Existence of a subgroup of order $pq$ 
Let $G$ be a finite group and assume $G$ has a single $p$-Sylow subgroup. Let $q\neq p$ be prime for which $q$ divides the order of $G$. Show that there exists a subgroup of $G$ with order $pq$.

Following are some of my attempts at solving this:
if $P<G$ is a $p$-Sylow group then $P\vartriangleleft G$ from Sylow's theorems. 
From Cauchy's theorem we have a $Q<G$ cyclic such that $\mid{Q}\mid = q$
Because $P$ is normal in $G$ I know that $PQ < G$, let's assume $\mid{PQ}\mid = p^{k}q$ for $k\geq 1$
My attempt mainly consisted from this point on taking a group of order $p$ from $PQ$, say $P'$ and showing that $P'Q<PQ$. I know that $\mid{P'Q}\mid = pq$ so that would conclude my proof (if it were the right approach, that is).
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 $QP=PQ$.
Am I even in the right direction?
I am not interested in a full proof but merely in a (good solid) hint.
Many thanks.
 A: You can't prove that there is a subgroup of order $pq$ in general, with just these hypotheses. For example, consider the case when $G$ is the alternating group $A_4$ of order $12,$ with $p=2,q = 3.$ Then $G$ has a single Sylow $2$-subgroup, but $G$ has no subgroup of order $6.$ Similar examples can be constructed for every choice of primes $p$ and $q$ such that $q$ does not divide $p-1.$
A: 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 $Q<G$, 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 $P'Q<PQ$. 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!
A: so first off, what makes you assume that the order of your q-sylow subgroup is of order q? Also you have that your sylow subgroups must have a non-trivial center.  This would be an abelian p-subgroup of P, so it must have a subgroup of order p.  The same goes for Q. Not sure if that leads to a proof, but it should at least be on the right track.  You seem to be on the right track though, looking at product of subgroups.  As for the last comment, thats definitely an important fact but does not constitute a proof, as you have not found subgroups of order p and q.
