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Let $P$ be a group of order p, on $S_p$ , How can I prove that the cardinality of normalizer of $P$ it's $p(p-1)$ ?

If I compute that the number of conjugates of the group P, it's $ \frac{{n!}} {{p\left( {p - 1} \right)}} $ then I'm done, since equals to the index of the normalizer. But I don't know how.

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up vote 2 down vote accepted

It's probably easier to calculate the normalizer directly. Since all subgroups of order $p$ are conjugate in $S_p$ (they are all Sylow $p$-subgroups), we can take the subgroup $H$ generated by $(123\ldots p)$. If we conjugate that generator by $\sigma \in S_p$, we get $(\sigma(1) \sigma(2) \ldots \sigma(p))$. Now how many ways can we choose $\sigma$ so that cycle is still in $H$?

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