= Solution
Use the <class equation> for the action of $G$ on itself by <conjugation>:
$$
|G|=|Z(G)|+\sum_j[G:C_G(g_j)],
$$
where the sum runs over noncentral <conjugacy classes>. By <Lagrange's theorem>, each noncentral class has size a positive power of $p$ greater than one. Thus $|Z(G)|\equiv |G|\equiv0\pmod p$. Since the <identity element> belongs to the <center of a group>,
$$
\boxed{|Z(G)|\geq p>1}.
$$
For $|G|=p^3$, the possible orders of the <center of a group> are $p,p^2,p^3$. Order $p^3$ would make $G$ an <abelian group>. Order $p^2$ would make the <quotient group> $G/Z(G)$ have prime order and hence be a <cyclic group>. But a <cyclic quotient by the center> forces $G$ to be <Abelian>: if $gZ(G)$ generates the quotient, any two elements are $g^a z$ and $g^b w$ with $z,w\in Z(G)$, and their products commute. Both larger orders contradict the hypothesis. Consequently
$$
\boxed{|Z(G)|=p}.
$$
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