From: Marston Conder 
Date: Thu, 14 May 1998 06:29:49 +1200 (NZST)
Subject: Re: Is SL(3,Z) a quotient of PSL(2,Z) ?

Dear everyone

The answer is no: SL(3,Z) is not a quotient of PSL(2,Z).

I'm very grateful to Chiara Tamburini for emailing me a copy of a section
from the PhD thesis of one of her students, P Zucca, showing that GL(3,Z)
is not (2,3)-generated.  A shortened and modified version of the argument
for SL(3,Z) is attached below, in English.

All the best
Marston Conder


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Assume the contrary, so SL(3,Z) can be generated by A and B such that
A^2 = B^3 = I.   Conjugating within GL(3,Z) using a matrix which has two
eigenvectors for A (corresp. to eigenvalue -1) and one eigenvector for B
(corresp. to eigenvalue 1) as its columns,  we may suppose
A = mat(-1,0,a:0,-1,b:0,0,1)   and   B = mat(*,*,0:*,*,0:*,*,1).

Now gcd(a,b) = 1, for if p were a prime divisor of both a and b, then under
reduction mod p the generators A and B would have (0,0,1)^T as a common
eigenvector, so could not generate SL(3,p), contradiction.  Hence there exist
integers c and d such that  ac + bd = 1,  and then conjugating further by
mat(a,-d,0:b,c,0:0,0,1)  we may suppose  A = mat(-1,0,1:0,-1,0:0,0,1)  and 
B = mat(r,s,0:u,v,0:x,y,1)  for some integers r,s,u,v,x,y.

As trace(B) = 0 and det(B) = 1 we have v = -(r+1) and su = rv-1 = -(r^2+r+1).
Also u is not divisible by any prime p, for otherwise (0,1,0)^T would be
a common eigenvector for the reductions of A^T and B^T mod p, so u = -1 or 1.

Replacing B by B^-1 if necessary, we may suppose u = 1, and then conjugating
by  mat(1,r,0:0,1,0:0,0,1)  we may take  A = mat(-1,0,1:0,-1,0:0,0,1)  and
B = mat(0,-1,0:1,-1,0:x,y,1)  for some x,y in Z.

Now assume that y = 0.  In this case if  X = mat(2,1,-x:1,2,0:-x,0,-2x)
then an easy calculation shows  AXA^T = X  and  BXB^T = X , and it follows
that  MXM^T = M  for every M in SL(3,Z), which is easily seen to be impossible.

Moreover, considering reduction of these matrices mod p, the same argument
shows y is not divisible by any prime p, and hence  y = -1 or 1.

But also  2x+y+6 = -1 or 1, for otherwise (2,x+2,-1)^T would be a common
eigenvector for A^T and B^T modulo any prime p dividing  2x+y+6.
As y = -1 or 1, it follows that  (x,y) = (-2,-1), (-3,-1), (-3,1) or (-4,1).
In the cases  (x,y) = (-2,-1) or (-3,1), however, (1,-xy,0)^T is a common
eigenvector for A and B mod 7, while in the other two cases (1,-xy,0)^T is a
common eigenvector for A and B mod 13, so all cases are eliminated.

Hence SL(3,Z) has no generators of the required form.


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