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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?

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2 Answers 2

up vote 10 down vote accepted

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.

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Thank you for pointing this out to me! And thank Jimmy Wales for Wikipedia. –  Sebastian Nov 27 '12 at 13:43

The case of finitely generated free groups

Let me reproduce the proof that Lee Mosher pointed out.

Claim:
Let $F=A\ast B$ be a free splitting of a finitely generated free group and $\phi\in Aut(F)$. If $\phi(A)\subseteq A$ then $\phi(A)=A$.

Proof:
If $A\ast B$ is a free splitting of $F$ then so is $\phi(A)\ast \phi(B)$. Therefore, $F=\phi(A)\ast \phi(B)$ is the fundamental group of a graph of groups $X$ that consists of a single edge with trivial edge group and vertex groups $\phi(A)$ and $\phi(B)$. Let $T$ be the Bass-Serre covering tree of $X$. By construction, the group $F$ acts on $T$ with trivial edge stabilizers and with vertex stabilizers conjugate to $\phi(A)$ and $\phi(B)$ respectively. In particular, there exists a vertex $v\in V(T)$ such that $F_v=\phi(A)$, where $F_v$ denotes the point stabilizer of $v$ under the action of $F$.
The subgroup $A$ of $F$ also acts on the tree $T$ and the edge stabilizers under this action are also trivial. Our assumption $\phi(A)\subseteq A$ implies that the vertex stabilizer $A_v=F_v\cap A=\phi(A)\cap A$ is in fact given by $\phi(A)$. Now choose a fundamental domain for the action of $A$ on $T$ accordingly such that the quotient graph of groups $Y$ with underlying topological graph $A\backslash T$ has $\phi(A)$ as one of its vertex groups. Bass-Serre theory tells us that $A$ is the fundamental group of $Y$, and since $Y$ has trivial edge groups it follows that $A$ contains $\phi(A)$ as a free factor.
We haven't yet made use of the fact that we are dealing with free groups ($A$ and $B$ are free, as all subgroups of free groups are free). Since $\phi$ restricted to $A$ is an isomorphism onto its image $\phi(A)$, the two free groups $A$ and $\phi(A)$ have the same finite rank. Therefore, $A$ cannot contain $\phi(A)$ as a proper free factor and hence $\phi(A)=A$.

Does this statement also hold for arbitrary finitely generated groups?

No, not in general.

In 'A Finitely Related Group with An Isomorphic Proper Factor Group', J. London Math. Soc. (1951) s1-26(1): 59-61, Graham Higman gives an example of a finitely presented group $G$ that is isomorphic to a proper free factor of itself. In other words, there exists a free splitting $G=G_1\ast G_2$ and an isomorphism $$\phi\colon G_1\ast G_2\stackrel{\cong}{\longrightarrow} G_1.$$ The free product of $\phi$ with its inverse $\phi^{-1}$ is then an automorphism $$\Phi=\phi\ast\phi^{-1}\colon (G_1\ast G_2)\ast {G_1}'\stackrel{\cong}{\longrightarrow} G_1\ast ({G_1}'\ast G_2)$$ which has the property that $\Phi(G_1\ast G_2)=G_1\subseteq G_1\ast G_2$, but equality does not hold.

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