The action on the first level defines a surjective homomorphism
$$
G\longrightarrow\langle(\mathbf0\,\mathbf1\,\mathbf2)\rangle\cong C_3
$$
that 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 $\operatorname{Stab}_G(1)$.

Apply the <Reidemeister–Schreier theorem> with transversal $\{1,a,a^{-1}\}$. The generators arising from $a$ are trivial, while those arising from $b$ are
$$
b,qquad aba^{-1},qquad a^{-1}ba.
$$
Consequently these three elements generate $\operatorname{Stab}_G(1)$.

Solution

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

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Given a presentation of a group and a Schreier transversal for a subgroup, the Reidemeister–Schreier theorem gives a presentation of the subgroup. Its generators are the Schreier generators $ts\overline{ts}^{-1}$, and its relators are rewritten conjugates of the original relators.

= Schreier lemma
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 Incoming links: Reidemeister–Schreier theorem