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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.

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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
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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
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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.

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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
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