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\begin{document}% \tiny \scriptsize \footnotesize \small \normalsize \large \Large \LARGE \huge \Huge 
\begin{center}{\Huge Sylow's Theorems}\bigskip\\ Pierre-Yves Gaillard\footnote{ORCID https://orcid.org/0000-0002-7960-1698} \end{center} 

\noi Let $G$ be a group of order $|G|=p^nm$ with $p$ prime and $m$ prime to $p$. A $p$-\emph{Sylow subgroup} of $G$ is, by definition, a subgroup of $G$ of order $p^n$. (For any set $S$ we write $|S|$ for the cardinality of $S$.) 

\begin{thm}\label{S1}
If $G$ is as above, then $G$ has a Sylow $p$-subgroup.
\end{thm} 

\begin{thm}\label{S2}
Let $G$ be as above, let $S$ be a Sylow $p$-subgroup, and let $P$ be a $p$-subgroup of $G$. Then there is a $g$ in $G$ such that $gPg^{-1}\subset S$. In particular, Sylow $p$-subgroup are conjugate. 
\end{thm} 

\begin{thm}\label{S3}
Let $G$ be as above, and let $s$ be the number of Sylow $p$-subgroups of $G$. Then 

\emph{(a)} $s$ divides $m$, 

\emph{(b)} $s\equiv1\bmod p$.
\end{thm} 

\begin{proof}[Proof of Theorems \ref{S1} and \ref{S3}(b)]\footnote{The proof below is taken from the text \emph{Groupes finis} by Jean-Pierre Serre \href{https://arxiv.org/abs/math/0503154}{https://arxiv.org/abs/math/0503154}.} It suffices to prove Theorem~\ref{S3}(b). Let $G$ act on the set $X$ of cardinality $p^n$ subsets of $G$ by $(g,A)\mapsto gA$, and, for $A\in X$ write $X_A$ for the $G$-orbit of $A$. Let $\mc S\subset X$ be the set of Sylow $p$-subgroups of $G$. We claim 

(c) $S,T\in\mc S$ and $S\ne T$ imply $X_S\ne X_T$, 

(d) $S\in\mc S$ implies $|X_S|=m$, 

(e) $A\in X$ and $X_A\cap\mc S=\varnothing$ imply that $p$ divides $|X_A|$. 

\noi Proof that (c), (d) and (e) imply Theorem~\ref{S3}(b): We have $|X|\equiv sm\bmod p$. In particular $s\bmod p$ depends only on the order of $G$. Since $\mathbb Z/|G|\mathbb Z$ is easily seen to have exactly one Sylow $p$-subgroup, we get $s\equiv1\bmod p$, that is, Theorem~\ref{S3}(b), and, \emph{a fortiori}, Theorem~\ref{S1}. 

\noi Proof of (c) and (d): This follows immediately from the fact that, for $S\in\mc S$, we have $X_S=\{gS\ |\ g\in G\}=G/S$. 

\noi Proof of (e): Set $B:=a^{-1}A\in X_A$ with $a\in A$. The stabilizer $H$ of $B$  acting freely on $B$, we have $|H|=p^k$ with $0\le k\le n$, hence $|X_A|=|G|/|H|=p^{n-k}m$, and it suffices to show $k\ne n$. Since $H\subset B$, the equality $k=n$ would imply $B=H\in X_A\cap\mc S$. 
\end{proof} 

\begin{proof}[Proof of Theorem \ref{S2}] 
Let $S$ be a Sylow $p$-subgroup of $G$ and $P$ a $p$-subgroup of $G$. Set $X:=G/S$ and note that $|X|=m$. The cardinality of any $P$ orbit on $X$ is of the form $p^k$ with $0\le k\le n$. It follows from Theorem~\ref{S3}(b) that $P$ has at least one fixed point on $X$. This means that there is a $g$ in $G$ such that $PgS=gS$, that is $(g^{-1}Pg)S=S$, that is $g^{-1}Pg\subset S$. This proves the theorem. 
\end{proof} 

\begin{proof}[Proof of Theorem \ref{S3}(a)] 
Let $G$ act by conjugacy on the set $\mc S$ of all Sylow $p$-subgroups of $G$. This action is transitive by Theorem~\ref{S2}. Let $S$ be in $\mc S$. Since the stabilizer $N$ of $S$ (which is its normalizer) contains $S$, we see that $|S|$ divides $|N|$, hence $s=|G/N|$ divides $|G/S|=m$. 
\end{proof} 

\end{document}