Let $G$ be a group, where $(ab)^3=a^3b^3$ and $(ab)^5=a^5b^5$. Prove that $G $ is an abelian group.

Question

Let G be a group, where $(ab)^3=(a^3)(b^3)$ and $(ab)^5=(a^5)(b^5)$. Prove that $G$ is abelian group.

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Answer 1

From $(ab)^3=a^3b^3$ and $(ab)^5=a^5b^5$ it follows that $(ba)^2=a^2b^2$, and $(ba)^4=a^4b^4$. This implies $(ba)^4=(ba)^2(ba)^2=a^2b^2a^2b^2=a^4b^4$, so $b^2a^2=a^2b^2$, hence all squares commute. Thus $(ba)^2=baba=a^2b^2=b^2a^2=bbaa$ and it follows that $ab=ba$.