I've been looking for some concrete examples of group homology calculations and have been struggling to find any so I thought I'd work through one myself that should be simple: $H_n(\mathbb{Z}, \mathbb{Z})$, but need some help making the last step.
I made a free $\mathbb{Z}[\mathbb{Z}]$-resolution of $\mathbb{Z}$: $$ \cdots\rightarrow\mathbb{Z}[\mathbb{Z}]\rightarrow\mathbb{Z}[\mathbb{Z}]\rightarrow\mathbb{Z}[\mathbb{Z}]\xrightarrow{\partial_0}\mathbb{Z}\rightarrow0 $$ With: $$ \partial_0:\mathbb{Z}[\mathbb{Z}]\rightarrow\mathbb{Z} $$ $$(a_0,a_1,\dots) \mapsto a_0$$ and the other maps created to fit in with exactness and $\ker{\partial_0}$. I then tensor over $\mathbb{Z}[\mathbb{Z}]$ with $\mathbb{Z}$: $$ \cdots\rightarrow\mathbb{Z}[\mathbb{Z}]\otimes_{\mathbb{Z}[\mathbb{Z}]}\mathbb{Z}\rightarrow\mathbb{Z}[\mathbb{Z}]\otimes_{\mathbb{Z}[\mathbb{Z}]}\mathbb{Z}\rightarrow\mathbb{Z}[\mathbb{Z}]\otimes_{\mathbb{Z}[\mathbb{Z}]}\mathbb{Z}\rightarrow\mathbb{Z}\otimes_{\mathbb{Z}[\mathbb{Z}]}\mathbb{Z}$$ Which I know becomes: $$\cdots \rightarrow \mathbb{Z} \rightarrow \mathbb{Z} \rightarrow \mathbb{Z}\rightarrow\mathbb{Z}\otimes_{\mathbb{Z}[\mathbb{Z}]}\mathbb{Z}$$ And it is at this point I know I need the homology of this sequences and that is where I am getting stuck. Can anyone point me in the right direction? and does anyone have any group homology calculation examples they know of online?
Here is a short free resolution of $\mathbb{Z}$ as a trivial $\mathbb{Z}[\mathbb{Z}]$-module: $$0 \to \mathbb{Z}[\mathbb{Z}] \xrightarrow{\partial} \mathbb{Z}[\mathbb{Z}] \xrightarrow{\epsilon} \mathbb{Z} \to 0.$$
The map $\epsilon: \mathbb{Z}[\mathbb{Z}] \to \mathbb{Z}$ is the usual augmentation, given by $[n] \mapsto 1$. The map $\partial: \mathbb{Z}[\mathbb{Z}] \to \mathbb{Z}[\mathbb{Z}]$ is the map $[n] \mapsto [n] - [n-1]$. You should verify that this complex is exact. Tensoring this resolution down with $\mathbb{Z}$, we get the complex $$0 \to \mathbb{Z} \xrightarrow{\bar{\partial}} \mathbb{Z} \to 0.$$ The map $\bar{\partial}$ is the zero map, so the homology of this chain complex is $\mathbb{Z}$ in degrees $0, 1$, and is trivial in other degrees. Hence $$H_n(\mathbb{Z}, \mathbb{Z}) \cong \begin{cases} \mathbb{Z}, & n = 0, 1 \\ 0, & \text{else}. \end{cases}$$ This agrees with the answer given in Michael's comment.