Let $F$ be a finite-rank free group, $g$ an element of $F$ and $\Phi\colon F \to F$ an automorphism. Consider the dynamical system $\psi_g\colon F \to F$ defined by $x \mapsto g\Phi(x)$. Say that $g$ is principal of period $k$ for $\Phi$ if the identity of $F$ is periodic with period $k$ under the dynamical system $\psi_g$. For example, if $\Phi$ inverts $g$, then $1 \mapsto g \mapsto g\Phi(g) = 1$, so such $g \ne 1$ are principal of period 2 for $\Phi$. Being principal with period 3, i.e. $g\Phi(g)\Phi^2(g) = 1$, already seems more mysterious.
I'd like to say that there is a uniform bound (depending on $\Phi$ but not $g$) to the period of principal elements of $F$. Any thoughts in this direction are most welcome!
Best Answer
Ahhhh, suppose $\psi_g^k(1) = 1$, i.e. that $g\Phi(g)\cdots\Phi^{k-1}(g) = 1$. Then $g = \psi^{k+1}_g(g) = g\Phi(g)\cdots \Phi^{k-1}(g)\Phi^k(g) = \Phi^k(g)$, so $g$ is $\Phi$-periodic with period dividing $k$. Therefore if there is a uniform bound to the period of $\Phi$-periodic elements, then there is a uniform bound (depending on $\Phi$) for the period of principal elements for $\Phi$.
This latter statement holds for $F$ a finite-rank free group, say of rank $n$. I claim the subgroup $P(\Phi)$ of $F$ comprising all $\Phi$-periodic elements has rank at most $n$. (Since the restriction of $\Phi$ to $P(\Phi)$ is periodic, this will show there is a bound on the period.) Indeed, suppose $x_0,\ldots,x_n$ are $n+1$ elements of a free basis for $P(\Phi)$. There exists $N$ such that each $x_i$ has period dividing $N$, thus they belong to the fixed subgroup of $\Phi^N$, which has rank at most $n$ by Bestvina–Handel's Theorem (the Scott Conjecture), a contradiction.
Indeed, the argument in the first paragraph shows that the result holds for a group $F$ provided that there is a bound on the period of $\Phi$-periodic elements.