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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 $A \cdot F$?

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    $\begingroup$ $M$ acts on $A$ by multiplication in the field $K$, i.e. for $m \in M$, $a \in A$, $m:a \mapsto ma$. And $H$ acts naturally on $A$, since $H = {\rm Aut}(K)$. Putting these actions together, we get an action of $F$ on $A$. In your example, the resulting semidirect product is just $S_4$. In general, the group $G$ is often denoted by ${\rm A \Gamma L}(r,2)$. $\endgroup$ – Derek Holt Nov 6 '15 at 13:30
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    $\begingroup$ By the way, these groups are defined in Section 2.8 of Dixon & Mortimer's book on Permutation Groups. $\endgroup$ – Derek Holt Nov 6 '15 at 13:48
  • $\begingroup$ Thank you! But guess you mean $H \le \mbox{Aut}(K)$, not equality. $\endgroup$ – StefanH Nov 6 '15 at 18:02
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    $\begingroup$ No I meant equality. The Galois group of any finite field over its prime subfield is equal to the full automorphism group of that field. $\endgroup$ – Derek Holt Nov 6 '15 at 20:17

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