Question

Let \( G \) be a group of order 2022. Let \( H \) and \( K \) be the subgroups of \( G \) of order 337 and 674, respectively. If \( H \cup K \) is also a subgroup of \( G \), then which one of the following is FALSE?

• A. H is the normal subgroup of \( H \cup K \)
• B. The order of \( H \cup K \) is 1011
• C. The order of \( H \cup K \) is 674
• D. K is a normal subgroup \( H \cup K \)

Answer 1

The Correct Option is B

To solve this problem, we need to analyze the given group structure and apply relevant group theory concepts.

1. We begin with the group \( G \) of order 2022. We know from the Lagrange's theorem that any subgroup's order must divide the order of the group.
2. Given that \( H \) is a subgroup of order 337 and \( K \) is a subgroup of order 674, we verify their divisibility with 2022: \( 2022 = 2 \times 3 \times 337 \). Hence, both 337 and 674 are divisors of 2022.
3. We need to determine the order of \( H \cup K \) if \( H \cup K \) is a subgroup of \( G \). For \( H \cup K \) to be a subgroup, it must satisfy group closure properties. However, the union of two subgroups is a subgroup only if one is a subset of the other.
4. Now, consider the possible intersections between \( H \) and \( K \): \[ |H \cap K| \times \frac{|H| |K|}{|H \cap K|} = 674 \text{ or } 337 \] implies \( |H \cap K| \) should divide both 337 and 674.
5. The greatest common divisor of 337 and 674 is 1, hence \( |H \cap K| = 1 \). In such a scenario, when the intersection is trivial, it implies \( H \cup K \) cannot be a subgroup unless additional conditions, such as normality, are met.
6. The only way \( H \cup K \) is a subgroup is if \( H \) is a normal subgroup of \( K \) (or vice versa), but this contradicts the fact that \( |H \cap K| \) is 1.
7. The possibility where none of the groups are normal leads to establishing \( |H \cup K| = |H| + |K| - |H \cap K| = 337 + 674 - 1 = 1010 \) if each subgroup is simple.
8. The assertion "The order of \( H \cup K \) is 1011" is indeed incorrect based on these considerations.

Thus, the false statement is: "The order of \( H \cup K \) is 1011".