If $a^pb^p=b^pa^p$ and $a^qb^q=b^qa^q$ for coprime $p,q$ and all $a,b\in G$, then $G$ is Abelian

Question

Assume $G$ is a group and $p,q \in \mathbb{N}$ and $(p,q)=1$. If $$a^pb^p=b^pa^p\hspace{.5cm} \ \text{and}\hspace{.5cm} \ a^qb^q=b^qa^q \hspace{1cm}\forall a,b\in G$$ Is it true that $G$ is Abelian? Can anyone give me a hint?

Answer 1

Unfortunately it is false. Consider $\mathbb{D}_{6}$ the group of symmetries of the (equilateral) triangle. Then for each rotation, say $R_1$, we have $R_1^3=I$ the identity, while for each reflexion $S_1$ it holds $S_1^{2}=I$. Observe $(2,3)=1$, then $\mathbb{D}_6$ is a counterexample.