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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

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.

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$\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.

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    Maybe I don't understand the notation. Does $\left \langle S \right \rangle$ represent the group generated by elements of $S$? – 2017-01-27
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    Sorry for taking so long to get back. The answer is indeed yes. – 2017-01-28
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    I still don't understand how I use these subgroups to show that $P$ is abelian. sure, $Q_k$ are abelian subgroups of a sylow $p$-subgroup. but how do I then conclude it is abelian? – 2017-01-28
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    The order of $Q_{k}$ is precisely $p^{k}$, so $Q_{k}$ is a Sylow $p$-subgroup. – 2017-01-28
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    For which $k$? it's a $p$-subgroup but not necessarily a Sylow $p$-subgroup. – 2017-01-28
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    Are we working with the same definitions here? A sylow $p$-subgroup is a subgroup of order $p^m$ for **maximal** $m$ – 2017-01-28
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    Yes. So when $n < p^{2}$, what is the ***maximal*** $m$ such that $p^{m}$ divides the order of $S_{n}$? – 2017-01-28
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    Note what I have written in my answer. If $n < p^{2}$, then a Sylow $p$-subgroup $P$ has order $p^{k}$, where $k$ is... Then I take $Q_{k}$ for the ***same*** $k$. – 2017-01-28
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    I understand now. I will try that – 2017-01-28
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    why does $k=\floor{\frac{n}{p}}$? – 2017-01-28
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    It is the integer part of $n/p$, the largest integer $s$ such that $s \le n/p$. – 2017-01-28
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    but why is it equal to $k$? if $k$ is defined as the power of the order of $P$, then how did you say that $k = \left \lfloor \frac{n}{p} \right \rfloor$? – 2017-01-28
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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

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    I don't understand the part where "(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)". First of all, what is maximimal? did you mean maximal? and I still don't see why from there it follows that $n – 2017-01-27
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    yup, if a non abelian p-subgroup exists it must be contained in a nonabelian sylow p-subgroup, which would make it non-abelian. – 2017-01-27
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    Oh so you meant that the sylow p-subgroup is also non-abelian. That's what i missed. I'm still having some trouble with the first part: to show that $n$S_{n^2}$ and $P$. – 2017-01-27
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    I meant that if $n\geq p^2$ then the $p$-sylow subgroups are not abelian. – 2017-01-27
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    I think I understand what you mean, but isn't this proving that **some** p-sylow subgroup is non-abelian? or does it immediately follow for all of them because they are conjugate? – 2017-01-27
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    it follows, conjugate subgroups are clearly isomorphic (conjugation is an injective homomorfism). And all the p-sylow subgroups are proven to be conjugate during the proof of the sylow theorems – 2017-01-27
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    and cleary if two groups are isomorphic it is not possible for only one to be abelian. – 2017-01-27