Let $G$ be a group. Can one always find a $G$-module $M$ such that the homology group $H_i(G,M)$ is non-zero for some $i\geq1$?
In other words, I'm wondering if there exists a group which has the most vanishing homology conceivable or, at least, how far it is known one can go in this direction.
Clearly, acyclic groups, namely groups whose classifying space has the homology of a point, are on the right path, and are the starting point of my inquiry. Notice that an acyclic group has to be perfect, as otherwise $H_1(G,\mathbb{Z})$ would not vanish.
The homological dimension of a group $G$, often denoted $\text{hd}(G)$, is defined to be the minimal $n$ so that for every $G$-module $M$ and $i > n$, $H_i(G,M) = 0$. With this language, it seems your question is "does there exist a group so that $\text{hd}(G) = 0$", which would mean that for $n \geq 1$ we must have $H_n(G,M) = 0$ for every $G$-module $M$.
It turns out that $\text{hd}(G) = 0 \iff G = 1$, so no nontrivial group will do what you want.
The cohomological version of this theorem is an exercise in Brown's Cohomology of Groups (chapter VIII.2), and the homological version is stated as a remark in this paper. I think some variant on the proof Brown is alluding to should still work, but I'm not seeing how to flesh out the details. As a sketch:
Say $\text{hd}(G) = 0$. Then in particular $H_1(G,\mathbb Z) = 0$, so $\mathbb{Z}$ is flat. Brown's hint is to use the fact that if $\mathbb{Z}$ is projective, then the augmentation map $\epsilon : \mathbb{Z}G \to \mathbb{Z}$ splits, and from here we can show $G = 1$.
Of course, flat and projective are different (since group rings are not always noetherian), so this proof as written doesn't go through. I'm not seeing how to fix this, unfortunately, and I also can't seem to find a reference. I might spend some more time thinking about this tomorrow, though.
I hope this helps ^_^