Group homology of the rationals

Question

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.

Answer 1

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 $n>1$