Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 133 / 2 / b

Let $g$ be an isometry of the tree $T$. After subdividing edges if necessary, it has no edge inversion. If $g$ fixes a vertex, it is an elliptic isometry of a tree. Otherwise choose a vertex $v$ minimizing the positive integer $d(v,gv)$. The geodesic segments $g^n[v,gv]$ concatenate without backtracking: any backtracking would produce a vertex with smaller displacement. Their union over $n\in \mathbb{Z}$ is therefore a bi-infinite geodesic, the axis of a tree isometry, and $g$ translates it by $d(v,gv)$. This proves the elliptic-hyperbolic classification of an isometry of a tree.

An element of finite order cannot translate a line through a positive distance, since its powers would have unbounded displacement. Every finite-order element of $S{L}_2(\mathbb{Z})$ therefore fixes a vertex of its Bass-Serre tree. Vertex stabilizers are conjugates of $⟨A⟩$ and $⟨B⟩$, so the element is conjugate to a power of $A$ or a power of $B$.