Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 3 / 2 / f

Use the field $F={\mathbb{F}}_2[\zeta ]/({\zeta }^3+\zeta +1)$. The defining polynomial has no root in ${\mathbb{F}}_2$, hence is irreducible. Label its elements bySince $F^{\ast }$ has prime order seven, $\zeta$ has order seven. Multiplication by $\zeta$ is exactly $b$. The identities ${\zeta }^3=\zeta +1$, ${\zeta }^6={\zeta }^2+1$ and ${\zeta }^5={\zeta }^4+1$ show that translation by one is exactly $a$.

Conjugating $a$ by powers of $b$ gives the translations $T_{{\zeta }^i}:z\mapsto z+{\zeta }^i$. The translations by $1,\zeta ,{\zeta }^2$ generate all eight translations. ConsequentlyFor distinct $u,v$ and distinct target points $u^{\prime },v^{\prime }$, the unique affine map has $s=(v^{\prime }-u^{\prime })/(v-u)$ and $t=u^{\prime }-su$. Thus the action is sharply two-transitive.

Its regular characteristic subgroup isTranslations act regularly, $K$ is normal, and its order eight makes it the unique Sylow two-subgroup, hence characteristic. This identifies the abstract subgroup and explicit generators in the original permutation notation.

Adjoin the label $9=\infty$. Inversion on the projective line, with zero and infinity interchanged, isThe 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 $x{M}_{s}x=M_{s^{-1}}$, so $x$ normalizes $H$.

For $g(z)=sz+t$ outside $H$, we have $t\ne 0$. On the projective line,The equality uses characteristic two and is valid as an equality of fractional linear transformations, including poles and infinity. ThusAll conditions of the double-coset criterion for a one-point extension hold. Therefore $G^{+}=⟨G,x⟩$ is a one-point extension, with $|G^{+}|=9\cdot 56=504$. Its stabilizer subgroup is sharply two-transitive, so its action on nine points is sharply three-transitive.

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.