# (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

### 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)