Sylow -subgroup
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Revision 2616
8/29/2026, 2:27:36 PM · Ancient Tree
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Let $p$ be a [[Prime number|prime number]]. Let also $G$ be a [[Finite group|finite group]] of [[Order of a finite group|order]] $|G|=p^{n}m$ with $p\nmid m$. A Sylow $p$-subgroup of $G$ is $p$-[[$p$-subgroup|subgroup]] of maximal order, that is, of order $p^{n}$.1
Let $p$ be a [[Prime number|prime number]]. Let also $G$ be a [[Finite group|finite group]] of [[Order of a finite group|order]] $|G|=p^{n}m$ with $p\nmid m$. A Sylow $p$-subgroup of $G$ is a $p$-[[$p$-subgroup|subgroup]] of maximal order, that is, of order $p^{n}$.Revision 1011
8/6/2026, 9:38:41 PM · Sequoia
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Revision 1010
8/6/2026, 9:38:36 PM · Sequoia
Updated text
Compare with revision 9642 changed lines
1
Let $G$ be a [[Finite group|finite group]] of [[Order of a finite group|order]] $|G|=p^{n}m$ with $p\nmid m$. A Sylow $p$-subgroup of $G$ is $p$-[[$p$-subgroup|subgroup]] of maximal order, that is, of order $p^{n}$.1
Let $p$ be a [[Prime number|prime number]]. Let also $G$ be a [[Finite group|finite group]] of [[Order of a finite group|order]] $|G|=p^{n}m$ with $p\nmid m$. A Sylow $p$-subgroup of $G$ is $p$-[[$p$-subgroup|subgroup]] of maximal order, that is, of order $p^{n}$.Revision 964
8/6/2026, 12:59:18 PM · Ancient Tree
Concept created
Let $G$ be a [[Finite group|finite group]] of [[Order of a finite group|order]] $|G|=p^{n}m$ with $p\nmid m$. A Sylow $p$-subgroup of $G$ is $p$-[[$p$-subgroup|subgroup]] of maximal order, that is, of order $p^{n}$.