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

Statement

Suppose $G$ is a group and $n$ is an integer such that the power map $x \mapsto x^{n-1}$ is an endomorphism of $G$ and $x^{n-1}$ is in the center of $G$ for all $x$ in $G$. Then, the map $x \mapsto x^n$ is also an endomorphism of $G$.

Related facts

Applications

• Frattini-in-center odd-order p-group implies (mp plus 1)-power map is automorphism

Converse

The precise converse is not true, but a partial converse is: nth power map is surjective endomorphism implies (n-1)th power map is endomorphism taking values in the center

Proof

Given: A group $G$, an integer $n$ such that the map $x \mapsto x^{n-1}$ is an endomorphism taking values in the center.

To prove: The map $x \mapsto x^n$ is an endomorphism of $G$, i.e., $(gh)^n = g^nh^n$ for all $g,h \in G$.

Proof: We have the following for all $g,h \in G$.

Step no.	Assertion/construction	Given data used	Previous steps used
1	$(gh)^n = (gh)(gh)^{n-1}$
2	$(gh)^{n-1} = g^{n-1}h^{n-1}$	$x \mapsto x^{n-1}$ is an endomorphism
3	$(gh)^n = ghg^{n-1}h^{n-1}$		Step (2) plugged into Step (1)
4	$(gh)^n = gg^{n-1}hh^{n-1}$	$g^{n-1}$ is in the center of $G$, so commutes with $h$	Step (3)
5	$(gh)^n = g^nh^n$		Step (4) (simplified)