= Solution
Use the convention $[g,h]=g^{-1}h^{-1}gh$. In $G/K$ the images of $a$ and $b$ commute and both have order three, so $G/K$ is a quotient of $C_3\times C_3$. Thus
$$
[G:K]\leq9.
$$
Put $d=a^{-1}ba$. From the preceding section calculations,
$$
\phi(d)=(1,b,a),qquad
\phi(x)=\phi([a,b])=(a,b^{-1},a^{-1}b).
$$
Since both elements fix the first level, their commutator is computed coordinatewise, and
$$
\phi([d,x])=(1,1,x).
$$
The element $[d,x]$ belongs to $K$. The third-coordinate projection of $\phi(\operatorname{Stab}_G(1))$ is onto $G$, so conjugating this element inside the stabilizer shows that $\phi(K)$ contains $(1,1,x^g)$ for every $g\in G$. Because the conjugates $x^g$ generate $K$, it contains $1\times1\times K$. Conjugation by $a$ cyclically permutes the coordinates; hence it also contains $K\times1\times1$ and $1\times K\times1$. These coordinate subgroups commute, giving
$$
\boxed{K\times K\times K\ leq\phi(K\cap\operatorname{Stab}_G(1)).}
$$
Injectivity of $\phi$ identifies its inverse image with a subgroup of $K$ isomorphic to $K^3$.
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