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It's very well known that the dodecahedral space is an example of topological manivold with the same homology of $S^3$ but with different fundamental group. I'm trying to compute these homology groups. However any referece is apreciated.

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Since the fundamental group is perfect, the description of $H_1$ as the abelianization of the fundamental group tells you that $H_1=0$. The space is an orientable manifold so Poincaré duality applies to it, and from that we see that $H_2$ is also zero. Finally, since the space is a compact and orientable 3-manifold $H_3$ is free of rank 1

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    Thanks for the answer :) Do you know any book which proves this without use the homology of the manifolds? – 2017-02-28
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    In principle, you could use the description of the space as a dodecahedron with faces identified, which gives yuo a CW-structure, and use cellular homology to compute. I have never tried. – 2017-02-28