# how can we compute the homology of these groups without using topology?

I'd like to know the homology of a free group and a free abelian group of rank 2. I know that they could be computed topologically, but I'm searching a proof purely algebraic, could you help me please?

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## 1 Answer

If you look in the book by Cartan-Eilenberg on Homological Algebra (and surely many others, but each of us has a favorite!) you will find explicit projective resolutions for both free groups and for free abelian groups. From that one can immediately compute homology and cohomology.

You can also use technology. For example, it is easy to show that the (co)homology of a free product of groups is (in positive degrees) the direct sum of the factors (you can find this in Hilton-Stammbach's book on homological algebra, for example) and this easily reduces the computation of the (co)homology of free groups to that of infinite cyclic groups. Likewise, there is a Künneth formula for the (co)homology of direct products of groups which reduces the computation of (co)homology of free abelian groups to —again­— that of infinite cyclic groups.

Infinite cyclic groups are easily handled by writing down an explicit resolution: if we write $G$ for the infinite cyclic group generated by $\sigma$, we can use $$\mathbb ZG\xrightarrow{d}\mathbb ZG\stackrel\varepsilon\twoheadrightarrow\mathbb Z$$ with $\varepsilon$ the usual augmentation and $d$ the unique $\mathbb ZG$-linear map such that $d(1)=\sigma-1$.

Alternatively, every textbook includes an explicit description of the functors $H^0(G,\mathord-)$ and $H^1(G,\mathord-)$, which in the case of the infinite cyclic group makes them immediately computable, and then you can check any problems that $H^1(G,\mathord-)$ is right exact. It follows that the higher $H^p$'s are zero, and that we have computed the whole cohomology. One can proceed in a similar way for homology.

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