Question

If p is a prime number and a group G is of the order p2, then G is:

• A. trivial
• B. non-abelian
• C. non-cyclic
• D. either cyclic of order p2 or isomorphic to the product of two cyclic groups of order p each

Hint:

For competitive exams, it's crucial to memorize the classifications of groups of small orders, especially for orders like \(p\), \(p^2\), \(p^3\), and \(2p\), where \(p\) is a prime. This can save a lot of time.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This question is about the classification of finite groups, specifically groups whose order is the square of a prime number. There is a fundamental theorem in group theory that addresses this exact case.

Step 2: Key Formula or Approach:
The theorem for groups of order \(p^2\), where \(p\) is a prime, states that any such group is abelian.
Furthermore, an abelian group of order \(p^2\) is isomorphic to either \(\mathbb{Z}_{p^2}\) (the cyclic group of order \(p^2\)) or \(\mathbb{Z}_p \times \mathbb{Z}_p\) (the direct product of two cyclic groups of order \(p\)).

Step 3: Detailed Explanation:
Let \(G\) be a group with order \(|G| = p^2\), where \(p\) is a prime.
\begin{enumerate}
A key result states that any group of order \(p^2\) must be abelian. This immediately rules out option (B) non-abelian.

A trivial group has only one element (the identity), so its order is 1. Since \(p\) is a prime, \(p^2 \ge 4\), so the group cannot be trivial. This rules out option (A).

Since the group could be cyclic of order \(p^2\) (isomorphic to \(\mathbb{Z}_{p^2}\)), option (C) which states it must be non-cyclic is incorrect.

By the structure theorem for finite abelian groups, or the specific classification for groups of order \(p^2\), \(G\) must be isomorphic to one of two possible structures:

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 directly corresponds to option (D).

Step 4: Final Answer:
Therefore, if \(p\) is a prime number and a group \(G\) is of the order \(p^2\), then \(G\) is either cyclic of order \(p^2\) or isomorphic to the product of two cyclic groups of order \(p\) each.