= Solution
Adjoin the label $9=\infty$. Inversion on the projective line, with zero and infinity interchanged, is
$$
\boxed{x=(1\ 9)(3\ 6)(5\ 8)(4\ 7).}
$$
The label $2=1\in F$ is fixed. Set $H=G_1=\{M_s:z\mapsto sz\}$, where the subscript one denotes the original point label, namely field zero. Then $x^2=1\in H$ and $xM_sx=M_{s^{-1}}$, so $x$ normalizes $H$.
For $g(z)=sz+t$ outside $H$, we have $t\ne0$. On the projective line,
$$
xgx(z)=\frac{z}{s+tz}
=\frac1t+\frac{s/t^2}{z+s/t}.
$$
The equality uses characteristic two and is valid as an equality of fractional linear transformations, including poles and infinity. Thus
$$
xgx=T_{1/t}M_{s/t^2}\,x\,T_{s/t}\in GxG.
$$
All conditions of the <double-coset criterion for a one-point extension> hold. Therefore \b[$G^+=\langle G,x\rangle$ is a one-point extension], with $|G^+|=9\cdot56=504$. Its <stabilizer subgroup> is sharply two-transitive, so its action on nine points is sharply three-transitive.
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