Exercise 14.4.3

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}}

Let $M_1$ and $M_2$ be the subgroups defined in the text, and assume that $p>3$. Prove that $M_2$ is not doubly transitive and not isomorphic to a subgroup of $M_1$.

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} By Theorem 14.3.4, if $M_2 \subset S_{p^2}$ was doubly transitive, then $$p^2(p^2-1) \mid |M_2| = 2p^2(p-1)^2.$$
Then $p+1 \mid 2(p-1)$, thus $p+1 \mid  2(p+1) - 2(p-1) = 4$, where $p$ is prime. The only solution is $p=3$. We can conclude:
\begin{center}
If $p>3$, then $M_2$ is not doubly transitive.
\end{center}
If $M_2$ was isomorphic to a subgroup of $M_1$, then, by Lagrange's Theorem,
$$
\frac{|M_1|}{|M_2|} = \frac{p+1}{p-1} \in \Z.
$$
In this case, $p-1 \mid p+1$, therefore $p-1 \mid (p+1)- (p-1) = 2$, which is only possible if $p = 2$ or $p=3$. Here $p>3$, so we can conclude:
\begin{center}
If $p>3$, then $M_2$ is not isomorphic to a subgroup of $M_1$.
\end{center}
\end{proof}

Exercise 14.4.3

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Exercise 14.4.3

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