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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.$ $ $ $ – Alex B. Dec 5 '12 at 8:51
up vote 3 down vote accepted

Sylow $p$-subgroups of symmetric groups are iterated wreath products. As long as $n<p^2$, 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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@suergin: In fact, if $G$ is a permutation group of order $p^k$ built on $n$ alphabets such that $n<p^2$ then $G$ is a $p$- primary abelian group. This is what you see through the first lines of the answer. +1 for it. – Babak S. Dec 5 '12 at 9:01
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 $0$). – Marc van Leeuwen Dec 5 '12 at 11:00
@Marc You are right, thank you! – Alex B. Dec 6 '12 at 9:34

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