Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|cite|improve this question
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. –  Andy Putman Sep 17 '13 at 15:23
Ah - that's a good point. I've edited my question. –  Nick Salter Sep 17 '13 at 15:31
To be clear, for you, a 'commutator relation' is a relation which is a product of commutators, right? –  HJRW Sep 17 '13 at 15:49
Yes, I mean a product of commutators. I'll edit the language to make it more precise. –  Nick Salter Sep 17 '13 at 16:58
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 $H_1$. –  HJRW Sep 18 '13 at 8:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.