# Periodic automorphisms of free groups

Hi, I am troubled with the following question: Does there exist a finite order automorphism of a free group, $f\in Aut(F)$, such that it fixes no non trivial conjugacy class and no non trivial centralizer, i.e. $f(g)$ is not conjugate to $g$ and $f(g)\neq g^{-1}$ for any $g\in F$ ? Can we find such, so the same holds for any of its non trivial powers $f^i$?

By inspection of the finite order automorphisms of $F_2$, this cannot be an automorphism of $F_2$.

Moreover, if there exists such an automorphism is it possible to find one with as large order as we want (possibly after moving to a larger free group)?

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What about $x\rightarrow y, y\rightarrow z, z\rightarrow w$ in $F_3 = \langle x, y, z\rangle? – Igor Rivin Apr 25 '12 at 20:46 Sorry, that was supposed to be$z\rightarrow x.$– Igor Rivin Apr 25 '12 at 20:55 Doesn't that give a general proof that such a thing does not exist (take an arbitrary$x,$then write the word$x f(x) f^2(x)...?$– Igor Rivin Apr 25 '12 at 22:12 @Igor: the word you write could be trivial. For the order 3 automorphism of$<x,y>$given by$x \to y$,$y \to y^{-1} x^{-1}$, the word$x f(x) f^2(x)$is trivial. This is an order 3 rotation of an egg beater. – Lee Mosher Apr 25 '12 at 22:19 But upon looking at your (@Lee's) answer, I see that I was not THAT far off... – Igor Rivin Apr 26 '12 at 0:36 ## 1 Answer No such automorphism exists. Every finite order automorphism of a finite rank free group has a nontrivial fixed conjugacy class. To see why, you can represent the free group automorphism as a simplicial automorphism$f : G \to G$of some finite connected graph having no vertices of valence 1. Take any vertex$p \in G$of valence$\ge 3$. Let$\gamma$be an immersed edge path with initial oriented edge$E_0$having initial vertex$p$, and terminal oriented edge$E_1$having terminal vertex$f(p)$, such that the initial direction of$f(E_0)$does not equal the terminal direction of$E_1$. The valence condition on$p$lets you make this choice; if you made a bad choice of$\gamma$then, since the valence of$f(p)$is$\ge 3$, you could concatenate with some immersed path from$f(p)$to$f(p)$so as to change your terminal direction. Letting$k$be the order of the simplicial automorphism$f$, it follows that$f$fixes the conjugacy class corresponding to the immersed loop$\gamma * f(\gamma) * ... * f^{k-1}(\gamma)\$.

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@Lee: In the second phrase "finite order" is missing? – Mark Sapir Apr 25 '12 at 21:28
Oops, yes, fixed that. – Lee Mosher Apr 25 '12 at 21:29