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Use the projective-line realization from part (ii). The subgroup $K$ of translations is abelian and normal in the stabilizer subgroup $G$ of infinity. We verify both remaining conditions of Iwasawa's simplicity lemma, instead of concluding simplicity from transitivity alone.

Let $R$ be generated by all conjugates of $K$ in $G^{+}$. It contains the matricesHere matrices act by fractional linear transformations. For $t\ne 0$,In particular $w(1)=x\in R$, and $w(t)w(1)=\mathrm{diag}(t,t^{-1})$ acts as multiplication by $t^2$. Squaring is a bijection of $F^{\ast }$, so every $M_s$ belongs to $R$. Hence $R$ contains $G$ and $x$, and $\boxed{R=G^{+}}$.

Choose $s\ne 1$. The commutator, with convention $[u,v]=uv{u}^{-1}{v}^{-1}$, isAs $t$ varies, this gives all translations. Thus $(G^{+}{)}^{\prime }$ contains $K$, and normality makes it contain every conjugate of $K$. Since those generate $G^{+}$, the group is a perfect group.

A sharply three-transitive action on nine points is a primitive group action, and this permutation action is a faithful group action. All Iwasawa hypotheses now hold, soIt may be identified with ${\mathrm{PSL}}_2(8)={\mathrm{PGL}}_2(8)$: the generators are fractional linear transformations and $|{\mathrm{PGL}}_2(8)|=8(8^2-1)=504$, while all nonzero field elements are squares. The simplicity proof above does not rely on assuming simplicity of that named family.