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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 $2$.
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    $\begingroup$ Are you assuming free-by-cyclic means {finitely generated free}-by-cyclic? I know this is usual, but a preprint of Kielak and Linton dropped last week which is very relevant to the more general setting. $\endgroup$
    – ADL
    Feb 28, 2023 at 16:58
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    $\begingroup$ @ADL no, the free kernel may be infinitely generated (eg. I include surface groups in the free-by-cyclic category). And thanks for the reference! In fact, the question was motivated by this very same paper, I was trying to understand if the class of (torsion-free-virtually-free-by-cyclic) is actually strictly bigger than simply all free-by-cyclic $\endgroup$
    – HASouza
    Feb 28, 2023 at 17:03

2 Answers 2

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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.
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  • $\begingroup$ There may be some words missing from the final bullet point. Which finite index subgroup is shown to be free-by-cyclic? (Presumably $\phi^{-1}\langle d\rangle$.) $\endgroup$
    – Mark Grant
    Mar 1, 2023 at 12:38
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    $\begingroup$ @MarkGrant Yes, $\phi^{-1}(d)$ works. The point is that $G$ is shown to be free-by-$D_{\infty}$, so the pre-image of any $\mathbb{Z}$ subgroup of $D_{\infty}$ will work. $\endgroup$
    – ADL
    Mar 1, 2023 at 13:23
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    $\begingroup$ OK, understood. Then maybe you meant to write "that the group is virtually free-by-cyclic" in the last bullet point. $\endgroup$
    – Mark Grant
    Mar 1, 2023 at 13:33
  • $\begingroup$ @MarkGrant Thanks for pointing this out. I've fixed it now, and also tried to made it all slightly clearer. $\endgroup$
    – ADL
    Mar 1, 2023 at 13:50
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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.

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    $\begingroup$ On reflection, the paper of Kielak and Linton mentioned in comments makes it possible to write down many more explicit examples. If $G=\langle a,b\mid w\rangle$ is a two-generator, one-relator group with abelianisation $\mathbb{Z}$, and if the Alexander polynomial of $w$ (which is easy to write down explicitly) is 1 then, for the same reason as the knot case, $[G,G]$ is perfect and $G$ can't be free-by-cyclic. On the other hand, if $w$ is $C'(1/6)$ small cancellation then $G$ is virtually free-by-cyclic by Kielak--Linton. These are both explicit criteria that are easy to fulfill. $\endgroup$
    – HJRW
    Mar 1, 2023 at 9:39
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    $\begingroup$ Maybe a stupid question, but why does the Alexander polynomial of $G = \langle a, b \mid w \rangle$ being $1$ imply that $[G, G]$ is perfect? (I don't quite know the reason in the knot case). $\endgroup$ Mar 1, 2023 at 14:06
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    $\begingroup$ Nevermind, sorry, it's easy, at least in the case of knots- I'll write it out for my own sanity. Since the abelianisation of $G$ is $\mathbb{Z}$ there is an infinite cyclic cover whose fundamental group is $[G, G]$. The first homology of this space, the Alexander module, is a module over $\mathbb{Z}[t, t^{-1}]$. Since the Alexander polynomial is $1$, the Alexander module (which is $\mathbb{Z}[t, t^{-1}]$ modulo the principal ideal generated by the polynomial) is $1$. So the first homology of $[G, G]$ is trivial i.e. $[G, G]$ is perfect. I guess this can now be translated to one-relator groups? $\endgroup$ Mar 1, 2023 at 14:59
  • $\begingroup$ @Carl-FredrikNybergBrodda: Yes, exactly. Furthemore, at least in the 2-generator 1-relator case, the Alexander polynomial can be read off the relator in an easy way (that I'm happy to explain if anyone is interested). $\endgroup$
    – HJRW
    Mar 1, 2023 at 15:06

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