Integral homology groups of free product of cyclic groups

Question

Please pardon my ignorance when it comes to group homology? Let $n$ and $m$ be nonnegative integers. I would like to compute $H_2(\mathbb{Z}/n * \mathbb{Z}/m; \mathbb{Z})$? I am interesed also in $H_2(\mathbb{Z}/m * \mathbb{Z} ; \mathbb{Z})$. If there is a general method of relating the homology of a free product to the homology of the components that would be useful.

Answer 1

I don't know how to answer this question using only the definition of group homology, but here is an alternative method.

Recall that $H_n(G; \mathbb{Z}) \cong H_n(K(G, 1); \mathbb{Z})$. Also note that $K(G_1, 1)\vee K(G_2, 1)$ is a $K(G_1\ast G_2, 1)$, so for $n > 0$

\begin{align*} H_n(G_1\ast G_2; \mathbb{Z}) &\cong H_n(K(G_1\ast G_2, 1); \mathbb{Z})\\ &\cong H_n(K(G_1, 1)\vee K(G_2, 1); \mathbb{Z})\\ &\cong H_n(K(G_1, 1); \mathbb{Z})\oplus H_n(K(G_2, 1); \mathbb{Z})\\ &\cong H_n(G_1; \mathbb{Z})\oplus H_n(G_2; \mathbb{Z}). \end{align*}

In particular

$$H_2(\mathbb{Z}/n\ast\mathbb{Z}/m; \mathbb{Z}) \cong H_2(\mathbb{Z}/2; \mathbb{Z})\oplus H_2(\mathbb{Z}/m; \mathbb{Z}) \cong 0\oplus 0 \cong 0$$

and

$$H_2(\mathbb{Z}/n\ast\mathbb{Z}; \mathbb{Z}) \cong H_2(\mathbb{Z}/2; \mathbb{Z})\oplus H_2(\mathbb{Z}; \mathbb{Z}) \cong 0\oplus 0 \cong 0.$$