For $|G|=pq$ ($p$ & $q$ are distinct primes), $H$ and $K$ are normal subgroups of order $p$ and $q$ resp., then, prove that $G$ is cyclic My Attempt:
Since $H$ and $K$ are groups of prime order, so, $H$ and $K$ are cyclic.
Let $H=\langle a\rangle $ and $K=\langle b\rangle $, for some $a$,$b$ in G. Then $o(a)=p$, $o(b)=q$
Now we consider the element $ab$ in $G$:
$$o(ab)=(1/p)/(q/pq)$$
If $o(ab)=1$, then, $ab=e \Rightarrow a=b^{-1} or\space b=a^{-1} \Rightarrow H=K \Rightarrow |H|=|K| \Rightarrow p=q$, a contradiction.
If $o(ab)=p$, then, $(ab)^p=e \implies a^pb^p=e \implies b^p=e$, a contradiction.
Similarly, $o(ab)=q \space\space$ gives a contradiction.
Therefore, $o(ab)=pq \implies G$ is cyclic.
But if this proof is correct then what is the use of the normal property?
Thank you for any help or hint.


Lemma 1 : Let $G$ be a group and $M \triangleleft G$, $N < G$. It holds
that $MN < G$

What's interesting is the following :

Lemma 2 : Let $G$ be a group and $M,N$ be normal subgroup such that $M
\cap N = \{e\}$. Then $mn = nm \forall n \in N, m \in M$.

As suggested in the commments this can be showed as follows : consider forall $n \in N, m \in M$ the element $mnm^{-1}n^{-1}$. This element can be seen both as $(mnm^{-1})n^{-1} \in N$ since N is closed by conjugancy being normal and as $m(nm^{-1})n^{-1}) = m((nmn^{-1})^{-1}) \in M$ with the same argument, then $mnm^{-1}n^{-1} \in M \cap N = \{e\}$ which proves what we're looking for.
At this point in you case $H \cap K$ must be trivial since by Lagrange theorem you'd have that $g \in H \cap K$ has $o(g) \mid o(H)$ and $o(g) \mid o(K) \Rightarrow o(g) \mid (p,q) = 1$, hence $g = e$.
At this point by Lemma $1$ $HK = G$ is a subgroup and by Lemma $2$ is abelian. Taking (in your notation) the element $ab$, by abelianity $o(ab) = lcm(o(a),o(b))= o(a)o(b) = pq = o(G)$.
In other words $G$ is cyclic of order $pq$ hence $G \simeq \mathbb{Z}_{pq} \simeq \mathbb{Z}_p \times \mathbb{Z}_q$
