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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 $B$.

Let $x,y\in \mathrm{Fix}(\varphi )$. A tree has a unique geodesic segment $[x,y]$. The isometry $\varphi$ sends this segment to the segment $[\varphi x,\varphi y]=[x,y]$. An isometry of a segment that fixes both endpoints fixes every point of it, soThus the fixed-point set of an elliptic isometry of a tree is a convex subtree, in particular it is path-connected.

Let $T^{Z(G)}$ be the common fixed subtree of the centre, which is nonempty by hypothesis and is $G$-invariant because $Z(G)$ is central. Restrict the action to this subtree. There $x$ and $y$ act pointwise trivially, so the action factors throughIn particular, the induced tree isometries satisfy

Assume, as usual for a combinatorial tree action, that edge inversions have been removed by barycentric subdivision. The finite-order elements $a,b,c$ are then elliptic. Moreoverso each pairwise product is elliptic. Serre lemma for tree actions implies that the fixed subtrees of each pair intersect. Convex subtrees of a tree have the Helly property, soSince $a,b,c$ generate $G/⟨x,y⟩$, $G$ fixes a vertex of $T^{Z(G)}$. Thus the action of $G$ on $T$ is trivial in the tree-action sense.