Let $p$ be an odd prime and let $H$ be the non-abelian $p$-group of order $p^3$ and exponent $p^2$. A presentation of $H$ is given by
$$H=\langle a,b \mid a^{p^2}=1,b^p=1,[a,b]=a^p \rangle. $$
How do I find a presentation with $2$ generators and $2$ relations (if it exists)? I played around with the relations and couldn't see how any two relations imply the third and I tried using different genrators.
Motivation: Let $G$ be a finite $p$-group and let $d(G)$ be the minimal number of generators of $G$. Let $R(G)$ and $r(G)$ be the minimal number of relations needed to present $G$ as an abstract and pro-$p$ group respectively. It is always true that $R(G)\geq r(G)$ and I think there is no known example for $G$ such that $R(G)>r(G)$. Since in this case I know that $r(H)=2$, I expect such a presentation to exist.
This seems to work: $$\langle a,b\mid a^{p^2}=b^p,[a,b]=a^p\rangle$$
Since $(1+p)^p\equiv 1+p^2\pmod{p^3}$ and $a=b^pab^{-p}=a^{(1+p)^p}$, we have $a^{p^2}=1$.