Final answer:
The proof shows that a finite group G is abelian by utilizing the property that representatives of its conjugacy classes commute pairwise, ultimately demonstrating that any two elements in G commute.
Step-by-step explanation:
The question asks to prove that a finite group G is abelian given a special property of its representatives of conjugacy classes: the representatives commute pairwise.
To prove that G is abelian, consider any two elements a and b in G. Because each element of G is conjugate to one of the representatives g₁, g₂, ..., gᵣ, there exist x, y in G such that xax⁻¹ = gᵢ and yby⁻¹ = gᵣ for some i and j. Since the representatives commute, we have gᵢgᵣ = gᵣgᵢ. By conjugating both sides by x⁻¹ and y⁻¹ respectively, we get ab = ba, showing that G is abelian.
Given a finite group G with representatives of conjugacy classes g₁, g₂, ..., gᵣ, that pairwise commute, we aim to show G is abelian. Leveraging the class equation, where
∣G∣ equals the sum of the sizes of the centralizers, we observe that since the representatives commute, the centralizers commute pairwise, resulting in their intersections being trivial. This leads to each element in G being uniquely represented, implying all elements commute.
Consequently, G is abelian. This concise proof demonstrates how the pairwise commutation of conjugacy class representatives manifests abelianness in the finite group G.