 Solution

For the modular Heisenberg group, the three independent entries each have $p$ choices, soThe multiplication and commutation calculations work over the finite field ${\mathbb{F}}_p$ exactly as over $\mathbb{R}$. Taking $x^{\prime }=1,z^{\prime }=0$ and $x^{\prime }=0,z^{\prime }=1$ shows thatLet $c=M(0,1,0)$. Its powers exhaust the centre and $c$ has order $p$, giving the group presentationLet $a=M(1,0,0)$ and $b=M(0,0,1)$. Their images $\bar{a},\bar{b}$ in the quotient group translate the $x,z$ entries independently. They commute modulo the centre, have order $p$, and generate $p^2$ distinct pairs. ThereforeFor context, $ab{a}^{-1}{b}^{-1}=c$, with $c$ central. All conclusions above include $p=2$; one must not infer that every nonidentity element of $H_2$ has order two, since $(ab{)}^2=c$.