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.

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 $g^{n-1}h^n = h^ng^{n-1}$ 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]
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]

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