Prove a group is equal to its commutator subgroup

Question

Let $G$ be the group generated by $a,b$ satisfying the relations $ba^3=a^2b, ab^2=b^3a$.

How can I prove $G = [G,G]$ which is the commutator group of $G$?

Answer 1

From $ab^2=b^3a$ we have: $$ \begin{align} [a,b^2]&=ab^2a^{-1}b^{-2}\\ &=b^3aa^{-1}b^{-2}\\ &=b \end{align}$$ so $b\in[G,G]$.

Also, from $ba^3=a^2b$ we have: $$\begin{align} [b^{-1},a^2]&=b^{-1}a^2ba^{-2} \\ &=b^{-1}ba^3a^{-2} \\ &=a \end{align}$$ so $a\in[G,G]$.

Thus, both $a,b\in [G,G]$, which implies $G=[G,G]$.