I was given the following problem:
Let $p$ be a prime, and $P$ be a Sylow p-subgroup of $S_n$. Prove that $P$ is abelian if and only if $n<p^2$
and I was also given the following hint:
find a subgroup of $S_{n^2}$ of order $p^p$, and show that no other element of $S_{p^2}$ commutes with all of it's elements
I do not understand how to tackle this problem, neither how the existence of such subgroup as mentioned in the hint helps.
On
To prove that $n\geq p^2$ then the $p$-sylow subgroup is not abelian it suffices to find a non-abelain $p$-subgroup of $s_{p^2}$. (this is because the $p$-sylow subgroups are the maximimal $p$-subgroups, so if a non-abelian $p$-subgroup exists it must be contained in a $p$-sylow subgroup).
An example of such a subgroup is the subgroup of permutations that permute the elements inside the subsets $\{1,2,\dots,p\},\{p+1,p+2,\dots,p+2\},\dots,((p-1)p+1,\dots,p^2\}$ cyclically and also permute the subsets among each other cyclically.
To see why it is not abelian notice that doing one outer rotation and then one inner rotation is not the same as doing both these things in different order.
To see why it is true when $n<p^2$ look at the construction for the $p$-sylow subgroup given in the other answer.
$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$I will give you a slightly different hint.
If $n < p^{2}$, then $\Size{P} = p^{k}$, where $$k = \left\lfloor \dfrac{n}{p} \right\rfloor \le \dfrac{n}{p} < p.$$ Consider the subgroup $$ Q_{k} = \Span{ (12\dots p), (p+1, p+2, \dots 2 p), \dots, ((k-1)p + 1, (k-1) p + 2, \dots, kp)}. $$
If $n \ge p^{2}$, consider the subgroup $Q_{p}$, and then another suitable element.