Classifying Groups of Order 15

Abstract Algebra · Axiom Academy

EXAMPLE Classifying Groups of Order 15 Using Sylow theorems to prove every group of order 15 is cyclic Excellent work! You've proven that every group of order 15 is cyclic. Here's what we established: Sylow's Third Theorem is powerful: The conditions n_p | m and n_p ≡ 1 (mod p) often force unique Sylow subgroups Unique Sylow subgroups are normal: If there's only one Sylow p-subgroup, it must be normal in G Coprime orders give trivial intersection: When gcd(|H|, |K|) = 1, we have H ∩ K = e Internal direct product structure: When H, K are normal with trivial intersection and HK = G, then G ≅ H × K Prime order implies cyclic: Every group of prime order is cyclic by Lagrange's theorem Direct product of coprime cyclic groups: ℤₘ × ℤₙ ≅ ℤₘₙ when gcd(m,n) = 1 This technique generalizes! Any group whose order is a product of distinct primes (called a square-free number) has a unique structure. For instance, every group of order 35 = 5 × 7 or order 33 = 3 × 11 is also cyclic. The method: use Sylow theorems to show all Sylow subgroups are unique and normal, then apply the internal direct product theorem!

This is the written version of the interactive lesson above. See the full Abstract Algebra course.