Date: Sun, 1 Oct 95 23:13:32 CDT From: rusin (Dave Rusin) To: sivak-d@cs.buffalo.edu Subject: Re: Analogue of Fermat for Matrix Groups? Newsgroups: sci.math [Was: can you generalize Fermat's little theorem to matrix groups] Sure. Fermat's little theorem is nothing but Lagrange's theorem specialized to the group Z_p^*. So what you want to do is ignore Fermat and listen to Lagrange: every element of GL_n(F_p) has order dividing the order of GL_n(F_p). At the level of your question I would say the order of this group would be considered hard or messy, but when n=2 it's not so bad; the conclusion is that for every 2 x 2 invertible matrix M over F_p, we have M^k = I if k=(p^2-1)*(p^2-p). Actually you can do a bit better with some matrix theory behind you: if the eigenvalues of M are distinct, then M is conjugate to (and thus has the same order as) a diagonal matrix. If D=diag(a,b) is a diagonal 2x2 matrix, its order is the lcm of the orders of a and b. By Fermat, this is at worst (p-1) if a and b lie in F_p. If they lie in an extension field, they'll have the same order and it will divide p^2-1. (Elements with precisely this order exist). If the eigenvalues of M are not distinct, M might still be diagonalizable, so this discussion applies. If not, it has some element b actually from F_p on its diagonal and a 1 above the diagonal (not M, I mean a conjugate matrix.). To get the diagonal elements equal to 1, you need to raise to the (p-1) power (at least for the worst b's you do); then to get rid of what's above the diagonal you will in general have to raise to the p-th power, giving elements of order dividing and including p^2-p. So the smallest exponent you could use for your theorem would be the lcm of these two numbers p^2-1=(p-1)(p+1) and (p^2-p)=p*(p-1), which is clearly p(p-1)(p+1). Naturally the situation gets even more cumbersome with matrices larger than 2 x 2. dave