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A comment on another question (linked below) states

"The group $PSL_2((\mathbb{Z}/p^n))$ is the automorphisms group of the $(p+1)$ regular tree of depth $n$, where at level $m$ of the tree you have the points of $\mathbb{P}(\mathbb{Z}/p^m)$."

I was unable to find a reference stating this. Is it true, and if so what is the correct reference?

Comment: Presentations of PSL(2, Z/p^n)

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  • $\begingroup$ Isn't this already false for $n=1$ where for $T$ the $(p+1)$-regular tree of depth 1, $Aut(T)$ is $S_{p+1}$ which is not isomorphic to $PSL(2,p)$ unless $p=2$? $\endgroup$
    – ARupinski
    Commented May 13, 2014 at 22:07
  • $\begingroup$ Yes you appear to be correct. I wonder if the person who wrote that was completely off base, or if something along these lines is true... $\endgroup$
    – user94741
    Commented May 13, 2014 at 22:49
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    $\begingroup$ Note that the finite group in the header would usually be interpreted as the points over a finite field of order $p^n$ rather than over the finite ring $\mathbb{Z}/p^n \mathbb{Z}$. A little more care with the notation is needed. $\endgroup$
    – Jim Humphreys
    Commented May 14, 2014 at 15:44

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As already noted in the comments, the group $G=PSL_2(\mathbb{Z}/p^n)$ is not the full automorphism group of the $(p+1)$-regular tree of depth $n$. These particular groups of automorphisms have quite some additional structure.

For example, if one fixes any leaf $x$, then the point-stabilizer $G_x$ contains a normal subgroup $U_x$ which works regular on the leaves which are not in the same branch as $x$ (i.e. the leaves $y$ for which the path fro $x$ to $y$ passes the root of the tree). In this particular case, the groups $U_x$ will be isomorphic to the additive group $\mathbb{Z}/p^n$.

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For much more than you ever wanted, see Serre's Trees, Chapter II.

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