Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
@MikeShulman: Agreed. I've also tried the "Theorem A (=Theorem 2.7)" "Theorem B (=Theorem 4.6)" approach in the past, which I still kind of like. I think changing to letters for the main results can help by not having two numbering systems, which could get confusing. (But still get the best of both worlds, as you suggest.)
@TimothyChow: If we only consider these on the category of groups & isomorphisms, then we can take $F_2$ to be contravariant as well. Yet even in that situation, I think the answers on the question linked to by Eric rule out a natural isomorphism of these functors.
For groups given by matrix generators over $\mathbb{Z}$ this is certainly true, but for groups given by matrix generators over a finite field (another common setting), certainly most questions are decidable, and the question then becomes one of whether there are provably efficient algorithms and/or practical implementations.
According to this, the part you cite is still due to Wedderburn. Malcev's contribution was to show that any semisimple $S'$ such that $A = Rad(A) \oplus S'$ must be conjugate to $S$ by an element of the form $(1+r)$ for $r \in Rad(A)$.
In particular, group cohomology can be seen as a topological invariant of a group. Although it has a purely algebraic definition, it is naturally isomorphic to the cohomology of the corresponding classifying space.
Sure, i see. But note that this analogy is off by a level of abstraction. Each graph has an independence poly counting the ind sets in that graph; what you're asking for is more like a formula for the number of graphs with a given property. While this may be given by a poly in q (or, more likely, a quasi-poly?), it seems hard to find and not so easy to work with. It is probably not much easier to work with than counting points on a general variety...
While these are interesting characterizations, I wonder whether checking for such special bases is any easier than just looking for a group basis directly? (by which I mean a basis of A which is closed under multiplication and consists of units) it may very well be, I just don't see either way yet.
@JohnShareshian: Indeed, there are many applications of CFSG in group theory, combinatorics, computer science, and beyond. Even more basic is that deciding isomorphism of simple groups can be done in poly(|G|) time (since they are generated by 2 elements, which seems to need CFSG), and the much deeper "effective recognition of black-box simple groups" by many authors. One small note on the state of the art for GI: I believe Pyber removed the dependence of Babai's result on CFSG in arxiv.org/abs/1605.08266.
@WillSawin: If you look naively, sure. But instead of a classification theorem you can ask for an efficient algorithm to decide isomorphism. There are also nice results like the classification of p-groups of small coclass. Also new results in (higher, i.e. beyond H^2 = extensions) group cohomology, such as Symond's result bounding the degrees of the gens & relations of the cohomology ring. There are also very cool relationships between cohomology & (modular) representation theory, see, e.g., Benson's 2-volume book "Representations & Cohomology".
@Will Sawin: my impression is that it significantly slowed progress in finite group theory in general. There was, and still is, much to be done regarding group cohomology and modular representation theory. The finite groups are the building blocks, but it remains to study how they can be glued together.
Could you give a reference for the case of Z coefficients? It's somewhat hard to search for online because the search terms are all so similar to the case of Q coefficients, or to GL(n,Z), etc.
@PaceNielsen: I don't think all right ideals are right principal in Lam's Exercise 12 (you mean Section 19, Chapter 7, I believe). In particular, $\bigcap_{i \geq 0} (rad R)^i \neq 0$, so by my answer this intersection can't be right principal. The exercise asks to prove that all right ideals are two-sided ideals, but says nothing about principal.
If M isn't a ring, what is the algebra structure on $H^*(G, M)$? I believe it is just a module over the algebra $H^*(G, R)$ where $M$ is an $R$-module (or even an $RG$-module, appropriately defined).
