Question

Consider the group \( \mathbb{Z}^2 = \{ (a, b) | a, b \in \mathbb{Z} \} \) under component-wise addition. Then which of the following is a subgroup of \( \mathbb{Z}^2 \)?

• A. \( \{ (a, b) \in \mathbb{Z}^2 | ab = 0 \} \)
• B. \( \{ (a, b) \in \mathbb{Z}^2 | 3a + 2b = 15 \} \)
• C. \( \{ (a, b) \in \mathbb{Z}^2 | 7 \text{ divides } ab \} \)
• D. \( \{ (a, b) \in \mathbb{Z}^2 | 2 \text{ divides } a \text{ and } 3 \text{ divides } b \} \)

Hint:

When checking subgroups, verify closure under addition, the presence of the identity element, and the existence of inverses.

Answer 1

The Correct Option is D

Step 1: Checking subgroup criteria. 
To be a subgroup, the set must satisfy the following: 
- It must be closed under addition. 
- It must contain the identity element (in this case, \( (0, 0) \). 
- It must contain inverses.

Step 2: Checking each option. 
(A) The set \( \{ (a, b) | ab = 0 \} \) is not closed under addition, as \( (1, 0) + (0, 1) = (1, 1) \) and \( 1 \times 1 \neq 0 \). 
(B) The set \( \{ (a, b) | 3a + 2b = 15 \} \) is not closed under addition because adding two such elements does not guarantee that the resulting vector satisfies the equation. 
(C) The set \( \{ (a, b) | 7 \text{ divides } ab \} \) is not closed under addition because adding two such vectors does not necessarily result in a vector where \( ab \) is divisible by 7. 
(D) The set \( \{ (a, b) | 2 \text{ divides } a \text{ and } 3 \text{ divides } b \} \) is closed under addition, contains the identity element \( (0, 0) \), and contains inverses, so it forms a subgroup. 
 

Step 3: Conclusion. 
The correct answer is (D) \( \{ (a, b) \in \mathbb{Z}^2 | 2 \text{ divides } a \text{ and } 3 \text{ divides } b \} \).