How does $(ab)^{3}=a^{3}b^{3}$ and $(ab)^{5}=a^{5}b^{5}$ for all $a,b \in G$ implies that $G$ is abelian? I know that the first criteria alone isn't enough because there's a counterexample. What could we say if in those equation instead 3 and 5 we had two other relatively prime numbers?
Asked
Active
Viewed 127 times
