Exercise 4.5.46 (Sylow $p$-subgroup of $S_{p^2}$)

Question

Find generators for a Sylow $p$-subgroup of $S_{p^2}$, where $p$ is a prime. Show that this is a non abelian group of order $p^{p+1}$.

Richard Ganaye · Accepted Answer

\begin{proof} 
By the Legendre's formula,
$$
u_p((p^2)!) = \left \lfloor \frac{p^2}{p} ight floor + \left \lfloor \frac{p^2}{p^2} ight floor = p+1,$$
so
$$|S_{p^2}| =(p^2)! =  p^{p+1} m ,\qquad p 
mid m.$$
This shows that any Sylow $p$-subgroup of $S_{p^2}$ has order $p^{p+1}$.

\bigskip
Consider the subgroup
$$P =\langle 	au, \sigma_0, \sigma_1, \cdots \sigma_{p-1} angle,$$
where
\begin{align*}
\sigma_ i &= (1 + ip\ \ 2 + ip\ \    3 + ip \cdots p + ip)\qquad \qquad (0 \leq i \leq p-1),\
	au &= (1 \ \ p+1 \ \ 2p+1 \cdots (p-1)p + 1)(2\ \ p+2 \ \ 2p+2 \cdots (p-1)p + 2)\cdots (p\ \ 2p\ \  3p \cdots  p^2).
\end{align*}
For instance, for $p = 5$,
\begin{align*}
\sigma_0 &= (1\ 2\ 3\ 4\ 5),\
\sigma_1 &=  (6\ 7\ 8\ 9\ 10),\
\sigma_2 &=  (11\ 12\ 13\ 14\ 15),\
\sigma_3 &=  (16\ 17\ 18\ 19\ 20),\
\sigma_4 &= (21\ 22\ 23\ 24\ 25),\
	au &= (1\ 6\ 11\ 16\ 21)(2\ 7\ 12\ 17\ 22)(3\ 8\ 13\ 18\ 23)(4\ 9\ 14\ 19\ 24)(5\ 10\ 15\ 20\ 25).
\end{align*}

Consider the subgroups of $P$:
\begin{align*}
H &= \langle  \sigma_0, \sigma_1, \cdots \sigma_{p-1} angle,\
K &= \langle 	au angle.
\end{align*}
Since $\sigma _i \sigma_j = \sigma_j \sigma_i$ for all $i,j$,  $H$ is abelian, and every element $\sigma\in H$ is of the form
$$\sigma = \sigma_0^{i_0}\sigma_1^{i_1}\cdots \sigma_{p-1}^{i_{p-1}}, \qquad 0 \leq i_0 <p, \cdots, 0 \leq i_{p-1} < p.$$
Consequently, every $\sigma \in H$ preserves the sets $\{1 + ip,  2 + ip,   3 + ip, \cdots, p + ip\}$ for all $i \in [\![0, p-1]\!]$. Moreover
$$|H| = p^p,\qquad |K| = p.$$

Note that 
$$	au \sigma_i 	au^{-1} = \sigma_{i+1}\ (0\leq i <p -1),\qquad 	au \sigma_{p-1} 	au^{-1} = \sigma_{0}.$$
This shows that $H \unlhd P$. Therefore $HK$ is a subgroup of $P$, where $P = \langle H  \cup K angle$, so
\begin{align}
P = HK.
\end{align}
We prove that $H \cap K = \{1\}$. If $\lambda \in H \cap K$, then $\lambda = 	au^{i}$ for some $i \in  [\![0, p-1]\!]$. Moreover $\lambda \in H$, therefore
$$ \lambda(1) = 	au^i (1) \in \{1, 2, 3,\ldots,p-1\} \cap \{1 \ \ p+1 \ \ 2p+1 \cdots (p-1)p + 1\},$$
thus $	au^i (1) = 1$, so $i = 0$ and $\lambda = 	au^i = () = 1$. This proves
\begin{align}
H \cap K = \{1\}.
\end{align}
By (1) and (2), 
$$|P| = |HK| = \frac{|H||K|}{|H \cap K|} = |H||K| = p^{p+1}.$$
This shows that $P$ is a Sylow $p$-subgroup of $S_{p^2}$.

(Since $	au \sigma_1 	au^{-1} = \sigma_2 
e \sigma_1$, $P$ is not abelian.)

\end{proof}

With Sagemath:
\begin{verbatim}
sage: G = SymmetricGroup(25)
sage: P = G.sylow_subgroup(5); P.gens()

[(21,22,23,24,25),
 (16,17,18,19,20),
 (11,12,13,14,15),
 (6,7,8,9,10),
 (1,2,3,4,5),
 (1,6,11,16,21)(2,7,12,17,22)(3,8,13,18,23)(4,9,14,19,24)(5,10,15,20,25)]
\end{verbatim}

Exercise 4.5.46 (Sylow $p$-subgroup of $S_{p^2}$)

Answers

Comments

Exercise 4.5.46 (Sylow $p$-subgroup of $S_{p^2}$)

Answers

Comments

Add answer