# 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

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.

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.