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?
An elementary fact about Sylow subgroup of Sym(n)
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group-theory
symmetric-groups
sylow-theory
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0Well, how many times does $p$ divide the order of the group? – 2012-12-04
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0The answer is $p^{[n/p]+[n/p^{2}]+...}$. Here [n/p] denotes the largest integer$\leq$n/p – 2012-12-04
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0Ok, now compare that to the number of times $p$ divides the order of the smaller group. – 2012-12-04
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0It also divides the order of the smaller group. – 2012-12-04
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1The way you have written it, it is not true. *Some* of the p-Sylows are subgroups of $S_{n-1}$. – 2012-12-04
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0@user I recently got scolded about the addition of the abstract algebra tag when a user tags as group theory. I was directed to a meta post on this topic. So I (who has added abstract algebra to many a questions about groups) was just acting on that reprimand. – 2013-11-10
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0@amWhy Well, then something must be wrong with these tags, since as it is described now any group theory (and not only!) question belong to abstract algebra. – 2013-11-10
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0@user I agree, wholeheartedly. It was Asaf who called me out on it. I think I'll continue adding it (and allow others to add it too), since I don't believe there is community consensus on the matter, and until the tag description changes (if it changes), I'll simply appeal to your logic, though I won't literally "point" to you. I'll just write out the tag description, as did you. – 2013-11-10