Suppose that we have a finite group $G$. Its classifying space $BG$ is not naturally pointed, so if we would like to consider it as a spectrum there are two possible approaches. The first is to take any point in the classifying space and let it be the basepoint when constructing the suspension spectrum; the second is to add a disjoint basepoint.

My question is: is it possible to reconstruct $G$ from these spectra? If so, how? What information do we know about $G$ if we are given one of these spectra?

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    $\begingroup$ 1) Yes. 2) Taking the fundamental group (of the nontrivial component in the second case. 3) I don't understand this question. $\endgroup$ Jul 1, 2013 at 20:35
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    $\begingroup$ Sorry, am I just being daft? I always thought $\mathrm B G$ was pointed for any group $G$. Take the canonical map of groups $1 \to G$ (where $1$ is the one-element group) and apply $\mathrm B$ to it. But $\mathrm B 1 = \mathrm{pt}$. $\endgroup$ Jul 1, 2013 at 20:44
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    $\begingroup$ @TheoJohnson-Freyd, it is pointed if you point it! One can decide not to :-) $\endgroup$ Jul 1, 2013 at 21:11
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    $\begingroup$ @FernandoMuro: if G is not abelian, then $\pi_1^s BG \neq G$. And for question 3, I want to know how much, if not all, information can we reconstruct from the space? (This assumes that the answer to the first question is "no.") $\endgroup$
    – Inna
    Jul 1, 2013 at 21:21
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    $\begingroup$ @MarianoSuárez-Alvarez, sure, you may always forget the pointing of a pointed space, or re-point it some other way. But OP had said "not naturally pointed", and I think that the functor $B: \mathrm{Groups} \to \mathrm{Spaces}$ does receive a canonical natural transformation from the constant functor $pt$. In any case, trying to turn it into a spectrum will, at least a priori, lose information. $\endgroup$ Jul 1, 2013 at 21:44

2 Answers 2


Let $X_+$ be a space $X$ together with a disjoint base-point. Since $\Sigma^\infty(BG_+)=\Sigma^\infty(BG)\vee S$, where $S$ is the sphere spectrum, there's not much difference between the two spectra you propose. In Example 5.2 of Martino–Priddy's 'Stable homotopy classification of $BG^{\hat{}}_p$' you can find non-isomorphic finite groups whose classifying spaces have the same suspension spectrum, so the answer to your first question in its current form is no. In that paper you can also find very interesting results concerning your last question.


I'm not sure exactly what the question is you are asking in terms of how much information you want to have about the spectrum. From my perspective, the answer to the first question is no, in the sense that if you are only considering the homotopy type of the spectrum, you cannot reconstruct the group from the spectrum.

To see this, consider any non-trivial acyclic group. Then the homology of BG is the same as that of a point, so if you choose a basepoint in BG the suspension spectrum of BG will be homotopy equivalent to the suspension spectrum of B(e) (the classifying space of the trivial group) even though G and (e) are not isomorphic groups.

The same problem arises with disjoint basepoints of course. The homology of $BG_{+}$ is the same as that of $B(e)_{+} \simeq S^0$.

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    $\begingroup$ The question asks about finite groups; there are no acyclic finite groups. $\endgroup$ Jul 1, 2013 at 22:01
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    $\begingroup$ Quite right. I missed the finite criterion somehow and answered a much simpler question. Fernando's answer is much better and gives a nice sensible reference too. $\endgroup$ Jul 3, 2013 at 18:46

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