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?