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

This article describes an easy-to-prove fact about basic notions in group theory, that is not very well-known or important in itself
View other elementary non-basic facts
VIEW FACTS USING THIS: directly | directly or indirectly, upto two steps | directly or indirectly, upto three steps|

## 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.

## 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.

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)

## 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.

## Proof

### 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:

Step no. Assertion Facts used Given data used Previous steps used Explanation
1 We get $h^ng^{n-1} = g^{n-1}h^n$ for all $g,h \in G$. In other words, every $n^{th}$ power commutes with every $(n - 1)^{th}$ power. Fact (2) $n^{th}$ power map is endomorphism Given+Fact direct
2 $G$ is abelian. Fact (3) Step (1) [SHOW MORE]