# 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.

## 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. PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]