I am trying to verify that the so-called Heisenberg group over the finite field ${\mathbb{F}}_p$ of order $p$ (with order $p^3$ and exponent $p$) has the following presentation: $$ G = \langle x, y ~:~ x^p, y^p, [y,x,y], [y,x,x] \rangle $$ given in a paper (Ref. https://link.springer.com/article/10.1007/s00605-016-0938-5 Page-687, Thm.4.3)
On a different look, it seems like that the commutator $[y,x]$ in $G$ is central and is of infinite order. Shouldn't we add another relation $[y,x]^p$ here?
Yes, that presentation is correct.
Since $x$ commutes with $[x,y]$, we have $[x,y]^p = [x^p,y] = 1$.
This follows from the standard commutator identities, which you can find here.