N-abelian implies every nth power and (n-1)th power commute

Statement

Suppose $G$ is a group and $n$ is an integer such that $G$ is n-abelian: the $n^{th}$ power map on $G$ is an endomorphism of $G$. Then, for every $g,h\in G$ we have $h^{n}g^{n-1}=g^{n-1}h^{n}$.

Related facts

Applications

• nth power map is automorphism implies (n-1)th power map is endomorphism taking values in the center
• nth power map is endomorphism iff abelian (if order is relatively prime to n(n-1))
• n-abelian implies n(n-1)-central

Other related facts

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.

• n-abelian iff (1-n)-abelian
• The set of $n$ for which $G$ is $n$-abelian is termed the exponent semigroup of $G$. It is a submonoid of the multiplicative monoid of integers.
• abelian implies n-abelian for all n
• n-abelian implies every nth power and (n-1)th power commute
• n-abelian implies n(n-1)-central
• n-abelian iff abelian (if order is relatively prime to n(n-1))
• nth power map is surjective endomorphism implies (n-1)th power map is endomorphism taking values in the center
• (n-1)th power map is endomorphism taking values in the center implies nth power map is endomorphism
• Frattini-in-center odd-order p-group implies p-power map is endomorphism
• Frattini-in-center odd-order p-group implies (mp plus 1)-power map is automorphism
• Characterization of exponent semigroup of a finite p-group
• Alperin's structure theorem for n-abelian groups

Value of $n$ (note that the condition for $n$ is the same as the condition for $1-n$)	Characterization of $n$-abelian groups	Proof	Other related facts
0	all groups	obvious
1	all groups	obvious
2	abelian groups only	2-abelian iff abelian	endomorphism sends more than three-fourths of elements to squares implies abelian
-1	abelian groups only	-1-abelian iff abelian
3	3-abelian group means: 2-Engel group and derived subgroup has exponent dividing three	Levi's characterization of 3-abelian groups	cube map is surjective endomorphism implies abelian, cube map is endomorphism iff abelian (if order is not a multiple of 3), cube map is endomorphism implies class three
-2	same as for 3-abelian	(based on n-abelian iff (1-n)-abelian)

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)

Facts used

1. Group acts as automorphisms by conjugation: For any $g\in G$, the map $c_{g}:x\mapsto gxg^{-1}$ is an automorphism of $G$.

Proof

From the power map being an endomorphism

Given: A group $G$ and a natural number $n$ such that $g\mapsto g^{n}$ is an endomorphism of $G$.

To prove: For every $g,h\in G$, $h^{n}g^{n-1}=g^{n-1}h^{n}$.

Proof: We denote by $c_{g}$ the map $x\mapsto gxg^{-1}$.

Step no.	Assertion	Facts used	Given data used	Previous steps used	Explanation
1	For $g,h\in G$, we have $gh^{n}g^{-1}=(ghg^{-1})^{n}$	Fact (1)			By fact (1), $c_{g}$ is an automorphism of $G$, so $c_{g}(h^{n})=(c_{g}(h))^{n}$. Applying the definition of $c_{g}$ gives the result.
2	We have $(ghg^{-1})^{n}=g^{n}h^{n}g^{-n}$ for all $g,h\in G$.		$n^{th}$ power map is endomorphism		Since the $n^{th}$ power map is an endomorphism, we have $(abc)^{n}=a^{n}b^{n}c^{n}$. Setting $a=g,b=h,c=g^{-1}$ gives the desired result.
3	We get $h^{n}g^{n-1}=g^{n-1}h^{n}$ for all $g,h\in G$. In other words, every $n^{th}$ power commutes with every $(n-1)^{th}$ power.			Steps (1), (2)	Combining steps (1) and (2) gives that $gh^{n}g^{-1}=g^{n}h^{n}g^{-n}$. Cancel a left-most $g$ and a right-most $g^{-1}$ on both sides, and get $h^{n}=g^{n-1}h^{n}g^{-(n-1)}$. Multiply both sides on the right by $g^{n-1}$ to get the desired result. Thus, every $n^{th}$ power commutes with every $(n-1)^{th}$ power.