Let $p,q$ be odd primes. Prove that a group of order $2 pq$ is solvable. I got a problem as below. 

Let $p,q$ be odd primes.  Prove that a group of order $2 pq$ is solvable. ($p$ and $q$ may or may not be distinct.)

Honestly, I even don't know where to start. Thanks for any help in advance! 
Definition of Solvability(From dummit and Foote)
A group $G$ is solvable if there is a chain of subgroups $$1=G_0\trianglelefteq G_1\trianglelefteq \cdots \trianglelefteq G_s=G$$such that $G_{i+1}/G_i$ is abelian for $i=0,1,\dots, s-1$. 
 A: Lemma: If $p,q,r$ are any three prime numbers then a group of order $pqr$ is not simple.
Proof: We'll split into cases. If $p=q=r$ then $|G|=p^3$ and it is well known a $p$-group can't be simple. (unless its order is $p$) 
If $p=q\ne r$ then $|G|=p^2r$. We'll look for Sylow subgroups. $n_p$ is either $1$ or $r$. If $n_p=1$ then $G$ has a normal $p$-Sylow subgroup and we are done. So let's suppose $n_p=r$. We also know that $r=n_p\equiv 1$(mod $p$) so $r>p$. Now again, if $n_r=1$ then we have nothing left to prove so let's suppose $n_r\ne 1$, which implies $n_r\in\{p,p^2\}$. But if $n_r=p$ then we'll get $p\equiv 1$(mod $r$) and hence $p>r$ which is a contradiction. Then the only option left is $n_r=p^2$. But each two $r$-Sylow subgroups intersect trivially and hence there are $p^2(r-1)$ non trivial elements in $r$-Sylow subgroups. So there are only enough elements for one $p$-Sylow subgroup which is a contradiction. (because we assumed $n_p=r$)
Now let's suppose that all three primes are distinct. Without loss of generality we'll suppose $p>q>r$. Like before, if there is a prime with one Sylow subgroup then it is a normal subgroup of $G$ and we have nothing left to prove. So we suppose there are no such primes. Because $p$ is the largest prime we conclude $n_p$ cannot be equal to $q$ or $r$. So $n_p=qr$. Also, $q>r$ so $n_q\ne r$ and hence $n_q\geq p$. And finally $n_r\geq q$. And because each two Sylow subgroups intersect trivially we get that together with the identity the number of elements in $G$ is at least:
$qr(p-1)+p(q-1)+q(r-1)+1=pqr+pq-p-q+1=pqr+(p-1)(q-1)\geq pqr+1$
Which is of course a contradiction. 
Conclusion: So now let $G$ be a group of order $2pq$. It is not simple, so it has a subgroup $N\triangleleft G$ which is not $G$ and not $\{e\}$. Hence $N$ either has a prime order or its order is a product of two primes, and the same thing can be said about $G/N$. And any group of such order is solvable. So $N$ and $G/N$ are both solvable which implies $G$ is solvable. 
A: Basic theorem says, there exists the p-Sylowsubgroup P and the q-Sylowsubgroup Q.
Since both of them are cyclic (Groups of prime order are cyclic) they only have the neutral element in common. You should also check that there is only one of each.
The product of P and Q $PQ = \{ab : a \in P, b \in Q\}$ with $|PQ| = pq$ is a subgroup, since P or Q is normal (both are in this case). For further reading look here.
Now we have $PQ \leq G$ and $[G:PQ] = 2$, so it follows that $PQ$ is normal in $G$. (here)
Thus $G/PQ$ is abelian (prime order of 2) and PQ is solvable (because Groups of order pq are solvable, here).
So it follows by definition that G is solvable.
