Exercise 14.3.23

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}}

Use Theorem 14.3.21 to show that $\mathrm{AGL}(2,\F_p)$ is not solvable for $p>3$.

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}
Let $p$ be a prime such that $p>3$. By Theorem 14.3.21, we know that $\mathrm{PSL}(2,\F_p)$ is simple, and non-Abelian, therefore $\mathrm{PSL}(2,\F_p)$ is not solvable.

Since  $$\mathrm{PSL}(2,\F_p) \subset \mathrm{PGL}(2,\F_p), $$
$\mathrm{PGL}(2,\F_p)$ is not solvable.

Moreover, $$\mathrm{PGL}(2,\F_p) = \mathrm{GL}(2,\F_p)/\F_p^* I_n.$$
Since $\F_p^* I_n$ is cyclic, therefore solvable, $\mathrm{GL}(2,\F_p)$ is not solvable.

But $\mathrm{AGL}(2,\F_p)$ contains the subgroup $\{\gamma_{A,0} \mid A \in \mathrm{GL}(2,\F_p)\} \simeq \mathrm{GL}(2,\F_p)$.

This proves that $\mathrm{AGL}(2,\F_p)$ is not solvable if $p>3$.
\end{proof}

Exercise 14.3.23

Answers

Comments

Exercise 14.3.23

Answers

Comments

Add answer