I had to prove that the cohomology with rational coefficients of any finite group is trivial. I tried to prove it by proving that $\mathbb Q$ is an injective $G$-module, however it doesn't seem to me to be the case.
Are there any easy to construct counter examples? Is it even true for any non trivial $G$?
The trivial representation $\mathbb{Q}$ is an injective $\mathbb{Z}[G]$-module for any finite group $G$. To prove this, suppose $N$ is a module with a submodule $M$ and $f:M\to \mathbb{Q}$ is a homomorphism; we wish to extend $f$ to a homomorphism $\tilde{f}:N\to \mathbb{Q}$. First, since $\mathbb{Q}$ is an injective $\mathbb{Z}$-module, we can extend $f$ to a $\mathbb{Z}$-homomorphism $g:N\to\mathbb{Q}$. Now the trick is that we can make $g$ become $G$-equivariant by averaging. That is, define $$\tilde{f}(x)=\frac{1}{|G|}\sum_{a\in G}g(ax).$$ Then $\tilde{f}$ is $G$-equivariant (for any $b\in G$, $\tilde{f}(bx)$ is the same as $f(x)$ since the summands just get permuted), and extends $f$ since $f$ was $G$-equivariant so all the summands are the same when $x\in M$.
More generally, a similar argument shows that any $\mathbb{Z}[G]$-module which is a $\mathbb{Q}$-vector space is injective (if $G$ acts nontrivially, you need to replace $g(ax)$ with $a^{-1}g(ax)$ in the sum above).