Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 104 / 3 / d / Solution

An element $g$ fixing the first level restricts to an automorphism $g_i$ on each rooted subtree, and composition is coordinatewise. Thereforeis a homomorphism. If all three sections are trivial, $g$ fixes every word, so $\varphi$ is injective.

Directly from the recursions,The generators found in the preceding part therefore have every section in $G$, so $\varphi ({\mathrm{Stab}}_G(1))\subseteq G^3$.

Moreover, every coordinate projection of this image contains both $a$ and $b$, and is therefore onto $G$. Since $a$ acts transitively on the first level, induction shows that $G$ acts transitively on every level of the rooted tree. The $n$th level has $3^n$ vertices, so the orders of these finite orbits are unbounded. Hence $G$ is infinite.