# Cohomological dimension of finitely presented group

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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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. –  Misha Jun 14 '12 at 16:22
Or any Baumslag--Solitar group (presentation $\langle a,b\mid ba^mb^{-1}=b^n\rangle$ ). –  HJRW Jun 14 '12 at 18:02
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. –  HJRW Jun 14 '12 at 18:06

The quotient of the free group of rank 2 by a random, long relator has cohomological dimension 2 and is not commutative.

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In fact, quotient any relator that isn't a proper power, or primitive, works as well. –  Steve D Jun 14 '12 at 17:38
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. –  Ian Agol Jun 14 '12 at 17:44
Lyndon, "Cohomology Theory of Groups with a Single Defining Relation". –  Steve D Jun 14 '12 at 17:55
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. –  Ian Agol Jun 14 '12 at 22:53
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. –  Lee Mosher Jun 15 '12 at 2:52
Here's a concrete example of this type, one of my favorite groups: the solvable Baumslag-Solitar group $BS(1,n) = <a,t | tat^{-1}=a^n>$. –  Lee Mosher Jun 14 '12 at 16:31