$Theorem$-If $G$ is a group of order $pq$ where $p$ & $q$ are prime , $p>q$ and $q$ does not divide $p-1$ then there is a normal subgroup $H$ in $G$ which is of order $q$.
$Proof$-By $Sylow's \ first \ theorem$ there exists a subgroup $H$ of order $q$ and it is a sylow-$q$ subgroup, since it is the largest subgroup whose order is of the form $ \ q^{n}$
By $Sylow's \ third \ theorem$ we know that number of such sylow $q$ subgroup is of the form $kq+1$ for some integer $k \geq 0$ and divides $|G|=pq.$ If $kq+1$ divides $pq$ then it either divides $p$ or $q$, since it divides $q$ only when $kq+1=1$ and if it divides $p$ then $kq+1=p \ or \ 1$ if $kq+1=p$ then $k=\frac{p-1}{q}$ but since $q$ doesnot divide $p-1,$ $k=\frac{p-1}{q}$ is not an integer hence $kq+1=1$ and hence number of sylow $q$ subgroups is $1$. Thus $H$ is the only subgroup of order $q$.
By $Sylow's \ second \ theorem$ all sylow $q$ subgroups are conjugate to each other and since $H$ is the only sylow $q$ subgroup. $xHx^{-1}=H \ \forall \ x \in G$. Thus H is a unique normal subgroup of order $q$
I am assuming this proof is correct. Please correct me if there are any mistakes.
My doubt here is that in the theorem we include the condition $q$ does not divide $p-1$ to insure $kq+1\neq p$ and we know that $kq+1\neq q $ for any integer $k\geq 0$. But $p$ and $q$ are not the only divisors of $pq, pq$ is its own divisor too, so don't we need to include the condition $q$ doesnot divide $pq-1$ to insure that $kq+1\neq pq$
Best Answer
$ q $ cannot be a divisor of $ pq - 1 $ since we know it is a divisor of $ pq $. Or to phrase it differently, $$ pq - 1 \equiv -1\ (\mathrm{mod}\ q) $$