Finite implies p-divisible iff p-powered

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Suppose G is a finite group and p is a prime number. Then, the following are equivalent:

  1. G is p-divisible, i.e., every element of G has a p^{th} root in G.
  2. G is p-powered, i.e., every element of G has a unique p^{th} root in G.

Related facts

It is also true that the above equivalent conditions hold if and only if p is relatively prime to the order of G, which in this case means that it does not divide the order of G (since it is prime). For a proof, see kth power map is bijective iff k is relatively prime to the order.


(2) implies (1)

This is obvious.

(1) implies (2)

(1) can be reinterpreted as saying that the p^{th} power map from G to itself is surjective. Since G is finite, this implies that the map is bijective, which is equivalent to (2).