==============================================================================
[It turns out what was wanted were groups with H^k(G,Z)=0 for many k;
the following shows H^k(G,Z/2Z)=0 is possible for many k -- a weaker statement.
--djr]
==============================================================================
Date: Sun, 25 Feb 96 23:38:31 CST
From: rusin (Dave Rusin)
To: adem@math.wisc.edu
Subject: Groups with lots of vanishing cohomology
Alex,
Did I understand you to ask for an example of the following type:
WANTED: a group G_k such that H^n(G_k, F_2) = 0 for n = 1, 2, ..., k
but not for n=k+1.
(You have such a group G_k for k= 0, 1, 2, 3 and want examples with
higher k.)
If that's what you want, you're in luck --- such examples exist with large
k, and in all likelihood exist for arbitrarily many different values of k.
One need only consider groups which are semidirect products of the form
G_p = (Z/2)^p . C, where C is cyclic of order q = 2^p - 1, and this number
is a (Mersenne) prime. The action I intend for C on (Z/2)^p is the
action of the Sylow q-subgroup of GL_n(F_2). A generator of such a
(cyclic) C can thus be represented by a p x p matrix over F_2. It's
not diagonalizable, but over the Galois field K=F_(2^p) it can be diagonalized
as diag(w, w^2, w^4, w^8, ..., w^(2^(p-1)) ), where w is a primitive
q-th root of 1. In other words, if you extend the action of C on
V = (F_2)^p to an action on W = V \tensor K, a basis of W
may be chosen so that C acts diagonally. This then makes it easy to compute
the action of C on the tensor powers of V: we know T^n(V) \tensor K =
T^n(W), and the action of C on the latter is of course diagonalizable --
the eigenvalues are the products of n eigenvalues of the action
of C on W. Could 1 be among the eigenvalues? Yes, surely, but since
the order of w is 2^p-1, Prod (w^(2^i))^e_i is equal to 1 iff
Sum (2^i*e_i) = 0 mod 2^p - 1. The solution to this diophantine equation
for which Sum(e_i) is smallest is the one with e_0=e_1=...=e_(p-1)=1,
in which Sum(e_i) = p. In other words, there is an eigenvector in
T^n(W) with eigenvalue 1 only if n >= p. It follows in particular
that T^n(V) has no fixed points under the action of C for n