arXiv:1611.09350v1 [] 26 Nov 2016

A NOTE ABOUT THE (2, 3)-GENERATION OF
SL12(q)
TSANKO RAYKOV GENCHEV
Abstract
In this note we provide explicit uniform type (2, 3)-generators for the special
linear group SL12 (q) for all q except for q = 2 or q = 4. Our considerations
are easily traceable, self-contained and based only on the known list of maximal
subgroups of this group.
Key words: (2,3)-generated group.
2010 Mathematics Subject Classification: 20F05, 20D06.

Now, when the problem concerning the (2, 3)-generation (especially) of the special
linear groups and their projective images is completely solved (see [4] and references
there), we give our contribution by discussing the last remaining group SL12 (q) in the
light of the method used in our works [2], [3] and [5]. It has been proved in [4] the
(2, 3)-generation of the same group over all finite fields in characteristic not equal to 5.
The author’s approach is different from ours in which we make an essential use of the
known list of maximal subgroups of SL12 (q).
The group G = SL12 (q), where q = pm and p is a prime, acts naturally on a twelvedimensional vector space V over the field F = GF (q). We identify V with the column
vectors of F 12 , and let v1 , . . . , v12 be the standard base of the space V . Set Q = q 11 − 1
if q 6= 3, 7 and Q = (q 11 − 1)/2 if q = 3, 7.
First of all, based only to the orders of the maximal subgroups of G (which exact
structure is given in Tables 8.76 and 8.77 in [1]), it can be easily seen that there is only
one class of such subgroups having an element of order Q. Namely, this is the class
of reducible on the space V subgroups; actually they are the stabilizers in G of one or
eleven-dimensional subspaces of V .
Now, let us choose an element ω of order Q in the multiplicative group of the field
GF (q 11 ) and set
2

3

4

5

6

7

8

9

f (t) = (t−ω)(t−ω q )(t−ω q )(t−ω q )(t−ω q )(t−ω q )(t−ω q )(t−ω q )(t−ω q )(t−ω q )(t−
10
ω q ) = t11 − α1 t10 + α2 t9 − α3 t8 + α4 t7 − α5 t6 + α6 t5 − α7 t4 + α8 t3 − α9 t2 + α10 t − α11 .
Then f (t) ∈ F [t] and the polynomial f (t) is irreducible over the field F . Note that
α11 = ω

q11 −1
q−1

has order q − 1 if q 6= 3, 7, α11 = 1 if q = 3, and α311 = 1 6= α11 if q = 7.

1

Set











x=










−1 0
0 −1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0

0
0
0
0
0
−1
0
0
0
0
0
0

0
 1

 0

 0

 0

 0
y=
 0

 0

 0

 0

 0
0

0
0
0
0
0
0
α6 α−1
11
0
0
0
0
0
0
α5 α−1
11
0
0 −1 0
0
0
α4 α−1
11
0
0
0
0 −1 0
α3 α−1
11
0 −1 0
0
0
0
α8 α−1
11
0
0
0
0
0
0
α7 α−1
11
0
0
0
0
0
0
α1 α−1
11
−1 0
0
0
0
0
α9 α−1
11
0
0
0
0
0 −1 α2 α−1
11
0
0
0
0
0
0
0
0
0
0 −1 0
0 α10 α−1
11
0
0
0
0
0
0
α−1
11

0 1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 

1 0 0 0 0 0 0 0 0 0 0 

0 0 0 0 1 0 0 0 0 0 0 

0 0 1 0 0 0 0 0 0 0 0 

0 0 0 1 0 0 0 0 0 0 0 
.
0 0 0 0 0 0 0 1 0 0 0 

0 0 0 0 0 1 0 0 0 0 0 

0 0 0 0 0 0 1 0 0 0 0 

0 0 0 0 0 0 0 0 0 0 1 

0 0 0 0 0 0 0 0 1 0 0 
0 0 0 0 0 0 0 0 0 1 0

0
α6
0
α5
0
α7
0
α9
0
α8
0
α4
−1 α10
0
α3
0
α2
0 α11
0
α1
0
0












,










Then x and y are elements of G of orders 2 and 3, respectively. Denote

0
0 −1 0
0
0
0
0
0
0
α6 α6 α−1
11
 −1 0
0
0
0
0
0
0
0
0
α5 α5 α−1
11

 0
0
0
0 −1 0
0
0
0
0
α7 α4 α−1
11

 0
0
0
0
0
0 −1 0
0
0
α9 α3 α−1
11

 0
0
0 −1 0
0
0
0
0
0
α8 α8 α−1
11

 0 −1 0
0
0
0
0
0
0
0
α4 α7 α−1
11
z = xy = 
 0
0
0
0
0
0
0
0
0 −1 α10 α1 α−1
11

 0
0
0
0
0 −1 0
0
0
0
α3 α9 α−1

11
 0
0
0
0
0
0
0 −1 0
0
α2 α2 α−1

11
 0
0
0
0
0
0
0
0
0
0 α11
0

 0
0
0
0
0
0
0
0 −1 0
α1 α10 α−1
11
0
0
0
0
0
0
0
0
0
0
0
α−1
11












.










The characteristic polynomial of z is fz (t) = (t − α−1
11 )f (t) and the characteristic roots
10
−1
q
q2
q3
q4
q5
q6
q7
q8
q9
α11 , ω, ω , ω , ω , ω , ω , ω , ω , ω , ω , and ω q of z are pairwise distinct.
2
3
4
5
−1
Then, in GL12 (q 11 ), z is conjugate to the matrix diag (α11
, ω, ω q , ω q , ω q , ω q , ω q ,
6
7
8
9
10
ω q , ω q , ω q , ω q , ω q ) and hence z is an element of SL12 (q) of order Q.
2

Let H = hx, yi, H ≤ G. We prove that H acts irreducibly on the space V . Indeed,
assume that W is an H-invariant subspace of V and k = dim W , 1 ≤ k ≤ 11.
Let first k = 1 and 0 6= w ∈ W . Then y(w) = λw where λ ∈ F and λ3 = 1. This
yields
w = µ1 (v1 +λ2 v2 +λv3 )+µ2 (v4 +λ2 v5 +λv6 )+µ3 (v7 +λ2 v8 +λv9 )+µ4 (v10 +λ2 v11 +λv12 ),
where µ1 , µ2 , µ3 , and µ4 are elements of the field F .
Now, we involve the action of x onto w: x(w) = νw where ν = ±1. This yields consecutively µ4 6= 0 , α11 = λ2 ν, and
(1)

µ3 = λ(α1 + να10 − λν)µ4 ,

(2)

νµ1 + µ2 = (να4 + α7 )µ4 ,

(3)

νµ2 + λ2 µ3 = λ(να3 + α9 )µ4 ,

(4)

(ν + 1)(µ1 − λα6 µ4 ) = 0,

(5)

(ν + 1)(µ1 − λ2 α5 µ4 ) = 0,

(6)

(ν + 1)(µ2 − λ2 α8 µ4 ) = 0,

(7)

(ν + 1)(µ3 − α2 µ4 ) = 0,

In particular, we have α311 = ν and α611 = 1. This is impossible if q = 5 or q > 7 since
then α11 has order q − 1. According to our assumption (q 6= 2, 4) only two possibilities
left: q = 3 (and α11 = 1) or q = 7 (and α311 = 1 6= α11 ). So ν = 1, α11 = λ2 and (1), (2),
(3), (4), (5), (6) and (7) produce α1 = λ2 α2 − α10 + λ, α6 = λα5 , α7 = λ2 α5 + λ2 α8 − α4
and α9 = λα2 + λα8 − α3 . Now f (−1) = −(1 + λ + λ2 )(1 + α2 + α5 + α8 ) = 0 both for
q = 3 and q = 7, an impossibility as f (t) is irreducible over the field F .
Now let 2 ≤ k ≤ 11. Then the characteristic polynomial of z|W has degree k and has
to divide fz (t) = (t − α−1
11 )f (t).The irreducibility of f (t) over F leads immediately to the
conclusion that this polynomial is f (t) and k = 11. Now the subspace U of V which is
generated by the vectors v1 , v2 , v3 , . . . , v11 is hzi-invariant. If W 6= U then U ∩ W is
hzi-invariant and dim (U ∩ W ) = 10. This means that the characteristic polynomial of
z|U∩W has degree 10 and must divide fz (t) = (t − α−1
11 )f (t) which is impossible. Thus
W = U but obviously U is not hyi-invariant, a contradiction.
Note that the above considerations fail if q = 2 or 4.
Now, as H = hx, yi acts irreducible on the space V and it has an element of order
Q, we conclude that H can not be contained in any maximal subgroup of G(= SL12 (q)).
Thus H = G and G = hx, yi is a (2, 3)-generated group.


3

References
[1] J. N. BRAY, D.F. HOLT, C. M. RONEY-DOUGAL. The Maximal Subgroups of
the Low - Dimensional Finite Classical Groups. London Math. Soc. Lecture Note
Series 407, Cambridge University Press (2013).
[2] TS. GENCHEV, E. GENCHEVA. (2, 3)-generation of the special linear groups of
dimension 8. Proceedings of the Forty Fourth Spring Conference of the Union of
Bulgarian Mathematicians, SOK ”Kamchia”, April 2-6 (2015), 167-173.
[3] E. GENCHEVA, TS. GENCHEV and K. TABAKOV. (2, 3)-generation of the special
linear groups of dimensions 9, 10 and 11. arXiv 1412.8631v5 (2016).
[4] M. A. PELLEGRINI. The (2, 3)-generation of the special linear groups over finite
fields. arXiv 1605.04276v1 (2016).
[5] K. TABAKOV, E. GENCHEVA and TS. GENCHEV. (2, 3)-generation of the special
linear groups of dimension 11. Proceedings of the Forty Fifth Spring Conference of
the Union of Bulgarian Mathematicians, Pleven, April 6-10 (2016), 146-151.
Ts. Genchev
e-mail: genchev57@yahoo.com
Department of Mathematics
Technical University
Varna, Bulgaria

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