Suppose $0<a<p$ where $p$ is an odd prime. Then how can I show that the Sylow $p$-subgroup of the symmetric group $S_{ap}$ is an elementary abelian $p$-group?
EDIT: I showed that the order of Sylow $p$-subgroups are $p^a$. Now if I want these Sylow $p$-subgroups to be elementary abelian, then all of their nonidentity elements should have order $p$. Elements of order $p$ in $S_{ap}$ look like either a $p$-cycle $(1 2 ... p)$ or a product of disjoint $p$-cycles. Now the group generated by $\langle (1 2 ...p) \rangle $ has order $p$. To get a group of order $p^a$, I believe I should take the direct product of $a$-many different subgroups each of which is generated by a single $p$-cycle, and all these generators are different. (Is this correct?..) However, would that even be a subgroup of $S_{ap}$? Or is that group would be isomorphic to a subgroup of $S_{ap}$?
$S_{ap}$ contains a product of $a$ cyclic groups of order $p$. In fact a set of $ap$ elements is a disjoint union of $a$ sets of $p$ elements. Therefore it contains a subgroup isomorphic to $S_p^a$, and as a sub-subgroup a product of product of $a$ cyclic groups of order $p$.