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

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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.

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