Homology of a vector space modulo a finite group I'm looking at a proof concerning homology of graph complexes.
One has a chain complex $(V, \partial)$ of $\mathbb{Q}$-vector spaces
$$\cdots \xrightarrow{\partial}V_n \xrightarrow{\partial} V_{n-1} \xrightarrow{\partial} \cdots$$
There is a finite group $G$ acting on all of these spaces $V_n$, so that we may consider the quotient spaces $V_n/G$ and the corresponding chain complex $(V/G, \partial)$.
Now in the proof it says: "over the rationals, finite groups have no homology, a fact which implies that the chain complexes $(V, \partial)$ and $(V/G, \partial)$ are rationally quasi-isomorphic."
I don't understand the following things in this statement:
What does it mean for a group to have no homology (what does one even mean by the "homology of a group")?
Why do finite groups have no homology over the rationals?
Why does $G$ having no homology imply that the two chain complexes are quasi-isomorphic?
For the third question, I would think that one can do something with the long exact sequence in homology. So, if one would have a short exact sequence
$$0 \rightarrow G \rightarrow V \rightarrow V/G \rightarrow 0$$
then the long exact sequence in homology and the fact that $G$ has no homology would imply the quasi-isomorphicness. But the first map in the above short exact sequence doesn't really make sense...
Thanks in advance for any help!
 A: Group homology is a construction that takes a topological group $G$ and outputs a vector space (assuming you're working over some base field). There are several equivalent definitions; one of them is that it is given by $H^*(BG; \Bbbk)$ where $BG$ is the  classifying space of $G$.
It is a theorem that if $G$ is a finite group, then $H^n(G; \mathbb{Q}) = 0$ for all $n \ge 1$; this is what is meant by the informal sentence "over the rationals, a finite group has no cohomology".
Now this is where it gets technical. There is the Cartan–Leray spectral sequence, a spectral sequence with $E^2$ page
$$E^2_pq = H_p(G; H_q(V))$$
and converging to $H_*(V/G)$, when $G$ acts freely on $V$. Over the rationals higher group homology is trivial, so we have
$$H_p(G; H_q(V)) = \begin{cases} H_q(V)/G, & \text{if } p = 0; \\ 0, & \text{if } p > 0. \end{cases}$$
So the $E^2$ page is concentrated on the first column, so the sequence collapses and thus $H_*(V)/G$ is isomorphic (still over $\mathbb{Q})$ to $H_*(V/G)$ as a graded vector space.
Now this isn't quite what you've written, but maybe with the precise reference there are more hypotheses on how $G$ acts on $V$ to get the result.
As it is the result as stated can't be true in such generality, take for example $V = \mathbb{Q} \oplus \mathbb{Q}$ concentrated in degree $0$ with zero differential and $G = \mathbb{Z}/2\mathbb{Z}$ acting by swapping the two factors. Then $V/G \cong \mathbb{Q}$ concentrated in degree zero: $V$ and $V/G$ don't even have the same homology, so they cannot possibly be quasi-isomorphic. (However we do have $H_*(V/G) \cong H_*(V)/G$, which agrees with what I've written above.)
