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Let $T$ be a regular tree, and suppose that $G \leq \mathrm{Aut}(T)$ has finite quotient graph, $T / G$. Is it true (in general) that $G$ will have trivial centralizer in the full automorphism group? If not, are there more specialized situations when this is true?

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Not sure that I completely understand the question. What if T is an infinite chain? It is regular, and it is a tree. Its automorphism group is $D_{\infty} = \langle a, b\rangle$ - an infinite dihedral group generated by a translation $a$ and a reflection $b$. Then if $G=\langle a^3 \rangle$, then $T/G$ is a triangle, and the centralizer of $G$ in $D_\infty$ is $\langle a \rangle$. Maybe I don't understand the terminology correctly? –  Dan Shved Nov 6 '12 at 17:35
    
Nope, that's a perfectly good counter-example! A result along these lines is used in a paper I'm reading, but it's not backed up, so perhaps it's just that there need to be more restraints in the statement. –  pck7 Nov 6 '12 at 18:08

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