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theorem:sylow_s_theorem

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Sylow theorems

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 $p^k$.

Remarks

  • $(n,p)$ denotes the greatest common divisor of $n$ and $p$.
  • $\textrm{Syl}_p(G)$ denotes the collection of Sylow $p$-subgroups of $G$.

LaTeX version

%%%%%%%%%%
% DEPENDENCIES
% --RequiredMacros: \DeclareMathOperator{\syl}{Syl} 
%%%%%%%%%%
\begin{theorem}[Sylow thereoms]
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}

theorem/sylow_s_theorem.1375966923.txt.gz · Last modified: 2013/08/08 09:02 by joshuawiscons