Group of prime power order is either trivial or of prime order or has outer automorphism class of same prime order

Statement
Suppose $$P$$ is a nontrivial group of prime power order, say with prime $$p$$. Further, suppose that the order of $$P$$ is strictly greater than $$p$$. Then, the following equivalent statements hold:


 * The order of the fact about::outer automorphism group of $$P$$ is divisible by $$p$$.
 * $$\operatorname{Out}(P)$$ has a non-identity element of order $$p$$.
 * There is an element in $$\operatorname{Aut}(P)$$ and outside $$\operatorname{Inn}(P)$$ whose order is a power of $$p$$.

The first two statements are equivalent by Cauchy's theorem, which says that there is an element of order $$p$$ in a finite group for every $$p$$ dividing the order of the group, and Lagrange's theorem, that guarantees the converse. The equivalence of the second and third statement is clear by considering images and inverse images under the quotient map $$\operatorname{Aut}(P) \to \operatorname{Aut}(P)/\operatorname{Inn}(P) = \operatorname{Out}(P)$$.

Stronger facts

 * Group of prime power order is either elementary abelian or extraspecial or its outer automorphism group has a nontrivial p-core

Open questions

 * Berkovich's question on whether a group of prime power order has an outer automorphism of the same prime order