Involutions with that property were constructed in Lemma 2.11: x = r_{b}α_{1}(1)α_{7}(a)α_{6}(1)α_{10}(b)α_{12}(c)α_{11}(c + 1)α_{9}(a)α_{5}(1)α_{4}(1)α_{8}(1)r_{a}r_{b}. Calculations in K now show that the element of S_{5} represented by y := α_{7}(1)α_{9}(1)r_{a}r_{0}α_{7}(1) × α_{9}(1)α_{3}(1) has the property that y·x has order two. Also, α_{7}(1)·α_{10}(1)·x should have order three. Together these two conditions yield, upon calculation, b = c = 1, a ∈ {0, 1}. The two possibilities for x, according to a = 0 or a = 1, are conjugate via α_{3}(1). Hence K is uniquely determined in G. The centralizer of K in G is trivial. The outer automorphism group of K has the structure 2^{2}, but from [5] one sees that neither 2_{1} nor 2_{2} can be realized in G, for in the first case there are elements of order 24, and in the second we get involutions τ with 15 | |C_{G}(τ)|. But in ^{2}F_{4}(2), the normalizer is already equal to L_{2}(25)^{.}2_{3}, which proves the proposition.

2.9 Subgroups of type ^{2}F_{4}(2)′

Proposition 2.13 : There is just one class of subgroups ^{2}F_{4}(2)′ in G, with normalizer 2F_{4}(2).

Proof : Assume K = ^{2}F_{4}(2)′ < G. Then there is a subgroup 3^{2}: 4 < A_{6} < K, where the A_{6} can be assumed to be the standard one of G by Proposition 2.10, and it has one class of subgroups 3^{2}: 4. In K, the normalizer of 3^{2} is equal to (q^{2} + 1)^{2}: GL_{2}(3) by Proposition 1.2. An element of order 2 in 3^{2}: 4 inverts all elements of 3^{2} and hence of (q^{2} + 1)^{2}, so its centralizer in N_{G}(3^{2}) is just a GL_{2}(3). The normalizer in K of 3^{2} is equal to 3^{2}: GL_{2}(3), and the GL_{2}(3) is now uniquely determined to be the centralizer of an inverting involution in 3^{2}: 4. Thus K is also unique as the compositum of the A_{6} and the 3^{2}: GL_{2}(3). It must be contained in the generic ^{2}F_{4}(2) (obtained by restricting the field of definition), and by the local part has trivial centralizer, so the assertion about the normalizer holds.

2.10 Subgroups of type L_{2}(13)

The only ‘surprise’ would have been the existence of subgroups L_{2}(13) in G for 3|(2n+1). But the following proposition shows that no such subgroups arise.

Proposition 2.14 : There is no subgroup L_{2}(13) in G.

Proof : We attempt to construct K = L_{2}(13) in G. The group K contains a subgroup D_{12} ≅ S_{3}×Z_{2} (the normalizer of a subgroup of order six), which may be assumed to be ⟨α_{3}(1), r_{a}⟩× ⟨α_{10}(1)⟩ in G, as we already know. The two involutions α_{3}(1) and α_{10}(1) must be conjugate in K, as the latter has only one class of involutions. The two involutions generate a Sylow 2subgroup V_{4} of K, and its normalizer is isomorphic to A_{4}. Consequently there has to exist an element ρ of order three in G with α_{10}(1)^{ρ} = α_{3}(1), and α_{3}(1)^{ρ} = α_{3}(1)α_{10}(1). Calculations show that for ρ to satisfy the first condition, we get ρ = α_{7}(1)α_{8}(a)α_{9}(b)α_{10}(b^{θ})α_{12}(c)r_{b}r_{a}r_{b}, a, b, c ∈ 𝔽_{q^{2}}. But then α_{3}(1)^{ρ} = α_{10}(1)α_{9}(1)α_{3}(1), violating the second condition. Hence the assumption L_{2}(13) < G leads to a contradiction and the proposition is proved.

2.11 Subgroups of type B_{2}(q_{0}), q_{0} ≥ 4

We use the fact that there exists exactly one class of subgroups S_{6} ≅ B_{2}(2) (by Proposition 2.10). The group K := B_{2}(q_{0}) contains a subgroup L_{2}(q_{0}^{2}), so by Proposition 2.6, q_{0} = 2^{m} with m|(2n + 1). Moreover K contains a unique class of B_{2}(2), and the latter has a unique class of subgroups Z_{3} × Z_{3}. In B_{2}(q_{0}) (resp. in G) this lies in exactly one maximal torus T_{0} (resp. T = T_{8}) of order (q_{0} + 1)^{2}, (q^{2} + 1)^{2} respectively. Thus the relevant group K contains the subgroup B_{2}(2) and the intersection T ∩ K, but these two already generate K. Consequently, there is just one class of groups B_{2}(q_{0}), which is therefore always contained in the generic B_{2}(q^{2}) exhibited above. Its normalizer is investigated as before, and found to contain, apart from K itself, only the graph automorphism. So all normalizers of groups B_{2}(q_{0}) are conjugate to subgroups of the generic B_{2}(q^{2}): 2.

2.12 Subgroups of type ^{2}F_{4}(q_{0}), q_{0} ≥ 8

By definition, q_{0} is an odd power of 2. We use the existence of only one class of subgroups 2F_{4}(2) in G. Now K = ^{2}F_{4}(q_{0}) contains a unique class of ^{2}F_{4}(2) and of subgroups Z_{3} × Z_{3}. By the same argument as above we get K to contain ^{2}F_{4}(2) and the intersection of the torus T_{8} with K, which together generate K. Thus K is conjugate to the generic subgroup 2F_{4}(q_{0}), defined over a subfield of 𝔽_{q^{2}}. The normalizer is trivial in every case (any field automorphism would centralize a ^{2}F_{4}(2)), and so K is maximal iff (2n + 1)/(2m + 1) ∈ ℙ, with q_{0} = 2^{2m+1}.

3 Conclusion

By the results of the first two sections, the maximal subgroups are among the groups of Propositions 1.1, 1.2, 1.3, 2.3, 2.6, 2.8, 2.9, 2.10, 2.12, 2.13 or those discussed in subsections 2.11 and 2.12. Those of 2.6 were shown to lie inside a B_{2}(q^{2}), and the groups of 2.9, 2.10 and 2.12 are already contained in ^{2}F_{4}(2) and hence cannot be maximal. The group S: 2 ≅ A_{1}(q^{2}) ≀ 2 also lies inside a B_{2}(q^{2}). From the groups in Proposition 1.3, (1), (4) and (5) are conjugate to subgroups of ^{2}B_{2}(q^{2}) ≀ 2 = R: 2. Furthermore, (2) and (3) occur inside S: 2 and (6) lies inside B_{2}(q^{2}): 2. For q^{2} = 8, N_{G}(T_{6}) ≅ (5 × 5): [96] is contained in 2F_{4}(2). For larger q^{2}, none of the cases (7) up to (10) of Proposition 1.3 are nonmaximal for order reasons.

From this, the Main Theorem follows immediately. As for the Corollary, it is clear from the description of the maximal subgroups with generators and relations that all maximal subgroups of G extend to Aut(G). Now assume that L < Aut(G) is a novelty, i.e., a maximal subgroup of Aut(G) with L_{0} = G ∩ L nonmaximal in G. Let M be a characteristically simple normal subgroup of L_{0}. If L_{0} is 2-local, then we get a contradiction as in the proof of Proposition 2.6 in [8]. In the other local cases, by Lemma 1.5.4 of [8] we have L_{0} = N_{G}(M), hence L_{0} is among the groups listed in Propositions 1.2 and 1.3. But all of these extend to Aut(G) inside the extension of a maximal subgroup of G, which shows that they can not give rise to novelties in Aut(G). The case of nonlocal L_{0} is also easily handled. Namely because all maximal subgroups of G extend to Aut(G), we are in case (e) of Proposition 2.3 in [21]. By the known lists of maximal subgroups of the simple groups involved in G (see [18,7]), all direct products of isomorphic nonabelian simple subgroups in G are ‘lone’ (in the terminology of [21]) inside all maximal subgroups of G which contain them. Hence no novelties can arise.

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