Let $r$ be a prime and let $K$ be a finite field of order $2^r$. Let $A$ denote the addtive group of $K$, let $M$ denote the multiplicative group of $K$ and let $H$ denote the Galois group of $K$ over its prime field. Let $F$ be the semidirect product of $M$ and $H$. Then the semidirect product $G := A\cdot F$ acting on the set $\Omega := G / F$ of right cosets is transitive, nonregular and each nontrivial element of $G$ fixes at most two elements.
I tried to make sense of the above statements and proof the claim. But I am unsure how $G$ looks, for a semi-direct product I need some way that $F$ acts on $A$, i.e. a map $\varphi : F \to \mbox{Aut}(A)$, but I do not see in what sense it could be choosen?
Could you help me to figure this out and see the claim?
I tried to construct an example, if $r = 2$ we have $|K| = 4$, then $M \cong C_3$ and $A \cong C_2 \times C_2$, see here. Also the Galois group $H = \mbox{Gal}(K / \mathbb F_2)$ is generated by the map $\varphi_2(t) = t^2$ on $K$, as could be read here, Theorem 4.1, and hence $H \cong C_2$. If $K = \{0,1,2,3\}$, then $M = \{1,2,3\}$ and a generator is $2$ and we have $$ \varphi_2(2) = 3 = 2^{-1} $$ (a little bit odd to write $2\cdot 2 = 3$ as this has not that much to do with ordinary multiplication) and hence $2^{\varphi_2} = 2^{-1}$ in the semidirect product $F$ and we have $F \cong D_6 \cong S_3$ as the action of the involution is like in a dihedral group. But now I do not know how to interpret the semidirect product $A \cdot F$?