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Revision difference : groups of order pq
Version 4 Version 3
We can use Sylow's theorems to examine any group $G$ of order $pq$, where $p$ and $q$ are prime and $p<q$. We can use Sylow's theorems to examine any group $G$ of order $pq$, with $p<q$.
We have $|Syl_q(G)|=1+kq$ for some $k$ and $|Syl_q(G)|\mid p$. But, since $p<q$, $k=0$, and so $|Syl_q(G)|=1$, and there is a unique Sylow $q$-subgroup. Since the Sylow $q$-subgroup is unique, it is normal (indeed, \PMlinkname{characteristic}{CharacteristicSubgroup}) in $G$. We have $|Syl_q(G)|=1+kq$ for some $k$ and $|Syl_q(G)|\mid p$. But, since $p<q$, $k=0$, and so $|Syl_q(G)|=1$, and there is a unique Sylow $q$-subgroup. Since the Sylow $q$-subgroup is unique, it is normal (indeed, \PMlinkname{characteristic}{CharacteristicSubgroup}) in $G$.