I have the following question regarding group actions on trees to which I suspect the answer to be "yes", but it could very well be that extra conditions are required (it is certainly true for free actions on trees):

Let $G$ be a finitely generated group acting by simplicial automorphisms on a minimal simplicial tree $T$ (here, minimal means that $T$ does not have a $G$-invariant proper subgraph, and it implies that $T$ is cocompact). We call $g\in G$ elliptic if it fixes a point in $T$ and hyperbolic if not.

Moreover, let $R\subset T$ be an infinite ray based at a vertex $x_0\in V(T)$. Since $T$ is cocompact and therefore has only finitely many orbits of vertices, when walking along $R$ we eventually find a vertex $y\in V(R)$ that is a $G$-translate of some 'earlier' vertex, i.e. $y=gy_0$ for some $y_0\in [x_0,y)\subset R$ and $g\in G$. However, we do not know whether $g$ is elliptic or hyperbolic.

Question: Does there exist $y\in V(R)$ such that $y=gy_0$ for some $y_0\in [x_0,y)\subset R$ and some hyperbolic $g\in G$?

I ended up considering two cases and I have sorted out one of the two. The other one, however, seems to break up into many other cases again that I cannot yet handle. I decided not to post my partial result in order not to lead people in the (possibly) wrong direction from the very beginning.

  • $\begingroup$ Sorry my mobile cut off my comment. I meant to say if R is a half ray of the axis and $y=g^2x_0$, then $y=gy_0$ with $y_0=gx_0$, so you must exclude that case. $\endgroup$ – Benjamin Steinberg Aug 7 '13 at 14:38
  • $\begingroup$ Can you use the fact that the action is cofinite on the oriented edges of T? $\endgroup$ – staylor Aug 7 '13 at 14:45

The answer is yes. For the proof, the ray $R$ contains three distinct points in order, $y_0,y_1,y_2$, such that $y_1 = g_1 y_0$ and $y_2 = g_2 y_1$ for some elements $g_1,g_2 \in G$. If either of $g_1,g_2$ is hyperbolic, you are done. Suppose that both of $g_1,g_2$ are elliptic. The midpoint $p_1$ of $[y_0,y_1]$ must be fixed by $g_1$, and so $g_1$ takes the segment $[y_0,p_1]$ to the segment $[y_2,p_1]$. Similarly, $g_2$ fixes the midpoint $p_2$ of $[y_1,y_2]$ and takes $[y_1,p_2]$ to $[y_2,p_2]$. Consider $g = g_2 g_1$ which takes $y_0$ to $y_2$, and note by construction that $g[y_0,y_2] \cap [y_0,y_2] = \lbrace y_2 \rbrace$, so $g$ is hyperbolic and $[y_0,y_2]$ is a fundamental domain for the axis of $g$.

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