N-abelian iff abelian (if order is relatively prime to n(n-1))

Statement
Suppose $$G$$ is a finite group and $$n$$ is an integer such that the order of $$G$$ is relatively prime to $$n(n-1)$$. Then, the $$n^{th}$$ power map on $$G$$ is an endomorphism (i.e., $$G$$ is a n-abelian group if and only if $$G$$ is an abelian group.

Facts about n-abelian groups
We say that a group is a n-abelian group if the $$n^{th}$$ power map is an endomorphism. Here are some related facts about $$n$$-abelian groups.

Analogues in other algebraic structures

 * Multiplication by n map is a derivation iff derived subring has exponent dividing n
 * Multiplication by n map is an endomorphism iff derived subring has exponent dividing n(n-1)

Tightness
It turns out that the condition of being relatively prime to $$n(n-1)$$ is fairly tight for $$n \ge 4$$, i.e., we can find non-abelian groups of order not relatively prime to $$n$$, as well as non-abelian groups of order not relatively prime to $$n - 1$$, where the $$n^{th}$$ power map is an endomorphism.


 * The $$n - 1$$ case: $$n - 1$$ must be divisible either by 4 or by an odd prime. If 4 divides $$n-1$$, the $$n^{th}$$ power map gives an automorphism for any non-abelian group of exponent 4, such as dihedral group:D8. If $$p$$ is an odd prime dividing $$n - 1$$, we can find a non-abelian Frattini-in-center $$p$$-group, where the $$p^{th}$$ power map is an endomorphism taking values in the center. Hence, the $$n^{th}$$ power map is an automorphism.
 * The $$n$$ case: $$n$$ must be divisible either by 4 or by an odd prime. If 4 divides $$n$$, the $$n^{th}$$ power map is an endomorphism of any non-abelian group of exponent 4. If $$p$$ is an odd prime dividing $$n$$, we can find a non-abelian Frattini-in-center $$p$$-group, where the $$p^{th}$$ power map is an endomorphism taking values in the center. Hence, the $$n^{th}$$ power map is an endomorphism.

Facts used

 * 1) uses::Abelian implies universal power map is endomorphism
 * 2) uses::nth power map is endomorphism implies every nth power and (n-1)th power commute
 * 3) uses::kth power map is bijective iff k is relatively prime to the order

From abelianness to the power map being an endomorphism
This follows from fact (1).

From the power map being an endomorphism to being Abelian
Given: A finite group $$G$$ and an integer $$n$$ such that the order of $$G$$ is relatively prime to $$n(n-1)$$. Further, $$g \mapsto g^n$$ is an endomorphism of $$G$$.

To prove: $$G$$ is abelian.

Proof: