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Version 4 |
| Let $G$ be a finite group with order $n$, and let $p$ be a prime integer. |
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 $n$). |
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. |
Any subgroup of $G$ whose order is $p^k$ is called a Sylow $p$-subgroup. |
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| 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 |
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 |
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| When $|G|=p^k$ we simply say that $G$ is a $p$-group. |
When $|G|=p^k$ we simply say that $G$ is a $p$-group. |
