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Assuming $p$ divides $n$, let $P$ be a Sylow $p$-subgroup of $S_n$ and let $z=(1,2,...,p)$. Why is $z$ in the center of $P$? Thanks!

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    It's not.$ $ $ $2012-12-05

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Sylow $p$-subgroups of symmetric groups are iterated wreath products. As long as $n

In fact, the centre of this group is easily seen to be of order $p$, generated by the product of the $p$ disjoint cycles $(ip+1,\ldots,ip+p-1)$, $i=0,\ldots,p-1$. Since conjugation preserves the cycle structure, and since all Sylow subgroups are conjugate, the centre of any other $p$-Sylow is also generated by a product of $p$ disjoint $p$-cycles.

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    @suergin: In fact, if $G$ is a permutation group of order $p^k$ built on $n$ alphabets such that $n$G$ is a $p$- primary abelian group. This is what you see through the first lines of the answer. +1 for it. – 2012-12-05
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    I think your element $g$ is not correctly specified (even after changing $2p+2$ to $2p+1$). It should be a product of $p$ disjoint $p$-cycles, each cycling the elements of a congruence class modulo $p$ of values, rather than just one such $p$-cycle (in other words it would be $x\mapsto (x+p)\bmod p^2$ if one would start numbering from $0$).2012-12-05
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    @Marc You are right, thank you!2012-12-06