# Finite implies p-divisible iff p-powered

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(Redirected from Finite and p-divisible implies p-powered)

## Statement

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

- is -divisible, i.e., every element of has a root in .
- is -powered, i.e., every element of has a
*unique*root in .

## Related facts

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

## Proof

### (2) implies (1)

This is obvious.

### (1) implies (2)

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