Let $G$ be a group. We define its standard resolution by setting
$$
C_n = \mathbb{Z}G^{n+1}
$$
to be the free $\mathbb{Z}$-module on $G^{n+1}$ for each $n \geq 0$, with a $\mathbb{Z}[G]$-module structure given by
$$
g(g_0,\dots,g_n) := (gg_0,\dots,gg_n)
$$
for each $g,g_0,\dots,g_n \in G$ and defining the differentials as
$$
\partial_n : (g_0, \dots,g_n) \in C_n \mapsto \sum_{i=0}^n(-1)^i(g_0,\dots,\widehat{g_i},\dots,g_n) \in C_{n-1}.
$$
This gives a projective resolution $(C_\bullet,\partial_\bullet) \xrightarrow{\varepsilon} \mathbb{Z} \to 0$ and so we can compute the group homology of $G$ as the homology of tensoring this complex with $\mathbb{Z}$,
$$
H_\bullet(G) = H_\bullet(\mathbb{Z} \otimes_{\mathbb{Z}G} C_n).
$$
From $G$ we can also form its classifying space $BG$, defined as the simplicial set
$$
BG_n = \{(g_0,\dots,g_n) : g_i \in G\}
$$
with face and degeneracy operators
$$
d_i(g_0,\dots,g_n) = (g_0,\dots,g_ig_{i+1},\dots,g_n),\\ s_i(g_0,\dots,g_n) = (g_0,\dots,g_i,1,g_{i+1},\dots,g_n).
$$
From here one can take the Moore complex of $BG$
$$
\dots \to \mathbb{Z}BG_n \xrightarrow{\partial} \mathbb{Z}BG_{n-1} \to \cdots \to \mathbb{Z}BG_0
$$
given by the free $\mathbb{Z}$-modules generated by each set $BG_n$ and the boundary operator given by $\partial = \sum_{i=0}^n (-1)^id_i$, that is,
$$
\partial(g_0, \dots,g_n) = \sum_{i=0}^n(-1)^i(g_0,\dots,g_ig_{i+1},\dots,g_n).
$$
The homology of $BG$ is then defined as the homology of this complex,
$$
H_\bullet(BG) := H(\mathbb{Z}BG_\bullet,\partial_\bullet).
$$
The constructions look suspiciously similar, so my question is:
What is the relationship between these two? Do we have $H_\bullet(G) = H_\bullet(BG)$, and if so, is there an isomorphism between the aforementioned complexes that I am missing?
Best Answer
Yes, the two are isomorphic, there are plenty of ways to see that. I'll begin by sketching a topological argument and then make it more concrete below. If you're not comfortable with algebraic topology you can skip the first part (except the definition of $EG$) up to "why yes there is"
If you're comfortable with algebraic topology, here's one way : note that what you wrote $C_\bullet$ is actually the singular chain complex of the simplicial set $EG$ which is defined in the obvious way, and has a free $G$-action. Moreover, this simplicial set $EG$ is contractible, as it is the nerve of a contractible groupoid (this groupoid is simply the groupoid that has $G$ as objects, and one arrow precisely between any two objects).
It follows that $C_\bullet$ is a projective resolution of the $G$-module $\mathbb Z$, so that what you called $H_\bullet(G)$ is in fact $\mathrm{Tor}^{\mathbb Z[G]}_\bullet(\mathbb Z, \mathbb Z)$
On the other hand, let's have a look at $H_\bullet(BG)$ : $BG$ is a simplicial set which is the nerve of the group $G$. It is therefore a Kan complex, and one can easily compute its homotopy groups to be $G$ for the fundamental group and $0$ for higher homotopy groups.
Therefore if you take a universal covering $\tilde EG\to |BG|$ (the notation is not a coincidence, I'll explain later - note in the meantime that such a universal covering has $\tilde EG/G \cong |BG|$), it has a free $G$-action and is contractible; so that its singular chain complex $C_\bullet(\tilde EG)$ is a projective resolution of $\mathbb Z$ as a $G$-module, therefore to compute $\mathrm{Tor}^{\mathbb Z[G]}_\bullet (\mathbb Z,\mathbb Z)$, one may use this resolution and take $\mathbb Z\otimes_{\mathbb Z[G]}C_\bullet(\tilde EG)$, which, by $\tilde EG/G\cong |BG|$ is nothing but $C_\bullet(|BG|)$, whose homology is $H_\bullet(|BG|)$, and this coincides with simplicial homology, i.e. $H_\bullet(BG)$.
So the two homologies are isomorphic, as they are isomorphic to $\mathrm{Tor}^{\mathbb Z[G]}_\bullet (\mathbb Z,\mathbb Z)$.
Now this is all very abstract for an actually very concrete thing : take the geometric realization of $EG$ as above : it is contractible and has a free action of $G$, the quotient $|EG|/G$ is therefore a space with exactly one nonzero homotopy group, and it's equal to $G$ : if you know a bit about homotopy theory, this tells you that $|EG|/G$ is homotopy equivalent to $|BG|$. This suggests to maybe look at something earlier in our chain of functors: is there actually something relating $EG/G$ and $BG$ ?
Why yes, there is : $EG$ is the nerve of a certain category, $BG$ of another one, can we get a functor between the categories ?
Well send any object $g\in EG$ to the single object $*$ of $BG$ (here I'm seeing them both as categories), and the unique morphism $g\to h$ in $EG$ can be sent to $h^{-1}g \in G$ (I had written $hg^{-1}$ initially, it works too, but it won't work later because of the side you chose for the action - this one will fit nicely with the side of the action). One checks easily that this yields a functor $EG\to BG$ and so a map of simplicial sets.
Moreover, at the level of categories, one can clearly see that this exhibits $BG$ as $EG/G$, so this works for simplicial sets too.
Note that the connection between the nerve of $EG$ and your construction is as follows: an $n$-simplex of said nerve is a composable string of arrows $g_0\to ...\to g_k$, but as there is exactly one arrow between any two objects, this amounts to the list $(g_0,...,g_k)$ of objects (i.e. elements of $G$), and the $i$th boundary map, which corresponds to composition for usual nerves, corresponds to just deleting $g_i$ from the list in our case : so your $C_\bullet$ was indeed the simplicial chain complex of $EG$.
Now what does our functor $EG\to BG$ look like on the level of nerves ? Well it sends $(g_0,...,g_k)$ to $g_0\to...\to g_k$ to $*\overset{g_1^{-1}g_0}\to ... \overset{g_k^{-1}g_{k-1}}\to *$, that is, the map on chain complexes is, very concretely :
$(g_0,...,g_k) \mapsto (g_1^{-1}g_0,...,g_k^{-1}g_{k-1})$
You can now check that this is $G$-equivariant (with the trivial action on the right), and that it is a map of chain complexes (simply because it was a map of simplicial sets to begin with and so it commutes with boundaries ! you can also see it more concretely if you wish : if you remove $g_i$ on the left, then you can multiply $g_{i+1}^{-1}g_i$ and $g_i^{-1}g_{i-1}$ on the right, and you get the correct thing)
Since it is $G$-equivariant, it factors through $(C_\bullet)_G \to \mathbb ZBG_\bullet$, and you can in fact now check that this is an isomorphism (it will come from the fact that $C_\bullet$ is a free $\mathbb Z[G]$-module, and so taking the coinvariants amounts to taking a quotient)
Finally you have to understand that for any $\mathbb ZG$-module $M$, $M_G \cong \mathbb Z\otimes_{\mathbb Z[G]}M$