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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, a $p$-Sylow is generated by disjoint $p$-cycles, and in particular is abelian. But for $n=p^2$, a $p$-Sylow is generated by $(1,2,\ldots,p)$, $(p+1,p+2,\ldots,2p),\ldots,(p^2-p+1,\ldots,p^2)$, and $g=(1,p+1,2p+2,\ldots,p^2-p+1)(2,p+2,\ldots,p^2-2+2)\cdots(p,2p,\ldots,p^2),$ and is isomorphic to $C_p\wr C_p=(\overbrace{C_p\times\ldots\times C_p}^{p\text{ times}})\rtimes C_p$. In particular, the cycle $(1,2,\ldots,p)$ is not central, but is conjugated to the next generator $(p+1,p+2,\ldots,2p)$ under $g$.

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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    @Marc You are right, thank you!2012-12-06