Is there an example of a group $G$ such that $H_n(G;\mathbb{Z})=0$ for all $n\in \mathbb{Z}_{>0}$ and $H_n \left( \prod_{k=1}^{\infty} G;\ \mathbb Z\right) \neq 0$ for all $n\in\mathbb{Z}_{>0}$?
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No, by Kunneth formula for finite products. – Moishe Kohan Dec 04 '13 at 04:33
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The answer is that such groups do not exist. Even more, if you have acyclic groups $G_k$ then their direct product is also acyclic. The proof is a direct application of the Kunneth formula (if you do not know it, check wikipedia or Hatcher's book). First of all, it suffices to work with the usual homology of cell complexes $K(G_k,1)$. Second, nonzero homology will be detected by a finite direct product. Thus, you can apply induction to compute with the Kunneth formula and the question reduces to the statement that the product of two acyclic spaces $X, Y$ is acyclic. This follows by examining the tensor product and Tor terms of the Kunneth formula. By your assumption, they both vanish, hence, the middle terms is also zero. You have to be a bit careful with the term $Tor(H_0(X), H_0(Y))$, it is zero since both homology groups are isomorphic to the infinite cyclic group.
Moishe Kohan
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Why is nonzero homology detected by a finite direct product? Maybe I am missing something simple but I don't see any obvious reason that should be true. – Eric Wofsey Dec 15 '21 at 04:33
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@EricWofsey: I'll take a look, the answer was written 8 years ago. – Moishe Kohan Dec 15 '21 at 15:06