Skip to main content
3 of 6
added 863 characters in body; Post Made Community Wiki
Gregory Arone
  • 10.9k
  • 1
  • 47
  • 66

Edit: I thought I had a counterexample, but it does not quite work. It answers a different (easier?) question. I will make it CommunityWiki, and leave it up in a modified form, just in case it is of some interest, or might help in the search for a complete answer.

If I am not mistaken, For some finite groups $G$, $BG^+$ is equivalent to the product of the $p$-completions $BG\hat{_p}$. For example, it is true if $G$ is perfect. More generally, it is true if the final term of the lower central series is a perfect subgroup, and the plus construction is taken with respect to this subgroup. So we may consider the related question whether non-isomorphic finite groups can have equivalent $p$-completed classifying spaces for all primes $p$ (in the same spirit as Matthias Wendt's answer).

By a theorem of Oliver, the $p$-completion of $BG$ is determined by the $p$-fusion system of $G$, where the $p$-fusion system of $G$ is the category whose objects are subgroups of a $p$-Sylow subgroup $P$, and where morphisms are all homomorphisms that are induced by conjugation by an element of $G$.

So this question is equivalent to whether there can be non-isomorphic finite groups whose $p$-fusion systems are equivalent for all primes $p$. Such examples are known to exist. I found the following one in a paper of Martino and Priddy (who attribute it to Minami): $Q_{4p}\times {\mathbb Z}/2$ and $D_{2p}\times {\mathbb Z}/4$. Here $Q_{4p}$ is the generalised Quaternion group of order $4p$, where $p$ is an odd prime. The point is that they have isomorphic $p$-Sylow subgroups at all primes, namely ${\mathbb Z}/2\times {\mathbb Z}/4$ and ${\mathbb Z}/p$, and moreover the $p$-fusion systems are equivalent: you get the trivial structure at the prime $2$, and the only non-trivial morphism at $p$ is the inverse homomorphism.

However, these groups are solvable but not nilpotent. So they do not have non-trivial perfect subgroups, and their lower central series do not terminate at the trivial group. So I don't believe their plus constructions are equivalent. Indeed, the plus construction does nothing to these groups. An example where the groups are nilpotent would answer the original question.

Gregory Arone
  • 10.9k
  • 1
  • 47
  • 66