Exercise 4.5.45 (Sylow $p$- subgroups of $S_{2p}$)

Question

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

Richard Ganaye · Accepted Answer

\begin{proof} 
Since
$$|S_{2p}| = (2p)! = (2p)(2p-1)\cdots p (p-1) \cdots 2,$$
then
$$|S_{2p}|  = p^2 m,\qquad 	ext{where } p 
mid m.$$
Therefore any Sylow $p$-subgroup of $S_{2p}$ has order $p^2$, thus is abelian, as any group of order $p^2$.

Consider the subgroup
$$P = \langle \sigma, 	auangle,$$
Where
$$\sigma = (1\ 2\ \cdots p),\qquad 	au = (p + 1\ p+2 \cdots 2p).$$
Then $\sigma$ and $	au$ commute, since their supports are disjoint. Thus
$$P =\{\sigma^i 	au ^j \mid 0 \leq i \leq p, 0 \leq j \leq p\}$$
has order $p^2$, so $P$ is a Sylow $p$-subgroup of $S_{2p}$. There are many distinct conjugates.

If $Q$ is any Sylow $p$-subgroup of $S_{2p}$, then $Q$ is conjugate to $P$, so
$$Q = \langle \sigma', 	au' angle,$$
where
$$\sigma' = (a_1\ a_2\cdots a_p),\qquad 	au' = (a_{p+1} \ a_{p+2} \cdots a_{2p}),\qquad [\![1, 2p]\!] = \{a_1, a_2, \ldots, a_{2p}\}.$$
\end{proof}
Example (with Sagemath):
\begin{verbatim}
sage: G = SymmetricGroup(10)
sage: P = G.sylow_subgroup(5); P.gens()
[(6,7,8,9,10), (1,2,3,4,5)]
sage: P.order()
25
\end{verbatim}

Exercise 4.5.45 (Sylow $p$- subgroups of $S_{2p}$)

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Exercise 4.5.45 (Sylow $p$- subgroups of $S_{2p}$)

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