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

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.