Group cohomology without G-modules (a.k.a. what does this bar construction compute?)

Question

Without any prior exposure to the cohomology of groups, one might naively proceed by replacing a group by a sort of resolution.

For instance, let's take $G = \mathbb{Z}^2$, and "resolve":
$$ 0 \to \mathbb{Z} \overset{1 \mapsto xyx^{-1}y^{-1}}\longrightarrow \mathrm{F}_2 \to \mathbb{Z}^2 \to 0. $$ This mirrors exactly the topology of the torus $\mathrm{T} = \mathbf{B} \mathbb{Z}^2$, as obtained by attaching two loops, and then a disc along the commutator.
The same works for $G = \mathbb{Z}^3$, as there is a relation between the relations, given by the Jacobi–Witt–Hall identity (this is not strictly true, as I need to allow some twisting to conjugate the relations to get a relation between the relations). The cohomology of the three torus, and its cell structure, is mirrored by a sequence looking somewhat like
$$ 0 \to \mathbb{Z} \to \mathrm{F}_3 \to \mathrm{F}_3 \to 0. $$

A different example would be to take $G = \mathbb{Z}/2\mathbb{Z}$: $$ 0 \to \mathbb{Z} \overset{\times 2}{\longrightarrow} \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0.$$ This is not so good... following this on the nose would get us to construct $\mathbb{RP}^2$ instead of $\mathbf{B} \mathbb{Z}/2\mathbb{Z} = \mathbb{RP}^\infty$. The problem is that the relation $x^2 = 1$ is not imposed once, but infinitely many times, once for each pair $(x^{2n}=1, x^{2n+1}=1)$. To account for this, one has to extend the sequence by another map $ \mathbb{Z} \overset{\times 2}{\longrightarrow} \mathbb{Z}$, and a similar argument gives us that we should in fact be looking at

$$ \cdots \overset{\times 2}{\longrightarrow} \mathbb{Z} \overset{\times 2}{\longrightarrow} \mathbb{Z} \overset{\times 2}{\longrightarrow} \cdots \overset{\times 2}{\longrightarrow} \mathbb{Z} \overset{\times 2}{\longrightarrow} \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0.$$ This of course mirrors the cell structure of $\mathbf{B} \mathbb{Z}/2\mathbb{Z} = \mathbb{RP}^\infty$.

With this in mind, suppose we now try to define group cohomology while staying in the category of groups (rather than using $\mathbb{Z}[G]$-modules for instance). We have a bar construction

$$ \cdots F^3 G \underset{\underset{\longrightarrow}{\longleftarrow}}{\overset{\overset{\longrightarrow}{\longleftarrow}}{\longrightarrow}} F^2 G \underset{\longrightarrow}{\overset{\longrightarrow}{\longleftarrow}} F G \longrightarrow G \to 0,$$

where $F$ is the functor (comonad) sending a group to the free group on the underlying set. This corresponds to looking at all possible elements of $G$ as generators, then imposing every equation $g_1 \ldots g_j = k$ as a relation, etc. Mimicking the previous, we only remember the outer evaluation maps and get a sequence

$$ \cdots \to F^3 G \to F^2 G \to FG \to G \to 0. $$

In the case of $G = \mathbb{Z}/2\mathbb{Z}$, after cleaning up by removing extraneous terms corresponding to the identity (for instance replacing $FG = \langle e_0, e_1 \rangle$ by $\langle e_1 \rangle$), we get precisely the sequence $$ \cdots \overset{\times 2}{\longrightarrow} \mathbb{Z} \overset{\times 2}{\longrightarrow} \mathbb{Z} \overset{\times 2}{\longrightarrow} \cdots \overset{\times 2}{\longrightarrow} \mathbb{Z} \overset{\times 2}{\longrightarrow} \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0.$$

Question: Is the sequence
$$ \cdots \to F^3 G \to F^2 G \to FG \to 0 $$ always the "correct" replacement for $G$? Can it be used to compute the group cohomology of $G$?

Note: This object needs to be twisted somewhat, to account that one might not have literal relations between relations, but have them once we allow to conjugate relations by elements of the group (and so on, higher up). So it's not literally correct as is, but hopefully someone who understands crossed complexes can fix this.

(On the whole, perhaps it would be wiser to apply a functor valued in some abelian category to the simplicial group obtained by the bar construction, and then use Dold–Kan to turn the entire thing into a genuine chain complex... but the sequence above for $\mathbb{RP}^\infty$ really does feel like the correct object to consider, which is why I'm insisting on working with groups alone.)

More generally, I would like to understand why we have run into the problems we have (e.g. those with $ G = \mathbb{Z}/2\mathbb{Z}$), whereas doing the analogous thing for the comonad associated to the adjunction between $\mathbb{Z}[G]$-modules and abelian groups does compute the group cohomology with no hiccups.
I understand the need to distinguish between groups and $\infty$-groups (where imposing the same relation multiple times is different from imposing it once), but somehow that problem seems to have vanished when considering $\mathbb{Z}[G]$-modules...?

Answer 1

The following does not answer your question, but shows that the object you have in mind does give useful information (notice I did not drop all but the outer evaluation maps but kept the whole simplicial group).

From the adjunction $U:\mathsf{Grp}\leftrightarrows\mathsf{Set}:F$ between the free group functor and the forgetful functor, one gets a monad $F\circ U:\mathsf{Grp}\to\mathsf{Grp}$, which we can write just $F$. Given a group $G$, there is an associated (augmented) simplical group $F^\sharp G$, the bar construction for the monad, which looks like the one you wrote.

If $M$ is a $G$-module, them $M$ is a $F^nG$-module for all $n$, by pulling back along the composition of maps going from $F^nG$ to $G$ in the simplical group. It makes sense to talk about derivations from the $F^nG$ to $M$, and in fact we get a cosimplicial abelian group $\operatorname{Der}(F^\sharp G,M)$. From it we can construct as usual a cochain complex (whose differentials are alternating sums of transposed face maps in the simplicial group $F^\sharp G$)

(I have not completely check the details but) the cohomology of the complex $\operatorname{Der}(F^\sharp G,M)$ should be isomorphic to the cohomology of $G$ with values in $G$ up to a shift and a twist: explicitly, $$H^i(\operatorname{Der}(F^\sharp G,M))\cong\begin{cases}Der(G,M), &\text{if $i=0$;}\\H^{i+1}(G,M), &\text{if $i>0$.}\end{cases}$$ This is the usual group cohomology of $G$, but we dropped the $0$th group, and instead of having derivations modulo the inner ones, we have just derivations. Most of the work needed to check this is of the general nonsense kind.

If you apply a more interesting functor $\mathsf{Grp}\to\mathsf{Ab}$ to $F^\sharp G$, you'll get other things, of course.

All the above works because free groups play particularly nice as the domain of derivations: they behave like projectives. AAndré-Quillen cohomology resolves general commutative rings with polynomial rings, because it wants to find derivatives of the functor of derivations (or of Kähler forms, in the homology case) and for these polynomial rings are "projective")

Answer 2

It is well known that the underlying simplicial set of a simplicial group is a Kan complex. However, the underlying simplicial set of the canonical simplicial resolution (call it $F_\bullet G$) you describe has an extra degeneracy, so it is really just a disjoint union of contractible Kan complexes. In particular, the homotopy groups of $F_\bullet G$ are all trivial, and we recover the original group as $\pi_0$.

But we haven't really used the fact that we have a simplicial group. To get back group homology, we apply the abelianisation functor to $F_\bullet G$. The homology of the associated chain complex is then $H_{* + 1} (G, \mathbb{Z})$ (this seems to be folklore), and the 0th homology is the abelianisation of $G$. I suppose this should be expected, since the abelianisation of $G$ is $G \otimes \mathbb{Z}$, where $\otimes$ here is defined to be universal with respect to bihomomorphisms.