= 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 by
$$
1\leftrightarrow0,\quad2\leftrightarrow1,\quad3\leftrightarrow\zeta,\quad
5\leftrightarrow\zeta^2,\quad4\leftrightarrow\zeta^3,\quad
7\leftrightarrow\zeta^4,\quad8\leftrightarrow\zeta^5,\quad6\leftrightarrow\zeta^6.
$$
Since $F^*$ 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. Consequently
$$
\boxed{G=\{z\mapsto sz+t:s\in F^*,\ t\in F\}\cong\operatorname{AGL}_1(8),\qquad |G|=56.}
$$
For distinct $u,v$ and distinct target points $u',v'$, the unique affine map has $s=(v'-u')/(v-u)$ and $t=u'-su$. Thus the action is sharply two-transitive.
Its regular <characteristic subgroup> is
$$
\boxed{K=\{T_t:t\in F\}\cong C_2^3
=\langle a,bab^{-1},b^2ab^{-2}\rangle.}
$$
Translations 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.
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