# Howson property of automorphism group of $F_2$ and of $F_3$

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.

• Not for $n\ge 4$, since it contains $Aut(F_2)^2$, which contains $F_2^2$. – YCor Mar 16 at 16:15
• @YCor That’s excellent, thanks! I suspected something like this might happen for larger $n$ (as it does with $\operatorname{SL}_4(\mathbb{Z})$). That just leaves $n=2,3$. – Carl-Fredrik Nyberg Brodda Mar 16 at 16:21
• It's fine for me if you edit the question to focus on $n=2,3$. – YCor Mar 16 at 16:23

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.)
• Excellent, thanks a lot. And even better, this means that (by the embedding mentioned in the question) $\operatorname{Out}(F_3)$ is not Howson either. – Carl-Fredrik Nyberg Brodda Mar 16 at 18:52