Step 1: Introduction
This problem concerns the classification of finite groups, specifically those with order equal to the square of a prime number. A fundamental theorem in group theory provides the solution.
Step 2: Core Theorem
For a group of order \(p^2\), where \(p\) is prime, the group is abelian. Furthermore, any abelian group of order \(p^2\) is isomorphic to either \(\mathbb{Z}_{p^2}\) or \(\mathbb{Z}_p \times \mathbb{Z}_p\).
Step 3: Detailed Reasoning
Let \(G\) be a group with \(|G| = p^2\), where \(p\) is prime.
\begin{enumerate}
Groups of order \(p^2\) must be abelian, eliminating option (B) (non-abelian).
A trivial group has order 1. Since \(p^2 \ge 4\), the group cannot be trivial, thus ruling out option (A).
Since \(G\) can be cyclic of order \(p^2\) (isomorphic to \(\mathbb{Z}_{p^2}\)), option (C), which claims the group *must* be non-cyclic, is incorrect.
According to the structure theorem for finite abelian groups (or the specific classification for groups of order \(p^2\)), \(G\) must be isomorphic to one of the following:
The cyclic group of order \(p^2\), denoted \(\mathbb{Z}_{p^2}\).
The direct product of two cyclic groups of order \(p\), denoted \(\mathbb{Z}_p \times \mathbb{Z}_p\).
\end{enumerate}
This supports option (D).
Step 4: Conclusion
If a group \(G\) has order \(p^2\), where \(p\) is prime, then \(G\) is either cyclic of order \(p^2\) or isomorphic to the direct product of two cyclic groups of order \(p\).