Cohomology of free group acting trivially

Question

I read here the formula for cohomology of a free group with trivial action: $$H^q(G, M) = \begin{cases} M &\text{for } q = 0\\ M^n &\text{for } q = 1\\ 0 &\text{for } q \geq 2\\ \end{cases}$$ where $G$ is free group with $n$ free generators. $\newcommand{\Z}{\mathbb{Z}}$

How does one prove that? (I know the cohomology for $q = 0$ and $q = 1$. But I have no idea how to handle $q \geq 2$. I have tried recursion approach: $$0 \rightarrow M \rightarrow Ind^G(M) \rightarrow M' \rightarrow 0$$ so that $H^q(G, M) = H^{q-1}(G, M')$ but there is no way to show that $M'$ is induced. ($M'$ can be identified with the subset $\{\phi : G \rightarrow M \;|\; \phi(1) = 0\}$ of $Ind^G(M)$ with the action $g \cdot \phi : x \mapsto \phi(xg) - \phi(g)$.) I have tried the direct approach via taking cohomology of the complex $Hom_G(P^\bullet, M)$ where $P^\bullet \rightarrow \Z \rightarrow 0$ is standard resolution but cannot find a way to construct the coboundary.

REMARK: As a consequence of $H^1$, the number of free generators of a free group is uniquely determined (even though a free group might have different free bases and can have free subgroups of any countable cardinality). That also means if we know a free group $G$ has $n$ free generators then any $n$ elements $g_1,...,g_n \in G$ that generates $G$ must be free generators.

Answer 1

Start with $q=2$. We actually have $H^2(G,A)=0$ for every $G$-module $A$. This follows from the fact that $H^2(G,A)$ is in bijection with the set of isomorphism classes of extensions

$$1 \to A \to E \to G \to 1.$$ But all such extensions are trivial (i.e. split) by the universal property of $G$. Now use dimension shifting as you suggested before. Prove by induction the stronger statement that $H^n(G,A)=0$ for all $n \geq 2$ and all $G$-modules $A$ using the exact sequence

$$0 \to A \to Ind^G(A) \to A' \to 0$$ as you mentioned in your question. The cohomology sequence gives $$\cdots \to H^{n-1}(G,A') \to H^n(G,A) \to H^n(G, Ind^G(A)) \to \cdots,$$ and both terms surrounding $H^n(G,A)$ are zero (the first is by inductive hypothesis, the second by Shapiro's lemma).

Answer 2

You can also show there is a length one resolution of the trivial $G$-module $\Bbb Z$ as follows. Let $\varepsilon : \Bbb ZG\longrightarrow \Bbb Z$ be the canonical augmentation that sends eveyr $g\in G\mapsto 1 \in\Bbb Z$, and let $K=\ker\varepsilon$. Show that $K$ is a free $\Bbb ZG$-module with basis $\{ x-1 : x\in X\}$ where $X$ is a basis of $G$, so there is a free resolution of length $1$ $$ 0\longrightarrow K\longrightarrow \mathbb ZG\longrightarrow \mathbb Z\longrightarrow 0$$

Thus $H^n(G;M)=H_n(G;M)=0$ if $n>1$, and you can handle the remaining two cases by hand. There are some details in Weibel's book.