Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
3  
@ 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! –  Somnath Basu Jan 19 '11 at 23:03
13  
@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. –  Bill Thurston Jan 19 '11 at 23:15
6  
@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. –  Bill Thurston Jan 20 '11 at 1:40
2  
Theorem B in the paper On the integral homology of finitely-presented groups G. Baumslag, E. Dyer, and C. F. Miller Bull. Amer. Math. Soc. 4 (1981), 321-324. says roughly that for any recursively enumerable sequence of abelian groups $A_i$ with $A_1$, $A_2$ finitely generated, there exists a finitely presented group $G$ in which $H_n(G)=A_n$. More precisely, each $A_i$ must have a presentation of form $(X:R)_{ab}$, where $R$ is a free basis of the subgroup of FreeAb$(X)$ that it generates, the whole sequence of presentations must be r.e. –  Derek Holt Jan 22 '11 at 13:15
3  
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. –  Ben Wieland Jan 26 '11 at 5:43
show 6 more comments

1 Answer

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.

share|improve this answer
    
Wow, welcome to MO, prof. Sullivan! I will look into this. –  Chris Gerig Feb 10 '13 at 23:50
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.