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
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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

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.

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]