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I was assigned to compute the group homology of $\mathbb{Z}^k$ with $\mathbb Z$ as coefficient ring(with the trivial action): $H_*(\mathbb{Z}^k, \mathbb{Z})$.

I know that $H_*(\mathbb{Z}^k, \mathbb{Z}) = {\rm Tor} _* ^{\mathbb{Z}(\mathbb{Z}^k)}(\mathbb{Z}, \mathbb{Z}) $, but I don't know how to compute this Tor. I think, first of all, I need to take a projective resolution of $\mathbb{Z}$, but even I am stuck with it. Is there any easy way to compute this?

I have more problems which I have to do. If someone help me on this, then I will try to solve the others by myself. Thank you for your help!

S. Ha
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  • Do you know the Kunneth formula? – Moishe Kohan Jun 15 '14 at 16:10
  • @studiosus Sorry for being late. I have heard the Kunneth formula, which is stated in my text. I think it is okay to use the Kunneth formula(honestly, I have not studied very hard these days...), but I am not sure. If there is an easy way to compute the group homology, then it will be an enough answer for my question! If it is possible to do with just definitions, then I will very thank for giving that answer! – S. Ha Jun 16 '14 at 05:36
  • In fact, I have an example in my note, $H_*(\mathbb{Z}/k\mathbb{Z}, \mathbb{Z})$, which is computed with just definitions. But I am not sure if it is possible to find a projective resolution for the assigned problems. How do I find a projective resolution of $\mathbb{Z}$ as a $\mathbb{Z}(\mathbb{Z}^k))$-module? Am I missing something really easy? How about a free group on $k$ generators? – S. Ha Jun 16 '14 at 05:38
  • If you look closely at the Kunneth formula, it allows you to compute $H_*(Z^k, Z)$ by using the decomposition $Z^k=Z^{k-1}\times Z$. Then you just induct on $k$. – Moishe Kohan Jun 16 '14 at 21:32
  • I already solved through the definition, but I was not sure if my answer is correct. So I will check my answer with the conclusion of the Kunneth formula. Thank you! – S. Ha Jun 17 '14 at 07:56
  • Yeah I was correct. It was actually not difficult! Thank you for giving me comments. – S. Ha Jun 17 '14 at 08:15

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