Presentation of a non-abelian group of order $pq$. What is the presentation of the non-abelian group of order $pq$ where $p$ and $q$ are primes and $q\mid(p-1)$?
Thanks in advance.
 A: Since $\mathbb{Z}_p$ is cyclic, $\operatorname{Aut}(\mathbb{Z}_p)$ is abelian of order $p-1$, and thus (since the converse of Lagrange's theorem holds for abelian groups) $\operatorname{Aut}(\mathbb{Z}_p)$ contains an element of order $q$.  Let $\sigma$ be such an automorphism and $\Sigma:\mathbb{Z}_q\rightarrow \operatorname{Aut}(\mathbb{Z}_p)$ map the generator of $\mathbb{Z}_q$ to $\sigma$.  Then $\Sigma$ completely determines an embedding of $\mathbb{Z}_q$ into $\operatorname{Aut}(\mathbb{Z}_p)$ (why?), so we may form $$\mathbb{Z}_p\rtimes_\Sigma \mathbb{Z}_q.$$
The presentation of this group is $$\langle a,b|a^p,b^q,a^b=a^m\rangle,$$
where $m$ is the integer satisfying $\sigma(x)=x^m$ - that is, the problem is reduced to finding a number of multiplicative order $q$ modulo $p$.  This can be somewhat difficult to compute, but one approach is to find a primitive element $\bmod \,\,p$ (call it $u$) then compute $m=u^{(p-1)/q}\mod{p}$.
A: One may be arisen thinking of $$G=\langle a,b|a^p=b^q=1,~~bab^{-1}=a^l\rangle$$ wherein $l^q\equiv1~(\text{mod}~ p),~(l,p)=1$.
