Diameter of classical groups generated by transvections

Let $G$ be a finite classical group generated by transvections, i.e., one of $\operatorname{SL}_n(q)$, $\operatorname{SU}_n(q)$, $\operatorname{Sp}_{2n}(q)$, or $\operatorname{O}^\pm_{2n}(q)$ ($q$ even), and let $X$ be a generating set for $G$ containing at least one transvection. Building on work of Garonzi, Halasi, and Somlai, we prove that the diameter of the Cayley graph $\operatorname{Cay}(G, X)$ is bounded by $(n \log q)^C$ for some constant $C$. This confirms Babai's conjecture on the diameter of finite simple groups in the case of generating sets containing a transvection. By combining this with a result of the author and Jezernik it follows that if $G$ is one of $\operatorname{SL}_n(q)$, $\operatorname{SU}_n(q)$, $\operatorname{Sp}_{2n}(q)$ and $X$ contains three random generators then with high probability the diameter $\operatorname{Cay}(G, X)$ is bounded by $n^{O(\log q)}$. This confirms Babai's conjecture for non-orthogonal classical simple groups over small fields and three random generators.


Introduction
This paper represents another step in the direction of the long-standing conjecture of Babai [BS92, Conjecture 1.7] on the diameter of finite simple groups with respect to arbitrary generating sets.Given g ∈ G = X , let ℓ X (g) denote the minimal word length of g with respect to X ∪ X −1 , and for S ⊆ G let ℓ X (S) = sup g∈S ℓ X (g).Babai's conjecture predicts that ℓ X (G) ≤ (log |G|) O(1) for any (nonabelian) finite simple group G and generating set X. We call ℓ X (G) the diameter of G with respect to X; it is equivalent to the diameter of the undirected Cayley graph Cay(G, X).
The past two decades have seen a slew of progress on Babai's conjecture.In particular, following the seminal work of Helfgott [Hel08], Breuillard-Green-Tao [BGT11], and Pyber-Szabó [PS16], the conjecture is now completely resolved in bounded Lie rank.In the case of alternating groups A n , while Babai's conjecture predicts that the diameter should be bounded by a polynomial in n, a result of Helfgott and Seress [HS14,Hel19] establishes a quasipolynomial bound.Recently Bajpai, Dona, and Helfgott [BDH21] proved a bound for classical Chevalley groups of large characteristic of the form (log |G|) O(r 5 ) , where r is the rank of G.
The author is supported by the Royal Society.
Much stronger results are known about diameter with respect to typical generators (as opposed to worst-case generators, as in Babai's conjecture).In this formulation we assume the generators are chosen uniformly at random and we are interested in results that fail with asymptotically negligible probability.Polynomial (in n) bounds for the diameter of A n with respect to typical generators were established in [BH05,SP12,HSZ15].More recently, it was proved in [EJ22] that if G = Cl n (q) is a classical group with defining module F n q (so log |G| ≈ cn 2 log q) then ℓ X (G) ≤ q 2 n O(1) with high probability provided that X includes at least q C random generators.
Pyber has proposed 1 that Babai's conjecture should be broken down into three parts.Since the conjecture is resolved in bounded rank, we may assume that G = A n or G = Cl n (q), q = p f , that is, one of the groups SL n (q), SU n ( √ q), Sp n (q), Ω ε n (q).The degree of an element g ∈ G is the size of its support if G = A n or the rank of g − 1 if G = Cl n (q).Now let X be an arbitrary generating set for G. Pyber's programme consists of the following subproblems: (1) Find a nontrivial element g ∈ G of polylogarithmic length over X whose degree is at most (1 − ǫ)n for some constant ǫ > 0.
(2) Assuming X includes an element of degree at most (1 − ǫ)n as above, find a nontrivial element g ∈ G of polylogarithmic length over X whose degree is minimal in G.
(3) Assuming X includes an element of minimal degree, show that every element of G has polylogarithmic length over X.Part (1) is likely the most difficult.Part (2) has been solved for G = A n for ǫ = 0.67 in [BBS04] and more recently for ǫ = 0.37 in [BGH + 14] (doing the same for an arbitrary constant ǫ > 0 is a fascinating open problem).Part (3) is actually trivial for G = A n , since the number of 3-cycles in A n is less than n 3 , but for classical groups it is not at all trivial and in fact the focus of this paper.
Excluding the case of Ω ǫ n (q) (which we leave to future work), the elements of minimal degree in a classical group G = Cl n (q) are the transvections.By recent work of Halasi and Garonzi-Halasi-Somlai [Hal21,GHS22] we have a suitable bound for ℓ X (G) for SL n (q), SU n ( √ q), Sp n (q) and any generating set X containing a transvection, excluding all cases of characteristic 2 and a few cases of characteristic 3.
Theorem 1.1 (Garonzi-Halasi-Somlai, [GHS22, Theorem 1.5]).Let G be one of the non-orthogonal classical groups SL n (q), SU n ( √ q) (q square), Sp n (q) (n even) with defining module F n q .Let X be a generating set for G containing a transvection.Then ℓ X (G) ≤ (n log q) O(1) provided that • q is odd, In this paper we deal with the omissions above.It is also natural to include the (imperfect) orthogonal groups in characteristic 2. Our main theorem is the following.
It follows that Babai's conjecture [BS92, Conjecture 1.7] holds for the simple groups PSL n (q), PSU n ( √ q), and PSp n (q), as well as the index-2 overgroup PO ± n (q) of the simple group PΩ ± n (q) for q even, for generating sets containing the image of a transvection.By combining with [EJ22, Theorem 1.1], we also get the following corollary, which confirms Babai's conjecture for all non-orthogonal classical groups over small fields with respect to three random generators.(As above, this is new only for p = 2 and a few p = 3 cases.)Corollary 1.3.Let G be in one of the following intervals: Assume log q < cn/ log 2 n for a sufficiently small constant c > 0. Let X = {x, y, z} where x, y, z ∈ G are uniformly random.Then with probability 1 − e −cn we have ℓ X (G) ≤ n O(log q) .
The proof of Theorem 1.2 does not consist merely in reviewing [GHS22] and dealing with technicalities.There is a much greater variety of irreducible subgroups generated by transvections in characteristic 2 than in odd characteristic, and these subgroups arise as obstructions in the proof.While Garonzi, Halasi, and Somlai relied on more elementary results of Dickson and Humphries, our main tool is the following deep theorem of Kantor, building on previous work of Wagner, Piper, and Pollatsek.The theorem classifies irreducible subgroups of SL n (q) generated by transvections (and thus in particular gives additional context to Theorem 1.2).
In the rest of the paper, when G is a irreducible linear group generated by transvections, we say G has linear, unitary, symplectic, orthogonal, monomial, or symmetric type according to cases (1)-( 6) respectively of Theorem 1.4.In the three remaining cases (which will not concern us) we say G has exceptional type.We refer to the linear, unitary, symplectic, and orthogonal cases collectively as classical.
In the course of the paper we build on many of the ideas and tools already developed in [GHS22,Hal21], particularly the transvection graph and its associated features such as symplectic and unitary cycles.An effort has been made however to keep the present paper reasonably self-contained and it should not be necessary to refer to [GHS22,Hal21] extensively.Where a proposition in the present paper closely parallels a corresponding proposition in [GHS22,Hal21], a brief note is made to that effect.Moreover, while we are free to assume p ∈ {2, 3} throughout, doing so would not simplify the paper in any way, and simplifications are obtained for p ≥ 3 (namely, we avoid many of the difficult calculations of [GHS22, Sections 4.2-4.3]).

Bounded rank
Babai's conjecture is known for simple groups of Lie type of bounded rank.This is a corollary of the independent work of Breuillard-Green-Tao and Pyber-Szabó.
Theorem 2.1 (Breuillard-Green-Tao [BGT11], Pyber-Szabó [PS16]).Let G be a finite simple group of Lie type of rank n and let X be a generating set for G. Then .
Theorem 1.2 in bounded rank follows immediately from this and standard tools, as detailed below.
Proposition 2.2.In the situation of Theorem 1.2 we have .
Proof.Assume X = X −1 and 1 ∈ X without loss of generality.Note that [G : G ′ ] = 2 in the orthogonal case and G = G ′ in the other cases.Therefore by Schreier's lemma In particular |Y m | ≥ |G ′ /Z(G ′ )|.Applying Freiman's 3/2 theorem (see [Tao15, Theorem 1.5.2 and Proposition 1.5.4])we have (for any k ≥ 1) either By Landazuri-Seitz [LS74] and [NP11, Corollary 2.5], the minimal degree of a nontrivial complex representation of G is at least |G| c/n > q c ′ n for some constants c, c This special case is used in the proof of Theorem 1.2 in two essentially different ways.First, it allows us to assume that n is larger than any fixed constant, which enables to avoid low-rank pathologies.Second, in the course of the proof of Theorem 1.2 we will apply Proposition 2.2 to a wide variety of bounded-rank subgroups.

The transvection graph
Following [Hal21,GHS22] we define a graph based on a collection of transvections.To motivate the definition we first recall the concept of the transposition graph (which we will also use).
If T ⊆ Sym(Ω) is a set of transpositions, the transposition graph Γ(T ) is the graph with vertex set Ω and edge set {ij : (ij) ∈ T }.Proposition 3.1.Let T ⊆ Sym(Ω) be a set of transpositions and let G = T .Then G is the direct product of the full symmetric groups on the components of Γ(T ).In particular the following are equivalent: (1) Γ(T ) is connected; (2) G is transitive; (3) G = Sym(Ω).
Proof.First assume Γ(T ) is connected.We will prove G = Sym(Ω) by induction on n, the result being obvious if n ≤ 1.We may assume Γ(T ) is a tree by replacing T with a subset.Let i be a leaf of Γ(T ), let t 0 = (ij) ∈ T , and let T 0 = T {t 0 }.Then Γ(T 0 ) can be identified with a tree with vertex set Ω 0 = Ω {i}, so by induction G 0 = T 0 = Sym(Ω 0 ), and it follows easily that G = G 0 , (ij) = Sym(Ω).
In general, let Ω 1 , . . ., Ω k be the components of Γ(T ) and let T i ⊆ T be the set of transpositions in T supported on Ω i .Then T = T 1 ∪ • • • ∪ T k and the elements of T i commute with the elements of Now let K be a field of characteristic p > 0 and let V ∼ = K n be a vector space over K of finite dimension n ≥ 2. A transvection is an element t ∈ SL(V ) such that t − 1 is a rank-one map.Let T ⊆ SL(V ) be the set of transvections.Each transvection t ∈ T has the form We make T into a directed graph Γ(T) by declaring there to be a directed edge t → s if and only if (t − 1)(s − 1) = 0. Put another way, there is an edge 1 + v ⊗ φ → 1 + u ⊗ ψ if and only if φ(u) = 0.For any subset T ⊆ T we denote by Γ(T ) the corresponding induced subgraph.We also define linear subspaces where G = T .Here [V, G] denotes the commutator of V and G as subgroups of V ⋊ G, and V G denotes the space of G-invariants in V .
Let T ⊆ T be an arbitrary set of transvections and let G = T .In the rest of this section we describe how important properties of G can be detected through careful examination of the transvection graph Γ(T ) as well as the subspaces V (T ) and V * (T ).We start with irreducibility of G.
) and (2) hold.Now suppose S is a nonempty subset of T with no incoming edges in Γ(T ).Then, for t ∈ T and s ∈ S, Let U = V (S).Then U is nonzero and S -invariant, and the above equation shows that T S acts trivially on U, so U is G-invariant, so U = V .Thus for every t ∈ T there is some s ∈ S such that (t − 1)(s − 1) = 0, whence t ∈ S by (3.1).Hence (3) holds.Sufficiency: Let U ≤ V be a nonzero G-invariant subspace.Let S ⊆ T be the set of s such that im(s − 1) ≤ U. Let u ∈ U {0}.By (2) there is at least one t ∈ T such that (t − 1)u = 0.By G-invariance of U it follows that im(t − 1) ≤ U, so t ∈ S. Hence S is nonempty, and the same calculation shows that S has no incoming edges, so by (3) we have S = T .Thus by (1) we have U = V .Hence G is irreducible.
We say G defined over a subfield L ≤ K if for some basis of V the elements of G are represented by matrices with entries in L. The defining field of G is the smallest field such that G is defined over L.
If T ⊆ T we denote by L k (T ) the subfield generated by the weights of the cycles of Γ(T ) of length at most k.Let L(T ) = ∞ k=1 L k (T ).Proposition 3.3 (cf.[GHS22, Section 3.2]).Assume G is irreducible.Then L(T ) is the defining field of G.
Proof.It follows from (3.2) (and induction on k) that L k (T ) is equal to the field generated by {tr(g) : ℓ T (g) ≤ k}, so L(T ) = ∞ k=1 L k (T ) is the field generated by {tr(g) : g ∈ G} (the trace field of G).It is well-known that the latter is precisely the defining field of G in positive characteristic (see [Isa94, Corollaries 9.5(c) and 9.23]).
Again suppose t 1 , . . ., t k ∈ T. We define and we call the tuple (t 1 , . . ., t k ) symplectic if d s = 0. Similarly, if θ is an involution of K, we define d θ (t 1 , . . ., t k ) = w(t 1 , . . ., t k ) − (−1) k w(t k , . . ., t 1 ) θ , and we call the tuple (t 1 , . . ., t k ) θ-unitary if d θ = 0. (If K is the field F q where q is a square then there is a unique involution of K and we may drop the "θ-" prefix.)Note that if (t 1 , . . ., t k ) is symplectic or θ-unitary then it is a cycle in Γ(T) if and only if it is a two-way cycle.
(1) There is a G-invariant symplectic form if and only if every cycle of Γ(T ) of length at most 2D + 1 is symplectic.(2) There is a G-invariant θ-unitary form if and only if every cycle of Γ(T ) of length at most 2D + 1 is θ-unitary.
Proof.We can prove both statements at once by seeking a G-invariant nondegenerate sesquilinear form f such that f (x, y) + f (y, x) θ = 0 identically, where θ is an automorphism of K of order at most 2. If θ = 1 and p = 2 then we additionally require f (x, x) = 0 identically.If θ is nontrivial we obtain a unitary form by multiplying f by a nonzero scalar λ such that λ θ = −λ.
We use the unitary language for both cases.
If G preserves such a form f then V and V * can be identified by defining v * (u) = f (u, v) for all u, v ∈ V , and the resulting map v → v * is a θsemilinear isomorphism of V with V * .The transvections in Aut(f ) are precisely the transvections of the form 1 + λv ⊗ v * with λ ∈ Fix(θ) and f (v, v) = 0. Thus the weight of a cycle t 1 , . . ., t k , where so every cycle is unitary.
Conversely suppose every cycle in Γ(T ) of length at most 2D +1 is unitary.For each t ∈ T fix a representation t = 1 + v t ⊗ φ t .We claim we can choose nonzero scalars λ t ∈ Fix(θ) for each t ∈ T such that Indeed, first note that every edge (t, s) in Γ(T ) is two-way since every edge is contained in a cycle of length at most D + 1, and every such cycle is unitary.Now for any path γ = (t 0 , t 1 , . . ., t k ) in Γ(T ) of length at most D + 1 define The cycle condition ensures that λ γ = λ θ γ ′ whenever γ and γ ′ have the same endpoints and combined length at most 2D + 1.In particular λ γ ∈ Fix(θ) if γ has length at most D. Choose t 0 ∈ T arbitrarily and define λ t to be the common value of λ γ for paths γ from t 0 to t of length at most D. Then (λ t ) is the required global solution to (3.3).Now define L : By (3.3) we have In particular, since V (T ) = V and V * (T ) = V * , we have L 1 (x) = 0 if and only if L 2 (x) = 0. Thus im(L) ⊆ V × V * is the graph of a θ-semilinear isomorphism * : V → V * such that v * t = λ t φ t for all t ∈ T .Define f (u, v) = v * (u) for u, v ∈ V .Then f is sesquilinear and nondegenerate, and by (3.3) we have , which implies that f is hermitian antisymmetric since im(T ) = V .Moreover for every t ∈ T we have Finally, we explain how to determine from the transvection graph whether G is contained in an orthogonal group.Note that this is only relevant in characteristic 2. Proposition 3.5.Assume K is a perfect field of characteristic 2. Assume G is irreducible and preserves a symplectic form f .Let v → v * be the isomorphism V → V * defined by v * (u) = f (u, v).Then for each t ∈ T there is a unique v ∈ V such that t = 1 + v ⊗ v * , and G preserves a quadratic from associated to f if and only if for every linear relation with m ≤ n + 1 we have the corresponding quadratic relation Proof.We noted in the proof of the previous proposition that each transvec- Suppose Q is a quadratic form associated to f .By a well-known direct calculation, a transvection Conversely, suppose (3.4) implies (3.5) identically.Since G is irreducible, V (T ) = V by Proposition 3.2, so there is a basis v 1 , . . ., v n among the vectors To this linear relation we can apply either (3.4) =⇒ (3.5) or we can apply the quadratic from Q.By comparing the results we find that

Some splitting results
Continue to assume K is a field of characteristic p > 0 and V ∼ = K n is a vector space of dimension n ≥ 2. As before T ⊆ SL(V ) denotes the set of transvections, T ⊆ T, and G = T .
By Proposition 3.2, G is irreducible if and only if Γ(T ) is strongly connected and V (T ) = V and V * (T ) = V * .In this section we consider the situation in which Γ(T ) is strongly connected but one or other of the latter conditions may fail.
Throughout this section, Observe that U and W are G-invariant subspaces, so G acts naturally on the section U/W .Let g → g denote the homomorphism G → SL(U/W ).
Proposition 4.1.Assume Γ(T ) is strongly connected.Then T is a set of transvections of U/W and the map t → t defines a graph homomorphism Γ(T ) → Γ(T ) that is surjective on both vertices and edges.Moreover, the weights of corresponding cycles are equal.In particular, G is irreducible.
We say T ⊆ T is nondegenerate if the natural pairing We say T is weakly nondegenerate if the natural pairing has either zero left kernel or zero right kernel.Observe that T is nondegenerate if and only if it is weakly nondegenerate and dim Proof.Choose a basis for V extending a basis for U. Since G preserves U = [V, G] and acts trivially on V /U, the elements of G take the form * * 0 I .
The elements of M = C G (U) have the form and M can be identified with a subspace of Hom(V /U, U) ∼ = U m , where m = dim(V /U).Moreover the conjugation action of G on M corresponds to the natural action of G on U m .Since dim U ≥ 4 > 1 and G = SL(U), G acts transitively on U {0}.In particular, U is an irreducible kG-module where k = F p is the prime subfield of K.
Suppose U 1 is an irreducible kG-submodule of U m .Then U 1 ∼ = U.Let f : U → U 1 be an isomorphism.By composing with any of the m coordinate projections π i : U m → U we get a map π i • f ∈ End kG (U).It is easy to check that End kG (U) = K using the fact that each α ∈ End kG (U) must commute with every transvection of U.2 Therefore there is some This shows that all irreducible kG-submodules of U m are equivalent up to the natural action of GL m (K).By induction on dimension, it follows that every kG-submodule of U m is equivalent to {0} m−d ⊕ U d for some d.
Since M can be identified with a kG-submodule of U m , it follows from the previous paragraph that we can choose the basis so that for some well-defined map f : SL(U) → U m−d .By direct calculation, f must be a crossed homomorphism.By a result of Higman [Hig62, Lemma 4], H 1 (SL(U), U) = 0 for dim U ≥ 4 (see also [Pol71,CPS75]).Therefore f (g) = gu − u for some u ∈ U m−d , and by conjugating appropriately we can arrange that f = 0. Lemma 4.4.Let H = SL(V ) ⋉ V d and let h → h be the natural map from H to H = SL(V ).Let X be a generating set for H. Then Proof.Assume without loss of generality X = X −1 and 1 ∈ X.We may also assume that ℓ , and the claimed bound holds by induction.
The last part follows from G ∼ = G ⋉ U d and the previous lemma.

Path shortening and other tricks
The method of this section is inspired by [GHS22, Lemma 4.1] and [Hal21, Lemma 2.4].
Given a vector u ∈ V and a transvection t This notation is compatible with the corresponding notation for the transvection graph and Γ(T ) is strongly connected (in fact Γ(T ) has directed diameter at most 2), so G = T is irreducible by Proposition 3.2.The following useful proposition is a partial converse.
Proposition 5.1.Let T ⊆ T be a symmetric set of transvections such that T is irreducible.Then T 2n−1 ∩ T is dense.
Proof.Let v ∈ V and φ ∈ V * be both nonzero.By Proposition 3.2, V (T ) = V , V * (T ) = V * , and Γ(T ) is strongly connected.It follows that there are transvections t 1 , . . ., t m ∈ T for some m ≥ 1 such that Write t i = 1+v i ⊗φ i for i = 1, . . ., m. Assume m is minimal for the existence of such t 1 , . . ., t m .If m = 1 we are done, so assume m > 1.By minimality of m, (1) Using properties (1) and (3) above, we have Similarly, by property (2), Proposition 5.2.Let T ⊆ T be a dense set of transvections.Let T 0 ⊆ T be a nonempty subset and assume Γ(T 0 ) has k strongly connected components.Then there is a set of transvections Proof.Let t 1 , . . ., t k ∈ T 0 be representatives for the k strongly connected components of Γ(T 0 ).By density of T there are transvections t ′ 1 , . . ., t ′ k ∈ T such that If G preserves a symplectic or unitary form then all edges are two-way and we can omit Proposition 5.3.Let T ⊆ T be a dense set of transvections.Let T 0 ⊆ T be a nonempty subset such that Γ(T 0 ) is strongly connected.Then there is a set of transvections Γ(T 1 ) is strongly connected, (4) T 1 is weakly nondegenerate.
Proof.Consider the pairing V (T 0 ) × V * (T 0 ) → K and assume both the left and right kernels are nonzero.Assume 0 . By density there is some ) is again strongly connected.Moreover since ψ ∈ V * (T 1 ) and V (T 1 ) ∩ ψ ⊥ = V (T 0 ) we have and by a symmetric argument we have Thus δ(T 1 ) = δ(T 0 ) − 1 and we are done by induction on defect.

Certification
The main new idea in the present work is to leverage Theorem 1.4 using "certificates".By a certificate for T we mean a bounded-size subset T 0 ⊆ T such that ℓ T (T 0 ) ≤ (n log q) O(1) and such that G 0 = T 0 acts irreducibly on V 0 = V (T 0 ) = [V, G 0 ] with the same defining field and same type as G itself.We may assume dim V 0 ≥ 10 to avoid pathologies.As a consequence of Kantor's classification theorem, it follows in particular that if T 0 ⊆ H ≤ G and H is irreducible then H = G.
Consider the containment diagram shown in Figure 1 between the types of irreducible groups generated by transvections with defining field F q .Here we have ignored the groups of exceptional type.This is harmless as we can always choose T 0 so that dim V 0 ≥ 10.Now our strategy is to consider each of the possibilities above in turn.For example, if G is not contained in a symplectic group, then we aim to find a subset T 0 of bounded size which is also not contained in a symplectic group: this is actually easy using symplectic cycles.Some of the other types will give us much more difficulty.
For the rest of the paper we assume K = F q is a finite field of characteristic p.As in previous sections, T ⊆ T and G = T .We assume G is irreducible.In consideration of Proposition 5.1, we assume T is dense.
Warning: The definition of unitary cycle depends on the field.In particular even if T 0 ⊆ T contains a cycle which is non-unitary with respect to F q , it is still quite possible that G 0 = T 0 is a unitary group with respect to the subfield L(T 0 ) < L(T ).6.2.Symmetric-type subgroups.Let us review the construction of the irreducible linear groups of symmetric type.Let G = S m and consider its standard permutation representation K m , where K = F 2 .The unique proper nontrivial submodules are ℓ = (K m ) G = e , where e = (1, . . ., 1), and is a group of symmetric type.We call G odd symmetric type if m is odd and even symmetric type if m is even.
The unique nonzero G-invariant alternating bilinear form on K m is given by f (x, y) = i =j x i y j , and one checks H ⊥ = ℓ, so f induces a G-invariant symplectic form on V .Thus G is conjugate to a subgroup of Sp n (2).One may also check that G preserves a quadratic form on V if and only if m ≡ 2 (mod 4).
For small m there are various coincidences with groups of classical type: i (e j ) = δ ij .Let g = (i, j) ∈ G. Then g acts as 1+(e i +e j )⊗(e * i +e * j ) on K m , which descends to a transvection on V , so ρ(g) is a transvection.Conversely, suppose ρ(g) is a transvection.Then g is an involution, so g has cycle type 2 k for some k, say g = (1, k + 1)(2, k + 2) • • • (k, 2k) without loss of generality.Then g acts as 1 + k i=1 (e i + e k+i ) ⊗ (e * i + e * k+i ) on K m .The subspace (g − 1)(H) consists of all vectors x ∈ K m such that x i = x k+i for 1 ≤ i ≤ k and such that x i = 0 for i > 2k, and moreover such that 2 and k is even.Thus (g − 1)|V has rank k unless k = m/2, in which case it has rank k − 1 if k is odd and k − 2 if k is even.Therefore in order that g acts a transvection on V it is necessary that k = 1.
In the rest of this subsection, we consider the following question.Let K = F 2 and V = K n , where n is even.Given T ⊆ T ∩ Sp n (2) such that G = T is irreducible, how can we certify that G is not contained in a group of symmetric type?Lemma 6.3.The following are equivalent: (1) G has odd symmetric type, (2) there is a G-invariant spanning set B ⊆ V of cardinality n + 1.
Moreover B is uniquely determined by G.
Γ 0 = Γ(T 0 ) be the transposition graph defined by T 0 .Then G 0 = T 0 is the direct product of the full symmetric groups on each component of Γ 0 , and V 0 is the image in V of the space W of all x ∈ K m such that i∈C x i = 0 for each component C of Γ 0 .If e ∈ W then every component of Γ 0 has even cardinality and |T 0 | ≥ (n + 2)/2 > 1000, contrary to hypothesis.Therefore W ∩ ℓ = 0 and V 0 ∼ = W as G 0 -modules.Since G 0 |V 0 is irreducible, it must be that Γ 0 has only one non-singleton component, and it must have odd order.Thus G 0 |V 0 has odd symmetric type.
(2) =⇒ (1): By hypothesis, G 0 |V 0 has odd symmetric type for every T 0 ∈ A 1000 .In this case by Lemma 6.3 there is a unique Throughout this proof we will refer to the subgroup G 0 ≤ G, the subspace V 0 ≤ V , and the subset B 0 ⊆ V 0 constructed from admissible sets T 0 in this way. Suppose and from our analysis of the implication (1) =⇒ (2) we know that B 0 is related to B 1 in the following rigid way: there is a G 0 -equivariant injection ι : Also G 0 acts trivially on B 1 ι(B 0 ).In particular b We claim that, for every v ∈ ∆, there is in fact some By consideration of the transposition graph defined by T 0 ⊆ G 0 ∼ = Sym(B 0 ), there are t, t ′ ∈ T 0 that move b 0 , b ′ 0 , respectively.By Proposition 5.3, {t, t ′ } is contained in some T 1 ∈ A 6 , and T 0 ∪ T 1 is contained in some T 2 ∈ A 808 (note that |T 0 ∪ T 1 | ≤ 404 and Γ(T 0 ∪ T 1 ) is strongly connected).Consider the corresponding maps ι 0 : B 0 → B 2 and ι 1 : B 1 → B 2 .Since t acts as a transposition on each of B 0 , B 1 , B 2 , it must be that ι 0 (b 0 ) = b 2 = ι 1 (b 1 ) for some b 2 ∈ B 2 and b 1 ∈ B 1 , and likewise ).This proves our claim.
In particular it follows that ∆ is G-invariant.Indeed, it suffices to prove that if v ∈ ∆ and t ∈ T then t(v) ∈ ∆.By the previous paragraph, v ∈ ∆(T 0 ) for some T 0 ∈ A 6 .By Propositions 5.2 and 5.3, T 0 ∪ {t} is contained in some Similarly, G is transitive on ∆.Consider v, v ′ ∈ ∆.Then v ∈ ∆(T 0 ) and v ′ ∈ ∆(T ′ 0 ) for T 0 , T ′ 0 ∈ A 6 .By Propositions 5.2 and 5.3, ) is transitive on ∆(T 1 ), it follows that some element of G 1 maps v to v ′ , and G 1 ≤ G.
Next we define a symmetric relation E on ∆ by Plainly E is G-invariant, so G acts on the graph (∆, E) by automorphisms.Moreover we can describe the relation E another way as follows.Suppose (u, v) ∈ ∆ 2 and suppose u, v ∈ ∆(T 0 ) for some Then w ∈ ∆(T 1 ) for some T 1 ∈ A 6 .By Propositions 5.2 and 5.3 there is a set Let Ω be the set of maximal cliques C of (∆, E) of cardinality |C| > 3, and note that Ω inherits a G-action from (∆, E).
We claim that each clique C ∈ Ω is determined by any two of its elements, i.e., |C ∩ . By Propositions 5.2 and 5.3, there is some . By our alternative characterization of E, each v i and v ′ i corresponds to a 2-subset of B 0 in such a way that adjacent vertices correspond to overlapping subsets.Since {v 1 , v 2 , v 3 , v 4 } is a clique, we must have , and since v 3 and v ′ 3 were arbitrary this implies that C ∪ C ′ is a clique, as claimed.
In fact we claim that |C ∩ By maximality and distinctness of C and C ′ , we may assume that , w} is a clique of cardinality at least 4 containing v 1 , v 2 , so it must be contained in C.This shows w ∈ C and symmetrically w ∈ C ′ , so w ∈ C ∩ C ′ .
We make one further observation.Let , there is some t ∈ G 0 transposing b 0 and b ′ 0 and fixing every other element of B 0 .Observe that ), and as above we must have 2 )} is a clique, which implies C ′′ = t(C ′′ ), and this is what we wanted to prove.
At last we can complete the proof.Consider the G-space K Ω and let H ≤ K Ω be the hyperplane {x ∈ K Ω : C∈Ω x C = 0}.Fix any C 0 ∈ Ω and observe that {e by sending e C 0 + e C to the unique element of C 0 ∩ C (recalling that the elements of Ω are by definition cliques of (∆, E)) and linearly extending to H.It follows from (6.1) that f sends e C +e C ′ to the unique element of C ∩C ′ for all C, C ′ ∈ Ω, C = C ′ .Hence f is a nonzero G-module homomorphism, and it must be surjective since V is irreducible.Thus G has symmetric type, as claimed.Remark 6.5.In the above proof, it follows a posteriori that |Ω| ∈ {n + 1, n + 2}, and G has odd type or even type according to these two cases.6.3.Monomial subgroups.Let K = F q be a finite field of characteristic 2 and let a be a divisor of q − 1.Let M n (a) be the subgroup of SL n (q) consisting of those elements which are represented in the standard basis by monomial matrices with all nonzero entries being ath roots of unity.Then M n (a) ∼ = C n−1 a ⋊S n and M n (a) is a subgroup of SL n (q) generated by transvections, irreducible for a > 1.The transvections in M n (a) are the elements represented as In general if G ≤ GL(V ), a G-invariant monomial structure on V is a set of lines L = {Kb 1 , . . ., Kb n } permuted by G such that b 1 , . . ., b n is a basis for V .Lemma 6.6.If a > 1 there is a unique M n (a)-invariant monomial structure on Note that M n (a) does not preserve a symplectic form on V if a > 1.Indeed, the only bilinear forms invariant on the diagonal subgroup of M n (a) are those proportional to n i=1 x i y i , and the only such form that is alternating is the zero form.By similar reasoning M n (a) preserves a unitary form if and only if a | q + 1. Proposition 6.7.Let T ⊆ T be a dense set of transvections in SL n (q).Assume that G does not preserve a symplectic form.Then the following are equivalent: (1) G has monomial type; (2) for every subset T 0 ⊆ T of cardinality |T 0 | ≤ 1000, (a) if T 0 is weakly nondegenerate then it is nondegenerate, and , so weak nondegeneracy implies nondegeneracy.Now assume G 0 |V 0 is irreducible.Since V 0 = i:d i >1 U i and the U i are G 0 -invariant, there must be exactly one d i > 1, say d 1 > 1 (i.e., Γ 0 is connected apart from isolated vertices).Moreover G 0 |V 0 is monomial type or symmetric type 1 according to whether b 1 > 1 or b 1 = 1, and in the latter case d 1 must be odd.
(2) =⇒ (1): Since G does not preserve a symplectic form, by Proposition 6.1 there is a set T 0 of cardinality |T 0 | ≤ 5 inducing a non-symplectic cycle.Let A t be the set of subsets T 1 ⊆ T of cardinality |T 1 | ≤ t containing T 0 such that Γ(T 1 ) is strongly connected and T 1 is weakly nondegenerate.By hypothesis, every does not have symmetric type, so it must have monomial type.By the previous lemma, there is a unique G 1 -invariant monomial structure L 1 on V 1 .Let L be the union of the sets L 1 over all T 1 ∈ A 100 .Note that L = ∅ since Proposition 5.3 implies T 0 is contained in some T 1 ∈ A 10 .
Let t ∈ T .We claim that t transposes some two lines Kb, Kb ′ ∈ L and acts trivially on every other line in L. By Propositions 5.2 and 5.3, T 0 ∪ {t} is contained in some T 1 ∈ A 16 .The group G 1 has monomial type, and therefore the transvections in G 1 act as transpositions on L 1 .Thus t ∈ G 1 transposes some two lines Kb, Kb ′ ∈ L 1 .Now let Kb ′′ ∈ L be any other line.Then Kb ′′ ∈ L 2 for some T 2 ∈ A 100 , and by Propositions 5.2 and 5.3 the set T 1 ∪ T 2 is contained in some T 3 ∈ A 236 .Now G 3 |V 3 is monomial type with unique monomial structure L 3 such that L 1 ∪ L 2 ⊆ L 3 .By the structure of transvections in monomial groups, t acts trivially on each line in L 3 {Kb, Kb ′ }, including Kb ′′ .
In particular, L is G-invariant.To complete the proof it suffices to show that the lines of L are linearly independent, i.e., if L = {Kb 1 , . . ., Kb m } then b 1 , . . ., b m are linearly independent.Indeed, since L is G-invariant it follows by irreducibility that m = n and L is a G-invariant monomial structure.This shows that G is a subgroup of a group of monomial type, and, by the argument we used to show (1) =⇒ (2), any irreducible subgroup generated by transvections of a group of monomial type not preserving a symplectic form itself must be of monomial type. Suppose Since G 1 has monomial type, G 1 contains some element g mapping b 1 → xb 1 (x = 1) and say b 2 → x −1 b 2 , and acting trivially on every other line of L 1 .Consider any other line Kb ∈ L. We have Kb ∈ L 2 for some T 2 ∈ A 100 and by Propositions 5.2 and 5.3 we have T 1 ∪ T 2 ⊆ T 3 for some T 3 ∈ A 404 .Since G 1 is a monomial-type subgroup of the monomial-type group G 3 , the element g acts trivially on Kb.Thus Since this argument works for each index, it follows that b 1 , . . ., b m are linearly independent, as required.
Summarizing this subsection and the previous one, if T ⊆ T is dense and G = T is classical, we can find a bounded-size certificate T 0 ⊆ T for the classicality of G.This is the content of the next proposition.Lemma 6.8 ([GHS22, Lemma 4.3]).If T is dense then L 5 (T ) = L(T ).6.4.Subfield subgroups.Now let K = F q be an arbitrary finite field, T ⊆ T, G = T ≤ SL n (q).Assume T is dense, G is classical, and K = L(T ).We aim to find a certificate T 0 ⊆ T (log q) O(1) of cardinality |T 0 | ≤ 3 such that L(T 0 ) = K.Lemma 6.10.Let G ≤ SL n (q) be a classical group generated by transvections with defining field F q .Then there is a subset Proof.Let λ ∈ F q be a primitive element.
Finally consider an orthogonal group G = O ± n (q), n ≥ 6, p = 2.The transvections in G are precisely those of the form 1 Let H ≤ V be a 4-dimensional hyperbolic subspace (see [Asc86,(21.2)]) and let e 1 , e 2 , e 3 , e 4 be a hyperbolic basis for H.With respect to this basis we have, for x, y ∈ H, Let u = e 1 + e 2 and v = λe 1 + e 3 + e 4 .Then Q(u) = Q(v) = 1 and f (u, v) = λ.Since finite fields are perfect, λ 2 generates F q , so it suffices to take Proposition 6.11.Let T ⊆ T be dense, G = T classical, L(T ) = F q , and assume n is sufficiently large.Then there is a subset S ⊆ T (log q) O(1) ∩ T of cardinality |S| ≤ 3 such that L(S) = F q .
Proof.First assume G = SL n (q).Then there is a G-invariant symplectic or unitary form f .The presence of this form simplifies the argument enough that it is worth keeping it separate from the SL n (q) case.
First, by Proposition 6.9 and Proposition 5.3, there is a nondegenerate subset T 0 ⊆ T of bounded cardinality such that G 0 = T 0 acts irreducibly on U 0 = [V, G 0 ] with quasisimple classical type.Let K 0 = L(T 0 ).Now for i > 0 proceed as follows.Assume T i−1 ⊆ T is a nondegenerate set of transvections of bounded cardinality such that be a quasisimple classical subgroup with the same defining field as G i−1 and of rank at most 6 (e.g., a copy of SL 2 , SU 3 , or O ± 6 over K i−1 ).By the previous lemma, there is a subset T 0 i ⊆ H i ∩ T of cardinality at most 10 generating H i .By Lemma 6.8, there is a cycle T 1 i in Γ(T ) of length at most 5 whose weight is not contained in K i−1 .Applying Propositions 5.2 and 5.3, i since T i is nondegenerate, and G i acts irreducibly on U i = [V, G i ] by Proposition 4.1.Since H i ≤ G i and H i has quasisimple classical type, it follows from Theorem 1.4 that G i also has quasisimple classical type on U i .Moreover, G i has defining field K i > K i−1 .
The process ends with some K r = F q .Then K 0 < • • • < K r = F q .For each i, by Proposition 2.2, we have Finally, by the previous lemma there is a subset S ⊆ G r ∩ T of cardinality |S| ≤ 3 such that L(S) = F q , as required.
Now suppose G = SL n (q).We argue similarly but because we cannot guarantee nondegeneracy we instead work with the section throughout the argument.First we define a bounded-cardinality starting set T 0 ⊆ T .Using Proposition 6.1 we may include in T 0 a non-symplectic cycle of length at most 5.If q is a square, we also include a non-unitary cycle, as well as a cycle whose weight is not in F √ q , using Lemma 6.8.Using Propositions 5.2, 5.3, and 6.9, we may assume T 0 is weakly nondegenerate, Γ(T 0 ) is strongly connected, and G 0 = T 0 acts irreducibly and classically on W 0 = [V, G 0 ]/[V, G 0 ] G 0 , and dim W 0 > 10.Because of the presence of non-symplectic and non-unitary cycles, G 0 |W 0 ∼ = SL dim W 0 (K 0 ) where K 0 = L(T 0 ).Now for i > 0 proceed as follows.Assume T i−1 ⊆ T is a weakly nondegenerate set of transvections of bounded cardinality such that . Otherwise, by the previous lemma and Proposition 4.5 there is a subset T 0 i ⊆ G i−1 ∩ T of cardinality at most 3 (actually 2) such that L(T 0 i ) = K i−1 .By Lemma 6.8, there is a cycle T 1 i in Γ(T ) of length at most 5 whose weight is not contained in K i−1 .Let t = 1 + v ⊗ φ ∈ T 0 i and let H i = 1 + K i−1 v ⊗ φ ∼ = K i−1 be the full transvection subgroup over K i−1 generated by t.It follows from Proposition 4.5 that H i ≤ G i−1 .Applying Propositions 5.2 and 5.3, T 0 ∪ T 0 i ∪ T 1 i is contained in a bounded-cardinality weakly nondegenerate set As before the process ends with some K r = F q .Then K 0 < • • • < K r = F q .Now we use Proposition 2.2 in combination with the diameter bound in Proposition 4.5.For each i we have 1) .Since H i ≤ G i−1 for each i we get ℓ T (G r ) ≤ O(log q) O(1) as before.Finally we apply the previous lemma in combination with Proposition 4.5 again to find S ⊆ G r ∩ T as required.6.5.Orthogonal subgroups.The last thing remaining is distinguish orthogonal groups from symplectic groups.Throughout this subsection assume K = F q is a finite field of characteristic 2. Lemma 6.12.Let (V, Q) be a nondegenerate quadratic space over a finite field K = F q .Let w + H ⊆ V be an affine subspace of codimension 2.
Then H 2 is nondegenerate and H 2 has codimension at most 6, so dim H 2 ≥ 4. It follows from the classification of quadratic forms over finite fields [Asc86,(21.2)] that Q(H 2 ) = K.Now assume T ⊆ T ∩ Sp n (q) is dense, L(T ) = K = F q , and G = T is classical.Proposition 6.13.There is a constant C such that the following are equivalent: (1) G is an orthogonal group, (2) every nondegenerate subset T 0 ⊆ T (n log q) O(1) of cardinality at most C generates a subgroup G 0 = T 0 that preserves an orthogonal structure on Proof.(1) =⇒ (2): Suppose G preserves an orthogonal form Q with corresponding symplectic form f .If T 0 ⊆ T is nondegenerate then (2) =⇒ (1): First, by Propositions 5.2, 5.3, 6.9, and 6.11, there is a bounded-cardinality nondegenerate subset T 0 ⊆ T (log q) O(1) such that G 0 = T 0 acts irreducibly and classically on V 0 = V (T 0 ) , dim V 0 ≥ 10, and L(T 0 ) = K.Note G 0 must be an orthogonal group.By Proposition 2.2, Thus by replacing T with T (log q) O(1) ∩T we may assume T 0 ⊆ T and ℓ T (G 0 ) ≤ O(1).Now we apply Proposition 3.5.For each t ∈ T there is a unique v t ∈ V such that t = 1 + v t ⊗ v * t .Consider some linear relation Suppose λ is supported on t 1 , . . ., t m , and we may assume m ≤ n + 1.We must show that Q(λ) = 0, where Define also We may think of λ as an element of K T , and Q defines an orthogonal form on K T with corresponding symplectic form f .We aim to reduce the support of λ as much as possible.Suppose m ≥ 10, i.e., the support of λ contains at least 10 transvections t 1 , . . ., t 10 .Applying Propositions 5.2 and 5.3, T 0 ∪ {t 1 , . . ., t 10 } is contained in a boundedcardinality nondegenerate set T 1 ⊆ T such that G 1 = T 1 acts irreducibly on V 1 = V (T 1 ).Since T 0 ⊆ T 1 , G 1 must act classically, and therefore G 1 must be an orthogonal group by hypothesis.Since G 0 ≤ G 1 , by Proposition 2.2 we have ℓ By Lemma 6.12, we can find a transvection t ∈ G 1 ⊆ T L (for some constant L) Hence we split the given linear relation into two shorter linear relations Repeating this argument for ⌊m/10⌋ disjoint subsets of {t 1 , . . ., t m } of size 10, we find that we can add ≤ m/10 transvections t i = 1+v i ⊗v * i ∈ T O(1) such that our original linear relation can be written as a sum of linear relations of support ≤ 11 and one of support at most ⌈9m/10⌉, i.e., λ = λ where each λ (j) has support size at most ⌈9m/10⌉ and t∈T L λ (j) t v t = 0 (j = 1, . . ., s).Now we repeat this argument for T L and each of the λ (j) , etc, log n times.The end result is that λ gets written as a sum of some µ (1) , . . ., µ (N ) where each vector µ has support size O(1), and Note that L log n = n O(1) .Finally we use We claim that each term on the right-hand side of (6.2) is zero.Indeed, first consider some Q(µ), µ = µ (j) .We know that µ is supported on a set of O(1) transvections t ∈ T L log n .By Propositions 5.2 and 5.3, the union of T 0 with the support of µ is contained in a bounded-cardinality nondegenerate set acts classically and therefore as an orthogonal group on V 1 .Applying the forward direction of Proposition 3.5, we find that Q(µ) = 0.By the same argument applied to the union of the supports of µ and µ ′ , µ ′ = µ (j ′ ) , we also have Q This proves the claim.Hence Q(λ) = 0 and the proof is complete.6.6.Conclusion.Theorem 6.14 (Certification).Let n be sufficient large.Let T ⊆ T ⊆ SL n (q) be a set of transvections such that G = T is an irreducible classical group.Then there is a subset T 0 ⊆ T (n log q) O(1) of cardinality O(1) such that, for any subset T 1 ⊆ G ∩ T containing T 0 such that Γ(T 1 ) is strongly connected, the group G 1 = T 1 acts on the section W = [V, G 1 ]/[V, G 1 ] G 1 as an irreducible classical group with the same type and defining field as G itself, and moreover dim W > 10.
We may assume G has defining field K = F q .By Proposition 5.1 and replacing T with T 2n−1 ∩ T, we may assume T is dense.By Proposition 6.11, there is a subset S 0 ⊆ T (log q) O(1) of cardinality |S 0 | ≤ 3 such that L(S 0 ) = K.By Proposition 6.1 there is a set S 1 ⊆ T of cardinality at most 10 such that S 1 contains a non-symplectic cycle if T does and S 1 contains a non-unitary cycle if T does.By Proposition 6.4, if q = 2 and G is contained in a symplectic group, there is a subset S 2 ⊆ T of cardinality |S 2 | ≤ 2000 that is not contained in a symmetric-type subgroup.By Proposition 6.7, if G is not contained in a symplectic group, there is a subset S 3 ⊆ T of cardinality |S 3 | ≤ 1000 that is not contained in a monomial subgroup.By Proposition 6.13, if G = Sp n (q) there is a subset S 4 ⊆ T (n log q) O(1) of cardinality O(1) that is not contained in an orthogonal subgroup.Finally, by density of T there is a bounded-cardinality set T 0 ⊆ T such that for each i (such that S i is defined) we have S i ⊆ T 0 , moreover the natural maps are injective, and moreover both spaces on the right have dimension at least 2000.
Let T 1 ⊆ T be a set containing T 0 such that Γ(T 1 ) is strongly connected.By Proposition 4.1, and let T 1 be the image of T 1 in G 1 .By Proposition 4.1, Γ(T 1 ) and Γ(T 1 ) are naturally identified and the weights of corresponding cycles are equal.In particular, L(T 1 ) = L(T 1 ) = K, so G 1 has defining field K. Also Γ(T 1 ) contains symplectic cycles if and only if Γ(T 1 ) does, and ditto for unitary cycles.We also have dim W ≥ 2000.The symmetric-type and monomial-type certificates ensure that G 1 has neither symmetric type nor monomial type.Thus G 1 must have the same type as G.

Bounding the diameter
Finally we turn to the proof of our main theorem, Theorem 1.2.Let G be one of the groups in scope and let X be a generating set for G containing a transvection t.Let and T m = G.This demonstrates that for the purpose of proving Theorem 1.2 it is sufficient to assume X ⊆ T (cf.[GHS22, Lemma 2.1]).Thus assume T ⊆ T is a generating set for G.In particular, T is irreducible.Now we apply Theorem 6.14 to obtain a certificate T 0 ⊆ T (n log q) O(1) of cardinality O(1).We may assume T 0 is weakly nondegenerate and Γ(T 0 ) is strongly connected.Let G 0 = T 0 .Unless G = SL(V ), T 0 is in fact nondegenerate and G 0 is a classical group with the same type as G, defining field K, and domain 1) .By replacing T with T (n log q) O(1) ∩ T we may thus assume T 0 ⊆ G 0 ∩ T ⊆ T .
If G = SL(V ), T 0 may only be weakly nondegenerate, but , and dim W 0 > 10.By Propositions 2.2 and 4.5 we have ℓ T 0 (G 0 ) ≤ O(log q) O(1) .Thus again by replacing T with T (n log q) O(1) ∩ T we may assume T 0 ⊆ G 0 ∩ T ⊆ T .
Let V = F n q be the natural module for G. Then V is a linear, unitary, symplectic, or characteristic-2 orthogonal space.Let us call a vector v ∈ V transvective if there is a transvection t ∈ G ∩ T such that [V, t] = v .If V is linear or symplectic, all vectors are transvective; if V is unitary, a vector is transvective if and only if it is singular; if V is orthogonal, a vector is transvective if and only if it is nonsingular.
Proof.Clearly we may assume V is unitary or orthogonal, as otherwise all vectors are transvective.
Since the trace map tr : F q → F √ q is surjective and f (v, v) ∈ F √ q , we can choose λ ∈ K so that f (v, v) = tr(λf (v i , v)), and it follows that v − λv i is singular, as required.
Orthogonal case.Observe that Since Q(v i ) = 0, there are at most two λ ∈ K for which this is zero, so we are done unless q = 2, Q(v) = 0, and f (v, v i ) = 1 for all i.In this case there must be some i, j such that f (v i , v j ) = 0, and so we are done.
Lemma 7.2.Let b 1 , . . ., b n be a basis for V with b 1 , . . ., b n transvective.For v = n i=1 x i b i ∈ V , let s(v) be the number of nonzero coefficients x i .Then any transvective vector v with s(v) = k is the sum of 4 transvective vectors v 1 , v 2 , v 3 , v 4 with s(v i ) ≤ k/2 + 2 for each i.
Proof.Case G = SL n (q).Assume V is a symplectic, unitary, or orthogonal space.Let f be the invariant form and for each v ∈ V let v * be the dual element associated to v by f .Recall that a vector v ∈ V is transvective if and only if there is a transvection t ∈ G ∩ T such that t = 1 + λv ⊗ v * for some λ ∈ K.By Proposition 3.2, there is a (transvective) basis b 1 , . . ., b n for V such that 1 + µ i b i ⊗ b * i ∈ T for some µ i ∈ K for each i = 1, . . ., n.Let s : V → {0, . . ., n} be the support function defined in Lemma 7.2 and for each k ≥ 1 let T (k) be the set of transvections t = 1 + λv ⊗ v * ∈ G ∩ T such that s(v) ≤ k.It suffices to prove that T (5) ⊆ T O(1) and T (2k) ⊆ (T (k+2) ) O(1) for each k ≥ 1, for then it follows that G ∩ T = T n ⊆ T O(1) ⌈log 2 n⌉ = T n O(1) .
Let t ∈ T (5) , say t = 1 + λv ⊗ v * and v = k i=1 x i b i , k ≤ 5.By Propositions 5.2 and 5.3, T 0 ∪ {1 + µ i b i ⊗ b * i : 1 ≤ i ≤ k} is contained in a nondegenerate set T 1 ⊆ T of cardinality O(1) such that Γ(T 1 ) is strongly connected.Then G 1 = T 1 is a classical group with the same type as G and defining field K and domain and t ∈ G 1 .By Proposition 2.2 and the fact that T contains G 0 ∩ T and log |G 0 ∩ T| ≥ log q, we have G 1 ⊆ T O(1) .Thus T (5) ⊆ T O(1) .
Case G = SL n (q).Now all vectors are transvective.Again by Proposition 3.2 there are elements 1 + b i ⊗ f i ∈ T (1 ≤ i ≤ n) such that b 1 , . . ., b n is a basis for V , as well as elements 1 + c i ⊗ g i ∈ T (1 ≤ i ≤ n) such that g 1 , . . ., g n is a basis for V * (T ).Define s(v) exactly as before with respect to the basis b 1 , . . ., b n and s(φ) analogously for φ ∈ V * with respect to the basis g 1 , . . ., g n .For each k ≥ 1 let T (k) be the set of all transvections t = 1 + v ⊗ φ ∈ T such that s(v) ≤ k and s(φ) ≤ k.It suffices to prove that T (1) ⊆ T O(1) and T (2k) ⊆ (T (k) ) O(1) for each k ≥ 1.
Let t ∈ T (1) , so without loss of generality t = 1 + λb i ⊗ g j for some indices i, j and λ ∈ K.By Propositions 5.2 and 5.3, T 0 ∪ {1 + b i ⊗ f i , 1 + c j ⊗ g j } is contained in a bounded-cardinality weakly nondegenerate set T 1 ⊆ T such that Γ(T 1 ) is strongly connected.Applying Proposition 4.5 and Proposition 2.2, t ∈ G 1 = T 1 and G 1 ⊆ T O(1) .Since t was arbitrary this proves T (1) ⊆ T O(1) .

Figure 1 .
Figure 1.Containment between types of irreducible groups generated by transvections with defining field F q It is therefore harmless to assume m ≥ 7, m = 8.Unless m ∈ {2, 4}, an element of G ∼ = S m acts on {1, . . ., m} as a transposition if and only if it acts on V as a transvection.Lemma 6.2.Let G = S m and let ρ : G → SL(V ) ∼ = SL n (2) be the representation defined above, where n = m − gcd(m, 2).Assume m ≥ 5.An element g ∈ G is a transposition if and only if ρ(g) is transvection.Proof.Let e 1 , . . ., e m be the standard basis of K m .Define e * i ∈ (K m ) * by e * w is the sum of two distinct elements of B 2 .Moreover the expressions for u and v are u = ι(b 1 )+ι(b 2 ) and v = ι(b 3 )+ι(b 4 ), where ι : B 0 → B 2 is the usual map.If b 1 , b 2 , b 3 , b 4 are all distinct then w = ι(b 1 ) + ι(b 2 ) + ι(b 3 ) + ι(b 4 ) is not the sum of two elements of B 2 , since the unique nontrivial linear relation among the elements of B 2 is b∈B 2 b = 0 and |B 2 | is odd.Therefore |{b 1 , b 2 } ∩ {b 3 , b 4 }| = 1, as claimed.
and w = b 0 + b ′ 0 , w ′ = b ′ 0 + b ′′ 0 , and w ′′ = b 0 + b ′′ 0 .Finally, we claim that G|Ω = Sym(Ω).It suffices to prove that for any two distinct C, C ′ ∈ Ω, there is an element of G transposing C and C ′ and fixing every other element of Ω. Exactly as above let v 1 The kernel C G (L) of the action of G on L is abelian and normal, so by the above argument it must be contained in the diagonal subgroup D∼ = C n−1 a of G. Moreover G/C G (L) is isomorphic to a subgroup of Sym(L) ∼ = S n ; it follows that C G (L) = D.Hence b 1 , . . ., b n are joint eigenvectors of D. However it is straightforward to check that the only joint eigenvectors of D are the scalar multiples of e 1 , . . ., e n , so {Kb 1 , . . ., Kb n } = {Ke 1 , . . ., Ke n }.
0 has monomial type or odd symmetric type.Proof.(1) =⇒ (2): Assume G = M n (a) ≤ SL n (q) and recall M n (a) ∼ = C n−1 a ⋊ S n .Let T 0 ⊆ T be a subset (of any cardinality).Observe that the image of T 0 under the natural map M n (a) → S n is a set of transpositions.Let Γ 0 be the corresponding transposition graph.Then by Proposition 3.1 the image of G 0 in S n is the direct product of the full symmetric groups on each component C of Γ 0 .The group G 0 itself is the direct product of monomial groups M d (b) for b ≥ 1, say G 0 = k i=1 M d i (b i ), where d i = n, d i ≥ 1, and we take b Proposition 7.4.ℓ T (G) ≤ n O(1) .Proof.By the previous proposition, ℓ T (G ∩ T) ≤ n O(1) .By the Liebeck-Shalev theorem [LS01, Theorem 1.1] we have ℓ G∩T (G) ≤ O(n).Thus ℓ T (G) ≤ ℓ T (G ∩ T) ℓ G∩T (G) ≤ n O(1) , as claimed.