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Let $\,n\in\mathbb{N}\,$ and $p$ a prime. Let $P$ be a Sylow $p$-subgroup of $\,S_n\,$. If $p$ does not divide $n$, then $\,P\leq S_{n-1}\,$. Why?

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Well, how many times does $p$ divide the order of the group? –  Tobias Kildetoft Dec 4 '12 at 18:51
The answer is $p^{[n/p]+[n/p^{2}]+...}$. Here [n/p] denotes the largest integer$\leq$n/p –  suergin Dec 4 '12 at 18:58
Ok, now compare that to the number of times $p$ divides the order of the smaller group. –  Tobias Kildetoft Dec 4 '12 at 19:00
It also divides the order of the smaller group. –  suergin Dec 4 '12 at 19:06
The way you have written it, it is not true. Some of the p-Sylows are subgroups of $S_{n-1}$. –  Phira Dec 4 '12 at 19:10

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