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

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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.

• 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)

Tightness

Further information: Frattini-in-center odd-order p-group implies p-power map is endomorphism, Frattini-in-center odd-order p-group implies (p plus 1)-power map is automorphism

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.

Facts used

1. Abelian implies universal power map is endomorphism
2. nth power map is endomorphism implies every nth power and (n-1)th power commute
3. kth power map is bijective iff k is relatively prime to the order

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] By fact (3), we now have that since the order of the group is relatively prime to $n(n-1)$, every element is a $n^{th}$ power and every element is a $(n-1)^{th}$ power, so from step (1), any two elements must commute.