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?

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

• Three answers? Edit your old ones... Apr 1 '14 at 23:26

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.

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.