Question about soluble and cyclic groups of order pq I'm trying to solve the following problem:
If $p>q$ are prime, show that a group $g$ of order $pq$ is soluble. If $q$ doesn't divide $p-1$, show that $pq$ is cyclic. Show that two non abelian subgroups of order $pq$ are isomorphic.
Progress:
Let's show that $G$ is soluble. Let $n_p$ be the number of $p$-Sylows of $G$ and $q$ be the number of $q$-Sylows of $G$ By the Sylow theorems, there is only one $p$-Sylow since $n_p\equiv 1 (mod\,p)$ and $n_p | q<p$. Therefore there exists only one $p$-Sylow normal subgroup $P$. Therefore we know that: $\{e\}\unlhd P \unlhd G$. $P/\{e\}\approx P$ is abelian since $P$ is cyclic (because $|P|=p$) and $G/P$ is abelian since $|G/P|=q$.
Now suppose $q$ does not divide $p-1$. We know that $q|(n_q-1)$ and that $n_q|p$, therefore $n_q=1$ or $p$. $n_q$ cannot be $p$ since $q$ doesn't divide $p-1$. Therefore there is only one $Q$ $q$-Sylow subgroup and $G\approx P \times Q \approx \mathbb Z_p\times \mathbb Z_q$. Since $\mathbb Z_{pq}$ is abelian it's the product of all it's Sylow subgroups $P'$ and $Q'$, therefore $\mathbb Z_{pq}=P'\times Q'\approx \mathbb Z_p\times \mathbb Z_q$. It follows that $G$ is cyclic.
It remains to show that two non abelian groups of order $pq$ are isomorphic. That's where I'm stuck.
Strategy: Let $G, H$ with $|G|=|H|=pq$ be two non abelian groups. By the preceding paragraph, it's easy to see that both have a single $p$-Sylow and $p$ $q$-Sylows. I tried to build an isomorphism by sending a generator of each of these subgroups of $G$ into a generator of a subgroup of $H$, but I'm not managing to prove it works.
Can someone help me with this last part and say if my solution for the preceeding parts is ok?
 A: Prove first that a group of order $pq$ is solvable.
Use the Sylow theorems. Let $|G| = pq $
Let $n$ = number of $p$-Sylow subgroups. Then, $n\mid q$ and $n = 1 \pmod p$. Since $p$ and $q$ are primes with $p > q$, we conclude that $n = 1$. 
 Thus, the $p$-Sylow subgroup is normal in $G$. 
This leads to the composition series $\{1\} < \Bbb{Z}_p < G$, and thus $G$ is solvable. 
A: If using semidirect products is ok for you, check that please:
3.1, etc
http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/cauchyapp.pdf
A: Let m be the number of $q$-Sylow subgroups of $G$ 
using the Sylow Theorems you can see  that $m = 1$ or $p$.
But since $q$ does not divide $p-1$, we see that $m = 1$, too.
Soboth the $p$ and $q$-Sylow subgroups are normal in $G$ and we also know that $p$ and $q$ are relatively prime... do these 2 subgroups  intersect? Yes. So your $G = \Bbb{Z}_p \times \Bbb{Z}_q$. 
Using the the Chinese Remainder Theorem (as $\gcd(p,q) = 1$), we conclude that $G = \Bbb{Z}_{pq}$, which is cyclic.
