Is the intersection of any two finitely generated subgroups of $\operatorname{Aut}(F\_2)$ (resp. $\operatorname{Aut}(F\_3)$) again finitely generated? That is, does $\operatorname{Aut}(F\_2)$ (resp. $\operatorname{Aut}(F\_3)$) have the Howson property? Here $F\_n$ is the free group of rank $n$.


For $n \geq 4$, it follows by @YCor’s comment that $\operatorname{Aut}(F\_n)$ does not have the Howson property.


This question is maybe very simple (apologies if so!), but I have not been able to track down any references for this. I suppose my question extends quite naturally to the same question for the outer automorphism group $\operatorname{Out}(F\_3)$ given the embedding $\operatorname{Aut}(F\_n) \hookrightarrow \operatorname{Out}(F\_{n+1})$. Note that $\operatorname{Out}(F\_2) \cong \operatorname{GL}\_2(\mathbb{Z})$, which is virtually free and hence has the Howson property.



Howson property of automorphism group of $F_2$ and of $F_3$
According to wikipedia (<https://en.wikipedia.org/wiki/Howson_property>), any group of the form $F\_2 \rtimes \mathbb{Z}$ fails to have the Howson property. Assuming we believe wikipedia, then since $Aut(F\_2)$ contains subgroups of this form, it also fails to have the Howson property. (For example, take the subgroup generated by $Inn(F\_2)\cong F\_2$ together with any $\alpha\in Aut(F\_2)$ whose image in $Out(F\_2)\cong GL\_2(\mathbb{Z})$ has infinite order.)


