Suppose that $|G| = pq$ where $p$ and $q$ are primes such that $p < q$ and $p$ does not divide $q − 1$. Prove that $G$ is a cyclic group.
A cyclic group is a group that has a unique generator element, so is the way to go with this to find that element? Or am I missing something?
A precedent question shows Prove that G has a normal Sylow p-subgroup that $G$ has a Sylow normal group $N$ of order $p$, Consider $m:G\rightarrow G/N$ it defines an extension $1\rightarrow N\rightarrow G\rightarrow G/N\rightarrow 1$ this extension splits since the theorem of Sylow implies that $G$ has a $q$-subgroup, but it is trivial since splitting extensions of $Z/p$ by $Z/q$ are classified by morphisms $Z/q\rightarrow Aut(Z/p)$ https://en.wikipedia.org/wiki/Group_extension#Classifying_split_extensions but any action of $Z/p$ on $Z/p$ is trivial since $p$ and $q$ are different prime numbers and $p$ does not divide $q-1$. This implies that $G=Z/p\times Z/q$.
In my answer here I showed that any such group must be abelian. Now it is a general fact that an abelian group of order $pq$ where $p \ne q$ are primes, then the group is cyclic. The goal is to produce elements of order $p$ and $q$ and then find the order of the product.
The advantage of this approach is that it can be done without Sylow Theorems.
