Let $p$ be a prime number. For a group $G$ and a positive integer $m$, let $[G,G]$ denote the commutator subgroup of $G$ and $G^{m}=\langle g^{m};~g\in G\rangle$.
It is well known that a finite $p$-group $G$ is called a powerful $p$-group if either $p\neq2$ and $[G,G]\leq G^{p}$ or $p=2$ and $[G,G]\leq G^{4}$. Clearly, all abelian finite $p$-groups are also powerful $p$-groups since $[G,G]=1$, but these are the trivial examples and then non interesting.
In the book "Analytic Pro-$p$ Groups, 2nd edition, is proven in the page 44 the following theorem:
Theorem. Let $G$ be a finite $p$-group of rank $r$. Then $G$ has a characteristic powerful subgroup whose index is bounded by a function of $p$ and $r$.
The subgroup constructed in the proof of this theorem is $V(G,r)$, if $p\neq2$ and $V(G,r)^2$ if $p=2$, where $V(G,r)$ is the intersecction of the kernels of all homomorphisms from $G$ to the General Linear Group $GL$$_{r}(\mathbb{F}_p)$. Of course, it is not immeadiate to found this subgroup and I think that this subgroup may be the trivial subgroup.
Then, how to obtain a finite non-abelian powerful $p$-group? Is there a known presentation of a non abelian powerful $p$-group for each prime $p$?