Question

Let \( G \) be a finite abelian group of order 10. Let \( x_0 \) be an element of order 2 in \( G \). If \( X = \{ x \in G : x^3 = x_0 \} \), then which one of the following is TRUE?

• A. \( X \) has exactly one element
• B. \( X \) has exactly two elements
• C. \( X \) has exactly three elements
• D. \( X \) is an empty set

Hint:

In a finite group, the number of solutions to an equation like \( x^n = a \) depends on the order of the element and the structure of the group.

Answer 1

The Correct Option is A

Step 1: Use the properties of the group.
Since \( G \) is a finite abelian group of order 10, by Lagrange's Theorem, the order of any element in \( G \) must divide 10. Step 2: Analyze the set \( X \).
We are given that \( x_0 \) has order 2, i.e., \( x_0^2 = e \) (where \( e \) is the identity element). We are asked to find the set \( X = \{ x \in G : x^3 = x_0 \} \). Step 3: Solve for elements in \( X \).
Let \( x^3 = x_0 \). Since \( x_0^2 = e \), we have: \[ (x^3)^2 = x_0^2 = e \] Thus, the order of \( x \) must be 3, so there is exactly 1 element in \( G \) that satisfies this condition. Final Answer: \[ \boxed{X \text{ has exactly one element}} \]