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