Is $\mathbb{Q}$ an injective $\mathbb{Z}[G]$-module Here $G$ is a finite group. I can show that $\mathbb{Q}$ (as a trivial $G$-module) is cohomologically trivial, since the inclusion $\mathbb{Q} \to \mathbb{Q}[G]$, $a\mapsto a(\sum_{g\in G} g)$ splits (via mapping $g\mapsto |G|^{-1}$) and $\mathbb{Q}[G]$ is an induced module.
I wonder that is $\mathbb{Q}$ injective as a $G$-module? Of course, I also wonder what happens when $G$ is an infinite group.
 A: $\newcommand{\Hom}{\operatorname{Hom}}$If $G$ is finite, then $\mathbb Q$ is an injective $\mathbb Z[G]$-module. Indeed, we have the following isomorphisms
$$
\Hom_{\mathbb Z[G]}(X,\mathbb Q) \cong \Hom_{\mathbb Q[G]}(\mathbb Q\otimes X, \mathbb Q) \cong \Hom_{\mathbb Q}\bigl((\mathbb Q\otimes X)_G, \mathbb Q\bigr)
$$
which are natural in the $\mathbb Z[G]$-module $X$. Here, for any $\mathbb Q[G]$-module $Y$, we denote $Y_G$ the largest quotient of $Y$ on which $G$ acts trivially; more explicitly:
$$
Y_G = Y/\langle gy-y\mid g\in G, y\in Y\rangle.
$$
Since $G$ is finite and $\mathbb Q$ is a field of characteristic $0$, every $\mathbb Q[G]$-representation is projective by Maschke's theorem, and hence every additive functor is exact. In particular, $Y\mapsto Y_G$ is exact.
It follows that the functor $X\mapsto\Hom_{\mathbb Z[G]}(X,\mathbb Q)$ is exact being the composition of the exact functors $X\mapsto \mathbb Q\otimes X$, $Y\mapsto Y_G$, and $Z\mapsto \Hom_{\mathbb Q}(Z,\mathbb Q)$. This means that $\mathbb Q$ is injective as a $\mathbb Z[G]$-module.
If $G$ is infinite, $\mathbb Q$ need not be injective. As an example, let $G$ be the infinite cyclic group on a generator $g$. Put $X = Y = \mathbb Z[G]$ and consider the $G$-equivariant injective map
$$
\varphi\colon X\longrightarrow Y,\quad x\longmapsto gx-x.
$$
On coinvariants, this is the zero map $\varphi_G\colon X_G \to Y_G$ (because the image of $\varphi$ is contained in the kernel of the projection $\pi_Y\colon Y\to Y_G$). Now, consider the inclusion $\iota\colon X_G \cong \mathbb Z \hookrightarrow \mathbb Q$. If there existed a map $\psi\colon Y\to \mathbb Q$ such that $\psi\circ\varphi = \iota\circ \pi_X$ (where $\pi_X\colon X\to X_G$ is the projection), then on $G$-coinvariants we would get $\psi_G\circ \varphi_G = \iota \neq 0$, but this contradicts $\varphi_G = 0$. Hence $\mathbb Q$ is not an injective $\mathbb Z[G]$-module.
