Suppose $G$ is an arbitrary group and $a,b,x,y\in G$. Is the following relationship true? \begin{gather*}ba(xyx^{-1}y^{-1})=ab\end{gather*}
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3Yes, if the group is commutative. No, if the group is necomuativ. – medicu Nov 22 '13 at 11:22
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Regarding @draks post, in $S_3=\{\text{id},~(2,3),~(1,2),~(1,2,3),~(1,3,2),~(1,3)\}$: $$(2,3)(1,2)[(1,3,2),(1,2,3)]\neq (1,2)(2,3)$$
Mikasa
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From the Commutator#Group_theory:
The commutator of two elements, $g$ and $h$, of a group $G$, is the element
$$ [g, h] = g^{−1}h^{−1}gh.$$
It is equal to the group's identity $e$ if and only if $g$ and $h$ commute...
In your words, if $(xyx^{-1}y^{-1})=a^{-1}b^{-1}ab$ for all $a,b,x,y\in G$ then $a^{-1}b^{-1}ab=e$, thus $G$ is commutative.
BTW: It is true for arbitrary groups that [t]he quotient $G/[G,G]$ is an abelian group called the abelianization of G or G made abelian [Abelianization]
