Use Sylow’s theorem. To review Sylow’s theorem, check out the post Sylow’s Theorem (summary). Read the corollary there as well to understand the proof below.

Proof.

Let $n_p, n_q$ be the number of $p$-Sylow subgroups and $q$-Sylow subgroups, respectively.
Then by Sylow’s theorem, we have
\[n_p\equiv 1 \pmod p, \qquad n_p|q \tag{*}\]
and
\[n_q \equiv 1 \pmod q, \qquad n_q|p. \tag{**}\]
Since $q \equiv 1 \pmod p$, we have $q>p$. Thus $n_q$ must be $1$ from (**).
Hence, $G$ has a unique $q$-Sylow subgroup, and it is normal.

From (*), the possibilities for $n_p$ are either $1$ or $q$.
We eliminate the possibility of $n_p=1$ as follows.
If $n_p=1$, then $G$ has a unique $p$-Sylow subgroup, and hence it is normal.
Let $P, Q$ be the unique normal $p$-Sylow subgroup and $q$-Sylow subgroup of $G$, respectively. Then $P, Q$ are normal subgroup of order $p$ and $q$, and hence $P \cap Q=\{e\}$, where $e$ is the identity element of $G$.
Since the order $|PQ|=pq=|G|$, these conditions imply that we have
\[G\cong P \times Q.\]
Since $P$ and $Q$ are groups of prime order, hence it is cyclic, in particular abelian.
The direct product of abelian group is abelian, hence $G$ is abelian.
This is a contradiction.

Group of Order $pq$ Has a Normal Sylow Subgroup and Solvable
Let $p, q$ be prime numbers such that $p>q$.
If a group $G$ has order $pq$, then show the followings.
(a) The group $G$ has a normal Sylow $p$-subgroup.
(b) The group $G$ is solvable.
Definition/Hint
For (a), apply Sylow's theorem. To review Sylow's theorem, […]

If a Sylow Subgroup is Normal in a Normal Subgroup, it is a Normal Subgroup
Let $G$ be a finite group. Suppose that $p$ is a prime number that divides the order of $G$.
Let $N$ be a normal subgroup of $G$ and let $P$ be a $p$-Sylow subgroup of $G$.
Show that if $P$ is normal in $N$, then $P$ is a normal subgroup of $G$.
Hint.
It follows from […]

Sylow Subgroups of a Group of Order 33 is Normal Subgroups
Prove that any $p$-Sylow subgroup of a group $G$ of order $33$ is a normal subgroup of $G$.
Hint.
We use Sylow's theorem. Review the basic terminologies and Sylow's theorem.
Recall that if there is only one $p$-Sylow subgroup $P$ of $G$ for a fixed prime $p$, then $P$ […]

A Group of Order $20$ is Solvable
Prove that a group of order $20$ is solvable.
Hint.
Show that a group of order $20$ has a unique normal $5$-Sylow subgroup by Sylow's theorem.
See the post summary of Sylow’s Theorem to review Sylow's theorem.
Proof.
Let $G$ be a group of order $20$. The […]

Every Group of Order 12 Has a Normal Subgroup of Order 3 or 4
Let $G$ be a group of order $12$. Prove that $G$ has a normal subgroup of order $3$ or $4$.
Hint.
Use Sylow's theorem.
(See Sylow’s Theorem (Summary) for a review of Sylow's theorem.)
Recall that if there is a unique Sylow $p$-subgroup in a group $GH$, then it is […]

Subgroup Containing All $p$-Sylow Subgroups of a Group
Suppose that $G$ is a finite group of order $p^an$, where $p$ is a prime number and $p$ does not divide $n$.
Let $N$ be a normal subgroup of $G$ such that the index $|G: N|$ is relatively prime to $p$.
Then show that $N$ contains all $p$-Sylow subgroups of […]

Every Group of Order 72 is Not a Simple Group
Prove that every finite group of order $72$ is not a simple group.
Definition.
A group $G$ is said to be simple if the only normal subgroups of $G$ are the trivial group $\{e\}$ or $G$ itself.
Hint.
Let $G$ be a group of order $72$.
Use the Sylow's theorem and determine […]

Abelian Group and Direct Product of Its Subgroups
Let $G$ be a finite abelian group of order $mn$, where $m$ and $n$ are relatively prime positive integers.
Then show that there exists unique subgroups $G_1$ of order $m$ and $G_2$ of order $n$ such that $G\cong G_1 \times G_2$.
Hint.
Consider […]

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