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If I have a divisible abelian group (i.e $\mathbb{Z}$-injective) $D$ and I take an arbitrary group $G$, and then I give $D$ the trivial $G$ action, then will $D$ be an injective $\mathbb{Z}[G]$-module?

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This statement is false. Let $T$ be the infinite cyclic group with generator $t\in T$. Then I claim that nontrivial abelian group $M$ with the trivial $T$-action cannot be an injective $T$-module. (By $T$-module I mean a $\mathbb{Z}[T]$ module.)

Let $M$ be a nontrivial abelian group with trivial $T$-action. Then $M$ is injective as a $T$ module if and only if the functor $\mathrm{Hom}_T(-,M)$ is an exact functor. To show that this is not the case, it is sufficient to show that $\mathrm{Ext}^1_T(\mathbb{Z},M)\not= 0$. This can be computed using the projective $T$-resolution $0\to \mathbb{Z}[T]\xrightarrow{t-1}\mathbb{Z}[T]\xrightarrow{\epsilon}\mathbb{Z}\to 0$ where $\epsilon$ is the augmentation map.

Using this, one easily shows that $\mathrm{Ext}^1_T(\mathbb{Z},M) \cong M_T$, where $M_T$ is the module of coinvariants. Since $M$ has the trivial action, $M_T\cong M$ and $M$ is nonzero by assumption. (This of course is the usual group cohomology $H^1(T,M)$.) Hence $M$ cannot be injective.

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