I was looking for groups with order $pq$ for primes $p$ and $q$, and found that it wasn't too hard to classify all of them up to isomorphism. In particular, if $p=q$ then we have $C_{p^2}$ or $C_p\times C_p$. Otherwise, w.l.o.g. $p<q$. Then if $p\not\mid q-1$ we have only $C_{pq}$, and finally if $p\mid q-1$ we have $C_{pq}$, and a single non-abelian case, $C_p \ltimes C_q$.
The non-abelian case can be presented as $\langle g,h\vert g^q=h^p=1, gh=hg^x\rangle$ where $x$ is some integer such that $x\not\equiv1\pmod q$ (else the group is abelian) and $x^p\equiv1\pmod q$ (otherwise $p\not\mid\text{ord}(x)=n$, so $h^n$ generates $\langle h\rangle$, but $gh^n=h^ng^{x^n}=h^ng$, contradicting that the group is not abelian).
By the multiplicative group modulo prime $q$ being cyclic, it is clear that there are exactly $p-1$ distinct choices of $x$ up to congruence modulo $q$ provided that $p\mid q-1$; they are the order $p$ elements of $\left({\Bbb Z}\over{q\Bbb Z}\right)^\times$, which form a cyclic subgroup. Moreover, there are obvious automorphisms of the non-abelian group which show that all such choices of $x$ give the same group up to isomorphism.
So I was then wondering if there was a way to construct a particular such integer $x$ given $p$ and $q$ (perhaps using some standard operations for a closed-form expression) which would give a "nice" presentation of the group, i.e. without needing the extra information that $x$ satisfies some congruences which aren't constructive and don't make it very clear that this is always the same group up to isomorphism.
I can only find a solution to this in the obvious special cases. Firstly, the trivial case $p=2$ (and hence $q$ an odd prime), where the only such $x$ is $x\equiv-1\pmod q$. But then we simply have the dihedral group $D_{2q}$ anyway, so this case isn't very interesting or helpful. Secondly, very special cases where we are given that $q$ is a prime factor of ${x^p-1}\over{x-1}$ (for example, $x=2$, $p$ is a mersenne prime exponent, and $q$ is the mersenne prime itself) in which case that particular $x$ is given, so this is also not very helpful for the general case.
For any other case (even simply $p=3$) the next best thing I could think of was just saying, for example, that $x$ is the least positive integer satisfying the condition, which you might write as $\min{\{n\in\Bbb N:n^p\equiv1\pmod q,1<n<q\}}$, but this still doesn't feel very useful or constructive, and certainly would look quite strange as the exponent of $g$.