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.


$\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.

  • $\begingroup$ Maybe I don't understand the notation. Does $\left \langle S \right \rangle$ represent the group generated by elements of $S$? $\endgroup$ – Bary12 Jan 27 '17 at 19:00
  • $\begingroup$ Sorry for taking so long to get back. The answer is indeed yes. $\endgroup$ – Andreas Caranti Jan 28 '17 at 8:52
  • $\begingroup$ 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? $\endgroup$ – Bary12 Jan 28 '17 at 19:27
  • $\begingroup$ The order of $Q_{k}$ is precisely $p^{k}$, so $Q_{k}$ is a Sylow $p$-subgroup. $\endgroup$ – Andreas Caranti Jan 28 '17 at 21:06
  • $\begingroup$ For which $k$? it's a $p$-subgroup but not necessarily a Sylow $p$-subgroup. $\endgroup$ – Bary12 Jan 28 '17 at 21:10

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.

  • $\begingroup$ 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 $n<p^2$ $\endgroup$ – Bary12 Jan 27 '17 at 22:18
  • $\begingroup$ yup, if a non abelian p-subgroup exists it must be contained in a nonabelian sylow p-subgroup, which would make it non-abelian. $\endgroup$ – Jorge Fernández Hidalgo Jan 27 '17 at 22:19
  • $\begingroup$ 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 $n<p^2\Rightarrow P$ is abelian. I don't understand the connection between the subgroup of $S_{n^2}$ and $P$. $\endgroup$ – Bary12 Jan 27 '17 at 22:26
  • $\begingroup$ I meant that if $n\geq p^2$ then the $p$-sylow subgroups are not abelian. $\endgroup$ – Jorge Fernández Hidalgo Jan 27 '17 at 22:30
  • $\begingroup$ 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? $\endgroup$ – Bary12 Jan 27 '17 at 23:29

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