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An alternative construction of such a group: let $G$ be the free product of two copies of $A_5$, i.e., $G=A_5*A_5$. This group $G$ is perfect and it is also virtually free, so that any torsion-free s …

By Venkov's proof of the Evens-Venkov theorem, if you take a faithful representation $G\rightarrow SO(m)$, then the cohomology $H^*(BG)$ is a finitely generated module for the image of $H^*(BSO(m))$. …

Inspired by YCor's remark I realized that every finite abelian $p$-group $G$ of rank three will have essential elements in $H^3(G;U(1))$. I think that this works for $p=2$ as well, but here's an argu …

I would like to draw your attention to an article by Nick Gill and (his then MMath student) Sam Hughes The character table of a sharply 5-transitive subgroup of $A_{12}$, that constructs the character …

Yes. There is a stronger result too, which predates Quillen's theorem: if $G$ is a finite group whose Sylow $p$-subgroup $P$ is abelian, then the restriction map $H^*(G;\mathbb{F}_p)\rightarrow H^*(P …

There is an algebraic result that is relevant, due to Benson and Carlson and stated as Corollary 4.5 in `Complexity and Multiple Complexes' Math. Z. vol 195 (1987) 221--238. Given a finite group $G$, …

Here are some comments, including an answer to (3). Firstly, if you want an actual explicit computation of the mod-2 cohomology of symmetric groups $S_n$ for as large an $n$ as possible, you should l …