Exercise 13.2.2

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}}

This exercise will consider some simple properties of $S_5$.
\be
\item[(a)] Prove that $\langle (1\,2\,3\,4\,5) angle$ is a 5-Sylow subgroup of $S_5$ and more generally is a $5$-Sylow subgroup of any subgroup $G \subset S_5$ containing $\langle (1\,2\,3\,4\,5) angle$.
\item[(b)] Prove that $S_5$ has twenty-four $5$-cycles.
\ee

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}
\be
\item[(a)] As $|S_5| = 5! = 5 \cdot 24$, where $\gcd(5,24) = 1$, any subgroup of $S_5$ with order 5 is a $5$-Sylow of $S_5$, so  $\langle (1\,2\,3\,4\,5) angle$ is a 5-Sylow of $S_5$.

Let $G$ be a subgroup of $S_5$ containing  $\langle (1\,2\,3\,4\,5) angle$. Then $5$ divides $ |G|$ and  $|G|$ divides $ 5! = 5 \cdot 24$, so $|G | = 5d$, where $d\mid 24$, thus $\gcd(5,d) = 1$. Therefore  $\langle (1\,2\,3\,4\,5) angle$ is a 5-Sylow of $G$.
 
 \item[(b)] There are $5!$ arrangements $(a_1,a_2,a_3, a_4,a_5)$, with distinct $a_i$ in $\{1,2,3,4,5\}$. The 5 arrangements $(a_1,a_2,a_3, a_4,a_5), (a_2,a_3, a_4,a_5,a_1), \ldots$ correspond to the same permutation $(a_1\, a_2 \, a_3\, a_4\,a_5)$, so there are $5!/5 = 24$ 5-cycles in $S_5$.
\ee
\end{proof}

Exercise 13.2.2

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Exercise 13.2.2

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