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I understand that if $aba^{-1}b^{-1} = e$ then $ab$ is commutative, but I don't see how having multiple commutators will prevent the group from being abelian

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    $\begingroup$ If $aba^{-1}b^{-1} = c \ne e$, then $ab=cba$, so $ab \ne ba$. $\endgroup$ – Derek Holt Feb 24 '14 at 16:10
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    $\begingroup$ such a simple connection that I couldn't make, thank you!!! $\endgroup$ – Rod Feb 24 '14 at 16:14
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It's easier to see the contrapositive: If a group is abelian, then $e$ is its only commutator:

$$aba^{-1}b^{-1} = aa^{-1}bb^{-1} = ee = e $$

Therefore, if the group has more than one commutator, at least one of them will be different from $e$, and so the group cannot be abelian.

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