General properties of free-by-cyclic groups

Question

I admit this is a very broad question, but I am looking for general properties of [finitely generated free]-by-[infinite cyclic] groups. More precisely, what are some properties that the groups $\{F_n\rtimes_\phi\mathbb{Z}\ |\ n\geq 2,\ \phi\in Aut(F_n)\}$ have in common?

These are some properties that not all of those groups have (even though some do):

1. They are not all hyperbolic.
2. They are not all $\mathrm{CAT}(0)$ groups.

These are some properties that the groups do have in common:

1. They all have solvable conjugacy problem.
2. They all satisfy a quadratic isoperimetric inequality.

Furthermore for $n=2$, all groups of the form $F_2\rtimes_\phi\mathbb{Z}$ are strongly poly-free. This will not be true in higher rank however, as for $n\geq 3$ not all automorphisms of $F_n$ are geometric.

I understand that the structure of $F_n\rtimes_\phi\mathbb{Z}$ highly depends on $\phi$, so fully understanding all common properties probably boils down to fully understanding $Aut(F_n)$.

Answer 1

Well, you can turn any theorem about these groups into a property. For instance: $F_n \rtimes_\phi \mathbb{Z}$ satisfies the property ``hyperbolic if and only if no $\mathbb{Z}\oplus\mathbb{Z}$ subgroups''. That's Brinkmann's theorem.

Answer 2

Feighn and Handel proved that they're all coherent in 'Mapping tori of free group automorphisms are coherent', Ann. of Math. (2) 149 (1999), no. 3, 1061--1077.

Answer 3

They all have cohomological dimension $2$.

The upper bound follows easily from the Serre spectral sequence, the lower bound follows from the fact that groups of cohomological dimension $1$ are free groups. This last fact can be proved without using this big theorem by computing $H^*(G,\mathbf ZG)$ : it is concentrated in dimension 2 and is a free abelian group ... in particular your groups are duality groups (but rarely Poincaré duality groups).

Edit : sorry, I answered too quickly ... the group $H^2(G,\mathbf ZG)$ does not need to be a free abelian group. Thus your groups need not be duality groups.

Answer 4

1. They are all residually finite.
2. They are not all subgroup separable/LERF.
3. They do not all have decidable submonoid membership problem.

Residual finiteness is a result which can be found in the (aptly named) [G. Baumslag, "Finitely generated cyclic extensions of free groups are residually finite" (Bull. Amer. Math. Soc., 5, 87-94, 1971)].

The facts on the submonoid membership problem and subgroup separability/LERF follow from the following example: the free-by-cyclic one-relator group $$G = \langle a, t \mid [a, tat^{-1}] = 1 \rangle \cong \langle a, b, t \mid a^t = ab, b^t = b \rangle$$ embeds the right-angled Artin group $A(P_4)$ (see this article), which is known to have undecidable submonoid membership problem. It is shown that $G$ is not subgroup separable/LERF in [R.G. Burns, A. Karrass, and D. Solitar, A note on groups with separable finitely generated subgroups, Bull. Aust. Math. Soc. 36 (1987), 153–160].