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Let $p$ be a prime number. Show that there is an abelian subgroup $P$ of order $p^p$ in $S_{p^2}$ such that every element in $S_{p^2}$ that isn't in $P$ does not commute with every element in $P$.

I've tries using sylow theorems first theorem to show that there is a p-sylow subgroup $Q$ in $S_{p^2}$ with order $p^{(p+1)}$ and it has a subgroup of order $p^p$ because it is a p-group. I would like a hint on how to show that $P$ is abelian.

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    $\begingroup$ Write down generators in cycle notation. Intersect the centralizers of each generator to get the result. $\endgroup$ – Jack Schmidt Jan 1 '14 at 16:55
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    $\begingroup$ For example, when $p=2$, you could take $P = \langle (1,2),(3,4) \rangle$. $\endgroup$ – Derek Holt Jan 1 '14 at 17:41
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Start with the set $\{1,\dots,p^2\}$ on which $S_{p^2}$ acts, and partition it into $p$ blocks of cardinality $p$ each. Think of the elements of each block as being arranged in a cycle of length $p$. Let $P$ consist of those permutations in $S_{p^2}$ that map each block to itself and preserve the cyclic ordering within each block. That group is abelian because it's isomorphic to the direct product of $p$ copies of the cyclic group of order $p$.

Also, if an element $g\in S_{p^2}$ commutes with all the permutations in $P$, then it is fairly easy to see that $g\in P$. Indeed, if $g$ moved an element of one block into a different block, then it would fail to commute with a permutation in $P$ that fixes the elements of the former block and mover the elements of the latter. And if $g\in S_{p^2}$ moved the elements in some block in a way that doesn't respect the cyclic order there, then it would fail to commute with any element of $P$ that moves the elements of that block.

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