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