Conjugate Subgroups

1. Sep 10, 2011

*FaerieLight*

1. The problem statement, all variables and given/known data
How would I go about proving that if K = gHg-1, for some g $\in$G, where K and H are both subgroups of G with a prime number of elements, then K = H?

2. Relevant 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 realised that H does not necessarily equal gHg-1 unless g is in the normaliser of H. :(

3. 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.

2. Sep 10, 2011

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.