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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? –  Yuval Filmus Jan 25 '12 at 14:23

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Other than not checking that cycles of length $p$ are the only order type that have order $p$ (which is fairly easy to check using the primality of $p$), your argument fails exactly when you conclude that the only possible group is the one generated by $(12...p)$.

Indeed, we can start with ANY cycle of length $p$ and create a cyclic subgroup of order $p$. For $p$ larger than 5, you can take $(132...)$ where "..." is any arrangement of the numbers $4 \to p$, and that should generate a different subgroup.

We can in fact count how many subgroups of order $p$ there are. The number of cycles of length $p$ is the number of ways of arranging $2 \to p$ in $(1...)$. Thus, there are $(p-1)!$ different elements of order $p$, and in each subgroup there are $p-1$ elements (it is easy to check that every pair of distinct subgroups of order $p$ share only the identity element), so that gives the number of elements as $\frac{1}{p-1}(p-1)!=(p-2)!$.

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