I am trying to compute the group homologies $H^n(G,M)$ on a $G$-module $M$ where $G$ is the infinite cyclic group. I think I managed to get the case $n = 0$ and $n > 1$, but I'm unsure about the remaining case. My work is as follows.
Let $x$ be a generator of $G$, then the augmentation ideal is given by $IG = \langle x-1 \rangle$. This creates a short exact sequence $0 \to IG \xrightarrow{\iota} \mathbb{Z}[G] \xrightarrow{\epsilon} \mathbb{Z} \to 0$ where $\iota$ is the inclusion, and $\epsilon$ the augmentation map. Since $IG$ is infinitely cyclic, it is a free group, therefore this sequence can be considered a free resolution of $\mathbb{Z}$. The Hom functor induces the following sequence on the respective deleted resolution.
\begin{equation} 0 \to \text{Hom}(\mathbb{Z}[G],M) \xrightarrow{\iota^\ast} \text{Hom}(IG,M) \to 0 \to 0 \to \cdots \end{equation}
Consider an arbitrary $f \in \text{Ker}(\iota^\ast)$, which means $f \circ \iota = 0$ per definition of the induced map. Since these maps are defined over the ring $\mathbb{Z}[G]$, it follows that $f(x - 1) = (x - 1) \cdot f(1) = 0$. This defines a unique map for every choice of $f(1) \in M[x - 1]$. If there are two maps $f, g \in \text{Hom}(\mathbb{Z}[G],M)$ such that $f(1) = g(1)$ then for arbitrary $z \in \mathbb{Z}[G]$ follows that $f(z) = z \cdot f(1) = z \cdot g(1) = g(z)$. Therefore a map is uniquely defined by its image $f(1)$. This concludes $H^0(G,M) = \text{Ker}(\iota^\ast) \cong M[x-1]$.
I'm a bit lost on what to do next. Clearly the only other (potentially) non-trivial homology is $H^1(G,M)$. I'm not mistaken this homology is equal to $\text{Hom}(IG,M)/\text{Img}(\iota^\ast)$ right? I would think any $f \in \text{Hom}(IG,M)$ can be written as $f = g \circ \iota$ for some $g \in \text{Hom}(\mathbb{Z}[G],M)$. Is that enough to conclude $H^1(G,M) = 0$ as well?
Edit: The thing that sort of makes me hesitant about my results is, if this is true, then I should be able to show any group extension $0 \to M \to E \to G \to 1$ is equivalent to the trivial one. But I'm not sure how to show this.