Suppose $G = F/R$ is a finitely presented group with $F = F_n$ a free group and $R$ the normal closure of words $W_1, \dots, W_p$, not all of which are products of commutators. An obvious necessary condition for $G$ to admit a presentation with relators contained in the commutator subgroup of $F$ is that the abelianization be torsion-free; then the number of generators in a commutator-relators presentation will be equal to the rank of $G^{ab}$. This certainly isn't sufficient, as a nontrivial perfect group shows. Is there a criterion for deciding whether a given presentation can be changed by Tietze moves into one with relators in the commutator subgroup? Is there an algorithm for doing so?

EDIT: Andy Putman observes that a general algorithm would solve the problem of identifying the trivial group, so there can't be such a criterion. Given this, are there any known results for special classes of presentations?

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    $\begingroup$ This is impossible. The trivial group is the only group with this property whose abelianization is trivial, so if you could solve this problem then you could decide whether a group presentation gave the trivial group. $\endgroup$ – Andy Putman Sep 17 '13 at 15:23
  • $\begingroup$ Ah - that's a good point. I've edited my question. $\endgroup$ – Nick Salter Sep 17 '13 at 15:31
  • $\begingroup$ To be clear, for you, a 'commutator relation' is a relation which is a product of commutators, right? $\endgroup$ – HJRW Sep 17 '13 at 15:49
  • $\begingroup$ Yes, I mean a product of commutators. I'll edit the language to make it more precise. $\endgroup$ – Nick Salter Sep 17 '13 at 16:58
  • $\begingroup$ Here's another necessary condition: if your group $G$ is not free, then $b_2(G)>0$. Stallings proved (Homology and central series of groups, J. Algebra) that, in a group with $H_2(G,\mathbb{Q})=0$, any linearly independent subset in $H_1$ generates a free subgroup. But the generating set in your presentation is linearly independent in $H_1$. $\endgroup$ – HJRW Sep 18 '13 at 8:23

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