<?xml version="1.0" encoding="UTF-8"?>

<record version="5" id="5184">
 <title>$p$-subgroup</title>
 <name>PGroup</name>
 <created>2003-10-15 01:29:47</created>
 <modified>2004-12-12 14:57:43</modified>
 <type>Definition</type>
 <creator id="3" name="drini"/>
 <author id="348" name="bbukh"/>
 <author id="3" name="drini"/>
 <author id="3284" name="apmxi"/>
 <classification>
	<category scheme="msc" code="20D20"/>
 </classification>
 <defines>
	<concept>Sylow $p$-subgroup</concept>
	<concept>Sylow p-subgroup</concept>
	<concept>$p$-group</concept>
	<concept>p-group</concept>
 </defines>
 <related>
	<object name="PGroup4"/>
 </related>
 <preamble>\usepackage{graphicx}
%\usepackage{xypic} 
\usepackage{bbm}
\newcommand{\Z}{\mathbbmss{Z}}
\newcommand{\C}{\mathbbmss{C}}
\newcommand{\R}{\mathbbmss{R}}
\newcommand{\Q}{\mathbbmss{Q}}
\newcommand{\mathbb}[1]{\mathbbmss{#1}}
\newcommand{\figura}[1]{\begin{center}\includegraphics{#1}\end{center}}
\newcommand{\figuraex}[2]{\begin{center}\includegraphics[#2]{#1}\end{center}}</preamble>
 <content>Let $G$ be a finite group with order $n$, and let $p$ be a prime integer.
We can write $n=p^k m$ for some $k,m$ integers, such that $k$ and $m$ are coprimes (that is, $p^k$ is  the highest power of $p$ that divides $n$).
Any subgroup of $G$ whose order is $p^k$ is called a Sylow $p$-subgroup.

While there is no reason for Sylow $p$-subgroups to exist for any finite group, the fact is that all groups have Sylow $p$-subgroups for every prime $p$ that divides $|G|$. This statement is the First Sylow theorem
	
When $|G|=p^k$ we simply say that $G$ is a $p$-group.</content>
</record>
