Can a Sylow Subgroup be Contained in a Sylow p-Subgroup?

[shakeydakey]

Homework Statement

If J is a subgroup of G whose order is a power of a prime p, verify that J must be contained in a Sylow p-subgroup of G. The problem says to refer to a lemma that given an action by a subgroup H on its own left cosets, h(xH)=hxH, H is a normal subgroup iff every orbit of the induced action of H on a coset contains just one coset. In the problem it says to take H in the lemma to be a Sylow p-subgroup and consider the induced action of J on X.

Homework Equations

The Attempt at a Solution

I'm really not sure how to do this. I've been trying to start with the case that there's only one Sylow p-subgroup (H). Then I know that it's normal, so I could ideally use the lemma. I don't know how though, and I'm totally baffled by the case where there's more than one Sylow p-subgroup. A fellow student was trying to use the theorem that any two Sylow p-subgroups are conjugate, to simplify the case where there's more than one. But to me it seems like you'd need the converse of that theorem to get anywhere with it.

 

[konthelion]

If p is a prime where the order of G is 2p, then G has a normal subgroup, J, with order p (if p is not equal to 2).

Since J is normal, then it is contained in the intersection of all Sylow p-subgroups in G (I thought this was by definition... I don't know). Thus, it must be contained in at least one Sylow p-group of G.

Perhaps someone else can help you with this. Cheers.