Proving Conjugate Subgroups K=H with Prime Number of Elements

[*FaerieLight*]

Homework Statement

How would I go about proving that if K = gHg^-1, for some g \inG, where K and H are both subgroups of G with a prime number of elements, then K = H?

Homework Equations

I've tried to prove it by saying that if K = gHg^-1 then Kg = gH, and since H = gHg^-1, then Hg = gH also, so Hg = Kg, and hence H = K. I don't think that this proof is valid, unfortunately. And I've just realized that H does not necessarily equal gHg^-1 unless g is in the normaliser of H. :(

The Attempt at a Solution

This is actually something which I am attempting to prove in order to prove something else, so I'm not even sure if what I'm trying to prove holds at all. I just need K = H for my proof to work.

 

[Dick]

Take G=S3, the symmetric group on three elements. Take K={e,(12)} and H={e,(13)}. The subgroups both have order 2 (a prime) and they are conjugate via g=(23), but they aren't equal.