Question

Given a prime number \( p \), let \( n_p(G) \) denote the number of \( p \)-Sylow subgroups of a finite group \( G \). Which one of the following is TRUE for every group \( G \) of order \( 2024 \)?

• A. \( n_{11}(G) = 1 \) and \( n_{23}(G) = 11 \)
• B. \( n_{11}(G) \in \{1, 23\} \) and \( n_{23}(G) = 1 \)
• C. \( n_{11}(G) = 23 \) and \( n_{23}(G) = 188 \)
• D. \( n_{11}(G) = 23 \) and \( n_{23}(G) = 11 \)

Hint:

For Sylow subgroup problems, calculate \( n_p(G) \) using congruence and divisibility constraints derived from the group's order.

Answer 1

The Correct Option is B

Step 1: Analyzing the group order. The order of \( G \) is \( 2024 = 2^3 \cdot 11 \cdot 23 \). The Sylow theorems provide constraints on the number of \( p \)-Sylow subgroups: \[ n_p(G) \equiv 1 \pmod{p} \quad {and} \quad n_p(G) { divides } \frac{|G|}{p^k}. \] Step 2: Applying Sylow theorems. - For \( p = 11 \): \( n_{11}(G) \equiv 1 \pmod{11} \) and \( n_{11}(G) \mid 184 \). Thus, \( n_{11}(G) \in \{1, 23\} \). - For \( p = 23 \): \( n_{23}(G) \equiv 1 \pmod{23} \) and \( n_{23}(G) \mid 88 \). Hence, \( n_{23}(G) = 1 \). Step 3: Conclusion. The correct statement is \( {(2)} \): \( n_{11}(G) \in \{1, 23\} \) and \( n_{23}(G) = 1 \).