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

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.