Question

If \(G\) is a group of order 30, then the number of Sylow 5-subgroups in \(G\) must be ____.

• A. 1 or 2
• B. 2 or 3
• C. 3 or 5
• D. 1 or 6

Hint:

Apply Sylow's theorem: \(n_5 \equiv 1 \pmod 5\) and \(n_5 \mid 6\).

Answer 1

The Correct Option is D

Step 1: Write \(|G| = 30 = 2 \times 3 \times 5\). Let \(n_5\) denote the number of Sylow 5-subgroups of \(G\).

Step 2: By Sylow's third theorem, \(n_5 \equiv 1 \pmod 5\) and \(n_5\) divides \(\dfrac{|G|}{5} = 6\).

Step 3: The divisors of 6 are 1, 2, 3, 6. Checking each modulo 5: \(1 \equiv 1\), \(2 \equiv 2\), \(3 \equiv 3\), \(6 \equiv 1 \pmod 5\).

Step 4: Only 1 and 6 leave remainder 1 mod 5, so \(n_5 \equiv 1 \pmod 5\) forces \(n_5\) to be one of these two divisors.

\[\boxed{n_5 = 1 \text{ or } 6}\]