Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $F$ be a finitely generated free group and let $A$ be a finite index subgroup of $F$. Does there exist a subgroup $B\subset A$ such that $F/B$ is (elementary) amenable and torsion-free?

A group $G$ which is amenable and torsion-free has (at least conjecturally) the following nice properties: $\Bbb{Z}[G]$ has conjecturally no zero divisors (this is known if $G$ is say locally indicable or left-orderable) and $\Bbb{Z}[G]$ embeds in its Ore localization.

I have a certain application for the above question in mind, where using an Ore localization plays a role. But I am also curious since my intuition fails me on that question. Note that if $F/A$ is solvable, then one could just take $F$ to be an iterated commutator subgroup. But if $F/A$ is non-solvable, then I don't know what to do.

share|improve this question
just a side remark: zero divisors conjecture is known for arbitrary elementary amenable torsion-free groups (follows from the work of P. Linnell on Atiyah conjecture) –  Łukasz Grabowski Nov 10 '11 at 13:22
Let $A'$ be the normal closure of $A$. If the finite group $F/A'$ is the image of a torsionfree elementary amenable group $H$, then one can lift the surjection $F \to F/A'$ to $H$ to obtain a subgroup of $H$ as a quotient of $F$. This subgroup of $H$ is still elementary amenable, and one can take $B$ to be the kernel of $F \to H$. This shows that the actual question is: Is every finite group the quotient of a torsionfree elementary amenable group? I believe this is true, but I do not know how to prove it. –  Andreas Thom Nov 10 '11 at 16:22
add comment

1 Answer

up vote 10 down vote accepted

If F is a free group and R is a normal subgroup of F, then I thought it was well-known that F/R' is torsion free (cannot find an explicit reference now, though this is stated just after Lemma 5 of [Farkas, Daniel R. Miscellany on Bieberbach group algebras. Pacific J. Math. 59 (1975), no. 2, 427–435]). Of course if F/R is finite, then F/R' is virtually abelian and certainly elementary amenable.

share|improve this answer
I think this follows immediately from Lemma 5. The point is that if one considers any $g\in F−R$, then the subgroup generated by $G=\langle g,R\rangle < F$ will be a finite-index subgroup of $F$ containing $R$, such that $G/R$ is cyclic. Then one applies Lemma 5 to see that $G/R′$ is torsion-free. Thus, $F/R′$ itself is torsion-free. –  Ian Agol Nov 11 '11 at 7:05
thanks, that's a great answer! Ideally I really would like a stronger statement: Let $N$ be a (say) hyperbolic 3-manifold, $A\subset \pi:=\pi_1(N)$ a finite index subgroup, does there exist subgroup $B\subset \pi$ which is contained in $A$ and such that $\pi/A$ is once again torsion-free and (elementary) amenable. The proof for free groups implies the result for fibered 3-manifold groups. A positive answer would imply the following: if $K$ and $J$ are hyperbolic knots and if there exists an epimorphism $\pi_1(S^3\setminus K)\to \pi_1(S^3\setminus J)$, then genus(K) $\geq$ genus(J). –  Stefan Friedl Nov 11 '11 at 15:23
add comment

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.