= Solution
For the <modular Heisenberg group>, the three independent entries each have $p$ choices, so
$$
\boxed{|H_p|=p^3.}
$$
The multiplication and commutation calculations work over the <finite field> $\mathbb F_p$ exactly as over $\mathbb R$. Taking $x'=1,z'=0$ and $x'=0,z'=1$ shows that
$$
Z(H_p)=\{M(0,y,0):y\in\mathbb F_p\}.
$$
Let $c=M(0,1,0)$. Its powers exhaust the centre and $c$ has order $p$, giving the <group presentation>
$$
\boxed{Z(H_p)=\langle c\mid c^p=e\rangle\cong C_p.}
$$
Let $a=M(1,0,0)$ 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. Therefore
$$
\boxed{H_p/Z(H_p)=\langle\bar a,\bar b\mid\bar a^p=\bar b^p=e,\ \bar a\bar b=\bar b\bar a\rangle\cong C_p\times C_p.}
$$
For context, $aba^{-1}b^{-1}=c$, with $c$ central. All conclusions above include $p=2$; one must not infer that every nonidentity element of $H_2$ has order two, since $(ab)^2=c$.
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