Abelian implies uniquely p-divisible iff pth power map is automorphism

Statement

Suppose $p$ is a prime number and $G$ is an abelian group. The following are equivalent:

1. $G$ is a uniquely $p$-divisible abelian group, i.e., for every element $g \in G$, there is a unique $h \in G$ such that $ph = g$.
2. The multiplication by $p$ map is an automorphism of $G$, i.e., it is a bijective endomorphism.

Moreover, in this case, the division by $p$ map is an automorphism of $G$, i.e., it is a bijective endomorphism. These automorphisms are inverses of each other.

Related facts

• kth power map is bijective iff k is relatively prime to the order: In particular, if $G$ is a finite abelian group and $p$ is a prime not dividing the order of $G$, then $G$ is uniquely $p$-divisible and the above conclusions hold.
• Abelian implies universal power map is endomorphism (but need not be an automorphism)
• Square map is endomorphism iff abelian
• Inverse map is automorphism iff abelian
• Cube map is automorphism implies abelian

Applications

• Characteristic subgroup of uniquely p-divisible abelian group is uniquely p-divisible