= Solution
Consider the <principal congruence subgroup>
$$
\Gamma(3)=\ker\left(SL_2(\mathbb Z)\longrightarrow SL_2(\mathbb F_3)\right).
$$
It has finite index because $SL_2(\mathbb F_3)$ 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 $SL_2(\mathbb Z)$ contains the free subgroup $\Gamma(3)$ of finite index; equivalently, it is a <virtually free group>.
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