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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!

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  • $\begingroup$ Isn;t F torsion free? Thus g in F has infinite order if g is not id. $\endgroup$
    – Paul Fabel
    Commented Jan 7, 2022 at 20:14
  • $\begingroup$ Ahhh, suppose $\psi_g^k(1) = 1$, i.e. that $g\Phi(g)\cdots\Phi^{k-1}(g) = 1$. Then $g = \psi_g^{k+1}(1) = g\Phi(g)\cdots\Phi^{k-1}(g)\Phi^k(g) = \Phi^k(g)$, 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 $\Phi$. I think this latter statement is true for automorphisms of free groups, but the actual argument eludes me ... $\endgroup$
    – Robbie Lyman
    Commented Jan 7, 2022 at 20:38
  • $\begingroup$ Do you know whether the analogous thing works for $\mathbb{Z}^n$ and $GL_n(\mathbb{Z})$? $\endgroup$
    – Matt Zaremsky
    Commented Jan 7, 2022 at 21:33
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    $\begingroup$ @MattZaremsky I believe so, although I don't off the top of my head know an argument showing that elements of $GL_n(\mathbb{Z})$ have periodic points of uniformly bounded period $\endgroup$
    – Robbie Lyman
    Commented Jan 7, 2022 at 21:47

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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.

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    $\begingroup$ I think "latter statement" should be "former statement" (?) And...wait, I'm kind of turned around, but... this fully answers the original question, right? Really you just needed $P(\Phi)$ to be finitely generated. (Also, I guess this means the result is also true for $GL_n(\mathbb{Z})$, just since every subgroup of $\mathbb{Z}^n$ is finitely generated.) $\endgroup$
    – Matt Zaremsky
    Commented Jan 9, 2022 at 12:54
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    $\begingroup$ I forgot to connect the dots: if $P(\Phi)$ is finitely generated, the restriction of $\Phi$ to $P(\Phi)$ is a periodic automorphism, and there is a bound to the period of $\Phi|_{P(\Phi)}$ depending on the rank of $P(\Phi)$. But yes, I think this answers the question completely. $\endgroup$
    – Robbie Lyman
    Commented Jan 9, 2022 at 20:39
  • $\begingroup$ To connect the dots still further just to make sure I understand it, your argument is that $P(\Phi)$ IS (not just is contained in) the fixed subgroup of $\Phi^N$, right? $\endgroup$
    – HJRW
    Commented Jan 10, 2022 at 11:59
  • $\begingroup$ Or are you also invoking the fact that, in a free group, every ascending sequence of subgroups of bounded rank terminates? $\endgroup$
    – HJRW
    Commented Jan 10, 2022 at 16:17
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    $\begingroup$ I think I’m just reiterating the proof that the rank of $P(\Phi)$ is indeed bounded, which I guess follows from the Scott conjecture combined with the fact in my second comment. :) $\endgroup$
    – HJRW
    Commented Jan 10, 2022 at 18:14
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Edit: This is a mostly useless nonanswer and best ignored. It does NOT answer the question. It is NOT the case that $g,g^2,..$ are necessarily the iterates of $id$.

$F$ is torsion-free, since if the first and last letters of the irreducible $w$ have different type, then $w$, $ww$, $www,\dots$ admits no cancellation.

If the first and last letters of $w$ have the same type, conjugate and strictly shorten, until the previous case happens, we don't get $\mathrm{id}$ since conjugation is an automorphism of $F$.

Edit: The word:since, replaced by the word: IF, in the following sentence:

Thus, IF the iterates of $\mathrm{id}$ are $g$, $gg$, $ggg,\dots$ under the dynamical system in question, we must have $g=\mathrm{id}$ if the above sequence is periodic.

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    $\begingroup$ The iterates of $id$ are $g$, $g\Phi(g)$, $g\Phi(g)\Phi^2(g)$ and so on, not as you've written! $\endgroup$
    – Robbie Lyman
    Commented Jan 7, 2022 at 21:45

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