To determine the nature of a group \( G \) of order \( p^2 \), where \( p \) is a prime number, we need to consider some key results from group theory.

1. The order of a group, by Lagrange's Theorem, is the number of elements in the group; here, it is \( p^2 \).
2. A group of prime power order has interesting properties. Specifically, consider the following:
   • If \( G \) is a group of order \( p^2 \), then it is a \( p \)-group.
   • By Burnside's Theorem, every such group has a non-trivial center \( Z(G) \).
   • The center \( Z(G) \) must also be a \( p \)-group. Since the order of \( G \) is \( p^2 \), the possible orders of \( Z(G) \) are either \( p \) or \( p^2 \).
3. If the order of \( Z(G) = p^2 \), \( G \) is abelian, hence cyclic, and therefore isomorphic to the cyclic group of order \( p^2 \).
4. If the order of \( Z(G) = p \), then by the class equation, \( G/Z(G) \) has order \( p \), which implies it is cyclic (as every group of prime order is cyclic).
5. In such cases, \( G \) is isomorphic to the product of two cyclic groups of order \( p \) each (this is \( \mathbb{Z}_p \times \mathbb{Z}_p \)).

Given these considerations, for a group \( G \) of order \( p^2 \):

• It is either cyclic of order \( p^2 \) or isomorphic to the product of two cyclic groups of order \( p \) each.
• The options trivial, non-abelian, and non-cyclic do not correctly describe the structural possibilities for groups of this order.

Thus, the correct answer is: either cyclic of order \( p^2 \) or isomorphic to the product of two cyclic groups of order \( p \) each.