From: Marston ConderDate: 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 ------------------------------------------------------------------------------ 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. ------------------------------------------------------------------------------