I have a group of cohomological dimension 2 generated by two elements. Is it possible to deduce that the group is commutative or, more generally, does $\mathrm{cd}\ G=2$ imply anything about the relations that $G$ must satisfy?
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7$\begingroup$ A concrete example: Take figure 8 knot (or any nontrivial 2-bridge knot). Then the fundamental group of the complement is 2-generated and has cohomological dimension 2. $\endgroup$– MishaCommented Jun 14, 2012 at 16:22
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$\begingroup$ Or any Baumslag--Solitar group (presentation $\langle a,b\mid ba^mb^{-1}=b^n\rangle$ ). $\endgroup$– HJRWCommented Jun 14, 2012 at 18:02
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$\begingroup$ Oh, I see that Lee Mosher wrote something similar in a comment below. The bottom line is that there is a huge bestiary of groups of cohomological dimension two. $\endgroup$– HJRWCommented Jun 14, 2012 at 18:06
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2 Answers
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The quotient of the free group of rank 2 by a random, long relator has cohomological dimension 2 and is not commutative.
answered Jun 14, 2012 at 14:23
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4$\begingroup$ In fact, quotient any relator that isn't a proper power, or primitive, works as well. $\endgroup$– Steve DCommented Jun 14, 2012 at 17:38
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$\begingroup$ Do you know an old reference? Gromov's random group theorem is overkill, one could just use that a random relator will be small-cancellation, but was this observation made before Gromov? Anyway, an answer is more useful if it provides at least some reference. $\endgroup$– Ian AgolCommented Jun 14, 2012 at 17:44
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3$\begingroup$ Lyndon, "Cohomology Theory of Groups with a Single Defining Relation". $\endgroup$– Steve DCommented Jun 14, 2012 at 17:55
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$\begingroup$ Ok, I guess it follows from the Magnus-Moldovanskii hierarchy. Wise's recent results also imply the virtual cohomological dimension is $\leq 2$ in the presence of torsion. $\endgroup$– Ian AgolCommented Jun 14, 2012 at 22:53
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$\begingroup$ I wasn't thinking so much about Gromov's random groups as "any old random thing you want to do", the point being that cd G = 2 implies just about nothing about relators, and that it's quite easy to construct oodles of examples with cd G = 2 and no particular pattern to the relators. But I quite forgot about one relator groups! Which really nails in the point. $\endgroup$ Commented Jun 15, 2012 at 2:52
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To complement @Lee's answer, in "Classification of soluble groups of cohomological dimension two", (Math Z, 1979), Dion Gildenhuys does exactly what he claims, so you can see what you get under very strong additional conditions.
answered Jun 14, 2012 at 15:45
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1$\begingroup$ Here's a concrete example of this type, one of my favorite groups: the solvable Baumslag-Solitar group $BS(1,n) = <a,t | tat^{-1}=a^n>$. $\endgroup$ Commented Jun 14, 2012 at 16:31