Nth power map is surjective endomorphism implies (n-1)th power map is endomorphism taking values in the center

Statement

The following statements are equivalent for a group $G$ and an integer $n$. Suppose the $n^{th}$ power map $g\mapsto g^{n}$ is a surjective endomorphism (such as an automorphism) of $G$.

Then, the $(n-1)^{th}$ power map $g\mapsto g^{n-1}$ is an endomorphism of $G$ and $g^{n-1}$ is in the center of $G$ for all $g\in G$.

Related facts

Converse

A precise converse does not hold, but the following does: (n-1)th power map is endomorphism taking values in the center implies nth power map is endomorphism. We cannot guarantee surjectivity in general.

Similar facts

• nth power map is endomorphism iff abelian (if order is relatively prime to n(n-1))
• nth power map is endomorphism implies every nth power and (n-1)th power commute
• Inverse map is automorphism iff abelian
• Square map is endomorphism iff abelian
• Cube map is endomorphism iff abelian (if order is not a multiple of 3)
• Cube map is automorphism implies abelian
• 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

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)

Facts used

1. nth power map is endomorphism implies every nth power and (n-1)th power commute

Proof

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Given: A group $G$ and an integer $n$ such that $\sigma =g\mapsto g^{n}$ is a surjective endomorphism of $G$.

To prove: $\tau =g\mapsto g^{n-1}$ is an endomorphism of $G$ and $g^{n-1}x=xg^{n-1}$ for all $g,x\in G$.

Proof

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$g^{n-1}h^{n}=h^{n}g^{n-1}$ for all $g,h\in G$.	Fact (1)	$n^{th}$ power map is endomorphism		Given+Fact direct
2	$g^{n-1}x=xg^{n-1}$ for all $g,x\in G$		$n^{th}$ power map is surjective	Step (1)	[SHOW MORE] By Step (1), $g^{n-1}$ commutes with every $n^{th}$ power. Since the $n^{th}$ power map is an automorphism, every element is a $n^{th}$ power, so $g^{n-1}$ commutes with every element.
3	For any $g,x\in G$, $(gx)^{n-1}=g^{n-1}x^{n-1}$, so the $(n-1)^{th}$ power map is an endomorphism		$n^{th}$ power map is endomorphism	Step (2)	[SHOW MORE] We have $(gx)^{n}=g^{n}x^{n}$ from the given. Thus, we get $(gx)(gx)^{n-1}=gg^{n-1}xx^{n-1}$. Using step (2), interchange the $g^{n-1}$ and $x$ within the right side, and get $(gx)(gx)^{n-1}=(gx)(g^{n-1}x^{n-1})$. Cancel $gx$ from the left ends of both sides to get the required conclusion.

Steps (2) and (3) complete the proof.