N-abelian iff (1-n)-abelian

Statement
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.

Related facts

 * 2-abelian iff abelian
 * -1-abelian iff abelian

Proof
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.