Affine linear Frobenius groups over finite fields

By [F2], each $f_{a,b}$ is a bijection $F\to F$, so $G\subseteq \mathrm{Sym}(F)$; $G$ contains ${\mathrm{id}}_F=f_{1,0}$, is closed under composition and under inverses by the formulas of [F2] (with $ac\in F^{\times }$ and $a^{-1}\in F^{\times }$), so $G$ is a subgroup of $\mathrm{Sym}(F)$ by [F3]. The map $F^{\times }\times F\to G$, $(a,b)\mapsto f_{a,b}$, is bijective: it is surjective by the definition of $G$, and if $f_{a,b}=f_{c,d}$ then evaluating at $0$ gives $b=d$ and then evaluating at $1$ gives $a=c$. Hence $\mathrm{\mid }G\mathrm{\mid }=\mathrm{\mid }{F}^{\times }\times F\mathrm{\mid }=(q-1)q$ by [F4]. This is assertion 1.

The maps $\tau :F^{+}\to K$, $\tau (b)=f_{1,b}$, and $\delta :F^{\times }\to H$, $\delta (a)=f_{a,0}$, are bijections; by the composition formula [F2], $f_{1,b}\circ f_{1,c}=f_{1,b+c}$ and $f_{a,0}\circ f_{c,0}=f_{ac,0}$, while $\tau (0)=f_{1,0}={\mathrm{id}}_F=\delta (1)$ is the common identity; so $\tau$ and $\delta$ are group isomorphisms onto $K$ and $H$, and $K\le G$, $H\le G$ are subgroups.

For $g=f_{a,c}\in G$ and $f_{1,b}\in K$ we compute, using [F2] twice, $g{f}_{1,b}{g}^{-1}=f_{a,ab+c}\circ f_{a^{-1},-a^{-1}c}=f_{1,a(-a^{-1}c)+ab+c}=f_{1,ab}$, which lies in $K$; since conjugation by $g$ is a bijection $G\to G$ and $K$ is a subgroup, $K^g=K$ for every $g\in G$, that is $K⊴G$ by [F5].

Also $K\cap H=\{{\mathrm{id}}_F\}$: if $f_{1,b}=f_{a,0}$, evaluating at $0$ gives $b=0$ and then evaluating at $1$ gives $1=a$, so $f_{1,b}=f_{1,0}$. Moreover $G=KH$: for $f_{a,b}\in G$ we have $f_{a,b}=f_{a,0}\circ f_{1,a^{-1}b}$ by [F2], and $f_{a,0}\in H$, $f_{1,a^{-1}b}\in K$. Finally $1<\mathrm{\mid }K\mathrm{\mid }=q$ and $1<\mathrm{\mid }H\mathrm{\mid }=q-1$, because $q>2$. So $G=K⋊H$ is an internal semidirect product by [F5], which completes assertion 2.

We verify that the conjugation action of $H$ on $K\setminus \{\mathrm{id}\}$ is free. Let $h=f_{a,0}\in H$ with $h\mathrm{\ne }{\mathrm{id}}_F$, so $a\mathrm{\ne }1$, since $f_{a,0}(1)=a$ while ${\mathrm{id}}_F(1)=1$; and let $k=f_{1,b}\in K$ with $k\mathrm{\ne }{\mathrm{id}}_F$, so $b\mathrm{\ne }0$. By step 1.3 with $c=0$, $hk{h}^{-1}=f_{1,ab}$, and $ab\mathrm{\ne }b$ because $(a-1)b\mathrm{\ne }0$ by [F1]; hence $hk{h}^{-1}\mathrm{\ne }k$, that is, $h$ fixes no nonidentity element of $K$.

By [F7] applied to the internal semidirect product $G=K⋊H$ of step 1.4 and the free action of step 2.1, the subgroup $H$ is a Frobenius complement of $G$; and by the uniqueness statement of [F7], applied to the normal subgroup $K$ with $KH=G$ and $K\cap H=\{1\}$, the Frobenius kernel of $G$ with respect to $H$ is $K$. This is assertion 3. ∎