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The (2, 3)-generation of the finite 8-dimensional orthogonal groups

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Published/Copyright: August 25, 2022

Abstract

We construct ( 2 , 3 ) -generators for the finite 8-dimensional orthogonal groups, proving the following results: the groups Ω 8 + ( q ) and P Ω 8 + ( q ) are ( 2 , 3 ) -generated if and only if q 4 ; the groups Ω 8 - ( q ) and P Ω 8 - ( q ) are ( 2 , 3 ) -generated for all q 2 .

1 Introduction

A group is said to be ( 2 , 3 ) -generated if it can be generated by an involution and an element of order 3, equivalently if it is an epimorphic image of

C 2 C 3 PSL 2 ( Z ) .

By a famous result of Liebeck and Shalev [8], the finite simple classical groups are ( 2 , 3 ) -generated, apart from the two infinite families PSp 4 ( q ) with q = 2 f , 3 f , and a finite list ℒ of exceptions. The problem of determining which are exactly the members of this list ℒ requires a lot of detailed analysis, especially in small dimensions. Thanks to the efforts of many authors (see in particular [10, 12, 13] and the references within), the list ℒ contains the groups of items ( 2 ) ( 8 ) below and, with this contribution, remains possibly open only for orthogonal groups of dimension at least 9.

According to [11], the following finite simple classical groups are not ( 2 , 3 ) -generated:

  1. PSp 4 ( q ) P Ω 5 ( q ) with q = 2 f , 3 f ;

  2. PSL 2 ( 9 ) P Ω 3 ( 9 ) P Ω 4 - ( 3 ) ;

  3. PSL 3 ( 4 ) ;

  4. PSL 4 ( 2 ) P Ω 6 + ( 2 ) ;

  5. PSU 3 ( q 2 ) with q = 3 , 5 ;

  6. PSU 4 ( q 2 ) P Ω 6 - ( q ) with q = 2 , 3 ;

  7. PSU 5 ( 2 2 ) ;

  8. P Ω 8 + ( q ) with q = 2 , 3 .

The isomorphisms are well known; see [4]. From this list, it turns out that most exceptions are realized by orthogonal groups. Also for this reason, they are the most difficult case to be studied with respect to the ( 2 , 3 ) -generation problem. The two exceptions P Ω 8 + ( 2 ) and P Ω 8 + ( 3 ) have been found by Vsemirnov [14], while in [9] it was proved that the groups Ω 7 ( q ) are ( 2 , 3 ) -generated for all odd 𝑞. Note that, for 𝑞 even, Ω 7 ( q ) Sp 6 ( q ) . It is worth mentioning that all the known exceptions are ( 2 , 5 ) -generated, apart from PSU 3 ( 3 2 ) which is ( 2 , 7 ) -generated. So they are all ( 2 , r ) -generated for some small prime 𝑟.

Here, given a finite 8-dimensional orthogonal group, we construct plenty of ( 2 , 3 ) -generating pairs x , y , where 𝑦 is fixed and 𝑥 depends on the characteristic and on one or two parameters. In particular, we prove the following theorem.

Theorem 1.1

The groups Ω 8 + ( q ) and P Ω 8 + ( q ) are ( 2 , 3 ) -generated if and only if q 4 , while the groups Ω 8 - ( q ) and P Ω 8 - ( q ) are ( 2 , 3 ) -generated for all q 2 .

2 Notation and preliminary results

Denote by 𝔽 the algebraic closure of the field F p of prime order 𝑝. For every q = p f , the group GL 8 ( q ) acts on the left on the space F q 8 , whose canonical basis is denoted by { e 1 , e 2 , e 3 , e 4 , e - 1 , e - 2 , e - 3 , e - 4 } . Denote by 𝜔 a primitive cube root of 1 in 𝔽.

Given g Mat m , n ( F ) , m n , we denote by g ( i 1 , , i n ) the square submatrix of 𝑔 consisting of rows i 1 , , i n . Moreover, if 𝜎 is a field automorphism, we write g σ for the matrix obtained from 𝑔 applying 𝜎 to its entries.

The lists of maximal subgroups of Ω 8 + ( q ) , obtained by Kleidman [6], and of Ω 8 - ( q ) are described in [1, Tables 8.50–8.53]: this will be our main reference for the subdivision in classes C 1 , , C 6 , S , and for the structure of these subgroups.

In the following lemma, 𝐻 may be any subgroup of GL n ( F ) .

Lemma 2.1

Let 𝑈 be an 𝐻-invariant subspace of F n . Then there exists an H T -invariant subspace U ¯ such that dim ( U ¯ ) = codim ( U ) and, if 𝜆 is an eigenvalue of some h H , then either h | U or ( h T ) | U ¯ has the eigenvalue 𝜆.

Proof

Let B 1 = { u 1 , , u k } be a basis of 𝑈 and B 2 = { u k + 1 , , u n } a basis of any complement 𝑊 of 𝑈. Consider the matrix 𝑔 having as columns the elements of B 1 B 2 . From g e i = u i , 1 i k , and H U = U , we get that g - 1 H g fixes S 1 = e 1 , , e k . So ( g - 1 H g ) T = g T H T g - T fixes S 2 = e k + 1 , , e n , and hence H T fixes U ¯ = g - T S 2 . Finally, the matrix of h | U with respect to B 1 is the same as the matrix of ( g - 1 h g ) | S 1 with respect to the canonical basis. Thus, if h | U does not have the eigenvalue 𝜆, the same applies to ( g - 1 h g ) | S 1 . In this case, ( g T h T g - T ) | S 2 must have the eigenvalue 𝜆. Our claim is proved by noting that, if v S 2 is an eigenvector of g T h T g - T , then g - T v U ¯ is an eigenvector of h T . ∎

Assume further that 𝐻 preserves a non-singular form 𝐽, namely that

J h J - 1 = h - T for all h H .

It follows that J U ¯ is an 𝐻-invariant subspace of dimension n - dim ( U ) . In particular, this applies to H Ω 2 m ± ( q ) , in which case we may always assume that dim ( U ) m .

Given an eigenvalue 𝜆 of a matrix g GL n ( F ) , with corresponding eigenspace V λ ( g ) , we define the 𝑔-invariant subspace E λ ( g ) = V λ ( g ) Im ( g - I n ) .

3 Generators for Ω 8 + ( q ) and the triality automorphism

Since Ω 8 + ( 2 ) and Ω 8 + ( 3 ) are not ( 2 , 3 ) -generated, in this section and in Section 4, we assume q 2 , 3 . Let a F q * be such that F p [ a ] = F q . We define our generators x = x ( a ) , y as follows:

x = ( 0 0 0 0 0 1 0 0 0 0 0 0 1 ( a + 1 ) 2 ( a + 1 ) 2 a 0 0 0 0 0 ( a + 1 ) 2 ( a + 1 ) 2 a 0 0 0 0 0 a a 1 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 a 0 0 0 0 0 0 a ( a + 1 ) 2 0 0 0 0 ) if p = 2 , x = ( 0 - 1 0 0 0 0 0 0 - 1 0 0 0 0 0 0 0 1 - 1 - 1 - a 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 - 1 1 0 0 0 0 - 1 - 1 0 - 1 0 0 0 0 - 2 0 0 - 1 0 1 - 1 - 2 - 2 a 0 0 - a 1 ) if p > 2 , y = ( 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 ) .

Then 𝑥 and 𝑦 have respective orders 2 and 3. Moreover, they belong to the group SO 8 + ( q ) defined by the quadratic form

Q ( i = 1 4 ( α i e i + α - i e - i ) ) = i = 1 4 α i α - i .

The group P Ω 8 + ( q ) has an outer automorphism 𝜏 of order 3, called triality, which arises from a symmetry of order 3 of the Dynkin diagram of type D 4 . It can also be read from the spin representation of the Clifford group (e.g., see [15, pages 78–81]). We describe 𝜏 in terms of the Steinberg generators of Ω 8 + ( q ) (see [3, page 185]), namely

x 1 , 2 ( α ) = I 8 + α ( E 1 , 2 - E - 2 , - 1 ) , x 2 , 1 ( α ) = ( x 1 , 2 ( α ) ) T , x 2 , 3 ( α ) = I 8 + α ( E 2 , 3 - E - 3 , - 2 ) , x 3 , 2 ( α ) = ( x 2 , 3 ( α ) ) T , x 3 , 4 ( α ) = I 8 + α ( E 3 , 4 - E - 4 , - 3 ) , x 4 , 3 ( α ) = ( x 3 , 4 ( α ) ) T , x 1 , - 2 ( α ) = I 8 + α ( E 1 , - 2 - E 2 , - 1 ) , x - 2 , 1 ( α ) = ( x 1 , - 2 ( α ) ) T .

Then 𝜏 is defined by the following permutation action [3, Proposition 12.2.3]:

(3.1) ( x 1 , 2 ( α ) , x 3 , 4 ( α ) , x - 2 , 1 ( α ) ) ( x 2 , 1 ( α ) , x 4 , 3 ( α ) , x 1 , - 2 ( α ) ) .

We need 𝜏 mainly because it moves certain maximal subgroups from class C 1 to some other class C j . So, to exclude the possibility that H := x , y is contained in M C j , we show that τ i ( H ) , i = 1 , 2 , is absolutely irreducible, hence it cannot be contained in any maximal subgroup of class C 1 .

In order to calculate τ ( x ) and τ ( y ) , we need to express x , y as products of the above generators, proving in particular that H Ω 8 + ( q ) . To this purpose, it is useful to consider the factorization P Ω 8 + ( q ) = B N B , where 𝐵 is the Borel subgroup and 𝑁 the monomial subgroup. As a special case of the formula

[ I n + α E i , j , I n + β E j , k ] = I n + α β E i , k

for i , j , k distinct, we get

x 2 , 4 ( α ) = [ x 2 , 3 ( α ) , x 3 , 4 ( 1 ) ] , x 1 , 3 ( α ) = [ x 1 , 2 ( α ) , x 2 , 3 ( 1 ) ] , x 1 , 4 ( α ) = [ x 1 , 3 ( α ) , x 3 , 4 ( 1 ) ] .

Now set

π 1 , 2 = x 2 , 1 ( - 1 ) x 1 , 2 ( 1 ) x 2 , 1 ( - 1 ) , π 2 , 3 = x 3 , 2 ( - 1 ) x 2 , 3 ( 1 ) x 3 , 2 ( - 1 ) , π 3 , 4 = x 4 , 3 ( - 1 ) x 3 , 4 ( 1 ) x 4 , 3 ( - 1 ) , π 1 , - 2 = x - 2 , 1 ( - 1 ) x 1 , - 2 ( 1 ) x - 2 , 1 ( - 1 ) , π 1 , 3 = π 1 , 2 - 1 π 2 , 3 π 1 , 2 , π 2 , 4 = π 2 , 3 - 1 π 3 , 4 π 2 , 3 ,

and then

ρ 1 , 2 = π 1 , 2 π 1 , - 2 , ρ 3 , 4 = π 1 , 3 π 2 , 4 ρ 1 , 2 π 1 , 3 π 2 , 4 , J = ρ 1 , 2 ρ 3 , 4 .

Finally, we define

x - 1 , 3 ( α ) = ( x - 2 , 1 ( α ) ) π 2 , 3 , x - 1 , 4 ( α ) = ( x - 1 , 3 ( α ) ) π 3 , 4 , x - 2 , 4 ( α ) = ( x - 1 , 4 ( α ) ) π 1 , 2 , x - 3 , 4 ( α ) = ( x - 2 , 4 ( α ) ) π 2 , 3 .

Now H Ω 8 + ( q ) since we can write

y = ρ 1 , 2 π 2 , 3 ρ 1 , 2 π 3 , 4 3

and

x = { x 2 , 3 ( 1 ) x 3 , 4 ( a ) π 1 , 2 x 4 , 3 ( a ) x 3 , 2 ( 1 ) J if p = 2 , x - 1 , 4 ( - 1 ) x - 2 , 4 ( 1 ) x - 3 , 4 ( 2 ) x 1 , 2 ( 1 ) x 2 , 3 ( - 1 ) x 2 , 4 ( - a ) x 1 , 3 ( 1 ) x 1 , 4 ( a ) π 2 , 3 2 π 1 , 3 - 1 x 1 , 2 ( - 1 ) x 2 , 3 ( 1 ) x 1 , 4 ( - a ) if p > 2 .

Applying (3.1), we get

τ ( y ) = ( 0 0 0 - 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 - 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 - 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 - 1 0 0 1 0 0 0 0 0 0 ) , τ 2 ( y ) = ( 0 0 0 0 0 0 0 1 - 1 0 0 0 0 0 0 0 0 0 - 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 - 1 0 0 0 0 0 0 0 0 0 - 1 0 0 1 0 0 0 0 0 0 ) .

Furthermore, τ ( x ) and τ 2 ( x ) are, respectively, the matrices

( 0 a a 0 1 0 0 0 a 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 a 2 + 1 0 0 0 0 a 0 0 0 a 2 + 1 a 2 + 1 0 a 0 0 0 0 a 2 + 1 a 2 + 1 1 a 0 0 0 0 0 1 0 0 0 0 0 ) , ( a a 2 + 1 a 2 + 1 0 0 0 0 0 1 a a 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 a 1 0 0 0 0 0 0 a 2 + 1 a 0 0 0 0 1 0 a 2 + 1 a 0 0 0 0 0 1 0 0 0 0 )

if p = 2 and

( 1 0 0 0 0 0 0 0 0 - 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 - a 0 0 1 0 0 0 - a 0 - 1 1 0 - 1 1 1 0 - 1 - 2 0 0 0 0 1 0 1 0 2 0 0 1 0 ) , ( 0 0 0 - 1 - a 1 - 1 0 0 0 0 0 1 0 0 0 0 - 1 - 1 0 - 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 a 0 - 1 0 0 - 1 0 0 0 0 - 2 0 0 - 1 0 0 - 1 - 2 0 - 1 0 0 1 )

if p > 2 .

Lemma 3.1

If p = 2 , the groups 𝐻, τ ( H ) and τ 2 ( H ) are absolutely irreducible.

Proof

We begin by proving the absolute irreducibility of 𝐻. The eigenspace V ω ( y ) is generated by v i = v i ( ω ) , i = 1 , 2 , where

v 1 = e 2 + ω - 1 e - 3 + ω e - 4 and v 2 = e 3 + ω - 1 e 4 + ω e - 2 .

Let { 0 } U and U ¯ be, respectively, an 𝐻- and an H T -invariant subspace of F 8 as in Lemma 2.1. We may assume dim U 4 . If U V ω ( y ) = { 0 } , then we have V ω ( y T ) U ¯ . From y T = y - 1 , it follows that V ω ( y T ) = v 1 - , v 2 - , where v 1 - = v 1 ( ω - 1 ) and v 2 - = v 2 ( ω - 1 ) . The matrix 𝑀 whose columns are

v 1 - , x T v 1 - , ( x y ) T v 1 - , ( x y x ) T v 1 - , v 2 - , x T v 2 - , ( x y ) T v 2 - , ( x y x ) T v 2 -

has determinant ( a + 1 ) 4 ( a 2 + a + ω ) 4 , which is nonzero unless a 2 = a + ω . In this case, take the matrix M ¯ obtained by replacing the last two columns of 𝑀 with ( ( x y ) 2 ) T v 1 - and ( ( x y ) 2 x ) T v 1 - . Then det ( M ¯ ) = ( a + 1 ) 3 0 . So, in any case, we obtain the contradiction U = { 0 } . It follows that, for some ( b 1 , b 2 ) ( 0 , 0 ) , v = b 1 v 1 + b 2 v 2 U .

Let 𝑁 be the matrix whose columns are the 𝑣, x v , y x v , y 2 x v and ( y x ) 2 v . If v = v 2 , then det ( N ( 1 , 2 , 3 , 4 , 8 ) ) = ( a + 1 ) 6 , and hence 𝑁 has rank 5, in contrast with dim U 4 . If v = v 1 , again v 2 U as

( a 2 + a ) v 2 = ( a 2 + ω 2 a + 1 ) v 1 + ω x v 1 + y x v 1 + ω 2 y 2 x v 1 .

So assume v = b 1 v 1 + v 2 with b 1 0 , and set

A = ( ( a + 1 ) b 1 + ω ) ( a b 1 + ω ( a + 1 ) ) .

If A 0 , again v 2 U as

A v 2 = ( ( a 2 + ω 2 a + 1 ) b 1 + ω 2 ( a + 1 ) ) v + ω b 1 x v + b 1 y x v + ω 2 b 1 y 2 x v .

If A = 0 , then

b 1 { ω a + 1 , ω ( a + 1 ) a } .

In both cases, take the matrix 𝐵 whose columns are 𝑣, x v , y x v , x y x v , ( y x ) 2 v . If b 1 = ω a + 1 , then det ( B ( 1 , 2 , 3 , 4 , 5 ) ) = d 1 ( a + 1 ) 5 and det ( B ( 1 , 2 , 3 , 4 , 7 ) ) = d 2 ( a + 1 ) 5 , where

d 1 = ω 2 a 13 + a 12 + ω 2 a 10 + ω a 9 + ω a 8 + a 7 + ω 2 a 6 + a 5 + ω 2 a 4 + ω a 3 + ω a 2 + a + ω 2 ,
d 2 = ω 2 a 15 + ω a 14 + ω a 13 + ω 2 a 12 + a 10 + a 8 + ω a 7 + a 6 + a 5 + ω 2 a 4 + a 3 + ω a 2 + a + ω .
If b 1 = ω ( a + 1 ) a , then det ( B ( 1 , 2 , 3 , 4 , 5 ) ) = d 1 a 5 and det ( B ( 1 , 2 , 3 , 4 , 7 ) ) = d 2 a 5 , where
d 1 = a 11 + ω 2 a 10 + a 9 + ω 2 a 8 + ω a 7 + ω 2 a 6 + ω a 5 + ω a 4 + ω a 3 + a + ω ,
d 2 = a 13 + ω 2 a 12 + a 10 + ω 2 a 8 + ω a 7 + ω 2 a 6 + ω a 4 + a 3 + ω a + 1 .
In both cases, d 1 and d 2 are coprime, whence the contradiction rk ( B ) = 5 .

Now, to prove the absolute irreducibility of the groups τ ( H ) , τ 2 ( H ) , we simply adapt the previous argument. For sake of brevity, we only describe the necessary modifications.

Consider τ ( H ) , and write 𝑥 for τ ( x ) and 𝑦 for τ ( y ) . The eigenspace V ω ( y ) is generated by v i = v i ( ω ) , i = 1 , 2 , where

v 1 = e 1 + ω e 4 + ω - 1 e - 2 and v 2 = e 2 + ω e - 1 + ω - 1 e - 4 .

Then we have det ( M ) = ω ( a + 1 ) 2 ( a 2 + a + ω ) 4 and, assuming a 2 = a + ω , det ( M ¯ ) = ω 2 ( a + 1 ) 2 0 . If v = v 2 , then det ( N ( 1 , 2 , 3 , 5 , 7 ) ) = ω a ( a + 1 ) 3 0 . Furthermore, ( a 2 + a ) v 2 = ω 2 v 1 + ω x v 1 + y x v 1 + ω 2 y 2 x v 1 and

A v 2 = ω 2 ( a 2 + a + b 1 ) v + ω b 1 x v + b 1 y x v + ω 2 b 1 y 2 x v ,

where A = ( ( a + 1 ) b 1 + ω a ) ( a b 1 + ω ( a + 1 ) ) .

If b 1 = ω a a + 1 , then det ( B ( 1 , 2 , 3 , 4 , 5 ) ) = d 1 ( a + 1 ) 5 and det ( B ( 1 , 2 , 3 , 4 , 6 ) ) = d 2 ( a + 1 ) 5 , where

d 1 = a 14 + ω 2 a 13 + a 12 + a 11 + ω a 9 + ω a 8 + ω 2 a 7 + ω 2 a 5 + a 4 + ω 2 a 3 + a + ω ,
d 2 = a 15 + ω a 14 + ω a 13 + a 12 + ω a 10 + ω a 9 + ω 2 a 7 + ω 2 a 6 + ω a 5 + ω 2 a 4 + ω 2 a 3 + ω a + 1 .
If b 1 = ω ( a + 1 ) a , then det ( B ( 1 , 2 , 3 , 4 , 5 ) ) = d 1 a 4 and det ( B ( 1 , 2 , 3 , 4 , 7 ) ) = d 2 a 4 , where
d 1 = a 13 + ω a 12 + ω a 11 + ω a 10 + a 8 + a 7 + ω a 5 + ω 2 a 4 + a 3 + ω 2 a 2 + a + ω ,
d 2 = a 14 + ω a 13 + ω 2 a 12 + a 11 + ω a 10 + a 8 + ω a 7 + ω 2 a 6 + ω 2 a 4 + ω a 3 + ω 2 a 2 + ω 2 a + ω .
In both cases, d 1 and d 2 are coprime, whence the contradiction rk ( B ) = 5 .

Finally, consider τ 2 ( H ) , and write 𝑥 for τ 2 ( x ) and 𝑦 for τ 2 ( y ) . The eigenspace V ω ( y ) is generated by v i = v i ( ω ) , i = 1 , 2 , where

v 1 = e 1 + ω - 1 e 2 + ω e - 4 and v 2 = e 4 + ω - 1 e - 1 + ω e - 2 .

Then we have det ( M ) = a 2 ( a + 1 ) 2 ( a 2 + a + ω ) 4 and, assuming a 2 = a + ω , det ( M ¯ ) = ( a + 1 ) 3 0 . Take as 𝑁 the matrix having columns 𝑣, x v , y x v , x y 2 x v and ( y x ) 2 v . If v = v 2 , then det ( N ( 1 , 2 , 3 , 4 , 7 ) ) = ω a 2 ( a + 1 ) 4 0 . Furthermore,

ω v 2 = ( ω a + 1 ) 2 v 1 + x v 1 + ω 2 y x v 1 + ω y 2 x v 1 , ( a 2 + ω b 1 ) v 2 = ( ω a + 1 ) 2 v + x v + ω 2 y x v + ω y 2 x v .

If b 1 = ( ω a ) 2 , then det ( N ( 1 , 2 , 3 , 4 , 5 ) ) = d 1 and det ( N ( 1 , 3 , 4 , 5 , 6 ) ) = d 2 , where

d 1 = a 13 + a 12 + a 11 + a 10 + a 9 + ω a 7 + a 6 + a 3 + ω 2 a + 1 , d 2 = ω a 12 + ω a 11 + a 10 + ω a 8 + a 6 + ω a 5 + ω a 3 + a 2 + ω 2 a + ω .

Since d 1 and d 2 are coprime, we obtain the contradiction rk ( N ) = 5 . ∎

Lemma 3.2

Let 𝑝 be odd. If a ± 2 , the group 𝐻 is absolutely irreducible and contains an element 𝜂 whose order is divisible by 𝑝.

Proof

The characteristic polynomial of η = ( x y 2 ) 2 ( x y ) 2 is χ η ( t ) = ( t - 1 ) 4 ψ ( t ) with

ψ ( t ) = t 4 - ( 4 a 3 + 12 a 2 - 14 ) t 3 + ( 4 a 4 - 8 a 3 - 44 a 2 + 51 ) t 2 - ( 4 a 3 + 12 a 2 - 14 ) t + 1 .

By direct computations, if a - 2 , the eigenspace V 1 ( η ) is generated by the vectors

v 1 = e 1 + e 2 + ( 1 - a ) e - 2 + e - 3 + a e - 4 and v 2 = e - 1 - e - 2 + e - 4 .

Note that η x = η - 1 and that v , x v = V 1 ( η ) for all 0 v in V 1 ( η ) unless 𝑣 is either a multiple of v 2 or a multiple of v ~ , where

v ~ = 2 ( e 1 + e 2 + e - 3 ) - a e - 1 + ( 2 - a ) e - 2 + a e - 4 .

Let 𝑈 and U ¯ be, respectively, an 𝐻- and an H T -invariant subspace of F 8 as in Lemma 2.1. We may assume dim U 4 . Suppose first that η | U has the eigenvalue 1. By what we observed above, we may assume that 𝑈 contains the vector v { v 2 , v ~ } . Let 𝑀 be the matrix whose columns are 𝑣, y v , y 2 v , x y v , x y 2 v . Then det ( M ( 1 , 2 , 3 , 4 , 5 ) ) = - a if v = v 2 and det ( M ( 1 , 2 , 3 , 6 , 8 ) ) = 2 ( a - 2 ) 4 if v = v ~ . So, in both cases, 𝑀 has rank 5 under our assumptions.

Now suppose that η | U does not have the eigenvalue 1. Then U ¯ contains V 1 ( η T ) and, in particular, contains the eigenvector u = e 1 - e 2 + e 4 . The matrix N 1 whose columns are the vectors

(3.2) u , y T u , ( y 2 ) T u , ( y x ) T u , ( y 2 x ) T u , ( y x y ) T u

and the vectors ( y x y 2 ) T u , ( ( y x ) 2 ) T u has determinant ( a - 2 ) d 1 , where

d 1 = ( a 2 - a - 3 ) ( a 2 - a + 1 ) .

The matrix N 2 whose columns are the vectors (3.2) and ( ( y x ) 3 ) T u , ( ( y x ) 4 ) T u has determinant ( a - 2 ) d 2 , where

d 2 = a 6 - 2 a 5 - 7 a 4 + 6 a 3 + 15 a 2 - 8 a + 6 .

If a 2 = a + 3 , then d 2 = - 16 a 0 ; if a 2 = a - 1 , then d 2 = 16 a 2 0 . We conclude that dim ( U ¯ ) = 8 .

Finally, since 1 is an eigenvalue of 𝜂 having algebraic multiplicity at least 4, but geometric multiplicity equal to 2, we conclude that the order of 𝜂 is divisible by 𝑝. ∎

Lemma 3.3

Let 𝑝 be odd. If a - 2 , the group τ ( H ) is absolutely irreducible.

Proof

We write 𝑥 for τ ( x ) and 𝑦 for - τ ( y ) . We have that E = E ω ( y ) is generated by v i = v i ( ω ) , i = 1 , 2 , where

v 1 ( ω ) = e 1 + ω e 4 - ω - 1 e - 2 and v 2 ( ω ) = e 2 - ω e - 1 - ω - 1 e - 4 .

Let U { 0 } and U ¯ be, respectively, a τ ( H ) - and a τ ( H ) T -invariant subspace of F 8 as in Lemma 2.1. If U E = { 0 } , then we have E ω ( y T ) = E ω - 1 ( y ) U ¯ . Taking v 1 - = v 1 ( ω - 1 ) and v 2 - = v 2 ( ω - 1 ) , let 𝑀 and 𝑁 be the matrices whose columns are, respectively,

v 1 - , x T v 1 - , ( x y ) T v 1 - , ( x y x ) T v 1 - , ( ( x y ) 2 ) T v 1 - , ( ( x y ) 2 x ) T v 1 - , v 2 - , x T v 2 -

and

v 1 - , x T v 1 - , ( x y ) T v 1 - , ( x y x ) T v 1 - , v 2 - , x T v 2 - , ( x y ) T v 2 - , ( x y x ) T v 2 - .

Then 𝑀 has determinant a ( a + 2 ) ( a - ω + 2 ) 2 ( a + ω ) 2 , which is nonzero unless a { - ω , ω - 2 } . If a = - ω , then det ( N ) = - 2 6 ( a + 2 ) ; if a = ω - 2 , then det ( N ) = - 2 4 ( 3 ω + 1 ) , which is nonzero unless p = 7 and ω = 2 , which gives the contradiction a = 0 . In any case, we get the contradiction U = { 0 } .

Hence, there exists 0 u E U . Let 𝐴 be the matrix having columns 𝑢, x u , y x u , x y x u , y x y 2 x u . If u = v 1 , then 𝐴 has rank 5 since det ( A ( 1 , 2 , 3 , 5 , 7 ) ) = - 8 ω a . Thus, we may assume u = λ v 1 + v 2 . In this case, det ( A ( 2 , 3 , 5 , 7 , 8 ) ) 0 , except in the following cases:

( 1 ) λ = a ω - 1 2 , ( 2 ) λ 2 + ( a + 1 ) ( ω 2 - 1 ) 2 λ - ω 2 2 = 0 .

Let 𝐵 be the matrix whose columns are 𝑢, x u , y x u , ( y x ) 2 u , ( y x ) 3 u .

Case (1).  If p = 3 , then det ( B ( 2 , 3 , 4 , 5 , 8 ) ) = - ( a + 2 ) 3 0 . So assume p 3 . Then

det ( B ( 2 , 3 , 4 , 5 , 8 ) ) = ( ω 2 - ω ) ( a + 2 ) 3 36 ( 3 a + ( 5 ω + 1 ) ) 3

is nonzero unless a = - 5 ω + 1 3 , in which case p 5 and

det ( B ( 1 , 2 , 4 , 6 , 8 ) ) = 2 6 5 3 4 0 .

It follows that rk ( B ) = 5 .

Case (2).  We may assume λ a ω - 1 2 by previous analysis, and clearly, λ 0 . If p = 3 , from (2), we get λ = ± ι , where 𝜄 has order 4. For λ { ι , - ι } , we have det ( B ( 1 , 2 , 6 , 7 , 8 ) ) = 0 if and only if a = ± ι , in which case det ( B ( 1 , 2 , 3 , 4 , 5 ) ) 0 . So we may assume p 3 and set a = 2 ( 1 - ω ) 3 λ - 1 + 2 + ω 3 λ . Then

det ( A ( 1 , 2 , 3 , 4 , 5 ) ) = 2 d 1 3 2 λ 2 , det ( A ( 1 , 2 , 3 , 4 , 6 ) ) = 2 d 2 3 3 λ 2 , det ( A ( 1 , 2 , 3 , 4 , 7 ) ) = 2 d 3 3 2 λ 2 ,

where d 1 , d 2 , d 3 can be viewed as polynomials in 𝜆. Setting δ = gcd ( d 2 , d 3 ) , we get δ = c 2 d 2 + c 3 d 3 , where

4 δ = 4 λ 6 + 2 ( 2 ω + 1 ) λ 5 - ( 2 ω + 1 ) λ 4 - ( ω - 1 ) λ 3 - ( ω + 5 ) λ 2 - ( 5 ω + 1 ) λ + ω ,
144 c 2 = 2 ( 7 ω + 11 ) λ 2 - ( 7 ω - 25 ) λ - 2 ( ω + 5 ) ,
48 c 3 = - 4 ( ω + 6 ) λ 4 - 2 ( 9 ω + 14 ) λ 3 - ( 19 ω + 32 ) λ 2 + 8 ( 2 ω - 1 ) λ + 4 ( 2 ω + 5 ) .
Finally, gcd ( d 1 , δ ) = 1 , whence the contradiction rk ( A ) = 5 .∎

Lemma 3.4

Let 𝑝 be odd. If a - 2 , the group τ 2 ( H ) is absolutely irreducible.

Proof

We write 𝑥 for τ 2 ( x ) and 𝑦 for - τ 2 ( y ) . The characteristic polynomial of η = [ x , y ] is χ η ( t ) = ( t + 1 ) 4 ψ ( t ) with

ψ ( t ) = t 4 + ( a 2 - 4 ) t 3 - ( a 2 - 6 ) t 2 + ( a 2 - 4 ) t + 1 .

By direct computations, the eigenspace V - 1 ( η ) is generated by the vectors

v 1 = a e 1 - e 2 + e 3 + a e 4 - e - 1 - a e - 3 + ( a 2 + a - 1 ) e - 4 , v 2 = e - 2 + e - 3 .

Let 𝑈 and U ¯ be, respectively, a τ 2 ( H ) - and a ( τ 2 ( H ) ) T -invariant subspace of F 8 as in Lemma 2.1. We may assume dim U 4 . Suppose first that η | U has the eigenvalue −1. Let 𝑀 be the matrix whose columns are the vectors 𝑢, y u , y 2 u , x y 2 u , y x y 2 u , where u U . If u = v 2 U , then det ( M ( 1 , 3 , 5 , 6 , 7 ) ) = a + 2 0 . So assume u = v 1 + λ v 2 U for some λ F . Since 1 2 ( u + x u ) = v 1 , we obtain that v 1 U . Taking u = v 1 , we have

det ( M ( 1 , 2 , 3 , 4 , 5 ) ) = - d 1 , det ( M ( 1 , 2 , 3 , 5 , 6 ) ) = ( a + 2 ) ( a + ω ) ( a + ω 2 ) d 2 ,

where

d 1 = 2 a 9 + 7 a 8 - 5 a 7 - 30 a 6 - 3 a 5 + 26 a 4 + 23 a 3 + 15 a 2 - 58 a + 16 , d 2 = 2 a 6 + 5 a 5 - 8 a 4 - 17 a 3 + 16 a 2 + 13 a - 8 .

If a - ω ± 1 , from gcd ( d 1 , d 2 ) = 1 , we conclude that the matrix 𝑀 has rank 5. For p = 3 and a = - 1 , we have det ( M ( 1 , 2 , 3 , 4 , 6 ) ) = 1 ; for p 3 and a = - ω ± 1 , we have d 1 { 2 6 ( ω - 1 ) , - 2 6 ( ω + 2 ) } , whence d 1 0 . So, also in these cases, rk ( M ) = 5 .

Now suppose η | U does not have the eigenvalue −1. Then U ¯ contains V - 1 ( η T ) and, in particular, the eigenvector u = e 2 + e 3 . The matrix whose columns are the vectors

u , y T u , ( y 2 ) T u , ( y x ) T u , ( y x y 2 ) T u , ( ( y x ) 2 ) T u , ( y x y ) T u , ( y η ) T u

has determinant 2 a ( a + 2 ) 4 0 . We conclude that dim ( U ¯ ) = 8 . ∎

Corollary 3.5

If p = 2 , then 𝐻 is not contained in any maximal subgroup Sp 6 ( q ) of class 𝒮. If 𝑝 is odd and a ± 2 , then 𝐻 is neither contained in any maximal subgroup of class C 4 nor in any maximal subgroup 2 Ω 7 ( q ) of class 𝒮.

Proof

Suppose that 𝐻 is contained in one of the maximal subgroups described in the statement. By [1, Table 8.50] and [6], τ i ( M ) belongs to the class C 1 for some i = 1 , 2 . In particular, τ i ( H ) is reducible, in contrast with Lemmas 3.1, 3.2, 3.3 and 3.4.∎

Applying the same argument, we can also prove that 𝐻 is neither contained in any maximal subgroup

SL 4 ( q ) . q - 1 gcd ( q - 1 , 2 ) .2 of class C 2

nor in any maximal subgroup

SU 4 ( q 2 ) . q + 1 gcd ( q - 1 , 2 ) .2 of class C 3 .

However, the first type of subgroup will be excluded by Lemma 4.1, while the second type has been already excluded by the absolute irreducibility of 𝐻.

4 The (2, 3)-generation of Ω 8 + ( q )

Lemma 4.1

Suppose a ± 2 if 𝑝 is odd. Then 𝐻 is neither contained in any maximal subgroup of classes C 2 , C 6 nor in any maximal subgroup 2 Ω 8 + ( 2 ) , 2 Sz ( 8 ) , 2 × Alt ( 10 ) , 2 Alt ( 10 ) of class 𝒮.

Proof

The characteristic polynomial of x y is

χ x y ( t ) = { t 8 + ( a 2 + a + 1 ) t 7 + a 2 t 6 + ( a + 1 ) t 5 + ( a + 1 ) 4 t 4 + ( a + 1 ) t 3 + a 2 t 2 + ( a 2 + a + 1 ) t + 1 if p = 2 , t 8 + ( a + 1 ) t 7 - a t 6 - 2 t 5 + a 2 t 4 - 2 t 3 - a t 2 + ( a + 1 ) t + 1 if p > 2 .

Suppose that our claim is false. Note that 𝐻 is absolutely irreducible by Lemmas 3.1 and 3.2.

Case 1. H M with M C 2 monomial, or 𝑀 one of the remaining maximal subgroups of the statement.

Then q = p is odd, and the set of the prime divisors of | M | is contained in { 2 , 3 , 5 , 7 } . The element η = ( x y 2 ) 2 ( x y ) 2 of Lemma 3.2 has order divisible by 𝑝, whence q { 5 , 7 } . So assume q = 5 . If a = 1 , the order of η y x y is 2 31 ; if a = - 1 , the order of 𝜂 is 5 13 . Finally, assume q = 7 . If a = 1 , then η y 2 x y has order 19; if a = ± 3 , the element y η 3 has order 9 19 ; if a = - 1 , the element y η 2 has order 43. In all these cases, we get a contradiction considering the order of 𝑀.

Case 2. H M C 2 fixes a decomposition F q 8 = v 1 , w 1 v 4 , w 4 .

Let ν : H Sym ( 4 ) be the corresponding homomorphism, and let 𝑟 be an odd prime dividing | ker ( ν ) | . Then 𝑟 divides q ± 1 . If 𝑝 is odd, consideration of 𝜂 as in Lemma 3.2 gives p = 3 . We may assume that 𝑦 acts as

ζ = ( v 1 , v 2 , v 3 ) ( w 1 , w 2 , w 3 )

and, considering its rational form, that y v 4 = v 4 , y w 4 = w 4 . Hence, 𝑥 must act either as ξ = ξ 1 or as ξ = ξ 2 , where for some A , B , C GL 2 ( q ) ,

ξ 1 = ( 0 0 0 A 0 B 0 0 0 0 C 0 A - 1 0 0 0 ) , ξ 2 = ( 0 0 0 A 0 0 B 0 0 B - 1 0 0 A - 1 0 0 0 ) , B = ( β 1 β 2 β 3 β 4 ) .

Setting Δ = det ( B ) , the characteristic polynomial of ξ ζ is, respectively,

p 1 ( t ) = t 8 + α t 4 + β ,
p 2 ( t ) = t 8 - ( β 1 + β 4 ) t 7 + Δ t 6 - β 1 + β 4 Δ t 5 + ( β 1 + β 4 ) 2 Δ t 4 - ( β 1 + β 4 ) t 3 + 1 Δ t 2 - β 1 + β 4 Δ t + 1 .
Comparison of p 1 ( t ) with χ x y ( t ) gives immediately a contradiction. Next we compare p 2 ( t ) with χ x y ( t ) . If p = 2 , considering the coefficients of t 6 and of t 2 , we get the contradiction Δ = a 2 = 1 Δ . If p = 3 , considering the coefficients of t 7 and of t 3 , we get a + 1 = - ( β 1 + β 4 ) = - 2 , whence the contradiction a = 0 .

Case 3. H M C 2 fixes a decomposition F q 8 = V 1 V 2 , with dim ( V i ) = 4 .

Clearly, we must have x V 1 = V 2 and y V i = V i , i = 1 , 2 . It follows that x y has trace 0. This implies a = ω ± 1 if p = 2 , and a = - 1 if 𝑝 is odd. If p = 2 , from a = ω ± 1 , we get the contradiction 0 = tr ( ( x y ) 3 ) = ω 1 . If 𝑝 is odd, from a = - 1 , we get the contradiction 0 = tr ( y ( x y ) 3 ) = 4 . ∎

Lemma 4.2

The group 𝐻 is not contained in any maximal subgroup of class C 5 .

Proof

Suppose the contrary. Then there exists g GL 8 ( F ) such that x g = ϑ 1 x 0 , y g = ϑ 2 y 0 , with x 0 , y 0 GL 8 ( q 0 ) and ϑ i F . If p = 2 , we have tr ( [ x , y ] ) = a 2 ; if 𝑝 is odd, we have tr ( [ x , y ] ) = 2 a . From tr ( [ x , y ] ) = tr ( [ x g , y g ] ) = tr ( [ x 0 , y 0 ] ) , it follows that a F q 0 . So F q = F p [ a ] F q 0 implies q 0 = q . ∎

Lemma 4.3

The group 𝐻 is not contained in any maximal subgroup of type Ω 8 - ( q ) of class 𝒮.

Proof

Suppose the contrary, and write q = q 0 2 . Then, for some i = 1 , 2 , there exists g GL 8 ( F ) such that

( τ i ( x ) ) g = ϑ 1 x 0 , ( τ i ( y ) ) g = ϑ 2 y 0 , with x 0 , y 0 GL 8 ( q 0 ) and ϑ i F .

First, assume p = 2 . We have tr ( [ τ ( x ) , τ ( y ) ] ) = a 4 and tr ( τ 2 ( x ) τ 2 ( y ) ) = a 2 . So, if i = 1 , we proceed as in Lemma 4.2. If i = 2 , then

a 2 = tr ( τ 2 ( x ) τ 2 ( y ) ) = tr ( ( τ 2 ( x ) ) g ( τ 2 ( y ) ) g ) = ϑ 1 ϑ 2 tr ( x 0 y 0 ) .

It follows that a F q 0 , whence the contradiction F q = F 2 [ a ] F q 0 .

A similar argument can be applied for 𝑝 odd. In this case, we have

tr ( [ τ ( x ) , τ ( y ) ] ) = - 2 a and tr ( ( τ 2 ( x ) τ 2 ( y ) ) 2 ) = 2 a + 1 .

When p 3 , the class 𝒮 contains maximal subgroups

M 1 = - I × PSL 3 ( q ) .3 and M 2 = - I × PSU 3 ( q 2 ) .3 .

The subgroup PSL 3 ( q ) arises from the conjugation action of SL 3 ( q ) on the 8-dimensional adjoint module N 1 = { A Mat 3 ( q ) tr ( A ) = 0 } . Similarly,

SU 3 ( q 2 ) = { g SL 3 ( q 2 ) g T g σ = I 3 }

acts by conjugation on M = { A Mat 3 ( q 2 ) A T = A σ } , giving rise to the 8-dimensional adjoint module N 2 = M F q I 3 . (Here g σ is obtained from 𝑔 raising its entries to the 𝑞-th power.) The modules N i , i = 1 , 2 , are described in [1, Section 5.4.1] with the relevant properties.

Lemma 4.4

Each g ¯ PSL 3 ( q ) M 1 admits the eigenvalue 1. The same is true for each g ¯ PSU 3 ( q 2 ) M 2 .

Proof

Consider first a preimage 𝑔 of g ¯ in SL 3 ( q ) . We must show that n g = n for some n N 1 . By the Frobenius formula [5, Theorem 3.16], the centralizer C g of 𝑔 in Mat 3 ( q ) has dimension at least 3. Since dim ( Mat 3 ( q ) ) = 9 , it follows that dim ( C g N 1 ) 2 (by the Grassmann formula), which proves our claim.

Next, consider a preimage 𝑔 of g ¯ in SU 3 ( q 2 ) . We must show that n g = n for some non-scalar n M . We have g - 1 = ( g σ ) T , so 𝑔 centralizes g + ( g σ ) T M . Hence, our claim is true unless g + ( g σ ) T = ρ I 3 for some ρ F q . In this case, g 2 - ρ g + I 3 = 0 , and we show that 𝑔 fixes a non-singular 1-dimensional space. Indeed, 𝑔 has minimal polynomial of degree 2, as we may assume 𝑔 non-scalar. It follows that the invariant factors of t I 3 - g are t - λ and ( t - λ ) ( t - μ ) for some λ , μ F q 2 (see [5, pages 192–201]). In particular, we have dim ( V λ ( g ) ) = 2 . Hence, there exist v , w V λ ( g ) such that v T w σ 0 . If v T v σ = w T w σ = 0 , then ( η v + w ) T ( η v + w ) σ 0 for a suitable η F q 2 . So we may suppose that g v = λ v with v T v σ 0 . It follows that 𝑔 fixes the non-singular complement v of v . As any unitary space admits an orthonormal basis [7, page 22], 𝑔 is conjugate, under U 3 ( q 2 ) , to a matrix of shape diag ( λ , L ) , with L GL 2 ( q 2 ) . Clearly, 𝑔 centralizes the non-scalar matrix diag ( 1 , 0 I 2 ) . ∎

Remark 4.5

The characteristic polynomial of g i = τ i ( [ x , y ] ) is

χ g 0 ( t ) = { t 8 + a 2 t 7 + a 2 t 6 + a 2 ( a + 1 ) 2 t 5 + ( a + 1 ) 8 t 4 + a 2 ( a + 1 ) 2 t 3 + a 2 t 2 + a 2 t + 1 if p = 2 , t 8 - 2 a t 7 + ( 3 a 2 - 4 ) t 6 - 2 a ( a 2 - 1 ) t 5 + ( a 4 - 2 a 2 + 6 ) t 4 - 2 a ( a 2 - 1 ) t 3 + ( 3 a 2 - 4 ) t 2 - 2 a t + 1 if p > 2 , χ g 1 ( t ) = { t 8 + a 4 t 7 + a 2 t 6 + a 2 ( a + 1 ) 2 t 5 + ( a + 1 ) 4 t 4 + a 2 ( a + 1 ) 2 t 3 + a 2 t 2 + a 4 t + 1 if p = 2 , t 8 + 2 a t 7 + ( 3 a 2 - 4 ) t 6 + 2 a ( a 2 - 1 ) t 5 + ( a 4 - 2 a 2 + 6 ) t 4 + 2 a ( a 2 - 1 ) t 3 + ( 3 a 2 - 4 ) t 2 + 2 a t + 1 if p > 2 , χ g 2 ( t ) = { t 8 + t 7 + a 2 t 6 + ( a 3 + 1 ) 2 t 5 + a 4 ( a + 1 ) 4 t 4 + ( a 3 + 1 ) 2 t 3 + a 2 t 2 + t + 1 if p = 2 , t 8 + a 2 t 7 + ( 3 a 2 - 4 ) t 6 + 3 a 2 t 5 + 2 ( a 2 + 3 ) t 4 + 3 a 2 t 3 + ( 3 a 2 - 4 ) t 2 + a 2 t + 1 if p > 2 .

Corollary 4.6

Suppose a ± 2 if 𝑝 is odd. Then 𝐻 is not contained in any maximal subgroup M 1 = - I × PSL 3 ( q ) .3 or M 2 = - I × PSU 3 ( q 2 ) .3 of class 𝒮.

Proof

Clearly, PSL 3 ( q ) = M 1 and PSU 3 ( q 2 ) = M 2 . Hence, if our claim is false, at least one of the commutators [ x , y ] , τ ( [ x , y ] ) , τ 2 ( [ x , y ] ) belongs to PSL 3 ( q ) or to PSU 3 ( q 2 ) . As a consequence, by Lemma 4.4, it must have the eigenvalue 1. By Remark 4.5, this possibility is excluded. Indeed, for p = 2 , we have

χ [ x , y ] ( 1 ) = ( a + 1 ) 8 , χ τ ( [ x , y ] ) ( 1 ) = ( a + 1 ) 4 , χ τ 2 ( [ x , y ] ) ( 1 ) = a 4 ( a + 1 ) 4 ,

and for p > 2 , we have

χ [ x , y ] ( 1 ) = a 2 ( a - 2 ) 2 , χ τ ( [ x , y ] ) ( 1 ) = a 2 ( a + 2 ) 2 , χ τ 2 ( [ x , y ] ) ( 1 ) = 16 a 2 .

Lemma 4.7

The group 𝐻 is not contained in any maximal subgroup of type D 4 3 ( q 3 ) of class 𝒮.

Proof

Let q = q 0 3 , and assume, for the sake of contradiction, that

H - I × D 4 3 ( q 0 ) .

Then there exists a non-singular matrix 𝑃 and a scalar λ h such that, for some i = 1 , 2 ,

P - 1 ( τ i ( h φ ) ) P = λ h h for all h H ,

where h φ denotes the matrix obtained by substituting each entry 𝛼 of ℎ with α q 0 . In particular, taking h = [ x , y ] , τ i ( h φ ) and ℎ have the same characteristic polynomial.

The characteristic polynomials of [ x , y ] , τ ( [ x , y ] ) and τ 2 ( [ x , y ] ) are described in Remark 4.5. If p = 2 , equating the coefficients of the term t 6 in χ [ x , y ] ( t ) and in χ τ i ( [ x , y ] φ ) ( t ) , we get the contradiction a 2 = ( a 2 ) q 0 . Similarly, for 𝑝 odd, equating the coefficients of the term t 7 in χ [ x , y ] ( t ) and in χ τ ( [ x , y ] φ ) ( t ) , we get a q 0 = - a , which gives the contradiction a q = - a . Finally, equating the coefficients of the terms t 7 and t 5 in χ [ x , y ] ( t ) and in χ τ 2 ( [ x , y ] φ ) ( t ) , we get a 2 q 0 = - 2 a and 3 a 2 q 0 = - 2 a ( a 2 - 1 ) , which gives the contradiction a = - 2 . ∎

Theorem 4.8

Suppose q 4 . Let a F q * be such that F q = F p [ a ] . If 𝑝 is odd, assume also that a ± 2 . Then H = Ω 8 + ( q ) . In particular, Ω 8 + ( q ) is ( 2 , 3 ) -generated for all q 4 .

Proof

By the considerations of Section 3, Lemma 3.1 and Lemma 3.2, 𝐻 is an absolutely irreducible subgroup of Ω 8 + ( q ) . By Lemmas 4.1 and 4.2 and by Corollary 3.5, it is primitive and tensor-indecomposable, and is not contained in any maximal subgroup of classes C 5 , C 6 . So either H = Ω 8 + ( q ) or 𝐻 is contained in one of the following maximal subgroups of class 𝒮:

  1. Sp 6 ( q ) if p = 2 , or 2 Ω 7 ( q ) if 𝑝 is odd;

  2. d Ω 8 - ( q 0 ) , where d = gcd ( q - 1 , 2 ) and q = q 0 2 ;

  3. - I × PSL 3 ( q ) .3 if q 1 ( mod 3 ) , or - I × PSU 3 ( q 2 ) .3 if q 2 ( mod 3 ) ;

  4. - I × D 4 3 ( q 0 ) , where q = q 0 3 ;

  5. 2 Ω 8 + ( 2 ) if q = p is odd;

  6. 2 Sz ( 8 ) , 2 × Alt ( 10 ) or 2 Alt ( 10 ) if q = 5 .

Now case (i) is excluded by Corollary 3.5; cases (ii), (iii) and (iv) are excluded, respectively, by Lemmas 4.3, 4.6 and 4.7; cases (v) and (vi) are excluded by Lemma 4.1. ∎

5 Generators for Ω 8 - ( q )

5.1 Fields of characteristic 2

We first deal with the case q = 2 .

Lemma 5.1

The group Ω 8 - ( 2 ) is ( 2 , 3 ) -generated.

Proof

Take x , y as follows:

x = ( 0 1 0 0 1 0 0 0 1 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 ) , y = ( 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 ) .

Then 𝑥 is an involution, 𝑦 has order 3, and det ( x ) = det ( y ) = 1 . Both the elements 𝑥 and 𝑦 fix the quadratic form

Q ( i = 1 4 ( α i e i + α - i e - i ) ) = α 1 2 + α - 1 2 + i = 1 4 α i α - i .

Proceeding as we will do for the general case, it is easy to prove that H = x , y is a subgroup of Ω 8 - ( 2 ) . Since the elements ( [ x , y ] x y ) 3 and [ x , y ] 5 x y x have, respectively, order 7 and 17, by [4, page 89], we conclude that H = Ω 8 - ( 2 ) . ∎

Assume q 4 even, and let a F q * be such that F 2 [ a ] = F q . This means, in particular, that a 0 , 1 . Define the following elements x = x ( a ) and 𝑦:

x = ( 0 a - 1 0 0 0 0 0 0 a 0 0 0 0 0 0 0 0 0 a 2 ( a + 1 ) 2 1 ( a + 1 ) 2 0 0 0 0 0 0 1 ( a + 1 ) 2 a 2 ( a + 1 ) 2 0 0 0 0 ( a + 1 ) 2 a - 1 0 0 0 a 0 0 a - 1 ( a + 1 ) 2 a 2 0 0 a - 1 0 0 0 0 0 0 0 0 0 a 2 ( a + 1 ) 2 1 ( a + 1 ) 2 0 0 0 0 0 0 1 ( a + 1 ) 2 a 2 ( a + 1 ) 2 ) , y = ( 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 ) .

Then 𝑥 is an involution, 𝑦 has order 3, and det ( x ) = det ( y ) = 1 . Both the elements 𝑥 and 𝑦 fix the quadratic form

Q ( i = 1 4 ( α i e i + α - i e - i ) ) = α 1 2 + α - 1 2 + a 2 i = 2 4 ( α i 2 + α - i 2 ) + i = 1 4 α i α - i ,

whose associated symmetric form has Gram matrix J = ( 0 I 4 I 4 0 ) .

Clearly, y Ω 8 ± ( q ) , having order 3. Since rk ( I 8 - x ) = 4 , the quasideterminant of 𝑥 is +1 by [1, Proposition 1.6.11 (i)], and hence x Ω 8 ± ( q ) . The space V = F q 8 contains a totally singular subspace W = a e 1 + e 2 , a e 1 + e 3 , a e 1 + e 4 of dimension 3. Suppose that there exists a vector w = μ 1 e 1 + i = 1 4 μ - i e - i such that W , w is totally singular of dimension 4. Imposing the condition that 𝑤 is orthogonal to 𝑊, we obtain μ - 2 = μ - 3 = μ - 4 = a μ - 1 . Now Q ( w ) = 0 gives μ - 1 0 . Hence, we may assume μ - 1 = 1 , obtaining the necessary condition μ 1 2 + μ 1 + ( a + 1 ) 4 = 0 . So, if the polynomial t 2 + t + ( a + 1 ) 4 is irreducible in F q [ t ] , then 𝑊 is not contained in a totally singular subspace of dimension 4. By Witt’s lemma [7, Corollary 2.1.7], the Witt index of 𝑉 is 3, and so 𝑄 is a quadratic form of sign - . Thus,

H = x , y Ω 8 - ( q ) if t 2 + t + ( a + 1 ) 4 is irreducible over F q .

Lemma 5.2

For all q = 2 f , there exists a F q satisfying F q = F 2 [ a ] and such that the polynomial p ( t ) = t 2 + t + ( a + 1 ) 4 is irreducible over F q .

Proof

By Carlitz’s formula [2], there exist 2 f - 1 elements 𝛼 in F q such that t 2 + t + α is irreducible. We claim that F 2 [ α ] = F q for at least one 𝛼. Indeed, the number of elements lying in proper subfields of F q is 0 if f = 1 , it is 2 if 𝑓 is a prime, and it is at most 2 + 2 2 + + 2 f - 2 = 2 f - 1 - 2 < 2 f - 1 otherwise. Now write α = ( a + 1 ) 4 : clearly,

F 2 [ a ] = F 2 [ a + 1 ] = F 2 [ α ] = F q .

The characteristic polynomial of x y is

(5.1) χ x y ( t ) = ( t + ( a 2 + 1 ) - 1 ) ( t + a 2 + 1 ) ( t 2 + ( a 2 + 1 ) - 1 t + 1 ) ( t 2 + t + 1 ) 2 .

Note that tr ( x y ) = ( a + 1 ) 2 and that χ x y ( 1 ) = a 4 ( a + 1 ) 4 0 .

Lemma 5.3

The group 𝐻 is absolutely irreducible.

Proof

Let 𝑈 and U ¯ be, respectively, an 𝐻- and an H T -invariant subspace of F 8 as in Lemma 2.1. Assume a 2 + a + 1 0 . In this case, the characteristic polynomial of x y admits ( a + 1 ) 2 as a simple root. The corresponding eigenspaces of x y and ( x y ) T are generated, respectively, by the vector s = a 2 e 2 + e 3 + ( a 3 + a ) e - 1 and s ¯ = e 3 + a 2 e 4 + ( a 5 + a 3 ) e - 1 + a 2 e - 3 + a 4 e - 4 . When a 2 + a + 1 = 0 , the eigenspaces of x y and ( x y ) T relative to ( a + 1 ) 2 are still one-dimensional and generated by the same vectors 𝑠 and s ¯ .

If the restriction ( x y ) | U has the eigenvalue ( a + 1 ) 2 , then the vector 𝑠 belongs to 𝑈. Take the matrix 𝑀 whose columns are the images of 𝑠 by

I 8 , y , y 2 , x y 2 , y x y 2 , x y x y 2 , ( x y 2 ) 2 , ( x y 2 ) 3 .

Since det ( M ) = ( a + 1 ) 18 a 8 0 , then U = F 8 . If the restriction ( x y ) | U does not have the eigenvalue ( a + 1 ) 2 , then U ¯ contains the vector s ¯ . Take the matrix M ¯ whose columns are the images of s ¯ by

I 8 , y T , ( y 2 ) T , ( y x y 2 ) T , ( x y 2 x y ) T , ( ( x y 2 ) 3 ) T , ( y ( x y 2 ) 3 ) T , ( y ( x y 2 ) 4 ) T .

Then det ( M ¯ ) = a 8 ( a + 1 ) 22 0 , and hence U ¯ = F 8 , that is, U = { 0 } . ∎

5.2 Fields of odd characteristic

Suppose 𝑞 odd. Let a , ξ F q * , where F p [ a ] = F q and 𝜉 is a non-square in F q * . Define the following elements x = x ( a , ξ ) and 𝑦:

x = ( 0 0 0 0 2 0 0 0 0 1 0 0 0 0 0 0 0 0 - 1 0 0 0 - a 2 ξ 2 - a ξ 0 1 0 - 1 0 0 0 0 1 2 0 0 0 0 0 0 0 0 - 1 2 0 1 0 1 0 0 0 0 0 0 0 0 - 1 0 0 0 0 0 0 0 a 1 ) , y = ( 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 ) .

Then 𝑥 is an involution, 𝑦 has order 3, and det ( x ) = det ( y ) = 1 . Both the elements 𝑥 and 𝑦 fix the symmetric form whose Gram matrix is

J = ( 0 0 I 3 0 0 1 0 0 I 3 0 0 0 0 0 0 - ξ )

of determinant 𝜉. By [1, Proposition 1.5.42], the subgroup H = x , y is contained in SO 8 - ( q ) as 𝜉 is a non-square. Furthermore, y Ω 8 - ( q ) having order 3. The spinor norm of 𝑥 with respect to 𝐽 is +1: it follows from [1, Proposition 1.6.11 (ii)] that 𝑥 belongs to Ω 8 - ( q ) . Hence, H = x , y Ω 8 - ( q ) .

The characteristic polynomial of x y is

(5.2) χ x y ( t ) = t 8 - t 6 + a 2 ξ - 4 4 t 5 + a 2 ξ + 4 2 t 4 + a 2 ξ - 4 4 t 3 - t 2 + 1 .

Note that χ x y ( 1 ) = a 2 ξ 0 .

Lemma 5.4

The group 𝐻 is absolutely irreducible.

Proof

By direct computations, E ω ( y ) is generated by the vectors v i = v i ( ω ) , i = 1 , 2 , where

v 1 = e 1 + ω - 1 e 2 + ω e 3 and v 2 = J v 1 = e - 1 + ω - 1 e - 2 + ω e - 3 .

Let 𝑈 and U ¯ be, respectively, an 𝐻- and an H T -invariant subspace of F 8 as in Lemma 2.1. We may assume dim U 4 .

If U E ω ( y ) = { 0 } , then

E ω ( y T ) = E ω ( y - 1 ) = E ω - 1 ( y ) U ¯ .

Writing v 1 - = v 1 ( ω - 1 ) and v 2 - = v 2 ( ω - 1 ) , the matrix whose columns are the vectors

v 1 - , x T v 1 - , ( x y ) T v 1 - , ( x y x ) T v 1 - , ( ( x y ) 2 ) T v 1 - , ( ( x y ) 2 x ) T v 1 - , v 2 - , x T v 2 -

has determinant 1 32 a ξ ( a 2 ξ - 4 ) 2 ( a 2 ξ - 16 ω 2 ) 2 0 . This means that U = { 0 } .

So, for some ( b 1 , b 2 ) ( 0 , 0 ) , we have v = b 1 v 1 + b 2 v 2 U . Let 𝑀 be the matrix whose columns are

v , x v , y x v , x y 2 x v , ( y x ) 2 v .

We will show that rk ( M ) = 5 , a contradiction as we are assuming dim ( U ) 4 .

Suppose first b 2 = 0 , b 1 = 1 . We have det ( M ( 1 , 2 , 5 , 7 , 8 ) ) = - a 32 ( a 2 ξ - 4 ) 0 . Now suppose b 2 = 1 . If p = 3 , then det ( M ( 1 , 2 , 3 , 4 , 8 ) ) = - a b 1 ( a 2 ξ - 1 ) 3 , which is nonzero unless b 1 = 0 . In this case, det ( M ( 3 , 4 , 6 , 7 , 8 ) ) = a 5 ξ 2 ( a 2 ξ - 1 ) 0 .

If p 3 , then

det ( M ( 4 , 5 , 6 , 7 , 8 ) ) = 2 ω + 1 16 a ( a 2 ξ - 4 ) ( b 1 + 2 ω ) ( 4 ( ω + 2 ) b 1 + a 2 ξ - 4 ) ,

which is nonzero unless either (i) b 1 = - 2 ω or (ii) b 1 = a 2 ξ - 4 4 ( ω 2 - 1 ) . In case (i), we have

det ( M ( 2 , 5 , 6 , 7 , 8 ) ) = 2 ( ω + 2 ) a ( a 2 ξ - 4 ) 0 .

In case (ii), we have

det ( M ( 1 , 2 , 3 , 4 , 8 ) ) = ω - 1 2 6 3 a ( a 2 ξ - 4 ) 2 ( a 2 ξ - 8 ω + 4 ) 2 ,

which is nonzero unless ξ = 4 a - 2 ( 2 ω - 1 ) . This gives

det ( M ( 2 , 5 , 6 , 7 , 8 ) ) = - 48 a 0 .

Lemma 5.5

Suppose that F p [ a 2 ξ ] = F q . Then 𝐻 is not contained in any maximal subgroup of class C 5 .

Proof

Suppose the contrary. Then there exists g GL 8 ( F ) such that x g = ϑ 1 x 0 , y g = ϑ 2 y 0 , with x 0 , y 0 GL 8 ( q 0 ) and ϑ i F . From

- 2 a 2 ξ - 6 = tr ( ( x y ) 4 ) = tr ( ( x g y g ) 4 ) = ϑ 1 ϑ 2 tr ( ( x 0 y 0 ) 4 ) ,

it follows that a 2 ξ F q 0 . So F q = F p [ a 2 ξ ] F q 0 implies q 0 = q . ∎

6 The (2, 3)-generation of Ω 8 - ( q )

We recall that the maximal subgroups of Ω 8 - ( q ) can be subdivided into the classes C 1 , C 3 , C 5 and 𝒮; see [1, Tables 8.52 and 8.53]. Furthermore, the maximal subgroups belonging to the classes C 1 and C 3 are not absolutely irreducible.

Lemma 6.1

The group 𝐻 is not contained in any maximal subgroup of class 𝒮.

Proof

The class 𝒮 is non-empty only when p 3 , in which case it only contains maximal subgroups isomorphic either to PSL 3 ( q ) or to PSU 3 ( q 2 ) . Arguing as in Lemma 4.6, we can exclude these subgroups since x y does not have the eigenvalue 1; see (5.1) and (5.2). ∎

Theorem 6.2

Let q 4 be even, and let a F q be such that F 2 [ a ] = F q . If the polynomial t 2 + t + ( a + 1 ) 4 is irreducible, then H = Ω 8 - ( q ) . In particular, Ω 8 - ( q ) is ( 2 , 3 ) -generated for all even q 2 .

Proof

By the considerations of Section 5.1 and Lemma 5.3, 𝐻 is an absolutely irreducible subgroup of Ω 8 - ( q ) . If our claim is false, there exists a maximal subgroup 𝑀 of Ω 8 - ( q ) which contains 𝐻. Then 𝑀 must belong either to the class C 5 or to the class 𝒮. The second possibility is excluded by Lemma 6.1. Hence, we get M = Ω 8 - ( q 0 ) for some q 0 such that q = q 0 r with 𝑟 an odd prime. However, also this possibility can be easily excluded as tr ( x y ) = a 2 + 1 and F 2 [ a ] = F q .

For the second part of the statement, if q = 2 , we apply Lemma 5.1. If q 4 , there exists an element a F q * satisfying all the requirements by Lemma 5.2. ∎

Theorem 6.3

Let q 3 be odd. Let 𝜉 be a non-square in F q * , and let a F q * be such that F p [ a 2 ξ ] = F q . Then H = Ω 8 - ( q ) . In particular, Ω 8 - ( q ) is ( 2 , 3 ) -generated for all odd q 3 .

Proof

Take a , ξ as in the statement (for instance, a = 1 and any generator 𝜉 of the cyclic group F q * ). By the considerations of Section 5.2 and Lemma 5.4, 𝐻 is an absolutely irreducible subgroup of Ω 8 - ( q ) . If our claim is false, there exists a maximal subgroup 𝑀 of Ω 8 - ( q ) which contains 𝐻. So 𝑀 must belong either to the class C 5 or to the class 𝒮. However, these possibilities are excluded by Lemma 5.5 and Lemma 6.1, respectively. ∎

  1. Communicated by: Timothy C. Burness

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Received: 2022-01-02
Revised: 2022-07-21
Published Online: 2022-08-25
Published in Print: 2023-03-01

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