Discussion
The number of 5-Sylow subgroup(s) in a group of order 45 is
Answer: A
A group of order 45 has exactly 1 Sylow 5-subgroup.
Write \(45 = 3^{2}\cdot 5\). By Sylow's theorems the number \(n_{5}\) of Sylow 5-subgroups divides 9 and satisfies \(n_{5} \equiv 1 \pmod 5\).
The divisors of 9 are 1, 3 and 9. Of these, 3 and 9 leave remainders 3 and 4 on division by 5, so \(n_{5} = 1\), and the Sylow 5-subgroup is normal.
2 and 4 do not divide 9, and 3 is not congruent to 1 mod 5, so all three other options are impossible. (The same argument gives \(n_{3} = 1\), so every group of order 45 is abelian.)
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