 Solution

The action on the first level defines a surjective homomorphismthat sends $a$ to the displayed cycle and $b$ to the identity. Its kernel is therefore the normal closure of $b$, proving that $\{b\}$ normally generates ${\mathrm{Stab}}_G(1)$.

Apply the Reidemeister–Schreier theorem with transversal $\{1,a,a^{-1}\}$. The generators arising from $a$ are trivial, while those arising from $b$ areConsequently these three elements generate ${\mathrm{Stab}}_G(1)$.