Let $F\_n=F\ast F'$ be a free splitting of the free group $F\_n$ and $\phi\in Aut(F\_n)$. The free factor $F$ is said to be *invariant under $\phi$* if $\phi(F)\subseteq F$.


I recently wondered if this already implies that $\phi(F)=F$, and I found a positive answer:


An exercise in Magnus-Karrass-Solitar's book on combinatorial group theory shows that if $\phi(F)\subseteq F$ then $\phi(F)$ is a free factor of $F$, and since $\phi(F)$ and $F$ both have the same rank they have to be equal. The exercise in Magnus-Karrass-Solitar's book is proved using the existence of Schreier systems.


Can anyone think of an even easier argument why $$\phi(F)\subseteq F\quad\Rightarrow\quad\phi(F)=F$$ that does not involve Schreier systems?



Invariant free factor of a free group
There is a proof attributed to Peter Scott in Lemma 6.0.6 of "The Tits alternative for Out(F\_n) I: Dynamics of exponentially growing automorphisms", MR1765705. The proof uses the Kurosh subgroup theorem, although the proof is then carried out more generally for any finite rank subgroup $F$ of $F\_n$ using the LERF property for $F$.


The gist of the application of the Kurosh subgroup theorem in this context is that if you have an inclusion of free factors $H < H'$ then $rank(H) \le rank(H')$ with equality if and only if $H=H'$. It may be, then, that what you are really asking for is a proof of the Kurosh subgroup theorem that does not use Schreier systems, in which case I would point you to the Wikipedia page on the Kurosh subgroup theorem which outlines and gives references for a proof using Bass-Serre theory.


