Question

Let \( H \) and \( K \) be subgroups of \( \mathbb{Z}_{144} \). If the order of \( H \) is 24 and the order of \( K \) is 36, then the order of the subgroup \( H \cap K \) is

• A. 3
• B. 4
• C. 6
• D. 12

Hint:

When calculating the order of the intersection of two subgroups, always use the greatest common divisor of their orders.

Answer 1

The Correct Option is D

Step 1: Use the formula for the order of the intersection of subgroups.
For two subgroups \( H \) and \( K \) of a finite group, the order of the intersection \( |H \cap K| \) is given by: \[ |H \cap K| = \frac{|H| \cdot |K|}{|H \cup K|}. \] However, we also know that the order of \( H \cap K \) divides the orders of both \( H \) and \( K \). Therefore, it must divide the greatest common divisor of the orders of \( H \) and \( K \).
Step 2: Compute the greatest common divisor.
The order of \( H \) is 24, and the order of \( K \) is 36. We compute the greatest common divisor of 24 and 36: \[ \gcd(24, 36) = 12. \] Thus, the order of \( H \cap K \) is 12, and the correct answer is \( \boxed{4} \).