Exercise 14.3.26

Question

\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}}

Compute the orders of the groups in (14.15).

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}}

$$(14.15)\qquad \F_p^2 = \F_{p^2}  \subset \mathrm{AGL}(1,\F_{p^2}) \subset \mathrm{A\Gamma L}(1,\F_{p^2}) \subset \mathrm{AGL}(2,\F_p) \subset S_{p^2}.$$
\begin{proof}
 \item[$\bullet$] $|\F_p^2| = |\F_{p^2}| = p^2.$
 \item[$\bullet$] Every element of $\mathrm{AGL}(1,\F_{p^2})$ is of the unique form $\gamma_{a,b},\  a \in \F_{p^2}^*, b \in \F_{p^2}$, thus
  $$|\mathrm{AGL}(1,\F_{p^2})| = p^2(p^2-1).$$
   \item[$\bullet$] By Exercise 3(a), $\mathrm{AGL}(1,\F_{p^2})$ has index 2 in  $\mathrm{A\Gamma L}(1,\F_{p^2})$, therefore
   $$| \mathrm{A\Gamma L}(1,\F_{p^2})| = 2p^2(p^2-1).$$
   \item[$\bullet$] Every element of $\mathrm{AGL}(2,\F_{p})$ is of the unique form $\gamma_{A,v},\  A \in \mathrm{GL}(2,\F_p), v \in \F_{p}^2$.
   Moreover, by Exercise 14(a), $|\mathrm{GL}(2,\F_p)| = p(p-1)(p^2-1)$, therefore
   $$|\mathrm{AGL}(2,\F_{p})| = p^3(p-1)(p^2-1).$$
   \item[$\bullet$]  To be complete, $|S_{p^2}| = (p^2)!$.
\end{proof}

Exercise 14.3.26

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Exercise 14.3.26

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