
psylow subgroup
Let $\displaystyle o(G)=pq$, where $\displaystyle p>q$ are primes.
Since $\displaystyle p$ divides $\displaystyle o(G)$, $\displaystyle G$ contains an element $\displaystyle b$ of order p.
Let $\displaystyle S=<b>$.
Since $\displaystyle S$ is a $\displaystyle p$sylow subgroup of $\displaystyle G$, the number of conjugates is $\displaystyle 1+kp$ for some $\displaystyle k \geqslant 0$.
But $\displaystyle 1+kp=[G:N(S)]$ which divides $\displaystyle o(G)=pq$.
Since $\displaystyle (1+kp,p)=1$, then $\displaystyle 1+kp$ divides $\displaystyle q$.
Since $\displaystyle q<p$, then $\displaystyle k=0 $ and $\displaystyle S$ is normal in $\displaystyle G$.
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.