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--- theorem:sylow_s_theorem [2013/08/08 06:14]
joshuawiscons created
+++ theorem:sylow_s_theorem [2013/08/13 11:29] (current)
joshuawiscons
@@ Line 1: / Line 1: @@
-====== Sylow's Theorem ======
+====== Sylow's theorem ======
-===== Statement =====
+$\DeclareMathOperator{\syl}{Syl}$
+**Theorem.** Let $G$ be a finite [[Definition:Group|group]] and $p$ a prime. Write $|G| = np^k$ with $(n,p) = 1$. Then
+  - $G$ acts [[Definition:Transitive Group Action|transitively]] on $\syl_p(G)$ by [[Definition:Conjugation|conjugation]] with $|\syl_p(G)| \equiv 1$ modulo $p$, and
+  - every $P \in \syl_p(G)$ has [[Definition:Order of a Group|order]] $p^k$.
+----
+==== Remarks ====
+  * $(n,p)$ denotes the greatest common divisor of $n$ and $p$.
+  * $\syl_p(G)$ denotes the collection of [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] of $G$.
+----
+==== $\LaTeX$ version ====
+<code>
+%%%%%%%%%%
+% DEPENDENCIES
+% RequiredMacros: \DeclareMathOperator{\syl}{Syl}
+%%%%%%%%%%
+\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}
+</code>
+----
+==== External links ====
+  * [[wp>Sylow_theorems]]

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