When (G:S) is finite, why is every injective G-module, when considered as an S-module, still an injective S-module?

Question

I encountered the statement that $H^*(G,\pi ^*_{S\rightarrow G}A)$ is a universal functor, when S is an open subgroup of $G$. Its statement is accompanied with the reasning that, as $S$ has finite index in $G$, $\pi ^*_{S\rightarrow G}A$ is injective whenever $A$ is. But how to derive this?
P.S. Some additional conditions were missing in this question. I apologize. Here $G$ is a profinite group, which acts on the $G$-module $A$. So $G$ is compact, which justifies the claim that $S$ has finite index in $G$.
Thanks for your attention.

Answer 1

The key reason it works is that in this case, $kG$ is (finitely generated) projective as a $kS$-module (i.e. $\pi^*_{S \to G} kG$ is finitely generated projective in your notation). Indeed $kG$ is a free $kS$-module with free generators a set of coset reps for $S$ in $G$. Here $k$ is whatever ring you are working over.

Injectivity of $\pi^*_{S \to G} A$ is equivalent to the vanishing of all groups $\operatorname{Ext}^n_{kS}(-, \pi^*_{S \to G} A)$ with $n>0$. But by the Eckmann-Shapiro Lemma,

$$\operatorname{Ext}^n_{kS}(M, \pi^*_{S \to G} A) \cong \operatorname{Ext}^n_{kG}( M \!\!\uparrow _S ^G, A) = 0 $$

because $A$ is injective.