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 $$g \in G$$.

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

Proof
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

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