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?
Best Answer
If using semidirect products is ok for you, check that please: 3.1, etc http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/cauchyapp.pdf