Discussion
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?
Answer: B
For every group of order 2024, \(n_{11}(G) \in \{1, 23\}\) and \(n_{23}(G) = 1\).
Factor \(2024 = 2^{3}\cdot 11\cdot 23\). By Sylow's theorems:
\(n_{23}\) divides 88 and \(n_{23} \equiv 1 \pmod{23}\). The divisors of 88 are 1, 2, 4, 8, 11, 22, 44, 88, and only 1 is congruent to 1 mod 23 (88 leaves remainder 19). So \(n_{23} = 1\).
\(n_{11}\) divides 184 and \(n_{11} \equiv 1 \pmod{11}\). The divisors of 184 are 1, 2, 4, 8, 23, 46, 92, 184, and those congruent to 1 mod 11 are 1 and 23. So \(n_{11} \in \{1, 23\}\).
The options with \(n_{23} = 11\) or \(n_{23} \in \{1, 88\}\) contradict the first count, and \(n_{11}\) cannot be fixed at a single value for every group: the cyclic group has \(n_{11} = 1\).
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