N-abelian iff (1-n)-abelian

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Suppose n is an integer and G is a group. Then, G is a n-abelian group (see n-abelian group) if and only if G is a (1-n)-abelian group.

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The idea is to show that the condition for being n-abelian on x,y \in G is equivalent to the condition for being (1-n)-abelian on x^{-1},y^{-1}. Since the inverse map is bijective, varying (x,y) over all of G \times G also varies (x^{-1},y^{-1}) over all of G \times G.

Given: A group G, elements x,y \in G such that (xy)^n= x^ny^n.

To prove: (x^{-1}y^{-1})^{1-n} = (x^{-1})^{1-n}(y^{-1})^{1-n}.

Proof: This is straightforward group element manipulation. PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]