# Sylow subgroup 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.

## 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.

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.