The cohomology of 2groups 

groups of order 8 groups of order 16 groups of order 32 groups of order 64 
The computations include StiefelWhitney classes, Chern classes and
Steenrod operations.

The results are available in SAGE format.  
How were the computations done ? Are they valid ?A paper is now available, explaining the methods employed. You can also have a look at the source code. So you can check for yourself that the results are valid. (You can also compare with the results which you happen to know already.)Among the results, one thing should be ignored though: the Steenrod operations given on generators other than StiefelWhitney classes are not correct (unless it is specified that I have manually added the answer). There are, on the whole, very few classes which are not StiefelWhitney (see group 6 of order 16 for the smallest example). Let me point out that my own calculations rely on those by Jon Carlson, so that you have to believe in his results in order to believe in mine. 

Some Steenrod operations seem to be missing...No. The trivial operations Sq^{n}(x) = x^{2} when n=deg(x) are not given, for concision. Also, recall that Sq^{m}(x)=0 for m> n. 

How are the representations numbered ?
Let us start with the complex representations  they appear together
with the information given on Chern classes, and you need to
understand them in order to understand the real ones.

There is an issue with GAP here. The GAP manual asserts that irr[1] is
often the trivial character, but not necessarily always. I have never
come accross an example of a group for which it isn't (while
working manually, that is). To be on the safe side, my programs look for the trivial character and move it to the beginning of the list. The numbering just explained happens then. I am pretty sure that the groups I'm considering are not concerned with this issue anyway. Also, I hope GAP does not change its numbering in the future. 
The relations seem more complicated than those given by other authors. There'd better be a very good reason !There is. It is best to look at the paper if you want to understand it.In the meanwhile, I can at least say that I've tried to keep the relations which have a "geometric", or "representationtheoretic", meaning. Let me give an example. Consider the group of quaternions, i.e. the 4th group of order 8. The cohomology is generated by w_{4}(r_{4}), w_{1}(r_{2}) and w_{1}(r_{3}). The relations are presented as
Well, for a start we see that the second equation is obtained by applying a Steenrod operation to the first one, so that the ideal of relations is generated by the first relation alone as a "Steenrod ideal". What is more, both relations are the direct translation in terms of
StiefelWhitney classes of the single relation:
I'm slightly cheating here, as I am also using the relation r_{3}=r_{1}r_{2}, which gives w_{1}(r_{3}) = w_{1}(r_{1}) + w_{1}(r_{2}), so there is a simple substitution to do in order to obtain the second equation. However, I see this equation as more meaningful than w_{1}(r_{2})^{3}=0. Finding out which relation between the representations has been translated into a given equation between the StiefelWhitney classes (if indeed there is one) is not so straightforward, unfortunately. It is also unclear how much time I will spend trying to resolve the situation. For the time being, if the
address of a group page is, say: 
You can also have a look at the .homlog files. They give extra
information on the computation. Both the .replog and the .homlog files
are a bit messy, and I apologize for this. Originally they were just
debugging files for my own use, not meant to be made public.

What is the "algebra of Milnor constants" ??Milnor has defined a sequence of derivations Q_0, Q_1, Q_2, ..., on any unstable algebra. (In fact Q_0 = Sq^1 and Q_{n+1} is the commutator of Q_n and Sq^{2^n}). The kernel of each derivation is a subalgebra, and the elements belonging to all these kernels simultaneously I call the Milnor constants.My interest in these elements stems from the following result: it can be shown that if a cohomology class is supported by an algebraic cycle, then it is a Milnor constant. So computing this subalgebra gives an upper bound for the ring generated by all algebraic cycles. One knows that Chern classes are (supported by) algebraic cycles. Therefore, if the algebra of Milnor constants happens to be generated by Chern classes, then we have completely determined which classes are algebraic. This was the original aim of all the computations on this page. 