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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
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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
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@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
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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
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1 Answer

up vote 18 down vote accepted

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
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