Proving Ind functor preserves injectives from tensor functor

Question

For a subgroup $H$ of $G$. I've known that the functor: $$\textrm{Ind}_H^G: \textrm{Mod}_H\to \textrm{Mod}_G$$ is actually equivalent to the functor: $$\mathbb{Z}G \otimes_{\mathbb{Z}H} -: \mathbb{Z}H\to \mathbb{Z}G$$ Is there any simple way to prove the above functor preserves injectives?

Answer 1

This is false. Take $H$ to be the trivial group and $G = \mathbb{Z}$. Then $\mathbb{Z}[H] \cong \mathbb{Z}$ and $\mathbb{Z}[G] \cong \mathbb{Z}[x, x^{-1}]$, and the extension of scalars of the injective module $\mathbb{Q}$ is the module $\mathbb{Q}[x, x^{-1}]$. Since $\mathbb{Z}[x, x^{-1}]$ is an integral domain, injective modules must be divisible, but $\mathbb{Q}[x, x^{-1}]$ is not divisible (e.g. the action of $x - 1$ is not invertible).

What is true in general is that coinduction $\text{Hom}_{\mathbb{Z}[H]}(\mathbb{Z}[G], -)$ preserves injectives. This is a straightforward application of the universal property: coinduction is right adjoint to restriction, and generally the right adjoint of a functor which preserves monomorphisms preserves injectives. Dually, the left adjoint of a functor which preserves epimorphisms preserves projectives (so induction preserves projectives).

Induction and coinduction are naturally isomorphic if $H$ and $G$ are both finite, and I think more generally if $H$ has finite index in $G$; did you mean to include this as a hypothesis?