# Let $P$ be a Sylow $p$-subgroup of $\operatorname{Sym}(n)$. If $p$ does not divide $n$, then $P\leq\operatorname{Sym}(n-1)$

Let $P$ be a Sylow $p$-subgroup of $\operatorname{Sym}(n)$. If $p$ does not divide $n$, then $P\leq\operatorname{Sym}(n-1)$.

I can compute the order of $P$ and see that there exists a Sylow $p$-subgroup $Q$ of $\operatorname{Sym}(n-1)$ which is order $|P|$. But how can we say that $P$ can be equal to $Q$? Thanks.

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All Sylow $p$-subgroups are conjugate and hence isomorphic. – Derek Holt Dec 16 '12 at 10:38
Did you mean $|P|\leq|\operatorname{Sym}(n-1)|=(n-1)!$ in the title? – Mario Carneiro Dec 16 '12 at 10:38
$\leq$ means subgroup – Ergin Suer Dec 16 '12 at 10:47
@Ergin, have you considered Derek's hint? Do you know what the conjugates of $Sym(n-1)$ (inside $Sym(n)$) look like? – Jyrki Lahtonen Dec 16 '12 at 10:48
There is no natural containment $\,S_n\subset S_{n-1}\,$ ,so what do you exactly mean by a subset of $\,S_n\,$ being also a subset of $\,S_{n-1}\,$? As Mario asked you, perhaps you meant an inequality in the orders? We already had this question some 1.5-2 weeks ago and I don't think any thing clear came out of it, either. – DonAntonio Dec 16 '12 at 10:50

I think the question should be interpreted in this way: Given the hypothesis show that $P$ is contained in a subgroup of $\mathrm{Sym}(n)$ isomorphic to $\mathrm{Sym}(n-1)$. And this is absolutely clear: indeed it's true a little bit more, $P$ is contained in a subgroup of $\mathrm{Sym}(n)$ isomorphic to $\mathrm{Sym}(n-r)$, where $r$ is a nonnegative integer less then $p$, equivalent to $n \pmod p$. Indeed $P$ is a $p$-group acting on a set, so it has a set of fixed point of cardinality equivalent to $n \pmod p$ (orbit stabilizer+n=sum of the cardinality of orbits). Then clearly there are at least $r$ points fixed by the whole $P$. So $P$ has a natural immersion in the stabilizer of these $r$ points, that is isomorphic to $\mathrm{Sym}(n-r)$.