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 $S_{ap}$?

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    $\begingroup$ Just by exhibiting one, I think. $\endgroup$
    – YCor
    Dec 29, 2019 at 22:55
  • $\begingroup$ Also, "odd" is not necessary. $\endgroup$
    – verret
    Dec 29, 2019 at 23:02
  • $\begingroup$ You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. $\endgroup$
    – Shaun
    Dec 29, 2019 at 23:03
  • $\begingroup$ The fact that I didn’t include details about what I’ve tried so far doesn’t mean I haven’t put in hours to do it. $\endgroup$ Dec 29, 2019 at 23:06
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    $\begingroup$ @Shaun I have edited the original post with details of what I have thought so far. $\endgroup$ Dec 30, 2019 at 6:59

1 Answer 1


$S_{ap}$ 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 $p$.

  • $\begingroup$ Why is the first sentence in your answer correct? $\endgroup$ Dec 30, 2019 at 17:35
  • $\begingroup$ The explanation is in the second and third sentences... $\endgroup$
    – Thomas
    Dec 31, 2019 at 7:35
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    $\begingroup$ I think I don't understand how the second sentence implies the third. Why "therefore"? $\endgroup$ Dec 31, 2019 at 7:49
  • $\begingroup$ If you have a set X which is the disjoint union of two sets A, B then the group of permutations of X contains the product of the group of permutations of A and that of B. Agree ? $\endgroup$
    – Thomas
    Dec 31, 2019 at 11:31
  • $\begingroup$ This is something I have not heard before. Are you saying $S_{5}$ contains $S_{2} \times S_{3}$? Why is that -- by Cayley's Theorem? $\endgroup$ Dec 31, 2019 at 18:30

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