1. Sylow theory

If P is a p-sylow subgroup of G abd B is a subgroup of G of order $\displaystyle p^k$, prove that if B is not contained in some conjugate of P, then the number of conjugates of P in G is a multiple of $\displaystyle p$.

2. Originally Posted by Chandru1
If P is a p-sylow subgroup of G abd B is a subgroup of G of order $\displaystyle p^k$, prove that if B is not contained in some conjugate of P, then the number of conjugates of P in G is a multiple of $\displaystyle p$.

Something's wrong here, or else I misunderstood something: by Sylow Theorems, EVERY p-sbgp. of G is contained in some Sylow p-sbgp. (SPS) of G, and according to the same theorems, if P is a SPS then all the SPS's are conjugate of P, so B MUST be contained in some conjugate of P.
Also, by the same theorem, the number of SPS's is 1 mod p, so it cannot be a multiple of p.

Unless, of course, the above is part of the proof of Sylow's theorem that state every p-sbgp. of G is contained in some SPS, and you already know that the number of SPS's is 1 mod p, so you're going to get a straighforward contradiction...but you didn't explain this, so...

Tonio