# Calculation of homology groups of chain complex $C_*(Y) \otimes_{\mathbb{Z[G]}} \mathbb{Z}$.

Let $$G=\mathbb{Z^2}$$ be a group acting on a set $$Y=\mathbb{Z}$$ defined as

$$n\cdot(a,b)=n+a+b, ~\textrm{for all}~ n \in Y \textrm{ and } (a,b)\in G .$$

Consider chain complexes $$(C_*(Y), \partial)$$ and $$(C_*(Y)_G : = C_*(Y) \otimes_{\mathbb{Z[G]}} \mathbb{Z}, \partial \otimes id_{\mathbb{Z}})$$, where $$C_n(Y)$$ is the free $$\mathbb{Z}$$-module generated by $$(n+1)$$-tuples $$(y_0, y_1, \ldots,y_n)$$ of elements of $$Y$$, $$\partial = \sum_i (-1)^i \partial_i$$ and $$\partial_i(y_0, y_1, \ldots,y_n)=(y_0, \ldots,y_{i-1}, y_{i+1}, \ldots, y_n)$$. I have to prove that $$H_i(C_*(Y)_G)=0$$ for all $$i \geq 2$$ and $$H_1(C_*(Y)_G)=\mathbb{Z}$$. I don't know how to proceed. Can someone suggest direction to solve this?

• In the tensor product defining $C_*(Y)_G$, does $\mathbb{Z}$ have the trivial $G$-module structure? Commented Feb 1, 2020 at 16:31
• @Lama:: Yes, there $\mathbb{Z}$ has trivial $G$-module structure. And $C_{*}(Y)$ has natural $G$-action induced from the action of $G$ on set $Y$.
– eyp
Commented Feb 2, 2020 at 3:17