General properties of free-by-cyclic groups

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 $Aut(F_n)$.

• If you are interested in analytic properties: they all have the Haagerup property. – Alain Valette Nov 14 '12 at 22:40
• As this question seems to be asking for a list, perhaps it should be community wiki. – HJRW Nov 15 '12 at 17:15
• I agree with HW, why don't you CW the question? – YCor Nov 15 '12 at 22:08
• Here's a related question (which I'm sure you've seen): mathoverflow.net/questions/53973/… – Peter Samuelson Nov 15 '12 at 22:58

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.
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.