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.

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    $\begingroup$ @ David - For a connected CW complex $X$ the classifying space of the based loop space of $X$, i.e., $B\Omega X$ is homotopy equivalent to $X$. One can therefore think of $X$ as a classifying space of some H-space and I believe Milnor gave a model which realizes $\Omega X$ as a topological group. This is what the OP may be implying although I'm not so sure! $\endgroup$ Jan 19, 2011 at 23:03
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    $\begingroup$ @David: Dan Kan and I once wrote a paper constructing, for any finite CW complex $X$, a space $Y = K(G,1)$, where $G$ is a finitely generated group, and a map $Y \rightarrow X$ which induces an isomorphism in homology over the fundamental group of $X$. So it can be done with homology of $G$ in the algebraic sense. But the construction actually depends on having a way to "cone off" the homology of any group within a bigger group. John Mather had proved that the group of compactly supported homeomorphisms of $\mathbb R^n$ is acyclic (in the algebraic sense)---same trick, I got it from that. $\endgroup$ Jan 19, 2011 at 23:15
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    $\begingroup$ @David: for a discrete group, $BG = K(G,1)$ (as homotopy types), and the homology of the group is the same as the homology of this space. For a topological group, if $G_\delta$ denotes $G$ with discrete topology, then $K(G,1) = BG_\delta$, and often has quite different homology from $BG$. For instance $BSL_2(\mathbb R)$ is homotopy equivalent to $BSO_1$, which is $\mathbb{CP}^\infty$. But $BSL_2(\mathbb R)_\delta$ has uncountably generated 2nd and 3rd homology. I guess even easier is $B\mathbb R$, which is trivial, vs $B\mathbb R_\delta$, with homology rank $2^\omega$ in every dimension. $\endgroup$ Jan 20, 2011 at 1:40
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    $\begingroup$ If a finite group has vanishing homology in all degrees, it is trivial. There should be an earlier reference, but Quillen's ICM address on group cohomology announces that for element $g\in G$, $H^*(G)\to H^*(<g>)$ is nontrivial. I very much doubt that there is a uniform $n$ so that if a finite group has vanishing cohomology in degrees less than $n$, that it vanishes, but I can't give you any examples of groups with large vanishing ranges. $\endgroup$ Jan 26, 2011 at 5:43
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    $\begingroup$ Just for the record: This question was posed ("an obvious question to ask") by A.Adem on page 810 of his 1997 paper here. $\endgroup$
    – Misha
    Mar 16, 2013 at 4:04

1 Answer 1


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.

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    $\begingroup$ Wow, welcome to MO, prof. Sullivan! I will look into this. $\endgroup$ Feb 10, 2013 at 23:50

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