We have the following classification for finite minimal nonabelian groups by Huppert.
Let $G$ be a finite minimal nonabelian group. Then (i) The order of $G$ has at most two distinct prime divisors, (ii) if $\vert G\vert$ is not a power of a prime then $G=PQ$, where $P$ is a cyclic $p$-sylow subgroup of $G$ and $Q$ is elementary abelian minimal normal $q$-sylow subgroup of $G$.
My question is about a special kind of such groups.
Let $G$ be a minimal nonabelian group of order $pq^\alpha$, where $p \nmid q-1$ and $Z(G)=1$. By the above theorem we have $G\cong (\underbrace{{\mathbb{Z}}_q \times \cdots \times {\mathbb{Z}}_q}_{\alpha}) \rtimes {\mathbb{Z}}_p $. I aslo know that this kind of groups aren't supersolvable.
I want to know if there is a peresentation for this kind of minimal nonabelian groups or is there any more information about such kind of groups?
The minimality condition implies that $\alpha$ is minimal such that $p|q^\alpha-1$. In other words, $q$ has order $\alpha$ mod $p$.
A nice way to represent these groups $G = Q \rtimes P$ is to consider the Sylow $q$-subgroup as the additive group of the finite field $K$ of order $q^\alpha$. Then the multiplicative group of $K$ is cyclic, and contains a unique subgroup $\langle g \rangle$ of order $p$. The action of a Sylow $p$-subgroup $P$ of $G$ on $Q$ can be defined by multiplication in $K$ of a generator $g$ of this subgroup.
You asked how to compute a presentation of these groups. I can describe an algorithm for doing that. Assume that $\alpha$ is minimal with $p|q^\alpha-1$. We need to find an element (i.e. matrix) $A = (a_{ij})$ of order $q$ in ${\rm GL}(\alpha,p)$. You could do that in GAP, for example. Then using generators $x_1,\ldots,x_\alpha$ of $P$ and $y$ of $Q$, there is a presentation $\langle X \mid R \rangle$ of $P \rtimes Q$, with $X= \{x_1,\ldots,x_\alpha,y\}$ and $R = R_1 \cup R_2 \cup R_3$, with $$R_1 = \{x_1^p,x_2^p,\ldots,x_\alpha^p,y^q\},$$ $$R_2 = \{ [x_i,x_j] : 1 \le i < j \le \alpha \},$$ $$R_3 = \{ y^{-1}x_iy = x_1^{a_{i1}}x_2^{a_{i2}}\cdots x_\alpha^{a_{i\alpha}} : 1 \le i \le \alpha \}.$$
In keeping with the description of this group in terms of the finite field $K$ of order $p^\alpha$, we can also use that description to define the matrix $A$. We can take $A$ to be the companion matrix of the minimal polynomial over the prime field of an element of multiplicative order $q$ in $K$.
For example, if $p=3$ and $q=5$, then $\alpha=4$, and an element of order $5$ in $K$ has minimal polynomial $x^4+x^3+x^2+x+1$, giving $$A = \left(\begin{array}{rrrr}0&1&0&0\\0&0&1&0\\0&0&0&1\\2&2&2&2\end{array}\right).$$