Q:

To prove $|G|=p^2$ $\implies$ $|Z(G)| \ne 1$ , without using Class equation

It is known that if $G$ is a finite group of order $p^2$ , where $p$ is prime , then the center of $G$ is non-trivial i.e. $Z(G) \ne${$e$} , $e$ being the identity element of $G$. Can this be proved without using class equation ?   

A:

To have this question not remain unanswered, here is a variation of the answer given by Arturo Magidin at Showing non-cyclic group with $p^2$ elements is Abelian
First, we can assume that the group is not cyclic, so all non-identoty elements have order $p$ and thus the subgroup generated by such an element has order $p$ and index $p$.
Since $p$ is the smallest prime diving the order of the group, any subgroup of index $p$ is normal (see Normal subgroup of prime index but beware that the proof by myself there would turn this entire thing into a cyclic argument). 
Now, pick two distinct such subgroups (so pick an element $h\neq 1$ and $k\not\in\langle h\rangle$).  Now $H = \langle h\rangle$ and $K = \langle k\rangle$ are normal subgroups of order $p$ intersecting trivially, so we have $G = HK$ and in fact $G = H\times K$ so $G$ is abelian.