 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

Put $d=a^{-1}ba$. From the preceding section calculations,Since both elements fix the first level, their commutator is computed coordinatewise, andThe element $[d,x]$ belongs to $K$. The third-coordinate projection of $\varphi ({\mathrm{Stab}}_G(1))$ is onto $G$, so conjugating this element inside the stabilizer shows that $\varphi (K)$ contains $(1,1,x^g)$ for every $g\in G$. Because the conjugates $x^g$ generate $K$, it contains $1\times 1\times K$. Conjugation by $a$ cyclically permutes the coordinates; hence it also contains $K\times 1\times 1$ and $1\times K\times 1$. These coordinate subgroups commute, givingInjectivity of $\varphi$ identifies its inverse image with a subgroup of $K$ isomorphic to $K^3$.