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A group $G$ is perfect if $G=[G,G]$. For perfect groups, we know that the first group homology $H_1(G, \mathbb{Z})=G/[G,G]=0$. A group $G$ is called acyclic if its group homology $H_i(G, \mathbb{Z})=0$ for $i \geq 1$ and $H_0(G, \mathbb{Z})=\mathbb{Z}$. It is well-known that the group $A_5$ is a perfect group which is not acyclic. However, I am interested in computing group homology with rational coefficients. I would like to know an example of a perfect group which has non-trivial rational group homology. It seems to me that such a group cannot be a finite perfect group, because one can use the universal coefficient theorem $$0 \rightarrow H_i(G,\mathbb{Z})\otimes \mathbb{Q}\rightarrow H_i(G, \mathbb{Q})→Tor(H_{i-1}(G,\mathbb{Z}),\mathbb{Q})→0$$ and deduce the rational group homology vanish for $i \geq 1$ inductively. So I am wondering if a perfect group is actually rationally acyclic or a counterexample to this exisit?

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You can find an example where second homology group of a perfect group is not zero in this paper of Milnor. p. 79.

http://www.maths.ed.ac.uk/~aar/papers/milnor7.pdf

Milnor consider the cohomology of the discrete group $PSL(2,R)$ (in his terminology $PSL(2,R)$ made discrete) and it shows that there is a class in $H^2$ which is not trivial. The duality implies that you have an homology class $\geq 1$ which is not trivial.

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  • $\begingroup$ Thank you for your fast reply. Sorry that I have one more question. In this example, it seems that $dim H_*(G, \mathbb{Q})=1$, since the square of the generator is mapped to zero in $H^4$. I am still wondering if there exists an example of perfect group such that $dim H_*(G, \mathbb{Q})>1$. In particular, can $H_*(G, \mathbb{Q})>1$ has arbitrary dimension? $\endgroup$ – awivil Feb 18 '16 at 4:53

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