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 \( 2024 \)?
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 \( 2024 \)?
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For Sylow subgroup problems, calculate \( n_p(G) \) using congruence and divisibility constraints derived from the group's order.
Updated On: Feb 1, 2025
- \( n_{11}(G) = 1 \) and \( n_{23}(G) = 11 \)
- \( n_{11}(G) \in \{1, 23\} \) and \( n_{23}(G) = 1 \)
- \( n_{11}(G) = 23 \) and \( n_{23}(G) = 188 \)
- \( n_{11}(G) = 23 \) and \( n_{23}(G) = 11 \)
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The Correct Option is B
Solution and Explanation
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 \( n_{23}(G) = 1 \).
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