
psylow subgroup
Let , where are primes.
Since divides , contains an element of order p.
Let .
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?
(Headbang)(Headbang)(Headbang)

Sylow's theorem explains that. You know that psylow groups are all conjugate, so if there is only one psylow S in G, then it is normal.
The reason is that for all x in G, xSx^(1) is a psylow too (his order is p), so it is S.