Group Cohomology as a Derived Functor

Question

Let $G$ be a (finite, say) group acting on module $M$. I've been trying to understand how the (standard?) construction of group cohomology $H^i(G,M)$ given by wikipedia relates to the general definition of a right derived functor in wikipedia. The former defines the cochain complex made out of groups of the form: $$C^n(G,M) = \{ \text{ functions from $G^n$ to $M \}$ }.$$ Whereas the latter defines a cochain complex $$0 \rightarrow F(I^1) \rightarrow F(I^2) \rightarrow \cdots$$ Where $0 \rightarrow M \rightarrow I^0 \rightarrow I^1 \rightarrow \cdots$ form an injective resolution of $M$ and where $F$ is the functor on $G$-modules given by $F(N) = N^G = $ $G$-invariant elements of $N$. In both cases taking cohomology of the complex gives $H^n(G,M)$. (At least, that's my understanding.)

Question: It seems like $C^n(G,M)$ is not of the form $F(I^\bullet)$, is that correct? Specifically, it seems like the elements of $C^n(G,M)$ couldn't be the image of $F$ on any module, because they aren't $G$-invariant. (I am assuming the action of $G$ on $C^n(G,M)$ is the one induced by $G$'s action on the image, $M$.)

Is this simply a case of multiple complexes having the same cohomology, and the $C^n(G,M)$ complex being easier to compute than the one coming directly from the definition of a derived functor? Or am I missing something and actually $C^n(G,M) = F(I^n)$ holds?

P.S. Later on in that first wikipedia page they note $H^n(G,M) = \mathrm{Ext}^n_{\mathbb Z[G]}(G,M)$ and explicitly construct it as a derived functor (see also). But that seems to be a 3rd distinct complex because they avoid using an injective sequence for $M$ when building it.

Answer 1

As explained in the comments, the cochain complex that computes group cohomology in terms of cocycles comes from a different source than an injective resolution of $M$.

As to your actual question, the $G$ action on $C^n(G,M)$ is a red herring. You should think of $F$ as a functor to abelian groups, as the trivial action doesn't carry information (one typically doesn't consider the group cohomology groups as $G$-modules either, which one should if we consider $F$ as a functor to $G$-modules)

So your argument does not show that an injective resolution $I$ with $F(I^\bullet)=C^n(G,M)$ does not exist. You can't just take any action on $C^n(G,M)$ and say that the elements are not invariant. After all, as I explained, there is no natural action on $F(I^*)$. You might try to argue that there is the trivial action, but the trivial action is defined for every abelian group. Just because we can embed abelian groups to $G$-modules via considering the trivial $G$-action does not mean that every abelian group should always be considered a $G$-module.

Nonetheless, it is true that generally $C^\bullet(G,M)$ is not of the form $F(I^\bullet)$ for any injective resolution $M \to I^\bullet$. To prove this, let $G$ be the trivial group, then $F$ is just isomorphic the identity functor. Thus, if $C^\bullet(G,M)$ was of the form $F(I^\bullet)$, then $C^n(G,M)$ would itself need to be injective for $n\geq 1$. But taking e.g. $M=\Bbb Z$ and considering $n=1$ shows that this is generally not the case.