# Showing Sylow $p$-subgroup is an elementary abelian $p$-group

Suppose $$0 where $$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 $$S_{ap}$$?

• Just by exhibiting one, I think.
– YCor
Dec 29, 2019 at 22:55
• Also, "odd" is not necessary. Dec 29, 2019 at 23:02
• 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. Dec 29, 2019 at 23:03
• 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. Dec 29, 2019 at 23:06
• @Shaun I have edited the original post with details of what I have thought so far. Dec 30, 2019 at 6:59

$$S_{ap}$$ 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 $$p$$.
• 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? Dec 31, 2019 at 18:30