Exercise 14.2.11

Question

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

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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}}

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

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

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

Given a prime $p$, let $C_p \subset S_p$ be the cyclic subgroup generated by the $p$-cycle $(1\,2\cdots p)$. As explained in the text, this gives the wreath product $C_p \wr C_p \subset S_{p^2}$. Prove that $C_p \wr C_p$ is a $p$-Sylow subgroup of $S_{p^2}$.

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 Lemma 14.2.8, we know that $C_p \wr C_p$ is a subgroup of $S_p \wr S_p$, which may be viewed as $S_{p^2}$.

Exercise 6 shows that $|C_p \wr C_p| = |C_p| |C_p|^p = p^{p+1}$.

Moreover $|S_{p^2}| = (p^2)!$, and, if $
u_p(n)$ is the exponent of $p$ in the prime factorization of $n!$, then
$$
u_p(n!) =\left \lfloor \frac{n}{p} ight floor +  \left \lfloor \frac{n}{p^2}ight  floor + \cdots + \left \lfloor \frac{n}{p^k} ight floor + \cdots.$$
(see for instance Ex. 2.6 in Ireland and Rosen)
 
Therefore\begin{align*}

u_p((p^2)!) &=\left \lfloor \frac{p^2}{p} ight floor +  \left \lfloor \frac{p^2 }{p^2}ight  floor + \cdots + \left \lfloor \frac{p^2}{p^k} ight floor + \cdots\
&=\left \lfloor \frac{p^2}{p} ight floor +  \left \lfloor \frac{p^2 }{p^2}ight  floor\
&=p + 1.
\end{align*}
Therefore $p^{p+1}$ is the maximal power of $p$ which divides $(p^2)! = S_{p^2}$, so that $C_p \wr C_p$ is a $p$-Sylow subgroup of $S_{p^2}$.
\end{proof}

Exercise 14.2.11

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Exercise 14.2.11

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