Exercise 14.3.25

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

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ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Prove that $\mathrm{AGL}(1,\F_4) \simeq A_4$ and $\mathrm{A\Gamma L}(1,\F_4) \simeq S_4$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
$\mathrm{AGL}(1,\F_4)$ acts on $\F_4$, via the action defined by $$\gamma_{a,b} \cdot i = \gamma_{a,b}(i) = ai +b,$$ where $a \in \F_4^*, b \in \F_4$.

Since $\gamma_{a,b}$ is bijective, $\mathrm{AGL}(1,\F_4) \subset S(\F_4) \simeq S_4$.

Moreover, $|\mathrm{AGL}(1,\F_4)| = 3 	imes 4 = 12$, and the only subgroup with $12$ elements of $S_4$ is $A_4$. Therefore 
$$\mathrm{AGL}(1,\F_4) \simeq A_4.$$

Similarly, $\mathrm{A\Gamma L}(1,\F_4)$ acts on $\F_4$, via the action defined by 
$$\gamma_{a,\sigma,b} \cdot i = \gamma_{a,\sigma,b}(i) = a \sigma(i) + b,$$
where $a \in \F_4^*, b \in F_4, \sigma \in \Gal(F_4/F_2) = \{e,F\}$, $F$ being the Frobenius isomorphism $i \mapsto i^2$.

Since $\gamma_{a,\sigma,b}$ is bijective, $\mathrm{AGL}(1,\F_4) \subset S(\F_4) \simeq S_4$.

By Exercise 3(a), $\mathrm{AGL}(2,\F_4)$ is a subgroup of $\mathrm{A\Gamma L}(2,\F_4)$ of index 2, therefore $|\mathrm{A\Gamma L}(2,\F_4) | = 24$. This proves that $\mathrm{A\Gamma L}(2,\F_4) = S(\F_4)$, thus
$$\mathrm{AGL}(2,\F_4) \simeq S_4.$$
\end{proof}

Exercise 14.3.25

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Exercise 14.3.25

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