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]