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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 $G$?

Shaun
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bt203
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    G is the trivial group, see https://math.stackexchange.com/questions/66573/a-particular-two-variable-system-in-a-group?noredirect=1&lq=1 – zwim Sep 11 '19 at 19:57
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    It's a lot easier to prove that $G = [G,G]$ than it is to prove that $G$ is trivail. – Derek Holt Sep 11 '19 at 21:18

1 Answers1

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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 $G=[G,G]$.

Shaun
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