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The paper The Tits alternative for $\operatorname{Out}(F_n)$ I by Bestvina, Feighn and Handel and the paper Automorphisms of free groups and Outer space by Vogtmann both state that $\operatorname{Aut}(F_n)$ and $\operatorname{Out}(F_n)$ are not linear groups for $n \geq 3$ respectively $n \geq 4$. They both cite The automorphism group of a free group is not linear by Formanek and Procesi. This text proves the claim about the automorphism group of a free group, but does not mention the claim about the outer automorphism group.

I was wondering if some properties about linear groups imply this result, but I couldn't figure out what property would. Any help would be appreciated.

Once again, not really sure if this is the right place to post this question. If not, I'll take it down (or if someone knows how to relocate it to mathstack, that would be great too)

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There is an embedding $\text{Aut}(F_{n-1}) \hookrightarrow \text{Out}(F_n)$ for any $n \ge 2$, as follows.

First one embeds $\text{Aut}(F_{n-1}) \hookrightarrow \text{Aut}(F_n)$ by extending any automorphism of $F_{n-1}$ to an automorphism of $$F_n \approx F_{n-1} * \mathbb{Z} $$ using the identity automorphism of $\mathbb{Z}$.

Next one notices that none of the resulting automorphisms of $F_n$ are inner, hence we get an embedding $\text{Aut}(F_{n-1}) \hookrightarrow \text{Out}(F_n)$.

It follows that if $\text{Out}(F_n)$ is linear then $\text{Aut}(F_{n-1})$ is linear.

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  • $\begingroup$ Thanks for the answer. I can see why each automorphism of $F_{n-1}$ defines an automorphism of $F_n$. In order to see that none of those are inner, is the following reasoning OK?: suppose that for some automorphism $\alpha$ of $F_{n-1}$, we have that $\alpha \ast \operatorname{Id}$ is conjugation by some $g \in F_n$, then we would have that $x_n = (\alpha \ast \operatorname{Id})(x_n) = gx_ng^{-1}$ and hence $g$ must be the identity element. But then looking at the images of the generators of $F_{n-1}$ shows that $\alpha$ must be the identity automorphism. $\endgroup$
    – Student
    May 17, 2018 at 21:13
  • $\begingroup$ Well, $x_n = g x_n g^{-1}$ does not immediately imply that $g$ is the identity element, it only implies that $g_n = x_n^i$ for some $i \in \mathbb{Z}$. $\endgroup$
    – Lee Mosher
    May 17, 2018 at 21:30
  • $\begingroup$ ah, indeed... I'll have to think about this tomorrow, thanks for your quick reply. $\endgroup$
    – Student
    May 17, 2018 at 22:30
  • $\begingroup$ But $alpha$ still maps $x_1, \ldotd, x_{n-1}$ onto some combination of those elements, so there does not change much, right? (Too late to edit my previous comment) $\endgroup$
    – Student
    May 17, 2018 at 22:36
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    $\begingroup$ Having proved that the automorphism $\alpha * \text{Id}$ is equal to the inner automorphism known as "conjugation by $x^i_n$" for some exponent $i \in \mathbb{Z}$, the only way that the group element $$\alpha(x_1) = (\alpha * \text{Id})(x_1) = (x_n^i) x_1 (x_n^i)^{-1}$$ can be equal to a word in the letters $x_1,...,x_{n-1}$ is if $i=0$, in which case it follows that $\alpha(x_1)=x_1$. Similarly for each $x_k$ when $k \in \{1,...,n-1\}$. $\endgroup$
    – Lee Mosher
    May 18, 2018 at 15:15

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