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If $p$ is a prime number, then am I right in thinking that there is only one order $p$ subgroup in the symmetric group $\operatorname{Sym}(p)$? My rationalization is as follows, please correct me if i am wrong! Or if there is a better proof, please tell me.

$p$ is a prime means that a subgroup of order p must be cyclic and each element in it apart from the identity of course is of order $p$. So we can only have the group generated by $(12 \dots p)$ that satisfies the criterion.

Thank you.

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    Take $p = 5$. How many "large" cycles? How many in each order $5$ subgroup?2012-01-25

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