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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.

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    $\begingroup$ Not for $n\ge 4$, since it contains $Aut(F_2)^2$, which contains $F_2^2$. $\endgroup$ – YCor Mar 16 at 16:15
  • $\begingroup$ @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 $n=2,3$. $\endgroup$ – Carl-Fredrik Nyberg Brodda Mar 16 at 16:21
  • $\begingroup$ It's fine for me if you edit the question to focus on $n=2,3$. $\endgroup$ – YCor Mar 16 at 16:23
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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.)

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    $\begingroup$ Wikipedia provides a reference for this fact: R. G. Burns and A. M. Brunner, Two remarks on the group property of Howson, Algebra i Logika 18 (1979), 513–522 $\endgroup$ – YCor Mar 16 at 17:58
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    $\begingroup$ 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. $\endgroup$ – Carl-Fredrik Nyberg Brodda Mar 16 at 18:52

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