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Suppose that $G$ is the group generated by $a,b$ satisfying the relations $ba=a^2b, ab=b^2a$, then how can I find the cardinality of this group?

I've tried to rearrange these two relations by doing some cancellation and substitution, but nothing worked out.

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    $\begingroup$ What you have described in a group presentation, not a group representation. $\endgroup$ – Shaun Sep 11 at 19:28
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$ab=b(ba)=ba^2b$ so $a=ba^2$, hence $ba=1$ and $b=a^{-1}$.

Putting this into $ba=a^2b$ gives us $1=a$, so $G$ is the trivial group.

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