TITLE: Nontrivial finite group with trivial group homologies?
QUESTION [70 upvotes]: I stumbled across this question in a seminar-paper a long time ago:
Does there exist a positive integer $N$ such that if $G$ is a finite group with $\bigoplus_{i=1}^NH_i(G)=0$ then $G=\lbrace 1\rbrace$?
I believe this to still be an open problem.  For $N=1$, any perfect group (ex: $A_5$) is a counterexample.  For $N=2$, the binary icosahedral group $SL_2(F_5)$ suffices (perfect group with periodic Tate cohomology).  And I found in one of Milgram's papers a result for $N=5$, the sporadic Mathieu group $M_{23}$.  Note that this question is answered for infinite groups, because we can always construct a topological space (hence a $BG$ for some discrete group $G$) with prescribed homologies.
Is there another known group with a larger $N\ge 5$ before homology becomes nontrivial?
Are there any classifications of obstructions in higher homology groups?
[[Edit]]:  Another view.   A group is $\textit{acyclic}$ if it has trivial integral homology.  There are no nontrivial finite acyclic groups.  Indeed, a result of Richard Swan says that a group with $p$-torsion has nontrivial mod-$p$ cohomology in infinitely many dimensions, hence nontrivial integral homology.

REPLY [26 votes]: Here is an approach to try to answer this question.
To show the  affirmative side 
[there is an N] look at the prime two and at swans argument that for a finite group with a two primary subgroup the cohomology mod two must be nonzero in infinitely many dimensions. I tried but didn't find it. if that results in a finite N with nonzero cohomology for all finte groups with a two primary part  the problem is solved. because if the two primary component is absent, the group is solvable and N=1 results in non zero homology.
On the other hand, to show the negative side, there is no such N. if swans argument does not yield such a concrete N for the prime two then one might well believe there is no such N, and that a string of examples  might be constructed by thinking about the proof of swan's theorem.