Sylow's theorem explains that. You know that p-sylow groups are all conjugate, so if there is only one p-sylow S in G, then it is normal.
The reason is that for all x in G, xSx^(-1) is a p-sylow too (his order is p), so it is S.
Let , where are primes.
Since divides , contains an element of order p.
Since is a -sylow subgroup of , the number of conjugates is for some .
But which divides .
Since , then divides .
Since , then and is normal in .
I don't understand why S is normal in G. If it is because [G:N(S)]=1, why?