Homomorphism given by $x \mapsto x^p$

Question

Let $p$ be an odd prime and let $G$ be a group of order $p^3$. Suppose we have already proved that the map $\varphi: G \rightarrow Z(G)$ given by $\varphi(x) = x^p$ is a homomorphism. Assume also that we know that if $G$ is not abelian, then ker $\varphi$ has order either $p^2$ or $p^3$. How can we deduce the same information about ker $\varphi$ in the case that $G$ is abelian but not cyclic? That is, why can't ker $\varphi$ have order 1 or $p$?