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?

6$\begingroup$ A concrete example: Take figure 8 knot (or any nontrivial 2bridge knot). Then the fundamental group of the complement is 2generated and has cohomological dimension 2. $\endgroup$ – Misha Jun 14 '12 at 16:22

$\begingroup$ Or any BaumslagSolitar group (presentation $\langle a,b\mid ba^mb^{1}=b^n\rangle$ ). $\endgroup$ – HJRW Jun 14 '12 at 18:02

$\begingroup$ 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. $\endgroup$ – 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.

4$\begingroup$ In fact, quotient any relator that isn't a proper power, or primitive, works as well. $\endgroup$ – Steve D Jun 14 '12 at 17:38

$\begingroup$ Do you know an old reference? Gromov's random group theorem is overkill, one could just use that a random relator will be smallcancellation, but was this observation made before Gromov? Anyway, an answer is more useful if it provides at least some reference. $\endgroup$ – Ian Agol Jun 14 '12 at 17:44

3$\begingroup$ Lyndon, "Cohomology Theory of Groups with a Single Defining Relation". $\endgroup$ – Steve D Jun 14 '12 at 17:55

$\begingroup$ Ok, I guess it follows from the MagnusMoldovanskii hierarchy. Wise's recent results also imply the virtual cohomological dimension is $\leq 2$ in the presence of torsion. $\endgroup$ – Ian Agol Jun 14 '12 at 22:53

$\begingroup$ 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. $\endgroup$ – Lee Mosher Jun 15 '12 at 2:52
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.

1$\begingroup$ Here's a concrete example of this type, one of my favorite groups: the solvable BaumslagSolitar group $BS(1,n) = <a,t  tat^{1}=a^n>$. $\endgroup$ – Lee Mosher Jun 14 '12 at 16:31