 Solution

The centre of a group is $Z(G)=\{z\in G:zg=gz\text{ for every }g\in G\}$. It is a normal subgroup. For $|G|=p^r$ with $r\ge 1$, let $G$ act on itself by conjugation. The orbit-stabilizer theorem makes the size of the conjugacy class of $g$ equal to $[G:C_G(g)]$, a power of $p$. A class has size one exactly when $g\in Z(G)$; every other class size is divisible by $p$. The class equation therefore givesso $p$ divides $|Z(G)|$. Since the identity belongs to the centre of a group, its size is a positive multiple of $p$. This proves the nontrivial center of a finite p-group.

If $G/Z(G)$ is cyclic, choose a representative $a$ of a generator. Every element is $a^{m}z$ for an integer $m$ and $z\in Z(G)$. Any two such elements commute, since their central factors commute with everything and their remaining factors are powers of the same element. Thus $G$ is an abelian group, $Z(G)=G$, and the cyclic quotient by the centre of a group is trivial.

For a non-abelian group of order $p^3$, the centre of a group has order $p$, $p^2$ or $p^3$ by the proved result and Lagrange's theorem. The last would make the group abelian. The middle would make $G/Z(G)$ have prime order and hence cyclic, also making $G$ abelian by the preceding argument. Therefore its centre has order $p$.

Finally write the given upper-unitriangular matrix as $M(a,b,c)$. Direct matrix multiplication givesIt commutes with every $M(a^{\prime },b^{\prime },c^{\prime })$ exactly when $a{c}^{\prime }=a^{\prime }c$ for all $a^{\prime },c^{\prime }\in F$. Taking $(a^{\prime },c^{\prime })=(0,1)$ forces $a=0$, and taking $(1,0)$ forces $c=0$; those conditions also suffice. Thus the centre of a group in this Heisenberg group over a prime field is