Let $\mathbb{Q}$ be the group of rational numbers. How to compute group homology $H_n(\mathbb{Q},\mathbb{Z})=H_n(B\mathbb{Q},\mathbb{Z})$?

I know that $H_0(\mathbb{Q},\mathbb{Z})=\mathbb{Z}$ and $H_1(\mathbb{Q},\mathbb{Z})=\mathbb{Q}_{ab}=\mathbb{Q}$ and I think that $H_n(\mathbb{Q},\mathbb{Z})=0$ for $n>1$, but I don't know how to prove it.


You can explicitly construct a model of $B\mathbb{Q}$ by taking the mapping telescope of the sequence $S^1\stackrel{1}\to S^1\stackrel{2}\to S^1\stackrel{3}\to S^1\stackrel{4}\to\dots$, where $\stackrel{n}\to$ denotes a degree $n$ map. Indeed, if $K$ is such a mapping telescope, we see that the homotopy groups are the colimit of the homotopy groups of $S^1$ under these maps, and so $\pi_1(K)=\mathbb{Q}$ and the other homotopy groups of $K$ are trivial.

Thus we can compute the group homology of $\mathbb{Q}$ as the homology of this space $K$. But the homology of $K$ is just the colimit of the homology of $S^1$ under the induced maps of the sequence, and so we find $H_0(K;\mathbb{Z})=\mathbb{Z}$, $H_1(K;\mathbb{Z})=\mathbb{Q})$, and $H_n(K;\mathbb{Z})=0$ for $n>1$

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  • $\begingroup$ Eric Wofsey, thanks! $\endgroup$ – user327401 Oct 15 '17 at 18:52
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    $\begingroup$ It's crucial here that one use the mapping telescope and not the colimit of the sequence. It's fun to think through what happens if you just take the colimit of the sequence. See mathoverflow.net/questions/16829/… $\endgroup$ – Dan Ramras Oct 15 '17 at 18:56

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