if $G$ only has one subgroup of order $p$ and one subgroup of order $q$, then $G$ is a cyclic group I'm having some trouble proving the following:

Let $G$ be a group with $|G|=pq$ where $p$ and $q$ are two distinct prime numbers. Show that if $G$ only has one subgroup of order $p$ and one subgroup of order $q$, then $G$ is a cyclic group.
To prove so use the following:

If $G$ is a group and if $H_1,H_2 \unlhd G$ such that $H_1 \cap H_2=\{1\}$ and $H_1H_2=G$, then:$$G\simeq H_1 \times H_2$$


If $P\leq G$ and $Q\leq G$ are groups such that $|P|=p$ and $|Q|=q$ then $$G\simeq P\times Q$$
proves that $G$ is cyclic. To prove this we just need to show that:

*

*$P,Q \unlhd G$

*$P \cap Q = \{1\}$

*$PQ=G$
But I'm not being able to prove any of these. How can this be done?
 A: Since conjugation by an element of $G$ is an automorphism of $G$, we have $\lvert gPg^{-1}\rvert=\lvert P\rvert$ and $\lvert gQg^{-1}\rvert=\lvert Q\rvert$ for all $g\in G$, so that $gPg^{-1}=P$ and $gQg^{-1}=Q$ for all $g\in G$ by uniqueness of orders. Hence $P,Q\unlhd G$.
It is well-known that $P\cap Q\le P$ and $P\cap Q\le Q$, so, by Lagrange's Theorem, the coprimality of $p$ and $q$ implies that $|P\cap Q|=1$; therefore, $P\cap Q=\{1\}$.
Since $p$ and $q$ are coprime, it is clear that $\lvert PQ\rvert=pq=|G|$; thus $PQ=G$ (as $PQ\subseteq G$).
To show $G$ is cyclic, observe that $P\cong\Bbb Z_p$ and $Q\cong\Bbb Z_q$. Then it is routine to show that
$$\begin{align}
\Bbb Z_{pq}&\cong \Bbb Z_p\times \Bbb Z_q\\
&\cong P\times Q\\
&\cong G
\end{align}$$
by the Chinese remainder theorem.
A: This can be proved without Sylow's theorem.
Let $H$ be subgroup of order $p$ and $K$ be subgroup of order $q$, with $p,q$ distinct primes.
Clearly $H\nsubseteq K$ and $K\nsubseteq H$. Hence $H\cup K$ is proper subset of $G$.
Take $x\in G\setminus (H\cup K)$. What can be order of $x$? It is a divisor of $pq$, and it is not $1$.
If $o(x)=p$, then $x$ should lie in $H$ (unique subgroup of order $p$), contradiction. Similarly $o(x)\neq q$.
Hence $o(x)=pq$, i.e. $G$ is cyclic.
