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

Direct multiplication gives $A^3=B^2=-I$, so $⟨A⟩$ has order $6$, $⟨B⟩$ has order $4$, and their intersection is $\{I,-I\}$ of order $2$. For the amalgamated free productthe Bass-Serre tree has verticesand edges $S{L}_2(\mathbb{Z})/\{\pm I\}$. Each edge $g\{\pm I\}$ joins $g⟨A⟩$ to $g⟨B⟩$. Thus it is an infinite bipartite tree in which the $⟨A⟩$-vertices have degree $3$ and the $⟨B⟩$-vertices have degree $2$:with a third branch leaving every ${\circ }_3$ vertex and the same pattern continuing on every branch.

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 $B$.

Consider the principal congruence subgroupIt has finite index because $S{L}_2({\mathbb{F}}_3)$ is a finite group. It is torsion-free: by part (b), a finite-order element is conjugate to $A^i$ or $B^j$, while the reductions of $A$ and $B$ modulo $3$ still have orders $6$ and $4$, respectively. Hence no nonidentity power in either vertex group reduces to the identity.

It follows that $\Gamma (3)$ intersects every conjugate of the two vertex stabilizers trivially, so its action on the Bass-Serre tree is free. A group with a free group action on a tree is a free group. Consequently $S{L}_2(\mathbb{Z})$ contains the free subgroup $\Gamma (3)$ of finite index; equivalently, it is a virtually free group.