Nontrivial finite group with trivial group homologies? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T12:54:12Z http://mathoverflow.net/feeds/question/52552 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/52552/nontrivial-finite-group-with-trivial-group-homologies Nontrivial finite group with trivial group homologies? Chris Gerig 2011-01-19T22:17:06Z 2013-02-10T22:44:43Z <p>I stumbled across this question in a seminar-paper a long time ago:</p> <p><em>Does there exist a positive integer</em> $N$ <em>such that if</em> $G$ <em>is a finite group with</em> $\bigoplus_{i=1}^NH_i(G)=0$ <em>then</em> $G=\lbrace 1\rbrace$?</p> <p>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.</p> <p><strong>Is there another known group with a larger $N\ge 5$ before homology becomes nontrivial?<br> Are there any classifications of obstructions in higher homology groups?</strong></p> <p>[]: 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.</p> http://mathoverflow.net/questions/52552/nontrivial-finite-group-with-trivial-group-homologies/121428#121428 Answer by Dennis Sullivan for Nontrivial finite group with trivial group homologies? Dennis Sullivan 2013-02-10T22:44:43Z 2013-02-10T22:44:43Z <p>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.</p>