You were pretty much on the right track. Except that you need to notice the following:
To Prove that $PQ$ is a subgroup of $G$:
Given $P\trianglelefteq G$ and $Q, We claim that $PQ=\{pq|p\in P; q \in Q\}$ is a subgroup of $G$.
There are two ways of looking at this: One way of doing is to prove that $PQ$ is a subgroup if and only if $PQ=QP$.(Try Proving this and then show that this is true here!) Another way is to show through direct methods, which I'll show here:
- Note that identity of $G$ exists here in $PQ$.
- If $pq, p_1q_1 \in PQ$, $pq(p_1q_1)^{-1}=pqq_1^{-1}p^{-1}=p\cdot qq_1^{-1}p^{-1}(qq_1^{-1})^{-1}\cdot qq_1^{-1}$ Note that the term sandwiched between the dots is an element in $P$, because, $P$ is a normal subgroup in G. So, this completes the proof, thanks to the subgroup Test.
If $P$ and $Q$ are subgroups of $G$, then the subgroup $PQ$ is of cardinality, $|PQ|=\dfrac{|P|\cdot |Q|}{|P \cap Q|}$
Since $P$ and $Q$ are groups with co-prime orders, they intersect trivially.
(Does this not require a proof? It certainly does! Consider an element $x$ from the intersection $P \cap Q$, Since $P\cap Q$ is both a subgroup of $P$ and $Q$, by Lagrange's Theorem, $o(x)|p$ and $o(x)|q$ This forces that $o(x)=1$ which necessarily means that $x$ is the trivial element.)
So, $|P \cap Q|=1$ and hence $|PQ|=|P|\cdot |Q|=pq$.
This completes the proof!
I would like to point a few glaring errors in your way of proving.
My attempt mainly consisted from this point on taking a group of order $p$ from $PQ$, say P' and showing that P'Q. I know that ∣P'Q∣=pq so that would conclude my proof (if it were the right approach, that is).
How will this prove that $k=1$. Are there not groups of order $p^2q$ with subgroups of order $pq$?
This approach didn't work for me because I can't show that if $h\in Q$ and g\in P' then gh=h'g'\in P'Q. I do know that gh=hg'\in QP because $P$ is also normal in $QP=PQ$.
Where are all these elements (\cdot)' coming from? It's hard to comprehend that! From the sentence that, gh=hg'\in QP, I believe g'=h^{-1}gh.
The rest is fine!