# 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.