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...?