G/[G,G] is ALWAYS abelian, no matter what G is, yet you are phrasing it as if it's an additional assumption. xyx^-1y^-1=1 is the same equation as xy = yx, just by multiplying on the right by y and then by x on both sides of the equation. To put it another way, x and y commute if and only if their commutator, xyx^-1y^-1 is equal to the identity.
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