Generating Sets of Conjugate Involutions of Groups PSL_n(9)

G. Malle, J. Saxl, and T. Weigel in [Geom. Ded., 49, No. 1, 85-116 (1994)] formulated the following problem: For every finite simple non-Abelian group G, find the minimum number n_c(G) of generators of conjugate involutions whose product equals 1. (See also Question 14.69c in [Unsolved Problems in Group Theory. The Kourovka Notebook, No. 20, E. I. Khukhro and V. D. Mazurov (Eds.), Sobolev Institute of Mathematics SO RAN, Novosibirsk (2022); https://alglog.org/20tkt.pdf].) J. M. Ward [PhD Thesis, Queen Mary College, Univ. London (2009)] solved this problem for sporadic, alternating, and projective special linear groups PSL_n(q) over a field of odd order q, except in the case q = 9 for n ≥ 4 and also in the case q ≡ 3 (mod 4) for n = 6. Here we lift the restriction q ≠ 9 for dimensions n ≥ 9 and n = 6.

INTRODUCTION

It is known that any simple finite non-Abelian group, except the three-dimensional unitary group over a field of nine elements PSU₃(9), is generated by three involutions. The proof of this fact was completed by G. Malle, J. Saxl, and T. Weigel in [1]. There, too, the following problem was formulated (see also [2, Question 14.69c]):

(A) For every finite simple non-Abelian group G, find n_c(G), the minimum number of generators of conjugate involutions whose product equals 1.

If a group G is simple and non-Abelian, then n_c(G) ≥ 5 (cf. [3, p. 13] or [4, Lemma 4]). On the other hand, it is obvious that if G is generated by three conjugate involutions, then n_c(G) ≤ 6. We say that a group is (2, 3)-generated if it is generated by an involution and an element of order 3. Every perfect (2, 3)-generated group is generated by three conjugate involutions [5, p. 221]. (A group is perfect if it coincides with its derived subgroup.) Thus, for a finite simple (2, 3)-generated group G, the number n_c(G) equals either 5 or 6, and problem (A) reduces to the following questio n: Is a group G generated by five conjugate involutions whose product is equal to unity? Note that there are simple finite groups G for which n_c(G) > 6 (see [4]), for instance n_c(PSU₃(9)) = 7.

J. M. Ward [3] solved problem (A) for sporadic, alternating, and projective special linear groups PSL_n(q) over a field of odd order q, except in the case q = 9 for n ≥ 4 and also in the case q ≡ 3 (mod 4) for n = 6. In 2021 I. Yu. Efimov and Ya. N. Nuzhin [6] lifted the restriction q ≠ 9 for n = 4, 5, 7, 8. In this paper we lift the restriction q ≠ 9 for dimensions n ≥ 9 and for the dimension n = 6. In all of these cases, it turns out that generating quintuples of conjugate involutions whose product equals unity (see [3]), will do also for q = 9.

The main result of the paper is the following:

THEOREM 1. (a) The special linear group SL_n(9) of dimension n ≥ 9 over a field of order 9 is generated by three conjugate involutions α, β, γ, and the product αβ is also an involution, which is conjugate to α.

(b) The projective special linear group PSL₆(9) of dimension 6 over a field of order 9 is generated by three conjugate involutions α, β, γ, and the product αβ is also an involution, which is conjugate to α.

We fix as a lemma one result useful for solving problem (A) (see also [4, Lemma 6]).

PROPOSITION 1 [3, Lemma 1.2.2]. The following statements are equivalent:

(a) a group G is generated by three conjugate involutions α, β, γ, and the product αβ is also an involution and it is conjugate to α;

(b) a group G is generated by conjugate involutions α, β, γ, δ, ε two of which coincide and αβγδε = 1.

In [3] Proposition 1 was used and triples of generating conjugate involutions α, β, γ were found the first two of which commute and the involutions α, αβ are conjugate. The proof that groups are generated by a given set of involutions in [3] made essential use of a well-known theorem due to L. Dickson, which says that a group generated by two opposite elementary transvections over a finite field of odd order q other than 9 coincides (up to conjugation by a diagonal element) with the group SL₂ over some subfield of the base field. In the case of a field of order 9, however, two opposite transvections do not always generate a group isomorphic to SL₂(q); in [3], therefore, this case is excluded.

Obviously, if $\overline{G }$ is a nontrivial homomorphic image of G, then n_c($\overline{G }$) ≤ n_c(G). In view of Proposition 1 and the fact that n_c(G) ≥ 5 for a simple group G, Theorem 1 entails

COROLLARY 1. (a) If n ≥ 9, then n_c(SL_n(9)) = n_c(PSL_n(9)) = 5.

(b) n_c(PSL₆(9)) = 5.

Note that dimension 6 stands out as a special case in Theorem 1 and Corollary 1 since the group SL₆(q) for odd q is not generated by even an arbitrary quintuple of (not necessarily conjugate) involutions whose product equals unity [7]. However, it is (2, 3)-generated [8,9,10], so the equality n_c(SL₆(q)) = 6 holds.

Finally, summing up the results obtained in [3, 6] and Theorem 1, we obtain that for groups PSL_n(q), q is odd, Question 14.69(c) remains unsettled only for n = 6 with q ≡ 3 (mod 4).

1. PRELIMINARY RESULTS

We fix some elements from the special linear group SL_n(F ) over a field F . The elementary transvections

$${t}_{r,s}\left(u\right)={E}_{n}+{ue}_{r,s},r,s=\text{1,2},\dots ,n, r\ne s,u\in F,$$

will be called simply transvections, where E_n is the identity matrix of degree n, and e_r,_s is the (n × n) matrix with 1 on position (r, s) and 0’s elsewhere. Hereinafter, the following abbreviations are used. For any nonempty subset M of some group, <M> denotes a subgroup generated by the set M; x^y = yxy⁻¹; [x, y] = xyx⁻¹y⁻¹.

In the proof of Theorem 1(b), the following obvious fact was used. Let ϕ: G → $\overline{G }$ be a surjective homomorphism, and let a group G be generated by a set M. Then the homomorphic image $\overline{G }$ is generated by the set (M ). Since the natural homomorphism ϕ: SL_n(F) → PSL_n(F) is surjective, it suffices to show that the group SL_n(F) is generated by a set {α, β, γ}, where ϕ(α), ϕ(β), ϕ(γ) are involutions in PSL_n(F) having the required properties.

The following is well known:

LEMMA 1.1 [11, p. 26]. The special linear group SL_n(F ) over a field F is generated by all its elementary transvections t_r,_s(u), u ∈ F , r, s = 1, 2,... , n.

Throughout what follows, on matrix positions i is an element of the field GF (9) such that i² = −1. The additive group of a field of order 9 is generated by the elements 1 and i, so the previous lemma implies

LEMMA 1.2. (a) The special linear group SL_n(3) over a field of order 3 is generated by transvections t_r,_s(1), r, s = 1, 2,..., n.

(b) The special linear group SL_n(9) over a field of order 9 is generated by transvections t_r,_s(1), t_r,_s(i), r, s = 1, 2,..., n.

In proofs on generation of groups SL_n(9) by a certain collection of involutions, use is made of the matrices

$$\mu =\left(\begin{array}{ccccc}0& 0& \dots & 0& {\left(-1\right)}^{n+1}\\ 1& 0& \dots & 0& 0\\ 0& 1& \dots & 0& 0\\ \dots & \dots & \dots & \dots & \dots \\ 0& 0& \dots & 1& 0\dots \end{array}\right),$$

where the plus sign is taken for n = 4m+1, n = 4m+2 and the minus sign is taken for n = 4m+3, n = 4m so that the matrix γ₁ be in SL_n(9). A group generated by the matrix μ acts transitively by conjugations on a set of transvections given by

$$T=\left\{{t}_{1,n}\left({\left(-1\right)}^{n+1}\right),{t}_{r+1,r}\left(1\right),r=\text{1,2},\dots ,n-1\right\}$$

and on its transpose

$${T}{\prime}=\left\{{t}_{n,1}\left({\left(-1\right)}^{n+1}\right),{t}_{r,r+1}\left(1\right),r=\text{1,2},\dots ,n-1\right\}.$$

Commuting with each other elements from the set T or from T′, we obtain all transvections t_r,_s(1), r, s = 1,... , n. In view of Lemma 1.2(a), therefore, each of the sets T and T′ generates the group SL_n(3). Moreover, the following is valid:

LEMMA 1.3. (a) The group SL_n(3) is generated by the matrix μ and one transvection from the set T or from the set T′.

(b) The group SL_n(9) is generated by the matrix μ in combination with one transvection from T or from T′ and any transvection t_r,_s(i).

In proving, we use an isomorphism between the symmetric group S_n and the factor group M/D of a monomial group M in SL_n(F ) with respect to D, a subgroup of its diagonal matrices. In correspondence with every monomial matrix ϵ with a nonzero element on position (j_k, i_k) and zeros elsewhere in this column is a permutation $\widetilde{\epsilon }$ mapping element i_k into element j_k, respectively, with k = 1, 2,..., n. And conversely, every permutation on n symbols is in correspondence with monomial matrices from SL_n(F). In particular, the matrix μ agrees with a cycle (1, 2,..., n) of length n, and any monomial matrix from SL_n(F), which agrees with a cycle of length n, is conjugate to μ in GL_n(F ) by monomial and diagonal matrices.

It is easy to see that the following is valid:

LEMMA 1.4. Let η be a monomial (0, 1)-matrix (permutation matrix) which agrees with an element of the symmetric group mapping the symbols l and k into r and s, respectively, with l ≠ k, r ≠ s, l, k, r, s ∈ {1, 2,..., n}. Then the equality t_l,_k(u)^η = t_r,_s(u) holds for any element u ∈ F.

2. PROOF OF THEOREM 1

The proof is constructive and insists on all generators being given in an explicit form. However, generating involutions are the same as in Ward’s thesis [3].

To prove that the given involutions α, β, γ, αβ are conjugate, we use K. Jordan’s famous theory. Since the characteristic of a field F is not 2, Jordan boxes of involutions, as well as of elements of order 4 in the case of dimension n = 6, are one-dimensional. Consequently, if the corresponding matrices have equal sets of eigenvalues, and all eigenvalues are in F, then they are conjugate in GL_n(F) and, hence, in SL_n(F).

The case of dimensions n = 4m + 1. Generating involutions have the forms

(In the matrix γ, the element i is at position (n, 2m), and the element −i is at position (n, 2m+1)). Hereinafter, we use the sign diag not only for diagonal matrices but also for block-diagonal ones. The elements α, β, γ, αβ are involutions and have eigenvalues {1^2m+1, (−1)^2m}, so they are conjugate in SL_n(9). We put M = <α, β, γ> and show that the subgroup M coincides with SL_n(9). We have

$$\begin{array}{c}\eta ={\left(\beta \gamma \right)}^{4}={t}_{n,2m}\left(4i\right){t}_{n,2m+1}\left(-4i\right)={t}_{n,2m}\left(i\right){t}_{n,2m+1}\left(-i\right),\\ {\gamma }_{1}={t}_{n,2m}\left(i\right){t}_{n,2m+1}\left(-i\right)\gamma .\end{array}$$

Monomial matrices that are contained in the group M comply with the following permutations

$$\begin{array}{c}\widetilde{\alpha }=\left(\text{1,2}m+1\right)\dots \left(m+1\right)\left(2m+\text{2,2}m+3\right)\dots \left(n-1,n\right),\\ {\widetilde{\gamma }}_{1}=\left(1,n-1\right)\dots \left(2m+2m+1\right)\left(n\right),\\ \widetilde{\theta }=\widetilde{\alpha }{\widetilde{\gamma }}_{1}=\left(\text{1,2}m,n-\text{2,3},\dots ,\text{2,2}m+1,n-1,n\right).\end{array}$$

The key step in proof is

LEMMA 2.1. The equality <θ, t₂,₁(1)> = SL_n(3) holds.

Proof. Permutations that comply with the matrices θ and μ are cycles of length n. Since all cycles of length n are conjugate in S_n, there exists a monomial matrix δ lying in GL_n(3) and mapping by conjugation θ into μ. Then

$$\widetilde{\delta }=\left(2m,2,n-3,\dots \right)\left(1\right)\left(n-1\right)\left(n\right).$$

Note that t_2,1(1)^δ = t_n−3,₁(±1), hence

$${\langle \theta ,{t}_{\text{2,1}}\left(1\right)\rangle }^{\delta }=\langle \mu ,{t}_{n-\text{3,1}}\left(1\right)\rangle .$$

Further, conjugating the transvections by matrix μ and commuting the results of conjugation, we obtain an element of the set T′. Hereinafter, Lemma 1.4 is used. We have

$$\begin{array}{c}{t}_{n-\text{3,1}}{\left(1\right)}^{{\mu }^{4}}={t}_{\text{1,5}}\left(1\right),\\ \left[{t}_{n-\text{3,1}}\left(1\right),{t}_{\text{1,5}}\left(1\right)\right]={t}_{n-\text{3,5}}\left(1\right),\\ \begin{array}{c}{t}_{n-\text{3,5}}{\left(1\right)}^{{\mu }^{4}}={t}_{\text{1,9}}\left(1\right),\\ \dots \\ \begin{array}{c}{t}_{n-\text{3,4}}\left(m-2\right)+1{\left(1\right)}^{{\mu }^{4}}={t}_{\text{1,4}m-3}\left(1\right)={t}_{1,n-4}\left(1\right),\\ \left[{t}_{n-\text{3,1}}\left(1\right),{t}_{1,n-4}\left(1\right)\right]={t}_{n-3,n-4}\left(1\right).\end{array}\end{array}\end{array}$$

In view of Lemma 1.3(a),

$$\langle \mu ,{t}_{n-3,n-4}\left(1\right)\rangle ={SL}_{n}\left(3\right),$$

hence

$$\langle \theta ,{t}_{\text{2,1}}\left(1\right)\rangle ={SL}_{n}{\left(3\right)}^{\delta -1}={SL}_{n}\left(3\right),$$

The lemma is proved.

Our computations show that

$$\begin{array}{c}{\eta }^{\alpha \gamma }={t}_{n-\text{1,1}}\left(i\right){t}_{n-\text{1,2}}\left(-i\right),\\ {\eta }^{\gamma \alpha }={t}_{1,n-2,}\left(i\right){t}_{1,n-1}\left(-i\right),\\ \lambda ={\left({\eta }^{{\left(\alpha \gamma \right)}^{2}}\right)}^{\alpha }.\end{array}$$

The element λ has the following representation:

Multiplying the last equality from the left by (η⁻¹)^αγ, we have zeros at positions (n − 1, 1) and (n − 1, 2). Furthermore,

$$\xi ={\left({\left({\eta }^{-1}\right)}^{\alpha \gamma }\lambda \right)}^{2}=diag\left(1,-\text{1,1},\dots ,1,-1\right){t}_{\text{2,1}}\left(1\right){t}_{n-\text{1,2}m-1}\left(1\right).$$

Commuting this matrix with a suitable diagonal element, we obtain a transvection. We have

$$\begin{array}{c}{\beta }^{{\theta }^{3}}=diag\left(1,{\underbrace{-1,\dots ,-1}_{2m-3}},1,{\underbrace{-1,\dots ,-1}_{2m-3}},-\text{1,1},1,-1\right),\\ \left[\xi ,{\beta }^{{\theta }^{3}}\right]={t}_{\text{2,1}}\left(1\right).\end{array}$$

Thus there is a transvection t₂,₁(1). Since t₂,₁(1),θ ∈ M , in view of Lemma 2.1, the subgroup M contains the group SL_n(3). It remains to show that the subgroup SL_n(3), together with the product of transvections t_n,_2m(i)t_n,_2m+1(−i), generates the group SL_n(9). For that, it suffices to note that the commutator

$$\left[diag{\underbrace{\left(-1,\dots ,-1\right.,}_{2m}}{\underbrace{\left.1,\dots ,1\right)}_{2m+1}},{t}_{n,2m}\left(i\right){t}_{n,2m+1}\left(-i\right)\right]$$

with diagonal matrix from SL_n(3) gives a transvection t_n,2m(i). Now Lemma 1.3(b) yields the generation of the group SL_n(9).

The case of dimensions n = 4m + 2. As generators we can take the involutions

The elements α, β, γ, αβ are involutions and have eigenvalues {1^2m+2, (−1)^2m}. So they are conjugate in SL_n(9). Put M = <α, β, γ>. Then

$$\begin{array}{c}\eta ={\left(\beta \gamma \right)}^{8}={t}_{n,3}\left(-i\right){t}_{n,n-3}\left(i\right){t}_{n,n-2}\left(i\right),\\ {\gamma }_{1}={\eta }^{2}\gamma ,\\ \theta ={\gamma }_{1}\alpha .\end{array}$$

LEMMA 2.2. We have <α, β, γ₁, t_n,₁(1)> = SL_n(3).

Proof. In a permutation representation, monomial (0, 1)-matrices α, γ₁, and θ have the forms

$$\begin{array}{c}\widetilde{\alpha }=\left(\text{1,2}\right)\dots \left(n-5,n-4\right)\left(n-3\right)\left(n-2\right)\left(n-1,n\right),\\ {\widetilde{\gamma }}_{1}=\left(1,n-1\right)\dots \left(2m,2m+2\right)\left(2m+1\right)\left(n\right),\\ \widetilde{\theta }=\left(1,n-\text{2,2},n-1,n\right)\left(3,n-\text{4,5},n-6,\dots 6,n-\text{5,4},n-3\right).\end{array}$$

Conjugating transvections by the matrix θ and commuting them between each other, we obtain the following equalities:

$$\begin{array}{c}{t}_{n,1}{\left(1\right)}^{\langle \theta \rangle }=\left\{{t}_{n,1}\left(1\right),{t}_{1,n-2}\left(1\right),{t}_{n-\text{2,2}}\left(1\right),{t}_{2,n-1}\left(1\right),{t}_{n-1,n}\left(1\right)\right\},\\ \left[{t}_{n-1,n}\left(1\right),{t}_{n,1}\left(1\right)\right]={t}_{n-\text{1,1}}\left(1\right),\\ {t}_{n,1}{\left(1\right)}^{\langle \theta \rangle }=\left\{{t}_{n,n-2}\left(1\right),{t}_{\text{1,2}}\left(1\right),{t}_{n-2,n-1}\left(1\right),{t}_{2,n}\left(1\right),{t}_{n-\text{1,1}}\left(1\right)\right\},\\ \begin{array}{c}\left[{t}_{1,n-2}\left(1\right),{t}_{n-2,n-1}\left(1\right)\right]={t}_{1,n-1}\left(1\right),\\ \left[{t}_{1,n-1}\left(1\right),{t}_{n-1,n}\left(1\right)\right]={t}_{1,n}\left(1\right),\\ \begin{array}{c}\left[{t}_{n,1}\left(1\right),{t}_{1,j}\left(1\right)\right]={t}_{n,j}\left(1\right),j=2,n-1,n-2,\\ \left[{t}_{n-\text{1,1}}\left(1\right),{t}_{1,j}\left(1\right)\right]={t}_{n-1,j}\left(1\right),j=2,n,n-2,\\ \begin{array}{c}\left[{t}_{2,n}\left(1\right),{t}_{n,1}\left(1\right)\right]={t}_{\text{2,1}}\left(1\right),\\ \left[{t}_{\text{2,1}}\left(1\right),{t}_{1,j}\left(1\right)\right]={t}_{2,j}\left(1\right),j=n-2,n-1,n,\\ \begin{array}{c}\left[{t}_{n-\text{2,2}}\left(1\right),{t}_{\text{2,1}}\left(1\right)\right]={t}_{n-\text{2,1}}\left(1\right),\\ \left[{t}_{n-\text{2,1}}\left(1\right),{t}_{1,j}\left(1\right)\right]={t}_{n-2,j}\left(1\right),j=2,n-1,n.\end{array}\end{array}\end{array}\end{array}\end{array}$$

Thus, there is a set of transvections {t_i,j(1), i,j = 1, 2,n − 2,n − 1, n}. Below is a chain of commutators where, for simplicity, computations are done without regard for the sign:

$$\begin{array}{c}\left[{t}_{1,j}{\left(1\right)}^{\beta },{t}_{j,k}\left(1\right)\right]={t}_{4,k}\left(\pm 1\right),j=n-2,n-1,n,k=\text{1,2},n-2,n-1,n,\\ \left[{t}_{2,j}{\left(1\right)}^{\beta },{t}_{j,k}\left(1\right)\right]={t}_{3,k}\left(\pm 1\right),j=n-2,n-1,n,k=\text{1,2},n-2,n-1,n,\\ \begin{array}{c}\left[{t}_{k,j}\left(1\right),{t}_{j,1}{\left(1\right)}^{\beta }\right]={t}_{k,4}\left(\pm 1\right),j=n-2,n-1,n,k=\text{1,2},n-2,n-1,n,\\ \left[{t}_{k,j}\left(1\right),{t}_{j,2}{\left(1\right)}^{\beta }\right]={t}_{k,3}\left(\pm 1\right),j=n-2,n-1,n,k=\text{1,2},n-2,n-1,n.\end{array}\end{array}$$

Conjugating the results of commutation by the matrix θ, we obtain

$$\begin{array}{c}{t}_{j,3}{\left(1\right)}^{\langle \theta \rangle }=\left\{{t}_{k,j}\left(1\right),j=\text{1,2},n-2,n-1,n,k=\text{3,4},\dots ,n-3\right\},\\ {t}_{3,j}{\left(1\right)}^{\langle \theta \rangle }=\left\{{t}_{k,j}\left(1\right),j=\text{1,2},n-2,n-1,n,k=\text{3,4},\dots ,n-3\right\}.\end{array}$$

Commuting the resulting transvections, we have

$$\left[{t}_{{k}_{1},j}\left(1\right),{t}_{j,{k}_{2}}\left(1\right)\right]={t}_{{k}_{1},{k}_{2}}\left(1\right),j=n-2,n-1,n,{k}_{1},{k}_{2}=\text{3,4},\dots ,n-3.$$

It remains to apply Lemma 1.2(a) to conclude that

$$\langle \alpha ,\beta ,{\gamma }_{1},{t}_{n,1}\left(1\right)\rangle =S{L}_{n}\left(3\right).$$

The lemma is proved.

We show that the transvection t_n,1(1) lies in the set M. We have the following conjugates of η:

$$\begin{array}{c}{\eta }^{\alpha }={t}_{n-\text{1,1}}\left(-i\right){t}_{n-\text{1,4}}\left(-i\right){t}_{n-1,n-3}\left(i\right){t}_{n-1,n-2}\left(i\right),\\ {\eta }^{\gamma \alpha }={t}_{\text{1,2}}\left(i\right){t}_{\text{1,3}}\left(i\right){t}_{1,n-4}\left(-i\right){t}_{1,n-1}\left(-i\right),\\ \begin{array}{c}{\eta }^{\beta \gamma \alpha }={t}_{\text{4,2}}\left(i\right){t}_{\text{4,3}}\left(i\right){t}_{4,n-4}\left(-i\right){t}_{4,n-1}\left(-i\right),\\ {\eta }^{\alpha \beta \gamma \alpha }={t}_{\text{3,1}}\left(i\right){t}_{\text{3,4}}\left(i\right){t}_{3,n-5}\left(-i\right){t}_{3,n}\left(-i\right),\\ {\eta }^{\gamma \alpha \beta \gamma \alpha }=\tau \left(\begin{array}{ccccccc}1& \dots & 0& 0& 0& 0& 0\\ & \ddots & & & & & \\ 0& \dots & 1& 0& 0& 0& 0\\ 0& \dots & 0& -1& 1& i& i\\ 0& \dots & 0& 0& 1& 0& 0\\ 0& \dots & 0& 0& 0& 1& 0\\ 0& \dots & 0& i& i& -1& 0\end{array}\right),\end{array}\end{array}$$

where

$$\begin{array}{l}\tau ={t}_{n-\text{3,2}}\left(-1\right){t}_{n-\text{3,3}}\left(-1\right){t}_{n-\text{3,5}}\left(-i\right){t}_{n-3,n-4}\left(i\right){t}_{n,2}\left(-i\right){t}_{n,3}\left(-i\right)\\ {t}_{n,5}\left(1\right){t}_{n,n-4}\left(-1\right).\end{array}$$

Let

$$\xi ={\left({\eta \eta }^{\gamma \alpha \beta \gamma \alpha }\right)}^{2}={t}_{n-\text{3,5}}\left(-i\right){t}_{n-3,n-4}\left(i\right){t}_{n-3,n-1}\left(i\right){t}_{n-3,n}\left(i\right).$$

Then

$${\left(\xi \alpha \right)}^{2}={t}_{n-\text{3,5}}\left(-i\right){t}_{n-\text{3,6}}\left(-i\right){t}_{n-3,n-5}\left(i\right){t}_{n-3,n-4}\left(i\right){t}_{n-3,n-1}\left(-i\right){t}_{n-3,n}\left(-i\right).$$

Further, we have two options: n = 10 or n > 10.

Subsequent computations in these two cases are different. For n = 10, the following equalities hold:

$$\begin{array}{c}{\left({\left(\xi \alpha \right)}^{2}\right)}^{{\gamma }_{1}\beta {\gamma }_{1}}={t}_{\text{8,6}}\left(-i\right){t}_{\text{8,10}}\left(i\right),\\ \vartheta ={\left({\left({\left({\left(\xi \alpha \right)}^{2}\right)}^{{\gamma }_{1}\beta {\gamma }_{1}}\beta \right)}^{2}\right)}^{\alpha }={t}_{\text{8,5}}\left(i\right),\\ {\vartheta }^{{\gamma }_{1}\beta {\gamma }_{1}}={t}_{\text{7,5}}\left(i\right)={t}_{n-3,n-5}\left(i\right).\end{array}$$

For n > 10, we have

$$\begin{array}{c}{\left({\left(\xi \alpha \right)}^{2}\right)}^{{\gamma }_{1}\beta {\gamma }_{1}}={t}_{n-\text{2,5}}\left(-i\right){t}_{n-\text{2,6}}\left(-i\right){t}_{n-2,n-5}\left(-i\right){t}_{n-2,n-4}\left(-i\right)\\ {t}_{n-2,n-1}\left(i\right){t}_{n-2,n}\left(i\right),\end{array}$$

$$\begin{array}{c}\vartheta ={\left({\left({\left(\xi \alpha \right)}^{2}\right)}^{{\gamma }_{1}\beta {\gamma }_{1}}\beta \right)}^{2}={t}_{n-2,n-5}\left(-i\right){t}_{n-2,n-4}\left(i\right),\\ {\vartheta }^{{\gamma }_{1}\beta {\gamma }_{1}}={t}_{n-3,n-5}\left(-i\right){t}_{n-3,n-1}\left(i\right),\\ \zeta ={\left({\vartheta }^{{\gamma }_{1}\beta {\gamma }_{1}}\beta \right)}^{2}={t}_{n-3,n-5}\left(i\right).\end{array}$$

In either case, the subgroup M contains a transvection t_n−3,n−5(i), and the following equalities hold:

$$\begin{array}{c}{t}_{n-3,n-5}{\left(i\right)}^{\gamma 1\alpha }\upsilon ={t}_{\text{3,4}}\left(i\right),\\ {t}_{\text{3,4}}{\left(i\right)}^{\beta }={t}_{\text{2,1}}\left(i\right),\\ \begin{array}{c}{t}_{\text{2,1}}{\left(i\right)}^{{\left(\gamma 1\alpha \right)}^{2}}={t}_{n,2}\left(i\right),\\ \left[{t}_{n,2}\left(i\right),{t}_{\text{2,1}}\left(-i\right)\right]={t}_{n,1}\left(i\right).\end{array}\end{array}$$

Lemma 2.2 implies that SL_n(3) ≤ M, but M also has, for instance, a transvection t_n−3,n−5(i).

Now, in view of Lemma 1.3(b), we obtain the equality M = SL_n(9).

The case of dimensions n = 4m + 3. We show that a triple of involutions

generates a group SL_n(9). The involutions α, β, γ, αβ have eigenvalues {1^2m+1, (−1)^2m+2}. So they are conjugate in SL_n(9). Put M = <α, β, γ>. We have

$$\begin{array}{c}\eta ={\left(\beta \gamma \right)}^{8}={t}_{n,3}\left(i\right){t}_{n,4}\left(i\right){t}_{n,n-4}\left(i\right){t}_{n,n-3}\left(i\right),\\ \begin{array}{c}{\gamma }_{1}={\eta }^{2}\gamma ,\\ \theta ={\gamma }_{1}\alpha .\end{array}\end{array}$$

LEMMA 2.3. The equality <θ, t_n,n−1(1)> = SL_n(3) holds.

Proof. Matrices α, γ₁, and θ comply with the following permutations:

$$\begin{array}{c}\widetilde{\alpha }=\left(1,n-3\right)\dots \left(2m,2m+1\right)\left(n-2\right)\left(n-1,n\right),\\ {\widetilde{\gamma }}_{1}=\left(1,n-1\right)\dots \left(2m+\text{1,2}m+2\right)\left(2\right),\\ \widetilde{\theta }=\left(\text{1,3},5,\dots ,n-\text{2,2},\text{4,6},\dots ,n-1,n\right).\end{array}$$

Since θ is a cycle of length n, there exists an element

$$\widetilde{\delta }=\left(1\right)\left(\text{3,2},\dots ,5\right)\left(n-1\right)\left(n\right)$$

such that

$${\widetilde{\theta }}^{\widetilde{\delta }}=\widetilde{\mu }=\left(\text{1,2},\dots ,n-1,n\right),$$

with

$${t}_{n,n-1}{\left(1\right)}^{\delta }={t}_{n,n-1}\left(\pm 1\right).$$

If we apply Lemma 1.3(a) we obtain

$$\begin{array}{c}\langle \mu ,{t}_{n,n-1}\left(1\right)\rangle ={SL}_{n}\left(3\right)={\langle \mu ,{t}_{n,n-1}\left(1\right)\rangle }^{\delta },\\ \langle \mu ,{t}_{n,n-1}\left(1\right)\rangle ={SL}_{n}{\left(3\right)}^{\delta -1}={SL}_{n}\left(3\right).\end{array}$$

The lemma is proved.

Thus, further proof is reduced to obtaining a transvection t_n,n−1(1). Similarly to the case of dimensions 4m + 2, we find the following conjugates of η:

$$\begin{array}{c}{\eta }^{\alpha }={t}_{n-\text{1,1}}\left(i\right){t}_{n-\text{1,2}}\left(i\right){t}_{n-1,n-6}\left(i\right){t}_{n-1,n-5}\left(i\right),\\ {\eta }^{\gamma \alpha }={t}_{\text{1,5}}\left(i\right){t}_{\text{1,6}}\left(i\right){t}_{1,n-2}\left(i\right){t}_{1,n-1}\left(i\right),\\ \begin{array}{c}{\eta }^{\beta \gamma \alpha }={t}_{\text{2,5}}\left(i\right){t}_{\text{2,6}}\left(i\right){t}_{2,n-2}\left(i\right){t}_{2,n-1}\left(-i\right),\\ {\eta }^{\alpha \beta \gamma \alpha }={t}_{n-4,n-8}\left(i\right){t}_{n-4,n-7}\left(i\right){t}_{n-4,n-2}\left(-i\right){t}_{n-4,n}\left(-i\right),\\ {\eta }^{\gamma \alpha \beta \gamma \alpha }=\tau \left(\begin{array}{cccccccc}1& 0& 0& 0& 0& \dots & 0& 0\\ 0& 1& 0& 0& 0& \dots & 0& 0\\ 0& 0& 1& 0& 0& \dots & 0& 0\\ 0& -i& 1& -1& 0& \dots & 0& i\\ & & & & & \ddots & & \\ 0& 0& 0& 0& 0& \dots & 1& 0\\ 0& 1& i& i& 0& \dots & 0& 0\end{array}\right)\end{array}\end{array}$$

where

$$\tau ={t}_{\text{4,7}}\left(i\right){t}_{\text{4,8}}\left(i\right){t}_{4,n-3}\left(1\right){t}_{4,n-4}\left(1\right){t}_{n,7}\left(-1\right){t}_{n,8}\left(-1\right){t}_{n,n-3}\left(i\right){t}_{n,n-4}\left(i\right).$$

Multiplying from the left by η and squaring the result, we arrive at a product of four transvections:

$$\xi ={\left({\eta \eta }^{\gamma \alpha \beta \gamma \alpha }\right)}^{2}={t}_{\text{4,2}}\left(-i\right){t}_{\text{4,7}}\left(i\right){t}_{\text{4,8}}\left(i\right){t}_{4,n}\left(i\right).$$

Next, we will conjugate the element ξ by the monomial matrix θ and commute the results of conjugation. Thus,

$$\begin{array}{c}{\xi }^{{\theta }^{2m}}={t}_{n,3}\left(i\right){t}_{n,6}\left(i\right){t}_{n,n-4}\left(-i\right){t}_{n,n-1}\left(i\right),\\ \left[{\xi ,\xi }^{{\theta }^{2m}}\right]={t}_{\text{4,3}}\left(-1\right){t}_{\text{4,6}}\left(-1\right){t}_{4,n-4}\left(1\right){t}_{4,n-1}\left(-1\right),\\ {\xi }_{1}={\left[{\xi ,\xi }^{{\theta }^{2m}}\right]}^{{\theta }^{2m-1}}={t}_{n-1,n-8}\left(1\right){t}_{n-1,n-7}\left(-1\right){t}_{n-1,n-2}\left(-1\right){t}_{n-1,n}\left(1\right),\end{array}$$

$$\begin{array}{c}{\xi }_{2}={\left[{{\xi }_{1},\xi }^{{\theta }^{2m}}\right]}^{3}\\ =diag\left(1,\dots ,1,-1,-1\right){t}_{n-\text{1,3},}\left(-1\right){t}_{n-\text{1,6}}\left(-1\right)\\ \begin{array}{c}{t}_{n-1,n-4}\left(1\right){t}_{n,n-8}\left(-1\right){t}_{n,n-7}\left(1\right){t}_{n,n-2}\left(1\right)\\ {\vartheta }_{1}={\left({\xi }_{2\gamma }\right)}^{4}\eta ={t}_{n,2}\left(1\right){t}_{n,7}\left(1\right){t}_{n,8}\left(-1\right){t}_{n,n-8}\left(-1\right){t}_{n,n-7}\left(1\right){t}_{n,n-2}\left(1\right).\end{array}\end{array}$$

Note that in the case m = 3 the element ϑ₁ is a product of two transvections. Our calculations yield

$$\begin{array}{c}{\vartheta }_{1}^{\beta }={t}_{n,1}\left(-1\right){t}_{n,7}\left(1\right){t}_{n,8}\left(-1\right){t}_{n,n-8}\left(-1\right){t}_{n,n-7}\left(1\right){t}_{n,n-2}\left(-1\right),\\ {\vartheta }_{2}={\vartheta }_{1}^{\beta }{\vartheta }_{1}={t}_{n,1}\left(-1\right){t}_{n,2}\left(1\right){t}_{n,7}\left(-1\right){t}_{n,8}\left(1\right){t}_{n,n-8}\left(1\right){t}_{n,n-7}\left(-1\right),\\ \begin{array}{c}{\vartheta }_{2}^{{\gamma }_{1}}={t}_{n,7}\left(1\right){t}_{n,8}\left(-1\right){t}_{n,n-8}\left(-1\right){t}_{n,n-7}\left(1\right){t}_{n,n-2}\left(-1\right){t}_{n,n-1}\left(1\right),\\ {\left({\vartheta }_{2}^{{\gamma }_{1}}\beta \right)}^{2}={t}_{n,7}\left(-1\right){t}_{n,8}\left(1\right){t}_{n,n-8}\left(1\right){t}_{n,n-7}\left(-1\right){t}_{n,n-1}\left(-1\right),\\ {\left({\left({\vartheta }_{2}^{{\gamma }_{1}}\beta \right)}^{2}\right)}^{{\gamma }_{1}}={t}_{n,1}\left(1\right){t}_{n,7}\left(1\right){t}_{n,8}\left(-1\right){t}_{n,n-8}\left(-1\right){t}_{n,n-7}\left(1\right).\end{array}\end{array}$$

Multiplying the last two equalities gives a product of two transvections:

$${\vartheta }_{3}={\left({\vartheta }_{2}^{{\gamma }_{1}}\beta \right)}^{2}{\left({\left({\vartheta }_{2}^{{\gamma }_{1}}\beta \right)}^{2}\right)}^{{\gamma }_{1}}={t}_{n,1}\left(1\right){t}_{n,n-1}\left(-1\right).$$

Furthermore,

$$\begin{array}{c}{\vartheta }_{3}^{\beta }={t}_{n,2}\left(-1\right){t}_{n,n-1}\left(-1\right),\\ {\vartheta }_{3}^{2}{\vartheta }_{3}^{\beta }={t}_{n,1}\left(-1\right){t}_{n,2}\left(-1\right),\\ \begin{array}{c}{\left({\vartheta }_{3}^{2}{\vartheta }_{3}^{\beta }\right)}^{{\gamma }_{1}}={t}_{n,n-1}\left(1\right){t}_{n,n-2}\left(1\right),\\ {\left({\left({\vartheta }_{3}^{2}{\vartheta }_{3}^{\beta }\right)}^{{\gamma }_{1}}\beta \right)}^{2}={t}_{n,n-1}\left(-1\right).\end{array}\end{array}$$

Consequently, SL_n(3) ≤ M in virtue of Lemma 2.3. Matrix d = diag(1, 1, −1, 1,... , 1, −1) lies in SL_n(3) and (d(βγ)⁸)² = t_n,3(−i) ∈ M , Hence, by Lemma 1.3(b), the involutions α, β, γ generate SL_n(9).

The case of dimensions n = 4m. The scheme of the proof is analogous to the previous cases.

We have

The elements α, β, γ, αβ are involutions and have eigenvalues {1^2m, (−1)^2m}. So they are conjugate in SL_n(9). As in the previous cases, we obtain in M = <α, β, γ> the monomial matrices

$$\begin{array}{c}\eta ={\left(\beta \gamma \right)}^{8}={t}_{n,2}\left(i\right){t}_{n,3}\left(i\right){t}_{n,n-3}\left(i\right){t}_{n,n-2}\left(i\right),\\ {\gamma }_{1}={\eta }^{2}\gamma \\ \theta =\alpha {\gamma }_{1}.\end{array}$$

LEMMA 2.4. The equality <θ, t_n,1(1)> = SL_n(3) holds.

Proof. We write the appropriate permutations

$$\begin{array}{c}{\widetilde{\gamma }}_{1}=\left(1,n-1\right)\left(2,n-2\right)\dots \left(2m-\text{1,2}m+1\right)\left(2m\right)\left(n\right),\\ \widetilde{\alpha }=\left(\text{1,2}\right)\left(\text{3,4}\right)\dots \left(n-1,n\right),\\ \widetilde{\theta }=\left(1,n-\text{2,3},n-4,\dots ,2,n-1,n\right).\end{array}$$

The element $\widetilde{\theta }$ is a cycle of length n. Consequently, there exists an element

$$\widetilde{\delta }=\left(1\right)\left(n-\text{2,2}\right)\left(3\right)\left(n-\text{4,4}\right)\dots \left(n-1\right)\left(n\right)$$

such that

$${\widetilde{\theta }}^{\widetilde{\delta }}=\widetilde{\mu }=\left(\text{1,2},3,\dots ,n-1,n\right),$$

with

$${t}_{n,1}{\left(1\right)}^{\delta }={t}_{n,1}\left(\pm 1\right).$$

Passing to the conjugate subgroup

$${\langle \theta ,{t}_{n,1}\left(1\right)\rangle }^{\delta }=\langle \mu ,{t}_{n,1}\left(1\right)\rangle $$

and applying Lemma 1.3(a), we obtain

$$\langle \theta ,{t}_{n,1}\left(1\right)\rangle ={\langle \mu ,{t}_{n,1}\left(1\right)\rangle }^{\delta -1}={SL}_{n}{\left(3\right)}^{\delta -1}={SL}_{n}\left(3\right).$$

The lemma is proved.

It remains to show that the transvection t_n,1(1) lies in the subgroup M. For that, as in the previous two cases, we conjugate the matrix η, thus obtaining

$$\begin{array}{c}{\eta }^{\alpha }={t}_{n-\text{1,1}}\left(i\right){t}_{n-\text{1,4}}\left(i\right){t}_{n-1,n-3}\left(i\right){t}_{n-1,n-2}\left(i\right),\\ {\eta }^{\gamma \alpha }={t}_{\text{1,2}}\left(i\right){t}_{\text{1,3}}\left(i\right){t}_{1,n-4}\left(i\right){t}_{1,n-1}\left(i\right),\\ \begin{array}{c}{\eta }^{\beta \gamma \alpha }={t}_{\text{4,2}}\left(i\right){t}_{\text{4,3}}\left(i\right){t}_{4,n-4}\left(i\right){t}_{4,n-1}\left(-i\right),\\ {\eta }^{\alpha \beta \gamma \alpha }={t}_{\text{3,1}}\left(i\right){t}_{\text{3,4}}\left(i\right){t}_{3,n-5}\left(i\right){t}_{3,n}\left(-i\right).\end{array}\end{array}$$

Let

$$\tau ={t}_{n-\text{3,2}}\left(1\right){t}_{n-\text{3,3}}\left(1\right){t}_{n-\text{3,5}}\left(i\right){t}_{n-3,n-4}\left(i\right){t}_{n,2}\left(i\right){t}_{n,3}\left(i\right){t}_{n,5}\left(-1\right){t}_{n,n-4}\left(-1\right).$$

Then

$$\begin{array}{c}{\eta }^{\alpha \beta \gamma \alpha }=\tau \left(\begin{array}{cccccccc}1& \dots & 0& 0& 0& 0& 0& 0\\ & \ddots & & & & & & \\ 0& \dots & 1& 0& 0& 0& 0& 0\\ 0& \dots & 0& -1& 1& i& i& i\\ 0& \dots & 0& 0& 1& 0& 0& 0\\ 0& \dots & 0& 0& 0& 1& 1& 0\\ 0& \dots & 0& i& i& -1& -1& 0\end{array}\right),\\ \xi ={\left({\eta \eta }^{\gamma \alpha \beta \gamma \alpha }\right)}^{2}={t}_{n-\text{3,5}}\left(i\right){t}_{n-3,n-4}\left(i\right){t}_{n-3,n-1}\left(i\right){t}_{n-3,n}\left(i\right),\\ {\left(\xi \beta \right)}^{2}={t}_{n-3,n-4}\left(-i\right).\end{array}$$

Now, having obtained a transvection, we will use conjugation by monomial matrices and the matrix γ. Our calculations show that

$$\begin{array}{c}{\left({\left(\xi \beta \right)}^{2}\right)}^{\gamma 1}={t}_{\text{3,4}}\left(-i\right),\\ {\left({\left(\xi \beta \right)}^{2}\right)}^{\gamma }={t}_{\text{3,4}}\left(-i\right){t}_{n,4}\left(1\right),\\ \begin{array}{c}{\left({\left(\xi \beta \right)}^{2}\right)}^{\gamma }{\left({\left({\left(\xi \beta \right)}^{2}\right)}^{\gamma }\right)}^{2}={t}_{n,4}\left(1\right),\\ {t}_{n,4}{\left(1\right)}^{{\theta }^{2}}={t}_{\text{2,6}}\left(-1\right).\end{array}\end{array}$$

Again using the elements γ and γ₁ for conjugation, we have

$$\begin{array}{c}{\left({t}_{n,4}{\left(1\right)}^{{\theta }^{2}}\right)}^{\gamma }={t}_{n-2,n-6}\left(-1\right){t}_{n,n-6}\left(-i\right)={\left({t}_{n,4}{\left(1\right)}^{{\theta }^{2}}\right)}^{\gamma 1}={t}_{n,n-6}\left(-i\right),\\ {\left({t}_{n,4}{\left(1\right)}^{{\theta }^{2}}\right)}^{\gamma 1}{\left({\left({t}_{n,4}{\left(1\right)}^{{\theta }^{2}}\right)}^{\gamma }\right)}^{2}={t}_{n,n-6}\left(i\right),\\ \begin{array}{c}{t}_{n,n-6}{\left(i\right)}^{\gamma 1}={t}_{n,-6}\left(-i\right),\\ {\left({\left(\xi \beta \right)}^{2}\right)}^{\theta 3}={t}_{6,1}\left(-i\right)\\ \left[{t}_{n,6}\left(-i\right),{t}_{6,1}\left(-i\right)\right]={t}_{n,1}\left(1\right).\end{array}\end{array}$$

Applying Lemmas 2.4 and 1.3(b), we arrive at the equality <θ, t_n,1(1), t₆,₁(i)> = SL_n(9).

The case of dimension n = 6. In [3] the following generating involutions for the group PSL₆(9) are given:

$$\begin{array}{c}\alpha =\left(\begin{array}{cccccc}0& 1& 0& 0& 0& 0\\ -1& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& -1& 0& 0& 0\\ 0& 0& 0& 0& -1& 1\\ 0& 0& 0& 0& 0& 0\end{array}\right),\\ \beta =\left(\begin{array}{cccccc}0& 0& 0& 1& 0& 0\\ 0& 0& 1& 0& 0& 0\\ 0& -1& 0& 0& 0& 0\\ -1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& i& 1\\ 0& 0& 0& 0& 0& -i\end{array}\right),\\ \gamma =\left(\begin{array}{cccccc}0& 1& 0& 0& 0& 0\\ -1& 0& 0& 0& 0& 0\\ 0& 0& -i& 0& 0& 0\\ 0& 0& 0& 0& 1& 0\\ 0& 0& 0& -1& 0& 0\\ a& ia& 0& a& ia& i\end{array}\right),\end{array}$$

where a is an element of the multiplicative group of the field GF (9) satisfying some restrictions.

The matrices α, β, γ, αβ, when squared, give a scalar matrix −E₆; i.e., being elements of the group SL₆(9), they have order 4 and eigenvalues {i³, (−i)³} and are therefore conjugate in SL₆(9). Consequently, their images are conjugate in the group PSL₆(9). The elements (βγ)⁵ and ((βγ)⁵)^α of a group M = <α, β, γ> are transvections [3], i.e., are conjugate in SL₆(9) to elementary transvections. It is known that for any pair of transvections (ξ, η) in SL_n(q) there is an element ε in GL_n(q) such that ξ^ε and η^ε are elementary transvections [12]; see also [3, Lemma 4.1.8; 13, Lemma 1]. Furthermore, we show that α, β, γ generate SL₆(9), with the proof for n = 6, as distinct from the previous cases, making essential use of computer calculations. We put a = i and pass to a conjugate subgroup M^τ, in which (β^τ γ^τ )⁵ and ((β^τ γ^τ )⁵)^ατ are elementary transvections.

Below a pair of indices in elementary transvections will not be separated by a comma since all indices are numbers from 1 to 6. Let

$$\begin{array}{l}\tau =\left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ -i& 1& 0& 0& 0& 0\\ 1& -i& 1& 0& 0& 0\\ i& -1& i& 0& 0& 0\\ 1& i& 1& -i& 1& 0\\ -i& 1& i& 1& 0& 1\end{array}\right)\\ \theta ={\alpha }^{\tau }{\gamma }^{\tau }.\end{array},$$

We show that M^τ = <α^τ , β^τ , γ^τ> = SL₆(9), whence the equality <α, β, γ> = SL₆(9). In subsequent calculations, we show that M^τ contains all elementary transvections. So

$$\left({\beta }^{\tau }{\gamma }^{\tau }\right)={t}_{65}\left(-i-1\right).$$

Conjugation by the matrix θ yields the equality

$${\left({t}_{65}{\left(-i-1\right)}^{{\alpha }^{\tau }}\right)}^{-1}={t}_{65}\left(-i-1\right).$$

Next, we define

$$\begin{array}{c}\xi ={\left({t}_{65}{\left(-i-1\right)}^{\theta -1}{t}_{65}\left(-i-1\right)\right)}^{2},\\ \vartheta ={\left({t}_{56}{\left(-i-1\right)}^{\theta }{t}_{56}\left(-i-1\right)\right)}^{2},\\ \zeta ={\left({\xi }^{3}{\vartheta }^{{\alpha }^{\tau }}\right)}^{2}={t}_{45}\left(-i\right){t}_{61}\left(i\right){t}_{62}\left(i\right){t}_{63}\left(i\right)diag\left(\text{1,1},\text{1,1},-1,-1,\right).\end{array}$$

After obtaining the element ζ, calculations will be more transparent. In the chain of equalities below, we show that the subgroup <θ, ζ> contains a transvection t₆₂(-i). In fact,

$$\begin{array}{c}{\xi }^{\theta }=diag\left(\text{1,1},\text{1,1},-1,-1\right),\\ {\xi }^{\theta }={t}_{36}\left(i\right){t}_{46}\left(-1\right){t}_{51}\left(-1\right){t}_{52}\left(-i-1\right){t}_{53}\left(-1\right){t}_{54}\left(-i\right)diag\left(\text{1,1},\text{1,1},-1,-1\right),\\ \begin{array}{c}{\xi }_{1}={\left(\xi {\xi }^{\theta 2}\right)}^{2}={t}_{31}\left(1\right){t}_{32}\left(1\right){t}_{42}\left(1\right)diag\left(\text{1,1},-1,-1,-1,-1\right),\\ {\xi }_{2}={\left({\left({\xi }^{\theta 3}\right)}^{\theta 2}{\xi }^{\theta }\right)}^{2}={t}_{35}\left(1\right){t}_{62}\left(1-i\right){t}_{64}\left(1\right),\\ \left[{\xi }_{1},{\xi }_{2}\right]={t}_{62}\left(-i\right).\end{array}\end{array}$$

Furthermore,

$$\begin{array}{c}\left[{\beta }^{\tau },{t}_{62}\left(-i\right)\right]={t}_{64}\left(-i\right),\\ {\left({t}_{64}\left(-i\right)\xi \right)}^{2}={t}_{65}\left(-1\right),\\ {\left({t}_{64}\left(-i\right)\right)}^{\alpha \tau }={t}_{56}\left(1\right).\end{array}$$

So the transvections t₅₆(1), t₆₅(1), t₅₆(−i − 1), t₆₅(−i − 1) lie in the set M^τ. Hence the sets t₅₆(GF (9)), t₆₅(GF (9)) also lie in M^τ. Let u be an arbitrary element of the field GF (9). Then

$$\begin{array}{c}\left[{t}_{56}\left(u\right),{t}_{64}\left(-i\right)\right]={t}_{54}\left(-iu\right),\\ \left[{t}_{56}\left(u\right),{t}_{62}\left(-i\right)\right]={t}_{52}\left(-iu\right).\end{array}$$

Obviously, t₅₄(u) and t₅₂(u) also lie in M^τ . The following hold:

$$\begin{array}{c}{{t}_{56}\left(u\right)}^{{\alpha }^{\tau }}={t}_{56}\left(-iu\right),\\ {{t}_{54}\left(u\right)}^{{\alpha }^{\tau }}={t}_{64}\left(-iu\right){t}_{62}\left(-iu\right),\\ \begin{array}{c}\left[{t}_{35}\left(1\right),{t}_{62}\left(1-i\right){t}_{64}\left(1\right),{t}_{54}\left(u\right)\right]={t}_{34}\left(u\right),\\ \left[{t}_{35}\left(1\right),{t}_{62}\left(1-i\right){t}_{64}\left(1\right),{t}_{52}\left(u\right)\right]={t}_{32}\left(u\right).\end{array}\end{array}$$

Thus, we have obtained elements t_k2(u) for k = 3, 5, 6. Furthermore,

$$\begin{array}{c}{t}_{35}{\left(u\right)}^{{\gamma }^{\tau }}={t}_{32}\left(u\right){t}_{42}\left(iu\right){t}_{62}\left(-iu\right),\\ \left[{t}_{34}\left(u\right),{\beta }^{\tau }\right]={t}_{14}\left(u\right),\\ {t}_{14}{\left(u\right)}^{{\alpha }^{\tau }}={t}_{34}\left(u\right){t}_{32}\left(-u\right){t}_{14}\left(u\right){t}_{12}\left(-u\right).\end{array}$$

From this chain of equalities, we conclude that, for any k ≠ 2, the set t_k2(GF (9)) lies in M^τ. Again subsequent calculations yield the equalities

$$\begin{array}{c}{\beta }_{1}={t}_{34}\left(-i\right){t}_{12}\left(i\right){t}_{32}\left(i\right){t}_{14}\left(-i\right){\beta }^{\tau }=diag\left(-i,-i,i,i,i,-i\right){t}_{13}\left(1\right){t}_{24}\left(1\right),\\ {{t}_{64}\left(-i\right)}^{\gamma \tau }{t}_{62}\left(-i\right){t}_{64}\left(i\right){t}_{65}\left(1\right)={t}_{61}\left(-i\right){t}_{63}\left(i\right),\\ \begin{array}{c}{\left({t}_{61}\left(-i\right){t}_{63}\left(-1\right)\right)}^{\beta \tau }{t}_{62}\left(i\right){t}_{64}\left(-i\right)={t}_{61}\left(-1\right),\\ {{t}_{61}\left(-1\right)}^{\alpha \tau }{t}_{52}\left(-1\right)={t}_{51}\left(i\right),\\ \begin{array}{c}{51\left(-i\right)}^{\gamma \tau }{t}_{52}\left(1\right){t}_{51}\left(i\right){t}_{42}\left(i\right)={t}_{41}\left(1\right),\\ {\left({\beta }_{1}{t}_{41}\left(1\right)\right)}^{2}={t}_{21}\left(1\right){t}_{23}\left(1\right){t}_{43}\left(-1\right),\\ \begin{array}{c}{\left({\left({\beta }_{1}{t}_{41}\left(1\right)\right)}^{2}{t}_{32}\left(-1\right){t}_{42}\left(-1\right){\xi }_{1}\right)}^{2}{t}_{41}\left(-1\right)={t}_{43}\left(1\right),\\ {\xi }_{2}{t}_{43}\left(i-1\right){t}_{64}\left(-1\right)={t}_{35}\left(1\right).\end{array}\end{array}\end{array}\end{array}$$

Previously, we have obtained transvections t₁₄(u) and t₁₂(u). Note that

$$\begin{array}{c}\left[{t}_{14}\left(u\right){t}_{43}\left(1\right)\right]={t}_{13}\left(u\right),\\ \left[{t}_{13}\left(u\right){t}_{35}\left(1\right)\right]={t}_{15}\left(u\right),\\ \left[{t}_{15}\left(u\right){t}_{56}\left(1\right)\right]={t}_{16}\left(u\right).\end{array}$$

Hence M^τ contains sets t_1k(GF (9)) for any k ≠ 1. Furthermore,

$$\left[{t}_{41}\left(1\right){t}_{1k}\left(GF\left(9\right)\right)\right]={t}_{4k}\left(GF\left(9\right)\right),k\ne \text{4,1}.$$

Transvection t₄₁(i) can be obtained, for instance, as t₄₁(i) = [t₄₂(i), (β₁t₄₁(1))²]t₄₃(−i). In other words, the sets t_4k(GF (9)), for any k ≠ 4, lie in M^τ . The following equality holds:

$$\left[{t}_{14}\left(1\right){t}_{4k}\left(GF\left(9\right)\right)\right]={t}_{lk}\left(GF\left(9\right)\right),k\ne l,l=\text{3,5},6.$$

Transvections t₃₄(u), t₅₄(u), t₆₄(u) have been obtained previously. It remains to obtain

2. We have

$$\begin{array}{c}\left[{t}_{21}\left(1\right){t}_{23}\left(1\right){t}_{43}\left(-1\right),{t}_{15}\left(u\right)\right]={t}_{25}\left(u\right),\\ \left[{t}_{25}\left(u\right){t}_{5k}\left(1\right)\right]={t}_{2k}\left(u\right),k\ne \text{2,5}.\end{array}$$

Thus M^τ contains all transvections t_lk(u) with k ≠ l and u ∈ GF (9). Applying Lemma 1.2(b), we complete the proof of the theorem.