Torsion-free virtually free-by-cyclic groups

Question

Is it known if there are any examples of a finitely generated group $G$ such that:

1. $G$ has a finite index subgroup $H$ which is free-by-cyclic

2. $G$ itself is not free-by-cyclic

3. $G$ is torsion-free

Since subgroups of free-by-cyclic groups are free-by-cyclic, one may strengthen (1) and ask that $H$ is normal in $G$. It is then fairly easy to construct groups that satisfy (1) and (2) by extending $H$ under any finite group. However, I couldn't come up yet with an example satisfying all three conditions. I've already know such a group must satisfy some properties:

• By a combination of Serre's and Stallings-Swan's theorems, such a group must have cohomological dimension 2.
• Since $H/[H,H]$ has a finite index image in $G/[G,G]$, $G$ must have infinite abelianization.
• In particular, $G$ admits homomorphisms onto $\mathbb{Z}$, all of whose kernels must have cohomological dimension exactly $2$. So $G$ must be a semidirect product $K \rtimes \mathbb{Z}$ for some group $K$ of cohomological dimension $2$.

Answer 1

The group $$G=\langle a, b, x, y\mid [a, b]^2=[x, y]^2\rangle$$ is a torsion-free group which is not free by cyclic. However, $G$ is free-by-$D_{\infty}$ and so virtually free-by-cyclic (containing an index-two subgroup which is free-by-cyclic).

This example is from the paper Baumslag, Fine, Miller and Troeger, Virtual properties of cyclically pinched one-relator groups. Int. J. Alg. Comp. (2009).

• Firstly, $G$ is torsion-free as it is a free product with amalgamation of two torsion-free groups.
• Secondly, $G$ is not free-by-cyclic. Both $[a, b]$ and $[x, y]$ are contained in the commutator subgroup. These elements are non-equal but their squares are equal. Hence, the commutator subgroup does not have unique roots, and so it not free. Hence, any map to $\mathbb{Z}$ has non-free kernel.
• Finally, $G$ is free-by-$D_{\infty}$ by Theorem 4 of the above-mentioned paper. The idea of the proof is to map $G$ onto $D_{\infty}=\langle c, d\mid c^2=1, c^{-1}dc=d^{-1}\rangle$ by $\phi(a)=c=\phi(x)$ and $\phi(b)=d=\phi(y)$, and then prove that $\phi$ has free kernel. The index-two subgroup $\phi^{-1}(d)$ of $G$ is therefore free-by-cyclic.

Answer 2

Many more examples, including ones where the free kernel is finitely generated, arise by looking at knot complements.

Let $K$ be any non-trivial knot with Alexander polynomial $\Delta_K(t)=1$, (apparently the (-3,5,7)-pretzel knot is an example), let $M_K$ be the knot complement and let $G_K$ be its fundamental group.

The fact that $\Delta_K(t)=1$ implies that the commutator subgroup $G_K$ is perfect. In particular, $G_K$ can't be free-by-cyclic. All knot groups $G_K$ are torsion-free.

On the other hand, by work of Agol and Przytycki--Wise, all knot complements virtually fibre. This means that $G_K$ has a finite-index subgroup that is (finitely-generated-free)-by-cyclic.