From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: Conjugacy Classes of GL(n,F)
Date: 11 Nov 1999 00:12:06 GMT
Newsgroups: sci.math
In article <3821A826.65533AD8@giemsch.de>,
Peer Giemsch wrote:
>Let S be a Sylow p-subgroup of GL(n,q), the full
>linear group of dimension n over the field of q elements, where
>q is a power of a prime p.
>My problem is to count the number of conjugacy classes of S or
>to give a lower/upper bound.
>Any idea ??
Count elements (and think about orders of elements) to show that
S is (conjugate to) the set of upper-triangular matrices (with
1's on the diagonal).
Now view S as being the set of all matrices M with the following
properties (F is the field GF(q), e_1, ..., e_n are the standard
basis vectors of R^n):
M acts trivially on the subspace F*e1
M acts trivially on the quotient space (F*e1 + F*e2)/(F*e1)
M acts trivially on the quotient space (F*e1+F*e2+F*e3))/(F*e1+F*e2))
etc.
In other words, we have a maximal chain of subspaces
(0) < F* e1 < (F*e1 + F*e2) < ... < F^n
with the property that M acts trivially on each quotient V_i / V_(i-1).
It is easy to check that the given chain is the only such chain on which
S acts in this way.
Next observe that conjugation in GL(n,q) amounts to a change of basis,
so that a subgroup T of GL(n,q) is conjugate to S iff there is
a maximal chain of subspaces (0) < V1 < V2 < ... < F^n for which T is
precisely the set of all matrices which fix each V_i setwise and act
trivially on each quotient V_i / V_(i-1).
So the tally of conjugates of S is the same as the number of such chains.
Well, there are (|V|-1)/(|F|-1) one-dimensional subspaces of a vector
space V, so I count (q^n-1)/(q-1) choices for V_1, then (repeating
the argument on V' = V/V_1) (q^(n-1)-1)/(q-1) choices for V_2, and so on.
So it looks to me like the total count of Sylow subgroups is
(q^n -1)(q^(n-1) -1)(...)(q^2 -1)(q-1) / (q-1)^n
Reality check: the number of these subgroups is supposed to be congruent
to 1 mod p, and it is. (It's even equal to 1 mod q.)
You might also note that the number of subgroups is supposed to equal the
index [G : N_G(S)]. Well, prove that the normalizer of S is precisely
the set of upper-triangular matrices (with arbitrary invertible diagonal
elements) and you'll be done after a great number of q's have
moved around on your paper.
dave
==============================================================================
[Remark: What is counted here is the number of conjugates of S, that is,
the number of Sylow subgroups. In retrospect, it seems more likely that the
original poster was asking for something else... --djr]