Sylow's theorem and uniqueness of normal supgroup

Question

Let $G$ be a finite group of order $pq,$ where $p$ and $q$ are primes such that $p < q.$ Then how to prove that $G$ has a unique normal subgroup of order $q?$

Answer 1

Since $G$ is a group of order $pq$ and $p|pq\ \&\ p^2 \nmid pq,\ G$ has a $p$-Sylow subgroup of order $p.$ Since the number of $p$-Sylow subgroups is of the form $1+kp,k\geq 0$, and which is a divisor of order of $G,$

We have $1+kp\ |\ pq.$

$\implies 1+kp|q,$ since $(1+kp,p)=1$

$\implies k=0$

Hence the $p$-Sylow subgroup,say $H,$ in the group of order $pq$ is unique.

Since $H$ and $xHx^{-1}$ are subgroups of same order and $H$ is the unique subgroup of order $p,$ we have $H=xHx^{-1},\ \forall x \in G.$

$ \ \ $ i.e., $Hx=xH \ \forall x \in G$. Therefore $H$ is normal.