N-abelian implies n(n-1)-central

Statement

Suppose $G$ is a group and $n$ is an integer other than 0 or 1. Suppose that $G$ is a n-abelian group, i.e., the map $x\mapsto x^{n}$ is an endomorphism of $G$ (hence, it is a universal power endomorphism). Then, $G$ is n(n-1)-central: the inner automorphism group of $G$ has exponent dividing $n(n-1)$.

Related facts

Converse

• 2-central implies 4-abelian
• 3-central implies 9-abelian
• 4-central implies 16-abelian
• 6-central implies 36-abelian

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.

• 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)

Facts used

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

Proof

Given: A group $G$ such that the map $x\mapsto x^{n}$ is an endomorphism of $G$. Elements $g,h\in G$ (possibly equal).

To prove: $g^{n(n-1)}$ commutes with $h$. This is sufficient because $h$ being arbitrary, it shows that $g^{n(n-1)}$ is central, and $g$ being arbitrary, it shows that $G/Z(G)$ has exponent dividing $n(n-1)$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$g^{n(n-1)}$ commutes with $h^{n}$.	Fact (1)	The map $x\mapsto x^{n}$ is an endomorphism of $G$.		By the given data and Fact (1), every $n^{th}$ power commutes with every $(n-1)^{th}$ power. Treating $g^{n(n-1)}=(g^{n})^{n-1}$ as a $(n-1)^{th}$ power and $h^{n}$ as a $n^{th}$ power, we obtain that $g^{n(n-1)}$ commutes with $h^{n}$.
2	$g^{n(n-1)}$ commutes with $h^{n-1}$.	Fact (1)	The map $x\mapsto x^{n}$ is an endomorphism of $G$.		By the given data and Fact (1), every $n^{th}$ power commutes with every $(n-1)^{th}$ power. Treating $g^{n(n-1)}=(g^{n-1})^{n}$ as a $n^{th}$ power and $h^{n-1}$ as a $(n-1)^{th}$ power, we obtain that $g^{n(n-1)}$ commutes with $h^{n-1}$.
3	The centralizer in $G$ of $g^{n(n-1)}$ contains the elements $h^{n}$ and $h^{n-1}$. Since the centralizer of any element is a subgroup, it also contains the quotient $h^{n}(h^{n-1})^{-1}=h$, and hence $g^{n(n-1)}$ commutes with $h$.			Steps (1), (2)	The first part follows from Steps (1) and (2). The rest is self-explanatory.