Theorem. Let $G$ be a finite group and $p$ a prime. Write $|G| = np^k$ with $(n,p) = 1$. Then

1. $G$ acts transitively on $\textrm{Syl}_p(G)$ by conjugation with $|\textrm{Syl}_p(G)| \equiv 1$ modulo $p$, and
2. every $P \in \textrm{Syl}_p(G)$ has order $p^k$.

• $\textrm{Syl}_p(G)$ denotes the collection of Sylow $p$-subgroups of $G$.

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% --RequiredMacros: \DeclareMathOperator{\syl}{Syl} 
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\begin{theorem}[Sylow's Thereom]
Let $G$ be a finite group and $p$ a prime. Write $|G| = np^k$ with $(n,p) = 1$. Then
\begin{enumerate}
\item $G$ acts transitively on $\syl_p(G)$ by conjugation with $|\syl_p(G)| \equiv 1$ modulo $p$, and
\item every $P \in \syl_p(G)$ has order $p^k$.
\end{enumerate}
\end{theorem}

• Sylow's Theorem on Wikipedia